Proof. Suppose P satisfies (AF), and let $\mathbf {v} \in \mathscr {K}$ and $\mathbf {x} \in \mathbb {R}^n$ . Since $\mathscr {K}$ is an open cone, $\mathbf {w}= \mathbf {x}+t\mathbf {v} \in \mathscr {K}$ for all $t>0$ sufficiently large. Applying (AF) to $\mathbf {v},\mathbf {w}$ yields (AF2) after a simple calculation.

Assume (AF2), and let $\mathbf {v} \in \mathscr {K}$ and $\mathbf {x} \in \mathbb {R}^n$ . Then

(2.2) $$ \begin{align} f(t\mathbf{v} - \mathbf{x}) = t^2 \cdot P(\mathbf{v},\mathbf{v}) - 2t \cdot P(\mathbf{v},\mathbf{x}) + P(\mathbf{x},\mathbf{x}), \end{align} $$

which is real-rooted by nonnegativity of the discriminant. Thus f is hyperbolic with hyperbolicity cone containing $\mathscr {K}$ . Note in fact that (AF2) and (Hyp) are equivalent, since the discriminant condition is equivalent to real-rootedness.

Assume (Hyp), and let $\mathbf {v} \in \mathscr {K}$ . Since (Hyp) implies (AF2), we have that P is negative semidefinite on the hyperplane $\{\mathbf {x} : P(\mathbf {v},\mathbf {x})=0\}$ . By Cauchy interlacing and Sylvester’s law of inertia, P has at most one positive eigenvalue. Since $P(\mathbf {v},\mathbf {v})> 0$ , P has exactly one positive eigenvalue. Suppose $\mathbf {x} \in \mathscr {K}$ . Then $D_{\mathbf {v}} D_{\mathbf {x}} f = 2 \cdot P(\mathbf {v},\mathbf {x})> 0$ , since the quadratic (2.2) has two negative zeros. Hence $f(\mathbf {x})$ satisfies (P). The fact that (Lor) implies (H) is immediate from Definition 2.1.

Assume (H), and let $\mathbf {v}, \mathbf {w} \in \mathscr {K}$ . If $\mathbf {v}$ and $\mathbf {w}$ are parallel, then (AF) holds. Otherwise extend $\mathbf {v},\mathbf {w}$ to a basis of $\mathbb {R}^n$ . In this basis, the top-left $2 \times 2$ matrix reads

$$ \begin{align*}\begin{pmatrix} P(\mathbf{v},\mathbf{v}) & P(\mathbf{v},\mathbf{w}) \\ P(\mathbf{v},\mathbf{w}) & P(\mathbf{w},\mathbf{w}) \end{pmatrix}. \end{align*} $$

Since $P(\mathbf {v},\mathbf {v})>0$ , Cauchy interlacing and Sylvester’s law of inertia implies that this matrix has exactly one positive eigenvalue. The condition (AF) now follows since the determinant of this matrix is nonpositive.

Proof. Given a cone $\mathscr {K}$ , Remark 2.4 implies that a polynomial is $\mathscr {K}$ -Lorentzian if and only if it satisfies (P) and (AF) with respect to $\mathscr {K}$ . Notice that

(2.4) $$ \begin{align} D_{\mathbf{v}_1}D_{\mathbf{v}_2}\cdots D_{\mathbf{v}_d} g = D_{A\mathbf{v}_1}D_{A\mathbf{v}_2}\cdots D_{A\mathbf{v}_d}f. \end{align} $$

Thus (P) and (AF) for g follow directly from (P) and (AF) for f.

Proof. If $\mathbf {v} \in \mathscr {K}$ and $\mathbf {w} \in L_{\mathscr {K}}$ , then

$$ \begin{align*}D_{\mathbf{w}} f(\mathbf{v})= \frac{D_{\mathbf{w}} D_{\mathbf{v}}^{d-1}f}{(d-1)!} \geq 0 \ \ \text{ and } \ \ -D_{\mathbf{w}} f(\mathbf{v})= \frac {D_{-\mathbf{w}}D_{\mathbf{v}}^{d-1}f}{(d-1)!} \geq 0. \end{align*} $$

Hence $D_{\mathbf {w}} f(\mathbf {v}) =0 $ for all $\mathbf {v} \in \mathscr {K}$ . Since $D_{\mathbf {w}} f$ is a polynomial which is $0$ on the open set $\mathscr {K}$ , we have that $D_{\mathbf {w}} f \equiv 0$ as desired.

Proof. If g is a d-homogeneous polynomial and $g(\mathbf {x})>0$ , then the following three statements are equivalent

1. (a) the Hessian of g at $\mathbf {x}$ has exactly one positive eigenvalue,

2. (b) the Hessian of $g^{1/d}$ is negative semidefinite at $\mathbf {x}$ ,

3. (c) the matrix $d\cdot g \cdot \nabla ^2g - (d-1)\cdot \nabla g (\nabla g)^T$ is negative semidefinite at $\mathbf {x}$ ,

see, for example, [Reference Brändén and Huh14, Prop. 2.33].

