# On the non-transversality of the hyperelliptic locus and the supersingular locus $g = 3$

Andreas Pieper

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Andreas Pieper. On the non-transversality of the hyperelliptic locus and the supersingular locus $g = 3$. 2024. hal-04374672v2

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Distributed under a Creative Commons Attribution 4.0 International License
ON THE NON-TRANSVERSALITY OF THE
HYPERELLIPTIC LOCUS AND THE SUPERSINGULAR
LOCUS FOR $g = 3$

ANDREAS PIEPER

ABSTRACT. This paper gives a criterion for a moduli point to be a point of non-transversal intersection of the hyperelliptic locus and the supersingular locus in the Siegel moduli stack $\mathcal{M}_3 \times \mathbb{F}_p$. It is shown that for infinitely many primes $p$ there exists such a point.

CONTENTS

<table><tr><td>1. Introduction</td><td>1</td></tr><tr><td>2. Preliminaries</td><td>3</td></tr><tr><td>2.1. Hyperelliptic curves</td><td>3</td></tr><tr><td>2.2. Dieudonné modules</td><td>4</td></tr><tr><td>2.3. Polarized flag type quotients</td><td>5</td></tr><tr><td>3. Deformation theory</td><td>6</td></tr><tr><td>3.1. Notation</td><td>6</td></tr><tr><td>3.2. Dieudonné modules</td><td>7</td></tr><tr><td>3.3. Kodaira-Spencer map</td><td>8</td></tr><tr><td>3.4. Irreducible components of formal neighborhoods</td><td>10</td></tr><tr><td>4. Non-transversality criteria</td><td>13</td></tr><tr><td>4.1. General case</td><td>13</td></tr><tr><td>4.2. Simplifications for $a = 1$</td><td>17</td></tr><tr><td>5. Examples</td><td>19</td></tr><tr><td>5.1. CM-examples ($a = 1$)</td><td>19</td></tr><tr><td>5.2. One example with $a = 3$</td><td>25</td></tr><tr><td>References</td><td>26</td></tr></table>

1. INTRODUCTION

Let $k$ be an algebraically closed field of characteristic $p > 2$. Denote by $\mathfrak{M}_g$ the Siegel moduli space parametrizing $g$-dimensional principally polarized abelian varieties over $k$. The study of the intersection $\mathfrak{H}_3 \cap \mathcal{S}_3$ of the locus $\mathfrak{H}_3 \subset \mathfrak{M}_3$ of Jacobians of smooth hyperelliptic curves and the supersingular locus $\mathcal{S}_3 \subset \mathfrak{M}_3$ was initiated by Oort's seminal article [13]. He showed that this intersection is equidimensional of dimension 1. The interest in this particular situation arises because it is one of the simplest instances of the
series of difficult questions around the intersection of Newton polygon strata and loci in $\mathfrak{A}_g$ defined by Jacobians of certain curves, e.g. the Torelli locus. We recommend the recent survey article by Pries [14] and the references therein for the readers interested in this circle of ideas.

As Pries observes in her survey, for each prime $p$ the Torelli locus and the supersingular locus intersect for infinitely many values of $g$. But this means that the intersection is non-transversal for infinitely many $g$ because the expected dimension of the intersection is

$$ \dim(\mathfrak{A}_g) - \dim(\mathcal{M}_g) - \dim(\mathcal{S}_g) = \frac{g(g+1)}{2} - (3g-3) - \lfloor \frac{g^2}{4} \rfloor $$

which is negative if $g \ge 9$.

A natural question is: What is the smallest $g$ for which a Newton polygon stratum intersects a locus defined via curve geometry non-transversely? The answer came quite as a surprise to the author: Already the simplest open case, i.e. the hyperelliptic locus for $g = 3$ exhibits this phenomenon.

Just to clear things up, we will view $\mathfrak{A}_3$ as a stack, in order to avoid complications with the quotient singularities appearing on the coarse moduli space. The reader who does not want to stick to stacks may instead assume that we add a level-$N$ structure, $N > 2$, $p \nmid N$, making the moduli functor representable by a scheme.

The main result of this article looks at an irreducible component $\mathcal{W}$ of the formal neighborhood of a moduli point in $\mathcal{S}_3$. Our theorem will give a necessary and sufficient condition when $\mathcal{W}$ meets $\mathcal{H}_3$ non-transversely.<fnref n="1" />

To a smooth hyperelliptic supersingular curve $C$ and an irreducible component $\mathcal{W}$ of the formal neighborhood of the moduli point $[\mathrm{Jac}(C)]$ in $\mathcal{S}_3$ we associate a certain geometric configuration in $\mathbb{P}^2$, to wit: We define at a conic $Q$, a line $l$ and a point $P$ on $l$. The theorem is as follows:

**Theorem A.** The following are equivalent:

i) The component $\mathcal{W}$ meets $\mathcal{H}_3$ non-transversally at $[\mathrm{Jac}(C)]$.

ii) The line $l$ touches $Q$ at the point $P$.

To explain the origin of this geometric configuration: $Q$ is the image of the canonical map $C \to \mathbb{P}^2$ of our hyperelliptic curve. $P$ and $l$ will be defined via the PFTQ theory of Li and Oort [10].

In the special case where $a = 1$, the theorem simplifies as then the irreducible component $\mathcal{W}$ will be unique. Furthermore, the point $P$ and the line $l$ have a convenient description in terms of the Cartier-Manin matrix of $C$. Then Theorem A becomes:

**Theorem B.** Let $C$ be a hyperelliptic supersingular curve of genus 3 with $a(\mathrm{Jac}(C)) = 1$. Then the following are equivalent:

i) The supersingular locus $\mathcal{S}_3$ meets $\mathcal{H}_3$ non-transversally at $[C]$.

<footnote n="1">¹There is one case where $\mathcal{W}$ can be singular, which will be excluded, so that the notion of non-transversality is well-defined.</footnote>
ii) There exists a point $P \in C$ such that for any hyperelliptic equation

$$y^2 = f(x)$$

for $C$ that satisfies $x(P) = 0$ the corresponding Cartier-Manin matrix (formed by extracting suitable coefficients of $f^{\frac{p-1}{2}}$) has the shape

$$\begin{pmatrix} 0 & 0 & 0 \\ * & 0 & 0 \\ * & * & 0 \end{pmatrix}$$

The last stage of the article will be an example section. Examples for curves satisfying the criterion of Theorem B will be constructed as CM-reductions. The respective CM curve needs to satisfy a special condition on the CM-action that has a similar look-alike as ii) in Theorem B.

To obtain a CM curve with this property we apply a special case of the construction of Tautz-Top-Verberkmoes [16] for curves with RM by $\mathbb{Q}(\zeta_l + \zeta_l^{-1})$. Putting $l = 7$ this can be used to get a genus 3 hyperelliptic curve with CM by $\mathbb{Q}(\zeta_7 + \zeta_7^{-1}, i)$.

The idea behind the choice of the field $\mathbb{Q}(\zeta_7 + \zeta_7^{-1}, i)$ is as follows: The most obvious CM curve $y^2 = x^7 - 1$ with CM by the cyclotomic field $\mathbb{Q}(\zeta_7)$ almost satisfies the required lower triagonality condition of Theorem B ii). But it falls short, in fact, the triagonality is only achieved after an illegal permutation of the rows and columns of the Cartier-Manin matrix.

For that reason the author searched for a curve with CM by a field which is not quite cyclotomic, but close. The field $\mathbb{Q}(\zeta_7 + \zeta_7^{-1}, i)$ is the first option and fortunately it satisfies our needs.

**Acknowledgments:** The author thanks Irene Bouw, David Lubicz, Laurent Moret-Bailly, Mathieu Romagny, Jeroen Sijsling, and Stefan Wewers for enlightening and encouraging discussions on the subject of this article, and Rachel Pries for a helpful E-mail correspondence.

## 2. PRELIMINARIES

### 2.1. Hyperelliptic curves.
In this section we recall some preliminaries around the deformation theory of curves.

Let us denote by $\mathfrak{H}_g$ the moduli stack of hyperelliptic genus $g$ curves over a field of characteristic $\neq 2$. Furthermore we look at the Torelli morphism

$$t : \mathfrak{H}_g \to \mathfrak{M}_g, [C] \mapsto [(\mathrm{Jac}(C), \Theta)].$$

The map induced by $t$ on tangent spaces has the following properties:

**Lemma 2.1.** Let $C$ be a hyperelliptic curve. The map $t$ induces on tangent spaces a map

$$T_{[C]}\mathfrak{H}_g \to T_{\mathrm{Jac}(C)}\mathfrak{M}_g \cong S^2 \mathrm{H}^0(C, \Omega_C)^{\vee}$$

which

i) is injective
ii) and its image equals the orthogonal complement of the kernel of the multiplication map

$$S^2 \mathrm{H}^0(C, \Omega_C) \to \mathrm{H}^0(C, \Omega_C^{\otimes 2}).$$

*Proof.* See [2, p. 223]. The authors of that book assume that the base field $\mathbb{C}$, however, it is not difficult to see that their proof works over any field of characteristic $\neq 2$.
□

The preceding lemma and the Torelli theorem together imply that the map $t: \mathfrak{H}_g \to \mathfrak{A}_g$ is a locally closed immersion. By abuse of notation we will identify $\mathfrak{H}_g$ with its image in $\mathfrak{A}_g$.

**2.2. Dieudonné modules.** In [4][6] (relative) Dieudonné modules are defined. The purpose of this section is to review these definitions and the important theorems. Indeed, let $A$ be a regular local ring essentially of finite type over $k$. Berthelot & Messing and de Jong consider much more general rings, but these assumptions suffice for our purposes.

Now [4, Proposition 1.1.7] implies the existence of a lift of $A$, i.e., a flat $\mathbb{Z}_p$-algebra $\tilde{A}$ which is $p$-adically complete and satisfies $\tilde{A}/p \cong A$. Furthermore, there exists a lift of Frobenius $\sigma: \tilde{A} \to \tilde{A}$ by [4, Corollaire 1.2.7]. Notice that $\tilde{A}$ is unique up to isomorphism, however, $\sigma$ is not unique. We choose $\sigma$ once and for all.

