2.1. Tautological classes

Under the assumption that $\mathrm {gcd}(d,\chi )=1$ , there exists a universal sheaf [Reference Huybrechts and LehnHL, Theorem 4.6.5] over the product $\mathbb {P}^2 \times M_{d,\chi }$ which we denote by $\mathbb {F}$ . For a stable sheaf $[\mathcal {F}]\in M_{d,\chi }$ , the restriction of $\mathbb {F}$ to the fiber $\mathbb {P}^2 \times [\mathcal {F}]$ recovers $\mathcal {F}$ .

Consider the two projection maps $\pi _P$ and $\pi _M$ from $\mathbb {P}^2 \times M_{d,\chi }$ to its first and second components. One way to obtain natural cohomology classes on $M_{d,\chi }$ is to take the Chern characters of $\mathbb {F}$ , intersect with classes pulled back from $\mathbb {P}^2$ and push forward to $M_{d,\chi }$ . The choice of $\mathbb {F}$ is, however, not unique: Any two universal sheaves differ by tensoring with a line bundle pulled back from $M_{d,\chi }$ . We thus conduct a normalization of $\mathbb {F}$ as follows.

For a universal sheaf $\mathbb {F}$ and a class

$$\begin{align*}\delta = \pi_P^* \delta_P + \pi_M^* \delta_M \in A^1(\mathbb{P}^2 \times M_{d,\chi}),\;\;\; \textrm{with \,} \delta_P \in A^1(\mathbb{P}^2), \;\; \delta_M \in A^1(M_{d,\chi}), \end{align*}$$

we consider the twisted Chern character

and denote by $\mathrm {ch}_k^\delta (\mathbb {F})$ its degree k-part. For any class $\gamma \in A^*(\mathbb {P}^2)$ , we write

It is proven in [Reference Pi and ShenPS, Proposition 1.2] that for a fixed $\mathbb {F}$ , there exists a unique class $\delta _0=\delta _0({\mathbb {F}})$ as above satisfying the normalizing conditions

(2.1) $$ \begin{align} \int_H \mathrm{ch}_2^{\delta_0} (\mathbb{F})=0,\;\;\; \int_{\mathbf{1}_{\mathbb{P}^2}} \mathrm{ch}^{\delta_0}_2(\mathbb{F})=0. \end{align} $$

This class is computed explicitly in [Reference Kononov, Pi and ShenKPS, Proposition 2.1]:

(2.2) $$ \begin{align} \delta_0=\left(\frac{3}{2}-\frac{\chi}{d}\right)\cdot H-\frac{1}{d}\left(\left(\frac{3}{2}-\frac{\chi}{d}\right)\int_{H^2}{\mathrm{ch}}_1({\mathbb{F}})+\int_H \mathrm{ch}_2(\mathbb{F})\right). \end{align} $$

Note that the twisted Chern character ${\mathrm {ch}}^{\delta _0}({\mathbb {F}})$ does not depend on the choice of ${\mathbb {F}}$ anymore. With this normalizing class, we define the tautological classes

(2.3)

We collect some basic properties of the tautological classes from [Reference Pi and ShenPS, Section 1.2].

The next theorem gives a first result on the structure of $A^*(M_{d,\chi })$ in terms of the tautological classes. Note that we change the coefficient from $\mathbb {Q}$ to $\mathbb {C}$ .

We briefly recall the proof of Theorem 2.2: (i) By the results of Beauville [Reference BeauvilleBea] or Markman [Reference MarkmanMar], the ring $A^*(M_{d,\chi })$ is generated by the tautological classes. (ii) Using the geometry of $M_{d,\chi }$ , we produce tautological relations that express any tautological class in terms of the $3d-7$ classes in Equation (2.4); see the paragraph after Proposition 2.6. (iii) Part (b) and (c) follow from comparing the Betti numbers of $M_{d,\chi }$ with those of Hilbert schemes of points on $\mathbb {P}^2$ ; this is explained in Yuan [Reference YuanY1, Reference YuanY2].

As we will see, the three relations in $A^d(M_{d,\chi })$ are the central characters in our proof of Theorem 1.2. By carefully investigating the $\chi $ -dependence of the three relations, we are able to deduce that $A^*({M_{d,\chi }})$ are nonisomorphic for different choices of $\chi $ unless they are related by the symmetries in Section 1.1.

2.2. Tautological relations

We recall in this section certain tautological relations [Reference Pi and ShenPS, Section 1.2] on the cohomology of $M_{d,\chi }$ . By the symmetry (a) in Section 1.1, we may assume that $0 < \chi < d$ so that any $\mathcal {F} \in M_{d,\chi }$ satisfies $0< \mu (\mathcal {F}) <1$ .