Suppose $\mathbf {x}$ and f are as in the hypotheses of the lemma. Then, by (c),

$$ \begin{align*}(d-1)\cdot \partial_i f \cdot \nabla^2 \partial_i f \preceq (d-2) \cdot \nabla \partial_i f (\nabla \partial_i f)^T. \end{align*} $$

Euler’s identity, $ d \cdot f(\mathbf {x}) = \sum _{i=1}^n x_i \cdot \partial _i f(\mathbf {x}), $ yields

$$ \begin{align*}(d-2)\cdot \nabla^2 f = \sum_{i=1}^n x_i \nabla^2 \partial_i f \preceq \sum_{i=1}^n \frac {x_i}{\partial_i f} \frac {(d-2)}{(d-1)} \nabla \partial_i f (\nabla \partial_i f)^T. \end{align*} $$

Rewrite the above inequality as $ (d-1) \cdot \nabla ^2 f \preceq (\nabla ^2 f) \Lambda (\nabla ^2 f), $ where $\Lambda $ is the diagonal matrix $\operatorname {\mathrm {diag}}({x_1}/{\partial _1 f}, \ldots , {x_n}/{\partial _n f})$ . For the matrix $B= \Lambda ^{1/2} (\nabla ^2 f) \Lambda ^{1/2}$ , this implies $ B^2 -(d-1)B \succeq 0. $ Hence no eigenvalue of B lies in the open interval $(0,d-1)$ . The matrix B is irreducible and has nonnegative off-diagonal entries, so the Perron-Frobenius theorem applies to B. Notice that $\Lambda ^{-1/2} \mathbf {x}$ is a positive eigenvector of B, and the corresponding eigenvalue is $d-1$ . Hence $d-1$ is the unique largest eigenvalue of B afforded by the Perron-Frobenius theorem. We conclude that B, and thus also $\nabla ^2 f(\mathbf {x})$ , has exactly one positive eigenvalue.

Proof. Let $\mathbf {v}_1, \ldots , \mathbf {v}_{d-3} \in \mathscr {K}$ , and consider the cubic $g= D_{\mathbf {v}_1}\cdots D_{\mathbf {v}_{d-3}} f$ . Since $\partial _i f$ is $\mathscr {K}$ -Lorentzian, it follows from Definition 2.1 that so is $\partial _i g$ . By choosing $\mathbf {v}_{d-2}=\mathbf {x} \in \mathscr {K}$ , it follows from (2) that the Hessian $\nabla ^2g(\mathbf {x})$ is irreducible and its off-diagonal entries are nonnegative.

By (1) and Proposition 2.7, $g(\mathbf {x}+\mathbf {w})=g(\mathbf {x})$ for all $\mathbf {x} \in \mathbb {R}^n$ and $\mathbf {w} \in L_{\mathscr {K}}$ . Since $\mathscr {K}$ is effective, we may assume $\mathbf {x} \in \mathscr {K} \cap \mathbb {R}_{>0}^n$ . Lemma 2.8 then implies that the Hessian of g at $\mathbf {x}$ has exactly one positive eigenvalue.

The theorem now follows since the Hessian of $D_{\mathbf {v}_1}\cdots D_{\mathbf {v}_{d-3}} f$ at $\mathbf {x}$ is equal to the Hessian of $D_{\mathbf {v}_1}\cdots D_{\mathbf {v}_{d-3}}D_{\mathbf {x}} f$ .

Proof. Clearly $L_f \subseteq L_h$ and $\Delta _h \subseteq \tau \Delta _f$ . Hence h is hereditary.

Proof. If $S \in \Delta _f$ has fewer than d elements, then $\partial ^Sf \not \equiv 0$ so that

$$ \begin{align*}0 \not \equiv \partial^Sf(\mathbf{t}) = f^S(\pi_S(\mathbf{t})), \end{align*} $$

by (3.3). But then $f^S \not \equiv 0$ is a weakly hereditary polynomial in a set of variables $t_j$ such that $j \notin S$ . If $t_j$ is such a variable, then

$$ \begin{align*}0 \not \equiv \partial_j f^T = \partial^{S \cup \{j\}}f\big|_{t_i=0, i\in S}. \end{align*} $$

Hence S is not maximal in $\Delta _f$ .

Proof. We start by proving sufficiency of (C) and (Q) by induction over $d \geq 2$ . Suppose $d=2$ , and let $i \in V$ . Then $ \partial _if(\mathbf {v})=f^{\{i\}}(\pi _{\{i\}}(\mathbf {v}))>0 $ for all $\mathbf {v} \in \mathscr {K}_f$ , since $\pi _{\{i\}}(\mathbf {v}) \in \mathscr {K}_{f^{\{i\}}}$ . Hence $f^{\{i\}}$ is $\mathscr {K}_{f^{\{i\}}}$ -Lorentzian. Also

$$ \begin{align*}f (\mathbf{v}) = \frac 1 2\sum_{i\in V} v_i \cdot \partial_if(\mathbf{v})>0, \end{align*} $$

since $\mathscr {K}_f$ is effective. By Lemma 2.3 and (Q), f is $\mathscr {K}_f$ -Lorentzian.