We consider the continuous differential forms

$$\hat{\Omega}_{\tilde{A}} = \lim_{\underset{n}{\overline{\hspace{2em}}}} \Omega_{A_n/\mathbb{Z}_p}$$

where $A_n = \tilde{A}/p^n$. By [4, Proposition 1.3.1], $\hat{\Omega}_{\tilde{A}}$ is a free $\tilde{A}$-module of rank $d = \dim(A)$. We will exhibit a basis below.

Now we make the following definition, essentially a special case of [6, Definition 2.3.4]:

**Definition 2.2.** A Dieudonné module over $\tilde{A}$ is a finite locally free $\tilde{A}$-module $M$ equipped with an integrable, topologically quasi-nilpotent connection

$$\nabla : M \to M \hat{\otimes}_{\tilde{A}} \hat{\Omega}_{\tilde{A}}$$

and $\nabla$-horizontal, $\tilde{A}$-linear maps

$$F : M \hat{\otimes}_{\sigma} \tilde{A} \to M, V : M \to M \hat{\otimes}_{\sigma} \tilde{A}$$

satisfying $F \circ V = V \circ F = p$.

Here topologically quasi-nilpotent means: For any derivation $\delta \in \mathrm{Hom}_{\tilde{A}}(\hat{\Omega}_{\tilde{A}}, \tilde{A})$ and any $m \in M$, there exists an $n \in \mathbb{N}$ such that $\nabla_{\delta}^{\circ n}(m) \in pM$.

We now recall the following theorem from [6]:

**Theorem 2.3.** There is a contravariant functor

$$\mathbb{D} : \{ \mathrm{p-divisible \ groups \ over \ Spec(A)} \} \to \{ \mathrm{Dieudonné \ modules \ over \ \tilde{A}} \}$$

*Proof.* Follows from [3, Proposition 1.3.3].
□
To distinguish this relative theory of Dieudonné modules from the story over $k$, we will denote the functor

$$ \{p\text{-divisible groups over Spec}(k)\} \to \{\text{Dieudonné modules over } W(k)\} $$
by $D$ and call it the classical Dieudonné module.

**2.3. Polarized flag type quotients.** In this section we recall the results from [10] we shall need. We will specialize to the case $g = 3$, although Li & Oort treat the case of general $g$.

Choose once and for all $E/k$, a supersingular elliptic curve defined over $\mathbb{F}_p$ such that the relative Frobenius satisfies the equation $F^2 + p = 0$. The theory of Li & Oort gives a bijection between the set of irreducible components of $S_3$ and polarizations $\eta$ on $E^3$ satisfying $\ker(\eta) = E^3[p]$. The bijection is obtained by writing down families of certain principally polarized quotients of $(E^3, \eta)$, so-called polarized flag-type quotients. The definition is as follows:

**Definition 2.4.** Let $\eta$ be as in the discussion preceding the definition. Let $S$ be a $k$-scheme. A polarized flag-type quotient (PFTQ for short) starting in $(E^3, \eta)$ is a sequence of isogenies of abelian $S$-schemes

$$ E^3 \times_k S = \mathcal{Y}_2 \xrightarrow{\rho_2} \mathcal{Y}_1 \xrightarrow{\rho_1} \mathcal{Y}_0 $$

such that:

i) $\ker(\rho_i)$ is an $\alpha$-group scheme of rank $i$.

ii) $\eta$ descends along the $\rho_i$ to polarizations $\eta_i$ on $\mathcal{Y}_i$ for $i = 0, 1$.

iii) $\ker(\eta_1) \subset \mathcal{Y}_1[F]$.

Two PFTQs

$$ E^3 \times_k S = \mathcal{Y}_2 \xrightarrow{\rho_2} \mathcal{Y}_1 \xrightarrow{\rho_1} \mathcal{Y}_0 $$

$$ E^3 \times_k S = \mathcal{Y}'_2 \xrightarrow{\rho'_2} \mathcal{Y}'_1 \xrightarrow{\rho'_1} \mathcal{Y}'_0 $$

are said to be isomorphic if there exist isomorphisms $\varphi_i : \mathcal{Y}_i \to \mathcal{Y}'_i$ with $i = 0, 1$ such that

$$ \begin{array}{ccc} \mathcal{Y}_2 & \xrightarrow{\rho_2} & \mathcal{Y}_1 \\ \mathcal{Y}'_2 & \xrightarrow{\rho'_2} & \mathcal{Y}'_1 \end{array} \xrightarrow{\varphi_1} \begin{array}{ccc} \mathcal{Y}_0 & & \\ & \ddots & \\ & & \mathcal{Y}_0 \end{array} $$

commutes.

Li & Oort proved the following representability result:

**Theorem 2.5.** Let $\eta$ be as above. The functor:

$$ k - \mathrm{Sch} \to \mathrm{Sets} $$

$$ S \mapsto \{ \text{PFTQs starting in } (E^3, \eta) \} / \mathrm{iso.} $$

is representable by a smooth projective integral $k$-scheme, denoted $\mathcal{P}_{3,\eta}$.

More precisely $\mathcal{P}_{3,\eta}$ admits the following description:
i) There is a map $\pi : \mathcal{P}_{3,\eta} \to \mathbb{P}^2$ (induced from forgetting $\mathcal{Y}_0$).

ii) The image of $\pi$ is $\mathrm{im}(\pi) = \mathcal{V}(X_0^{p+1}+X_1^{p+1}+X_2^{p+1})$, i.e., the standard Hermitian curve. We will denote it by $\mathcal{C}_H$.

iii) The map $\pi$ is a $\mathbb{P}^1$-bundle. More precisely, there is an isomorphism

$$ \mathcal{P}_{3,\eta} \cong \mathbb{P}_{\mathcal{C}_H}(\mathcal{O}(1) \oplus \mathcal{O}(-1)) $$

*Proof.* See [10, Lemma 3.7] for the representability. The explicit description is [10, Section 9.4]. □

By the Yoneda lemma there is a universal family $\mathcal{Y}_2 \to \mathcal{Y}_1 \to \mathcal{Y}_0$ of PFTQs over $\mathcal{P}_{3,\eta}$. The Abelian scheme $\mathcal{Y}_0$ carries a principal polarization (obtained from descending $\eta$). We denote by

$$ \varphi_{\mathrm{LO}} : \mathcal{P}_{3,\eta} \to \mathfrak{A}_3 $$

the corresponding map to the Siegel moduli stack. Clearly the image is contained in the supersingular locus. Now Li & Oort have proven the following theorem:

**Theorem 2.6.** i) The map

$$ \varphi_{\mathrm{LO}} : \mathcal{P}_{3,\eta} \to \mathcal{S}_3 $$

contracts a curve $T$ and is quasi-finite on the complement of $T$.

ii) If we take the disjoint union over the (finitely many) equivalence classes of $\eta$ modulo automorphisms of $E^3$, the obtained map

$$ \coprod_{\eta} \mathcal{P}_{3,\eta} \to \mathcal{S}_3 $$

is surjective.

*Proof.* The curve $T$ is defined to be the image of the section

$$ s : \mathcal{C}_H \to \mathcal{P}_{3,\eta} $$

of $\pi$ corresponding to the map

$$ \mathrm{pr}_2 : \mathcal{O}(1) \oplus \mathcal{O}(-1) \to \mathcal{O}(-1). $$

See [10, pp. 58-59] for the details. The other statements are [10, Corollary 4.1]. □

### 3. DEFORMATION THEORY

**3.1. Notation.** We are going to fix the following notation for the rest of the article: $p > 2$ is a prime number and $k$ is an algebraically closed field of characteristic $p$.
## 3.2. Dieudonné modules.
In this section we describe the variation of the crystalline cohomology in a Li-Oort family. This is a technical tool we need for computing the Kodaira-Spencer map in the next section.

Indeed, let $\mathcal{Y}_0 \to \mathcal{P}_{3,\eta}$ be the principally polarized abelian scheme constructed by Li & Oort and $\xi \in \mathcal{P}_{3,\eta}$ be a closed point. Assume that $\xi \notin T$. Our goal is to describe the relative crystalline cohomology of $\mathcal{Y}_0$ in an open neighborhood of $\xi$. To this end, consider the stalk $\mathcal{O}_{\mathcal{P}_{3,\eta},\xi}$ and denote it by $A$. Choose $\tilde{A}$, $\sigma$, a lift of $A$ together with a lift of Frobenius as in Section 2.2. Our goal is to describe the Dieudonné module of the Abelian scheme $\mathcal{Y}_0 \times_{\mathcal{P}_{3,\eta}} \text{Spec}(A)$. Intuitively this means that we want to understand the variation of the crystalline cohomology in an open neighborhood of the point $\xi$.

To start with, recall that we are given a supersingular elliptic curve $E$ and a polarization $\eta$ on $E^3$ satisfying $\ker(\eta) = E^3[p]$. Denote by $M_2$ the Dieudonné module of $E^3$ (in the classical sense). Then $\eta$ induces an alternating pairing

$$ \langle \cdot, \cdot \rangle : M_2^t \times M_2^t \to W(k) $$

where $M_2^t$ is the dual Dieudonné module.

By [10, Lemma 6.1] there exists a $W(k)$-basis $m_0, Fm_0, m_1, Fm_1, m_2, Fm_2$ of $M_2$ such that for all $i = 0, 1, 2$

$$ F \cdot m_i = Fm_i, \quad F \cdot Fm_i = -p, $$

and such that the pairing $\langle \cdot, \cdot \rangle$ is given by the matrix $\begin{pmatrix} 0 & p \\ -p & 0 \end{pmatrix}^{\oplus 3}$ with respect to the basis dual to $m_0, \dots, Fm_2$.

Now consider the constant Abelian scheme $E_A^3 = E^3 \times \text{Spec}(A)$. We have

$$ \mathbb{D}(E_A^3) = M_2 \hat{\otimes}_{W(k)} \tilde{A} $$

where the map $W(k) \to \tilde{A}$ is the unique map lifting the inclusion $k \hookrightarrow A$ (which exists because $k$ is perfect and $\tilde{A}$ is $p$-adically complete). Denote $M_2 = \mathbb{D}(E_A^3)$. It carries a connection

$$ \nabla : M_2 \to M_2 \hat{\otimes} \hat{\Omega}_{\tilde{A}}, $$

which equals the trivial connection since $E_A^3$ is a constant family.