Consider the triple product , where $\check {\mathbb {P}}^2$ is the dual projective space parametrizing lines in $\mathbb {P}^2$ . Let $\pi _R : Y \to \check {\mathbb {P}}^2$ be the third projection. We write $p= \pi _P \times \pi _M, \, q = \pi _P \times \pi _R$ and $r = \pi _M \times \pi _R$ :

Let $Z\subset \mathbb {P}^2\times \check {\mathbb {P}}^2$ be the incidence subscheme, and let $\mathcal {O}_Z$ be its structure sheaf. For a fixed universal sheaf $\mathbb {F}$ on $M_{d,\chi }$ , we consider the complex

(2.5)

The projection $r: Y \to M_{d,\chi } \times \check {\mathbb {P}}^2$ is a trivial $\mathbb {P}^2$ -bundle, so the derived push-forward

$$\begin{align*}Rr_*\mathcal{H}(n)\in D^b(M_{d,\chi} \times \check{\mathbb{P}}^2) \end{align*}$$

admits a three-term resolution $0\to K^0 \to K^1 \to K^2 \to 0$ by vector bundles.

Proof. This is [Reference Pi and ShenPS, Lemma 2.4]; we briefly recall the proof here. Take a point

$$\begin{align*}P=([\mathcal{F}],[L])\in M_{d,\chi}\times \check{\mathbb{P}}^2, \end{align*}$$

where $[L]\in \check {{\mathbb {P}}}^2$ corresponds to the line $L\subseteq {\mathbb {P}}^2$ . Over the point P, cohomology of the complex $K^0(P) \to K^1(P)\to K^2(P)$ computes the extension groups

$$\begin{align*}\mathrm{Ext}^i(\mathcal{F}, \mathcal{O}_{L}(-n)), \quad i=1,2,3. \end{align*}$$

Note that $\mu (\mathcal {O}_{L}(-n))=1-n$ . By stability and Serre duality, one checks that

$$\begin{align*}\mathrm{Ext}^0({\mathcal F}, \mathcal{O}_{L}(-n))=0, \quad \mathrm{Ext}^2(\mathcal{F}, {\mathcal O}_{L}(-n))=0 \end{align*}$$

as long as $n\in \{1,2,3\}$ . This implies $Rr_*\mathcal {H}(n)$ can be represented by a single vector bundle $K^1$ concentrated in degree $1$ , whose rank is d by a Hirzebruch–Riemann–Roch calculation.

It follows that $-Rr_* \mathcal {H}(n)$ is a rank d vector bundle on $M_{d,\chi } \times \check {\mathbb {P}}^2$ , whence

Corollary 2.5. For $\ell \geq d+1$ and $n\in \{1,2,3\}$ , we have

(2.6) $$ \begin{align} c_\ell(-Rr_*\mathcal{H}(n))=0 \in A^{\ell}(M_{d,\chi} \times \check{{\mathbb{P}}}^2). \end{align} $$

Recall that $\mathcal {H}(n)$ is defined by Equation (2.5) in terms of a fixed universal sheaf $\mathbb {F}$ . If we replace $\mathbb {F}$ by

with L a fractional line bundle over $M_{d,\chi }$ , then according to Equation (2.5) and the projection formula, the vector bundle $-Rr_*\mathcal {H}(n)$ is replaced by its tensor with the dual line bundle $\pi _M^* L^\vee $ . Using a formal argument involving Chern roots, one checks that the identity (2.6) also holds for $\mathbb {F}'$ .

In particular, we choose the fractional line bundle L to have first Chern classFootnote ⁵

$$\begin{align*}c_1(L)=-\frac{1}{d}\left(\left(\frac{3}{2}-\frac{\chi}{d}\right)c_0(2)+\int_H \mathrm{ch}_2(\mathbb{F})\right) \end{align*}$$

so that the normalizing class (2.2) for $\mathbb {F}'$ is given by $\delta _0({\mathbb {F}}')=\left (\frac {3}{2}-\frac {\chi }{d}\right )\cdot H$ . Now, we apply Equation (2.6) to $\mathbb {F}'$ . The Chern classes in Equation (2.6) can be expressed in terms of the Chern character $\mathrm {ch}(Rr_*\mathcal {H}(n))$ , which is computed by the Grothendieck–Riemann–Roch theorem:

$$ \begin{align*} \mathrm{ch}(Rr_*\mathcal{H}(n))&=r_*(\mathrm{ch}(\mathcal{H}(n))\cdot \mathrm{td}({\mathbb{P}}^2)) \\ &=r_*(\mathrm{ch}(p^*{\mathbb{F}'}^\vee)\cdot \mathrm{ch}(q^*\mathcal{O}_Z\otimes \pi_P^* \mathcal{O}_{{\mathbb{P}}^2}(-n))\cdot \mathrm{td}({\mathbb{P}}^2)). \end{align*} $$

We denote by $\beta $ the class of a line on $\check {\mathbb {P}}^2$ . Expanding the right-hand side of the above equation, Corollary 2.5 then gives relations in $A^*(M_{d,\chi } \times \check {\mathbb {P}}^2)$ :

Proposition 2.6. For every $\ell \geq d+1$ and $n\in \{1,2,3\}$ , the following identity holds:

(2.7) $$ \begin{align} \sum_{\mathbf{m}} \prod_{s=1}^\ell \frac{((s-1)!)^{m_s}}{(m_s)!}\left( \pi_M^* A_{s}- \sum_{0\leq i\leq 2}\frac{\pi_R^*\beta^i}{i!}(-1)^i \pi_M^* B_{s-i}\right)^{m_s}=0. \end{align} $$

Here, the first sum is over all $\ell $ -tuple of nonnegative integers $\mathbf {m}=(m_1,m_2,\ldots , m_\ell )$ such that $m_1+2m_2+\cdots +\ell m_\ell =\ell $ , and writing , the terms $A_s, B_s$ are given by

To obtain relations in $A^*(M_{d,\chi })$ , we first take the identity (2.7) for some $\ell \geq d+1$ ; this gives a relation in $A^\ell (M_{d,\chi } \times \check {\mathbb {P}}^2)$ . Then we multiply it by $\pi _R^* (\beta ^j)$ and push forward to $M_{d,\chi }$ , where $0\leq j \leq 2$ . This produces a relation in $A^{\ell +j-2}(M_{d,\chi })$ among the tautological classes (2.3). The procedure is explained in detail in [Reference Pi and ShenPS, Section 2.3] and the paragraph before it. Indeed, as a key step in the proof of Theorem 2.2(a), it is shown in [Reference Pi and ShenPS, Section 2.3] that the relations produced by Proposition 2.6 express all the tautological classes (2.3) in terms of the $3d-7$ generators (2.4).

2.3. Three relations in $A^{d}(M_{d,\chi })$

The goal of this section is to prove the following result:

We first introduce some notations for the rest of the paper. Let $T^k$ be the linear space spanned by tautological classes in degree k, that is,

$$\begin{align*}T^1=\text{span}_{\mathbb{C}}\{c_2(0), c_0(2)\}\,,\quad T^k=\text{span}_{\mathbb{C}}\{c_{k+1}(0),c_{k}(1), c_{k-1}(2)\}\,.\end{align*}$$

Let

be the span of the $3d-7$ tautological generators in Equation (2.4), and let

We denote by $\mathbb {C}[T]$ the symmetric algebra in T (i.e., the free algebra generated by the $3d-7$ tautological generators), and by $\mathbb {C}[T]^k$ its graded piece of degree k. Similarly, we write ${\mathbb {C}}[T^+]$ for the symmetric algebra on $T^+$ .

Furthermore, we endow the basis of $T^+$ with a total ordering $\prec $ as follows: First order the basis by cohomological degrees; within each degree, we order by the Chern grading of the basis. In a precise way, this means $c_k(j) \prec c_{k'}(j')$ if and only if $k+j-1<k'+j'-1$ , or $k+j-1=k'+j'-1$ and $k<k'$ . Finally, we endow monomials in each degree of $\mathbb {C}[T^+]$ the lexicographical order induced by the total ordering $\prec $ on $T^+$ . In this way, we can talk about the leading term of a polynomial in $\mathbb {C}[T^+]$ , as well as in $\mathbb {C}[T]$ .

We have graded algebra homomorphisms

$$\begin{align*}{\mathbb{C}}[T]\hookrightarrow {\mathbb{C}}[T^+]\to A^\ast(M_{d,\chi})\,.\end{align*}$$

Theorem 2.2 states that the homomorphism ${\mathbb {C}}[T]\to A^\ast (M_{d,\chi })$ is surjective, that it is an isomorphism up to degree $d-1$ and that $\ker \!\big ({\mathbb {C}}[T]^d\to A^d(M_{d,\chi })\big )$ is 3-dimensional. A relation is by definition an element in the kernel of ${\mathbb {C}}[T]\to A^\ast (M_{d,\chi })$ or ${\mathbb {C}}[T^+]\to A^\ast (M_{d,\chi })$ .

With the above setup, Theorem 2.2 (b) and (c) are equivalent to the following.Footnote ⁶

Proposition 2.8. For $d\geq 5$ , we have

$$\begin{align*}\dim A^k(M_{d,\chi}) = \begin{cases} \dim \mathbb{C}[T^+]^k, \quad & k \leq d-2,\\ \dim \mathbb{C}[T^+]^k-3, \quad &k = d-1,\\ \dim \mathbb{C}[T^+]^k -12, \;\; &k=d. \end{cases} \end{align*}$$

Since $\dim {\mathbb {C}}[T^+]^d=\dim {\mathbb {C}}[T]^d+9$ , to prove Proposition 2.7 it suffices to produce 12 linearly independent relations in $\mathbb {C}[T^+]^d$ from Proposition 2.6. These 12 relations in degree d are produced in the following three ways:

1. (a) Take $\ell = d+1$ in Equation (2.7), we get the vanishing of a cohomology class in $M_{d, \chi }\times \check {{\mathbb {P}}}^2$ . Pushing forward to $M_{d, \chi }$ yields a relation $R_{{(a)}}^n\in {\mathbb {C}}[T^+]^{d-1}$ for $n=1,2,3$ . Each of these three relations produces two relations in degree d, namely $c_2(0)R_{(a)}^n$ and $c_0(2)R_{(a)}^n$ .