Suppose $d \geq 3$ . Then $f^S$ satisfies (C) and (Q) for all nonempty $S \in \tau \Delta _f$ . By induction, $f^S$ is $\mathscr {K}_{f^S}$ -Lorentzian for all nonempty $S \in \tau \Delta _f$ . It remains to prove that f is $\mathscr {K}_f$ -Lorentzian. To do this we apply Theorem 2.10. To verify (1) of Theorem 2.10, we prove that for all $\mathbf {v}_1, \ldots , \mathbf {v}_d$ in $\mathscr {K}_f$ the polynomial $ f(s_1\mathbf {v}_1+ \cdots +s_d\mathbf {v}_d) $ has positive coefficients. Since $\mathscr {K}_f$ is effective we may assume $\mathbf {v}_1, \ldots , \mathbf {v}_d \in \mathscr {K}_f \cap \mathbb {R}_{>0}^n$ . Let $\mathbf {y}=s_1\mathbf {v}_1+ \cdots +s_d\mathbf {v}_d$ . By Euler’s formula,

$$ \begin{align*}d \cdot f(\mathbf{y})= \sum_{i \in V} y_i \cdot f^{\{i\}} \big( s_1\pi_{\{i\}}(\mathbf{v}_1)+\cdots+ s_d\pi_{\{i\}}(\mathbf{v}_d)\big). \end{align*} $$

By assumption all coefficients of $y_i$ are positive for each $i \in V$ , and by induction all coefficients of $ f^{\{i\}} \big ( s_1\pi _{\{i\}}(\mathbf {v}_1)+\cdots + s_d\pi _{\{i\}}(\mathbf {v}_d)\big ) $ are positive since $\pi _{\{i\}}(\mathbf {v}_j) \in \mathscr {K}_{f^{\{i\}}}$ for all j. This proves the claim.

Next we claim that for each $\{i,j \} \in \tau \Delta _f$ and all $\mathbf {v}_1, \ldots , \mathbf {v}_{d-2}$ in $\mathscr {K}_f$ , the polynomial

$$ \begin{align*}(\partial_i\partial_j f)(s_1\mathbf{v}_1+ \cdots+s_{d-2}\mathbf{v}_{d-2})=f^{\{i,j\}} \big(s_1\pi_{\{i,j\}}(\mathbf{v}_1)+\cdots+ s_{d-2}\pi_{\{i,j\}}(\mathbf{v}_{d-2})\big) \end{align*} $$

has positive coefficients. The claim will prove the nonnegativity of the off-diagonal entries of the Hessian in Theorem 2.10 (2). The claim follows by induction as above, since $\pi _{\{i,j\}}(\mathbf {v}_k) \in \mathscr {K}_{f^{\{i,j\}}}$ for each k. By H-connectivity of $\tau \Delta _f$ , the Hessian of $D_{\mathbf {v}_1}\cdots D_{\mathbf {v}_{d-2}}f$ is irreducible. Condition (3) of Theorem 2.10 follows by induction since $ \partial _i f(\mathbf {t}) = f^{\{i\}} ( \pi _{\{i\}}(\mathbf {t})) $ is $\mathscr {K}_f$ -Lorentzian for each $i \in V$ by Proposition 2.6 and (3.4). Hence sufficiency follows from Theorem 2.10.

For the converse, we need to prove that for a weakly hereditary Lorentzian polynomial f, $\tau \Delta _f$ is H-connected. For $d=2$ , (C) is obvious, since $\tau \Delta _f$ is zero-dimensional. Suppose $d\geq 3$ . It suffices to prove that $ \tau \Delta _f $ is connected. Let $\mathbf {v} \in \mathscr {K}_f$ . Then $g=D_{\mathbf {v}}^{d-2}f$ is a $\mathscr {K}_f$ -Lorentzian polynomial. Moreover if $\{i,j\} \in \tau \Delta _f$ , then

$$ \begin{align*}\partial_i \partial_j g= D_{\mathbf{v}}^{d-2} \partial_i \partial_j f = (d-2)!f^{\{i,j\}}(\pi_{\{i,j\}}(\mathbf{v}))>0, \end{align*} $$

since $f^{\{i,j\}}$ is $\pi _{\{i,j\}}(\mathscr {K}_f)$ -Lorentzian. If $\tau \Delta _f$ is not connected, then we may write the Hessian of g as a block matrix $A\oplus B$ with nonnegative off-diagonal entries, where A has exactly one positive eigenvalue and $B=(b_{ij})$ is a negative semidefinite irreducible matrix for which $b_{ij}>0$ for some $\{i,j\} \in \tau \Delta _f$ . By Perron-Frobenius the largest eigenvalue of B is simple. Hence the kernel of B has dimension at most one. Since $\{i,j\} \in \tau \Delta _f$ and f is weakly hereditary, for each $(x, y) \in \mathbb {R}^2$ there exists $\ell \in L_f \subseteq L_g$ such that $\ell _i=x$ and $\ell _j=y$ . Hence the kernel of B is at least two-dimensional. Hence $\tau \Delta _f$ is connected.

Proof. If $\mathscr {K}$ satisfies the hypothesis in the statement, then so does $\mathscr {K}+L_f$ . Hence we may assume $L_f \subseteq \overline {\mathscr {K}}$ . Since f is weakly hereditary and $\mathscr {K}$ is open and nonempty, $\pi _S(\mathscr {K})$ is nonempty for all $S \in \tau \Delta _f$ . By induction we deduce $\pi _S(\mathscr {K}) \subseteq \mathscr {K}_{f^S}$ for all nonempty $S \in \tau \Delta _f$ . If $\mathbf {v} \in \mathscr {K}$ , then $\mathbf {v} \in \mathbb {R}_{>0}^V+ L_f$ , since $\mathscr {K}$ is effective. Moreover $\pi _{\{i\}}(\mathbf {v}) \in \pi _{\{i\}}(\mathscr {K}) \subseteq \mathscr {K}_{f^{\{i\}}}$ . Hence $\mathbf {v} \in \mathscr {K}_f$ by the definition of $\mathscr {K}_f$ . As in the proof of Theorem 3.10, (C) follows from that $f^S$ is $\pi _S(\mathscr {K})$ -Lorentzian. Also (Q) holds since the quadratics are $\pi _S(\mathscr {K})$ -Lorentzian. Hence the theorem follows from Theorem 3.10.