Before we proceed to determine the Dieudonné module of $\mathcal{Y}_0$, we digress and describe a certain affine open subset of $\mathcal{P}_{3,\eta}$. Indeed, recall that we have a $\mathbb{P}^1$-bundle

$$ \pi : \mathcal{P}_{3,\eta} \to \mathcal{C}_H $$

where $\mathcal{C}_H = \mathcal{V}(X_0^{p+1} + X_1^{p+1} + X_2^{p+1}) \subset \mathbb{P}^2$. Let us denote by $U_0 \subset \mathcal{C}_H$ the standard affine open where $X_0$ does not vanish. After relabeling the coordinates we can assume without loss that $\xi \in U_0$.

We put $x_i = \frac{X_i}{X_0}$. Now the $\mathbb{P}^1$-bundle $\pi$ has a trivialization over $U_0$ given by the standard trivialization of the vector bundle $\mathcal{O}(1) \oplus \mathcal{O}(-1)$ on $U_0$.
This determines an isomorphism

$$ \pi^{-1}(U_0) \setminus T \cong U_0 \times \mathbb{A}^1. $$

We denote the latter affine open by $U \subset \mathcal{P}_{3,\eta}$. By assumption, we have $\xi \in U$. Let us denote by $t$ the function in the coordinate ring of $U$ corresponding to the coordinate on the second factor of $U_0 \times \mathbb{A}^1$. By localization we get three elements in $A = \mathcal{O}_{\mathcal{P}_{3,\eta},\xi}$ which we also denote $x_1, x_2, t$ by slight abuse of notation.

We get to the main lemma of this section.

**Lemma 3.1.** The Dieudonné module $\mathbb{D}(\mathcal{Y}_0 \times \text{Spec}(A))$ is given by the $\tilde{A}$-submodule of $\mathbb{M}_2$ generated by

$$ m_0 + \tilde{x}_1^p m_1 + \tilde{x}_2^p m_2 + \tilde{t}^p Fm_0, -\tilde{x}_1^p Fm_0 + Fm_1, -\tilde{x}_2^p Fm_0 + Fm_2, p\mathbb{M}_2 $$
where $\tilde{x}_1, \tilde{x}_2, \tilde{t} \in \tilde{A}$ are lifts of $x_1, x_2, t$ respectively.

*Proof.* Let us denote by $\mathbb{M}_0$ the $\tilde{A}$-submodule of $\mathbb{M}_2$ generated by

$$ m_0 + \tilde{x}_1^p m_1 + \tilde{x}_2^p m_2 + \tilde{t}^p Fm_0, -\tilde{x}_1^p Fm_0 + Fm_1, -\tilde{x}_2^p Fm_0 + Fm_2, p\mathbb{M}_2. $$

First we notice that $\mathbb{M}_0$ is clearly independent of the choice of the lifts $\tilde{x}_1, \tilde{x}_2, \tilde{t}$.

To prove that $\mathbb{M}_0$ is a Dieudonné submodule, we must prove that it is finite locally free and closed under $\nabla, F, V$. The former assertion follows easily from the short exact sequence

$$ 0 \to \mathbb{M}_0 \to \mathbb{M}_2 \to \mathbb{M}_2/\mathbb{M}_0 \to 0 $$

and the observation that $\mathbb{M}_2/\mathbb{M}_0$ is a free $\tilde{A}/p$-module with basis $m_1, m_2, Fm_0$.

For the closedness under $\nabla$ we compute

$$ \nabla(m_0 + \tilde{x}_1^p m_1 + \tilde{x}_2^p m_2 + \tilde{t}^p Fm_0) = p \tilde{x}_1^{p-1} m_1 \otimes d \tilde{x}_1 + p \tilde{x}_2^{p-1} m_2 \otimes d \tilde{x}_2 + p \tilde{t}^{p-1} Fm_0 \otimes d \tilde{t} $$

which is in $p\mathbb{M}_2 \hat{\otimes} \hat{\Omega}_{\tilde{A}}$ and similarly for the other generators. Thus we see that the $p$-th powers in the definition of the generators are essential for the closedness under $\nabla$.

The closedness under $F, V$ is an easy verification left to the reader.

We conclude that $\mathbb{M}_0$ is a Dieudonné submodule of $\mathbb{M}_2$. The fact that it is equal to the image of

$$ \mathbb{D}(\mathcal{Y}_0 \times \text{Spec}(A)) \to \mathbb{D}(E_A^3) = \mathbb{M}_2 $$

follows from the explicit description of the $g = 3$ Li-Oort family in [10, Section 9.4]. $\square$

**3.3. Kodaira-Spencer map.** Let $\mathcal{Y}_0 \to \mathcal{P}_{3,\eta}$ be a Li-Oort family and $\xi \in \mathcal{P}_{3,\eta}$ be a closed point as in the previous section. The Kodaira-Spencer map

$$ \kappa : T_{\xi} \mathcal{P}_{3,\eta} \to \text{Hom}_{\text{sym}}(\text{H}^0(\mathcal{Y}_{0,\xi}, \Omega_{\mathcal{Y}_{0,\xi}}) \to \text{H}^1(\mathcal{Y}_{0,\xi}, \mathcal{O}_{\mathcal{Y}_{0,\xi}})) $$

first introduced in the algebraic setting by Illusie [9, 2.1.5.7], may be viewed as the differential of the morphism induced by the universal property of the
Siegel moduli stack. In this section we will compute the Kodaira-Spencer map.

We begin by choosing a basis of $T_\xi \mathcal{P}_{3,\eta}$. Indeed, the first basis element is $\delta_1 = \frac{\partial}{\partial t}$. The one-dimensional vector space generated by $\frac{\partial}{\partial t}$ is canonically defined because it is the direction parallel to the fibers of the $\mathbb{P}^1$-bundle. The second basis element we will choose is not canonically defined as it will depend on our choice of a trivialization of this $\mathbb{P}^1$-bundle. Let us take $\delta_2 = -x_2^p \frac{\partial}{\partial x_1} + x_1^p \frac{\partial}{\partial x_2}$. The reader will have no difficulties to see that this formula defines a derivation on the affine open $U \subset \mathcal{P}_{3,\eta}$ that does not vanish at $\xi$.

We are now ready to compute the Kodaira-Spencer map. Indeed, consider the $A$-module with connection $\mathcal{H}_{\mathrm{dR}}^1 = (\mathbb{M}_0/p, \nabla)$. Then $\mathcal{H}_{\mathrm{dR}}^1$ is isomorphic to the relative de Rham cohomology of $\mathcal{Y}_0 \times_{\mathcal{P}_{3,\eta}} \mathrm{Spec}(A)$ over $\mathrm{Spec}(A)$.

Therefore, it only remains to identify the submodule of differentials. For that purpose we look at the exact sequence

$$ \mathcal{H}_{\mathrm{dR}}^1 \xrightarrow{v} (\mathcal{H}_{\mathrm{dR}}^1)^{(p)} \xrightarrow{f} \mathcal{H}_{\mathrm{dR}}^1, $$

where $(\mathcal{H}_{\mathrm{dR}}^1)^{(p)} = (\mathcal{H}_{\mathrm{dR}}^1) \otimes_{A, x \mapsto x^p} A$ and $f$ (resp. $v$) is the map induced by $F$ (resp. $V$). We will use the following Lemma from de Jong's [6]:

**Lemma 3.2.** There is a locally free, locally direct summand $\omega \subset \mathcal{H}_{\mathrm{dR}}^1$ such that $\omega^{(p)} \subset (\mathcal{H}_{\mathrm{dR}}^1)^{(p)}$ is equal to $\mathrm{im}(v) = \ker(f)$. It is uniquely determined by these conditions.

$$ \textit{Furthermore } \omega = \mathrm{H}^0(\mathcal{Y}_0 \times_{\mathcal{P}_{3,\eta}} \mathrm{Spec}(A), \Omega_{\mathcal{Y}_0 \times_{\mathcal{P}_{3,\eta}} \mathrm{Spec}(A)}). $$

*Proof.* The first assertion is [6, Proposition 2.5.2]. The second assertion follows from [3, Proposition 4.3.10]. $\square$

In our case, $\omega$ is generated by the four classes

$$ Fm_0 + \tilde{x}_1 Fm_1 + \tilde{x}_2 Fm_2 - p\tilde{t}m_0, p\tilde{x}_1 m_0 - pm_1, p\tilde{x}_2 m_0 - pm_2, pFm_0 $$

We will now assume that $x_2$ does not vanish at $\xi$ which we can, without loss, after switching $x_1, x_2$ if necessary. With this assumption it is easy to see that $p\tilde{x}_2 m_0 - pm_2$ is a linear combination of the other classes in $\omega$. Therefore,

$$ g_1 = Fm_0 + \tilde{x}_1 Fm_1 + \tilde{x}_2 Fm_2 - p\tilde{t}m_0, \quad g_2 = p\tilde{x}_1 m_0 - pm_1, \quad g_3 = pFm_0 $$

is a basis for $\omega$ as an $A$-module.

