1 Introduction
 The Ehrhart function 
 $i_P \colon {\mathbb Z}_{\ge 0} \rightarrow {\mathbb Z}_{\ge 0}$
 of a polytope
$i_P \colon {\mathbb Z}_{\ge 0} \rightarrow {\mathbb Z}_{\ge 0}$
 of a polytope 
 $P \subseteq {\mathbb R}^n$
 counts the number of lattice points
$P \subseteq {\mathbb R}^n$
 counts the number of lattice points 
 $i_P(d) = |dP \cap {\mathbb Z}^n|$
 in its d-th dilation, for
$i_P(d) = |dP \cap {\mathbb Z}^n|$
 in its d-th dilation, for 
 $d \ge 1$
. By convention, we take
$d \ge 1$
. By convention, we take 
 $i_P(0) = 1$
. If the vertices of P are lattice points, then
$i_P(0) = 1$
. If the vertices of P are lattice points, then 
 $i_P(d) \in \mathbb Q[d]$
 is a polynomial called the Ehrhart polynomial. More generally, if the vertices of P have rational coordinates, then
$i_P(d) \in \mathbb Q[d]$
 is a polynomial called the Ehrhart polynomial. More generally, if the vertices of P have rational coordinates, then 
 $i_P$
 is the Ehrhart quasi-polynomial, which is a polynomial whose coefficients are periodic functions from
$i_P$
 is the Ehrhart quasi-polynomial, which is a polynomial whose coefficients are periodic functions from 
 ${\mathbb Z}$
 to
${\mathbb Z}$
 to 
 $\mathbb Q$
. We can associate a generating function to
$\mathbb Q$
. We can associate a generating function to 
 $i_P$
 which we call the Ehrhart series denoted
$i_P$
 which we call the Ehrhart series denoted 
 $\mathrm {ES}_P[t] \in {\mathbb Z}[[t]]$
. If P is an n-dimensional lattice polytope, then the Ehrhart series takes the form
$\mathrm {ES}_P[t] \in {\mathbb Z}[[t]]$
. If P is an n-dimensional lattice polytope, then the Ehrhart series takes the form 
 $$\begin{align*}\mathrm{ES}_P[t] = \sum_{d \ge 0} i_P(d) t^d = \frac{h^*_P(t)}{(1-t)^{n+1}} \end{align*}$$
$$\begin{align*}\mathrm{ES}_P[t] = \sum_{d \ge 0} i_P(d) t^d = \frac{h^*_P(t)}{(1-t)^{n+1}} \end{align*}$$
for some polynomial 
 $h^*_P(t) \in {\mathbb Z}[t]$
 of degree
$h^*_P(t) \in {\mathbb Z}[t]$
 of degree 
 $\leq d$
 called the
$\leq d$
 called the 
 $h^*$
-polynomial of P. Stanley [Reference Stanley19] showed that the coefficients of
$h^*$
-polynomial of P. Stanley [Reference Stanley19] showed that the coefficients of 
 $h^*_P$
 are all non-negative. Understanding the relationship between properties of polytopes and properties of their
$h^*_P$
 are all non-negative. Understanding the relationship between properties of polytopes and properties of their 
 $h^*$
-coefficients is one of the central themes in Ehrhart theory [Reference Stapledon21, Reference Higashitani11, Reference Davis5, Reference Hofscheier, Katthän and Nill12, Reference Braun3].
$h^*$
-coefficients is one of the central themes in Ehrhart theory [Reference Stapledon21, Reference Higashitani11, Reference Davis5, Reference Hofscheier, Katthän and Nill12, Reference Braun3].
 Given positive integers 
 $k \leq n$
, the
$k \leq n$
, the 
 $(k,n)$
-hypersimplex is defined as
$(k,n)$
-hypersimplex is defined as 
 $$\begin{align*}\Delta_{k,n} = \{(x_1, x_2, \dots, x_n) \in {\mathbb R}^n : 0 \le x_i \le 1 \text{ for all } 1 \le i \le n,\ x_1 + x_2 + \dots + x_n = k\}. \end{align*}$$
$$\begin{align*}\Delta_{k,n} = \{(x_1, x_2, \dots, x_n) \in {\mathbb R}^n : 0 \le x_i \le 1 \text{ for all } 1 \le i \le n,\ x_1 + x_2 + \dots + x_n = k\}. \end{align*}$$
Equivalently, 
 $\Delta _{k,n}$
 is the convex hull of the zero-one vectors in
$\Delta _{k,n}$
 is the convex hull of the zero-one vectors in 
 ${\mathbb R}^n$
 with exactly k ones. Stanley [Reference Stanley20, Chapter 4 Exercise 62(e)(h)] gave two open problems about the hypersimplex, which were posed as tricky exercises. The first was to prove Ehrhart positivity (the property that all coefficients of the Ehrhart polynomial are non-negative) for the hypersimplex, which was accomplished by Ferroni [Reference Ferroni10] using a computation of Katzman [Reference Katzman13]. The second open problem was to give a combinatorial interpretation of the coefficients of
${\mathbb R}^n$
 with exactly k ones. Stanley [Reference Stanley20, Chapter 4 Exercise 62(e)(h)] gave two open problems about the hypersimplex, which were posed as tricky exercises. The first was to prove Ehrhart positivity (the property that all coefficients of the Ehrhart polynomial are non-negative) for the hypersimplex, which was accomplished by Ferroni [Reference Ferroni10] using a computation of Katzman [Reference Katzman13]. The second open problem was to give a combinatorial interpretation of the coefficients of 
 $h^*_{\Delta _{k,n}}(t)$
. In [Reference Katzman13], Katzman studied so-called Veronese type algebras, which includes the Ehrhart ring of the hypersimplex. He showed
$h^*_{\Delta _{k,n}}(t)$
. In [Reference Katzman13], Katzman studied so-called Veronese type algebras, which includes the Ehrhart ring of the hypersimplex. He showed 
 $$\begin{align*}i_{\Delta_{k,n}}(d) = \sum_{s = 0}^{k-1} (-1)^s \binom{n}{s} \binom{d(k-s) - s + n -1}{n - 1} \end{align*}$$
$$\begin{align*}i_{\Delta_{k,n}}(d) = \sum_{s = 0}^{k-1} (-1)^s \binom{n}{s} \binom{d(k-s) - s + n -1}{n - 1} \end{align*}$$
and, from this, derived a formula for the 
 $h^*$
-polynomial [Reference Katzman13, Corollary 2.9]. In [Reference Li17], Li gave the first result towards a combinatorial interpretation of the coefficients of
$h^*$
-polynomial [Reference Katzman13, Corollary 2.9]. In [Reference Li17], Li gave the first result towards a combinatorial interpretation of the coefficients of 
 $h^*_{\Delta _{k,n}}$
 by proving two combinatorial formulae for the
$h^*_{\Delta _{k,n}}$
 by proving two combinatorial formulae for the 
 $h^*$
-polynomial of the half-open hypersimplex. To do this, Li gave two different shelling orders of a well-known unimodular triangulation of the hypersimplex. This unimodular triangulation was originally a consequence of a result by Stanley [Reference Stanley18] and separately arose in the study of Gröbner bases associated to the hypersimplex [Reference Sturmfels25]. Lam and Postnikov [Reference Lam and Postnikov15] showed that these two triangulations coincide and gave two equivalent triangulations called the alcove triangulation and the circuit triangulation. The number of simplices of a unimodular triangulation of a polytope P is equal to its normalised volume, which equals
$h^*$
-polynomial of the half-open hypersimplex. To do this, Li gave two different shelling orders of a well-known unimodular triangulation of the hypersimplex. This unimodular triangulation was originally a consequence of a result by Stanley [Reference Stanley18] and separately arose in the study of Gröbner bases associated to the hypersimplex [Reference Sturmfels25]. Lam and Postnikov [Reference Lam and Postnikov15] showed that these two triangulations coincide and gave two equivalent triangulations called the alcove triangulation and the circuit triangulation. The number of simplices of a unimodular triangulation of a polytope P is equal to its normalised volume, which equals 
 $h^*_P(1)$
. For
$h^*_P(1)$
. For 
 $\Delta _{k,n}$
, the normalised volume is equal to the Eulerian number
$\Delta _{k,n}$
, the normalised volume is equal to the Eulerian number 
 $A(n-1, k-1)$
 [Reference Stanley18], which is the number of permutations of
$A(n-1, k-1)$
 [Reference Stanley18], which is the number of permutations of 
 $1, \dots , n-1$
 with exactly
$1, \dots , n-1$
 with exactly 
 $k-1$
 ascents, a fact that was implicit in work of Laplace [Reference Laplace16, p. 257].
$k-1$
 ascents, a fact that was implicit in work of Laplace [Reference Laplace16, p. 257].
 The Worpitzky identity [Reference Worpitzky26] gives a relationship between Eulerian numbers and binomial coefficients. In his thesis [Reference Early6, Reference Early7], Early generalised this identity by viewing its terms as dimensions of certain 
 $\mathbb C[S_n]$
-modules, which arise from simplices, hypersimplices and so-called q-plates, for q a root of unity. In the special case
$\mathbb C[S_n]$
-modules, which arise from simplices, hypersimplices and so-called q-plates, for q a root of unity. In the special case 
 $q=1$
, the plates have a combinatorial description, which inspired Early to conjecture in [Reference Early8] that the coefficients of the
$q=1$
, the plates have a combinatorial description, which inspired Early to conjecture in [Reference Early8] that the coefficients of the 
 $h^*$
-polynomial are given in terms of decorated ordered set partitions or DOSPs.
$h^*$
-polynomial are given in terms of decorated ordered set partitions or DOSPs.
Definition 1.1. A 
 $(k,n)$
-DOSP is an ordered partition
$(k,n)$
-DOSP is an ordered partition 
 $(L_1, \dots , L_r)$
 of
$(L_1, \dots , L_r)$
 of 
 $\{1,2, \dots , n\}$
 together with a sequence of positive integers
$\{1,2, \dots , n\}$
 together with a sequence of positive integers 
 $(\ell _1, \dots , \ell _r)$
 such that
$(\ell _1, \dots , \ell _r)$
 such that 
 $\ell _1 + \ell _2 + \dots + \ell _r = k$
. We write a DOSP as a sequence of pairs
$\ell _1 + \ell _2 + \dots + \ell _r = k$
. We write a DOSP as a sequence of pairs 
 $D = ((L_1, \ell _1), \dots , (L_r, \ell _r))$
. A DOSP is defined up to cyclic permutation of its pairs
$D = ((L_1, \ell _1), \dots , (L_r, \ell _r))$
. A DOSP is defined up to cyclic permutation of its pairs 
 $(L_i, \ell _i)$
. So, for example, we have that
$(L_i, \ell _i)$
. So, for example, we have that 
 $((L_2, \ell _2), \dots , (L_r, \ell _r), (L_1, \ell _1))$
 is equal to D. We say D is hypersimplicial if
$((L_2, \ell _2), \dots , (L_r, \ell _r), (L_1, \ell _1))$
 is equal to D. We say D is hypersimplicial if 
 $|L_i|> \ell _i$
 for each
$|L_i|> \ell _i$
 for each 
 $i \in \{1,\dots , r\}$
.
$i \in \{1,\dots , r\}$
.
 For every DOSP D, Early defines the winding number 
 $w(D) \in \{0,1, \dots , n-1\}$
, see Definition 4.2, and conjectured that the
$w(D) \in \{0,1, \dots , n-1\}$
, see Definition 4.2, and conjectured that the 
 $h^*$
-polynomial is given by
$h^*$
-polynomial is given by 
 $$\begin{align*}h^*_{\Delta_{k,n}}(t) = \sum_{D} t^{w(D)} \end{align*}$$
$$\begin{align*}h^*_{\Delta_{k,n}}(t) = \sum_{D} t^{w(D)} \end{align*}$$
where the sum is taken over all hypersimplicial 
 $(k,n)$
-DOSPs. This conjecture was proved by Kim [Reference Kim14] and gives a satisfying answer to Stanley’s second problem. We now recount the story of Ehrhart theory for hypersimplices in the equivariant setting.
$(k,n)$
-DOSPs. This conjecture was proved by Kim [Reference Kim14] and gives a satisfying answer to Stanley’s second problem. We now recount the story of Ehrhart theory for hypersimplices in the equivariant setting.
 
Equivariant Ehrhart theory. The symmetric group 
 $S_n$
 acts by coordinate permutation on
$S_n$
 acts by coordinate permutation on 
 ${\mathbb R}^n$
 and, for each
${\mathbb R}^n$
 and, for each 
 $\sigma \in S_n$
, we have
$\sigma \in S_n$
, we have 
 $\sigma (\Delta _{k,n}) = \Delta _{k,n}$
. Thus,
$\sigma (\Delta _{k,n}) = \Delta _{k,n}$
. Thus, 
 $\Delta _{k,n}$
 is an
$\Delta _{k,n}$
 is an 
 $S_n$
-invariant polytope. Moreover, the vertices of the hypersimplex are integral and lie in a single
$S_n$
-invariant polytope. Moreover, the vertices of the hypersimplex are integral and lie in a single 
 $S_n$
-orbit. Such polytopes are known as orbit polytopes [Reference Elia, Kim and Supina9, Section 3.4.1].
$S_n$
-orbit. Such polytopes are known as orbit polytopes [Reference Elia, Kim and Supina9, Section 3.4.1].
 Motivated by the study of finite groups acting on nondegenerate toric hypersurfaces [Reference Stapledon23], Stapledon introduced an equivariant generalisation of Ehrhart theory [Reference Stapledon22]. We provide a detailed description of the setup in Section 2.1, which defines the equivariant analogues of the Ehrhart series and 
 $h^*$
-polynomial. Intuitively, if P is a G-invariant polytope, then the equivariant analogues are obtained by replacing the value of the Ehrhart function
$h^*$
-polynomial. Intuitively, if P is a G-invariant polytope, then the equivariant analogues are obtained by replacing the value of the Ehrhart function 
 $i_P(d)$
 with the permutation representation of G acting on the lattice points of the dilation
$i_P(d)$
 with the permutation representation of G acting on the lattice points of the dilation 
 $dP$
. The equivariant analogue of the
$dP$
. The equivariant analogue of the 
 $h^*$
-polynomial is the equivariant
$h^*$
-polynomial is the equivariant 
 $H^*$
-series denoted
$H^*$
-series denoted 
 $H^*(P; G)[t] \in \mathcal R_{\mathbb C}(G)[[t]]$
, which is a power series with coefficients in the representation ring of G over
$H^*(P; G)[t] \in \mathcal R_{\mathbb C}(G)[[t]]$
, which is a power series with coefficients in the representation ring of G over 
 $\mathbb C$
.
$\mathbb C$
.
 One of the central open problems in equivariant Ehrhart theory is to determine when the 
 $H^*$
-series is a polynomial. Stapledon [Reference Stapledon22, Theorem 7.7] showed that if the toric variety corresponding to P admits a nondegenerate G-invariant hypersurface defined by a G-invariant equation, then the
$H^*$
-series is a polynomial. Stapledon [Reference Stapledon22, Theorem 7.7] showed that if the toric variety corresponding to P admits a nondegenerate G-invariant hypersurface defined by a G-invariant equation, then the 
 $H^*$
-series is a polynomial and, moreover, each of its coefficients is effective, that is, each coefficient is a genuine representation. See Definition 2.5. This result led to the following conjecture.
$H^*$
-series is a polynomial and, moreover, each of its coefficients is effective, that is, each coefficient is a genuine representation. See Definition 2.5. This result led to the following conjecture.
Conjecture 1.2 [Reference Stapledon22, Conjecture 12.1].
Suppose P is a G-invariant lattice polytope and Y is the corresponding toric variety with torus-invariant line bundle L. Then the following are equivalent:
- 
(1)  $(Y, L)$
 admits a nondegenerate G-invariant hypersurface defined by a G-invariant equation, $(Y, L)$
 admits a nondegenerate G-invariant hypersurface defined by a G-invariant equation,
- 
(2) The coefficients of the equivariant  $H^*$
-series are effective, $H^*$
-series are effective,
- 
(3) The equivariant  $H^*$
-series is a polynomial. $H^*$
-series is a polynomial.
 The result 
 $(1) \Rightarrow (2) \Rightarrow (3)$
 follows from [Reference Stapledon22, Theorem 7.7]. Unfortunately, in [Reference Elia, Kim and Supina9, Theorem 1.2], the authors report that Stapledon and Santos had discovered a counterexample showing
$(1) \Rightarrow (2) \Rightarrow (3)$
 follows from [Reference Stapledon22, Theorem 7.7]. Unfortunately, in [Reference Elia, Kim and Supina9, Theorem 1.2], the authors report that Stapledon and Santos had discovered a counterexample showing 
 $(2) \nRightarrow (1)$
. However, the conjecture
$(2) \nRightarrow (1)$
. However, the conjecture 
 $(3) \Rightarrow (2)$
, which we refer to as the effectiveness conjecture, is open and verified only in a handful of cases.
$(3) \Rightarrow (2)$
, which we refer to as the effectiveness conjecture, is open and verified only in a handful of cases.
 We give a brief overview of the known cases for the effectiveness conjecture. Stapledon [Reference Stapledon22, Proposition 6.1] showed that the coefficients of the 
 $H^*$
-polynomial of the simplex, under any invariant group action, are permutation representations. In the same paper, the effectiveness conjecture is also proved for: all two-dimensional polytopes under any invariant action, the zero-one hypercube under the action of coordinate permutations, and centrally symmetric polytopes under the action of
$H^*$
-polynomial of the simplex, under any invariant group action, are permutation representations. In the same paper, the effectiveness conjecture is also proved for: all two-dimensional polytopes under any invariant action, the zero-one hypercube under the action of coordinate permutations, and centrally symmetric polytopes under the action of 
 ${\mathbb Z}/2{\mathbb Z}$
 generated by the diagonal matrix
${\mathbb Z}/2{\mathbb Z}$
 generated by the diagonal matrix 
 $\operatorname {\mathrm {diag}}(-1, -1, \dots , -1)$
. In [Reference Ardila, Supina and Vindas-Meléndez2], the authors use results from [Reference Ardila, Schindler and Vindas-Meléndez1] to show that the permutohedron, which is the orbit polytope of the point
$\operatorname {\mathrm {diag}}(-1, -1, \dots , -1)$
. In [Reference Ardila, Supina and Vindas-Meléndez2], the authors use results from [Reference Ardila, Schindler and Vindas-Meléndez1] to show that the permutohedron, which is the orbit polytope of the point 
 $(1,2,\dots ,n) \in {\mathbb R}^n$
 under the action of
$(1,2,\dots ,n) \in {\mathbb R}^n$
 under the action of 
 $S_n$
, satisfies the effectiveness conjecture. The effectiveness conjecture is known to hold for the hypersimplex
$S_n$
, satisfies the effectiveness conjecture. The effectiveness conjecture is known to hold for the hypersimplex 
 $\Delta _{k,n}$
 under the action of
$\Delta _{k,n}$
 under the action of 
 $S_n$
 because its toric variety admits a nondegenerate
$S_n$
 because its toric variety admits a nondegenerate 
 $S_n$
-invariant hypersurface defined by a G-invariant equation [Reference Elia, Kim and Supina9, Theorem 3.60]. There are also constructive approaches to the effectiveness conjecture using G-invariant subdivisions to compute
$S_n$
-invariant hypersurface defined by a G-invariant equation [Reference Elia, Kim and Supina9, Theorem 3.60]. There are also constructive approaches to the effectiveness conjecture using G-invariant subdivisions to compute 
 $H^*$
-series [Reference Elia, Kim and Supina9, Reference Stapledon24]. In [Reference Clarke, Higashitani and Kölbl4], it is shown that the conjecture is false if P is allowed to be a rational polytope.
$H^*$
-series [Reference Elia, Kim and Supina9, Reference Stapledon24]. In [Reference Clarke, Higashitani and Kölbl4], it is shown that the conjecture is false if P is allowed to be a rational polytope.
 Suppose that P is a G-invariant polytope. For each 
 $g \in G$
, the fixed polytope
$g \in G$
, the fixed polytope 
 $P_g = \{x \in P : g(x) = x\}$
 is the set of points in P fixed by g. Recall that
$P_g = \{x \in P : g(x) = x\}$
 is the set of points in P fixed by g. Recall that 
 $h^*_P(1)$
 measures the lattice volume of P, in particular, it is a non-negative integer. The equivariant analogue
$h^*_P(1)$
 measures the lattice volume of P, in particular, it is a non-negative integer. The equivariant analogue 
 $H^*(P; G)[1]$
 is the equivariant volume of P. For example, see [Reference Ardila, Schindler and Vindas-Meléndez1] for the permutohedron. In [Reference Early6, Reference Early7] Early uses q-plates to study generalisations of equivariant volumes of simplices and hypersimplices; in particular [Reference Early7, Theorem 9] gives a relationship between these equivariant volumes. Stapledon conjectured the following for equivariant volumes.
$H^*(P; G)[1]$
 is the equivariant volume of P. For example, see [Reference Ardila, Schindler and Vindas-Meléndez1] for the permutohedron. In [Reference Early6, Reference Early7] Early uses q-plates to study generalisations of equivariant volumes of simplices and hypersimplices; in particular [Reference Early7, Theorem 9] gives a relationship between these equivariant volumes. Stapledon conjectured the following for equivariant volumes.
Conjecture 1.3 [Reference Stapledon22, Conjectures 12.2, 12.3].
 Let P be a G-invariant lattice polytope. For each 
 $g \in G$
, we have that
$g \in G$
, we have that 
 $H^*(P; G)[1](g)$
 is a non-negative integer. If the
$H^*(P; G)[1](g)$
 is a non-negative integer. If the 
 $H^*$
-series is effective (and polynomial), then
$H^*$
-series is effective (and polynomial), then 
 $H^*(P; G)[1]$
 is a permutation representation.
$H^*(P; G)[1]$
 is a permutation representation.
 Part of the strength of the analogy between 
 $h^*_P$
 and
$h^*_P$
 and 
 $H^*(P; G)$
 lies in the fact that the evaluation
$H^*(P; G)$
 lies in the fact that the evaluation 
 $H^*(P; G)[1](g)$
 is directly related to the lattice volume of
$H^*(P; G)[1](g)$
 is directly related to the lattice volume of 
 $P_g$
. Currently, one of the few methods to tackle Conjecture 1.3 is to compute all coefficients of the
$P_g$
. Currently, one of the few methods to tackle Conjecture 1.3 is to compute all coefficients of the 
 $H^*$
-series. For instance, if each coefficient
$H^*$
-series. For instance, if each coefficient 
 $H^*_i$
 of the
$H^*_i$
 of the 
 $H^*$
-series is a permutation representation, then
$H^*$
-series is a permutation representation, then 
 $H^*(P; G)[1] = \sum _i H^*_i$
 is also a permutation representation. The conjecture has been verified for a few examples, including the simplex and zero-one hypercube [Reference Stapledon22], and the permutohedron [Reference Ardila, Supina and Vindas-Meléndez2].
$H^*(P; G)[1] = \sum _i H^*_i$
 is also a permutation representation. The conjecture has been verified for a few examples, including the simplex and zero-one hypercube [Reference Stapledon22], and the permutohedron [Reference Ardila, Supina and Vindas-Meléndez2].
 In the case of the hypersimplex [Reference Elia, Kim and Supina9, Section 3.3], the coefficients of the 
 $H^*$
-polynomial have been computed under the action of
$H^*$
-polynomial have been computed under the action of 
 $C_n \le S_n$
, a cyclic subgroup of order n.
$C_n \le S_n$
, a cyclic subgroup of order n.
Theorem 1.4 [Reference Elia, Kim and Supina9, Theorem 3.3].
 The coefficient of 
 $t^i$
 in
$t^i$
 in 
 $H^*(\Delta _{k,n}, C_n)[t]$
 is the permutation representation of
$H^*(\Delta _{k,n}, C_n)[t]$
 is the permutation representation of 
 $C_n$
 acting on the set of hypersimplicial
$C_n$
 acting on the set of hypersimplicial 
 $(k,n)$
-DOSPs with winding number i. Hence, Conjecture 1.3 holds for
$(k,n)$
-DOSPs with winding number i. Hence, Conjecture 1.3 holds for 
 $\Delta _{k,n}$
 under the action of
$\Delta _{k,n}$
 under the action of 
 $C_n$
.
$C_n$
.
 In the above result, the action of 
 $C_n$
 on the set of hypersimplicial
$C_n$
 on the set of hypersimplicial 
 $(k,n)$
-DOSPs arises as the restriction of the natural action of
$(k,n)$
-DOSPs arises as the restriction of the natural action of 
 $S_n$
 on the set of DOSPs given by
$S_n$
 on the set of DOSPs given by 
 $$\begin{align*}\sigma((L_1, \ell_1), \dots, (L_r, \ell_r)) = ((\sigma(L_1), \ell_1), \dots, (\sigma(L_r), \ell_r)), \end{align*}$$
$$\begin{align*}\sigma((L_1, \ell_1), \dots, (L_r, \ell_r)) = ((\sigma(L_1), \ell_1), \dots, (\sigma(L_r), \ell_r)), \end{align*}$$
for each 
 $\sigma \in S_n$
. For each
$\sigma \in S_n$
. For each 
 $1 \le i \le r$
, observe that
$1 \le i \le r$
, observe that 
 $|\sigma (L_i)| = |L_i|$
. So, the action of
$|\sigma (L_i)| = |L_i|$
. So, the action of 
 $S_n$
 leaves the set of hypersimplicial DOSPs invariant. By restricting the action to a particular cyclic subgroup
$S_n$
 leaves the set of hypersimplicial DOSPs invariant. By restricting the action to a particular cyclic subgroup 
 $C_n \le S_n$
, it is straightforward to show that the set of hypersimplicial DOSPs with a fixed winding number is invariant under
$C_n \le S_n$
, it is straightforward to show that the set of hypersimplicial DOSPs with a fixed winding number is invariant under 
 $C_n$
. However, the set of hypersimplicial DOSPs with a fixed winding number is in general not invariant under the action of
$C_n$
. However, the set of hypersimplicial DOSPs with a fixed winding number is in general not invariant under the action of 
 $S_n$
. For instance, the
$S_n$
. For instance, the 
 $(2,4)$
-DOSP
$(2,4)$
-DOSP 
 $D = ((12,1), (34, 1))$
 has winding number
$D = ((12,1), (34, 1))$
 has winding number 
 $w(D) = 1$
. But, the transposition
$w(D) = 1$
. But, the transposition 
 $\sigma = (2\ 3) \in S_n$
 sends D to the DOSP
$\sigma = (2\ 3) \in S_n$
 sends D to the DOSP 
 $\sigma (D) = ((13,1), (24,1))$
, which has winding number
$\sigma (D) = ((13,1), (24,1))$
, which has winding number 
 $w(\sigma (D)) = 2$
. Therefore, Theorem 1.4 does not hold, or even make sense, if we replace
$w(\sigma (D)) = 2$
. Therefore, Theorem 1.4 does not hold, or even make sense, if we replace 
 $C_n$
 with
$C_n$
 with 
 $S_n$
. The aim of this paper is to understand what we can say about the hypersimplex under the action of the full symmetric group.
$S_n$
. The aim of this paper is to understand what we can say about the hypersimplex under the action of the full symmetric group.
 