Proof. By Euler’s identity,

$$ \begin{align*} d \partial_j f &= \partial_j(t_jQ_j) -t_j\partial_jQ_j+ \sum_{i=1}^n t_i\partial_jQ_i \\ &= Q_j + \sum_{i=1}^n t_i\partial_iQ_j = Q_j+(d-1)Q_j= dQ_j.\\[-42pt] \end{align*} $$

Proof. That there is at most one weakly hereditary polynomial with these properties follows from the above discussion, since a weakly hereditary polynomial f is determined by $\partial ^S f$ for all $S \subset V$ for which $|S| = d$ . Further, if such a weakly hereditary polynomial f exists, then for each facet F of $\tau \Delta $ ,

$$ \begin{align*}f^F(\mathbf{t}) = \sum_{i \not \in F} w(F\cup\{i\})t_i \end{align*} $$

is identically zero on $\pi _F(L_f) \supseteq \pi _F(L) = L_F$ .

It remains to prove that the condition on linear forms is enough to guarantee the existence of the weakly hereditary polynomial f. To this end, we define a family of polynomials $g^S \in \mathbb {R}[t_i : i \in V_S]$ recursively by

(3.6) $$ \begin{align} (d-|S|)\cdot g^S(\mathbf{t})= \sum_{i \in V_S}t_i \cdot g^{S\cup\{i\}}(\pi_{\{i\}}^S(\mathbf{t})), \ \ \ S \in \Delta, \end{align} $$

where $ g^F=w(F) $ for each facet F of $\Delta $ .

We prove by induction over $d-|S|$ that $g^{S}$ is weakly hereditary and that $L_S \subseteq L_{g^{S}}$ . The case when $|S|=d-1$ is clear since then $L_S {\subseteq } L_{g^{S}}$ by hypothesis. Suppose $S \in \Delta $ with $|S|< d-1$ , and that $g^T$ is weakly hereditary with $L_{g^T}\supseteq L_T$ for all $T \in \Delta $ of size greater than $|S|$ . We claim $\partial _i g^S(\mathbf {t})= g^{S \cup \{i\}}(\pi ^S_{\{i\}}(\mathbf {t}))$ for each $i \in V_S$ . By Lemma 3.14 and Euler’s formula this holds if and only if

$$ \begin{align*}\partial_j \left(g^{S \cup\{i\}}(\pi_{\{i\}}^S(\mathbf{t}))\right) = \partial_i \left(g^{S \cup\{j\}}(\pi_{\{j\}}^S(\mathbf{t}))\right) \end{align*} $$

for each $\{i,j\} \in \Delta _S$ . For $|S|=d-2$ this is true by hypothesis. For $|S|<d-2$ , we get by induction

$$ \begin{align*}\partial_j \left(g^{S \cup\{i\}}(\pi_{\{i\}}^S(\mathbf{t}))\right)=g^{S \cup\{i,j\}}(\pi_{\{j\}}^{S\cup\{i\}}\pi_{\{i\}}^S(\mathbf{t})). \end{align*} $$

By the induction hypothesis it follows that $g^{S \cup \{i\}}(\pi _{\{i\}}^S(\mathbf {t}))$ does not depend on the choice of the vector $\ell ^{(i)}$ in the definition of $\pi _{\{i\}}^S$ . Hence we may assume $\ell ^{(i)}_j=0$ . Then $\pi _{\{j\}}^{S\cup \{i\}}\pi _{\{i\}}^S= \pi _{\{i,j\}}^S$ , and again by induction, $g^{S \cup \{i,j\}}(\pi _{\{i,j\}}^S(\mathbf {t}))$ does not depend on the particular choice of vectors in the definition of $\pi _{\{i,j\}}^S$ . Hence

$$ \begin{align*}\partial_j \left(g^{S \cup\{i\}}(\pi_{\{i\}}^S(\mathbf{t}))\right) = \partial_i \left(g^{S \cup\{j\}}(\pi_{\{j\}}^S(\mathbf{t}))\right)= g^{S \cup\{i,j\}}(\pi_{\{i,j\}}^S(\mathbf{t})), \end{align*} $$

which proves the claim.

Next we prove $L_S \subseteq L_{g^{S}}$ . Let $\mathbf {y} \in L_S$ and $i \in V_S$ . Since $\partial _i g^S(\mathbf {t})= g^{S \cup \{i\}}(\pi _{\{i\}}^S(\mathbf {t}))$ for all $i \in V_S$ , it follows by induction, (3.4), and Euler’s formula that

$$\begin{align*}\partial^\alpha g^S(\mathbf{y}) = \frac{1}{d-|S|-|\alpha|} \sum_{i \in V_S} t_i \cdot \partial^\alpha g^{S \cup \{i\}}(\pi_{\{i\}}^S(\mathbf{y})) = 0 \end{align*}$$

unless $|\alpha |=d-|S|$ . By Taylor expanding $g^S(\mathbf {t}+\mathbf {y})$ ,

$$ \begin{align*}g^S(\mathbf{t}+\mathbf{y}) =\sum_{\alpha} \frac{\partial^\alpha g^S(\mathbf{y})}{\alpha!} \mathbf{t}^\alpha, \end{align*} $$

we see that $g^S(\mathbf {t}+\mathbf {y})=g^S(\mathbf {t})$ as claimed. This finishes the proof of the theorem for the weakly hereditary case by induction. Finally, since $L \subseteq L_f$ , if $(\Delta ,L)$ is hereditary, then f is hereditary.