Similarly we get the basis

$$ c_1 = pm_0, c_2 = \tilde{x}_1^p Fm_0 - Fm_1, c_3 = m_0 + \tilde{x}_1^p m_1 + \tilde{x}_2^p m_2 + \tilde{t} Fm_0 $$

for $(\mathcal{H}_{\mathrm{dR}}^1)/\omega$. The Kodaira-Spencer map is now calculated as follows:

**Lemma 3.3.**

$$ \kappa(\delta_1)(g_1) = -pm_0 = -c_1, \kappa(\delta_1)(g_2) = \kappa(\delta_1)(g_3) = 0 $$

$$ \kappa(\delta_2)(g_1) = \tilde{x}_2^p Fm_1 - \tilde{x}_1^p Fm_2 \in \mathrm{span}(c_1, c_2), \kappa(\delta_2)(g_2) = \tilde{x}_2^p pm_0, \kappa(\delta_2)(g_3) = 0 $$
*Proof.* The Kodaira-Spencer map

$$ \kappa : T_{\xi} \mathcal{P}_{3,\eta} \to \mathrm{Hom}^{\mathrm{sym}}(\mathrm{H}^{0}(\mathcal{Y}_{0,\xi}, \Omega_{\mathcal{Y}_{0,\xi}}) \to \mathrm{H}^{1}(\mathcal{Y}_{0,\xi}, \mathcal{O}_{\mathcal{Y}_{0,\xi}})) $$

is computed as follows: For a derivation $\delta$ we look at the map

$$ \omega \to \mathcal{H}_{\mathrm{dR}}^{1}/\omega $$

induced from $\nabla_{\delta}$ and restrict it to the fiber at $\xi$. This gives the desired map

$$ \kappa(\delta) \in \mathrm{Hom}^{\mathrm{sym}}(\mathrm{H}^{0}(\mathcal{Y}_{0,\xi}, \Omega_{\mathcal{Y}_{0,\xi}}) \to \mathrm{H}^{1}(\mathcal{Y}_{0,\xi}, \mathcal{O}_{\mathcal{Y}_{0,\xi}})). $$

The calculation is elementary and left to the reader.

The assertion that $\tilde{x}_2^p Fm_1 - \tilde{x}_1^p Fm_2 \in \mathrm{span}(c_1, c_2)$ also requires a small verification. $\square$

**Corollary 3.4.** *The map*

$$ \varphi_{LO} : \mathcal{P}_{3,\eta} \to \mathcal{S}_3 $$

is unramified on the open set $\mathcal{P}_{3,\eta} \setminus T$.

*Proof.* The differential of $\varphi_{LO}$ at the point $\xi$ is given by the Kodaira-Spencer map which we computed in the previous lemma. The point $\xi$ was assumed to be in $U$ and satisfied $x_2 \neq 0$. It is not hard to verify that the two Kodaira-Spencer classes are linearly independent.

This implies that $\varphi_{LO}$ is unramified on the open set $U \cap \{x_2 \neq 0\}$ and similarly for the open sets we obtain by permuting coordinates. Since these open sets cover $\mathcal{P}_{3,\eta} \setminus T$, the lemma follows. $\square$

**3.4. Irreducible components of formal neighborhoods.** Let $Y$ be a supersingular p.p.a.v. In this section we study the formal completion of $\mathcal{S}_3$ at the moduli point $[Y]$. Let us denote this formal completion by $\hat{\mathcal{S}}_{3,[Y]}$. We show that every irreducible component of $\hat{\mathcal{S}}_{3,[Y]}$ is smooth with one exception when $a(Y) = 3$. All the results in this section are well-known to the experts, but we include them here as the author could not find proofs in the literature.

**Lemma 3.5.** Let $Y$ be as above. There is a bijection

$$ \left\{ \begin{array}{l} \text{Irreducible components} \\ \text{of } \hat{\mathcal{S}}_{3,[Y]} \end{array} \right\} \longleftrightarrow \left\{ \begin{array}{l} \text{Isogenies } E^3 \to Y \text{ that} \\ \text{admit a PFTQ structure} \end{array} \right\}_{\sim} $$

where the equivalence relation $\sim$ is defined as follows: $\pi_1, \pi_2 : E^3 \to Y$ are called equivalent if there exists an automorphism $\varphi$ of $E^3$ such that

<figure><img src="image_2.png" /></figure>

commutes.
*Proof.* We define a map

$$f: \left\{ \begin{array}{l} \text{Isogenies } E^3 \to Y \text{ that} \\ \text{admit a PFTQ structure} \end{array} \right\}_{/\sim} \to \left\{ \begin{array}{l} \text{Irreducible components} \\ \text{of } \hat{S}_{3,[Y]} \end{array} \right\}$$

as follows: Let $\pi: E^3 \to Y$ be an isogeny that admits a PFTQ structure. Denote by $\eta$ the pullback of the principal polarization on $Y$ via $\pi$. Then, by assumption, there exists a PFTQ starting in $(E^3, \eta)$

$$E^3 \to Y_1 \to Y_0 = Y$$

that composes to $\pi$. As a consequence of [10, Equation 9.4.11] and the properties of the notion of rigid PFTQs in (loc. cit.) this PFTQ is unique unless $\ker(\pi) = E^3[F]$. In the former case one gets a unique point $\xi \in \mathcal{P}_{3,\eta}$ such that $\varphi_{\text{LO}}(\xi) = [Y]$ under the map

$$\varphi_{\text{LO}}: \mathcal{P}_{3,\eta} \to \mathcal{S}_3$$

We define $f(E^3 \to Y)$ to be the irreducible component $\hat{S}_{3,[Y]}$ given by the image of the map induced by $\varphi_{\text{LO}}$ on the formal completion at $\xi$.

The latter case $\pi = F$ can only happen if $a(Y) = 3$. In this case the map

$$\varphi_{\text{LO}}: \mathcal{P}_{3,\eta} \to \mathcal{S}_3$$

contracts the curve $T$ to the point $[Y]$. We define $\Xi$ to be the irreducible component of $\hat{S}_{3,[Y]}$ given by the image of the map induced by $\varphi_{\text{LO}}$ on the formal completion along $T$. Furthermore we put $f(E^3 \to Y) = \Xi$. It is not hard to verify that the map $f$ is well-defined.

We have to show that $f$ is bijective. For that purpose we equip $\mathcal{S}_3$ with a level $N$ structure for some $N \ge 3$, $p \nmid N$ and denote the corresponding moduli space by $\mathcal{S}_{3,N}$. Then, $\mathcal{S}_{3,N}$ is a closed subscheme of the Siegel moduli scheme $\mathcal{A}_{g,N}$ (equipped with a level $N$ structure). Furthermore, one can show that the Li-Oort construction can also be equipped with a level structure yielding a map

$$\varphi: \coprod \mathcal{P}_{3,\eta} \to \mathcal{S}_{3,N}$$

with the same properties as in Theorem 2.6. The disjoint union is taken over the finite set $\Lambda_N$ of all triples $(l_N, \eta)$ where $l_N$ is a level $N$ structure on $E^3$ and $\eta$ is a polarization on $E^3$ satisfying $\ker(\eta) = E^3[p]$ (taken modulo automorphisms of $E^3$).

Furthermore $\varphi$ is birational by [10, Remark 6.4]. Then, since $\mathcal{S}_{3,N} \to \mathcal{S}_3$ is étale and $k$ is algebraically closed, it induces an isomorphism on completions. We choose $l_N$, a level $N$ structure on $Y$ and it remains to prove the claim about the irreducible components for the completion of $\mathcal{S}_{3,N}$ at $(Y, l_N)$. Indeed, assume first that $a(Y) = 3$. Then there is an isomorphism $Y \cong E^3$ as abelian varieties and the composition $\pi: E^3 \xrightarrow{F} E^3 \cong Y$ admits the structure of a PFTQ. This leads to the irreducible component $\Xi$ defined above. As that component is kind of problematic, we want to remove $\Xi$ and prove that $f$ restricts to a bijection between the remaining sets.
To this end we consider $l'_N$ the level $N$ structure on $E^3$ obtained from $l_N$ by pulling back along the isomorphism

$$E^3[N] \to Y[N]$$

induced by $\pi$. Furthermore denote by $\eta$ the polarization on $E^3$ defined as the pullback of the principal polarization on $Y$ via $\pi$. Consider the restricted map

$$\coprod_{\Lambda_N \setminus \{(l'_N, \eta)\}} \mathcal{P}_{3,\eta} \to \mathcal{S}_{3,N}$$

and denote its image by $X$. The closed subscheme $X \subset \mathcal{S}_{3,N}$ has one irreducible component less than $\mathcal{S}_{3,N}$. As a last bit of notation we put $\Lambda'_N = \Lambda_N \setminus \{(l'_N, \eta)\}$.

If instead $a(Y) \le 2$, then we do not have to remove an irreducible component and we put $\Lambda'_N = \Lambda_N$ and $X = \mathcal{S}_{3,N}$.

In any case we have a map, also denoted $\varphi$ by slight abuse of notation

$$\varphi : \coprod_{\Lambda'_N} \mathcal{P}_{3,\eta} \to X.$$

The map $\varphi$ is birational, proper (because $\mathcal{P}_{3,\eta}$ is projective). Now we choose $U \subset X$, a Zariski open neighborhood of $(Y, l_N)$ small enough so that $U$ does not contain any points with $a = 3$ except maybe $(Y, l_N)$ itself. Then we look at the cartesian diagram

$$\begin{tikzcd}
U' \arrow[r] \arrow[d] & \coprod_{\Lambda'_N} \mathcal{P}_{3,\eta} \arrow[d] \\
\coprod_{\Lambda'_N} \mathcal{P}_{3,\eta} \arrow[r, "\varphi'", "\varphi"] & X
\arrow[ur] \arrow[ul] \\
U \arrow[r] & X
\arrow[ur] \arrow[ul] \arrow[dl] \arrow[dr]
\end{tikzcd}$$

The map $\varphi'$ is proper. It is also quasi-finite by Theorem 2.6 because $U$ does not contain any points with $a = 3$ except maybe $(Y, l_N)$. Indeed, by construction we removed the only component of $\coprod_{\Lambda_N} \mathcal{P}_{3,\eta}$ containing the unique positive dimensional component of the fiber $\varphi^{-1}(Y, l_N)$. As a consequence we conclude that $\varphi'$ is proper, quasi-finite and thus finite.

On the other hand, $U'$ is an open subscheme of the smooth $k$-scheme $\coprod_{\Lambda'_N} \mathcal{P}_{3,\eta}$ and in particular $U'$ is normal. In summary, we have that $\varphi'$ is a finite birational map with normal source. By [15, Lemma 035Q, (3)] this implies that $\varphi'$ is the normalization of $U$.