2. (b) Take $\ell = d+1$ in Equation (2.7), multiply by $\pi _R^* (\beta )$ and push forward to $M_{d, \chi }$ . This produces three relations $R_{(b)}^n\in {\mathbb {C}}[T^+]^{d}$ for $n=1,2,3$ .

3. (c) Take $\ell = d+2$ in Equation (2.7) and push forward to $M_{d, \chi }$ . This produces three relations $R_{(c)}^n\in {\mathbb {C}}[T^+]^{d}$ for $n=1,2,3$ .

We now proceed to argue that these 12 relations are linearly independent. It can be checked, cf. det1 in the Mathematica file and [Reference Pi and ShenPS, Section 2.3], that under the assumption $\mathrm {gcd}(d,\chi )=1$ , the $3\times 3$ matrix of coefficients of $c_{d}(0), c_{d-1}(1), c_{d-2}(2)$ in the three relations $R_{(a)}^1, R_{(a)}^2, R_{(a)}^3$ is nonsingular; in fact, it has determinant

$$\begin{align*}\texttt{det1}= \frac{(-1)^{d}(d - 2)^4(d-1) }{4}\chi(d - \chi)(d-2\chi)\,. \end{align*}$$

In particular, equations $R_{(a)}^n$ can be used to express $c_{d}(0), c_{d-1}(1), c_{d-2}(2)$ in terms of the tautological generators (2.4). Multiplying these equations with $c_2(0), c_0(2)$ , we obtain six relations in $\mathbb {C}[T^+]^{d}$ . These are still linearly independent since the $6\times 6$ matrix given by the coefficients of the monomials

$$\begin{align*}\mathsf{Mon}_1=\{c_{d}(0)c_2(0), c_{d}(0)c_0(2), c_{d-1}(1)c_2(0), c_{d-1}(1)c_0(2), c_{d-2}(2)c_2(0), c_{d-2}(2)c_0(2)\}\, \end{align*}$$

is invertible with determinant $\texttt {det1}^2$ .

We consider now the six relations $R_{(b)}^n, R_{(c)}^n$ , for $n=1,2,3$ and the set of monomials

$$\begin{align*}\mathsf{Mon}_2=\{c_{d+1}(0), c_{d}(1), c_{d-1}(2), c_{d-1}(0)c_3(0), c_{d-1}(0)c_2(1),c_{d-1}(0)c_1(2)\}\,.\end{align*}$$

The $6\times 6$ matrix of coefficients of $\mathsf {Mon}_2$ in these six relations is nonsingular. Indeed, a direct computation (cf. det2 in the Mathematica file) gives the determinant

$$\begin{align*}\texttt{det2}=4(d-2)^6(d-1)^3 d^4\neq 0\,. \end{align*}$$

It follows that the six relations $R_{(b)}^n$ and $R_{(c)}^n$ with $n\in \{1,2,3\}$ are linearly independent.

Finally, we note that the monomials in $\mathsf {Mon}_2$ do not appear in the relations of the form $c_2(0)R_{(a)}^n, c_0(2)R_{(a)}^n$ , so the $12\times 12$ matrix of coefficients of monomials $\mathsf {Mon}_1\cup \mathsf {Mon}_2$ in the 12 relations obtained by (a), (b), (c) is also nonsingular. We conclude that these 12 relations are all linearly independent, completing the proof of Proposition 2.7.

To explicitly obtain the three relations in ${\mathbb {C}}[T]^d$ , we proceed as follows. We use the relations $R_{(a)}^n$ to express $c_{d}(0), c_{d-1}(1), c_{d-2}(2)$ in terms of lower degree generators in the image of ${\mathbb {C}}[T]$ ; we then use $R_{(b)}^1, R_{(c)}^1, R_{(c)}^2$ to write $c_{d+1}(0), c_{d}(1), c_{d-1}(2)$ in terms of the tautological generators. Finally, plugging these into the other three relations

$$\begin{align*}R_{{(b)}}^2,\;R_{{(b)}}^3,\; R_{{(c)}}^3 \end{align*}$$

yields the three relations in ${\mathbb {C}}[T]^d$ . By the nonvanishing of det2, the row-echelon form (with respect to the ordering $\prec $ of monomials) of this system produces three relations $R_1, R_2, R_3$ such that

(2.8) $$ \begin{align} R_i=c_{d-1}(0)c_{4-i}(i-1)+\Big(\text{sum of monomials}\prec c_{d-1}(0)c_1(2)\Big)\quad \!\! \in {\mathbb{C}}[T]^d\,. \end{align} $$

A truncated version (see Remark 3.2) of these relations is implemented in the Mathematica file as TruncRelations.