Overview and results. In Section 2, we fix our notation and explain the setup for equivariant Ehrhart theory. In Section 3, we prove our first main result, Theorem 3.3, which gives a formula for the coefficients of the 
 $H^*$
-polynomial. In particular, this theorem allows us to give a completely combinatorial proof that the equivariant
$H^*$
-polynomial. In particular, this theorem allows us to give a completely combinatorial proof that the equivariant 
 $H^*$
-series is a polynomial. See Corollary 3.4. In Section 4, we study hypersimplicial DOSPs under the action of
$H^*$
-series is a polynomial. See Corollary 3.4. In Section 4, we study hypersimplicial DOSPs under the action of 
 $S_n$
 and verify Conjecture 1.3 as follows.
$S_n$
 and verify Conjecture 1.3 as follows.
Theorem (Theorem 4.1).
 We have that 
 $H^*(\Delta _{k,n}; S_n)[1]$
 is equal to the permutation representation of
$H^*(\Delta _{k,n}; S_n)[1]$
 is equal to the permutation representation of 
 $S_n$
 acting on the set of hypersimplicial
$S_n$
 acting on the set of hypersimplicial 
 $(k,n)$
-DOSPs. Hence, Conjecture 1.3 holds for
$(k,n)$
-DOSPs. Hence, Conjecture 1.3 holds for 
 $\Delta _{k,n}$
 under the action of
$\Delta _{k,n}$
 under the action of 
 $S_n$
.
$S_n$
.
 The full statement of Theorem 4.1 gives a formula for the character 
 $H^*(\Delta _{k,n}; S_n)[1]$
. Evaluating this formula at the identity gives the well-known formula for Eulerian numbers
$H^*(\Delta _{k,n}; S_n)[1]$
. Evaluating this formula at the identity gives the well-known formula for Eulerian numbers 
 $$\begin{align*}H^*(\Delta_{k,n}; S_n)[1](id) = \sum_{h = 0}^{k-1}(-1)^h \binom{n}{h}(k-h)^{n-1} = A(n-1, k-1). \end{align*}$$
$$\begin{align*}H^*(\Delta_{k,n}; S_n)[1](id) = \sum_{h = 0}^{k-1}(-1)^h \binom{n}{h}(k-h)^{n-1} = A(n-1, k-1). \end{align*}$$
Similar to the Eulerian numbers, we show that the values 
 $H^*(\Delta _{k,n}; S_n)[1](\sigma )$
 satisfy a certain recurrence relation in Section 4.4.
$H^*(\Delta _{k,n}; S_n)[1](\sigma )$
 satisfy a certain recurrence relation in Section 4.4.
 In Section 5, we apply Theorem 3.3 to the second hypersimplex 
 $\Delta _{2,n}$
. The main result of this section is Theorem 5.1, which gives a complete description of the coefficients of the
$\Delta _{2,n}$
. The main result of this section is Theorem 5.1, which gives a complete description of the coefficients of the 
 $H^*$
-polynomial in simple terms. From this, we observe two facts. Firstly, in Corollary 5.4, we have a combinatorial proof that the coefficients of the
$H^*$
-polynomial in simple terms. From this, we observe two facts. Firstly, in Corollary 5.4, we have a combinatorial proof that the coefficients of the 
 $H^*$
-polynomial are effective. Secondly, in Corollary 5.5, we observe that there is exactly one coefficient of the
$H^*$
-polynomial are effective. Secondly, in Corollary 5.5, we observe that there is exactly one coefficient of the 
 $H^*$
-polynomial that is not a permutation representation of
$H^*$
-polynomial that is not a permutation representation of 
 $S_n$
. We do this by showing that the coefficient of t in
$S_n$
. We do this by showing that the coefficient of t in 
 $H^*(\Delta _{2,n}, S_n)[t]$
 contains no copies of the trivial representation, which provides a family of counterexamples to the following conjecture.
$H^*(\Delta _{2,n}, S_n)[t]$
 contains no copies of the trivial representation, which provides a family of counterexamples to the following conjecture.
Conjecture 1.5 [Reference Stapledon22, Conjecture 12.4].
 Let P be a G-invariant polytope. Assume 
 $H^*(P; G)[t]$
 is an effective polynomial. If the coefficient of
$H^*(P; G)[t]$
 is an effective polynomial. If the coefficient of 
 $t^m$
 in
$t^m$
 in 
 $h^*_P(t)$
 is positive, then the coefficient of
$h^*_P(t)$
 is positive, then the coefficient of 
 $t^m$
 in
$t^m$
 in 
 $H^*(P; G)[t]$
 contains the trivial representation with nonzero multiplicity.
$H^*(P; G)[t]$
 contains the trivial representation with nonzero multiplicity.
Example 1.6. Let 
 $n \ge 4$
 and let
$n \ge 4$
 and let 
 $H^*_1$
 be the coefficient of t in
$H^*_1$
 be the coefficient of t in 
 $H^*(\Delta _{2,n}; S_n)(t)$
. By Corollary 5.5, the multiplicity of the trivial representation in
$H^*(\Delta _{2,n}; S_n)(t)$
. By Corollary 5.5, the multiplicity of the trivial representation in 
 $H^*_1$
 is zero. On the other hand, let
$H^*_1$
 is zero. On the other hand, let 
 $h^*_1$
 be the coefficient of t in
$h^*_1$
 be the coefficient of t in 
 $h^*_{\Delta _{2,n}}(t)$
, which is equal to the number of hypersimplicial
$h^*_{\Delta _{2,n}}(t)$
, which is equal to the number of hypersimplicial 
 $(2,n)$
-DOSPs with winding number one. In particular, the hypersimplicial DOSP
$(2,n)$
-DOSPs with winding number one. In particular, the hypersimplicial DOSP 
 $ ((12,1), (34\cdots n,1)) $
 has winding number one, hence
$ ((12,1), (34\cdots n,1)) $
 has winding number one, hence 
 $h^*_1> 0$
. Therefore, the hypersimplex
$h^*_1> 0$
. Therefore, the hypersimplex 
 $\Delta _{2,n}$
 is a counterexample to Conjecture 1.5.
$\Delta _{2,n}$
 is a counterexample to Conjecture 1.5.
 We note that Stapledon has already given a counterexample to Conjecture 1.5 [Reference Stapledon24, Example 4.35] and asked whether the conjecture holds if we further assume that P exhibits a G-invariant lattice triangulation. In the case of the hypersimplex 
 $\Delta _{2,n}$
, we observe, at the very end of the paper, that it does not admit an
$\Delta _{2,n}$
, we observe, at the very end of the paper, that it does not admit an 
 $S_n$
-invariant lattice triangulation.
$S_n$
-invariant lattice triangulation.
 In Section 6, we discuss questions and future directions based on the results of the paper. We ask for which subgroups 
 $G \le S_n$
, the hypersimplex
$G \le S_n$
, the hypersimplex 
 $\Delta _{k,n}$
 admits a G-invariant triangulation. We conjecture that the well-known unimodular triangulation of
$\Delta _{k,n}$
 admits a G-invariant triangulation. We conjecture that the well-known unimodular triangulation of 
 $\Delta _{k,n}$
 is invariant under the dihedral group
$\Delta _{k,n}$
 is invariant under the dihedral group 
 $D_{2n}$
 and this is the largest subgroup for which the triangulation is invariant.
$D_{2n}$
 and this is the largest subgroup for which the triangulation is invariant.
2 Preliminaries and notation
In this section, we give the general setup for equivariant Ehrhart theory and fix our notation.
 
General notation. Let n be a positive integer. We write 
 $[n] := \{1,2,\dots ,n\}$
 and
$[n] := \{1,2,\dots ,n\}$
 and 
 $S_n$
 for the symmetric group on
$S_n$
 for the symmetric group on 
 $[n]$
. Given a finite set X, we write
$[n]$
. Given a finite set X, we write 
 $\binom {X}{n}$
 for the collection of n-subsets of X. Group actions are left actions and group representations are over the complex numbers
$\binom {X}{n}$
 for the collection of n-subsets of X. Group actions are left actions and group representations are over the complex numbers 
 $\mathbb C$
. Given a group G, its representation ring
$\mathbb C$
. Given a group G, its representation ring 
 $\mathcal R(G)$
 is the ring of formal differences of isomorphism classes of representations of G. Addition in
$\mathcal R(G)$
 is the ring of formal differences of isomorphism classes of representations of G. Addition in 
 $\mathcal R(G)$
 is given by direct sums and multiplication by tensor products. Since we work with finite groups and representations over
$\mathcal R(G)$
 is given by direct sums and multiplication by tensor products. Since we work with finite groups and representations over 
 $\mathbb C$
, by a slight abuse of notation, we identify representations with their characters. A class function of G is any function
$\mathbb C$
, by a slight abuse of notation, we identify representations with their characters. A class function of G is any function 
 $G \rightarrow \mathbb C$
 that is constant in conjugacy classes of G. Given an element
$G \rightarrow \mathbb C$
 that is constant in conjugacy classes of G. Given an element 
 $g \in G$
, we denote by
$g \in G$
, we denote by 
 $o(g)$
 the order of g.
$o(g)$
 the order of g.
Definition 2.1. Each permutation 
 $\sigma \in S_n$
 is a product of disjoint cycles. The cycle type of
$\sigma \in S_n$
 is a product of disjoint cycles. The cycle type of 
 $\sigma $
 is the multiset
$\sigma $
 is the multiset 
 $(s_1, s_2, \dots , s_r)$
 of cycle lengths. The indices of the disjoint cycles of
$(s_1, s_2, \dots , s_r)$
 of cycle lengths. The indices of the disjoint cycles of 
 $\sigma $
 are the cycle sets
$\sigma $
 are the cycle sets 
 $C_1, \dots , C_r$
, which partition
$C_1, \dots , C_r$
, which partition 
 $[n]$
. By convention, we assume that
$[n]$
. By convention, we assume that 
 $|C_i| = s_i$
 for each
$|C_i| = s_i$
 for each 
 $i \in [r]$
. For example, the permutation
$i \in [r]$
. For example, the permutation 
 $\sigma = (1) (2 \ 3) (4 \ 5)$
 has cycle type
$\sigma = (1) (2 \ 3) (4 \ 5)$
 has cycle type 
 $(1,2,2)$
 and cycle sets
$(1,2,2)$
 and cycle sets 
 $C_1 = \{1\}$
,
$C_1 = \{1\}$
, 
 $C_2 = \{2,3\}$
,
$C_2 = \{2,3\}$
, 
 $C_3 = \{4,5\}$
.
$C_3 = \{4,5\}$
.
 Suppose that G is a finite group that acts on a finite set X. The permutation representation of the action is the vector space 
 $V = \mathbb C^{X}$
, with basis
$V = \mathbb C^{X}$
, with basis 
 $\{e_x : x \in X\} \subseteq V$
, together with an action of G defined by
$\{e_x : x \in X\} \subseteq V$
, together with an action of G defined by 
 $g \cdot e_x = e_{g(x)}$
 for all
$g \cdot e_x = e_{g(x)}$
 for all 
 $g \in G$
 and
$g \in G$
 and 
 $x \in X$
. The group G acts on V by permutation matrices, hence its character
$x \in X$
. The group G acts on V by permutation matrices, hence its character  is given by
 is given by  for each
 for each 
 $g \in G$
.
$g \in G$
.
2.1 Equivariant Ehrhart theory
 The setup of equivariant Ehrhart theory requires two ingredients: a group G acting on a lattice and a G-invariant polytope P that lies in a codimension-one subspace. From this, we define the equivariant 
 $H^*$
-series denoted
$H^*$
-series denoted 
 $H^*(P; G)$
, which encodes the
$H^*(P; G)$
, which encodes the 
 $h^*$
-polynomials of all fixed polytopes of P. If G is the trivial group, then
$h^*$
-polynomials of all fixed polytopes of P. If G is the trivial group, then 
 $H^*(P; G)$
 is naturally equivalent to the
$H^*(P; G)$
 is naturally equivalent to the 
 $h^*$
-polynomial of P.
$h^*$
-polynomial of P.
 
Group setup. Throughout this section, let 
 $G \le \operatorname {\mathrm {GL}}_{n+1}({\mathbb Z})$
 be a finite subgroup that naturally acts on the lattice
$G \le \operatorname {\mathrm {GL}}_{n+1}({\mathbb Z})$
 be a finite subgroup that naturally acts on the lattice 
 ${\mathbb Z}^{n+1}$
 and, by extension, the vector space
${\mathbb Z}^{n+1}$
 and, by extension, the vector space 
 ${\mathbb R}^{n+1} = {\mathbb R} {\otimes }_{{\mathbb Z}} {\mathbb Z}^{n+1}$
. Assume there exists a G-fixed one-dimensional subspace
${\mathbb R}^{n+1} = {\mathbb R} {\otimes }_{{\mathbb Z}} {\mathbb Z}^{n+1}$
. Assume there exists a G-fixed one-dimensional subspace 
 $A \subseteq {\mathbb R}^{n+1}$
, that is, for all
$A \subseteq {\mathbb R}^{n+1}$
, that is, for all 
 $a \in A$
 and
$a \in A$
 and 
 $g \in G$
 we have
$g \in G$
 we have 
 $g(a) = a$
. Assume there exists a primitive lattice point
$g(a) = a$
. Assume there exists a primitive lattice point 
 $a \in {\mathbb Z}^{n+1}$
 that spans A. Let
$a \in {\mathbb Z}^{n+1}$
 that spans A. Let 
 $A^{\perp } \cong {\mathbb R}^n$
 be the orthogonal space to A. Let
$A^{\perp } \cong {\mathbb R}^n$
 be the orthogonal space to A. Let 
 $N = (A^\perp \cap {\mathbb Z}^{n+1}) + \langle a \rangle _{\mathbb Z} \subseteq {\mathbb Z}^{n+1}$
 be the rank
$N = (A^\perp \cap {\mathbb Z}^{n+1}) + \langle a \rangle _{\mathbb Z} \subseteq {\mathbb Z}^{n+1}$
 be the rank 
 $n+1$
 sub-lattice generated by the lattice points of
$n+1$
 sub-lattice generated by the lattice points of 
 $A^\perp $
 and a and define
$A^\perp $
 and a and define 
 $d = [{\mathbb Z}^{n+1} \colon N]$
 the index of N in
$d = [{\mathbb Z}^{n+1} \colon N]$
 the index of N in 
 ${{\mathbb Z}}^{n+1}$
. Fix a positive integer k. Let M be the n-dimensional affine lattice at height k along A, that is,
${{\mathbb Z}}^{n+1}$
. Fix a positive integer k. Let M be the n-dimensional affine lattice at height k along A, that is, 
 $M = (\frac kd a + A^\perp ) \cap {{\mathbb Z}}^{n+1}$
, and
$M = (\frac kd a + A^\perp ) \cap {{\mathbb Z}}^{n+1}$
, and 
 $M_{{\mathbb R}} = \frac kd a + A^\perp $
 its corresponding affine space. Throughout, we identify
$M_{{\mathbb R}} = \frac kd a + A^\perp $
 its corresponding affine space. Throughout, we identify 
 ${\mathbb R}^n$
 with
${\mathbb R}^n$
 with 
 $M_{1{\mathbb R}}$
 and we identify
$M_{1{\mathbb R}}$
 and we identify 
 ${{\mathbb Z}}^n$
 with the lattice M.
${{\mathbb Z}}^n$
 with the lattice M.
Example 2.2. Let 
 $n = 1$
 and let
$n = 1$
 and let 
 $G \cong S_2$
 be the subgroup of
$G \cong S_2$
 be the subgroup of 
 $\operatorname {\mathrm {GL}}_2({\mathbb Z})$
 of permutation matrices:
$\operatorname {\mathrm {GL}}_2({\mathbb Z})$
 of permutation matrices: 
 $$\begin{align*}G = \left\{ e := \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix},\, \sigma := \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\}. \end{align*}$$
$$\begin{align*}G = \left\{ e := \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix},\, \sigma := \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\}. \end{align*}$$
Let us compute the G-invariant affine subspace of 
 ${\mathbb R}^2$
 at height one (
${\mathbb R}^2$
 at height one (
 $k = 1$
) described above and depicted in Figure 1a.
$k = 1$
) described above and depicted in Figure 1a.
 There is a unique one-dimensional G-fixed subspace 
 $A = \{(x,x) \in {\mathbb R}^2 : x \in {\mathbb R}\}$
. We take
$A = \{(x,x) \in {\mathbb R}^2 : x \in {\mathbb R}\}$
. We take 
 $a = (1,1)$
 a primitive lattice point in A. The orthogonal space
$a = (1,1)$
 a primitive lattice point in A. The orthogonal space 
 $A^\perp = \{(x,-x) \in {\mathbb R}^2 : x \in {\mathbb R}\}$
 is one dimensional with lattice basis
$A^\perp = \{(x,-x) \in {\mathbb R}^2 : x \in {\mathbb R}\}$
 is one dimensional with lattice basis 
 $b = (1,-1)$
. The sublattice
$b = (1,-1)$
. The sublattice 
 $N \subseteq {\mathbb Z}^2$
 is freely generated by
$N \subseteq {\mathbb Z}^2$
 is freely generated by 
 $\{a,b\}$
, so we have
$\{a,b\}$
, so we have 
 $$\begin{align*}d = [{\mathbb Z}^2 : N] = \left\lvert \det\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \right\rvert = 2. \end{align*}$$
$$\begin{align*}d = [{\mathbb Z}^2 : N] = \left\lvert \det\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \right\rvert = 2. \end{align*}$$
We define the affine lattice 
 $M = \frac {1}{2}e + A^\perp $
 and identify
$M = \frac {1}{2}e + A^\perp $
 and identify 
 ${\mathbb R}^n$
 with
${\mathbb R}^n$
 with 
 $M_{\mathbb R}$
.
$M_{\mathbb R}$
.
The next example shows the predominant setup in the rest of the paper.
Example 2.3. Fix positive integers 
 $k < n$
. Fix
$k < n$
. Fix 
 $G \le \operatorname {\mathrm {GL}}_n({\mathbb Z})$
 the set of permutation matrices, which form a subgroup isomorphic to
$G \le \operatorname {\mathrm {GL}}_n({\mathbb Z})$
 the set of permutation matrices, which form a subgroup isomorphic to 
 $S_n$
. Let
$S_n$
. Let 
 $A \subseteq {\mathbb R}^n$
 be the G-fixed subspace given by the span of the primitive lattice point
$A \subseteq {\mathbb R}^n$
 be the G-fixed subspace given by the span of the primitive lattice point 
 $a_1 = (1,1, \dots , 1)$
. Let
$a_1 = (1,1, \dots , 1)$
. Let 
 $e_1, \dots , e_n$
 be the standard unit vectors of
$e_1, \dots , e_n$
 be the standard unit vectors of 
 ${\mathbb R}^n$
. The lattice of the orthogonal space
${\mathbb R}^n$
. The lattice of the orthogonal space 
 $A^\perp \cap {\mathbb Z}^n$
 is generated by the vectors
$A^\perp \cap {\mathbb Z}^n$
 is generated by the vectors 
 $\{a_i := e_1 - e_i : 2 \le i \le n\}$
. The lattice N generated by
$\{a_i := e_1 - e_i : 2 \le i \le n\}$
. The lattice N generated by 
 $a_1, a_2, \dots , a_n$
 has index
$a_1, a_2, \dots , a_n$
 has index 
 $$\begin{align*}d = [{\mathbb Z}^n : N] = \left\lvert \det \begin{pmatrix} 1 & 1 & 1 & \dots & 1 \\ 1 & -1 & 0 & \dots & 0 \\ 1 & 0 &-1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \dots & -1 \end{pmatrix} \right\rvert = n. \end{align*}$$
$$\begin{align*}d = [{\mathbb Z}^n : N] = \left\lvert \det \begin{pmatrix} 1 & 1 & 1 & \dots & 1 \\ 1 & -1 & 0 & \dots & 0 \\ 1 & 0 &-1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \dots & -1 \end{pmatrix} \right\rvert = n. \end{align*}$$
We identify 
 ${\mathbb R}^{n-1}$
 with the affine space
${\mathbb R}^{n-1}$
 with the affine space 
 $M_{\mathbb R} = \frac {k}{n}a_1 +A^\perp \subseteq {\mathbb R}^n$
, which contains the affine lattice
$M_{\mathbb R} = \frac {k}{n}a_1 +A^\perp \subseteq {\mathbb R}^n$
, which contains the affine lattice 
 $M = \{x \in {\mathbb Z}^n : x_1 +\dots + x_n = k\}$
.
$M = \{x \in {\mathbb Z}^n : x_1 +\dots + x_n = k\}$
.
 
Polytope setup. Assume the group setup with 
 $G\le \operatorname {\mathrm {GL}}_{n+1}({\mathbb Z})$
 a finite group. Since
$G\le \operatorname {\mathrm {GL}}_{n+1}({\mathbb Z})$
 a finite group. Since 
 $\operatorname {\mathrm {GL}}_{n+1}({\mathbb Z}) \subseteq \operatorname {\mathrm {GL}}_{n+1}(\mathbb C)$
 there is a natural
$\operatorname {\mathrm {GL}}_{n+1}({\mathbb Z}) \subseteq \operatorname {\mathrm {GL}}_{n+1}(\mathbb C)$
 there is a natural 
 $(n+1)$
-dimensional representation of G given by the inclusion map
$(n+1)$
-dimensional representation of G given by the inclusion map 
 $\rho : \operatorname {\mathrm {GL}}_{n+1}({\mathbb Z}) \rightarrow \operatorname {\mathrm {GL}}_{n+1}(\mathbb C)$
. Let
$\rho : \operatorname {\mathrm {GL}}_{n+1}({\mathbb Z}) \rightarrow \operatorname {\mathrm {GL}}_{n+1}(\mathbb C)$
. Let 
 $P \subseteq {\mathbb R}^n$
 be a polytope. We say that P is G-invariant if for every
$P \subseteq {\mathbb R}^n$
 be a polytope. We say that P is G-invariant if for every 
 $g \in G$
 we have
$g \in G$
 we have 
 $g(P) = P$
. Suppose P is G-invariant. The group G acts by permutation on the lattice points of P and the lattice points of each of its dilates
$g(P) = P$
. Suppose P is G-invariant. The group G acts by permutation on the lattice points of P and the lattice points of each of its dilates 
 $mP \subseteq mM_{\mathbb R} \subseteq {\mathbb R}^{n+1}$
 for
$mP \subseteq mM_{\mathbb R} \subseteq {\mathbb R}^{n+1}$
 for 
 $m \in {\mathbb Z}_{\ge 0}$
. For each
$m \in {\mathbb Z}_{\ge 0}$
. For each 
 $m \in {\mathbb Z}_{\ge 0}$
, we denote by
$m \in {\mathbb Z}_{\ge 0}$
, we denote by  the character of this permutation representation. By convention, we define
 the character of this permutation representation. By convention, we define  to be the trivial character. The equivariant Ehrhart series
 to be the trivial character. The equivariant Ehrhart series 
 $\mathrm {EES_P}$
 and equivariant
$\mathrm {EES_P}$
 and equivariant 
 $H^*$
-series
$H^*$
-series 
 $H^*(P; G)$
 are elements of
$H^*(P; G)$
 are elements of 
 $\mathcal R(G)[[t]]$
, the power series ring with coefficients in
$\mathcal R(G)[[t]]$
, the power series ring with coefficients in 
 $\mathcal R(G)$
 defined by
$\mathcal R(G)$
 defined by 

where 
 $I_{n+1}$
 is the identity matrix of size
$I_{n+1}$
 is the identity matrix of size 
 $n+1$
. The function
$n+1$
. The function 
 $1/\det (I_{n+1} - t\rho )$
 is the class function of G whose evaluation at
$1/\det (I_{n+1} - t\rho )$
 is the class function of G whose evaluation at 
 $g \in G$
 is
$g \in G$
 is 
 $1/\det (I_{n+1} - t\rho (g))$
. We may also think of this as the ‘character’ of a certain representation. Let
$1/\det (I_{n+1} - t\rho (g))$
. We may also think of this as the ‘character’ of a certain representation. Let 
 $V = \mathbb C^{n+1}$
 be the
$V = \mathbb C^{n+1}$
 be the 
 $\mathbb C[G]$
-module associated to
$\mathbb C[G]$
-module associated to 
 $\rho $
; then, by [Reference Stapledon22, Lemma 3.1], we have
$\rho $
; then, by [Reference Stapledon22, Lemma 3.1], we have 
 $$\begin{align*}\sum_{m \ge 0} \mathrm{Sym}^m V t^m = \frac{1}{1 - Vt + \bigwedge^2 Vt^2 - \dots + (-1)^{n+1} \bigwedge^{n+1} V t^{n+1}}, \end{align*}$$
$$\begin{align*}\sum_{m \ge 0} \mathrm{Sym}^m V t^m = \frac{1}{1 - Vt + \bigwedge^2 Vt^2 - \dots + (-1)^{n+1} \bigwedge^{n+1} V t^{n+1}}, \end{align*}$$
where 
 $\mathrm {Sym}^m V$
 is the m-th symmetric power of V and
$\mathrm {Sym}^m V$
 is the m-th symmetric power of V and 
 $\bigwedge ^m V$
 is the m-th exterior power of V. Moreover, the character of the above representation is
$\bigwedge ^m V$
 is the m-th exterior power of V. Moreover, the character of the above representation is 
 $1/\det (I_{n+1} - t\rho )$
.
$1/\det (I_{n+1} - t\rho )$
.
 We often write 
 $H^*$
 for
$H^*$
 for 
 $H^*(P; G)$
 if P and G are clear from context and we write
$H^*(P; G)$
 if P and G are clear from context and we write 
 $H^*_m$
 for the coefficient of
$H^*_m$
 for the coefficient of 
 $t^m$
 in
$t^m$
 in 
 $H^*$
. For each
$H^*$
. For each 
 $g \in G$
, we have that
$g \in G$
, we have that  is the Ehrhart series of the fixed polytope
 is the Ehrhart series of the fixed polytope 
 $P_g = \{x \in P : g(x) = x\}$
. If the coefficients of
$P_g = \{x \in P : g(x) = x\}$
. If the coefficients of 
 $H^*(P; G)$
 are eventually zero, then we call it the equivariant
$H^*(P; G)$
 are eventually zero, then we call it the equivariant 
 $H^*$
-polynomial of P with respect to the action of G.
$H^*$
-polynomial of P with respect to the action of G.
Example 2.4. Assume the setup in Example 2.2 and let 
 $P = \mathrm {Conv}\{(1,0), (0,1)\}$
 be a line segment. Let us compute the equivariant
$P = \mathrm {Conv}\{(1,0), (0,1)\}$
 be a line segment. Let us compute the equivariant 
 $H^*$
-series of P under the action of
$H^*$
-series of P under the action of 
 $G = \{e, \sigma \}$
. The denominator of the equivariant Ehrhart series is given by
$G = \{e, \sigma \}$
. The denominator of the equivariant Ehrhart series is given by 
 $$\begin{align*}\det(I_2 - t\rho(e)) = (1-t)^2 \text{ and } \det(I_2 - t\rho(\sigma)) = \det\begin{pmatrix} 1 & -t \\ -t & 1 \end{pmatrix} = 1-t^2. \end{align*}$$
$$\begin{align*}\det(I_2 - t\rho(e)) = (1-t)^2 \text{ and } \det(I_2 - t\rho(\sigma)) = \det\begin{pmatrix} 1 & -t \\ -t & 1 \end{pmatrix} = 1-t^2. \end{align*}$$
The polytope P and its dilations are shown in Figure 1b. In particular, the 
 $\sigma $
-fixed lattice points are precisely those on the dotted line. The Ehrhart series of
$\sigma $
-fixed lattice points are precisely those on the dotted line. The Ehrhart series of 
 $P_e = P$
 and
$P_e = P$
 and 
 $P_\sigma = \{(1/2, 1/2)\}$
 are given by
$P_\sigma = \{(1/2, 1/2)\}$
 are given by 
 $$ \begin{align*} \mathrm{ES}_{P_e}[t] &= 1 + 2t + 3t^2 + \dots = 1/(1-t)^2 &= 1/\det(I_2 - t\rho(e)),\\ \mathrm{ES}_{P_\sigma}[t] &= 1 + t^2 + t^4 + \dots = 1/(1-t^2) &= 1/\det(I_2 - t\rho(\sigma)). \end{align*} $$
$$ \begin{align*} \mathrm{ES}_{P_e}[t] &= 1 + 2t + 3t^2 + \dots = 1/(1-t)^2 &= 1/\det(I_2 - t\rho(e)),\\ \mathrm{ES}_{P_\sigma}[t] &= 1 + t^2 + t^4 + \dots = 1/(1-t^2) &= 1/\det(I_2 - t\rho(\sigma)). \end{align*} $$
So we have  is the trivial representation.
 is the trivial representation.
Definition 2.5. Fix a group G and a G-invariant polytope P. An element of 
 $\mathcal R(G)$
 is effective if it is a representation, or in other words, it can be expressed as a non-negative integer sum of irreducible representations. We say that the
$\mathcal R(G)$
 is effective if it is a representation, or in other words, it can be expressed as a non-negative integer sum of irreducible representations. We say that the 
 $H^*$
-series
$H^*$
-series 
 $H^* = \sum _{m \ge 0} H^*_m t^m$
 is effective if
$H^* = \sum _{m \ge 0} H^*_m t^m$
 is effective if 
 $H^*_m$
 is effective for each
$H^*_m$
 is effective for each 
 $m \ge 0$
.
$m \ge 0$
.
2.2 Hypersimplices
 Fix 
 $0 < k \le n$
. We denote by
$0 < k \le n$
. We denote by 
 $e_i$
 with
$e_i$
 with 
 $i \in [n]$
 the standard basis vectors of
$i \in [n]$
 the standard basis vectors of 
 ${\mathbb R}^n$
. For each subset
${\mathbb R}^n$
. For each subset 
 $I \in \binom {[n]}{k}$
, we write
$I \in \binom {[n]}{k}$
, we write 
 $e_I = \sum _{i \in I} e_i$
. The hypersimplex
$e_I = \sum _{i \in I} e_i$
. The hypersimplex 
 $\Delta _{k,n} \subseteq {\mathbb R}^n$
 is the convex hull
$\Delta _{k,n} \subseteq {\mathbb R}^n$
 is the convex hull 
 $$\begin{align*}\Delta_{k,n} = \mathrm{Conv}\left\{e_I : I \in \binom{[n]}{k}\right\} \end{align*}$$
$$\begin{align*}\Delta_{k,n} = \mathrm{Conv}\left\{e_I : I \in \binom{[n]}{k}\right\} \end{align*}$$
with vertices given by 
 $e_I$
 for each k-subset I of
$e_I$
 for each k-subset I of 
 $[n]$
. The group
$[n]$
. The group 
 $S_n$
 acts on
$S_n$
 acts on 
 ${\mathbb R}^n$
 by permutation of its coordinates. The action of
${\mathbb R}^n$
 by permutation of its coordinates. The action of 
 $S_n$
 satisfies
$S_n$
 satisfies 
 $\sigma (e_I) = e_{\sigma (I)}$
, hence the action leaves the polytope
$\sigma (e_I) = e_{\sigma (I)}$
, hence the action leaves the polytope 
 $\Delta _{k,n}$
 invariant. Following the setup in the previous section and Example 2.3, the
$\Delta _{k,n}$
 invariant. Following the setup in the previous section and Example 2.3, the 
 $S_n$
-fixed space A is the linear span of
$S_n$
-fixed space A is the linear span of 
 $a = (1,1,\dots , 1)$
. The hypersimplex lies in the affine space
$a = (1,1,\dots , 1)$
. The hypersimplex lies in the affine space 
 ${\mathbb R}^{n-1} := \frac kn a +A^\perp $
. We recall, by [Reference Elia, Kim and Supina9, Theorem 3.60], that
${\mathbb R}^{n-1} := \frac kn a +A^\perp $
. We recall, by [Reference Elia, Kim and Supina9, Theorem 3.60], that 
 $H^*(\Delta _{k,n}; S_n)$
 is effective and polynomial.
$H^*(\Delta _{k,n}; S_n)$
 is effective and polynomial.