Now since $U$ is excellent, [15, Lemma 0C23] implies that normalization commutes with completion. Finally [15, Lemma 035Q, (2)] gives the desired bijection for the irreducible components of the completion. $\square$

**Corollary 3.6.** In the notation as above.

i) If $a(Y) \le 2$, then all the irreducible components of $\hat{S}_{3,[Y]}$ are smooth.

ii) If $a(Y) = 3$, then all the irreducible components of $\hat{S}_{3,[Y]}$ are smooth, except one.
*Proof.* Let $\mathcal{W}$ be an irreducible component of $\hat{\mathcal{S}}_{3,[Y]}$. Assume that either $a(Y) \le 2$ or $a(Y) = 3$ and $\mathcal{W}$ is not the singular component discussed in the proof of the previous lemma.

Then we know that there exists a polarization $\eta$ on $E^3$ satisfying $\ker(\eta) = E^3[p]$ and a point $\xi \in \mathcal{P}_{3,\eta}$ such that the formal completion $\hat{\mathcal{P}}_{3,\eta,\xi}$ dominates $\mathcal{W}$ via the map

$$ \varphi_{LO}: \mathcal{P}_{3,\eta} \to \mathcal{S}_3. $$

Let us denote $\hat{\mathcal{P}}_{3,\eta,\xi}$ by $\mathcal{W}'$ and the induced map $\mathcal{W}' \to \mathcal{W}$ by $\varphi$.

Then, by our assumptions on $\mathcal{W}$, we have $\xi \notin T$ and thus $\varphi$ is unramified (Corollary 3.4). Furthermore $\mathcal{W}$ is integral and geometrically unbranched being an irreducible component of the spectrum of a reduced complete local algebra over an algebraically closed field. Therefore, [8, Théorème 18.10.1] implies that $\varphi$ is étale. Since $\mathcal{W}'$ is smooth, it follows that $\mathcal{W}$ is smooth as well. $\square$

## 4. NON-TRANSVERSALITY CRITERIA

**4.1 General case.** Let $C$ be a hyperelliptic supersingular curve of genus 3. Let $\mathcal{W}$ be an irreducible component of the completion of $\mathcal{S}_3$ at the moduli point $[C]$. Assume that $\mathcal{W}$ is smooth at $[C]$. In this section we derive a necessary and sufficient criterion for the non-transversality of the intersection of $\mathcal{W}$ and $\mathcal{H}_3$ at $[C]$. We first describe the criterion geometrically. Later we specialize to the case $a=1$, where the criterion becomes equivalent to a condition on the Cartier-Manin matrix.

We begin with a lemma on PFTQs:

**Lemma 4.1.** Let $E^3 \to Y_0$ be a PFTQ (we omit $Y_1$ from the notation). Then

i) $E^3 \to Y_0$ factors through multiplication by $p$. Denote by

$$ \pi: Y_0 \to E^3 $$

the factored map.

ii) The images of the two maps

$$ \pi^*: \mathrm{H}^0(E^3, \Omega_{E^3}) \to \mathrm{H}^0(Y_0, \Omega_{Y_0}) $$

$$ \pi^*: \mathrm{H}^1(E^3, \mathcal{O}_{E^3}) \to \mathrm{H}^1(Y_0, \mathcal{O}_{Y_0}) $$

are both 1-dimensional, unless $\pi = F$. (In the latter case they are both 0.)

iii) Furthermore, the two images in ii) are orthogonal with respect to the natural pairing induced by the principal polarization on $Y_0$.

*Proof.* The first assertion follows trivially from the fact that the polarization $\eta$ in Definition 2.4 satisfies $\ker(\eta) = E^3[p]$. For the assertion ii) we use the explicit description of the Dieudonné modules. Recall that $M_2$ was the Dieudonné of $E^3$ with the additional structure as in Section 2.2. We will
denote by $M_0$ the Dieudonné module of $Y_0$ viewed as a submodule of $M_2$ via the map induced by $E^3 \to Y_0$.

Then the map $Y_0 \to E^3$ induces on Dieudonné modules the inclusion $pM_2 \subset M_0$. The map

$$ \pi^* : \mathrm{H}^0(E^3, \Omega_{E^3}) \to \mathrm{H}^0(Y_0, \Omega_{Y_0}) $$

is identified with

$$ \frac{V(pM_2)}{p(pM_2)} \to \frac{VM_0}{pM_0} $$

under the isomorphism from [12, Corollary 5.11]. From the explicit description of the Dieudonné modules in Section 2.2 (which uses the assumption that $\pi \neq F$) one readily sees that the image is one-dimensional with basis $g_3 = pFm_0$. Similarly the map

$$ \pi^* : \mathrm{H}^1(E^3, \mathcal{O}_{E^3}) \to \mathrm{H}^1(Y_0, \mathcal{O}_{Y_0}) $$

equals the map

$$ \frac{(pM_2)}{V(pM_2)} \to \frac{M_0}{VM_0}. $$

whose image has the basis $c_1 = pm_0$. This shows ii).

For iii) we need to make the pairing between $\mathrm{H}^0(Y_0, \Omega_{Y_0})$ and $\mathrm{H}^1(Y_0, \mathcal{O}_{Y_0})$ explicit. Indeed, the polarization on $\eta$ induces an alternating non-degenerate pairing

$$ M_2^t \times M_2^t \to W(k). $$

This gives an isomorphism $M_2^t[\frac{1}{p}] \simeq M_2[\frac{1}{p}]$. Therefore, we get a pairing

$$ M_2 \times M_2 \to W(k)[\frac{1}{p}]. $$

The restriction to $M_0$ has again values in $W(k)$ because $\eta$ descends to a principal polarization on $Y_0$. Thus we get a pairing

$$ \langle \cdot, \cdot \rangle : M_0 \times M_0 \to W(k). $$

From this and the explicit description of the pairing $\langle \cdot, \cdot \rangle$ in Section 2.2 one computes $\langle pm_0, pFm_0 \rangle = p$. Claim iii) follows. $\square$

Now let $C$ be as in the head of this chapter. The choice of an irreducible component $\mathcal{W}$ of the completion of $S_3$ at the moduli point $[C]$ is equivalent to the choice of a PFTQ ending in Jac$(C)$ by Lemma 3.5. This gives us a map

$$ \pi : \mathrm{Jac}(C) \to E^3 $$

as in Lemma 4.1. Recall that we assumed that $\mathcal{W}$ was non-singular at $[C]$, which by Corollary 3.6 is equivalent to $\pi \neq F$.

**Definition 4.2.** We define the following filtration on $V = \mathrm{H}^0(C, \Omega_C)$:

* $V_0 = V$.
* $V_1$ is defined to be the orthogonal complement of
  $\mathrm{im}(\pi^* : \mathrm{H}^1(E^3, \mathcal{O}_{E^3}) \to \mathrm{H}^1(\mathrm{Jac}(C), \mathcal{O}_{\mathrm{Jac}(C)}) \cong \mathrm{H}^1(C, \mathcal{O}_C))$
  with respect to the Serre duality pairing.

* $V_2$ is defined to be
  $V_2 = \mathrm{im}(\pi^* : \mathrm{H}^0(E^3, \Omega_{E^3}) \to \mathrm{H}^0(\mathrm{Jac}(C), \Omega_{\mathrm{Jac}(C)}) \cong \mathrm{H}^0(C, \Omega_C))$

Notice that Lemma 4.1 implies that $V_2 \subset V_1$ and $\dim(V_i) = 3 - i, i = 0, 1, 2$.

Consider now the canonical map $\varphi_{\mathrm{can}} : C \to \mathbb{P}(\mathrm{H}^0(C, \Omega_C)) \cong \mathbb{P}^2$. The image of $\varphi_{\mathrm{can}}$ is a conic.

On the other hand, the filtration $V_2 \subset V_1 \subset V_0 = \mathrm{H}^0(C, \Omega_C)$ defines a point $P$ and a line $l$ containing $P$ in $\mathbb{P}(\mathrm{H}^0(C, \Omega_C))$:

<figure><img src="image_3.png" /><figcaption>FIGURE 1. Generic picture</figcaption></figure>

We are now ready to state Theorem A from the introduction. Let $\mathcal{W}$ be as above. By Lemma 3.5 there is a PFTQ $\pi : E^3 \to \mathrm{Jac}(C)$ corresponding to $\mathcal{W}$. Furthermore the smoothness assumption on $\mathcal{W}$ is equivalent to saying that $\pi \neq F$. We can thus define the line $l$ and the point $P$ via the previous definition.

**Theorem 4.3.** The following are equivalent:

i) The component $\mathcal{W}$ meets $\mathfrak{H}_3$ non-transversally at $[C]$.

ii) The line $l$ touches $\mathrm{im}(\varphi_{\mathrm{can}})$ at the point $P$.

*Proof.* The tangent space of $\mathcal{W}$ at the point $[C]$ is two-dimensional with basis given by the two Kodaira-Spencer classes $\kappa(\delta_1), \kappa(\delta_2)$ as in Lemma 3.3. We can interpret $\kappa(\delta_1), \kappa(\delta_2)$ as elements in $S^2 \mathrm{H}^0(C, \Omega_C)^{\vee}$. By Lemma 2.1, the component $\mathcal{W}$ meets $\mathfrak{H}_3$ non-transversally at $[C]$ if and only if $\kappa(\delta_1), \kappa(\delta_2)$ kill the kernel of

$$S^2 \mathrm{H}^0(C, \Omega_C) \to \mathrm{H}^0(C, \Omega_C^{\otimes 2}).$$

Since $g = 3$ this kernel is 1-dimensional and generated by the quadratic form cutting out the conic $\mathrm{im}(\varphi_{\mathrm{can}})$.

Thus we need to show that the orthogonality of this quadratic form to the Kodaira-Spencer classes $\kappa(\delta_1), \kappa(\delta_2)$ has the desired geometric interpretation.

Indeed, recall that we computed in Lemma 3.3

$$\begin{align*}
\kappa(\delta_1)(g_1) &= -pm_0 = -c_1, \quad \kappa(\delta_1)(g_2) = \kappa(\delta_1)(g_3) = 0 \\
\kappa(\delta_2)(g_1) \in \mathrm{span}(c_1, c_2), \quad \kappa(\delta_2)(g_2) &= \tilde{x}_2^p pm_0, \quad \kappa(\delta_2)(g_3) = 0
\end{align*}$$
under the additional assumption that $x_2$ is not zero at $\xi$ which we can assume without loss of generality.