Step 1: truncating relations

We start by truncating the relations $R_i$ by looking only at the terms which are obtained as a product of a generator of degree $d-2$ with a generator of degree $2$ . In other words, we consider the projection of $R_i$ to $T^2\otimes T^{d-2}\subseteq {\mathbb {C}}[T]^d$ and regard it as a $3\times 3$ matrix $M_i$ by identifying

$$\begin{align*}M_i\in T^2\otimes T^{d-2}\cong \operatorname{Hom}((T^{d-2})^\vee, T^2)\cong M_{3\times 3}\,.\end{align*}$$

This is implemented in the Mathematica file as TopRelations[d,chi]. The identification with a $3\times 3$ matrix uses the ordered basis for $T^2, T^{d-2}$ given by the generators from Section 2.3.

More concretely,

$$\begin{align*}M_i=\Big([c_{3-s}(s)c_{d-1-t}(t)]R_i\Big)_{0\leq s,\,t\leq 2},\end{align*}$$

where $[c_{3-s}(s)c_{d-1-t}(t)]$ means reading off the corresponding coefficient with respect to the monomial basis. The matrices $M_1, M_2, M_3$ can be explicitly calculated as follows:

$$ \begin{align*} M_1&=\begin{bmatrix}1& 0& 0\\ 0& 0& \frac{(d - 2) (d-2\chi) \chi (d - \chi)}{8 d^3}\\[3pt] \frac{d - 4}{8 d - 16}& \frac{(d-2\chi) \chi (\chi-d)}{2 (d - 2) d^3}& \frac{(-6 \chi - 1) d^5 + (6 \chi^2 + 6 \chi + 3) d^4 + 18 \chi^2 d^3 - (48 \chi^3 + 48 \chi^2) d^2 + (24 \chi^4 + 96 \chi^3) d - 48 \chi^4}{32 (d - 2) d^4}\end{bmatrix},\\M_2&=\begin{bmatrix}0& 1& 0\\ 1& 0& \frac{(-6 \chi - 1) d^2 + (6 \chi^2 + 12 \chi) d - 12 \chi^2}{8 d^2}\\[3pt] 0&\frac{24 \chi^2-24 \chi d-d^3+2 d^2}{8 (d-2) d^2}&\frac{\chi \left(d^2+8 d-16\right) (2 \chi-d) (d-\chi)}{8 (d-2) d^3}\end{bmatrix}, \quad \: M_3=\begin{bmatrix}0& 0& 1\\ 0& 0& 0\\ \frac{2}{d-2}&0&\frac{12 \chi^2 - 12 \chi d - d^2}{8 (d-2) d}\end{bmatrix}. \end{align*} $$

Remark 3.2. To compute these matrices and also the extended matrices that will appear later in Section 3.3, we follow the strategy described in the previous section. Note that the truncation of the relations $R_{(a)}^n, R_{(b)}^n, R_{(c)}^n$ can be computed since only finitely many terms of Equation (2.7) contribute. For instance, for $\ell =d+1$ the only contributing terms are the ones corresponding to the partitions

$$\begin{align*}(d + 1),\, (d, 1),\, (d - 1, 1, 1),\, (d - 1, 2),\, (d - 2, 2, 1),\, (d - 2, 3),\, (d - 2, 1, 1, 1)\,.\end{align*}$$

Similarly, we define $M_i'$ with $\chi $ replaced by $\chi '$ . Let $A, B$ be the invertible linear maps

They are implemented in the Mathematica file as AutA and AutB, respectively. By considering the standard basis for $T^2, T^{d-2}$ we identify again $A, B$ with $3\times 3$ matrices. More concretely, $A=(a_{st})_{0\leq s,\,t\leq 2}, B=(b_{st})_{0\leq s,\,t\leq 2}$ , where

$$ \begin{align*} \phi\big(c_{d-1-s}(s)\big)&=\sum_{t=0}^2 a_{st}c_{d-1-t}(t)+ \text{(lower terms in }\mathbb{C}[T]^{d-2})\,,\\ \phi\big(c_{3-s}(s)\big)&=\sum_{t=0}^2 b_{st}c_{3-t}(t)+ \text{(lower terms in }\mathbb{C}[T]^{2})\,. \end{align*} $$

Then the truncation of Equation (3.1) to $T^2\otimes T^{d-2}$ can be written as an equality between $3\times 3$ matrices

(3.2) $$ \begin{align} A^{\mathsf{T}} M_i B=\sum_{j=1}^3 s_{ij} M_j'\,,\quad i=1,2,3\,. \end{align} $$

Step 2: finding $s_{ij}$

We begin with solving $s_{ij}$ for which there are matrices $A, B$ satisfying Equation (3.2). By scaling both $A, B$ (hence also S), we may assume that $\det (A)=\det (B)=1$ . Then taking a linear combination of Equation (3.2) and applying the determinant it follows that

(3.3) $$ \begin{align} \det\left(\sum_{i=1}^3x_i M_i\right)=\det\left(\sum_{i,j=1}^3 x_i s_{ij} M_j'\right)\,,\end{align} $$

as an equality between two homogeneous cubic polynomials in $x_1, x_2, x_3$ . We let $E\subseteq {\mathbb {P}}^2$ be the cubic curve defined by

$$\begin{align*}E=\left\{[x_1: x_2:x_3]\colon \det\left(\sum_{i=1}^3x_i M_i\right)\right\}\subseteq {\mathbb{P}}^2_{x_1, x_2, x_3}\,.\end{align*}$$

Similarly, define $E'\subseteq {\mathbb {P}}^2$ . The matrix S defines an automorphism of ${\mathbb {P}}^2_{x_1, x_2, x_3}$ sending E to $E'$ .