Figure 2 Fixed polytopes of the hypersimplex under the action of 
 $S_4$
 in Example 2.6.
$S_4$
 in Example 2.6.
Example 2.6. Let 
 $S_4$
 act by coordinate permutation on
$S_4$
 act by coordinate permutation on 
 ${\mathbb R}^4$
 and consider the
${\mathbb R}^4$
 and consider the 
 $S_4$
-invariant hypersimplex
$S_4$
-invariant hypersimplex 
 $\Delta _{2,4}$
, which has vertices
$\Delta _{2,4}$
, which has vertices 
 $$ \begin{align*} e_{12} = (1,1,0,0),\, e_{13} = (1,0,1,0),\, e_{14} = (1,0,0,1),\\ e_{23} = (0,1,1,0),\, e_{24} = (0,1,0,1),\, e_{34} = (0,0,1,1), \end{align*} $$
$$ \begin{align*} e_{12} = (1,1,0,0),\, e_{13} = (1,0,1,0),\, e_{14} = (1,0,0,1),\\ e_{23} = (0,1,1,0),\, e_{24} = (0,1,0,1),\, e_{34} = (0,0,1,1), \end{align*} $$
as shown in Figure 2a. The hypersimplex 
 $\Delta _{2,4}$
 is a three-dimensional polytope that lies in the
$\Delta _{2,4}$
 is a three-dimensional polytope that lies in the 
 $S_4$
-invariant affine subspace
$S_4$
-invariant affine subspace 
 $M_{\mathbb R} = \{x \in {\mathbb R}^4 : x_1 + x_2 + x_3 + x_4 = 2\} \cong {\mathbb R}^3$
. For each element
$M_{\mathbb R} = \{x \in {\mathbb R}^4 : x_1 + x_2 + x_3 + x_4 = 2\} \cong {\mathbb R}^3$
. For each element 
 $\sigma \in S_4$
, we obtain a fixed polytope of
$\sigma \in S_4$
, we obtain a fixed polytope of 
 $\Delta _{2,4}$
, which is its intersection with the space of
$\Delta _{2,4}$
, which is its intersection with the space of 
 $\sigma $
-fixed points. For example, if
$\sigma $
-fixed points. For example, if 
 $\sigma = (1 \ 2)$
, then the fixed polytope
$\sigma = (1 \ 2)$
, then the fixed polytope 
 $(\Delta _{2,4})_{\sigma } = \{x \in \Delta _{2,4} : \sigma (x) = x \}$
 is the convex hull of the points
$(\Delta _{2,4})_{\sigma } = \{x \in \Delta _{2,4} : \sigma (x) = x \}$
 is the convex hull of the points 
 $$\begin{align*}e_{12} = (1,1,0,0),\, e_{34} = (0,0,1,1),\, \frac 12(e_{13} + e_{23}) = \left(\frac 12, \frac 12, 1, 0\right),\, \frac 12(e_{14} + e_{24}) = \left(\frac 12, \frac 12, 0, 1\right), \end{align*}$$
$$\begin{align*}e_{12} = (1,1,0,0),\, e_{34} = (0,0,1,1),\, \frac 12(e_{13} + e_{23}) = \left(\frac 12, \frac 12, 1, 0\right),\, \frac 12(e_{14} + e_{24}) = \left(\frac 12, \frac 12, 0, 1\right), \end{align*}$$
see Figure 2b. The group 
 $S_4$
 has five conjugacy classes indexed by cycle types. The equivariant
$S_4$
 has five conjugacy classes indexed by cycle types. The equivariant 
 $H^*$
-series is a quadratic polynomial
$H^*$
-series is a quadratic polynomial 
 $H^* = H^*_0 + H^*_1 t + H^*_2 t^2$
 with coefficients shown in Table 1. Theorem 5.1 shows that
$H^* = H^*_0 + H^*_1 t + H^*_2 t^2$
 with coefficients shown in Table 1. Theorem 5.1 shows that 
 $H^*_1 = \rho _2 - \rho _1$
 where
$H^*_1 = \rho _2 - \rho _1$
 where 
 $\rho _2$
 is the permutation action of
$\rho _2$
 is the permutation action of 
 $S_4$
 on the
$S_4$
 on the 
 $2$
-subsets of
$2$
-subsets of 
 $[4]$
 and
$[4]$
 and 
 $\rho _1$
 is the natural representation given by the action of
$\rho _1$
 is the natural representation given by the action of 
 $S_4$
 on
$S_4$
 on 
 $[4]$
.
$[4]$
.
Table 1 Coefficients of 
 $H^*({\Delta _{2,4}, S_4)}$
 and characters in Example 2.6.
$H^*({\Delta _{2,4}, S_4)}$
 and characters in Example 2.6.

 Let 
 $\sigma = (1 \ 2) \in S_4$
 be a transposition, which has cycle type
$\sigma = (1 \ 2) \in S_4$
 be a transposition, which has cycle type 
 $(2,1,1)$
. Let
$(2,1,1)$
. Let 
 $\rho $
 denote the representation of
$\rho $
 denote the representation of 
 $S_4$
 by permutation matrices. The evaluation
$S_4$
 by permutation matrices. The evaluation 
 $H^*(\sigma ) = 1 + t^2$
 tells us that the Ehrhart series of the fixed polytope
$H^*(\sigma ) = 1 + t^2$
 tells us that the Ehrhart series of the fixed polytope 
 $(\Delta _{2,4})_\sigma $
, see Figure Figure 2b, is given by
$(\Delta _{2,4})_\sigma $
, see Figure Figure 2b, is given by 
 $$\begin{align*}\mathrm{ES}_{(\Delta_{2,4})_\sigma}[t] = \mathrm{EES}_{\Delta_{2,4}}[t](\sigma) = \frac{1 + t^2}{\det(I_4 - t\rho(\sigma))} = \frac{1 + t^2}{(1-t)^2(1 - t^2)} = \frac{1+2t+2t^2+2t^3+t^4}{(1-t^2)^3}. \end{align*}$$
$$\begin{align*}\mathrm{ES}_{(\Delta_{2,4})_\sigma}[t] = \mathrm{EES}_{\Delta_{2,4}}[t](\sigma) = \frac{1 + t^2}{\det(I_4 - t\rho(\sigma))} = \frac{1 + t^2}{(1-t)^2(1 - t^2)} = \frac{1+2t+2t^2+2t^3+t^4}{(1-t^2)^3}. \end{align*}$$
2.3 Stirling numbers of the second kind
 In Section 4.2, we count families of DOSPs. For this, we require Stirling numbers of the second kind, which, for brevity, we refer to simply as Stirling numbers. Given non-negative integers n and k, the Stirling number  is the number of partitions of the set
 is the number of partitions of the set 
 $[n]$
 into k nonempty unlabelled parts. For example, the set
$[n]$
 into k nonempty unlabelled parts. For example, the set 
 $[3]$
 is partitioned into two nonempty parts in three ways:
$[3]$
 is partitioned into two nonempty parts in three ways: 
 $ \{1,23\},\, \{2,13\},\, \{3,12\}. $
 Therefore
$ \{1,23\},\, \{2,13\},\, \{3,12\}. $
 Therefore  . For further background on Stirling numbers, we refer the reader to [Reference Stanley20].
. For further background on Stirling numbers, we refer the reader to [Reference Stanley20].
Proposition 2.7 [Reference Stanley20, Section 1.9].
The Stirling numbers satisfy the relationship

for all x, where 
 $(x)_k=\prod _{i=0}^{k-1} (x-i)$
 denotes the k-th falling factorial.
$(x)_k=\prod _{i=0}^{k-1} (x-i)$
 denotes the k-th falling factorial.
3 Coefficients of the equivariant 
 $H^*$
-polynomial
$H^*$
-polynomial
 The main result of this section is Theorem 3.3, which is a formula for the coefficients of the equivariant 
 $H^*$
-polynomial of the hypersimplex
$H^*$
-polynomial of the hypersimplex 
 $\Delta _{k,n}$
 under the action of
$\Delta _{k,n}$
 under the action of 
 $S_n$
. The formula is given in terms of the following combinatorial families of objects.
$S_n$
. The formula is given in terms of the following combinatorial families of objects.
Definition 3.1. Fix 
 $\sigma \in S_n$
 a permutation with cycle type
$\sigma \in S_n$
 a permutation with cycle type 
 $(s_1, \dots , s_r)$
. For each
$(s_1, \dots , s_r)$
. For each 
 $0 < k < n$
 and
$0 < k < n$
 and 
 $m \ge 0$
, we define the set of functions
$m \ge 0$
, we define the set of functions 
 $$\begin{align*}{\Phi}_k(\sigma, m) = \left\{ f \colon [r] \rightarrow \{0,1,\dots,k-1\} : \sum_{i = 1}^r f(i)s_i = m \right\}. \end{align*}$$
$$\begin{align*}{\Phi}_k(\sigma, m) = \left\{ f \colon [r] \rightarrow \{0,1,\dots,k-1\} : \sum_{i = 1}^r f(i)s_i = m \right\}. \end{align*}$$
By convention, we define 
 ${\Phi }_k(\sigma , m) = \emptyset $
 for all
${\Phi }_k(\sigma , m) = \emptyset $
 for all 
 $m < 0$
. For each
$m < 0$
. For each 
 $h \ge 0$
, we define the set
$h \ge 0$
, we define the set 
 $$\begin{align*}\mathcal I_h = \left\{ I = (I_1,I_2, \dots, I_{k-1}) \in {\mathbb Z}^{k-1}_{\ge 0} : \sum_{i = 1}^{k-1} i \cdot I_i = h \right\}. \end{align*}$$
$$\begin{align*}\mathcal I_h = \left\{ I = (I_1,I_2, \dots, I_{k-1}) \in {\mathbb Z}^{k-1}_{\ge 0} : \sum_{i = 1}^{k-1} i \cdot I_i = h \right\}. \end{align*}$$
For each 
 $I \in \mathcal I_h$
, we write
$I \in \mathcal I_h$
, we write 
 $|I| = I_1 + I_2 + \dots + I_{k-1}$
.
$|I| = I_1 + I_2 + \dots + I_{k-1}$
.
 We observe that the map 
 $\sigma \mapsto |{\Phi }_k(\sigma , m)|$
 is a permutation character of
$\sigma \mapsto |{\Phi }_k(\sigma , m)|$
 is a permutation character of 
 $S_n$
.
$S_n$
.
Proposition 3.2. Fix 
 $0 < k < n$
 and
$0 < k < n$
 and 
 $m \ge 0$
. Let
$m \ge 0$
. Let  be the permutation character of
 be the permutation character of 
 $S_n$
 acting on the set of functions
$S_n$
 acting on the set of functions 
 $\{f \colon [n] \rightarrow \{0,1, \dots , k-1\} : \sum _{i = 1}^n f(i) = m\}$
 by
$\{f \colon [n] \rightarrow \{0,1, \dots , k-1\} : \sum _{i = 1}^n f(i) = m\}$
 by 
 $(\sigma \cdot f)(i) = f(\sigma ^{-1}(i))$
. Then
$(\sigma \cdot f)(i) = f(\sigma ^{-1}(i))$
. Then  .
.
Proof. Follows immediately from the definition.
 The main result of this section is the following formula for the coefficients of 
 $H^*(\Delta _{k,n}; S_n)$
 in terms of the permutation characters
$H^*(\Delta _{k,n}; S_n)$
 in terms of the permutation characters 
 $|{\Phi }_k(\sigma , m)|$
 and the elements of
$|{\Phi }_k(\sigma , m)|$
 and the elements of 
 $\mathcal I_h$
.
$\mathcal I_h$
.
Theorem 3.3. Fix 
 $0 < k < n$
 and
$0 < k < n$
 and 
 $\sigma \in S_n$
. For each
$\sigma \in S_n$
. For each 
 $i \in [n]$
, denote by
$i \in [n]$
, denote by 
 $\lambda _i$
 the number of cycles of
$\lambda _i$
 the number of cycles of 
 $\sigma $
 of length i. For each
$\sigma $
 of length i. For each 
 $m \ge 0$
, let
$m \ge 0$
, let 
 $H^*_m$
 be the coefficient of
$H^*_m$
 be the coefficient of 
 $t^m$
 in
$t^m$
 in 
 $H^*(\Delta _{k,n}; S_n)[t]$
. Then
$H^*(\Delta _{k,n}; S_n)[t]$
. Then 
 $$\begin{align*}H^*_m(\sigma) = \sum_{h = 0}^{k-1}\left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right)|{\Phi}_{k-h}(\sigma, m(k-h) - h)|. \end{align*}$$
$$\begin{align*}H^*_m(\sigma) = \sum_{h = 0}^{k-1}\left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right)|{\Phi}_{k-h}(\sigma, m(k-h) - h)|. \end{align*}$$
 We will give a proof of the theorem at the end of the section after we have introduced notation and developed the necessary tools. For now, an immediate consequence of this theorem is a combinatorial proof that the 
 $H^*$
-series
$H^*$
-series 
 $H^*(\Delta _{k,n}; S_n)$
 is a polynomial and, moreover, we can write down its degree.
$H^*(\Delta _{k,n}; S_n)$
 is a polynomial and, moreover, we can write down its degree.
Corollary 3.4. Fix 
 $0 < k < n$
. We have that
$0 < k < n$
. We have that 
 $H^*(\Delta _{k,n}; S_n)$
 is a polynomial of degree
$H^*(\Delta _{k,n}; S_n)$
 is a polynomial of degree 
 $\lfloor (k-1)n/k \rfloor $
.
$\lfloor (k-1)n/k \rfloor $
.
Proof. Let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1, \dots , s_r)$
 and fix
$(s_1, \dots , s_r)$
 and fix 
 $h \ge 0$
. For any function
$h \ge 0$
. For any function 
 $f \colon [r] \rightarrow \{0,1, \dots , k-h-1\}$
, we have
$f \colon [r] \rightarrow \{0,1, \dots , k-h-1\}$
, we have 
 $\sum _{i = 1}^r f(i)s_i \le (k-h-1)n$
. If m satisfies
$\sum _{i = 1}^r f(i)s_i \le (k-h-1)n$
. If m satisfies 
 $(k-1)n < km$
, then we have
$(k-1)n < km$
, then we have 
 $$\begin{align*}\sum_{i = 1}^r f(i)s_i \le (k-h-1)n < \frac{(k-h-1)km}{k-1} = km - \frac{khm}{k-1} = m(k - h) - \frac{hm}{k-1} < m(k-h) -h, \end{align*}$$
$$\begin{align*}\sum_{i = 1}^r f(i)s_i \le (k-h-1)n < \frac{(k-h-1)km}{k-1} = km - \frac{khm}{k-1} = m(k - h) - \frac{hm}{k-1} < m(k-h) -h, \end{align*}$$
hence the set 
 ${\Phi }_k(\sigma , m(k-h)-h)$
 is empty. So, by Theorem 3.3, the coefficient
${\Phi }_k(\sigma , m(k-h)-h)$
 is empty. So, by Theorem 3.3, the coefficient 
 $H^*_m(\sigma ) = 0$
 for all
$H^*_m(\sigma ) = 0$
 for all 
 $m> (k-1)n/k$
. In particular, the
$m> (k-1)n/k$
. In particular, the 
 $H^*$
-series is a polynomial.
$H^*$
-series is a polynomial.
 On the other hand, let 
 $e \in S_n$
 be the identity. The set
$e \in S_n$
 be the identity. The set 
 ${\Phi }_{k}(e, mk)$
 is nonempty if and only if m satisfies
${\Phi }_{k}(e, mk)$
 is nonempty if and only if m satisfies 
 $(k-1)n \ge k m$
. Therefore, the degree of the
$(k-1)n \ge k m$
. Therefore, the degree of the 
 $H^*$
-polynomial is at least
$H^*$
-polynomial is at least 
 $\lfloor (k-1)n/k \rfloor $
, which concludes the proof of the result.
$\lfloor (k-1)n/k \rfloor $
, which concludes the proof of the result.
3.1 Katzman’s method
 In this section, we apply the method used by Katzman [Reference Katzman13] to obtain a formula for the coefficients of the equivariant 
 $H^*$
-series. The main result of this section is Lemma 3.5, which is the first step towards Theorem 3.3. We introduce two pieces of useful notation. Given
$H^*$
-series. The main result of this section is Lemma 3.5, which is the first step towards Theorem 3.3. We introduce two pieces of useful notation. Given 
 $\sigma \in S_n$
 with cycle type
$\sigma \in S_n$
 with cycle type 
 $(s_1, \dots , s_r)$
, we define the formal power series
$(s_1, \dots , s_r)$
, we define the formal power series 
 $$\begin{align*}u^\sigma = \sum_{i \ge 0} u^\sigma_i t^i := \prod_{i = 1}^k (1 + t^{s_i} + t^{2s_i} + \dots) = \prod_{i = 1}^k \frac{1}{1 - t^{s_i}} \in {\mathbb Z}[[t]]. \end{align*}$$
$$\begin{align*}u^\sigma = \sum_{i \ge 0} u^\sigma_i t^i := \prod_{i = 1}^k (1 + t^{s_i} + t^{2s_i} + \dots) = \prod_{i = 1}^k \frac{1}{1 - t^{s_i}} \in {\mathbb Z}[[t]]. \end{align*}$$
If 
 $\sigma $
 is clear from context, then we write u for
$\sigma $
 is clear from context, then we write u for 
 $u^\sigma $
 and
$u^\sigma $
 and 
 $u_i$
 for
$u_i$
 for 
 $u^\sigma _i$
. For each subset
$u^\sigma _i$
. For each subset 
 $S \subseteq [r]$
, we write
$S \subseteq [r]$
, we write 
 $\Sigma ^\sigma S = \sum _{i \in S} s_i$
. If the permutation
$\Sigma ^\sigma S = \sum _{i \in S} s_i$
. If the permutation 
 $\sigma $
 is clear from context then we write
$\sigma $
 is clear from context then we write 
 $\Sigma S$
 for
$\Sigma S$
 for 
 $\Sigma ^\sigma S$
.
$\Sigma ^\sigma S$
.
Lemma 3.5. Fix 
 $0 < k < n$
 and
$0 < k < n$
 and 
 $\sigma \in S_n$
 with cycle type
$\sigma \in S_n$
 with cycle type 
 $(s_1, \dots , s_r)$
. For each
$(s_1, \dots , s_r)$
. For each 
 $i \in [n]$
, let
$i \in [n]$
, let 
 $\lambda _i$
 be the number of length i cycles of
$\lambda _i$
 be the number of length i cycles of 
 $\sigma $
. For each
$\sigma $
. For each 
 $m \ge 0$
, the coefficient of
$m \ge 0$
, the coefficient of 
 $t^m$
 in
$t^m$
 in 
 $H^*(\Delta _{k,n}; S_n)[t]$
 is
$H^*(\Delta _{k,n}; S_n)[t]$
 is 
 $$\begin{align*}H^*_m(\sigma) = \sum_{S \subseteq [r]} (-1)^{|S|} \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) u_{(m-\Sigma S)(k-h) - h}. \end{align*}$$
$$\begin{align*}H^*_m(\sigma) = \sum_{S \subseteq [r]} (-1)^{|S|} \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) u_{(m-\Sigma S)(k-h) - h}. \end{align*}$$
Proof. By conjugating 
 $\sigma $
, we may assume without loss of generality that
$\sigma $
, we may assume without loss of generality that 
 $$\begin{align*}\sigma = (1 \ 2 \ \dots \ s_1)(s_1+1 \ s_1+2 \ \dots \ s_1+s_2 ) \dots (n-s_r+1 \ n-s_r+2 \ \dots \ n). \end{align*}$$
$$\begin{align*}\sigma = (1 \ 2 \ \dots \ s_1)(s_1+1 \ s_1+2 \ \dots \ s_1+s_2 ) \dots (n-s_r+1 \ n-s_r+2 \ \dots \ n). \end{align*}$$
Fix 
 $d \ge 0$
. We have that
$d \ge 0$
. We have that 
 $(d\Delta _{k,n})_\sigma \cap {\mathbb Z}^n$
 is equal to
$(d\Delta _{k,n})_\sigma \cap {\mathbb Z}^n$
 is equal to 
 $$\begin{align*}\left\{ ( \underbrace{x_1, \dots, x_1}_{s_1}, \underbrace{x_2, \dots, x_2}_{s_2}, \dots, \underbrace{x_r, \dots, x_r}_{s_r}) \in {\mathbb Z}^n : \sum_{i = 1}^r x_i s_i = k d,\, 0 \le x_i \le d \text{ for all } i \in [r] \right\}. \end{align*}$$
$$\begin{align*}\left\{ ( \underbrace{x_1, \dots, x_1}_{s_1}, \underbrace{x_2, \dots, x_2}_{s_2}, \dots, \underbrace{x_r, \dots, x_r}_{s_r}) \in {\mathbb Z}^n : \sum_{i = 1}^r x_i s_i = k d,\, 0 \le x_i \le d \text{ for all } i \in [r] \right\}. \end{align*}$$
So there is a bijection between the set solutions 
 $(x_1, x_2, \dots , x_r) \in \{0,1, \dots , d \}^r$
 to the equation
$(x_1, x_2, \dots , x_r) \in \{0,1, \dots , d \}^r$
 to the equation 
 $\sum _{i = 1}^r x_i s_i = kd$
 and lattice points
$\sum _{i = 1}^r x_i s_i = kd$
 and lattice points 
 $(d \Delta _{k,n})_\sigma \cap {\mathbb Z}^n$
. Consider the polynomial
$(d \Delta _{k,n})_\sigma \cap {\mathbb Z}^n$
. Consider the polynomial 
 $$\begin{align*}f_d(t) = \prod_{i = 1}^r (1 + t^{s_i} + t^{2s_i} + \dots + t^{ds_i}) = \prod_{i = 1}^r \frac{1 - t^{(d+1)s_i}}{1 - t^{s_i}}. \end{align*}$$
$$\begin{align*}f_d(t) = \prod_{i = 1}^r (1 + t^{s_i} + t^{2s_i} + \dots + t^{ds_i}) = \prod_{i = 1}^r \frac{1 - t^{(d+1)s_i}}{1 - t^{s_i}}. \end{align*}$$
For each solution 
 $(x_1, \dots , x_r)$
 to the above equation, we have a term
$(x_1, \dots , x_r)$
 to the above equation, we have a term 
 $t^{kd} = t^{x_1 s_1} t^{x_2 s_2} \dots t^{x_r s_r}$
 in the expansion of
$t^{kd} = t^{x_1 s_1} t^{x_2 s_2} \dots t^{x_r s_r}$
 in the expansion of 
 $f_d(t)$
. Moreover, each term
$f_d(t)$
. Moreover, each term 
 $t^{kd}$
 in the expansion of
$t^{kd}$
 in the expansion of 
 $f_d(t)$
 arises from such a solution. So we have that
$f_d(t)$
 arises from such a solution. So we have that 
 $|(d \Delta _{k,n})_\sigma \cap {\mathbb Z}^n|$
 is equal to the coefficient of
$|(d \Delta _{k,n})_\sigma \cap {\mathbb Z}^n|$
 is equal to the coefficient of 
 $t^{k d}$
 in
$t^{k d}$
 in 
 $f_d(t)$
, which we denote by
$f_d(t)$
, which we denote by 
 $[f_d]_{k d}$
.
$[f_d]_{k d}$
.
 For each 
 $s_i \ge k$
, we have that
$s_i \ge k$
, we have that 
 $t^{(d+1)s_i}$
 does not divide
$t^{(d+1)s_i}$
 does not divide 
 $t^{k d}$
. It follows that
$t^{k d}$
. It follows that 
 $[f_d]_{k d}$
 is equal to the coefficient of
$[f_d]_{k d}$
 is equal to the coefficient of 
 $t^{k d}$
 in the formal power series:
$t^{k d}$
 in the formal power series: 
 $$ \begin{align*} [f_d]_{k d} &= \left[ \prod_{j = 1}^{k - 1} (1 - t^{(d+1)j})^{\lambda_j} \prod_{i = 1}^r \frac{1}{1 - t^{s_i}} \right]_{k d} \\ &= \left[ \prod_{j = 1}^{k - 1} \sum_{h = 0}^{\lambda_j} (-1)^h \binom{\lambda_j}{h} t^{(d+1)jh} \cdot u \right]_{k d}\\ &= \left[ \sum_{h = 0}^{k - 1} \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \binom{\lambda_2}{I_2} \cdots \binom{\lambda_{\ell-1}}{I_{k-1}} t^{(d+1)h} \cdot u \right]_{k d} \\ &= \sum_{h = 0}^{k - 1} \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \binom{\lambda_2}{I_2} \cdots \binom{\lambda_{\ell-1}}{I_{k-1}} u_{kd - (d+1)h}. \end{align*} $$
$$ \begin{align*} [f_d]_{k d} &= \left[ \prod_{j = 1}^{k - 1} (1 - t^{(d+1)j})^{\lambda_j} \prod_{i = 1}^r \frac{1}{1 - t^{s_i}} \right]_{k d} \\ &= \left[ \prod_{j = 1}^{k - 1} \sum_{h = 0}^{\lambda_j} (-1)^h \binom{\lambda_j}{h} t^{(d+1)jh} \cdot u \right]_{k d}\\ &= \left[ \sum_{h = 0}^{k - 1} \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \binom{\lambda_2}{I_2} \cdots \binom{\lambda_{\ell-1}}{I_{k-1}} t^{(d+1)h} \cdot u \right]_{k d} \\ &= \sum_{h = 0}^{k - 1} \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \binom{\lambda_2}{I_2} \cdots \binom{\lambda_{\ell-1}}{I_{k-1}} u_{kd - (d+1)h}. \end{align*} $$
So the Ehrhart series of 
 $(\Delta _{k,n})_\sigma $
 is given by
$(\Delta _{k,n})_\sigma $
 is given by 
 $$\begin{align*}\sum_{d \ge 0} [f_d]_{kd} t^d = \sum_{d \ge 0} \left( \sum_{h = 0}^{k - 1} \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} u_{k d - h(d+1)} \right) t^d = \frac{H^*[t](\sigma)}{\prod_{i = 1}^r(1 - t^{s_i})}, \end{align*}$$
$$\begin{align*}\sum_{d \ge 0} [f_d]_{kd} t^d = \sum_{d \ge 0} \left( \sum_{h = 0}^{k - 1} \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} u_{k d - h(d+1)} \right) t^d = \frac{H^*[t](\sigma)}{\prod_{i = 1}^r(1 - t^{s_i})}, \end{align*}$$
where the right-most equality follows from definition of the equivariant 
 $H^*$
-series and the denominator comes from
$H^*$
-series and the denominator comes from 
 $\det (I_n - t\rho (\sigma ))$
 where
$\det (I_n - t\rho (\sigma ))$
 where 
 $\rho (\sigma )$
 is the permutation matrix of
$\rho (\sigma )$
 is the permutation matrix of 
 $\sigma $
 and
$\sigma $
 and 
 $I_n$
 is the identity matrix of size n. So, by clearing the denominator, we obtain a formula for the coefficients of equivariant
$I_n$
 is the identity matrix of size n. So, by clearing the denominator, we obtain a formula for the coefficients of equivariant 
 $H^*$
-series. For each
$H^*$
-series. For each 
 $m \ge 0$
, we have
$m \ge 0$
, we have 
 $$ \begin{align*} H^*_m(\sigma) &= \left[ \prod_{i = 1}^r (1 - t^{s_i}) \sum_{d \ge 0} \left( \sum_{h = 0}^{k - 1} \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} u_{kd - h(d+1)} \right) t^d \right]_m \\ &= \sum_{S \subseteq [r]} (-1)^{|S|} \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{i_{\ell-1}} \right) u_{(m-\Sigma S)(k-h) - h}. \end{align*} $$
$$ \begin{align*} H^*_m(\sigma) &= \left[ \prod_{i = 1}^r (1 - t^{s_i}) \sum_{d \ge 0} \left( \sum_{h = 0}^{k - 1} \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} u_{kd - h(d+1)} \right) t^d \right]_m \\ &= \sum_{S \subseteq [r]} (-1)^{|S|} \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{i_{\ell-1}} \right) u_{(m-\Sigma S)(k-h) - h}. \end{align*} $$
This concludes the proof of the result.
We illustrate some of the main steps of the above proof with the following example.
Example 3.6. Let 
 $k = 3$
. In this case, we consider the sets
$k = 3$
. In this case, we consider the sets 
 $\mathcal I_0 = \{(0,0)\}$
,
$\mathcal I_0 = \{(0,0)\}$
, 
 $\mathcal I_1 = \{(1,0)\}$
, and
$\mathcal I_1 = \{(1,0)\}$
, and 
 $\mathcal I_2 = \{(2,0), (0,1)\}$
. So, we have
$\mathcal I_2 = \{(2,0), (0,1)\}$
. So, we have 
 $$\begin{align*}[f_d]_{3d} = \left[ \left( 1 - \lambda_1 t^{d+1} + \left( \binom{\lambda_1}{2} - \lambda_2 \right) t^{2(d+1)} \right)u^\sigma \right]_{3 d} = u_{3d} - \lambda_1 u_{2d-1} + \left( \binom{\lambda_1}{2} - \lambda_2 \right) u_{d-2}. \end{align*}$$
$$\begin{align*}[f_d]_{3d} = \left[ \left( 1 - \lambda_1 t^{d+1} + \left( \binom{\lambda_1}{2} - \lambda_2 \right) t^{2(d+1)} \right)u^\sigma \right]_{3 d} = u_{3d} - \lambda_1 u_{2d-1} + \left( \binom{\lambda_1}{2} - \lambda_2 \right) u_{d-2}. \end{align*}$$
The Ehrhart series of 
 $(\Delta _{3,n})_\sigma $
 is given by
$(\Delta _{3,n})_\sigma $
 is given by 
 $$\begin{align*}\frac{H^*[t](\sigma)}{\prod_{i = 1}^r (1-t^{s_i})} = \sum_{d \ge 0} \left( u_{3d} - \lambda_1 u_{2d - 1} + \left( \binom{\lambda_1}{2} - \lambda_2 \right) u_{d-2} \right) t^d. \end{align*}$$
$$\begin{align*}\frac{H^*[t](\sigma)}{\prod_{i = 1}^r (1-t^{s_i})} = \sum_{d \ge 0} \left( u_{3d} - \lambda_1 u_{2d - 1} + \left( \binom{\lambda_1}{2} - \lambda_2 \right) u_{d-2} \right) t^d. \end{align*}$$
The coefficient of 
 $t^m$
 in the
$t^m$
 in the 
 $H^*$
-series is given by
$H^*$
-series is given by 
 $$\begin{align*}H^*_m(\sigma) = \sum_{S \subseteq [r]} (-1)^{|S|} \left( u_{3(m - \Sigma S)} - \lambda_1 u_{2(m - \Sigma S) - 1} + \left( \binom{\lambda_1}{2} - \lambda_2 \right) u_{m - \Sigma S - 2} \right). \end{align*}$$
$$\begin{align*}H^*_m(\sigma) = \sum_{S \subseteq [r]} (-1)^{|S|} \left( u_{3(m - \Sigma S)} - \lambda_1 u_{2(m - \Sigma S) - 1} + \left( \binom{\lambda_1}{2} - \lambda_2 \right) u_{m - \Sigma S - 2} \right). \end{align*}$$
3.2 Permutation-representation interpretation
 In this section we prove Theorem 3.3. We recall the sets of functions 
 ${\Phi }_{k}(\sigma , m)$
 from Definition 3.1. We use these sets of functions to give an interpretation of terms appearing in formula for
${\Phi }_{k}(\sigma , m)$
 from Definition 3.1. We use these sets of functions to give an interpretation of terms appearing in formula for 
 $H^*_m$
 in Lemma 3.5.
$H^*_m$
 in Lemma 3.5.
Lemma 3.7. Fix 
 $0 < k < n$
 and
$0 < k < n$
 and 
 $\sigma \in S_n$
 with cycle type
$\sigma \in S_n$
 with cycle type 
 $(s_1, \dots , s_r)$
. With our usual notation, we have
$(s_1, \dots , s_r)$
. With our usual notation, we have 
 $$\begin{align*}\sum_{S \subseteq [r]} (-1)^{|S|} u_{m - k \Sigma S} = |{\Phi}_k(\sigma, m)|. \end{align*}$$
$$\begin{align*}\sum_{S \subseteq [r]} (-1)^{|S|} u_{m - k \Sigma S} = |{\Phi}_k(\sigma, m)|. \end{align*}$$
Proof. We prove the result by induction on r. For the base case, assume 
 $r = 1$
, that is,
$r = 1$
, that is, 
 $\sigma $
 is an n-cycle. We have
$\sigma $
 is an n-cycle. We have 
 $\Sigma \emptyset = 0$
 and
$\Sigma \emptyset = 0$
 and 
 $\Sigma \{1\} = n$
, and
$\Sigma \{1\} = n$
, and 
 $ u = 1 + t^n + t^{2n} + \dots = 1/(1-t^n). $
 Therefore the left-hand sum is given by
$ u = 1 + t^n + t^{2n} + \dots = 1/(1-t^n). $
 Therefore the left-hand sum is given by 
 $$\begin{align*}\sum_{S \subseteq [r]} (-1)^{|S|} u_{m - k \Sigma S} = u_m - u_{m - k n} = \begin{cases} 1 & \text{if } m \in \{0, n, 2n, \dots, (k-1)n \}, \\ 0 & \text{otherwise}. \end{cases} \end{align*}$$
$$\begin{align*}\sum_{S \subseteq [r]} (-1)^{|S|} u_{m - k \Sigma S} = u_m - u_{m - k n} = \begin{cases} 1 & \text{if } m \in \{0, n, 2n, \dots, (k-1)n \}, \\ 0 & \text{otherwise}. \end{cases} \end{align*}$$
On the other hand, there are exactly k functions 
 $f \colon [r] \rightarrow \{0,1,\dots , k-1 \}$
 and any such function f satisfies
$f \colon [r] \rightarrow \{0,1,\dots , k-1 \}$
 and any such function f satisfies 
 $\sum _{i} f(i)s_i = f(1)n$
. Therefore
$\sum _{i} f(i)s_i = f(1)n$
. Therefore 
 $$\begin{align*}|{\Phi}_k(\sigma, m)| = \begin{cases} 1 & \text{if } m \in \{0, n, 2n, \dots, (k-1)n \} \\ 0 & \text{otherwise} \end{cases} = \sum_{S \subseteq [r]} (-1)^{|S|} u_{m - k \Sigma S}, \end{align*}$$
$$\begin{align*}|{\Phi}_k(\sigma, m)| = \begin{cases} 1 & \text{if } m \in \{0, n, 2n, \dots, (k-1)n \} \\ 0 & \text{otherwise} \end{cases} = \sum_{S \subseteq [r]} (-1)^{|S|} u_{m - k \Sigma S}, \end{align*}$$
and we are done with the base case.
 For the induction step, let 
 $\sigma $
 be a permutation with cycle type
$\sigma $
 be a permutation with cycle type 
 $(s_1, s_2, \dots , s_{r+1})$
 and assume that the result holds for any permutation with r disjoint cycles. Let
$(s_1, s_2, \dots , s_{r+1})$
 and assume that the result holds for any permutation with r disjoint cycles. Let 
 $\tau $
 be a permutation with cycle type
$\tau $
 be a permutation with cycle type 
 $(s_1, \dots , s_r)$
. For ease of notation, we define
$(s_1, \dots , s_r)$
. For ease of notation, we define 
 $s := s_{r+1}$
. We note that
$s := s_{r+1}$
. We note that 
 $u^\sigma = u^\tau (1 + t^s + t^{2s} + \dots )$
, so it follows that
$u^\sigma = u^\tau (1 + t^s + t^{2s} + \dots )$
, so it follows that 
 $u^\sigma _i = \sum _{j \ge 0} u^\tau _{i - sj}$
. Then we have the following chain of equalities:
$u^\sigma _i = \sum _{j \ge 0} u^\tau _{i - sj}$
. Then we have the following chain of equalities: 
 $$ \begin{align*} \sum_{S \subseteq [r+1]} (-1)^{|S|} u^\sigma_{m - k \Sigma^\sigma S} &= \sum_{S \subseteq [r]} (-1)^{|S|} \left( u^\sigma_{m - k \Sigma^\sigma S} - u^\sigma_{m - k \Sigma^\sigma S - k s} \right)\\&= \sum_{S \subseteq [r]} (-1)^{|S|} \left( u^\sigma_{m - k \Sigma^\tau S} - u^\sigma_{m - k \Sigma^\tau S - k s} \right) \\&= \sum_{S \subseteq [r]} (-1)^{|S|} \sum_{j \ge 0} \left( u^\tau_{m - k \Sigma^\tau S - sj} - u^\tau_{m - k \Sigma^\tau S - s(j + k)} \right) \\&= \sum_{S \subseteq [k]} (-1)^{|S|} \sum_{j = 0}^{k-1} u^\tau_{m - k \Sigma^\tau S - sj} = \sum_{j = 0}^{k-1} |{\Phi}_k(\tau, m - sj)|. \end{align*} $$
$$ \begin{align*} \sum_{S \subseteq [r+1]} (-1)^{|S|} u^\sigma_{m - k \Sigma^\sigma S} &= \sum_{S \subseteq [r]} (-1)^{|S|} \left( u^\sigma_{m - k \Sigma^\sigma S} - u^\sigma_{m - k \Sigma^\sigma S - k s} \right)\\&= \sum_{S \subseteq [r]} (-1)^{|S|} \left( u^\sigma_{m - k \Sigma^\tau S} - u^\sigma_{m - k \Sigma^\tau S - k s} \right) \\&= \sum_{S \subseteq [r]} (-1)^{|S|} \sum_{j \ge 0} \left( u^\tau_{m - k \Sigma^\tau S - sj} - u^\tau_{m - k \Sigma^\tau S - s(j + k)} \right) \\&= \sum_{S \subseteq [k]} (-1)^{|S|} \sum_{j = 0}^{k-1} u^\tau_{m - k \Sigma^\tau S - sj} = \sum_{j = 0}^{k-1} |{\Phi}_k(\tau, m - sj)|. \end{align*} $$
To conclude the proof, we note that there is a natural bijection between the sets