We will now compute the Serre duality pairing

$$ \langle \cdot, \cdot \rangle : \mathrm{H}^0(C, \Omega_C) \times \mathrm{H}^1(C, \mathcal{O}_C) \to k $$

on some of the given basis elements $g_1, g_2, g_3$ resp. $c_1, c_2, c_3$. Indeed this pairing equals the pairing induced by the principal polarization on $Y_0 \cong \mathrm{Jac}(C)$. This has the explicit description given in the proof of Lemma 4.1.

Thus we can compute

$$ \langle g_2, c_1 \rangle = \langle g_3, c_1 \rangle = \langle g_3, c_2 \rangle = 0. $$

This implies that $\mathrm{span}(c_1)^\perp = \mathrm{span}(g_2, g_3)$. But that means that $\kappa(\delta_1)$ is the unique map (up to scalar)

$$ \mathrm{H}^0(C, \Omega_C) \to \mathrm{H}^1(C, \mathcal{O}_C) $$

that kills the two-dimensional vector space $\mathrm{span}(c_1)^\perp$ and has image $\mathrm{span}(c_1)$.

It is not difficult to see that this means that the quadratic forms in $S^2 \mathrm{H}^0(C, \Omega_C)$ orthogonal to $\kappa(\delta_1)$ are exactly those vanishing at the point in

$$ \mathbb{P}(\mathrm{H}^0(C, \Omega_C)) $$

corresponding to $\mathrm{span}(c_1)$. But that point is equal to $P$ by its definition. Therefore, we conclude that $\kappa(\delta_1)$ represents a tangent direction to the hyperelliptic locus if and only if $P \in \mathrm{im}(\varphi_{\mathrm{can}})$.

Next, we analyze $\kappa(\delta_2)$. Indeed, we showed above

$$ \mathrm{span}(c_1, c_2)^\perp = \mathrm{span}(g_3). $$

Therefore

$$ \begin{gathered} \mathrm{im}(\kappa(\delta_2)) \subseteq \mathrm{span}(c_1, c_2), \\ \kappa(\delta_2)(\mathrm{span}(c_1)^\perp) \subseteq \mathrm{span}(c_1), \\ \kappa(\delta_2)(\mathrm{span}(c_1, c_2)^\perp) = 0. \end{gathered} $$

Let us denote by $\Upsilon \subset \mathrm{Hom}^{\mathrm{sym}}(\mathrm{H}^0(C, \Omega_C) \to \mathrm{H}^1(C, \mathcal{O}_C))$ the linear subspace of maps satisfying these conditions. One has $\mathrm{dim}(\Upsilon) = 2$ because $\Upsilon$ is equal to the vector space of maps whose matrix is of the form

$$ X = \begin{pmatrix} * & * & 0 \\ * & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, X = X^T $$

with respect to the basis $c_1, c_2, c_3$ on $\mathrm{H}^1(C, \mathcal{O}_C)$ and its dual basis on $\mathrm{H}^0(C, \Omega_C)$.

But we also have $\kappa(\delta_1) \in \Upsilon$. Since $\kappa(\delta_1), \kappa(\delta_2)$ are linearly independent, we conclude that $\Upsilon = \mathrm{span}(\kappa(\delta_1), \kappa(\delta_2))$.

Now let $Q \in S^2 \mathrm{H}^0(C, \Omega_C)$ be arbitrary. Then $Q$ is contained in the orthogonal complement of $\Upsilon$ if and only if $\mathcal{V}(Q)$ is a conic touching the line $\mathcal{V}(g_3)$ at the point $P$. (This can be seen from the matrix representation above.) But that line equals $l$ by definition.
Thus we may conclude: The component $\mathcal{W}$ meets $\mathfrak{H}_3$ non-transversally at $[C]$ if and only of the line $l$ touches $\mathrm{im}(\varphi_{\mathrm{can}})$ at the point $P$. The theorem follows. $\square$

4.2. **Simplifications for $a = 1$.** We will now specialize to the case $a = 1$. By [10, Remark on p. 38] this is equivalent to saying that there is a unique component of $\mathcal{S}_3$ passing through our moduli point.

Our first lemma will describe the filtration $V_2 \subset V_1 \subset V = \mathrm{H}^0(C, \Omega_C)$ in terms of the Cartier operator

$$ \mathcal{C} : \mathrm{H}^0(C, \Omega_C) \to \mathrm{H}^0(C, \Omega_C) $$

as follows:

**Lemma 4.4.** Assume further that $a = 1$. Let $E^3 \to \mathrm{Jac}(C)$ be the unique PFTQ ending at $\mathrm{Jac}(C)$. Let further $V_2 \subset V_1 \subset V = \mathrm{H}^0(C, \Omega_C)$ be as in Definition 4.2.

$$ \text{Then } V_2 = \ker(\mathcal{C}) \text{ and } V_1 = \ker(\mathcal{C}^2). $$

*Proof.* The assumption $a = 1$ implies that $\dim(\ker(\mathcal{C})) = 1$. Therefore, it suffices to show that $V_2 \subseteq \ker(\mathcal{C})$. This follows from

$$ V_2 = \mathrm{im}(\pi^* : \mathrm{H}^0(E^3, \Omega_{E^3}) \to \mathrm{H}^0(\mathrm{Jac}(C), \Omega_{\mathrm{Jac}(C)})) \cong \mathrm{H}^0(C, \Omega_C) $$

and the naturality of the Cartier operator.

To show that $V_1 = \ker(\mathcal{C}^2)$ it suffices again to prove $V_1 \subseteq \ker(\mathcal{C}^2)$. Recall that $V_1$ is defined to be the orthogonal complement of

$$ \mathrm{im}(\pi^* : \mathrm{H}^1(E^3, \mathcal{O}_{E^3}) \to \mathrm{H}^1(\mathrm{Jac}(C), \mathcal{O}_{\mathrm{Jac}(C)})) \cong \mathrm{H}^1(C, \mathcal{O}_C) $$

with respect to the Serre duality pairing. But since $F$ vanishes on this image, the identity

$$ \langle Fx, y \rangle = \langle x, \mathcal{C}(y) \rangle^p $$

shows that $V_1$ is closed under $\mathcal{C}$. Now $\mathcal{C}$ is nilpotent as a consequence of the supersingularity of $\mathcal{C}$. Therefore,

$$ V_1 \subseteq \mathrm{ker}(\mathcal{C}^{\dim(V_1)}) = \mathrm{ker}(\mathcal{C}^2). $$

This proves the lemma. $\square$

We can now state a reformulation of Theorem 4.3 in the special case $a = 1$ leading to Theorem B from the introduction:

**Theorem 4.5.** Let $C$ be a hyperelliptic supersingular curve of genus 3 with $a(\mathrm{Jac}(C)) = 1$. Denote by $\tau : C \to C$ the hyperelliptic involution. Then the following are equivalent:

i) The supersingular locus $\mathcal{S}_3$ meets $\mathfrak{H}_3$ non-transversally at $[C]$.

ii) There exists a point $P \in C$ such that the filtration $V_2 \subset V_1 \subset V_0 = \mathrm{H}^0(C, \Omega_C)$ of Lemma 4.4 agrees with the filtration

$$ \mathrm{H}^0(C, \Omega_C(-2P - 2\tau(P))) \subset \mathrm{H}^0(C, \Omega_C(-P - \tau(P))) \subset \mathrm{H}^0(C, \Omega_C) $$
iii) There exists a point $P \in C$ such that for any hyperelliptic equation

$$y^2 = f(x)$$

for $C$ that satisfies $x(P) = 0$ the corresponding Cartier-Manin matrix (formed by extracting suitable coefficients of $f^{\frac{p-1}{2}}$) has the shape

$$\begin{pmatrix} 0 & 0 & 0 \\ * & 0 & 0 \\ * & * & 0 \end{pmatrix}$$

*Proof.* In i) we use that $a = 1$ implies that there is unique component of $S_3$ passing through $[C]$ [10, p. 38] which furthermore is smooth at $[C]$ (Corollary 3.4). The implication “(i) $\Leftrightarrow$ (ii)” is a reformulation of Theorem 4.3.

Assume now that ii) is satisfied. We will show that iii) holds true with the same point $P \in C$. Indeed let

$$y^2 = f(x)$$

be an arbitrary hyperelliptic equation for $C$ satisfying $x(P) = 0$. Then the filtration

$$H^0(C, \Omega_C(-2P - 2\tau(P))) \subset H^0(C, \Omega_C(-P - \tau(P))) \subset H^0(C, \Omega_C)$$

is explicitly given by

$$\operatorname{span}\left(x^2 \frac{dx}{y}\right) \subset \operatorname{span}\left(x \frac{dx}{y}, x^2 \frac{dx}{y}\right) \subset \operatorname{span}\left(\frac{dx}{y}, x \frac{dx}{y}, x^2 \frac{dx}{y}\right).$$

By assumption and Lemma 4.4 one then has

$$\operatorname{span}\left(x^2 \frac{dx}{y}\right) = \ker(\mathcal{C}), \operatorname{span}\left(x \frac{dx}{y}, x^2 \frac{dx}{y}\right) = \ker(\mathcal{C}^2).$$

Together with the nilpotence of $\mathcal{C}$ this implies that the matrix of $\mathcal{C}$ with respect to the basis $\frac{dx}{y}, x \frac{dx}{y}, x^2 \frac{dx}{y}$, i.e., the Cartier-Manin matrix has the shape

$$\begin{pmatrix} 0 & 0 & 0 \\ * & 0 & 0 \\ * & * & 0 \end{pmatrix}$$

proving iii).

The implication iii) $\Rightarrow$ ii) is shown by reversing the previous argument. This concludes the proof of the theorem. $\square$

The utility of part iii) of the Theorem is the fact that it is easy to check whether the criterion is satisfied for a given hyperelliptic equation $y^2 = f(x)$.