Proof. We can directly compute the equation defining E. In the chart

$$\begin{align*}{\mathbb{C}}^2_{x_1, x_2}=\{[x_1: x_2: 1]\}\subseteq {\mathbb{P}}^2_{x_1, x_2, x_3},\end{align*}$$

the cubic polynomial defining E has the form

$$ \begin{align*}-\frac{\chi(d-\chi)(d-2\chi)}{4(d-2)d^2}x_1x_2+(\text{cubic terms in } x_1, x_2).\end{align*} $$

It follows that $[0:0:1]$ is indeed a node and the two branches of E at $[0:0:1]$ are tangent to the lines $x_1=0$ and $x_2=0$ . A cubic with a node either has exactly one node or has two nodes and is the union of a line and a conic. If the latter were the case, the line would necessarily be $x_1=0$ or $x_2=0$ , but this is not possible since $\det (M_2)\neq 0$ and $\det (M_1)\neq 0$ .

Finally, we note that an automorphism of ${\mathbb {P}}^2_{x_1, x_2, x_3}$ sending E to $E'$ must preserve the common node $[0:0:1]$ , so it follows that $s_{31}=s_{32}=0$ .

We now use Equation (3.3) to determine the remaining entries of S. Given a triple $(u,v,w)$ of nonnegative integers such that $u+v+w=3$ , we get an equation among the entries of S by comparing the $x_1^ux_2^vx_3^w$ coefficient of both sides of Equation (3.3); this is implemented in the Mathematica file as coeff[u,v,w].Footnote ⁷ The triples $(0,2,1)$ and $(2,0,1)$ give, respectively,

$$\begin{align*}s_{21}s_{22}s_{33}=0\;\;\text{ and }\;\;s_{12}s_{11}s_{33}=0\,.\end{align*}$$

On the other hand, the equation for $(1,1,1)$ implies that $s_{33}\neq 0$ and $s_{12}s_{21}+s_{11}s_{22}\neq 0$ . It follows that solutions must have either:

1. I. $s_{11}=s_{22}=0$ . We call the solutions of this form Type I solutions.

2. II. $s_{12}=s_{21}=0$ . We call the solutions of this form Type II solutions.

From now on, we divide the analysis of Equations (3.2) and (3.3) according to the type of the solutions.

3.1. Type I

Suppose that S is a solution of Type I to Equation (3.3) so that we have already the vanishing of the entries $s_{31}=s_{32}=s_{11}=s_{22}=0$ . We now obtain the remaining entries. By looking at the $(0,3,0)$ and $(3,0,0)$ equations, we obtain $s_{21}, s_{12}$ up to a choice of a cubic root of unity:

$$ \begin{align*} s_{21}^3&=-d^3\frac{2 \chi (d-\chi) (d-2\chi) (d^2+8 d -16)}{ \chi^{\prime2} (d-\chi^{\prime})^2 (d-2\chi^{\prime})^{2} (d-2)}\,,\\ s_{12}^3&=-\frac{1}{d^3}\frac{ \chi^2 (d-\chi)^2 (d-2\chi)^2 (d-2)}{2 \chi^{\prime} (d-\chi^{\prime}) (d-2\chi^{\prime}) (d^2+8 d -16)}\,. \end{align*} $$

Equation $(1,1,1)$ then writes $s_{33}$ in terms of $s_{12}$ and $s_{21}$ and we can conclude that $s_{33}^3=1$ . By simultaneously scaling the matrices B and S, if necessary, by a cubic root of unityFootnote ⁸ we may assume without loss of generality that $s_{33}=1$ . Finally, we obtain $s_{23}$ using $(1,2,0)$ and $s_{13}$ using $(2,1,0)$ ; note that these are uniquely determined once we choose which cubic root we take for $s_{21}$ .

After having obtained S, the next step now is to solve Equation (3.2) for the entries of A and B. The equation is quadratic in the entries of A and B; however, by rewriting it as

(3.4) $$ \begin{align} A^{\mathsf{T}} M_i =\left(\sum_{j=1}^3 s_{ij} M_j'\right) B^{-1}\,,\quad i=1,2,3\,, \end{align} $$

it becomes a linear (homogeneous) system in the entries $a_{st}$ of A and in the entries $\tilde b_{st}$ (written as bb[s,t] in the Mathematica file) of $B^{-1}$ . It turns out that the linear system only admits the trivial solution $a_{st}=\tilde b_{st}=0$ . Indeed, we can express all $a_{st}, \tilde {b}_{st}$ as a multiple of $a_{33}$ , and one additional constraint, denoted by Diff[d,chi1,chi2][[1]][[1,2]] in the Mathematica file, imposes that

$$\begin{align*}-\frac{2 \chi (d-4) (d-\chi)(d-2\chi) a_{33}}{d ((d-2) d^2+ 24 d{\chi'}-24 {\chi'}^2)}=0, \end{align*}$$

whence $a_{33}=0$ . Note that the denominator is always positive, thus nonzero. This gives a contradiction since A and B are invertible matrices; thus, there are no solutions to Equation (3.1) with S of Type I.