Proof. The fact that $f_1f_2$ is hereditary follows from $\Delta _{f_1f_2} = \Delta _1 \times \Delta _2$ and $L_1 \oplus L_2 \subseteq L_{f_1f_2}$ . Further, $\partial ^{S_1 \cup S_2} (f_1f_2) = \partial ^{S_1} f_1 \partial ^{S_2} f_2 = w_1(S_1) w_2(S_2)$ for all facets $S_1 \in \Delta _1$ and $S_2 \in \Delta _2$ (and equals 0 otherwise), which shows that $f_1f_2$ is a weakly hereditary polynomial corresponding to $\Delta _1\times \Delta _2$ , $L_1\oplus L_2$ and $w_1w_2$ . The uniqueness then follows from Theorem 3.15.

Proof. We apply Theorem 3.10. Clearly $\Delta _{fg}= \Delta _f \times \Delta _g$ , $L_{fg}=L_f \oplus L_g$ and $\mathscr {K}_{fg}=\mathscr {K}_f \times \mathscr {K}_g$ . Hence $fg$ is hereditary and $\mathscr {K}_{fg}$ is nonempty. The Cartesian product of two nonempty simplicial complexes is automatically connected. Hence, the H-connectedness of $\Delta _{fg}$ follows from the H-connectedness of $\Delta _f$ and $\Delta _g$ . This verifies (C).

The quadratics of $fg$ appearing in (Q) are either products of two linear polynomials with positive coefficients, or are positive constants times the quadratics appearing in (Q) for f or g. This verifies (Q) for $fg$ .

Proof. We first prove that $\mathscr {S}_{\mathscr {L}}$ is effective. Notice that $\mathbf {v} = (|F \setminus K| \cdot |E \setminus F|)_{K < F < E}$ is a vector in $\mathscr {S}_{\mathscr {L}}$ with positive entries. Suppose $\mathbf {y} \in \mathscr {S}_{\mathscr {L}}$ . Then $\mathbf {z} := \mathbf {y} - \epsilon \mathbf {v} \in \mathscr {S}_{\mathscr {L}}$ for $\epsilon> 0$ sufficiently small. By, for example, [Reference Murota25, Prop. 4.4], there exists $\mathbf {w} \in L(\mathscr {L})$ such that $\mathbf {z} + \mathbf {w} \in \mathbb {R}_{\geq 0}^{\underline {\mathscr {L}}}$ . Thus $\mathbf {y} + \mathbf {w} = \mathbf {z} + \mathbf {w} + \epsilon \mathbf {v} \in \mathbb {R}_{> 0}^{\underline {\mathscr {L}}}$ as desired.

We now prove $\mathscr {S}_{\mathscr {L}} \subseteq \mathscr {K}_{\operatorname {\mathrm {pol}}_{\mathscr {L}}}$ by induction on the degree of $\operatorname {\mathrm {pol}}_{\mathscr {L}}$ , where $\mathscr {L} = [K, E]$ . First, the inclusion is clear when $\operatorname {\mathrm {pol}}_{\mathscr {L}}$ is linear. To complete the proof, we need to prove

$$ \begin{align*}\pi_{\{F\}}(\mathbf{y}) \in \mathscr{K}_{\operatorname{\mathrm{pol}}_{\mathscr{L}}^{\{F\}}},\end{align*} $$

for all $F \in \underline {\mathscr {L}}$ and $\mathbf {y} \in \mathscr {S}_{\mathscr {L}}$ . Recall that $\operatorname {\mathrm {pol}}_{\mathscr {L}}^{\{F\}} = \operatorname {\mathrm {pol}}_{[K,F]} \cdot \operatorname {\mathrm {pol}}_{[F,E]}$ . Further, $\pi _{\{F\}}(\mathbf {y}) \in \mathscr {S}_{[K,F]} \oplus \mathscr {S}_{[F,E]}$ by the definition of $\pi _{\{F\}}$ , since the vectors in $L(\mathscr {L})$ are modular. Thus by induction $\pi _{\{F\}}(\mathbf {y}) \in \mathscr {S}_{[K,F]} \oplus \mathscr {S}_{[F,E]} \subseteq \mathscr {K}_{\operatorname {\mathrm {pol}}_{[K,F]}} \oplus \mathscr {K}_{\operatorname {\mathrm {pol}}_{[F,E]}} = \mathscr {K}_{\operatorname {\mathrm {pol}}_{\mathscr {L}}^{\{F\}}}$ , as desired.

Proof. By Lemma 4.2, $\mathscr {K}_{\operatorname {\mathrm {pol}}_{\mathscr {L}}}$ is nonempty. By Remark 3.12, Proposition 3.17, (4.1) and induction on the rank of $\mathscr {L}$ , it suffices to prove that $\Delta (\mathscr {L})$ is connected for any $\mathscr {L}$ , and that the Hessian of $\operatorname {\mathrm {pol}}_{\mathscr {L}}(\mathbf {t})$ has at most one positive eigenvalue whenever the rank of $\mathscr {L}$ is three. The latter follows immediately from Example 4.1.