It follows that 
 $|{\Phi }_k(\sigma , m)| = \sum _{j = 0}^{k-1} |{\Phi }_k(\tau , m-sj)| = \sum _{S \subseteq [r+1]} (-1)^{|S|} u^\sigma _{m - k \Sigma ^\sigma S}$
. This concludes the proof of the result.
$|{\Phi }_k(\sigma , m)| = \sum _{j = 0}^{k-1} |{\Phi }_k(\tau , m-sj)| = \sum _{S \subseteq [r+1]} (-1)^{|S|} u^\sigma _{m - k \Sigma ^\sigma S}$
. This concludes the proof of the result.
With this result, we give a proof of Theorem 3.3.
Proof of Theorem 3.3.
By Lemma 3.7, we have
 $$\begin{align*}\sum_{S \subseteq [r]} (-1)^{|S|} u_{m(k-h)-h -(k-h)\Sigma S} = |{\Phi}_{k-h}(\sigma, m(k-h)-h)|. \end{align*}$$
$$\begin{align*}\sum_{S \subseteq [r]} (-1)^{|S|} u_{m(k-h)-h -(k-h)\Sigma S} = |{\Phi}_{k-h}(\sigma, m(k-h)-h)|. \end{align*}$$
By Lemma 3.5, we have
 $$ \begin{align*} H^*_m(\sigma) &= \sum_{S \subseteq [r]} (-1)^{|S|} \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) u_{(m-\Sigma S)(k-h) - h}\\ &= \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) \sum_{S \subseteq [r]} (-1)^{|S|} u_{m(k-h)-h -(k-h)\Sigma S}\\ &= \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) |{\Phi}_{k-h}(\sigma, m(k-h)-h)|. \end{align*} $$
$$ \begin{align*} H^*_m(\sigma) &= \sum_{S \subseteq [r]} (-1)^{|S|} \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) u_{(m-\Sigma S)(k-h) - h}\\ &= \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) \sum_{S \subseteq [r]} (-1)^{|S|} u_{m(k-h)-h -(k-h)\Sigma S}\\ &= \sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) |{\Phi}_{k-h}(\sigma, m(k-h)-h)|. \end{align*} $$
This concludes the proof.
4 Decorated ordered set partitions
 In this section, we show that 
 $H^*(\Delta _{k,n}; S_n)[1]$
 is the permutation character of
$H^*(\Delta _{k,n}; S_n)[1]$
 is the permutation character of 
 $S_n$
 acting naturally on the set of hypersimplicial
$S_n$
 acting naturally on the set of hypersimplicial 
 $(k,n)$
-DOSPs. From Definition 3.1, we recall the definition of the set
$(k,n)$
-DOSPs. From Definition 3.1, we recall the definition of the set 
 $\mathcal I_h$
. Our main result gives a formula for the number of hypersimplicial
$\mathcal I_h$
. Our main result gives a formula for the number of hypersimplicial 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSPs.
$(k,n)$
-DOSPs.
Theorem 4.1. Let 
 $2 \le k < n$
 and denote
$2 \le k < n$
 and denote 
 $H^* = H^*(\Delta _{k,n}; S_n)$
. Then
$H^* = H^*(\Delta _{k,n}; S_n)$
. Then 
 $H^*[1]$
 is equal to the permutation character of the action of
$H^*[1]$
 is equal to the permutation character of the action of 
 $S_n$
 on hypersimplicial
$S_n$
 on hypersimplicial 
 $(k,n)$
-DOSPs. Let
$(k,n)$
-DOSPs. Let 
 $\sigma \in S_n$
 be a permutation with r disjoint cycles, and, for each
$\sigma \in S_n$
 be a permutation with r disjoint cycles, and, for each 
 $i \in [n]$
, write
$i \in [n]$
, write 
 $\lambda _i$
 for the number of length i cycles of
$\lambda _i$
 for the number of length i cycles of 
 $\sigma $
. Then the number of
$\sigma $
. Then the number of 
 $\sigma $
-fixed hypersimplicial
$\sigma $
-fixed hypersimplicial 
 $(k,n)$
-DOSPs is
$(k,n)$
-DOSPs is 
 $$\begin{align*}H^*[1](\sigma) = g \sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_{h}} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r - 1} \end{align*}$$
$$\begin{align*}H^*[1](\sigma) = g \sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_{h}} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r - 1} \end{align*}$$
where 
 $g = \gcd (\{k\} \cup \{i \in [n] : \lambda _i \ge 1\})$
.
$g = \gcd (\{k\} \cup \{i \in [n] : \lambda _i \ge 1\})$
.
 Let us outline the results of the next two sections, which we use to prove Theorem 4.1. In Section 4.1, we use Theorem 3.3 to show that the above formula for 
 $H^*[1]$
 holds (Corollary 4.12). We show that the number of
$H^*[1]$
 holds (Corollary 4.12). We show that the number of 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSPs, including non-hypersimplicial DOSPs, is equal to
$(k,n)$
-DOSPs, including non-hypersimplicial DOSPs, is equal to 
 $gk^{r-1}$
 (Corollary 4.8) and observe that this is equal to the
$gk^{r-1}$
 (Corollary 4.8) and observe that this is equal to the 
 $h=0$
 term in the above sum. In Section 4.2, we give an explicit formula for the number of
$h=0$
 term in the above sum. In Section 4.2, we give an explicit formula for the number of 
 $\sigma $
-fixed non-hypersimplicial DOSPs in Lemma 4.15. The proof of Theorem 4.1 then follows by simplifying the formula using a modified version of the falling factorial identity for Stirling numbers. In the remainder of this section, we give an alternative but equivalent definition for DOSPs under the action of
$\sigma $
-fixed non-hypersimplicial DOSPs in Lemma 4.15. The proof of Theorem 4.1 then follows by simplifying the formula using a modified version of the falling factorial identity for Stirling numbers. In the remainder of this section, we give an alternative but equivalent definition for DOSPs under the action of 
 $S_n$
 and define a notion of directed distance within a DOSP.
$S_n$
 and define a notion of directed distance within a DOSP.
 
Alternative DOSP definition. Fix 
 $k < n$
. Let
$k < n$
. Let 
 $\Psi = \{{f\colon [n]\to {\mathbb Z}/k{\mathbb Z}}\}$
 be the set of functions modulo the equivalence relation
$\Psi = \{{f\colon [n]\to {\mathbb Z}/k{\mathbb Z}}\}$
 be the set of functions modulo the equivalence relation 
 $f\sim g$
 if and only if
$f\sim g$
 if and only if 
 $f-g$
 is constant. There is an action of
$f-g$
 is constant. There is an action of 
 $S_n$
 on
$S_n$
 on 
 $\Psi $
 given by
$\Psi $
 given by 
 $(\sigma \cdot f)(i) = f(\sigma ^{-1}(i))$
 for each
$(\sigma \cdot f)(i) = f(\sigma ^{-1}(i))$
 for each 
 $\sigma \in S_n$
 and
$\sigma \in S_n$
 and 
 $f\in \Psi $
. We now describe the natural
$f\in \Psi $
. We now describe the natural 
 $S_n$
-set isomorphism between
$S_n$
-set isomorphism between 
 $\Psi $
 and
$\Psi $
 and 
 $(k,n)$
-DOSPs. Given a DOSP
$(k,n)$
-DOSPs. Given a DOSP 
 $D=((L_1,\ell _1),(L_2,\ell _2),\ldots ,(L_t,\ell _t))$
, its corresponding function is
$D=((L_1,\ell _1),(L_2,\ell _2),\ldots ,(L_t,\ell _t))$
, its corresponding function is 
 $f_D$
 such that
$f_D$
 such that 
 $f_D(i)=0$
 if
$f_D(i)=0$
 if 
 $i \in L_1$
 and
$i \in L_1$
 and 
 $f_D(i)=\ell _1+\ell _2+\cdots +\ell _{j-1}$
 if
$f_D(i)=\ell _1+\ell _2+\cdots +\ell _{j-1}$
 if 
 $i\in L_j$
 with
$i\in L_j$
 with 
 $j \ge 2$
. It is straightforward to check that the map
$j \ge 2$
. It is straightforward to check that the map 
 $D\mapsto f_D$
 is an isomorphism of
$D\mapsto f_D$
 is an isomorphism of 
 $S_n$
-sets.
$S_n$
-sets.
Definition 4.2. Let 
 $D = ((L_1, \ell _1), \dots , (L_r, \ell _r))$
 be a
$D = ((L_1, \ell _1), \dots , (L_r, \ell _r))$
 be a 
 $(k,n)$
-DOSP. Let
$(k,n)$
-DOSP. Let 
 $i, j \in [n]$
. We define the directed distance
$i, j \in [n]$
. We define the directed distance 
 $d_D(i,j)$
 from i to j in D as follows. Without loss of generality, we may assume
$d_D(i,j)$
 from i to j in D as follows. Without loss of generality, we may assume 
 $i \in L_1$
. Suppose that
$i \in L_1$
. Suppose that 
 $j \in L_u$
 for some
$j \in L_u$
 for some 
 $u \in [r]$
. Then
$u \in [r]$
. Then 
 $d_D(i,j) := \ell _1 + \ell _2 + \dots + \ell _{u-1} \in \{0,1, \dots , k-1\}$
. The winding number of D is
$d_D(i,j) := \ell _1 + \ell _2 + \dots + \ell _{u-1} \in \{0,1, \dots , k-1\}$
. The winding number of D is 
 $w(D) = (d_D(1,2) + d_D(2,3) + \dots + d_D(n-1, n) + d_D(n,1)) / k$
. If the DOSP is given as a function
$w(D) = (d_D(1,2) + d_D(2,3) + \dots + d_D(n-1, n) + d_D(n,1)) / k$
. If the DOSP is given as a function 
 $f: [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
, then, for each
$f: [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
, then, for each 
 $i,j \in [n]$
, the directed distance is
$i,j \in [n]$
, the directed distance is 
 $d_f(i,j) = f(j) - f(i)$
 where we take the representative of
$d_f(i,j) = f(j) - f(i)$
 where we take the representative of 
 $f(j) - f(i)$
 in
$f(j) - f(i)$
 in 
 $\{0,1, \dots , k-1\}$
.
$\{0,1, \dots , k-1\}$
.
 Fix a permutation 
 $\sigma \in S_n$
. Given a
$\sigma \in S_n$
. Given a 
 $\sigma $
-fixed DOSP
$\sigma $
-fixed DOSP 
 $f \colon [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
, we define the turning number
$f \colon [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
, we define the turning number 
 $\tau \in {\mathbb Z}/k{\mathbb Z}$
 of f with respect to
$\tau \in {\mathbb Z}/k{\mathbb Z}$
 of f with respect to 
 $\sigma $
 such that
$\sigma $
 such that 
 $\tau + f(i) = (\sigma \cdot f)(i)$
 for any
$\tau + f(i) = (\sigma \cdot f)(i)$
 for any 
 $i\in [n]$
. This notion is well-defined, that is,
$i\in [n]$
. This notion is well-defined, that is, 
 $\tau $
 does not depend on i, since
$\tau $
 does not depend on i, since 
 $\sigma $
 fixes f.
$\sigma $
 fixes f.
Example 4.3. Let 
 $k = 6$
,
$k = 6$
, 
 $n = 10$
, and
$n = 10$
, and 
 $\sigma = (1 \ 2 \ 3 \ 4 \ 5 \ 6)(7 \ 8 \ 9 \ 10) \in S_{10}$
. Let us compute the turning number of the DOSP
$\sigma = (1 \ 2 \ 3 \ 4 \ 5 \ 6)(7 \ 8 \ 9 \ 10) \in S_{10}$
. Let us compute the turning number of the DOSP 
 $D = ((\{{1,3,5}\}, 1), (\{{7,9}\}, 2), (\{{2,4,6}\}, 1), (\{{8,10}\}, 2))$
 under
$D = ((\{{1,3,5}\}, 1), (\{{7,9}\}, 2), (\{{2,4,6}\}, 1), (\{{8,10}\}, 2))$
 under 
 $\sigma $
, which is depicted in Figure 3. If we apply
$\sigma $
, which is depicted in Figure 3. If we apply 
 $\sigma $
 to D then we obtain
$\sigma $
 to D then we obtain 
 $\sigma (D) = ((\{{2, 4, 6}\}, 1), (\{{8, 10}\}, 2), (\{{1, 3, 5}, 1)\}, (\{{7, 9}\}, 2))$
, which is equivalent to D. The equivalence of
$\sigma (D) = ((\{{2, 4, 6}\}, 1), (\{{8, 10}\}, 2), (\{{1, 3, 5}, 1)\}, (\{{7, 9}\}, 2))$
, which is equivalent to D. The equivalence of 
 $\sigma (D)$
 and D can be observed in the figure: the right DOSP is obtained by turning the left DOSP clockwise by
$\sigma (D)$
 and D can be observed in the figure: the right DOSP is obtained by turning the left DOSP clockwise by 
 $3$
 spaces. In general, the turning number is the number of spaces we turn D to obtain
$3$
 spaces. In general, the turning number is the number of spaces we turn D to obtain 
 $\sigma (D)$
. So, the turning number of D with respect to
$\sigma (D)$
. So, the turning number of D with respect to 
 $\sigma $
 is
$\sigma $
 is 
 $3$
.
$3$
.

