On the de-Rham cohomology of hyperelliptic curves

Abstract

For any hyperelliptic curve X, we give an explicit basis of the first de-Rham cohomology of X in terms of Čech cohomology. We use this to produce a family of curves in characteristic $p>2$ for which the Hodge-de-Rham short exact sequence does not split equivariantly; this generalises a result of Hortsch. Further, we use our basis to show that the hyperelliptic involution acts on the first de-Rham cohomology by multiplication by $-1$, i.e., acts as the identity when $p=2$.

1 Introduction

Recall that the de-Rham cohomology $H^*_\mathrm {dR}(X/k)$ of a smooth projective curve X over an algebraically closed field k is defined as the hypercohomology of the de-Rham complex

$$\begin{aligned} {\mathcal {O}}_X \xrightarrow {d} \Omega _X \end{aligned}$$

where d denotes the usual differential map $f \mapsto df$. In particular, we have a long exact sequence relating $H^*_\mathrm {dR}(X/k)$ to ordinary cohomology of the structure sheaf ${\mathcal {O}}_X$ and of the sheaf $\Omega _X$ of differentials on X. The very general and famous fact that the Hodge-de-Rham spectral sequence degenerates at $E_1$ (e.g., see [13]) means for our curve X that the following main part of that long sequence is a short exact sequence, see Proposition 3.1:

$$\begin{aligned} 0 \rightarrow H^0(X, \Omega _X) \rightarrow H^1_\mathrm {dR}(X/k) \rightarrow H^1(X, {\mathcal {O}}_X) \rightarrow 0. \end{aligned}$$

We call this sequence the Hodge-de-Rham short exact sequence. In particular, the vector space $H^1_\mathrm {dR}(X/k)$ is the direct sum of the vector spaces $H^0(X, \Omega _X)$ and $H^1(X, {\mathcal {O}}_X)$ over k.

We now assume furthermore that a finite group G acts on our curve X. If $p:= \mathrm {char}(k)$ does not divide the order of G, Maschke’s Theorem implies that the Hodge-de-Rham short exact sequence also splits as a sequence of modules over the group ring k[G].

However, the latter fact fails to be true in general when $p>0$ does divide $\mathrm {ord}(G)$. A counterexample has been constructed in the recent paper [4] by Hortsch. The main goal of this paper is to generalise that counterexample. More precisely, we will prove the following theorem, see Theorem 4.3 and Example 4.4.

Theorem

Let $p \ge 3$ and let $q(z) \in k[z]$ be a monic polynomial of odd degree without repeated roots. Let X denote the hyperelliptic curve over k defined by the equation $y^2=q(x^p-x)$ and let G denote the subgroup of $\mathrm {Aut}(X)$ generated by the automorphism $\tau$ given by $(x,y) \mapsto (x+1, y)$. Then the Hodge-de-Rham short exact sequence does not split as a sequence of k[G]-modules.

We remark that the hyperelliptic curves considered in this theorem are exactly those hyperelliptic curves $y^2=f(x)$ which allow an automorphism that maps x to $x+1$ and for which f(x) is of odd degree, see Example 4.4 and Proposition 4.6.

When $q(z)= z$, the theorem above becomes the main theorem of [4]. Beyond [4], our theorem shows (see Remark 4.5) that, for every algebraically closed field k of characteristic $p\ge 3$, there exist infinitely many $g \ge 2$ and hyperelliptic curves X over k of genus g for which the Hodge-de-Rham short exact sequence does not split equivariantly. It also shows that, for every $g \ge 2$, there exists a prime $p\ge 3$ and hyperelliptic curves in characteristic p of genus g for which the Hodge-de-Rham short exact sequence does not split equivariantly.

In Example 4.8 and Remark 4.9 we show, using the modular curve $X_0(22)$ for $p=3$, that, without assuming the degree of q(x) to be odd, this theorem may be false.

To prove our main theorem, we follow the same broad strategy as in [4]: we give an explicit basis of $H^1_\mathrm {dR}(X/k)$ in terms of Čech cohomology (in fact for an arbitrary hyperelliptic curve X), see Theorem 3.2, and study the action of $\tau$ on that basis. The actual computations towards the end however do not generalise those in [4], see Remarks 3.5 and 4.7.