3.2. Type II

Suppose that S is a solution of Type II to Equation (3.3). We already have the vanishing of the entries $s_{31}=s_{32}=s_{12}=s_{21}=0$ . The remaining entries are obtained exactly as in the Type I case. The equations $(3,0,0)$ and $(0,3,0)$ determine, respectively, $s_{11}$ and $s_{22}$ up to a choice of a cubic root:

$$\begin{align*}s_{11}^3=\left(\frac{\chi(d-\chi)(d-2\chi)}{\chi'(d-\chi')(d-2\chi')}\right)^2\,,\quad s_{22}^3=\frac{\chi(d-\chi)(d-2\chi)}{\chi'(d-\chi')(d-2\chi')}\,.\end{align*}$$

Equation $(1,1,1)$ then writes $s_{33}$ in terms of $s_{11}$ and $s_{22}$ and we can conclude that $s_{33}^3=1$ . As we did for Type I, we may assume without loss of generality that $s_{33}=1$ . This fixes $s_{11}=s_{22}^2$ . Finally, we obtain $s_{13}$ using $(2,1,0)$ and $s_{23}$ using $(1,2,0)$ ; note that these are uniquely determined once we choose which cubic root we take for $s_{22}$ .

After obtaining S, Equation (3.4) is again linear in the entries $a_{st}$ of A and in the entries $\tilde b_{st}$ of $B^{-1}$ . The linear system can be solved explicitly, and it turns out that it has a one-dimensional space of solutions. By writing all the variables in terms of $a_{11}$ , the normalization imposed on the determinant $\det (A)=1$ gives

$$\begin{align*}a_{11}^3=\frac{\chi(d-\chi)(d-2\chi)}{\chi'(d-\chi')(d-2\chi')}=s_{22}^3\,.\end{align*}$$

By further scaling A by a cubic root of unity and scaling B by its inverse hence leaving S unchanged, we may assume without loss of generality that $a_{11}=s_{22}$ . This completely solves all the matrices $S, A, B$ once we choose which cubic root we take for $s_{11}$ . Solutions are implemented in the Mathematica file as solS[d,chi1,chi2], solA[d,chi1,chi2], solB[d,chi1,chi2], respectively.

Remark 3.4. When $\chi =\chi '$ , the unique solution of Type II is given by $S=A=B=\text {Id}_{3\times 3}$ , as expected. When $\chi +\chi '=d$ , the solution is given by simple sign matrices

$$ \begin{align*}A=B=\begin{bmatrix}-1& 0& 0\\ 0& 1&0\\ 0&0&-1\end{bmatrix},\quad S=\begin{bmatrix}1& 0& 0\\ 0& -1& 0\\ 0&0&1\end{bmatrix}. \end{align*} $$

In either case, these truncated solutions extend to a solution of Equation (3.1) since the moduli spaces $M_{d, \chi }$ and $M_{d, \chi '}$ are isomorphic. Indeed, the solution matrices are compatible with Proposition 2.1(c) which describes how the tautological generators are mapped under the two types of symmetries.

3.3. Type II – extended matrices

Recall that we have used only a part of the full Equation (3.1) truncated via the projection ${\mathbb {C}}[T]^d\twoheadrightarrow T^2\otimes T^{d-2}$ . This was sufficient to determine $A, B$ and S.

Step 3: bigger truncation

We proceed by considering a bigger truncation via the projection

$$ \begin{align*}{\mathbb{C}}[T]^d\twoheadrightarrow \Big(T^2\oplus {\text{Sym}}^2(T^1)\Big)\otimes T^{d-2}. \end{align*} $$

We can regard the truncated relations as a block $3\times (3\mid 3)$ matrix by identifying

$$\begin{align*}\begin{bmatrix}M_i&N_i\end{bmatrix} \in \Big(T^2\oplus {\text{Sym}}^2(T^1)\Big)\otimes T^{d-2}\cong \operatorname{Hom}((T^{d-2})^\vee, T^2\oplus {\text{Sym}}^2(T^1))\cong M_{3\times 6}\,,\end{align*}$$

where $M_i$ is as before and $N_i$ is implemented in the Mathematica file as ExtRelations[d,chi]. Similarly, we define $\begin {bmatrix}M_i'&N_i'\end {bmatrix}$ with $\chi $ replaced by $\chi '$ . As before, $\phi $ induces a linear map

where $U:T^2\rightarrow {\text {Sym}}^2(T^1)$ and $V:{\text {Sym}}^2(T^1)\rightarrow {\text {Sym}}^2(T^1)$ . They are implemented in the Mathematica file as AutU and AutV, respectively. Note that the lower left block is zero because $\phi $ is a graded ring isomorphism. Then the truncation of Equation (3.1) to $\Big (T^2\oplus {\text {Sym}}^2(T^1)\Big )\otimes T^{d-2}$ can be written as an equality between $3\times 6$ matrices

$$ \begin{align*} A^{\mathsf{T}} \begin{bmatrix}M_i&N_i\end{bmatrix} \begin{bmatrix}B& U\\ 0& V\\\end{bmatrix}=\sum_{j=1}^3 s_{ij} \begin{bmatrix}M_j'&N_j'\end{bmatrix}\,,\quad i=1,2,3\,. \end{align*} $$