Figure 3 The DOSPs in Example 4.3. The DOSP 
 $\sigma (D)$
 (right) is obtained from D (left) by turning it three spaces clockwise, hence the turning number of D with respect to
$\sigma (D)$
 (right) is obtained from D (left) by turning it three spaces clockwise, hence the turning number of D with respect to 
 $\sigma $
 is
$\sigma $
 is 
 $3$
.
$3$
.
Lemma 4.4. Fix 
 $2\leq k < n$
 and let
$2\leq k < n$
 and let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1,s_2,\dots ,s_r)$
. Let
$(s_1,s_2,\dots ,s_r)$
. Let 
 $g=\gcd (s_1,\ldots ,s_r,k)$
. If D is a
$g=\gcd (s_1,\ldots ,s_r,k)$
. If D is a 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSP with turning number
$(k,n)$
-DOSP with turning number 
 $\tau $
, then
$\tau $
, then 
 $g \tau =0$
.
$g \tau =0$
.
Proof. Suppose that 
 $f \colon [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
 is a
$f \colon [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
 is a 
 $\sigma $
-fixed DOSP with nonzero turning number
$\sigma $
-fixed DOSP with nonzero turning number 
 $\tau $
. If
$\tau $
. If 
 $C_i\subset {\mathbb Z}/k{\mathbb Z}$
 denotes the cycle with cycle type
$C_i\subset {\mathbb Z}/k{\mathbb Z}$
 denotes the cycle with cycle type 
 $s_i$
, choose an element
$s_i$
, choose an element 
 $q_i\in C_i$
. Notice that
$q_i\in C_i$
. Notice that 
 $(\sigma ^{s_i} \cdot f)(q_i)=f(q_i)$
, so we have
$(\sigma ^{s_i} \cdot f)(q_i)=f(q_i)$
, so we have 
 $s_i \tau = 0$
 for every
$s_i \tau = 0$
 for every 
 $1\leq i\leq r$
. Since
$1\leq i\leq r$
. Since 
 $\tau \in {\mathbb Z}/k{\mathbb Z}$
 we have
$\tau \in {\mathbb Z}/k{\mathbb Z}$
 we have 
 $k \tau =0$
, and it follows that
$k \tau =0$
, and it follows that 
 $g \tau = 0$
.
$g \tau = 0$
.
4.1 Interpreting terms with DOSPs
 Throughout, we fix 
 $k < n$
 and write
$k < n$
 and write 
 $H^*[t] = H^*(\Delta _{k,n}; S_n)[t]$
 for the equivariant
$H^*[t] = H^*(\Delta _{k,n}; S_n)[t]$
 for the equivariant 
 $H^*$
-polynomial. By Corollary 3.4, let
$H^*$
-polynomial. By Corollary 3.4, let 
 $d = \lfloor (k - 1)n/k \rfloor $
 be the degree of
$d = \lfloor (k - 1)n/k \rfloor $
 be the degree of 
 $H^*$
. By Theorem 3.3, we have that
$H^*$
. By Theorem 3.3, we have that 
 $$\begin{align*}H^*[1](\sigma) = \sum_{m = 0}^{d}\sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) |{\Phi}_{k-h}(\sigma, m(k-h) - h)|. \end{align*}$$
$$\begin{align*}H^*[1](\sigma) = \sum_{m = 0}^{d}\sum_{h = 0}^{k - 1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) |{\Phi}_{k-h}(\sigma, m(k-h) - h)|. \end{align*}$$
The 
 $h = 0$
 term in the above sum is
$h = 0$
 term in the above sum is 
 $ \sum _{m = 0}^{d} |{\Phi }_\ell (\sigma , m \ell )|, $
 which we will show corresponds to the number of
$ \sum _{m = 0}^{d} |{\Phi }_\ell (\sigma , m \ell )|, $
 which we will show corresponds to the number of 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSPs. In subsequent sections, we show that the remaining terms count the non-hypersimplicial
$(k,n)$
-DOSPs. In subsequent sections, we show that the remaining terms count the non-hypersimplicial 
 $(k,n)$
-DOSPs. Hence, the overall value
$(k,n)$
-DOSPs. Hence, the overall value 
 $H^*[1]$
 is the number of
$H^*[1]$
 is the number of 
 $\sigma $
-fixed hypersimplicial
$\sigma $
-fixed hypersimplicial 
 $(k,n)$
-DOSPs.
$(k,n)$
-DOSPs.
 For ease of notation, let us write functions as tuples. The function 
 $f \colon [n] \rightarrow \{0,\dots , k-1\}$
 is written as
$f \colon [n] \rightarrow \{0,\dots , k-1\}$
 is written as 
 $(f(1), f(2), \dots , f(n)) \in \{0, \dots , k-1\}^n$
.
$(f(1), f(2), \dots , f(n)) \in \{0, \dots , k-1\}^n$
.
Example 4.5. Consider the case 
 $n = 6$
,
$n = 6$
, 
 $k = 3$
, and take the permutation
$k = 3$
, and take the permutation 
 $\sigma = (1\ 2\ 3\ 4)(5\ 6)$
. In this case we have the functions
$\sigma = (1\ 2\ 3\ 4)(5\ 6)$
. In this case we have the functions 
 $$\begin{align*}\bigsqcup_{m = 0}^4 {\Phi}_3(\sigma, 3m) = \{(0,0,0,0,0,0), (1,1,1,1,1,1), (2,2,2,2,2,2) \}. \end{align*}$$
$$\begin{align*}\bigsqcup_{m = 0}^4 {\Phi}_3(\sigma, 3m) = \{(0,0,0,0,0,0), (1,1,1,1,1,1), (2,2,2,2,2,2) \}. \end{align*}$$
Indeed there are three 
 $\sigma $
-fixed DOSPs, which are given by
$\sigma $
-fixed DOSPs, which are given by 
 $$\begin{align*}D_1 = ((123456,3)), \quad D_2 = ((1234,2), (56,1)), \quad D_3 = ((1234,1), (56,2)). \end{align*}$$
$$\begin{align*}D_1 = ((123456,3)), \quad D_2 = ((1234,2), (56,1)), \quad D_3 = ((1234,1), (56,2)). \end{align*}$$
Lemma 4.6. Fix 
 $0 < k < n$
. Let
$0 < k < n$
. Let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1, \dots , s_r)$
 and define
$(s_1, \dots , s_r)$
 and define 
 $g = \gcd (k, s_1, s_2, \dots , s_r)$
. Then the number of
$g = \gcd (k, s_1, s_2, \dots , s_r)$
. Then the number of 
 $\sigma $
-fixed DOSPs is
$\sigma $
-fixed DOSPs is 
 $g k^{r-1}$
. In particular, there is a bijection between the set of
$g k^{r-1}$
. In particular, there is a bijection between the set of 
 $\sigma $
-fixed DOSPs and the set
$\sigma $
-fixed DOSPs and the set 
 $$\begin{align*}\{(\alpha_1, \alpha_2, \dots, \alpha_r) : 0 \le \alpha_1 \le g-1,\, 0 \le \alpha_i \le k - 1,\, 2 \le i \le r\}. \end{align*}$$
$$\begin{align*}\{(\alpha_1, \alpha_2, \dots, \alpha_r) : 0 \le \alpha_1 \le g-1,\, 0 \le \alpha_i \le k - 1,\, 2 \le i \le r\}. \end{align*}$$
Proof. Throughout, we consider DOSPs as functions 
 $f \colon [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
 up to equivalence. If
$f \colon [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
 up to equivalence. If 
 $\sigma $
 is the identity permutation, then every DOSP is fixed by
$\sigma $
 is the identity permutation, then every DOSP is fixed by 
 $\sigma $
. The number of
$\sigma $
. The number of 
 $\sigma $
-fixed DOSPs is
$\sigma $
-fixed DOSPs is 
 $k^{n-1}$
, because we may take
$k^{n-1}$
, because we may take 
 $f(1) = 0$
 and freely choose the values
$f(1) = 0$
 and freely choose the values 
 $f(i) \in {\mathbb Z}/k{\mathbb Z}$
 for each
$f(i) \in {\mathbb Z}/k{\mathbb Z}$
 for each 
 $i \in \{2,3, \dots , n\}$
. Note that each such choice gives a distinct DOSP.
$i \in \{2,3, \dots , n\}$
. Note that each such choice gives a distinct DOSP.
 Now suppose that 
 $\sigma $
 is not the identity. Let
$\sigma $
 is not the identity. Let 
 $C_1, C_2, \dots , C_r$
 be the cycle sets of
$C_1, C_2, \dots , C_r$
 be the cycle sets of 
 $\sigma $
. Without loss of generality, we may assume that
$\sigma $
. Without loss of generality, we may assume that 
 $s_1> 1$
 and
$s_1> 1$
 and 
 $1 \in C_1$
. We define
$1 \in C_1$
. We define 
 $q_1 = \sigma (1)$
 and, for each
$q_1 = \sigma (1)$
 and, for each 
 $i \in \{2, \dots , r\}$
, let us fix a distinguished element
$i \in \{2, \dots , r\}$
, let us fix a distinguished element 
 $q_i \in C_i$
. Let
$q_i \in C_i$
. Let 
 $f \colon [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
 be a
$f \colon [n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
 be a 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSP. Without loss of generality we assume that
$(k,n)$
-DOSP. Without loss of generality we assume that 
 $f(1) = 0$
. We will show that the sequence of integers
$f(1) = 0$
. We will show that the sequence of integers 
 $$\begin{align*}\alpha = (\alpha_1, \alpha_2, \dots, \alpha_r)= (d_f(1,q_1),\, d_f(1,q_2),\, \dots,\, d_f(1,q_r)) \end{align*}$$
$$\begin{align*}\alpha = (\alpha_1, \alpha_2, \dots, \alpha_r)= (d_f(1,q_1),\, d_f(1,q_2),\, \dots,\, d_f(1,q_r)) \end{align*}$$
uniquely determines the DOSP. By assumption 
 $f(1) = 0$
. By definition, we have
$f(1) = 0$
. By definition, we have 
 $f(q_1) = d_f(1,q_1)$
. Since D is invariant under
$f(q_1) = d_f(1,q_1)$
. Since D is invariant under 
 $\sigma $
, it follows that
$\sigma $
, it follows that 
 $$\begin{align*}d_D(1,\sigma(1)) = d_D(\sigma(1),\sigma^2(1)) = \dots = d_D(\sigma^{s_1-1}(1), \sigma^{s_1}(1)) = d_D(1,q_1). \end{align*}$$
$$\begin{align*}d_D(1,\sigma(1)) = d_D(\sigma(1),\sigma^2(1)) = \dots = d_D(\sigma^{s_1-1}(1), \sigma^{s_1}(1)) = d_D(1,q_1). \end{align*}$$
So the value 
 $f(\sigma ^i(1))$
 for each element of
$f(\sigma ^i(1))$
 for each element of 
 $C_1 = \{1, \sigma (1), \sigma ^2(1), \dots , \sigma ^{s_1 - 1}(1) \}$
 is determined by
$C_1 = \{1, \sigma (1), \sigma ^2(1), \dots , \sigma ^{s_1 - 1}(1) \}$
 is determined by 
 $d_f(1,q_1)$
. Explicitly, we have
$d_f(1,q_1)$
. Explicitly, we have 
 $f(\sigma ^i(1)) = i \cdot d_f(1,q_1)\ \mod {k}$
 for all
$f(\sigma ^i(1)) = i \cdot d_f(1,q_1)\ \mod {k}$
 for all 
 $i \ge 0$
. By a similar argument, the value
$i \ge 0$
. By a similar argument, the value 
 $f(\sigma ^i(q_2))$
 for each element of
$f(\sigma ^i(q_2))$
 for each element of 
 $C_2 = \{q_2, \sigma (q_2), \sigma ^2(q_2), \dots , \sigma ^{s_2 - 1}(q_2)\}$
 is determined by
$C_2 = \{q_2, \sigma (q_2), \sigma ^2(q_2), \dots , \sigma ^{s_2 - 1}(q_2)\}$
 is determined by 
 $d_f(1, q_2)$
. To see this, observe that
$d_f(1, q_2)$
. To see this, observe that 
 $f(q_2) = d_f(1, q_2)$
 and, since f is invariant under
$f(q_2) = d_f(1, q_2)$
 and, since f is invariant under 
 $\sigma $
, it follows that
$\sigma $
, it follows that 
 $d_f(q_2, \sigma (q_2)) = d_f(1,\sigma (1))$
. We deduce that the DOSP f is uniquely determined by
$d_f(q_2, \sigma (q_2)) = d_f(1,\sigma (1))$
. We deduce that the DOSP f is uniquely determined by 
 $\alpha $
.
$\alpha $
.
 We now consider the possible vectors 
 $\alpha $
. By definition, we have that
$\alpha $
. By definition, we have that 
 $d_f(1,q_1) = d_f(1, \sigma (1))$
 is the turning number of f. By Lemma 4.4 we have that
$d_f(1,q_1) = d_f(1, \sigma (1))$
 is the turning number of f. By Lemma 4.4 we have that 
 $g \cdot d_f(1, q_1) \equiv 0\ \mod k$
. The possible values for
$g \cdot d_f(1, q_1) \equiv 0\ \mod k$
. The possible values for 
 $d_f(1,q_1) \cdot g$
 are
$d_f(1,q_1) \cdot g$
 are 
 $ \beta \cdot k $
 for each
$ \beta \cdot k $
 for each 
 $\beta \in \{0,1, \dots , g - 1\}$
. Hence, the possible values for
$\beta \in \{0,1, \dots , g - 1\}$
. Hence, the possible values for 
 $d_f(1,q_1)$
 are
$d_f(1,q_1)$
 are 
 $\beta k / g$
 for each
$\beta k / g$
 for each 
 $0 \le \beta < g$
. For each such value of
$0 \le \beta < g$
. For each such value of 
 $\alpha _1 = d_f(1,q_1)$
, we may freely choose the values
$\alpha _1 = d_f(1,q_1)$
, we may freely choose the values 
 $\alpha _2, \dots , \alpha _k$
 in
$\alpha _2, \dots , \alpha _k$
 in 
 $\{0,1,\dots , k-1\}$
. Each choice gives a distinct DOSP and every
$\{0,1,\dots , k-1\}$
. Each choice gives a distinct DOSP and every 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSP arises in this way. So the number of DOSPs is
$(k,n)$
-DOSP arises in this way. So the number of DOSPs is 
 $gk^{r-1}$
.
$gk^{r-1}$
.
Proposition 4.7. Let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1, s_2, \dots , s_r)$
, and fix
$(s_1, s_2, \dots , s_r)$
, and fix 
 $k \in [n]$
. Define
$k \in [n]$
. Define 
 $$\begin{align*}\Phi = \left\{(f_1, f_2, \dots, f_r) \in \{0,1, \dots, k-1\}^r : \sum_{i = 1}^r f_i s_i \equiv 0 \quad\mod k \right\}, \end{align*}$$
$$\begin{align*}\Phi = \left\{(f_1, f_2, \dots, f_r) \in \{0,1, \dots, k-1\}^r : \sum_{i = 1}^r f_i s_i \equiv 0 \quad\mod k \right\}, \end{align*}$$
and let 
 $g := \gcd (s_1, s_2, \dots , s_r, k)$
. Then
$g := \gcd (s_1, s_2, \dots , s_r, k)$
. Then 
 $|\Phi | = g k^{r-1}$
.
$|\Phi | = g k^{r-1}$
.
Proof. Consider the homomorphism of abelian groups
 $$\begin{align*}\varphi \colon ({\mathbb Z}/k{\mathbb Z})^r \rightarrow {\mathbb Z}/k{\mathbb Z}, \quad (f_1, f_2, \dots, f_r) \mapsto \sum_{i = 1}^r f_i s_i. \end{align*}$$
$$\begin{align*}\varphi \colon ({\mathbb Z}/k{\mathbb Z})^r \rightarrow {\mathbb Z}/k{\mathbb Z}, \quad (f_1, f_2, \dots, f_r) \mapsto \sum_{i = 1}^r f_i s_i. \end{align*}$$
The image of 
 $\varphi $
 is the subgroup of
$\varphi $
 is the subgroup of 
 ${\mathbb Z}/k{\mathbb Z}$
 generated by
${\mathbb Z}/k{\mathbb Z}$
 generated by 
 $s_1, s_2, \dots , s_r$
. By Bezout’s identity
$s_1, s_2, \dots , s_r$
. By Bezout’s identity 
 $$\begin{align*}\langle s_1, s_2, \dots, s_k \rangle = \langle \gcd(s_1, s_2, \dots, s_r, k) \rangle = \langle g \rangle \subseteq {\mathbb Z}/k{\mathbb Z}. \end{align*}$$
$$\begin{align*}\langle s_1, s_2, \dots, s_k \rangle = \langle \gcd(s_1, s_2, \dots, s_r, k) \rangle = \langle g \rangle \subseteq {\mathbb Z}/k{\mathbb Z}. \end{align*}$$
So we have 
 $|\operatorname {\mathrm {Im}}(\varphi )| = k / g$
. Therefore
$|\operatorname {\mathrm {Im}}(\varphi )| = k / g$
. Therefore 
 $ |\Phi | = |\ker (\varphi )| = \frac {k^r}{k / g} = g k^{r-1} $
 and we are done.
$ |\Phi | = |\ker (\varphi )| = \frac {k^r}{k / g} = g k^{r-1} $
 and we are done.
The two results above, give us the following.
Corollary 4.8. Fix 
 $0 < k < n$
. Let
$0 < k < n$
. Let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $s_1, \dots , s_r$
 and define
$s_1, \dots , s_r$
 and define 
 $g = \gcd (k, s_1, s_2, \dots , s_r)$
. Recall the set
$g = \gcd (k, s_1, s_2, \dots , s_r)$
. Recall the set 
 $\Phi $
 from the statement of Proposition 4.7. There is a bijection between the set of all
$\Phi $
 from the statement of Proposition 4.7. There is a bijection between the set of all 
 $\sigma $
-fixed DOSPs and the set
$\sigma $
-fixed DOSPs and the set 
 $$\begin{align*}\Phi = \bigsqcup_{m = 0}^{\lfloor (k-1)n/k \rfloor } {\Phi}_k(\sigma, mk). \quad \text{ Therefore} \quad \sum_{m \ge 0} |{\Phi}_k(\sigma, mk)| = g k^{r-1}. \end{align*}$$
$$\begin{align*}\Phi = \bigsqcup_{m = 0}^{\lfloor (k-1)n/k \rfloor } {\Phi}_k(\sigma, mk). \quad \text{ Therefore} \quad \sum_{m \ge 0} |{\Phi}_k(\sigma, mk)| = g k^{r-1}. \end{align*}$$
Proof. Suppose that 
 $\sigma $
 has cycle type
$\sigma $
 has cycle type 
 $(s_1, s_2, \dots , s_r)$
. By Lemma 4.6, the number of
$(s_1, s_2, \dots , s_r)$
. By Lemma 4.6, the number of 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSPs is
$(k,n)$
-DOSPs is 
 $g k^{r-1}$
. By Proposition 4.7, we have that
$g k^{r-1}$
. By Proposition 4.7, we have that 
 $|\Phi | = g k^{r-1}$
, and we are done.
$|\Phi | = g k^{r-1}$
, and we are done.
Lemma 4.9. Let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1, s_2, \dots , s_r)$
. Fix
$(s_1, s_2, \dots , s_r)$
. Fix 
 $h \in \{0,1,\dots , k-1\}$
. Define
$h \in \{0,1,\dots , k-1\}$
. Define 
 $g' = \gcd (s_1, s_2, \dots , s_r, k-h)$
. Then we have
$g' = \gcd (s_1, s_2, \dots , s_r, k-h)$
. Then we have 
 $$\begin{align*}\sum_{m \ge 0} |{\Phi}_{k-h}(\sigma, m(k-h)-h)| = \begin{cases} g' (k-h)^{r-1} & \text{if } g' \text{ divides } h, \\ 0 & \text{otherwise.} \end{cases} \end{align*}$$
$$\begin{align*}\sum_{m \ge 0} |{\Phi}_{k-h}(\sigma, m(k-h)-h)| = \begin{cases} g' (k-h)^{r-1} & \text{if } g' \text{ divides } h, \\ 0 & \text{otherwise.} \end{cases} \end{align*}$$
Proof. Consider the homomorphism of abelian groups
 $$\begin{align*}\varphi \colon ({\mathbb Z}/(k-h){\mathbb Z})^r \rightarrow {\mathbb Z}/(k-h){\mathbb Z}, \quad (f_1, f_2, \dots, f_r) \mapsto \sum_{i = 1}^r f_i s_i. \end{align*}$$
$$\begin{align*}\varphi \colon ({\mathbb Z}/(k-h){\mathbb Z})^r \rightarrow {\mathbb Z}/(k-h){\mathbb Z}, \quad (f_1, f_2, \dots, f_r) \mapsto \sum_{i = 1}^r f_i s_i. \end{align*}$$
Observe that there is a natural bijection between 
 $\bigsqcup _{m \ge 0} {\Phi }_{k-h}(\sigma , m(k-h)-h)$
 and the set
$\bigsqcup _{m \ge 0} {\Phi }_{k-h}(\sigma , m(k-h)-h)$
 and the set 
 $\Phi ' := \{f \in ({\mathbb Z}/(k-h){\mathbb Z})^r : \varphi (f) = -h \ \mod (k-h)\}$
. By Bezout’s identity, it follows that the image of
$\Phi ' := \{f \in ({\mathbb Z}/(k-h){\mathbb Z})^r : \varphi (f) = -h \ \mod (k-h)\}$
. By Bezout’s identity, it follows that the image of 
 $\varphi $
 is principally generated by
$\varphi $
 is principally generated by 
 $g'$
. It follows that
$g'$
. It follows that 
 $$\begin{align*}|\Phi'| = \begin{cases} |\ker(\varphi)| = g'(k-h)^{r-1} & \text{if } -h \in \langle g' \rangle \subseteq {\mathbb Z}/(k-h){\mathbb Z}, \\ 0 & \text{otherwise.} \end{cases} \end{align*}$$
$$\begin{align*}|\Phi'| = \begin{cases} |\ker(\varphi)| = g'(k-h)^{r-1} & \text{if } -h \in \langle g' \rangle \subseteq {\mathbb Z}/(k-h){\mathbb Z}, \\ 0 & \text{otherwise.} \end{cases} \end{align*}$$
In the first case, we have that 
 $-h \in \langle g' \rangle $
 if and only if
$-h \in \langle g' \rangle $
 if and only if 
 $g'$
 divides h.
$g'$
 divides h.
Proposition 4.10. Fix 
 $0 \le h < k$
 and some positive integers
$0 \le h < k$
 and some positive integers 
 $s_1, \dots , s_r$
. For each
$s_1, \dots , s_r$
. For each 
 $0 \le i < k$
 define
$0 \le i < k$
 define 
 $g_i = \gcd (k-i, s_1, \dots , s_r)$
. Then
$g_i = \gcd (k-i, s_1, \dots , s_r)$
. Then 
 $g_h | h$
 if and only if
$g_h | h$
 if and only if 
 $g_0 | h$
.
$g_0 | h$
.
Proof. Define 
 $\tilde g = \gcd (s_1, \dots , s_r)$
, so
$\tilde g = \gcd (s_1, \dots , s_r)$
, so 
 $g_h = \gcd (k-h, \tilde g)$
 and
$g_h = \gcd (k-h, \tilde g)$
 and 
 $g_0 = \gcd (k, \tilde g)$
. We have
$g_0 = \gcd (k, \tilde g)$
. We have 
 $g_h | h$
 if and only if
$g_h | h$
 if and only if 
 $(k - h) | h$
 and
$(k - h) | h$
 and 
 $\tilde g | h$
. On the other hand
$\tilde g | h$
. On the other hand 
 $g_0 | h$
 if and only if
$g_0 | h$
 if and only if 
 $k | h$
 and
$k | h$
 and 
 $\tilde g | h$
. So it suffices to show that
$\tilde g | h$
. So it suffices to show that 
 $(k - h) | h$
 if and only if
$(k - h) | h$
 if and only if 
 $k | h$
, which easily follows from the assumption that
$k | h$
, which easily follows from the assumption that 
 $k> h \ge 0$
.
$k> h \ge 0$
.
Proposition 4.11. Let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1, \dots , s_r)$
. For each
$(s_1, \dots , s_r)$
. For each 
 $h \ge 0$
, define
$h \ge 0$
, define 
 $g := \gcd (k, s_1, \dots , s_r)$
 and
$g := \gcd (k, s_1, \dots , s_r)$
 and 
 $g_h := \gcd (k-h, s_1, \dots , s_r)$
. We have
$g_h := \gcd (k-h, s_1, \dots , s_r)$
. We have 
 $$\begin{align*}H^*[1](\sigma) = \sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) g_h (k - h)^{r-1} d(g, h), \end{align*}$$
$$\begin{align*}H^*[1](\sigma) = \sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) g_h (k - h)^{r-1} d(g, h), \end{align*}$$
where 
 $d(g, h) = 1$
 if g divides h and
$d(g, h) = 1$
 if g divides h and 
 $d(g, h) = 0$
 otherwise.
$d(g, h) = 0$
 otherwise.
Corollary 4.12. Let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1, \dots , s_r)$
. For each
$(s_1, \dots , s_r)$
. For each 
 $h \ge 0$
, define
$h \ge 0$
, define 
 $g := \gcd (k, s_1, \dots , s_r)$
. We have
$g := \gcd (k, s_1, \dots , s_r)$
. We have 
 $$\begin{align*}H^*[1](\sigma) = g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1}. \end{align*}$$
$$\begin{align*}H^*[1](\sigma) = g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1}. \end{align*}$$
Proof. For each 
 $h \ge 0$
, define
$h \ge 0$
, define 
 $g_h := \gcd (k-h, s_1, \dots , s_r)$
. Consider the formula in Proposition 4.11. Let
$g_h := \gcd (k-h, s_1, \dots , s_r)$
. Consider the formula in Proposition 4.11. Let 
 $h \ge 0$
 and
$h \ge 0$
 and 
 $I \in \mathcal I_h$
. Assume that the product of binomials
$I \in \mathcal I_h$
. Assume that the product of binomials 
 $\binom {\lambda _1}{I_1} \cdots \binom {\lambda _{k-1}}{I_{k-1}}$
 is nonzero. For each
$\binom {\lambda _1}{I_1} \cdots \binom {\lambda _{k-1}}{I_{k-1}}$
 is nonzero. For each 
 $s \in [k-1]$
 such that
$s \in [k-1]$
 such that 
 $\lambda _s \ge 1$
, we have that
$\lambda _s \ge 1$
, we have that 
 $g | s$
. Therefore each nonzero term of
$g | s$
. Therefore each nonzero term of 
 $1 \cdot I_1 + \dots + (k-1) \cdot I_{k-1}$
 is divisible by g, hence g divides h and so
$1 \cdot I_1 + \dots + (k-1) \cdot I_{k-1}$
 is divisible by g, hence g divides h and so 
 $d(g,h) = 1$
. Since g divides h, we have that
$d(g,h) = 1$
. Since g divides h, we have that 
 $g_h = \gcd (k-h,s_1, \dots , s_r) = \gcd (k, s_1, \dots , s_r) = g$
. The result immediately follows.
$g_h = \gcd (k-h,s_1, \dots , s_r) = \gcd (k, s_1, \dots , s_r) = g$
. The result immediately follows.
4.2 Counting non-hypersimplicial DOSPs
 In this section we count the non-hypersimplicial 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSPs. Throughout this section, we will frequently make use of the alternative definition of DOSPs in terms of functions
$(k,n)$
-DOSPs. Throughout this section, we will frequently make use of the alternative definition of DOSPs in terms of functions 
 $[n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
. See the beginning of Section 4 and Definition 4.2.
$[n] \rightarrow {\mathbb Z}/k{\mathbb Z}$
. See the beginning of Section 4 and Definition 4.2.
 
Setup for counting non-hypersimplicial DOSPs. Fix 
 $k < n$
 and
$k < n$
 and 
 $\sigma \in S_n$
. We denote by
$\sigma \in S_n$
. We denote by 
 $\mathcal D$
 the set of
$\mathcal D$
 the set of 
 $\sigma $
-fixed non-hypersimplicial DOSPs and by
$\sigma $
-fixed non-hypersimplicial DOSPs and by 
 $\mathcal D^\tau \subseteq \mathcal D$
 the subset of DOSPs with turning number
$\mathcal D^\tau \subseteq \mathcal D$
 the subset of DOSPs with turning number 
 $\tau \in {\mathbb Z}/k{\mathbb Z}$
. We define the set
$\tau \in {\mathbb Z}/k{\mathbb Z}$
. We define the set 
 $\Lambda $
 of nonempty unions of cycles in
$\Lambda $
 of nonempty unions of cycles in 
 $\sigma $
:
$\sigma $
: 