**Definition 4.6.** • Let $P \in C(k)$ be arbitrary. We define the filtration associated to $P$ to be the filtration in Theorem 4.5 ii):

$$H^0(C, \Omega_C(-2P - 2\tau(P))) \subset H^0(C, \Omega_C(-P - \tau(P))) \subset H^0(C, \Omega_C)$$
• When $a(\mathrm{Jac}(C)) = 1$ and one (and hence both) of conditions ii) or iii) of Theorem 4.5 are satisfied, we call $P$ a *touchpoint for the Cartier operator*.

It is easy to see that a touchpoint for the Cartier operator (if it exists) is unique up to replacing $P$ by $\tau(P)$.

## 5. EXAMPLES

**5.1. CM-examples** ($a = 1$). In this section we will construct an example of a curve $C$ over $\mathbb{Q}$ with the property that the reduction at any prime $p$ with $p \equiv \pm 2 \pmod 7, p \equiv 3 \pmod 4$ satisfies the criterion of Theorem 4.5.

To setup our notation consider a hyperelliptic genus 3 curve $C$ over $\mathbb{C}$. Suppose that $\mathrm{Jac}(C)$ has CM by a field $K$. Let us look at the differential forms $V = \mathrm{H}^0(C, \Omega_C)$. By CM-theory there is a decomposition

$$V = \bigoplus_{\iota \in \Phi^c} V_\iota$$

into one dimensional $\mathbb{C}$-vector spaces. Here $\Phi \subset \mathrm{Hom}(K, \mathbb{C})$ is the CM-type and $\Phi^c$ its complement. An element $a \in K$ acts on $V_\iota$ via multiplication by $\iota(a)$.

Furthermore we denote by $L$ the Galois closure of $K$ over $\mathbb{Q}$ and by $G = \mathrm{Gal}(L/\mathbb{Q})$ the Galois group.

**Definition 5.1.** Let $C$ be as above and let $P \in C(\mathbb{C})$ be a point. We say that $P$ is a *touchpoint for the CM-action* if the following condition holds true:

i) Denote by $V = V_0 \supset V_1 \supset V_2$ the filtration associated to $P$. Then we demand that there is an ordering $\iota_0, \iota_1, \iota_2$ of $\Phi^c$ such that

$$V_{\iota_i} \subseteq V_i.$$

ii) The choice of the ordering in i) gives an ordering $\iota_0, \iota_1, \iota_2, \overline{\iota_0}, \overline{\iota_1}, \overline{\iota_2}$ of all the embeddings of $K$ into $\mathbb{C}$. This gives a map

$$\mathrm{per} : G \hookrightarrow S_6$$

induced from the action of $G$ on the complex embeddings of $K$. The second condition is that

$$(123456) \in \mathrm{im}(\mathrm{per}).$$

**Remark 5.2.** The motivation for the terminology is that we can draw the following picture: The decomposition $V = \bigoplus_{\iota \in \Phi^c} V_\iota$ gives 3 points $P_0, P_1, P_2$ in $\mathbb{P}^2$ and condition i) is equivalent to $P_0 = \varphi_{\mathrm{can}}(P)$ and the line $P_0P_1$ should be tangent to $\mathrm{im}(\varphi_{\mathrm{can}})$:
<figure><img src="image_4.png" /><figcaption>FIGURE 2. CM touchpoint</figcaption></figure>

Notice that the existence of a touchpoint $P$ is very restrictive. Most of the CM-curves do not have any. Contrary to touchpoints for the Cartier operator there could exist more than one touchpoint (modulo the hyperelliptic involution) because in the picture above it might happen that $P_2$ is on $\text{im}(\varphi_{\text{can}})$ with tangent line $P_1P_2$

A touchpoint can also be a Weierstrass point. We will now discuss an example where that happens.

**Example 5.3.** Consider the hyperelliptic curve $C_{14}$ over $\mathbb{C}$ given by the affine equation

$$y^2 = x(x^{14} - 1).$$

It is known [5] that the automorphism group of $C_{14}$ is isomorphic to the group $U_{14}$, where

$$U_n = \langle a, b \mid a^2 = b^{2n} = abab^{n+1} = 1 \rangle.$$

The action is given by

$$a : x \mapsto \frac{-1}{x}, \quad b : x \mapsto -\zeta_7 x \\ y \mapsto \frac{y}{x^8}, \quad y \mapsto i\zeta_7^4 y$$

where $\zeta_7 = e^{\frac{2\pi i}{7}} \in \mathbb{C}$.

We now define $C$ to be the quotient $C_{14}/a$. Denote by

$$\pi : C_{14} \to C$$

the quotient map. We define a point $P$ on $C$ given by the image of $(0,0) \in C_{14}$ under $\pi$. The situation is summarized by the following diagram.

<figure><img src="image_5.png" /></figure>

To facilitate the reading of the proof of the following proposition, we remark that we do not choose coordinates on $C$. Thus the coordinates $x, y$ always refer to $C_{14}$.

**Proposition 5.4.** In the notation from above.

i) The curve $C$ is hyperelliptic of genus 3.

ii) Jac($C$) has CM by $K = \mathbb{Q}(\zeta_7 + \zeta_7^{-1}, i)$.

iii) The point $P$ is a touchpoint of the CM action.
(as it is given by the differential forms fixed by $a$). Moreover, one can compute

$$i^*(\omega_0) = -i\omega_0, \alpha^*(\omega_0) = (\zeta_7^3 + \zeta_7^{-3})\omega_0$$

$$i^*(\omega_1) = i\omega_1, \alpha^*(\omega_1) = (\zeta_7^2 + \zeta_7^{-2})\omega_1$$

$$i^*(\omega_2) = -i\omega_2, \alpha^*(\omega_2) = (\zeta_7 + \zeta_7^{-1})\omega_2.$$

This gives us the desired description of the action of $K$ on the differential forms on $C$. It is now an elementary verification to show that the point $P = \pi((0,0))$ is a touchpoint for the CM-action. □

**Remark 5.5.** The construction for the curve $C$ can be seen as a special case of Tautz-Top-Verberkmoes [16] in the case where (in their notation) $p = 7, t = 0$.

This explains the rationale behind our choice for $C$: We use the Tautz-Top-Verberkmoes construction to get RM by $\mathbb{Q}(\zeta_7 + \zeta_7^{-1})$. By implementing the additional order 4 automorphism $b$ into the covering curve $C_{14}$ we can achieve CM by $\mathbb{Q}(\zeta_7 + \zeta_7^{-1}, i)$.

The idea to use CM fields containing $i$ to force the curve to be hyperelliptic goes back to Weng [17]. In fact, our curve $C$ happens to be isomorphic to the first curve that appears in her list (p. 18 loc. cit.).

For the next theorem we use the following notation: Let $C$ be a hyperelliptic genus 3 curve over $\mathbb{C}$ such that Jac$(C)$ has CM by a field $K$. We denote by $L$ the Galois closure of $K/\mathbb{Q}$ and by $G = \text{Gal}(L/\mathbb{Q})$ the Galois group. Assume, furthermore, that there exists a point $P \in C$ which is a touchpoint for the CM-action. Let us denote by

$$\text{per} : G \hookrightarrow S_6$$

the group homomorphism introduced in Definition 5.1.ii)

By CM-theory $C$ is defined over a number field, say $M$. We assume that $M$ is large enough, such that $M$ contains $L$, all endomorphisms of Jac$(C)$ are defined over $M$, and such that $P$ is defined over $M$ (it is easy to see that $P$ must be defined over $\overline{\mathbb{Q}}$).

**Theorem 5.6.** Let $\mathfrak{P}$ be a prime ideal of $M$ such that $C$ has good reduction at $\mathfrak{P}$. Assume that $p$ does not divide $\text{disc}(R)$ where $R = K \cap \text{End}(\text{Jac}(C))$ is the CM-order and $p$ is the residue characteristic of $\mathfrak{P}$.

Denote by $\overline{C}$ the reduction of $C$ and by $\overline{P}$ the reduction of the point $P$ and by $\mathfrak{p}$ the intersection $\mathfrak{P} \cap L$.

$$\text{Assume that } \text{per}(\text{Frob}_{\mathfrak{p}}) = (123456)^{-1}.$$

Then the curve $\overline{C}$ is supersingular and the pair $(\overline{C}, \overline{P})$ satisfies the criterion of Theorem 4.5, i.e., $\overline{P}$ is a touchpoint for the Cartier operator on $\overline{C}$.

*Proof.* In order to prove the theorem we need to describe the action of the Cartier operator on $H^0(\overline{C}, \Omega_{\overline{C}})$ in terms of CM-theory. By [12, Corollary 5.11] it suffices for that purpose to understand the Verschiebung operator
on the Dieudonné module $M = D(\mathrm{Jac}(\bar{C}))$. Now the description of the Dieudonné module of the reduction of a CM abelian variety is known (see e.g. [7] cited after [11] or look at [1] for a more general result in $p$-adic Hodge theory). Nevertheless, we will reproduce a proof for the special case at our hand because we need the inner mechanics of the identification for showing the other claims of the theorem.

Indeed, choose $k$, an algebraically closed field containing the residue field of $\mathfrak{P}$. By slight abuse of notation we will denote the base change of $\bar{C}$ to $k$ by the same letter.