Recall that lattices of flats of matroids are graded and semimodular, see [Reference Oxley26, Chapter 1.7]. Also, by [Reference Stanley30, Prop. 3.3.2] a finite lattice $\mathscr {L}$ is semimodular if and only if for all $a,b \in \mathscr {L}$ ,

$$\begin{align*}\text{if } a \text{ and } b \text{ cover } a \wedge b, \text{ then } a \vee b \text{ covers } a \text{ and } b. \end{align*}$$

Connectedness of $\Delta (\mathscr {L})$ now follows from semimodularity.

Proof. It suffices to prove (3.1) for facets F of $\Delta _S$ .

If $0\not \in F$ , then $F \in \Delta $ so that (3.1) for $(\Delta _S, L^{\mathbf {c}})$ follows from (3.1) for $(\Delta , L)$ . Otherwise $0 \in F$ , and then F contains all but one element of S, and $(F \setminus \{0\}) \cup S \in \Delta $ . In this case (3.1) for F follows from (3.1) for $(F \setminus \{0\}) \cup S \in \Delta $ .

Proof. The lemma is trivially true if $T\cup \{0\} \notin \Delta _S$ . Suppose $T\cup \{0\} \in \Delta _S$ . The polynomial $\partial _0^k g^T$ is hereditary by Lemma 3.3. Hence

$$ \begin{align*}\partial_0^{k+1}\partial^T g \cdot t_0 + \sum_{i \in S\setminus T} \partial_0^k\partial^T \partial_i g\cdot t_i =0, \end{align*} $$

for all $\mathbf {t} \in \mathbb {R}^{S \cup \{0\}}$ such that $t_0 = \sum _{i\in S\setminus T} c_it_i$ . The lemma follows.

Proof. Let $g \in \mathscr {P}^d(\Delta _S, L^{\mathbf {c}})$ and $f=\operatorname {\mathrm {weld}}_S^{\mathbf {c}}(g)$ . Then $L \subseteq L_f$ . Suppose $T \notin \Delta $ . If $S \cap T=\varnothing $ , then $\partial ^T f \equiv 0$ . If $T=S_0 \cup R$ , where $S_0=T \cap S \neq \varnothing $ , then $\{0\} \cup R \notin \Delta _S$ . By the chain rule, $\partial ^Tf \equiv 0$ . Hence $f \in \mathscr {P}^d(\Delta ,L)$ .

Let $T=S'\cup R$ , where $S'=T\cap S$ , be a facet of $\Delta $ . Then, by the chain rule,

$$ \begin{align*}\partial^T f = \prod_{i \in S'}(c_i\partial_0+\partial_i)\partial^R g. \end{align*} $$

If $|S'|=0$ , then $\partial ^T f= \partial ^Tg$ . If $|S'|>0$ , then by using Lemma 5.2,

$$ \begin{align*}\partial^T f = \prod_{i \in S'}(c_i\partial_0+\partial_i)\partial^R g = (c_j\partial_0+\partial_j)\partial^{S'\setminus\{j\}} \partial^Rg, \end{align*} $$

where $j \in S'$ . If $S'=S$ , then $\partial ^T f= c_j\partial _0 \partial ^{T\setminus \{j\}}g$ . If $S' \neq S$ , then $\partial ^T f = \partial ^T g$ . This proves (5.1) and (5.2).

From Theorem 3.15 and (5.1), (5.2) it follows that $\operatorname {\mathrm {weld}}_S^{\mathbf {c}}$ is injective.

Proof. We first prove $\operatorname {\mathrm {sub}}_S^{\mathbf {c}} : \mathscr {P}^d(\Delta ,L) \longrightarrow \mathscr {P}^d(\Delta _S,L^{\mathbf {c}})$ . Let $f \in \mathscr {P}^d(\Delta ,L)$ , and let $g=\operatorname {\mathrm {sub}}_S^{\mathbf {c}}(f)$ . It is straightforward to see that $L^{\mathbf {c}} \subseteq L_g$ . We claim $\Delta _g \subseteq \Delta _S$ . By the definition of $\Delta _S$ , for any $T \subseteq V \cup \{0\}$ ,

$$ \begin{align*}(S\cup T) \setminus \{0\} \in \Delta \ \ \Longrightarrow \ \ T \in \Delta_S \text{ or } S \subseteq T. \end{align*} $$

Assume $T \notin \Delta _S$ and $S \not \subseteq T$ . Then $(S \cup T) \setminus \{0\} \notin \Delta $ , so that $\partial ^{S\cup T}f \equiv 0$ , from which it follows $\partial ^Tg = \partial ^Tf$ . If $0 \in T$ , then $\partial ^Tf \equiv 0$ . If $0 \notin T$ , then $T \notin \Delta $ so that $\partial ^Tf \equiv 0$ . Hence $T \not \in \Delta _g$ . To prove the claim it remains to prove $\partial ^Sg \equiv 0$ . If $e_j(\bar {\partial })$ denotes the elementary symmetric polynomial of degree j in the variables $\bar {\partial }_i$ , $i \in S$ , then

$$\begin{align*}\begin{aligned} \bar{\partial}^Sg &= \bar{\partial}^S f - (-1)^{s} \sum_{n=s}^\infty\sum_{j=0}^{s} (-1)^{s-j} \frac{z^{n-s+j}}{(n-s+j)!} \cdot e_{j}(\bar{\partial})\, h_{n-s}(\bar{\partial}) \, \bar{\partial}^S f \\ &= \bar{\partial}^S f - \sum_{n=0}^\infty \frac{z^{n}}{n!} \left(\sum_{j=0}^{n} (-1)^{j} e_{j}(\bar{\partial}) \, h_{n-j}(\bar{\partial})\right) \bar{\partial}^S f =0, \end{aligned} \end{align*}$$

since $\sum _{j=0}^{n} (-1)^{j} e_{j} \, h_{n-j}=\delta _{0n}$ .