Since $A, B$ and S satisfy Equation (3.2), this reduces to

(3.5) $$ \begin{align} A^{\mathsf{T}}M_iU+A^{\mathsf{T}}N_iV=\sum_{j=1}^3 s_{ij} N_j'\,,\quad i=1,2,3\,, \end{align} $$

which is a linear system in the entries $u_{st}$ and $v_{st}$ of U and V, respectively.

We show that this linear system, referred to as ExtDiff[d,chi1,chi2] in the Mathematica file, has a solution only if $\chi =\chi '$ or $\chi +\chi '=d$ . By using certain parts of the linear system (3.5), as explained in the Mathematica file, we can express $u_{11}, u_{21}, u_{31}, v_{11}, v_{21}$ in terms of $v_{31}$ . Once this is done, we further use two of the remaining linear system (3.5), referred to as ExtDiff[d,chi1,chi2][[1]][[1,1]] and ExtDiff[d,chi1,chi2][[1]][[2,1]], which take the form

$$ \begin{align*} \texttt{AA[d,chi1,chi2]}v_{31}+\texttt{BB[d,chi1,chi2]}&=0\,,\\ \texttt{CC[d,chi1,chi2]}v_{31}+\texttt{DD[d,chi1,chi2]}&=0\,, \end{align*} $$

following the notations from the code. Existence of the solution $v_{31}$ implies an equation which is purely in terms of d, $\chi $ and $\chi '$ :

This expression admits a factorization

$$ \begin{align*}\texttt{Constraint[d,chi1,chi2]}=\texttt{P1[d,chi1]}\cdot \texttt{P2[d,chi1,chi2]}\cdot\texttt{(other terms)} \end{align*} $$

with $\texttt {(other terms)}$ being clearly nonzero; see the Mathematica file. Note that the roles of $\chi $ and $\chi '$ are symmetric; we conclude that

(3.6) $$ \begin{align} P_1(d,\chi)\cdot P_2(d,\chi,\chi')=0\quad \text{and}\quad P_1(d,\chi')\cdot P_2(d,\chi',\chi)=0. \end{align} $$

We are left to prove that Equation (3.6) implies $\chi =\chi '$ or $\chi +\chi '=d$ . Suppose for the contradiction that $\chi \neq \chi '$ and $\chi +\chi '\neq d$ .

We first show that $P_2(d,\chi ,\chi ')$ and $P_2(d,\chi ',\chi )$ are nonzero. From the formula in the file, it is straightforward to check that $P_2(d,\chi ,\chi ')=0$ if and only if $P_2(d,\chi ',\chi )=0$ , if and only if

(3.7) $$ \begin{align} d^2 (\chi-\chi') (\chi+\chi'-d)\left(216\chi^4{\chi'}^4-432\chi^3{\chi'}^3(\chi+\chi')d+d^2\cdot f(d,\chi,\chi')\right)=0 \end{align} $$

for some explicit integral polynomial $f(d,\chi ,\chi ')$ . Since we assumed that $\chi \neq \chi '$ and $\chi +\chi '\neq d$ , the last term must vanish. On the other hand, $\gcd (d,\chi )=\gcd (d,\chi ')=1$ implies that d must divide $216$ . This further implies that $d^2$ divides $216$ by looking at the linear term in d, so we are left with $d=1,2,3,6$ . But there are no nontrivial pairs $\chi $ and $\chi '$ for such d.

Therefore, we may assume $P_1(d,\chi )=P_1(d,\chi ')=0$ . We can check that $P_1(d,x)$ is a degree four polynomial in x, with the symmetry

$$ \begin{align*}P_1(d,x)=P_1(d,d-x). \end{align*} $$

This implies that $P_1(d,x)$ is a degree four polynomial with four distinct roots

$$ \begin{align*}x=\chi,\; \chi',\; d-\chi,\; d-\chi',\end{align*} $$

all within the interval $[1,d-1]$ . On the other hand, explicit computation shows that

$$ \begin{align*}P_1(d,0)>0,\quad P_1(d,1)<0\end{align*} $$

as long as $d\geq 5$ . This forces $P_1(d,x)$ to have an additional root in the interval $(0,1)$ , which contradicts that it is a degree four polynomial. It follows that Equation (3.6) implies $\chi =\chi '$ or $\chi +\chi '=d$ , hence completing the proof.