We will now show that ii) implies i). Indeed, assuming ii) we claim that the Dieudonné module $M = D(\mathrm{Jac}(\bar{C}))$ has a $W(k)$ basis $m_1, \dots, m_6$ such that

$$Vm_1 = m_2, Vm_2 = m_3, Vm_3 = pm_4, Vm_4 = pm_5, Vm_5 = pm_6, Vm_6 = m_1.$$

Indeed $M$ naturally carries the structure of an $R \otimes_Z W(k)$-module. By assumption $R \otimes_Z \mathbb{Z}_p$ is an étale $\mathbb{Z}_p$-algebra and therefore

$$R \otimes_Z W(k) \cong \bigoplus_{i=1}^{6} W(k)$$

where the copies of $W(k)$ are indexed by $\mathrm{Hom}(R, W(k))$. Now under our assumptions we claim that there is a bijection $\mathrm{Hom}(K, \mathbb{C}) \xrightarrow{\sim} \mathrm{Hom}(R, W(k))$. (Here we see why it is awkward to define CM-types as embeddings into $\mathbb{C}$, but anyway.) Indeed, we assumed that $C$ is defined over the number field $M \subset \mathbb{C}$. Furthermore $M$ was assumed to be large enough to contain $L$, a Galois closure of $K$. This gives an embedding $L \hookrightarrow \mathbb{C}$. Now every embedding $\iota: K \hookrightarrow \mathbb{C}$ must factor through $L$. We define the map $\mathrm{Hom}(K, \mathbb{C}) \to \mathrm{Hom}(R, W(k))$ to be the composition

$$\mathrm{Hom}(K, \mathbb{C}) \to \mathrm{Hom}(K, L) \to \mathrm{Hom}(R, \mathcal{O}_L) \to \mathrm{Hom}(R, \mathcal{O}_L/\mathfrak{p})$$

$$\iota \mapsto \iota|_R$$

followed by the composition $\mathrm{Hom}(R, \mathcal{O}_L/\mathfrak{p}) \to \mathrm{Hom}(R, k) \to \mathrm{Hom}(R, W(k))$ where the first map is induced from the natural inclusion

$$\mathcal{O}_L/\mathfrak{p} \hookrightarrow \mathcal{O}_M/\mathfrak{P} \hookrightarrow k$$

and the second map exists because of the identification $\mathrm{Hom}(R, W(k)) \xrightarrow{\sim} \mathrm{Hom}(R \otimes_Z \mathbb{Z}_p, W(k))$ (a consequence of $p$-adic completeness, the assumption that $R \otimes_Z \mathbb{Z}_p$ is an étale $\mathbb{Z}_p$-algebra, and general properties of the Witt ring). This gives us the map $\mathrm{Hom}(K, \mathbb{C}) \to \mathrm{Hom}(R, W(k))$. It is not too hard to verify that this map is a bijection, e.g. by writing down a chain of maps inverting all the arrows.

We are now allowed to index the direct sum decomposition

$$R \otimes_Z W(k) \cong \bigoplus_{i=1}^{6} W(k)$$
with the ordering corresponding to the ordering $\iota_0, \iota_1, \iota_2, \bar{\iota}_0, \bar{\iota}_1, \bar{\iota}_2$ of $\operatorname{Hom}(K, \mathbb{C})$.

This gives a direct sum decomposition

$$M = \bigoplus_{i=1}^{6} M_i$$

as $W(k)$-modules such that $R$ acts on $M_i$ via multiplication by a scalar. More precisely, if $\psi_1, \dots, \psi_6$ are the maps $R \to W(k)$ corresponding to $\iota_0, \iota_1, \iota_2, \bar{\iota}_0, \bar{\iota}_1, \bar{\iota}_2$, then $a \in R$ acts on $M_i$ via multiplication by $\psi_i(a)$.

Choose at first an arbitrary generator $m_1 \in M_1$. (We will later adjust $m_1$.) Then for any $a \in R$ we have

$$a \cdot Vm_1 = V(a \cdot m_1) = V(\psi(a)m_1) = (\sigma^{-1} \circ \psi_1)(a)Vm_1$$

where $\sigma : W(k) \to W(k)$ is the lift of Frobenius. Now the assumption ii) implies that $\sigma^{-1} \circ \psi_1 = \psi_2$ and thus $m_2 \in M_2$. We claim that $m_2$ generates $M_2$. Indeed, we look at the isomorphism

$$M/VM \cong H^1(\bar{C}, \mathcal{O}_{\bar{C}}).$$

The image must be generated by the image of $M_4, M_5, M_6$, because the maps $\psi_4, \psi_5, \psi_6$ correspond to the embeddings contained in the CM-type $\Phi$, i.e. $\bar{\iota}_0, \bar{\iota}_1, \bar{\iota}_2$. Therefore $M_2$ is contained in $VM$ and this proves the claim.

Similarly we can define $m_3 = Vm_2$ and $m_3$ must generate $M_3$. Now consider $Vm_3$. We have $Vm_3 \in M_4$. But $Vm_3$ does not generate $M_4$ because $M_4/pM_4$ injects into $H^1(\bar{C}, \mathcal{O}_{\bar{C}})$ by the argument above. Therefore, $Vm_3 \in pM_4$ and we define $m_4 = \frac{1}{p}Vm_3$. Since this implies that $Fm_4 = m_3$, we see that $m_4$ must be a generator of $M_4$.

Along the same lines of reasoning we obtain generators $m_5$, (resp. $m_6$) of $M_5$ (resp. $M_6$) satisfying the relations $Vm_4 = pm_5, Vm_5 = pm_6$.

Then, $Vm_6$ will also be a generator of $M_1$, say $Vm_6 = um_1$ for some unit $u \in W(k)^\times$. Now, since $k$ is algebraically closed, one can show that there exists a unit $\tilde{u} \in W(k)^\times$ such that $\sigma^{-6}(\tilde{u})u = \tilde{u}$. After replacing $m_1$ by $\tilde{u}m_1$ and redefining $m_2, \dots, m_6$ by the relations above, we see that $Vm_6 = m_1$ holds true. This gives the desired description of the Dieudonné module $M$.

Then it follows from that description of $M$ that $\bar{C}$ is supersingular<fnref n="2" /> and $a(\mathrm{Jac}(\bar{C})) = 1$.

We claim that $\bar{P}$ is a touchpoint for the Cartier operator. Indeed, by [12, Corollary 5.11] there is an isomorphism

$$\frac{VM}{pM} \cong H^0(\bar{C}, \Omega_{\bar{C}}^1)$$

that identifies $V$ on the left with the Cartier operator on the right. Therefore, the filtration

$$\ker(\mathcal{C}) \subset \ker(\mathcal{C}^2) \subset H^0(\bar{C}, \Omega_{\bar{C}}^1)$$

<footnote n="2">²Alternatively one can use the Shimura-Taniyama formula to compute the Newton polygon, but in fact Deligne [7] proves that formula by giving the same description of the Dieudonné module</footnote>
is given by

$$ \mathrm{span}(m_3) \subset \mathrm{span}(m_2, m_3) \subset \mathrm{span}(m_1, m_2, m_3). $$

On the other hand, let us look at the filtration $V_2 \subset V_1 \subset V_0$ associated to $\bar{P}$, i.e.

$$ H^0(\bar{C}, \Omega_{\bar{C}}(-2\bar{P} - 2\tau(\bar{P}))) \subset H^0(\bar{C}, \Omega_{\bar{C}}(-\bar{P} - \tau(\bar{P}))) \subset H^0(\bar{C}, \Omega_{\bar{C}}). $$

Definition 5.1 implies that $V_2$ contains the eigenspace where $R$ acts via the map correponding to the embedding $\iota_2$ (a priori we have such an inclusion on the generic fiber, but it passes to the specialization). Therefore, we see that the class of $m_3$ must be contained in $V_2$ because $m_3$ generates that eigenspace. Similarly we get $m_2 \in V_1$. This shows that $\bar{P}$ is a touchpoint for the Cartier operator on $\bar{C}$. $\square$

**Corollary 5.7.** For every prime number $p$ such that

$$ p \equiv \pm 2 \mod 7, p \equiv 3 \mod 4 $$

the hyperelliptic locus and the supersingular locus intersect non-transversally in $\mathfrak{A}_3 \times \mathbb{F}_p$.

*Proof.* The curve $C$ from Example 5.3 is defined over $\mathbb{Q}$ and has good reduction at every prime different from 2, 7. Furthermore Jac$(C)$ has CM by $K = \mathbb{Q}(\zeta_7 + \zeta_7^{-1}, i)$. Let $R = \mathrm{End}(\mathrm{Jac}(C)) \cap K$ be the CM-order. It follows from the proof of Proposition 5.4 that the index of $R$ in $\mathcal{O}_K$ must be a power of 2. In fact, with a bit more work one can prove that $R = \mathcal{O}_K$, but we shall not need this.

Now Proposition 5.4 implies that $C$ satisfies the assumptions of Theorem 5.6. Let $p$ be a prime that satisfies the congruences $p \equiv \pm 2 \mod 7, p \equiv 3 \mod 4$. Then $p$ is unramified in $L$, which equals $K$ in our case. Furthermore $\mathrm{per}(\mathrm{Frob}_p) = (123456)^{-1}$ for any prime $p$ of $L$ lying over $p$. Thus with Theorem 5.6 and Theorem 4.5 we conclude that $\mathfrak{H}_3$ and $\mathcal{S}_3$ intersect non-transversely in the moduli point $[\bar{C}]$. $\square$

It is worthwile to remark that the curve $C$ from Example 5.3 also has good supersingular reduction at the primes satisfying

$$ p \equiv \pm 3 \mod 7, p \equiv 3 \mod 4. $$

However, in this case the criterion of Theorem 4.5 is not satisfied because one has $\mathrm{per}(\mathrm{Frob}_p) = (123456)$ instead of $(123456)^{-1}$.

5.2. **One example with $a = 3$.** The curve $C: y^2 = x^8 - 1$ over $\mathbb{F}_7$ is supersingular and satisfies $a(\mathrm{Jac}(C)) = 3$ (this can be checked by computing the Hasse-Witt matrix, for example). As a consequence of Lemma 3.5 the formal completion of $\mathcal{S}_3$ at $[C]$ has $p^5 + p^2 + 1 = 16857$ irreducible components. One can show that $p^2 = 49$ of these have the property that they intersect $\mathfrak{H}_3$ non-transversely.

To prove this, it does not suffice to compute the Cartier-Manin matrix as one would do in the $a = 1$ case. One rather must know the matrix of
Frobenius on crystalline cohomology mod $p^2$, i.e. $H_{\text{crys}}^1(C/W_2)$. But, if $C$ is defined over $\mathbb{F}_{p^2}$, as in our example, it suffices to know the matrix of Frobenius acting on $H_{\text{dr}}^1$.

In any case, these matrices can be computed with Kedlaya's algorithm.

The details of the verification of the claims in this section are not hard, but tedious, and thus will be omitted.

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