For each 
 $u=C_{i_1}\sqcup C_{i_2}\sqcup \cdots \sqcup C_{i_s} \in \Lambda $
, we will denote the corresponding set
$u=C_{i_1}\sqcup C_{i_2}\sqcup \cdots \sqcup C_{i_s} \in \Lambda $
, we will denote the corresponding set 
 $\{{i_1, i_2,\ldots , i_s}\}\subseteq [r]$
 by
$\{{i_1, i_2,\ldots , i_s}\}\subseteq [r]$
 by 
 $\operatorname {\mathrm {ind}}(u)$
. Furthermore, we define the subset
$\operatorname {\mathrm {ind}}(u)$
. Furthermore, we define the subset 
 $\mathcal D_u^{\tau }\subseteq \mathcal D^{\tau }$
 of DOSPs containing a tuple
$\mathcal D_u^{\tau }\subseteq \mathcal D^{\tau }$
 of DOSPs containing a tuple 
 $(L,\ell )$
 such that:
$(L,\ell )$
 such that: 
- 
•  $ \left \lvert {L} \right \rvert \leq \ell $
 (we call such L a bad set), $ \left \lvert {L} \right \rvert \leq \ell $
 (we call such L a bad set),
- 
• L completely lies in u and 
- 
• for every  $C_i\subseteq u$
, $C_i\subseteq u$
, $L\cap C_i$
 is nonempty. $L\cap C_i$
 is nonempty.
In other words, 
 $\mathcal D^\tau _u$
 is the set of all
$\mathcal D^\tau _u$
 is the set of all 
 $\sigma $
-fixed DOSPs D such that u is a disjoint union of bad sets of D, and those bad sets form a single
$\sigma $
-fixed DOSPs D such that u is a disjoint union of bad sets of D, and those bad sets form a single 
 $\sigma $
 orbit. Note, for any
$\sigma $
 orbit. Note, for any 
 $D\in \mathcal D^\tau $
 there exists
$D\in \mathcal D^\tau $
 there exists 
 $u \in \Lambda $
 such that
$u \in \Lambda $
 such that 
 $D\in \mathcal D_u^{\tau }$
. Explicitly, D is non-hypersimplicial so it has a bad set, say
$D\in \mathcal D_u^{\tau }$
. Explicitly, D is non-hypersimplicial so it has a bad set, say 
 $(L, \ell )$
, then
$(L, \ell )$
, then 
 $D \in \mathcal D_u^\tau $
 where
$D \in \mathcal D_u^\tau $
 where 
 $u = L \cup \sigma (L) \cup \sigma ^2(L) \cup \dots \cup \sigma ^{o(\sigma )-1}(L)$
 is the
$u = L \cup \sigma (L) \cup \sigma ^2(L) \cup \dots \cup \sigma ^{o(\sigma )-1}(L)$
 is the 
 $\sigma $
-orbit of L.
$\sigma $
-orbit of L.
Lemma 4.13. Fix 
 $2\leq k < n$
 and let
$2\leq k < n$
 and let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1,s_2,\dots ,s_r)$
. Define
$(s_1,s_2,\dots ,s_r)$
. Define 
 $g=\gcd (s_1,\ldots ,s_r,k)$
 and let
$g=\gcd (s_1,\ldots ,s_r,k)$
 and let 
 $\tau \in {\mathbb Z}/k{\mathbb Z}$
 such that
$\tau \in {\mathbb Z}/k{\mathbb Z}$
 such that 
 $g\cdot \tau = 0$
. Fix
$g\cdot \tau = 0$
. Fix 
 $h \in [k-1]$
 and let
$h \in [k-1]$
 and let 
 $J\in \binom {\Lambda }{h}$
 be a nonempty subset of
$J\in \binom {\Lambda }{h}$
 be a nonempty subset of 
 $\Lambda $
 such that the elements of J are pairwise disjoint. Then
$\Lambda $
 such that the elements of J are pairwise disjoint. Then 
 $$\begin{align*}\left\lvert {\bigcap_{u\in J}\mathcal D^\tau_u} \right\rvert = \frac{((k-i)/o(\tau)+h-1)!}{((k-i)/o(\tau))!} o(\tau)^{j-1} (k-i)^{r-j} \end{align*}$$
$$\begin{align*}\left\lvert {\bigcap_{u\in J}\mathcal D^\tau_u} \right\rvert = \frac{((k-i)/o(\tau)+h-1)!}{((k-i)/o(\tau))!} o(\tau)^{j-1} (k-i)^{r-j} \end{align*}$$
where 
 $o(\tau )$
 denotes the order of
$o(\tau )$
 denotes the order of 
 $\tau $
, j is the number of elements in
$\tau $
, j is the number of elements in 
 $\bigsqcup _{u\in J}\operatorname {\mathrm {ind}}(u)$
, and i is the number of elements in
$\bigsqcup _{u\in J}\operatorname {\mathrm {ind}}(u)$
, and i is the number of elements in 
 $\bigsqcup _{u\in J} $
u.
$\bigsqcup _{u\in J} $
u.
Before giving the proof, we will outline the concepts in the proof with an example.
Example 4.14. Fix 
 $k=12$
,
$k=12$
, 
 $n=24$
, and
$n=24$
, and 
 $\sigma \in S_n$
 with cycle type
$\sigma \in S_n$
 with cycle type 
 $(3,3,6,3,9)$
. This means that
$(3,3,6,3,9)$
. This means that 
 $r=5$
 and
$r=5$
 and 
 $g=3$
. For simplicity, we assume that the cycle sets are
$g=3$
. For simplicity, we assume that the cycle sets are 
 $$\begin{align*}C_1=\{{1,2,3}\},\ C_2=\{{4,5,6}\},\ C_3=\{{7,8,\ldots,12}\},\ C_4=\{{13,14,15}\},\ C_5=\{{16,17,\ldots,24}\}. \end{align*}$$
$$\begin{align*}C_1=\{{1,2,3}\},\ C_2=\{{4,5,6}\},\ C_3=\{{7,8,\ldots,12}\},\ C_4=\{{13,14,15}\},\ C_5=\{{16,17,\ldots,24}\}. \end{align*}$$
For each cycle 
 $C_i$
, we fix a distinguished element
$C_i$
, we fix a distinguished element 
 $q_i \in C_i$
. Explicitly, we choose
$q_i \in C_i$
. Explicitly, we choose 
 $q_i$
 to be the smallest element:
$q_i$
 to be the smallest element: 
 $q_1=1$
,
$q_1=1$
, 
 $q_2=4$
,
$q_2=4$
, 
 $q_3=7$
, etc. Let us fix a subset
$q_3=7$
, etc. Let us fix a subset 
 $J = \{ u_1, u_2\}$
 where
$J = \{ u_1, u_2\}$
 where 
 $u_1=C_1\sqcup C_2$
 and
$u_1=C_1\sqcup C_2$
 and 
 $u_2=C_4$
. This gives us
$u_2=C_4$
. This gives us 
 $$\begin{align*}\bigsqcup_{u \in J} u = \{1,2,3, 4, 5, 6, 13, 14, 15\} \quad \text{and} \quad \bigsqcup_{u \in J} \operatorname{\mathrm{ind}}(u) = \{1,2,4\}, \end{align*}$$
$$\begin{align*}\bigsqcup_{u \in J} u = \{1,2,3, 4, 5, 6, 13, 14, 15\} \quad \text{and} \quad \bigsqcup_{u \in J} \operatorname{\mathrm{ind}}(u) = \{1,2,4\}, \end{align*}$$
hence 
 $i=9$
 and
$i=9$
 and 
 $j=3$
. Lastly we fix
$j=3$
. Lastly we fix 
 $\tau =8 \in {\mathbb Z}/12{\mathbb Z}$
, meaning that
$\tau =8 \in {\mathbb Z}/12{\mathbb Z}$
, meaning that 
 $o(\tau )=3$
. We will give an overview of the proof of Lemma 4.13, which counts the number of non-hypersimplicial DOSPs in
$o(\tau )=3$
. We will give an overview of the proof of Lemma 4.13, which counts the number of non-hypersimplicial DOSPs in 
 $\mathcal D^\tau _{u_1} \cap \mathcal D^\tau _{u_2}$
. To do this, we construct DOSPs in this set. We imagine starting with an empty DOSP
$\mathcal D^\tau _{u_1} \cap \mathcal D^\tau _{u_2}$
. To do this, we construct DOSPs in this set. We imagine starting with an empty DOSP 
 $(L_1 = \{ \}, \dots , L_{12} = \{ \})$
 of
$(L_1 = \{ \}, \dots , L_{12} = \{ \})$
 of 
 $k = 12$
 empty sets. We then consider the possible ways to place the cycles into the DOSP. Note that the turning number
$k = 12$
 empty sets. We then consider the possible ways to place the cycles into the DOSP. Note that the turning number 
 $\tau = 8$
 is fixed, so each
$\tau = 8$
 is fixed, so each 
 $\sigma $
-orbit consists of
$\sigma $
-orbit consists of 
 $o(\tau ) = 3$
 sets of the DOSP. We will place the cycles into the DOSP with three steps.
$o(\tau ) = 3$
 sets of the DOSP. We will place the cycles into the DOSP with three steps.
 Our first step is to distribute 
 $u_1$
 across a single
$u_1$
 across a single 
 $\sigma $
-orbit of
$\sigma $
-orbit of 
 $o(\tau )=3$
 sets of
$o(\tau )=3$
 sets of 
 $2$
 elements each and to adorn each of these three sets with a decoration
$2$
 elements each and to adorn each of these three sets with a decoration 
 $\ell _i \geq 2$
 so that each set is a bad set of the resulting DOSP. Our second step is to distribute
$\ell _i \geq 2$
 so that each set is a bad set of the resulting DOSP. Our second step is to distribute 
 $u_2$
 across
$u_2$
 across 
 $3$
 singletons, which we note always results in bad sets in the final DOSP. The third step is to put the rest of the elements into the remaining spaces. See Figure 4 for a specific instance.
$3$
 singletons, which we note always results in bad sets in the final DOSP. The third step is to put the rest of the elements into the remaining spaces. See Figure 4 for a specific instance.

Figure 4 A DOSP with the setup of Example 4.14, the choices for the 
 $q_i$
 are
$q_i$
 are 
 $f_D(1)=0$
,
$f_D(1)=0$
, 
 $f_D(4)=4$
,
$f_D(4)=4$
, 
 $f_D(7)=10 f_D(13)=7$
,
$f_D(7)=10 f_D(13)=7$
, 
 $f_D(16)=2$
, are the underlined elements. The bold sets are the bad sets corresponding to
$f_D(16)=2$
, are the underlined elements. The bold sets are the bad sets corresponding to 
 $u_1$
 and
$u_1$
 and 
 $u_2$
.
$u_2$
.
 
Step 1. A 
 $\sigma $
-fixed DOSP D with turning number
$\sigma $
-fixed DOSP D with turning number 
 $\tau $
 is completely determined by the values the function
$\tau $
 is completely determined by the values the function 
 $f_D$
 takes on the distinguished elements
$f_D$
 takes on the distinguished elements 
 $q_i$
. In Figure 4, the
$q_i$
. In Figure 4, the 
 $q_i$
 are the underlined elements. As a starting point, we will assume that
$q_i$
 are the underlined elements. As a starting point, we will assume that 
 $f_D(q_1) = 0$
, or in other words
$f_D(q_1) = 0$
, or in other words 
 $1 \in L_1$
. This choice fixes the positions of the elements in
$1 \in L_1$
. This choice fixes the positions of the elements in 
 $C_1 = \{1,2,3\}$
. Since
$C_1 = \{1,2,3\}$
. Since 
 $\tau = 8$
, it follows that
$\tau = 8$
, it follows that 
 $f_D(u_1)=\{{0,4,8}\}$
, meaning that we have
$f_D(u_1)=\{{0,4,8}\}$
, meaning that we have 
 $o(\tau )=3$
 choices for the position of
$o(\tau )=3$
 choices for the position of 
 $4=q_2\in u_1$
. Once we have placed
$4=q_2\in u_1$
. Once we have placed 
 $u = C_1 \sqcup C_2$
, we mark the positions
$u = C_1 \sqcup C_2$
, we mark the positions 
 $0,1, 4,5, 8,9$
 as filled, this guarantees that each set in the resulting DOSP containing elements of
$0,1, 4,5, 8,9$
 as filled, this guarantees that each set in the resulting DOSP containing elements of 
 $u_1$
 are bad sets. In Figure 4, these filled sets include the white circles.
$u_1$
 are bad sets. In Figure 4, these filled sets include the white circles.
 
Step 2. The placement of the element 
 $13=q_4\in u_2$
 is restricted to the locations
$13=q_4\in u_2$
 is restricted to the locations 
 $\{{2,6,10}\}$
 and
$\{{2,6,10}\}$
 and 
 $\{{3,7,11}$
 because
$\{{3,7,11}$
 because 
 $\{{1,5,9}\}$
 (the white spaces in Figure 4) need to remain clear. This gives us
$\{{1,5,9}\}$
 (the white spaces in Figure 4) need to remain clear. This gives us 
 $6$
 choices for
$6$
 choices for 
 $q_4$
. Notice that we count possible locations for a
$q_4$
. Notice that we count possible locations for a 
 $q_i$
 in sets of
$q_i$
 in sets of 
 $3$
. This corresponds to the factor
$3$
. This corresponds to the factor 
 $o(\tau )^{j-1}$
 in the formula in the lemma.
$o(\tau )^{j-1}$
 in the formula in the lemma.
 
Step 3. Finally, we must choose placements for the remaining cycles 
 $C_3$
 and
$C_3$
 and 
 $C_5$
. After having placed
$C_5$
. After having placed 
 $u_1$
 and
$u_1$
 and 
 $u_2$
, there are only
$u_2$
, there are only 
 $3$
 spaced left in the DOSP. So, we have
$3$
 spaced left in the DOSP. So, we have 
 $3$
 choices for
$3$
 choices for 
 $q_3=7$
 and
$q_3=7$
 and 
 $q_5=16$
, which corresponds to
$q_5=16$
, which corresponds to 
 $9$
 choices to finish off the DOSP. This part corresponds to the right-most factor
$9$
 choices to finish off the DOSP. This part corresponds to the right-most factor 
 $(k-i)^{r-j}$
 in the formula from the lemma.
$(k-i)^{r-j}$
 in the formula from the lemma.
Proof of Lemma 4.13.
 For the purpose of this proof, we will endow the set 
 ${\mathbb Z}/k{\mathbb Z}$
 with a total ordering induced by identifying it with the set
${\mathbb Z}/k{\mathbb Z}$
 with a total ordering induced by identifying it with the set 
 $\{{0,1,\ldots ,k-1}\}$
. We write the elements of J as
$\{{0,1,\ldots ,k-1}\}$
. We write the elements of J as 
 $u_1,u_2,\ldots ,u_h$
. Let
$u_1,u_2,\ldots ,u_h$
. Let 
 $$\begin{align*}D = ((L_1, \ell_1), (L_2, \ell_2), \dots, (L_t, \ell_t)) \in \bigcap_{u \in J} \mathcal D^\tau_u \end{align*}$$
$$\begin{align*}D = ((L_1, \ell_1), (L_2, \ell_2), \dots, (L_t, \ell_t)) \in \bigcap_{u \in J} \mathcal D^\tau_u \end{align*}$$
be a DOSP. Without loss of generality, we may assume that 
 $L_1 \subseteq u_1$
. Since the turning number of D is
$L_1 \subseteq u_1$
. Since the turning number of D is 
 $\tau $
, we may assume that
$\tau $
, we may assume that 
 $f_D(L_1) = 0$
 and
$f_D(L_1) = 0$
 and 
 $f_D(\sigma (L_1)) = \tau $
. For each
$f_D(\sigma (L_1)) = \tau $
. For each 
 $u_a$
 with
$u_a$
 with 
 $a \in \{2, \dots h\}$
, there exists a set L of D such that
$a \in \{2, \dots h\}$
, there exists a set L of D such that 
 $0 < f_D(L) < \tau $
 with
$0 < f_D(L) < \tau $
 with 
 $L \subseteq u_a $
 and we write
$L \subseteq u_a $
 and we write 
 $p_a = f_D(L)$
 for the value of this function on L. We also fix
$p_a = f_D(L)$
 for the value of this function on L. We also fix 
 $p_1=0$
. Let us count the DOSPs D, as above, such that
$p_1=0$
. Let us count the DOSPs D, as above, such that 
 $p_2 < \dots < p_h$
.
$p_2 < \dots < p_h$
.
 Suppose that we are given the values of function 
 $f_D(L)$
 for each set
$f_D(L)$
 for each set 
 $L \subseteq u_a$
 over all
$L \subseteq u_a$
 over all 
 $u_a$
. Let us count the ways to distribute the elements of the
$u_a$
. Let us count the ways to distribute the elements of the 
 $u_a$
 into the DOSP if
$u_a$
 into the DOSP if 
 $u_a$
 is already placed. For each
$u_a$
 is already placed. For each 
 $b \in \operatorname {\mathrm {ind}}(u_a)$
, there are
$b \in \operatorname {\mathrm {ind}}(u_a)$
, there are 
 $k/o(\tau )$
 different possible values for
$k/o(\tau )$
 different possible values for 
 $f_D(q_b)$
. So, in total, there are
$f_D(q_b)$
. So, in total, there are 
 $o(\tau )^{j-1}$
 possible choices for the positions of the q’s, where the first q is put in the first position and the other q’s, of which there are
$o(\tau )^{j-1}$
 possible choices for the positions of the q’s, where the first q is put in the first position and the other q’s, of which there are 
 $j-1$
, are placed relatively to the first.
$j-1$
, are placed relatively to the first.
 Now let us count the ways to position the sets 
 $u_1, \dots , u_h$
 in the DOSP. The position of
$u_1, \dots , u_h$
 in the DOSP. The position of 
 $u_1$
 and its corresponding sets of the DOSP is fixed. Then we use a stars and bars argument to count the placements of the remaining spaces between the
$u_1$
 and its corresponding sets of the DOSP is fixed. Then we use a stars and bars argument to count the placements of the remaining spaces between the 
 $u_a$
’s:
$u_a$
’s: 
 $$\begin{align*}(L(u_1)) \ \square \ (L(u_2)) \ \square \ (L(u_3)) \ \square \ \dots \ \square \ (L(u_h)) \ \square, \end{align*}$$
$$\begin{align*}(L(u_1)) \ \square \ (L(u_2)) \ \square \ (L(u_3)) \ \square \ \dots \ \square \ (L(u_h)) \ \square, \end{align*}$$
where 
 $L(u_a)$
 is the set of the DOSP with
$L(u_a)$
 is the set of the DOSP with 
 $f_D(L(u_a))=p_a$
 and the boxes represent some number of spaces. Since each set
$f_D(L(u_a))=p_a$
 and the boxes represent some number of spaces. Since each set 
 $L(u_a)$
 is a bad set, it takes up at least
$L(u_a)$
 is a bad set, it takes up at least 
 $|L(u_a)|$
 spaces of the DOSP. Since
$|L(u_a)|$
 spaces of the DOSP. Since 
 $u_a$
 is a
$u_a$
 is a 
 $\sigma $
-orbit of bad sets, we have that each set that partitions
$\sigma $
-orbit of bad sets, we have that each set that partitions 
 $u_a$
 is bad and together they take up
$u_a$
 is bad and together they take up 
 $|u_a|$
 spaces of the DOSP. So the bad sets, whose union is
$|u_a|$
 spaces of the DOSP. So the bad sets, whose union is 
 $u_1 \sqcup \dots \sqcup u_h$
, take up i spaces of the DOSP. Hence, in total there are
$u_1 \sqcup \dots \sqcup u_h$
, take up i spaces of the DOSP. Hence, in total there are 
 $k-i$
 spaces to place between the
$k-i$
 spaces to place between the 
 $u_a$
’s. Note that whenever we place one space, its
$u_a$
’s. Note that whenever we place one space, its 
 $\sigma $
-orbit has length
$\sigma $
-orbit has length 
 $o(\tau )$
. So we are free to choose the positions of
$o(\tau )$
. So we are free to choose the positions of 
 $(k-i)/o(\tau )$
 spaces and the rest are determined by their
$(k-i)/o(\tau )$
 spaces and the rest are determined by their 
 $\sigma $
-orbits. So by stars and bars there are
$\sigma $
-orbits. So by stars and bars there are 
 $\binom {(k-i)/o(\tau ) + h-1}{(k-i)/o(\tau )} = \binom {(k-i)/o(\tau ) + h-1}{h-1}$
 many ways place the spaces.
$\binom {(k-i)/o(\tau ) + h-1}{(k-i)/o(\tau )} = \binom {(k-i)/o(\tau ) + h-1}{h-1}$
 many ways place the spaces.
 So far, we have fixed the position of all cycles of 
 $\sigma $
 in
$\sigma $
 in 
 $u_1, \dots , u_h$
, that is, we have placed j cycles into the DOSP. There are
$u_1, \dots , u_h$
, that is, we have placed j cycles into the DOSP. There are 
 $r-j$
 remaining cycles. Placing a cycle
$r-j$
 remaining cycles. Placing a cycle 
 $C_c$
 into the DOSP is equivalent to choosing the position of
$C_c$
 into the DOSP is equivalent to choosing the position of 
 $q_c$
. Each
$q_c$
. Each 
 $q_c$
 can be placed into the
$q_c$
 can be placed into the 
 $k-i$
 spaces. Hence, there are
$k-i$
 spaces. Hence, there are 
 $(k-i)^{r-j}$
 DOSPs with the given
$(k-i)^{r-j}$
 DOSPs with the given 
 $u_a$
 positions.
$u_a$
 positions.
 Finally, there are 
 $(h-1)!$
 different total orderings of
$(h-1)!$
 different total orderings of 
 $p_2, \dots , p_h$
 and each gives the same number of DOSPs. So, the number of
$p_2, \dots , p_h$
 and each gives the same number of DOSPs. So, the number of 
 $\sigma $
-fixed non-hypersimplicial DOSPs with turning number
$\sigma $
-fixed non-hypersimplicial DOSPs with turning number 
 $\tau $
 is:
$\tau $
 is: 
 $$\begin{align*}\kern-24pt \binom{(k-i)/o(\tau)+h-1}{h-1}(h-1)!o(\tau)^{j-1}(k-i)^{r-j} = \frac{((k-i)/o(\tau)+h-1)!}{((k-i)/o(\tau))!}o(\tau)^{j-1}(k-i)^{r-j}.\\[-45pt] \end{align*}$$
$$\begin{align*}\kern-24pt \binom{(k-i)/o(\tau)+h-1}{h-1}(h-1)!o(\tau)^{j-1}(k-i)^{r-j} = \frac{((k-i)/o(\tau)+h-1)!}{((k-i)/o(\tau))!}o(\tau)^{j-1}(k-i)^{r-j}.\\[-45pt] \end{align*}$$
 We proceed to count the 
 $\sigma $
-fixed non-hypersimplicial DOSPs.
$\sigma $
-fixed non-hypersimplicial DOSPs.
Lemma 4.15. Fix 
 $2\leq k < n$
 and let
$2\leq k < n$
 and let 
 $\sigma \in S_n$
 be a permutation with cycle type
$\sigma \in S_n$
 be a permutation with cycle type 
 $(s_1,s_2,\dots ,s_r)$
. For each
$(s_1,s_2,\dots ,s_r)$
. For each 
 $i\in [n]$
, let
$i\in [n]$
, let 
 $\lambda _i$
 be the number of length i-cycles of
$\lambda _i$
 be the number of length i-cycles of 
 $\sigma $
. Let
$\sigma $
. Let 
 $g=\gcd (s_1,\ldots ,s_r,k)$
 and define the set
$g=\gcd (s_1,\ldots ,s_r,k)$
 and define the set 
 $T=\{{\tau \in {\mathbb Z}/k{\mathbb Z}\colon g\cdot \tau =0}\}$
. The number of
$T=\{{\tau \in {\mathbb Z}/k{\mathbb Z}\colon g\cdot \tau =0}\}$
. The number of 
 $\sigma $
-fixed non-hypersimplicial DOSPs is
$\sigma $
-fixed non-hypersimplicial DOSPs is 

Proof. Let 
 $\tau \in {\mathbb Z}/k{\mathbb Z}$
 satisfy
$\tau \in {\mathbb Z}/k{\mathbb Z}$
 satisfy 
 $g\cdot \tau = 0$
 and let
$g\cdot \tau = 0$
 and let 
 $\mathcal D^{\tau }$
 be the set of non-hypersimplicial
$\mathcal D^{\tau }$
 be the set of non-hypersimplicial 
 $\sigma $
-fixed DOSPs with turning number
$\sigma $
-fixed DOSPs with turning number 
 $\tau $
. The number of
$\tau $
. The number of 
 $\sigma $
-fixed non-hypersimplicial DOSPs is
$\sigma $
-fixed non-hypersimplicial DOSPs is 
 $$\begin{align*}\sum_{\tau\in T} \left\lvert {\mathcal D^{\tau}} \right\rvert. \end{align*}$$
$$\begin{align*}\sum_{\tau\in T} \left\lvert {\mathcal D^{\tau}} \right\rvert. \end{align*}$$
By the inclusion-exclusion principle, we have
 $$\begin{align*}\left\lvert {\mathcal D^{\tau}} \right\rvert = \left\lvert {\bigcup_{u\in\Lambda}\mathcal D_u^{\tau}} \right\rvert = \sum_{h\geq 1} (-1)^{h+1} \sum_{J\in\binom{\Lambda}{h}} \left\lvert {\bigcap_{u\in J}\mathcal D_u^{\tau}} \right\rvert. \end{align*}$$
$$\begin{align*}\left\lvert {\mathcal D^{\tau}} \right\rvert = \left\lvert {\bigcup_{u\in\Lambda}\mathcal D_u^{\tau}} \right\rvert = \sum_{h\geq 1} (-1)^{h+1} \sum_{J\in\binom{\Lambda}{h}} \left\lvert {\bigcap_{u\in J}\mathcal D_u^{\tau}} \right\rvert. \end{align*}$$
Given a non-empty subset 
 $J\subseteq \Lambda $
, suppose we have
$J\subseteq \Lambda $
, suppose we have 
 $u_1,u_2\in J$
. For a DOSP D to lie both in
$u_1,u_2\in J$
. For a DOSP D to lie both in 
 $\mathcal D_{u_1}$
 and
$\mathcal D_{u_1}$
 and 
 $\mathcal D_{u_2}$
, it means there exist (not necessarily distinct) sets
$\mathcal D_{u_2}$
, it means there exist (not necessarily distinct) sets 
 $L_1$
 and
$L_1$
 and 
 $L_2$
 whose
$L_2$
 whose 
 $\sigma $
-orbits are
$\sigma $
-orbits are 
 $u_1$
 and
$u_1$
 and 
 $u_2$
 respectively. In particular, if
$u_2$
 respectively. In particular, if 
 $u_1\cap u_2\neq \emptyset $
, the
$u_1\cap u_2\neq \emptyset $
, the 
 $\sigma $
-orbits must be the same and
$\sigma $
-orbits must be the same and 
 $u_1=u_2$
. Hence, we may always assume that the sets
$u_1=u_2$
. Hence, we may always assume that the sets 
 $u_i$
 contained in J are disjoint. It also follows that h is bounded by the number of sets
$u_i$
 contained in J are disjoint. It also follows that h is bounded by the number of sets 
 $L_i$
, which is k. In the case
$L_i$
, which is k. In the case 
 $h=k$
, every
$h=k$
, every 
 $L_i$
 satisfies
$L_i$
 satisfies 
 $ \left \lvert {L_i} \right \rvert = \ell _i = 1$
, which means that
$ \left \lvert {L_i} \right \rvert = \ell _i = 1$
, which means that 
 $n=k$
, a contradiction. Thus
$n=k$
, a contradiction. Thus 
 $1\leq h\leq k-1$
. We define the set
$1\leq h\leq k-1$
. We define the set 
 $$\begin{align*}\Lambda(h,i,j) = \{{ J\in\binom{\Lambda}{h} : \left\lvert {\bigcup_{u\in J}u} \right\rvert = i\text{ and } \left\lvert {\bigcup_{u\in J}\operatorname{\mathrm{ind}}(u)} \right\rvert = j }\} \end{align*}$$
$$\begin{align*}\Lambda(h,i,j) = \{{ J\in\binom{\Lambda}{h} : \left\lvert {\bigcup_{u\in J}u} \right\rvert = i\text{ and } \left\lvert {\bigcup_{u\in J}\operatorname{\mathrm{ind}}(u)} \right\rvert = j }\} \end{align*}$$
This is the collection of all h-subsets of 
 $\Lambda $
 involving exactly j distinct cycles that contain a total of i elements across all cycles. The cardinality of
$\Lambda $
 involving exactly j distinct cycles that contain a total of i elements across all cycles. The cardinality of 
 $\Lambda (h,i,j)$
 is exactly
$\Lambda (h,i,j)$
 is exactly 

which follows from the argument that there are  ways to partition j distinct cycles into h sets and that we choose
 ways to partition j distinct cycles into h sets and that we choose 
 $I_1$
 many fixed points of
$I_1$
 many fixed points of 
 $\sigma $
,
$\sigma $
, 
 $I_2$
 many
$I_2$
 many 
 $2$
-cycles of
$2$
-cycles of 
 $\sigma $
 and so on, such that
$\sigma $
 and so on, such that 
 $ \left \lvert {I} \right \rvert = j$
 and
$ \left \lvert {I} \right \rvert = j$
 and 
 $1\cdot I_1 + 2\cdot I_2 +\cdots + (k-1)\cdot I_{k-1} = i$
. With this, we apply Lemma 4.13 and rearrange the previous formula:
$1\cdot I_1 + 2\cdot I_2 +\cdots + (k-1)\cdot I_{k-1} = i$
. With this, we apply Lemma 4.13 and rearrange the previous formula: 

4.3 Proof of main result
 In this section, we simplify the formula in Lemma 4.15 to prove Theorem 4.1. We use the following identity derived from the falling factorial identity for Stirling numbers. For each 
 $j \ge 1$
, define the Laurent polynomial
$j \ge 1$
, define the Laurent polynomial 
 $F_j(y) \in \mathbb Q[y, y^{-1}]$
 by
$F_j(y) \in \mathbb Q[y, y^{-1}]$
 by 

These Laurent polynomials are, in fact, constants.
Lemma 4.16. We have 
 $F_j(y) = (-1)^{j+1}$
.
$F_j(y) = (-1)^{j+1}$
.
Proof. For each 
 $h \ge 0$
, notice that
$h \ge 0$
, notice that 
 $ (-1)^{h+1} (y+1)(y+2)\cdots (y+h-1) = \frac {(-y)_h}{y}. $
 By Proposition 2.7, with
$ (-1)^{h+1} (y+1)(y+2)\cdots (y+h-1) = \frac {(-y)_h}{y}. $
 By Proposition 2.7, with 
 $x = -y$
, we obtain
$x = -y$
, we obtain 
 $F_j(y)=\frac {(-y)^{j-1}}{y^{j-1}}$
, which concludes the proof.
$F_j(y)=\frac {(-y)^{j-1}}{y^{j-1}}$
, which concludes the proof.
Proof of Theorem 4.1.
By Corollary 4.12, we have
 $$\begin{align*}g k^{r-1} - H^*[1](\sigma) = -g\sum_{h = 1}^{k-1} \left( \sum_{I \in \mathcal I_i} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - i)^{r-1}. \end{align*}$$
$$\begin{align*}g k^{r-1} - H^*[1](\sigma) = -g\sum_{h = 1}^{k-1} \left( \sum_{I \in \mathcal I_i} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - i)^{r-1}. \end{align*}$$
We will show that the expression above is equal to the number of 
 $\sigma $
-fixed non-hypersimplicial
$\sigma $
-fixed non-hypersimplicial 
 $(k,n)$
-DOSPs. Define the set
$(k,n)$
-DOSPs. Define the set 
 $T = \{{\tau \in {\mathbb Z}/k{\mathbb Z} : g \tau = 0}\}$
 and note that
$T = \{{\tau \in {\mathbb Z}/k{\mathbb Z} : g \tau = 0}\}$
 and note that 
 $|T| = g$
. By Lemma 4.15, we have that the number of
$|T| = g$
. By Lemma 4.15, we have that the number of 
 $\sigma $
-fixed non-hypersimplicial
$\sigma $
-fixed non-hypersimplicial 
 $(k,n)$
-DOSPs is
$(k,n)$
-DOSPs is 