Since $(\operatorname {\mathrm {weld}}_S^{\mathbf {c}} \circ \operatorname {\mathrm {sub}}_S^{\mathbf {c}})(f) =f$ for all f by definition, $\operatorname {\mathrm {sub}}_S^{\mathbf {c}}$ is a bijection. The final statement follows from Lemma 5.3.

Proof. The case when $i \not \in S$ follows immediately from the definition of $\operatorname {\mathrm {sub}}_S^{\mathbf {c}}$ . The case when $S = \{i\}$ follows from the fact that $\operatorname {\mathrm {weld}}_S$ is the inverse of $\operatorname {\mathrm {sub}}_S$ and that $\{i\} \not \in \Delta _S$ . For the case of $i \in S$ , we compute

$$\begin{align*}\begin{aligned} \bar{\partial}_i \operatorname{\mathrm{sub}}_S^{\mathbf{c}}(f) &= \bar{\partial}_i f - (-1)^{s-1} \sum_{n=s-1}^\infty \frac{z^n}{n!} \left[h_{n+1-s}(\bar{\partial}) - h_{n-s}(\bar{\partial}) \bar{\partial}_i\right] \bar{\partial}^S f \\ &= \bar{\partial}_i f - (-1)^{s-1} \sum_{n=s-1}^\infty \frac{z^n}{n!} \cdot h_{n-(s-1)}\big((\bar{\partial}_j)_{j \in S \setminus \{i\}}\big) \, \bar{\partial}^{S \setminus\{i\}} \bar{\partial}_i f \end{aligned} \end{align*}$$

where $h_{-1} = 0$ . The result then follows by scaling by $c_i$ and setting $t_i = 0$ .

Proof. We may assume that $|S| \geq 2$ . It is straightforward to see that $\operatorname {\mathrm {lk}}_{\Delta _S}(\{0\})$ is always connected.

By Lemma 5.6 and induction on the dimension of $\Delta $ , we now only need to show that $\Delta $ is connected if and only if $\Delta _S$ is connected, which is standard and clear by definition.

Proof. Consider the property (Q) restricted to positive hereditary polynomials f of degree d:

1. (Q) The Hessian of $f^T$ has at most one positive eigenvalue for each $T \in \Delta _f$ of size $d-2$ .

Let $g = \operatorname {\mathrm {sub}}_S^{\mathbf {c}}(f)$ . We prove by induction over d that f satisfies (Q) if and only if g satisfies (Q). If $s=1$ , then this follows from Lemma 5.5, and if $d < s$ , then this follows from the fact that $g = f$ . Assume $2 \leq s \leq d$ .

If $d=2$ , then $ g = f-z^2 \bar {\partial }^S f/2 $ . If f satisfies (Q), then so does g by the Cauchy interlacing theorem, since $\bar {\partial }^S f> 0$ . If g satisfies (Q), then so does $f=\operatorname {\mathrm {weld}}_S^{\mathbf {c}}(g)$ by Sylvester’s law of inertia and Cauchy interlacing. This verifies the statement for $d=2$ .

If $d=3$ , then $g^{\{0\}}$ is either identically zero or the product of two linearly independent linear polynomials. Hence $g^{\{0\}}$ has at most one positive eigenvalue. The statement for $d=3$ follows from Lemma 5.5 for $T = \{i\}$ , $i \neq 0$ , by induction.

Finally if $d \geq 4$ , then the statement follows from Lemma 5.5 by induction.

Suppose $f \sim g$ , and that $\mathscr {K}_f$ and $\mathscr {K}_g$ are nonempty. By Lemma 5.7, $\Delta _f$ is H-connected if and only if $\Delta _g$ is H-connected. Also, f satisfies (Q) if and only if g satisfies (Q). The theorem now follows from Theorem 3.10.

Proof. Let $\mathbf {t} \in \mathscr {K}_P$ be the support numbers of a polytope Q which contains the origin in its interior, and is such that the ray along $\rho _i$ intersects the relative interior of the facet $Q_i$ . It is straightforward that $\partial _i \operatorname {\mathrm {pol}}_P(\mathbf {t}) = \operatorname {\mathrm {vol}}(Q_i)$ . The polytope $Q' = Q - t_i \rho _i$ contains the origin in the relative interior of $Q_i'$ . The support numbers of $Q'$ are then

$$\begin{align*}t_j' = t_j - t_i \langle\rho_i, \rho_j\rangle =t_j - t_i \cos(\theta_{ij}). \end{align*}$$

Suppose $P_j$ is adjacent to $P_i$ . The projection of the unit normal vector $\rho _j$ onto $\mathcal {V}_i$ is given by $\sin (\theta _{ij}) \cdot \rho _j'$ , where $\rho _j' \in \mathcal {V}_i$ is the corresponding unit normal vector of $P_i$ . This implies the $1/\sin (\theta _{ij})$ factors.