We provide a basis of $H^1_\mathrm {dR}(X/k)$ for any hyperelliptic curve X also when $p=2$ and use this to show that the hyperelliptic involution acts trivially on $H^1_\mathrm {dR}(X/k)$ when $p=2$. In fact, the hyperelliptic involution acts on $H^1_\mathrm {dR}(X/k)$ by multiplication by $-1$ for all p, see Theorem 4.1.

If $p=2$, Elkin and Pries construct a subtler basis of $H^1_\mathrm {dR}(X/k)$ in [2] which is suitable to study the action of Frobenius and Verschiebung and, finally, to determine the Ekedahl–Oort type.

2 Preliminaries

In this section, we introduce assumptions and notations used throughout this paper and collect and prove some auxiliary results.

We assume that k is an algebraically closed field of characteristic $p \ge 0$ and that X is a hyperelliptic curve over k of genus $g \ge 2$. We recall that a curve (always assumed to be smooth, projective and irreducible in this paper) is hyperelliptic if there exists a finite, separable morphism of degree two from the curve to ${\mathbb {P}}_k^1$. We fix such a map

$$\begin{aligned} \pi :X \rightarrow {\mathbb {P}}_k^1, \end{aligned}$$

which is unique up to automorphisms of X and of ${\mathbb {P}}_k^1$ (see [7, Remark 7.4.30]). Let $K(X)/K({\mathbb P}^1_k)= k(x)$ denote the extension of function fields corresponding to $\pi$. According to [7, Proposition 7.4.24 and Remark 7.4.25], we may and will furthermore assume the following concrete description of K(X).

If $p \not =2$, then $K(X)=k(x,y)$ where y satisfies

$$\begin{aligned} y^2 = f(x) \end{aligned}$$

(1)

for some monic polynomial $f(x) \in k[x]$ which has no repeated roots; moreover, f(x) is of degree $2g +1$ if $\infty \in {\mathbb P}^1_k$ is a branch point of $\pi$ and of degree $2g+2$ otherwise. The branch points of $\pi$ are then the roots of f(x), together with $\infty \in {\mathbb P}^1_k$ if $\mathrm {deg}(f(x)) = 2g+1$.

If $p=2$, then $K(X) = k(x,y)$ where y satisfies

$$\begin{aligned} y^2 - h(x)y = f(x) \end{aligned}$$

(2)

for some polynomials $h(x), f(x) \in k[x]$ such that $h'(x)^2 f(x) + f'(x)^2$ and h(x) have no common roots in k; moreover, we have $d:=\mathrm {deg}(h(x)) \le g+1$, with equality if and only if $\infty$ is not a branch point of $\pi$. The branch points of $\pi$ are the roots of h(x), together with $\infty \in {\mathbb P}^1_k$ if $d<g+1$.

The following estimate for the order of y above $\infty$ is true for both $p \not =2$ and $p=2$.

Lemma 2.1

Let $P \in \pi ^{-1}(\infty )$. Then we have:

$$\begin{aligned}\mathrm {ord}_P(y) \ge \left\{ \begin{aligned}&-(g+1)&\quad \, if \, \pi \,is\, unramified\, at \,P\\&-2(g+1)&\quad \,if \,\pi \, is\, ramified\, at \,P. \end{aligned}\right. \end{aligned}$$

Proof

This is [5, Inequality (5.2)]. $\square$

Lemma 2.2

If $p \not =2$, let $\omega := \frac{dx}{y}$ and, if $p=2$, let $\omega := \frac{dx}{h(x)}$. Then the differentials $\omega , x\omega , \ldots , x^{g-1}\omega$ form a basis of the k-vector space $H^0(X, \Omega _X)$ of global holomorphic differentials on X.

Proof

This is [7, Proposition 7.4.26]. $\square$

Remark 2.3

A different basis of $H^0(X,\Omega _X)$ is given in [8, Lemma 5]. The action of the Cartier operator on $H^0(X,\Omega _X)$ is studied in [10] and [14].