We reorder the sums in this expression to obtain

Next, we apply Lemma 4.16 to the above expression by setting 
 $y = (k-i)/o(\tau )$
. Note that
$y = (k-i)/o(\tau )$
. Note that 
 $o(\tau )^{j-1} = (k-i)^{j-1} (1/y)^{j-1}$
. So, the above expression is equal to the following
$o(\tau )^{j-1} = (k-i)^{j-1} (1/y)^{j-1}$
. So, the above expression is equal to the following 
 $$ \begin{align*} & \sum_{\tau \in T} \sum_{i = 1}^{k-1} \sum_{j = 1}^{i} \left( \sum_{\substack{I \in \mathcal I_i \\ |I| = j}} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k-i)^{r-j} (k-i)^{j-1} F_j((k-i)/o(\tau))\\& \quad = \sum_{\tau \in T} \sum_{i = 1}^{k-1} \sum_{j = 1}^{i} \left( \sum_{\substack{I \in \mathcal I_i \\ |I| = j}} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k-i)^{r-1} (-1)^{j+1}\\& \quad = -g \sum_{i = 1}^{k-1} \left( \sum_{I \in \mathcal I_i} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k-i)^{r-1} = gk^{r-1} - H^*[1](\sigma). \end{align*} $$
$$ \begin{align*} & \sum_{\tau \in T} \sum_{i = 1}^{k-1} \sum_{j = 1}^{i} \left( \sum_{\substack{I \in \mathcal I_i \\ |I| = j}} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k-i)^{r-j} (k-i)^{j-1} F_j((k-i)/o(\tau))\\& \quad = \sum_{\tau \in T} \sum_{i = 1}^{k-1} \sum_{j = 1}^{i} \left( \sum_{\substack{I \in \mathcal I_i \\ |I| = j}} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k-i)^{r-1} (-1)^{j+1}\\& \quad = -g \sum_{i = 1}^{k-1} \left( \sum_{I \in \mathcal I_i} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k-i)^{r-1} = gk^{r-1} - H^*[1](\sigma). \end{align*} $$
So, the number of 
 $\sigma $
-fixed non-hypersimplicial DOSPs is equal to
$\sigma $
-fixed non-hypersimplicial DOSPs is equal to 
 $gk^{r-1} - H^*[1](\sigma )$
. By Corollary 4.8, the number of
$gk^{r-1} - H^*[1](\sigma )$
. By Corollary 4.8, the number of 
 $\sigma $
-fixed
$\sigma $
-fixed 
 $(k,n)$
-DOSPs is
$(k,n)$
-DOSPs is 
 $gk^{r-1}$
. Therefore
$gk^{r-1}$
. Therefore 
 $H^*[1](\sigma )$
 is equal to the number of
$H^*[1](\sigma )$
 is equal to the number of 
 $\sigma $
-fixed hypersimplicial
$\sigma $
-fixed hypersimplicial 
 $(k,n)$
-DOSPs. This completes the proof.
$(k,n)$
-DOSPs. This completes the proof.
4.4 Recurrence relation
 In this section, we show that 
 $H^*(\Delta _{k,n}; S_n)[1](\sigma )$
 satisfies a recurrence relation similar to that for Eulerian numbers. Given
$H^*(\Delta _{k,n}; S_n)[1](\sigma )$
 satisfies a recurrence relation similar to that for Eulerian numbers. Given 
 $k \in {\mathbb Z}$
, a tuple
$k \in {\mathbb Z}$
, a tuple 
 $\lambda = (\lambda _1, \lambda _2, \dots , \lambda _{n}) \in {\mathbb Z}^{n}_{\ge 0}$
, and
$\lambda = (\lambda _1, \lambda _2, \dots , \lambda _{n}) \in {\mathbb Z}^{n}_{\ge 0}$
, and 
 $r \ge 1$
, we define
$r \ge 1$
, we define 
 $$\begin{align*}B(k, \lambda, r) = g(k, \lambda) \sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1} \end{align*}$$
$$\begin{align*}B(k, \lambda, r) = g(k, \lambda) \sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1} \end{align*}$$
where 
 $g(k, \lambda ) = \gcd (\{k\} \cup \{i \in [n] : \lambda _i \ge 1\})$
.
$g(k, \lambda ) = \gcd (\{k\} \cup \{i \in [n] : \lambda _i \ge 1\})$
.
 Suppose that 
 $\sigma \in S_n$
 has cycle type
$\sigma \in S_n$
 has cycle type 
 $(s_1, \dots , s_r)$
 and for each
$(s_1, \dots , s_r)$
 and for each 
 $i \in [n]$
 we denote by
$i \in [n]$
 we denote by 
 $\lambda _i$
 the number of cycles of
$\lambda _i$
 the number of cycles of 
 $\sigma $
 of length i. Then, by Theorem 4.1, we have
$\sigma $
 of length i. Then, by Theorem 4.1, we have 
 $H^*(\Delta _{k,n}; S_n)[1](\sigma ) = B(k, \lambda , r)$
.
$H^*(\Delta _{k,n}; S_n)[1](\sigma ) = B(k, \lambda , r)$
.
Proposition 4.17. We have 
 $B(k, \lambda , r) = 0$
 if
$B(k, \lambda , r) = 0$
 if 
 $k < 1$
,
$k < 1$
, 
 $B(1, \lambda , r) = gk^{r-1}$
, and
$B(1, \lambda , r) = gk^{r-1}$
, and 
 $B(k, \lambda , r) = gk^{r-1}$
 if
$B(k, \lambda , r) = gk^{r-1}$
 if 
 $\lambda _1 = \cdots = \lambda _{k-1} = 0$
. Suppose there exists
$\lambda _1 = \cdots = \lambda _{k-1} = 0$
. Suppose there exists 
 $a \in [k-1]$
 such that
$a \in [k-1]$
 such that 
 $\lambda _a \ge 1$
. Define
$\lambda _a \ge 1$
. Define 
 $$\begin{align*}\lambda' = (\lambda_1, \dots, \lambda_{a-1}, \lambda_a - 1, \lambda_{a+1}, \dots, \lambda_n). \end{align*}$$
$$\begin{align*}\lambda' = (\lambda_1, \dots, \lambda_{a-1}, \lambda_a - 1, \lambda_{a+1}, \dots, \lambda_n). \end{align*}$$
Then, we have the recurrence relation
 $$\begin{align*}B(k, \lambda, r) = \frac{g(k, \lambda)}{g(k, \lambda')} B(k, \lambda', r) - \frac{g(k, \lambda)}{g(k-a,\lambda')} B(k-a, \lambda', r). \end{align*}$$
$$\begin{align*}B(k, \lambda, r) = \frac{g(k, \lambda)}{g(k, \lambda')} B(k, \lambda', r) - \frac{g(k, \lambda)}{g(k-a,\lambda')} B(k-a, \lambda', r). \end{align*}$$
Proof. The first part of the result follows easily from the definition of 
 $B(k, \lambda , r)$
. For the recurrence relation, fix
$B(k, \lambda , r)$
. For the recurrence relation, fix 
 $a \in [k-1]$
 such that
$a \in [k-1]$
 such that 
 $\lambda _a \ge 1$
. To simplify notation, we write
$\lambda _a \ge 1$
. To simplify notation, we write 
 $g = g(k, \lambda )$
,
$g = g(k, \lambda )$
, 
 $g' = g(k, \lambda ')$
, and
$g' = g(k, \lambda ')$
, and 
 $g" = g(k-a, \lambda ')$
. First, we apply Pascal’s identity
$g" = g(k-a, \lambda ')$
. First, we apply Pascal’s identity 
 $$ \begin{align*} B(k, \lambda, r) &= g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1} \\ &= g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1}\cdots \binom{\lambda_a - 1}{I_a} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1} \\ & \quad +g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1}\cdots \binom{\lambda_a - 1}{I_a - 1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1}. \end{align*} $$
$$ \begin{align*} B(k, \lambda, r) &= g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1} \\ &= g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1}\cdots \binom{\lambda_a - 1}{I_a} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1} \\ & \quad +g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1}\cdots \binom{\lambda_a - 1}{I_a - 1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1}. \end{align*} $$
 The first sum coincides with 
 $(g/g') B(k, \lambda ', r)$
. For the second sum, we re-index as follows
$(g/g') B(k, \lambda ', r)$
. For the second sum, we re-index as follows 
 $$ \begin{align*} B(k, \lambda, r) - \frac{g}{g'} B(k, \lambda', r) &= g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1}\cdots \binom{\lambda_a - 1}{I_a - 1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1} \\ &= g\sum_{h = 0}^{k-1-a} \left( \sum_{I \in \mathcal I_{h+a}} (-1)^{|I|} \binom{\lambda'_1}{I_1}\cdots \binom{\lambda'_a}{I_a - 1} \cdots \binom{\lambda'_{k-1}}{I_{k-1}} \right) (k - h - a)^{r-1} \\ &= g\sum_{h = 0}^{(k-a)-1} \left( \sum_{I \in \mathcal I_{h}} (-1)^{|I|+1} \binom{\lambda'_1}{I_1}\cdots \binom{\lambda'_{k-1}}{I_{k-1}} \right) ((k - a) - h)^{r-1} \\ &= - \frac{g}{g"} B(k-a, \lambda', r). \end{align*} $$
$$ \begin{align*} B(k, \lambda, r) - \frac{g}{g'} B(k, \lambda', r) &= g\sum_{h = 0}^{k-1} \left( \sum_{I \in \mathcal I_h} (-1)^{|I|} \binom{\lambda_1}{I_1}\cdots \binom{\lambda_a - 1}{I_a - 1} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \right) (k - h)^{r-1} \\ &= g\sum_{h = 0}^{k-1-a} \left( \sum_{I \in \mathcal I_{h+a}} (-1)^{|I|} \binom{\lambda'_1}{I_1}\cdots \binom{\lambda'_a}{I_a - 1} \cdots \binom{\lambda'_{k-1}}{I_{k-1}} \right) (k - h - a)^{r-1} \\ &= g\sum_{h = 0}^{(k-a)-1} \left( \sum_{I \in \mathcal I_{h}} (-1)^{|I|+1} \binom{\lambda'_1}{I_1}\cdots \binom{\lambda'_{k-1}}{I_{k-1}} \right) ((k - a) - h)^{r-1} \\ &= - \frac{g}{g"} B(k-a, \lambda', r). \end{align*} $$
So we have shown that B satisfies the recursive relation, which concludes the proof.
Remark 4.18. The evaluation of 
 $H^*(\Delta _{k,n}; S_n)[1]$
 at the identity is equal to Eulerian number
$H^*(\Delta _{k,n}; S_n)[1]$
 at the identity is equal to Eulerian number 
 ${A(n-1, k-1)}$
, which is equal to
${A(n-1, k-1)}$
, which is equal to 
 $B(k, (n,0, \dots , 0), n)$
. The standard recurrence relation for Eulerian numbers is
$B(k, (n,0, \dots , 0), n)$
. The standard recurrence relation for Eulerian numbers is 
 $$\begin{align*}A(n,k) = (k+1)A(n-1,k) + (n-k)A(n-1,k-1). \end{align*}$$
$$\begin{align*}A(n,k) = (k+1)A(n-1,k) + (n-k)A(n-1,k-1). \end{align*}$$
The equation that we obtain, using Proposition 4.17, by starting with an Eulerian number is
 $$\begin{align*}B(k,(n,0, \dots, 0), n) = B(k, (n-1,0, \dots, 0), n) - B(k-1, (n-1, 0, \dots, 0), n). \end{align*}$$
$$\begin{align*}B(k,(n,0, \dots, 0), n) = B(k, (n-1,0, \dots, 0), n) - B(k-1, (n-1, 0, \dots, 0), n). \end{align*}$$
This relation differs a little from the standard recurrence as the terms appearing on the right-hand side are, in general, not Eulerian numbers.
5 The second hypersimplex
 In this section, we give a complete description of the coefficients of the 
 $H^*$
-polynomial for the second hypersimplex
$H^*$
-polynomial for the second hypersimplex 
 $\Delta _{2,n}$
. Throughout this section, we denote
$\Delta _{2,n}$
. Throughout this section, we denote 
 $H^* = H^*(\Delta _{2,n}, S_n)$
. We interpret the coefficients
$H^* = H^*(\Delta _{2,n}, S_n)$
. We interpret the coefficients 
 $H^*_i$
 of
$H^*_i$
 of 
 $H^*$
 in terms of DOSPs, see Definition 1.1, as well as actions of
$H^*$
 in terms of DOSPs, see Definition 1.1, as well as actions of 
 $S_n$
 on subsets and partitions of
$S_n$
 on subsets and partitions of 
 $[n]$
.
$[n]$
.
 
Notation. For each 
 $m \in [n]$
, we denote by
$m \in [n]$
, we denote by 
 $\rho _m$
 the character of the permutation representation of
$\rho _m$
 the character of the permutation representation of 
 $S_n$
 acting on
$S_n$
 acting on 
 $\binom {[n]}{m}$
. So, we have
$\binom {[n]}{m}$
. So, we have 
 $\rho _m(\sigma ) = |\{S \subseteq [n] : |S| = m,\, \sigma (S) = S\}|$
 for each
$\rho _m(\sigma ) = |\{S \subseteq [n] : |S| = m,\, \sigma (S) = S\}|$
 for each 
 $\sigma \in S_n$
. By taking complements, we have
$\sigma \in S_n$
. By taking complements, we have 
 $\rho _{n-m} = \rho _m$
 for each m. For each
$\rho _{n-m} = \rho _m$
 for each m. For each 
 $\sigma \in S_n$
 and
$\sigma \in S_n$
 and 
 $i \in [n]$
, we write
$i \in [n]$
, we write 
 $\lambda _i$
 for the number of cycles of
$\lambda _i$
 for the number of cycles of 
 $\sigma $
 of length i. So, we have
$\sigma $
 of length i. So, we have  is the natural representation whose value at
 is the natural representation whose value at 
 $\sigma $
 is the number of fixed points of
$\sigma $
 is the number of fixed points of 
 $\sigma $
 and
$\sigma $
 and  is the trivial character. We define
 is the trivial character. We define 
 $\tau _m$
 to be the character of the permutation representation of
$\tau _m$
 to be the character of the permutation representation of 
 $S_n$
 acting on the set of partitions of
$S_n$
 acting on the set of partitions of 
 $[n]$
 into two parts: one of size m and the other of size
$[n]$
 into two parts: one of size m and the other of size 
 $n-m$
. Note that, unless n is even and
$n-m$
. Note that, unless n is even and 
 $m = n/2$
, we have that
$m = n/2$
, we have that 
 $\rho _m = \tau _m$
.
$\rho _m = \tau _m$
.
Theorem 5.1. Let 
 $n> 2$
. The coefficients of the
$n> 2$
. The coefficients of the 
 $H^*$
-polynomial
$H^*$
-polynomial 
 $H^*(\Delta _{2,n}; S_n)$
 are:
$H^*(\Delta _{2,n}; S_n)$
 are: 
- 
•  the trivial character, the trivial character,
- 
•  $H^*_1 = \rho _2 - \rho _1$
, $H^*_1 = \rho _2 - \rho _1$
,
- 
•  $H^*_m = \rho _{2m}$
 for each $H^*_m = \rho _{2m}$
 for each $2 \le m \le \lfloor n/2 \rfloor $
. $2 \le m \le \lfloor n/2 \rfloor $
.
The evaluation of 
 $H^*$
 at one is given by
$H^*$
 at one is given by  , which is a permutation character. If n is odd, then the leading coefficient is
, which is a permutation character. If n is odd, then the leading coefficient is 
 $H^*_{(n-1)/2} = \rho _1$
. Otherwise, if n is even, then the leading coefficient is
$H^*_{(n-1)/2} = \rho _1$
. Otherwise, if n is even, then the leading coefficient is  .
.
 Before we prove the theorem, we consider the formula for the coefficients 
 $H^*_m$
 in Theorem 3.3. We note that the final term, indexed by
$H^*_m$
 in Theorem 3.3. We note that the final term, indexed by 
 $h = k-1$
, has a simple description.
$h = k-1$
, has a simple description.
Proposition 5.2. Let 
 $\sigma \in S_n$
 and
$\sigma \in S_n$
 and 
 $m \ge 0$
. We have
$m \ge 0$
. We have 
 $$\begin{align*}|{\Phi}_1(\sigma, m)| = \begin{cases} 1 & \text{if } m = 0, \\ 0 & \text{otherwise}. \end{cases} \end{align*}$$
$$\begin{align*}|{\Phi}_1(\sigma, m)| = \begin{cases} 1 & \text{if } m = 0, \\ 0 & \text{otherwise}. \end{cases} \end{align*}$$
Proof. The set 
 ${\Phi }_1(\sigma , m)$
 consists of functions
${\Phi }_1(\sigma , m)$
 consists of functions 
 $f \colon [r] \rightarrow \{0\}$
 such that
$f \colon [r] \rightarrow \{0\}$
 such that 
 $\sum _{i = 1}^r f(i) s_i = m$
. There is only one such function, which belongs to the set
$\sum _{i = 1}^r f(i) s_i = m$
. There is only one such function, which belongs to the set 
 ${\Phi }_1(\sigma , 0)$
.
${\Phi }_1(\sigma , 0)$
.
 To prove Theorem 5.1, we require the following result about 
 $S_n$
-representations.
$S_n$
-representations.
Lemma 5.3. Let n be even. The following equation of 
 $S_n$
-representations holds
$S_n$
-representations holds 
 $$\begin{align*}\sum_{m = 0}^{n/2} \rho_{2m} = \sum_{m = 0}^{n/2} \tau_m. \end{align*}$$
$$\begin{align*}\sum_{m = 0}^{n/2} \rho_{2m} = \sum_{m = 0}^{n/2} \tau_m. \end{align*}$$
Proof. Fix a permutation 
 $\sigma $
 with cycle type
$\sigma $
 with cycle type 
 $(s_1, s_2, \dots , s_r)$
. It suffices to show that the number of subsets of
$(s_1, s_2, \dots , s_r)$
. It suffices to show that the number of subsets of 
 $[n]$
 with even size that are fixed by
$[n]$
 with even size that are fixed by 
 $\sigma $
 is equal to the number of two-part partitions of
$\sigma $
 is equal to the number of two-part partitions of 
 $[n]$
 that are fixed by
$[n]$
 that are fixed by 
 $\sigma $
. Suppose that
$\sigma $
. Suppose that 
 $A \sqcup B = [n]$
 is a two-part partition, then we write
$A \sqcup B = [n]$
 is a two-part partition, then we write 
 $AB := \{A, B\}$
 for the partition of
$AB := \{A, B\}$
 for the partition of 
 $[n]$
 into A and B. Given a partition
$[n]$
 into A and B. Given a partition 
 $AB$
 of
$AB$
 of 
 $[n]$
, we write
$[n]$
, we write 
 $\sigma (AB)$
 for the partition of
$\sigma (AB)$
 for the partition of 
 $[n]$
 with parts
$[n]$
 with parts 
 $\sigma (A)$
 and
$\sigma (A)$
 and 
 $\sigma (B)$
. We define the sets
$\sigma (B)$
. We define the sets 
 $$\begin{align*}L^\sigma = \{A \subseteq [n] : |A| \text{ is even},\, \sigma(A) = A \} \text{ and } R^\sigma = \{AB : A \sqcup B = [n],\, \sigma(AB) = AB \}.\end{align*}$$
$$\begin{align*}L^\sigma = \{A \subseteq [n] : |A| \text{ is even},\, \sigma(A) = A \} \text{ and } R^\sigma = \{AB : A \sqcup B = [n],\, \sigma(AB) = AB \}.\end{align*}$$
We will show that 
 $|L^\sigma | = |R^\sigma |$
.
$|L^\sigma | = |R^\sigma |$
.
 We first consider the special case where each cycle of 
 $\sigma $
 has odd length. Each element
$\sigma $
 has odd length. Each element 
 $A \in L^\sigma $
 is the union of the supports of cycles of
$A \in L^\sigma $
 is the union of the supports of cycles of 
 $\sigma $
. Since
$\sigma $
. Since 
 $|A|$
 is even and each cycle has odd length, it follow that A is the union of an even number of cycle supports. The indices of the cycles whose supports form A uniquely determine A, and any even subset of cycles forms a unique set A. So we have
$|A|$
 is even and each cycle has odd length, it follow that A is the union of an even number of cycle supports. The indices of the cycles whose supports form A uniquely determine A, and any even subset of cycles forms a unique set A. So we have 
 $$ \begin{align*} |L^\sigma| &= |\{S \subseteq [r] : \Sigma S \text{ is even} \}| \\ &= |\{S \subseteq [r] : |S| \text{ is even} \}|\\ &= 2^{r-1}. \end{align*} $$
$$ \begin{align*} |L^\sigma| &= |\{S \subseteq [r] : \Sigma S \text{ is even} \}| \\ &= |\{S \subseteq [r] : |S| \text{ is even} \}|\\ &= 2^{r-1}. \end{align*} $$
On the other hand, for each subset 
 $S \subset [r]$
, we obtain a partition
$S \subset [r]$
, we obtain a partition 
 $TU$
 where T is the union of supports of the cycles in
$TU$
 where T is the union of supports of the cycles in 
 $\sigma $
 indexed by S and
$\sigma $
 indexed by S and  is the complement of T. Observe that every
 is the complement of T. Observe that every 
 $\sigma $
-invariant partition arises in this way because each cycle has odd length. Moreover, each partition arises from a subset
$\sigma $
-invariant partition arises in this way because each cycle has odd length. Moreover, each partition arises from a subset 
 $S \subseteq [k]$
 or its complement
$S \subseteq [k]$
 or its complement  . So we conclude that
. So we conclude that 
 $|R^\sigma | = 2^{k-1} = |L^\sigma |$
. This concludes the special case.
$|R^\sigma | = 2^{k-1} = |L^\sigma |$
. This concludes the special case.
 Suppose that 
 $\sigma $
 contains a cycle of even length. Without loss of generality we assume that
$\sigma $
 contains a cycle of even length. Without loss of generality we assume that 
 $s := s_r$
 is even. We prove
$s := s_r$
 is even. We prove 
 $|L^\sigma | = |R^\sigma |$
 by induction on r. For the base case with
$|L^\sigma | = |R^\sigma |$
 by induction on r. For the base case with 
 $r = 1$
, we have that
$r = 1$
, we have that 
 $\sigma = (\sigma _1 \ \sigma _2 \ \dots \ \sigma _n)$
 is an n-cycle where n is even. It is easy to see that
$\sigma = (\sigma _1 \ \sigma _2 \ \dots \ \sigma _n)$
 is an n-cycle where n is even. It is easy to see that 
 $$\begin{align*}L^\sigma = \{\emptyset,\, [n]\} \quad \text{and} \quad R^\sigma = \{\{\emptyset,[n]\},\, \{\sigma_1\sigma_3\dots\sigma_{n-1}, \sigma_2\sigma_4\dots\sigma_n\} \}. \end{align*}$$
$$\begin{align*}L^\sigma = \{\emptyset,\, [n]\} \quad \text{and} \quad R^\sigma = \{\{\emptyset,[n]\},\, \{\sigma_1\sigma_3\dots\sigma_{n-1}, \sigma_2\sigma_4\dots\sigma_n\} \}. \end{align*}$$
So we have 
 $|L^\sigma | = 2 = |R^\sigma |$
.
$|L^\sigma | = 2 = |R^\sigma |$
.
 For the induction step, assume that 
 $r> 1$
 and consider a permutation
$r> 1$
 and consider a permutation 
 $\tau $
 that has cycle type
$\tau $
 that has cycle type 
 $s_1,s_2, \dots , s_{r-1}$
. Without loss of generality, let us assume that
$s_1,s_2, \dots , s_{r-1}$
. Without loss of generality, let us assume that 
 $\tau $
 is equal to the permutation
$\tau $
 is equal to the permutation 
 $\sigma $
 restricted to
$\sigma $
 restricted to 
 $[n-s_r]$
. Define the set
$[n-s_r]$
. Define the set  . It is easy to see that
. It is easy to see that 
 $$\begin{align*}L^\sigma = L^\tau \sqcup \{A \sqcup S : A \in L^\tau\} \end{align*}$$
$$\begin{align*}L^\sigma = L^\tau \sqcup \{A \sqcup S : A \in L^\tau\} \end{align*}$$
and so we have 
 $|L^\sigma | = 2|L^\tau |$
. On the other hand, let us consider a partition
$|L^\sigma | = 2|L^\tau |$
. On the other hand, let us consider a partition 
 $AB \in R^\tau $
. If
$AB \in R^\tau $
. If 
 $\sigma (A) = A$
, then we have that the partitions
$\sigma (A) = A$
, then we have that the partitions 
 $(A \sqcup S)B$
 and
$(A \sqcup S)B$
 and 
 $A(B \sqcup S)$
 lie in
$A(B \sqcup S)$
 lie in 
 $R^\sigma $
. On the other hand, if
$R^\sigma $
. On the other hand, if 
 $\sigma (A) = B$
, then write
$\sigma (A) = B$
, then write 
 $(c_1, c_2, \dots , c_{s_r})$
 for the cycle of
$(c_1, c_2, \dots , c_{s_r})$
 for the cycle of 
 $\sigma $
 supported on S. Then we have
$\sigma $
 supported on S. Then we have 
 $$\begin{align*}(A \sqcup \{c_1, c_3, \dots, c_{s_k-1} \})(B \sqcup \{c_2, c_4, \dots, c_{s_k} \}) \text{ and } (A \sqcup \{c_2, c_4, \dots, c_{s_k} \})(B \sqcup \{c_1, c_3, \dots, c_{s_k-1} \}) \end{align*}$$
$$\begin{align*}(A \sqcup \{c_1, c_3, \dots, c_{s_k-1} \})(B \sqcup \{c_2, c_4, \dots, c_{s_k} \}) \text{ and } (A \sqcup \{c_2, c_4, \dots, c_{s_k} \})(B \sqcup \{c_1, c_3, \dots, c_{s_k-1} \}) \end{align*}$$
are elements of 
 $R^\sigma $
. Every element of
$R^\sigma $
. Every element of 
 $R^\sigma $
 arises uniquely in one of the ways described above. So it follows that
$R^\sigma $
 arises uniquely in one of the ways described above. So it follows that 
 $|R^\sigma | = 2|R^\tau |$
. By induction, we have
$|R^\sigma | = 2|R^\tau |$
. By induction, we have 
 $|L^\tau | = |R^\tau |$
 and so we deduce that
$|L^\tau | = |R^\tau |$
 and so we deduce that 
 $|L^\sigma | = |R^\sigma |$
 and we are done.
$|L^\sigma | = |R^\sigma |$
 and we are done.
Proof of Theorem 5.1.
 Fix 
 $\sigma \in S_n$
 with cycle type
$\sigma \in S_n$
 with cycle type 
 $(s_1, \dots , s_r)$
 and denote by
$(s_1, \dots , s_r)$
 and denote by 
 $C_1, \dots , C_r$
 the cycle sets of
$C_1, \dots , C_r$
 the cycle sets of 
 $\sigma $
 such that
$\sigma $
 such that 
 $|C_i| = s_i$
 for each
$|C_i| = s_i$
 for each 
 $i \in [r]$
. Let us consider the coefficients given by Theorem 3.3 and Proposition 5.2 for the second hypersimplex
$i \in [r]$
. Let us consider the coefficients given by Theorem 3.3 and Proposition 5.2 for the second hypersimplex 
 $\Delta _{2,n}$
. We have
$\Delta _{2,n}$
. We have 
 $$\begin{align*}H^*_m(\sigma) = |{\Phi}_2(\sigma, 2m)| - \lambda_1 |{\Phi}_1(\sigma, m-1)| = \begin{cases} |{\Phi}_2(\sigma, 2m)| & \text{if } m \neq 1, \\ |{\Phi}_2(\sigma, 2m)| - \lambda_1 & \text{if } m = 1. \end{cases} \end{align*}$$
$$\begin{align*}H^*_m(\sigma) = |{\Phi}_2(\sigma, 2m)| - \lambda_1 |{\Phi}_1(\sigma, m-1)| = \begin{cases} |{\Phi}_2(\sigma, 2m)| & \text{if } m \neq 1, \\ |{\Phi}_2(\sigma, 2m)| - \lambda_1 & \text{if } m = 1. \end{cases} \end{align*}$$
The value 
 $\lambda _1$
 is equal to the number of fixed points of
$\lambda _1$
 is equal to the number of fixed points of 
 $\sigma $
. So
$\sigma $
. So 
 $\lambda _1(\sigma ) = \rho _1 (\sigma )$
 is the character of the natural representation of
$\lambda _1(\sigma ) = \rho _1 (\sigma )$
 is the character of the natural representation of 
 $S_n$
. On the other hand, the set
$S_n$
. On the other hand, the set 
 ${\Phi }_2(\sigma , 2m)$
 contains all functions
${\Phi }_2(\sigma , 2m)$
 contains all functions 
 $f \colon [r] \rightarrow \{0,1\}$
 such that
$f \colon [r] \rightarrow \{0,1\}$
 such that 
 $\sum _{i = 1}^r f(i) s_i = 2m$
. There is a natural correspondence between
$\sum _{i = 1}^r f(i) s_i = 2m$
. There is a natural correspondence between 
 $f \in {\Phi }_2(\sigma , 2m)$
 and subsets
$f \in {\Phi }_2(\sigma , 2m)$
 and subsets 
 $F \subseteq [n]$
 with
$F \subseteq [n]$
 with 
 $|F| = 2m$
 and
$|F| = 2m$
 and 
 $\sigma (F) = F$
, which is given by
$\sigma (F) = F$
, which is given by 
 $$\begin{align*}f \mapsto F = \bigcup_{i \in f^{-1}(1)} C_i. \end{align*}$$
$$\begin{align*}f \mapsto F = \bigcup_{i \in f^{-1}(1)} C_i. \end{align*}$$
So 
 $|{\Phi }_2(\sigma , 2m)| = \rho _{2m}(\sigma )$
 is equal to the permutation character of
$|{\Phi }_2(\sigma , 2m)| = \rho _{2m}(\sigma )$
 is equal to the permutation character of 
 $S_n$
 acting on
$S_n$
 acting on 
 $\binom {[n]}{2m}$
. This proves that
$\binom {[n]}{2m}$
. This proves that  ,
, 
 $H^*_1 = \rho _2 - \rho _1$
, and
$H^*_1 = \rho _2 - \rho _1$
, and 
 $H^*_m = \rho _{2m}$
 for each
$H^*_m = \rho _{2m}$
 for each 
 $2 \le m \le \lfloor n/2 \rfloor $
. By Corollary 3.4, if n is odd then the leading coefficient is
$2 \le m \le \lfloor n/2 \rfloor $
. By Corollary 3.4, if n is odd then the leading coefficient is 
 $H^*_{(n-1)/2} = \rho _{n-1} = \rho _1$
, otherwise if n is even then the leading coefficient is
$H^*_{(n-1)/2} = \rho _{n-1} = \rho _1$
, otherwise if n is even then the leading coefficient is  .
.
 It remains to show that  . If n is odd, then the result follows from the above and the fact that
. If n is odd, then the result follows from the above and the fact that 
 $\rho _m = \tau _m$
 for each
$\rho _m = \tau _m$
 for each 
 $m \in [n]$
. On the other hand, if n is even, then result follows from Lemma 5.3.
$m \in [n]$
. On the other hand, if n is even, then result follows from Lemma 5.3.
 Theorem 5.1 allows us to give a complete combinatorial proof of effectiveness for the 
 $H^*$
-polynomial.
$H^*$
-polynomial.
Corollary 5.4. Fix n. Each coefficient of 
 $H^*(\Delta _{2,n}, S_n)$
 is an effective representation.
$H^*(\Delta _{2,n}, S_n)$
 is an effective representation.
Proof. Let 
 $m \in \{0,1,2, \dots , \lfloor n/2 \rfloor \}$
 and consider the
$m \in \{0,1,2, \dots , \lfloor n/2 \rfloor \}$
 and consider the 
 $t^m$
 coefficient
$t^m$
 coefficient 
 $H^*_m$
 of
$H^*_m$
 of 
 $H^*(\Delta _{2,n}, S_n)$
. If
$H^*(\Delta _{2,n}, S_n)$
. If 
 $m \neq 1$
, then
$m \neq 1$
, then 
 $H^*_m$
 is a permutation character, hence it is effective. Otherwise, if
$H^*_m$
 is a permutation character, hence it is effective. Otherwise, if 
 $m = 1$
 then let V and W be
$m = 1$
 then let V and W be 
 $\mathbb C S_n$
-modules with characters
$\mathbb C S_n$
-modules with characters 
 $\rho _1$
 and
$\rho _1$
 and 
 $\rho _2$
 respectively. Explicitly, we assume V has basis
$\rho _2$
 respectively. Explicitly, we assume V has basis 
 $e_i$
 with
$e_i$
 with 
 $i \in [n]$
 and action
$i \in [n]$
 and action 
 $\sigma (e_i) = e_{\sigma (i)}$
 and W has basis
$\sigma (e_i) = e_{\sigma (i)}$
 and W has basis 
 $f_I$
 with
$f_I$
 with 
 $I \in \binom {[n]}{2}$
 and action
$I \in \binom {[n]}{2}$
 and action 
 $\sigma (f_I) = f_{\sigma (I)}$
. Define the map
$\sigma (f_I) = f_{\sigma (I)}$
. Define the map 
 $ \varphi \colon V \rightarrow W$
 given by
$ \varphi \colon V \rightarrow W$
 given by 
 $ \varphi (e_i) = \sum _{j \neq i} f_{ij} $
. It is straightforward to show that
$ \varphi (e_i) = \sum _{j \neq i} f_{ij} $
. It is straightforward to show that 
 $\varphi $
 is an injective
$\varphi $
 is an injective 
 $\mathbb C S_n$
-module homomorphism, hence
$\mathbb C S_n$
-module homomorphism, hence 
 $H^*_1 = \rho _2 - \rho _1$
 is effective.
$H^*_1 = \rho _2 - \rho _1$
 is effective.
 So, the above corollary shows that the coefficient 
 $H^*_1$
 is a representation. However, as we will now show,
$H^*_1$
 is a representation. However, as we will now show, 
 $H^*_1$
 is special in that the trivial character does not appear as a direct summand.
$H^*_1$
 is special in that the trivial character does not appear as a direct summand.
Corollary 5.5. Fix 
 $n \ge 4$
 and let
$n \ge 4$
 and let 
 $0 \le m \le \lfloor n/2 \rfloor $
. Then the coefficient
$0 \le m \le \lfloor n/2 \rfloor $
. Then the coefficient 
 $H^*_m$
 of the equivariant
$H^*_m$
 of the equivariant 
 $H^*$
-polynomial of
$H^*$
-polynomial of 
 $\Delta _{2,n}$
 is a permutation character if and only if
$\Delta _{2,n}$
 is a permutation character if and only if 
 $m \neq 1$
. Moreover, the trivial character does not appear in
$m \neq 1$
. Moreover, the trivial character does not appear in 
 $H^*_1$
.
$H^*_1$
.
Proof. Suppose 
 $m \neq 1$
. By Theorem 5.1, we have that
$m \neq 1$
. By Theorem 5.1, we have that 
 $H^*_m$
 is the permutation character
$H^*_m$
 is the permutation character 
 $\rho _{2m}$
 if
$\rho _{2m}$
 if 
 $m> 1$
 and
$m> 1$
 and  if
 if 
 $m = 0$
. Otherwise, let
$m = 0$
. Otherwise, let 
 $m = 1$
 and assume by contradiction that
$m = 1$
 and assume by contradiction that 
 $H^*_1$
 is a permutation character. For any permutation character
$H^*_1$
 is a permutation character. For any permutation character 
 $\rho $
, a consequence of the Orbit-Stabiliser Theorem is that
$\rho $
, a consequence of the Orbit-Stabiliser Theorem is that  is equal to the number of orbits of the action. By Theorem 5.1, we have
 is equal to the number of orbits of the action. By Theorem 5.1, we have 
 $H^*_1 = \rho _2 - \rho _1$
. Since
$H^*_1 = \rho _2 - \rho _1$
. Since 
 $n \ge 4$
, we have
$n \ge 4$
, we have 
 $H^*_1(1_{S_n}) = \binom n2 - n \neq 0$
, hence
$H^*_1(1_{S_n}) = \binom n2 - n \neq 0$
, hence 
 $H^*_1 \neq 0$
. Since
$H^*_1 \neq 0$
. Since 
 $S_n$
 acts transitively on
$S_n$
 acts transitively on 
 $[n]$
 and the
$[n]$
 and the 
 $2$
-subsets of
$2$
-subsets of 
 $[n]$
, the action associated to
$[n]$
, the action associated to 
 $H^*_1$
 has
$H^*_1$
 has  orbits, a contradiction. This completes the proof.
 orbits, a contradiction. This completes the proof.
Remark 5.6. Each coefficient of the 
 $h^*$
-polynomial of the hypersimplex has a combinatorial interpretation in terms of DOSPs. Explicitly
$h^*$
-polynomial of the hypersimplex has a combinatorial interpretation in terms of DOSPs. Explicitly 
 $h^*_m$
 is the number of hypersimplicial
$h^*_m$
 is the number of hypersimplicial 
 $(k,n)$
-DOSPs with winding number m, see Definition 4.2. We note that the winding number is not invariant under the action of
$(k,n)$
-DOSPs with winding number m, see Definition 4.2. We note that the winding number is not invariant under the action of 
 $S_n$
, so the same interpretation does not apply in the most general setting. However, the winding number is invariant under the cyclic group
$S_n$
, so the same interpretation does not apply in the most general setting. However, the winding number is invariant under the cyclic group 
 $C_n \le S_n$
. In [Reference Elia, Kim and Supina9], it is shown that the coefficient of
$C_n \le S_n$
. In [Reference Elia, Kim and Supina9], it is shown that the coefficient of 
 $t^m$
 in
$t^m$
 in 
 $H^*(\Delta _{k,n}; C_n)[t]$
 is the number of
$H^*(\Delta _{k,n}; C_n)[t]$
 is the number of 
 $\sigma $
-fixed hypersimplicial
$\sigma $
-fixed hypersimplicial 
 $(k,n)$
-DOSPs with winding number m. In the case
$(k,n)$
-DOSPs with winding number m. In the case 
 $k= 2$
, this result can be deduced from Theorem 5.1 as follows. If
$k= 2$
, this result can be deduced from Theorem 5.1 as follows. If 
 $D = ((A, 1), (B, 1))$
 is a DOSP, then we define the set
$D = ((A, 1), (B, 1))$
 is a DOSP, then we define the set 
 $J(D)$
 of jumping points to be the set of
$J(D)$
 of jumping points to be the set of 
 $i \in [n]$
 such that i and
$i \in [n]$
 such that i and 
 $i+1$
 belong to different sets of D. Since
$i+1$
 belong to different sets of D. Since 
 $k = 2$
, the winding number of D is equal to half the number of jumping points. For
$k = 2$
, the winding number of D is equal to half the number of jumping points. For 
 $m = 0$
 and
$m = 0$
 and 
 $m \ge 2$
, the restriction
$m \ge 2$
, the restriction 
 $\operatorname {\mathrm {Res}}^{S_n}_{C_n}\rho _{2m} (\sigma )$
 counts the number of
$\operatorname {\mathrm {Res}}^{S_n}_{C_n}\rho _{2m} (\sigma )$
 counts the number of 
 $\sigma $
-fixed partitions
$\sigma $
-fixed partitions 
 $\{A, B\}$
 of
$\{A, B\}$
 of 
 $[n]$
 with
$[n]$
 with 
 $|A| = 2m$
. For each such partition there is a unique DOSP with jumping points A. This DOSP is
$|A| = 2m$
. For each such partition there is a unique DOSP with jumping points A. This DOSP is 
 $\sigma $
-fixed and has winding number m. Every such DOSP arises in this way and so
$\sigma $
-fixed and has winding number m. Every such DOSP arises in this way and so 
 $H^*_m$
 is the number of
$H^*_m$
 is the number of 
 $\sigma $
-fixed DOSPs with winding number m. In the case
$\sigma $
-fixed DOSPs with winding number m. In the case 
 $m = 1$
, we have that
$m = 1$
, we have that 
 $\operatorname {\mathrm {Res}}^{S_n}_{C_n} (\rho _2 - \rho _1)$
 is isomorphic to the permutation representation of
$\operatorname {\mathrm {Res}}^{S_n}_{C_n} (\rho _2 - \rho _1)$
 is isomorphic to the permutation representation of 
 $C_n$
 acting on the set of
$C_n$
 acting on the set of 
 $2$
-subsets
$2$
-subsets 
 $ij \in \binom {[n]}{2}$
 such that
$ij \in \binom {[n]}{2}$
 such that 
 $|i - j|> 1$
. For each such
$|i - j|> 1$
. For each such 
 $2$
-subset, we obtain a
$2$
-subset, we obtain a 
 $\sigma $
-fixed hypersimplicial DOSP, which concludes the proof.
$\sigma $
-fixed hypersimplicial DOSP, which concludes the proof.
6 Discussion and open problems
In this section, we discuss various open questions that arise from our investigation.
 