Proof. If $S \in \Delta _P$ , then $\rho _i, i \in S$ are linearly independent since the dual of P is simplicial. Hence $(\Delta _P, L_P)$ is hereditary. Let $S \in \Delta _P$ and $Q \in \mathscr {K}_P$ , and let F be the face of Q corresponding to S. Then we may translate by $\mathbf {y} \in \mathcal {V}$ so that the origin is in the relative interior of F. Then $t_i(Q+\mathbf {y})=0$ for all $i \in S$ and $t_j(Q+\mathbf {y})>0$ for all j in the link of S. Hence $Q \in \mathscr {K}_{\operatorname {\mathrm {pol}}_P}$ .

Proof. Since P is a simplex, there is a (unique up to positive scalar) $\mathbf {v} \in \mathbb {R}_{>0}^{d+1}$ such that $v_1 \rho _1 + \cdots + v_{d+1} \rho _{d+1} = 0$ . Clearly $f(\mathbf {t})$ for this $\mathbf {v}$ is in $\mathscr {P}^d(\Delta _P,L_P)$ . On the other hand if $g \in \mathscr {P}^d(\Delta _P,L_P)$ , then $L_g$ contains $\{ \mathbf {t} \in \mathbb {R}^{d+1} : v_1 t_1 + \cdots + v_{d+1} t_{d+1}=0\}$ , from which it follows that g is a constant multiple of f.

Proof. The cone $\mathscr {K}_{\operatorname {\mathrm {vol}}_P}$ is nonempty by Lemma 6.2. Moreover $\Delta _P$ is H-connected. By Proposition 6.1, it remains to prove that $\operatorname {\mathrm {vol}}_P$ has at most one positive eigenvalue for $d=2$ . By performing suitable edge subdivisions and inverse edge subdivisions to $\operatorname {\mathrm {vol}}_P$ , it follows that $\operatorname {\mathrm {vol}}_P \sim f$ for some polynomial f such that $\Delta _f = \Delta _Q$ and $L_Q \subseteq L_f$ , where Q is a triangle. Since $\mathscr {P}^2(\Delta _Q,L_Q)$ is one-dimensional by Lemma 6.4, it follows that f is a positive constant multiple of $\operatorname {\mathrm {vol}}_Q$ . The theorem now follows from Lemma 6.4 and Theorem 5.8.

Proof. If $K_1, \ldots , K_n$ are simple polytopes in $\mathscr {K}_P$ , then using Proposition 2.6, the Alexandrov-Fenchel inequalities follow directly from Theorem 6.5 and Remark 2.4. By a simple approximation argument, the general case reduces to this case, see [Reference Schneider28, Theorem 2.4.15].

Proof. Follows from the fact that any k-dimensional cone in $\Sigma $ is spanned by k linearly independent ray vectors.

Proof. If $a \in I(\Delta )+J(L)$ and $f \in \mathscr {P}^k(\Delta ,L)$ , then $a(\partial )f=0$ . Hence the linear map $ \phi : \mathscr {P}^k(\Delta ,L) \to A^k(\Delta , L)^* $ is well-defined by $ \phi (f)(a)= a(\partial )f. $ The inverse of $\phi $ is given by $ \alpha \longmapsto \hat {\alpha } $ .

Proof. If $\mathbf {x}^T \in I(\operatorname {\mathrm {lk}}_\Delta (S))$ , then $\mathbf {x}^S \mathbf {x}^T \in I(\Delta )$ . If $(\ell _i)_{i \in V_S} \in L_S$ , then

$$ \begin{align*}\mathbf{x}^S \sum_{i \in V_S}\ell_i x_i = \mathbf{x}^S \sum_{i \in V}\ell_i x_i - \sum_{j : S\cup \{j\} \notin \Delta} \ell_jx_j\mathbf{x}^S \in I(\Delta) + J(\Delta). \end{align*} $$

Hence $\phi _S$ is well-defined. Moreover, (7.2) follows from (7.1).

Proof. Suppose

$$ \begin{align*}x_{i_1}^{\alpha_1} \cdots x_{i_s}^{\alpha_s} \in A^k(\Delta, L)\setminus \{0\}, \end{align*} $$

where $\alpha _j>0$ for all $1\leq j \leq s$ . Then $S=\{i_1,\ldots , i_s\} \in \Delta $ . If $s=k$ we have nothing to prove. Hence we may assume $\alpha _j>1$ for some j. Without loss of generality we may assume $j=1$ and $i_1=1$ . There is an element $\ell \in L$ such that $\ell _1=1$ and $\ell _i=0$ for all $i \in S \setminus \{1\}$ . Then

$$ \begin{align*}x_{i_1}^{\alpha_1} \cdots x_{i_s}^{\alpha_s}= \left(x_1-\sum_{i \in V}\ell_i x_i\right)x_{i_1}^{\alpha_1-1}x_{i_2}^{\alpha_2} \cdots x_{i_s}^{\alpha_s}. \end{align*} $$

By induction over $\sum _{j=1}^s(\alpha _j-1)$ this proves the lemma.

Proof. Both statements follow from the fact that $(\Delta ', L')$ being weakly hereditary is weaker than $(\Delta , L)$ being weakly hereditary.