Lemma 2.4

Let $p=2$ and let $P \in \pi ^{-1}(\infty )$. Then we have:

$$\begin{aligned} \mathrm {ord}_{P}(dx) = \left\{ \begin{aligned}&-2&\quad if \,\pi \, is\, unramified \,at \,P\\&2(g-1-d)&\quad if \,\pi \, is\, ramified\, at \,P. \end{aligned} \right. \end{aligned}$$

(3)

Proof

By the Riemann–Hurwitz formula [9, Theorem 3.4.6] we have

$$\begin{aligned} \mathrm {ord}_P(dx) = e_P \cdot \mathrm {ord}_\infty (dx) + \delta _P \end{aligned}$$

where $e_P$ denotes the ramification index of $\pi$ at P and $\delta _P$ denotes the order of the ramification divisor of $\pi$ at P. It is easy to see that $\mathrm {ord}_\infty (dx) = -2$. Therefore $\mathrm {ord}_P(dx) = -2$ if $\pi$ is unramified at P. On the other hand, if $\pi$ is ramified at P, we have $\delta _P = 2(g+1-d)$ by [5, Equation (5.3)] and hence

$$\begin{aligned} \mathrm {ord}_P(dx) = 2 \cdot (-2) + 2(g+1-d) = 2(g-1-d), \end{aligned}$$

as claimed. $\square$

We define $U_a = X \backslash \pi ^{-1}(a)$ for any $a \in {\mathbb {P}}_k^1$ and let $\mathcal{U}$ be the affine cover of X formed by $U_0$ and $U_\infty$. Given any sheaf $\mathcal F$ on X we have the Čech differential $\check{d}:\mathcal{F}(U_0) \times \mathcal{F} (U_\infty ) \rightarrow \mathcal{F}(U_0 \cap U_\infty )$, defined by $(f_0,f_\infty ) \mapsto f_0|_{U_0 \cap U_\infty } - f_\infty |_{U_0 \cap U_\infty }$. In general we will suppress the notation denoting the restriction map. The first cohomology group $\frac{ {\mathcal {O}}_X(U_0 \cap U_\infty )}{\mathrm{Im}(\check{d})}$ of the cochain complex

$$\begin{aligned} 0 \rightarrow {\mathcal {O}}_X(U_0)\times {\mathcal {O}}_X(U_\infty ) \xrightarrow {\check{d}} {\mathcal {O}}_X(U_0 \cap U_\infty ) \rightarrow 0. \end{aligned}$$

is the first Čech cohomology group $\check{H}^1({{\mathcal {U}}}, {\mathcal {O}}_X)$. By Leray’s theorem [7, Theorem 5.2.12] and Serre’s affineness criterion [7, Theorem 5.2.23] we therefore have

$$\begin{aligned} H^1(X, {\mathcal {O}}_X)\cong \frac{ {\mathcal {O}}_X(U_0 \cap U_\infty )}{\{f_0 - f_\infty \mid f_0 \in {\mathcal {O}}_X(U_0), \, f_\infty \in {\mathcal {O}}_X(U_\infty ) \}}. \end{aligned}$$

(4)

When describing elements of $H^1(X, {\mathcal {O}}_X)$ using this isomorphisms we will denote the residue class of $f \in {\mathcal {O}}_X(U_0 \cap U_\infty )$ by [f].

Proposition 2.5

The elements $\frac{y}{x}, \ldots , \frac{y}{x^g} \in K(X)$ are regular on $U_0 \cap U_\infty$, and their residue classes $\left[ \frac{y}{x} \right] , \ldots , \left[ \frac{y}{x^g} \right]$ form a basis of $H^1(X, {\mathcal {O}}_X)$.

Proof

By [7, Proposition 7.4.24(b)], we may identify ${\mathcal {O}}_X(U_\infty )$ with the k-algebra k[x, y] defined by the relation given in (1) or (2). Then ${\mathcal {O}}_X(U_0 \cap U_\infty )$ is $k[x^{\pm 1},y]$. As the relations in (1) and (2) are quadratic in y, the elements $\ldots , \frac{1}{x^2}, \frac{1}{x}, 1, x, x^2, \ldots$ and $\ldots , \frac{y}{x^2}, \frac{y}{x}, y, xy, x^2y, \ldots$ form a k-basis of $k[x^{\pm 1},y]$. The elements $1, x, x^2, \ldots$ and $y, xy, x^2y, \ldots$ obviously form a basis of the image of ${\mathcal {O}}_X(U_\infty )$ in ${\mathcal {O}}_X(U_0\cap U_\infty )$. By [7, Proposition 7.4.24(b)], the image of ${\mathcal {O}}_X(U_0)$ in ${\mathcal {O}}_X(U_0\cap U_\infty )$ consists of elements of the form $g\left( \frac{1}{x}, \frac{y}{x^{g+1}}\right)$ where $g \in k[s,t]$. Hence, the elements $\ldots , \frac{1}{x^2}, \frac{1}{x}, 1$ and $\ldots \frac{y}{x^{g+3}}, \frac{y}{x^{g+2}}, \frac{y}{x^{g+1}}$ form a basis of that image. We conclude that the residue classes $\left[ \frac{y}{x} \right] , \ldots , \left[ \frac{y}{x^g} \right]$ form a basis of $H^1(X, {\mathcal {O}}_X)$, as was to be shown. $\square$