Formula for 
 $\boldsymbol {H}^{\boldsymbol {*}}$
 coefficients. First, let us consider the formula for the number of
$\boldsymbol {H}^{\boldsymbol {*}}$
 coefficients. First, let us consider the formula for the number of 
 $\sigma $
-fixed hypersimplicial DOSPs in Theorem 4.1. Our proof is quite technical but leads to a significant amount of cancellation and simplification. So we ask whether there is a simpler or direct combinatorial proof of this result.
$\sigma $
-fixed hypersimplicial DOSPs in Theorem 4.1. Our proof is quite technical but leads to a significant amount of cancellation and simplification. So we ask whether there is a simpler or direct combinatorial proof of this result.
 Next, consider the formula for the coefficients of 
 $H^*(\Delta _{k,n}; S_n)$
 in Theorem 3.3. By Proposition 3.2, the map
$H^*(\Delta _{k,n}; S_n)$
 in Theorem 3.3. By Proposition 3.2, the map 
 $\sigma \mapsto |{\Phi }_{k-h}(\sigma , m(k-h)-h)|$
 is a permutation character of
$\sigma \mapsto |{\Phi }_{k-h}(\sigma , m(k-h)-h)|$
 is a permutation character of 
 $S_n$
. However, for a given
$S_n$
. However, for a given 
 $h \in \{0,1, \dots , k-1\}$
 and
$h \in \{0,1, \dots , k-1\}$
 and 
 $I \in \mathcal I_h$
 the map
$I \in \mathcal I_h$
 the map 
 $$\begin{align*}\sigma \mapsto \binom{\lambda_1}{I_1} \binom{\lambda_2}{I_2} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \end{align*}$$
$$\begin{align*}\sigma \mapsto \binom{\lambda_1}{I_1} \binom{\lambda_2}{I_2} \cdots \binom{\lambda_{k-1}}{I_{k-1}} \end{align*}$$
is not generally the character of an 
 $S_n$
-representation, where
$S_n$
-representation, where 
 $\lambda _i$
 is the number of cycles of
$\lambda _i$
 is the number of cycles of 
 $\sigma $
 of length i. If
$\sigma $
 of length i. If 
 $k=2$
 then the only nontrivial map is
$k=2$
 then the only nontrivial map is 
 $\sigma \mapsto \lambda _1$
, which is the character of the natural representation. In Section 5, we use this observation to give a simple description of the coefficients of the
$\sigma \mapsto \lambda _1$
, which is the character of the natural representation. In Section 5, we use this observation to give a simple description of the coefficients of the 
 $H^*$
-polynomial. So, for
$H^*$
-polynomial. So, for 
 $k> 2$
, we ask if the above map is the character of an
$k> 2$
, we ask if the above map is the character of an 
 $S_n$
 representation. More generally we ask what are the irreducible representations appearing in
$S_n$
 representation. More generally we ask what are the irreducible representations appearing in 
 $H^*_m$
.
$H^*_m$
.
Question 6.1. For each 
 $m \ge 0$
, what is a multiplicity of each irreducible representation of
$m \ge 0$
, what is a multiplicity of each irreducible representation of 
 $S_n$
 in the coefficient
$S_n$
 in the coefficient 
 $H^*_m$
 of
$H^*_m$
 of 
 $H^*(\Delta _{k,n}; S_n)$
?
$H^*(\Delta _{k,n}; S_n)$
?
 
Symmetric triangulations. Our next questions concern the well-known unimodular triangulation of the hypersimplex [Reference Lam and Postnikov15]. We have seen in Corollary 5.5 that the trivial character does not appear in the coefficient of t in 
 $H^*(\Delta _{2,n}; S_n)$
, for each
$H^*(\Delta _{2,n}; S_n)$
, for each 
 $n \ge 4$
. Therefore, these hypersimplicies provide a family of counterexamples to Conjecture 1.5; see Example 1.6. In [Reference Stapledon24], Stapledon asks whether Conjecture 1.5 holds if we assume that P admits a G-invariant lattice triangulation. It would be interesting to know whether the hypersimplex supports this conjecture. In particular, we ask for which subgroups G of
$n \ge 4$
. Therefore, these hypersimplicies provide a family of counterexamples to Conjecture 1.5; see Example 1.6. In [Reference Stapledon24], Stapledon asks whether Conjecture 1.5 holds if we assume that P admits a G-invariant lattice triangulation. It would be interesting to know whether the hypersimplex supports this conjecture. In particular, we ask for which subgroups G of 
 $S_n$
, the hypersimplex
$S_n$
, the hypersimplex 
 $\Delta _{k,n}$
 admits a G-invariant lattice triangulation. And, in each of these cases, whether the coefficients of the
$\Delta _{k,n}$
 admits a G-invariant lattice triangulation. And, in each of these cases, whether the coefficients of the 
 $H^*$
-polynomial are permutation representations.
$H^*$
-polynomial are permutation representations.
Example 6.2. Consider the hypersimplex 
 $\Delta _{2,4} \subseteq {\mathbb R}^4$
. We write
$\Delta _{2,4} \subseteq {\mathbb R}^4$
. We write 
 $[I_1, I_2, I_3, I_4]$
 for the convex hull of the points
$[I_1, I_2, I_3, I_4]$
 for the convex hull of the points 
 $\{e_{I_1}, e_{I_2}, e_{I_3}, e_{I_4}\} \subseteq {\mathbb R}^4$
, where
$\{e_{I_1}, e_{I_2}, e_{I_3}, e_{I_4}\} \subseteq {\mathbb R}^4$
, where 
 $I_i$
 is a
$I_i$
 is a 
 $2$
-subsets of
$2$
-subsets of 
 $[4]$
 for each
$[4]$
 for each 
 $i \in [4]$
. The well-known lattice triangulation of the hypersimplex is given by the simplices
$i \in [4]$
. The well-known lattice triangulation of the hypersimplex is given by the simplices 
 $$\begin{align*}\nabla_1 = [12, 13, 14, 24],\quad \nabla_2 = [23, 24, 12, 13],\quad \nabla_3 = [34, 13, 23, 24],\quad \nabla_4 = [14, 24, 34, 13]. \end{align*}$$
$$\begin{align*}\nabla_1 = [12, 13, 14, 24],\quad \nabla_2 = [23, 24, 12, 13],\quad \nabla_3 = [34, 13, 23, 24],\quad \nabla_4 = [14, 24, 34, 13]. \end{align*}$$
Let 
 $a := (1 \ 2 \ 3 \ 4) \in S_n$
 be a
$a := (1 \ 2 \ 3 \ 4) \in S_n$
 be a 
 $4$
-cycle. Observe that cyclic group
$4$
-cycle. Observe that cyclic group 
 $C_4$
 generated by a acts on the set of simplices by permutation. For instance, we have
$C_4$
 generated by a acts on the set of simplices by permutation. For instance, we have 
 $$\begin{align*}a(\nabla_1) := [a(12), a(13), a(14), a(24)] = [23, 24, 12, 13] = \nabla_2 \end{align*}$$
$$\begin{align*}a(\nabla_1) := [a(12), a(13), a(14), a(24)] = [23, 24, 12, 13] = \nabla_2 \end{align*}$$
and, in general, we have 
 $a(\nabla _i) = \nabla _{a(i)}$
 for each
$a(\nabla _i) = \nabla _{a(i)}$
 for each 
 $i \in [4]$
. By Theorem 1.4, the coefficients of the
$i \in [4]$
. By Theorem 1.4, the coefficients of the 
 $H^*$
-polynomial
$H^*$
-polynomial 
 $H^*(\Delta _{2,4}; C_4)$
 are permutation representations, hence they satisfy Conjecture 1.5.
$H^*(\Delta _{2,4}; C_4)$
 are permutation representations, hence they satisfy Conjecture 1.5.
 Observe that the simplices are also invariant under the transposition 
 $b := (1 \ 3) \in S_4$
. Explicitly, we have
$b := (1 \ 3) \in S_4$
. Explicitly, we have 
 $b(\nabla _i) = \nabla _{b(i)}$
 for all
$b(\nabla _i) = \nabla _{b(i)}$
 for all 
 $i \in [4]$
. So, the dihedral group
$i \in [4]$
. So, the dihedral group 
 $D_8 := \langle a, b \rangle \le S_4$
 acts naturally on the lattice triangulation. By Theorem 5.1, to show that Conjecture 1.5 holds for
$D_8 := \langle a, b \rangle \le S_4$
 acts naturally on the lattice triangulation. By Theorem 5.1, to show that Conjecture 1.5 holds for 
 $H^*(\Delta _{2,4}; D_8)$
, it suffices to consider the coefficient of t, which is given by the character
$H^*(\Delta _{2,4}; D_8)$
, it suffices to consider the coefficient of t, which is given by the character 
 $H^*_1 = \rho _2 - \rho _1$
. One easily verifies this character is the permutation character of
$H^*_1 = \rho _2 - \rho _1$
. One easily verifies this character is the permutation character of 
 $D_8$
 acting on the set
$D_8$
 acting on the set 
 $\{13, 24\}$
, hence Conjecture 1.5 holds. We also observe that the set of simplices is not invariant under the transposition
$\{13, 24\}$
, hence Conjecture 1.5 holds. We also observe that the set of simplices is not invariant under the transposition 
 $(1 \ 2) \in S_n$
. Therefore, the subgroup
$(1 \ 2) \in S_n$
. Therefore, the subgroup 
 $D_8$
 is the maximal subgroup of
$D_8$
 is the maximal subgroup of 
 $S_4$
 that leaves the triangulation invariant.
$S_4$
 that leaves the triangulation invariant.
 Our computations suggest that the above example generalises to all hypersimplices. For each hypersimplex 
 $\Delta _{k,n}$
 with
$\Delta _{k,n}$
 with 
 $(k,n)$
 in the set
$(k,n)$
 in the set 
 $$\begin{align*}\{(2,4),\, (2,5),\, \dots,\, (2,10),\, (3,6),\, (3,7),\, (3,8)\} \end{align*}$$
$$\begin{align*}\{(2,4),\, (2,5),\, \dots,\, (2,10),\, (3,6),\, (3,7),\, (3,8)\} \end{align*}$$
we have verified that the symmetry group of the well-known triangulation is the dihedral group 
 $D_{2n} \le S_n$
. So, we make the following conjecture.
$D_{2n} \le S_n$
. So, we make the following conjecture.
Conjecture 6.3. For all 
 $0 < k < n$
, the well-known lattice triangulation of
$0 < k < n$
, the well-known lattice triangulation of 
 $\Delta _{k,n}$
 is invariant under the dihedral group
$\Delta _{k,n}$
 is invariant under the dihedral group 
 $\langle a, b\rangle = D_{2n} \le S_n$
 generated by the permutations
$\langle a, b\rangle = D_{2n} \le S_n$
 generated by the permutations 
 $$\begin{align*}a = (1 \ 2 \ \cdots \ n) \quad \text{and} \quad b = \begin{cases} (1 \ n)(2 \ n-1)(3 \ n-2) \cdots ((n-1)/2 \ (n+3)/2) & \text{if } n \text{ is odd},\\ (1 \ n)(2 \ n-1)(3 \ n-2) \cdots (n/2 \ n/2+1) & \text{if } n \text{ is even}. \end{cases} \end{align*}$$
$$\begin{align*}a = (1 \ 2 \ \cdots \ n) \quad \text{and} \quad b = \begin{cases} (1 \ n)(2 \ n-1)(3 \ n-2) \cdots ((n-1)/2 \ (n+3)/2) & \text{if } n \text{ is odd},\\ (1 \ n)(2 \ n-1)(3 \ n-2) \cdots (n/2 \ n/2+1) & \text{if } n \text{ is even}. \end{cases} \end{align*}$$
Moreover, 
 $D_{2n}$
 is the maximal subgroup of
$D_{2n}$
 is the maximal subgroup of 
 $S_n$
 for which the lattice triangulation is invariant.
$S_n$
 for which the lattice triangulation is invariant.
 We note that by Theorem 5.1, the coefficient 
 $H^*_1$
 of t in
$H^*_1$
 of t in 
 $H^*(\Delta _{2,n}; D_{2n})$
 is a permutation character. Explicitly, it is straightforward to show that
$H^*(\Delta _{2,n}; D_{2n})$
 is a permutation character. Explicitly, it is straightforward to show that 
 $H^*_1$
 is the permutation character of
$H^*_1$
 is the permutation character of 
 $D_{2n}$
 acting on the set
$D_{2n}$
 acting on the set 
 $\{\{i,j\} \subseteq [n] : |i-j| \ge 2 \}$
, which can be thought of as the set of diagonals of an n-gon. It would be interesting to know if this generalises to all hypersimplices
$\{\{i,j\} \subseteq [n] : |i-j| \ge 2 \}$
, which can be thought of as the set of diagonals of an n-gon. It would be interesting to know if this generalises to all hypersimplices 
 $\Delta _{k,n}$
 with
$\Delta _{k,n}$
 with 
 $k> 2$
.
$k> 2$
.
Question 6.4. Let 
 $k < n$
. For which groups
$k < n$
. For which groups 
 $G \le S_n$
 does there exist a G-invariant lattice triangulation of
$G \le S_n$
 does there exist a G-invariant lattice triangulation of 
 $\Delta _{k,n}$
? Given such G, does Conjecture 1.5 hold for
$\Delta _{k,n}$
? Given such G, does Conjecture 1.5 hold for 
 $H^*(\Delta _{k,n}; G)$
? Does Theorem 1.4 hold if we replace the cyclic group
$H^*(\Delta _{k,n}; G)$
? Does Theorem 1.4 hold if we replace the cyclic group 
 $C_n$
 with the dihedral group
$C_n$
 with the dihedral group 
 $D_{2n}$
 (as defined in Conjecture 6.3)?
$D_{2n}$
 (as defined in Conjecture 6.3)?
 We remark that the final question is well defined as the winding number of a DOSP is invariant under the action of the dihedral group. We also note that 
 $\Delta _{2,n}$
 does not admit a
$\Delta _{2,n}$
 does not admit a 
 $S_n$
-invariant lattice triangulation for any
$S_n$
-invariant lattice triangulation for any 
 $n \ge 4$
. To see this, we write
$n \ge 4$
. To see this, we write 
 $e_{ij} := e_i + e_j \in {\mathbb R}^n$
 for each
$e_{ij} := e_i + e_j \in {\mathbb R}^n$
 for each 
 $i,j \in [n]$
 and observe that any triangulation of
$i,j \in [n]$
 and observe that any triangulation of 
 $\Delta _{2,n}$
 must include a simplex
$\Delta _{2,n}$
 must include a simplex 
 $\nabla = \mathrm {Conv}\{e_{12}, e_{13},\ldots ,e_{1n},e_{ab}\}$
 with
$\nabla = \mathrm {Conv}\{e_{12}, e_{13},\ldots ,e_{1n},e_{ab}\}$
 with 
 $2 \le a < b \le n$
. This follows from the fact that
$2 \le a < b \le n$
. This follows from the fact that 
 $\Delta _{2,n} \cap \{x \in {\mathbb R}^n : x_1 = 1\}$
 is a facet of
$\Delta _{2,n} \cap \{x \in {\mathbb R}^n : x_1 = 1\}$
 is a facet of 
 $\Delta _{2,n}$
 given by the convex hull of the vertices
$\Delta _{2,n}$
 given by the convex hull of the vertices 
 $e_{1j}$
 with
$e_{1j}$
 with 
 $2 \le j \le n$
. Since
$2 \le j \le n$
. Since 
 $n \ge 4$
, there exists
$n \ge 4$
, there exists  . Let
. Let 
 $\sigma $
 be the
$\sigma $
 be the 
 $3$
-cycle
$3$
-cycle 
 $(a \ b \ c) \in S_n$
. The relative interior of the simplex
$(a \ b \ c) \in S_n$
. The relative interior of the simplex 
 $\sigma (\nabla ) = \mathrm {Conv}\{e_{12}, \dots , e_{1n}, e_{bc}\}$
 intersects the relative interior of
$\sigma (\nabla ) = \mathrm {Conv}\{e_{12}, \dots , e_{1n}, e_{bc}\}$
 intersects the relative interior of 
 $\nabla $
, hence they cannot belong to the same triangulation. So the original triangulation is not
$\nabla $
, hence they cannot belong to the same triangulation. So the original triangulation is not 
 $S_n$
-invariant.
$S_n$
-invariant.
Acknowledgements
The authors would like to thank Nick Early for explaining the motivation behind DOSPs and the connection to equivariant volumes of hypersimplices.
Competing interests
The authors have no competing interests to declare.
Funding statement
The second author was supported by a Grant-in-Aid for JSPS Research Fellows (Grant Number: 24KJ1590) under the JSPS DC2 programme.
 
 







 
 
 

 
 
 
 
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