Remark 2.6

Let $\omega _j := \frac{x^{j-1}}{y} {dx}$ when $p \not =2$ and let $\omega _j =\frac{ x^{j-1}}{h(x)}{dx}$ when $p=2$. Then, by Lemma 2.2, the elements $\omega _j$, $j=1, \ldots , g$, form a k-basis of $H^0(X,\Omega _X)$. Let $\langle , \rangle :H^0(X, \Omega _X) \times H^1(X, {\mathcal {O}}_X) \rightarrow k$ denote the Serre duality pairing. Then $\langle \omega _j,\left[ \frac{y}{x^i} \right] \rangle$ vanishes if $j \not = i$ and is non-zero if $j=i$, see the proof of [12, Theorem 4.2.1]. In other words, up to multiplication by scalars, the basis $\left[ \frac{y}{x^i} \right]$, $i=1, \ldots , g$, of $H^1(X, {\mathcal {O}}_X)$, given in Proposition 2.5, is dual to the basis $\omega _j$, $j=1, \ldots , g$, with respect to Serre duality.

Remark 2.7

Different bases of $H^1(X, {\mathcal {O}}_X)$ are described in [11, Lemma 6] and [8, Lemma 6]. The action of Frobenius on $H^1(X, {\mathcal {O}}_X)$ is studied in [1].

3 Bases of $H^1_\text {dR}(X/k)$

The object of this section is to give an explicit k-basis for the first de-Rham cohomology group $H^1_\mathrm {dR}(X/k)$ using Čech cohomology. If $p \not = 2$, we will moreover refine our result when another open subset is added to our standard open cover of X.

The algebraic de-Rham cohomology of X is defined to be the hypercohomology of the de-Rham complex

$$\begin{aligned} 0 \rightarrow {\mathcal {O}}_X \xrightarrow {d} \Omega _X \rightarrow 0 \end{aligned}$$

(5)

where d denotes the usual differential map $f \mapsto df$. We use the cover $\mathcal U$ and the Čech differentials defined in the previous section to obtain the Čech bicomplex of (5):

(6)

By a generalisation of Leray’s theorem [3, Corollaire 12.4.7] and Serre’s affineness criterion [7, Theorem 5.2.23], the first de-Rham cohomology of X is isomorphic to the first cohomology of the total complex of (6). Thus, $H^1_\mathrm {dR}(X/k)$ is isomorphic to the quotient of the space

$$\begin{aligned} \begin{aligned}&\big \{(\omega _0, \omega _\infty , f_{0\infty }) \in \Omega _X(U_0)\times \Omega _X (U_\infty ) \times {\mathcal {O}}_X(U_0 \cap U_\infty )\mid \\&df_{0\infty } = \omega _0|_{U_0\cap U_\infty } - \omega _\infty |_{U_0\cap U_\infty } \big \} \end{aligned} \end{aligned}$$

(7)

by the subspace

$$\begin{aligned} \left\{ (df_0, df_\infty , f_0|_{U_0 \cap U_\infty } -f_\infty |_{U_0 \cap U_\infty } )|f_0 \in {\mathcal {O}}_X(U_0), f_\infty \in {\mathcal {O}}_X(U_\infty ) \right\} . \end{aligned}$$

(8)

Via this representation of $H^1_\mathrm {dR}(X/k)$ and the isomorphism (4) we obtain the canonical maps

$$\begin{aligned} i :H^0(X, \Omega _X)\rightarrow H^1_\mathrm {dR}(X/k), \qquad \omega \mapsto [(\omega |_{U_0}, \omega |_{U_\infty }, 0)] \end{aligned}$$

(9)

and

$$\begin{aligned} p :H^1_\mathrm {dR}(X/k)\rightarrow H^1(X, {\mathcal {O}}_X), \qquad [(\omega _0, \omega _\infty , f_{0 \infty })] \mapsto [f_{0 \infty }]. \end{aligned}$$

(10)

The following proposition is equivalent to the more familiar and fancier sounding statement that the Hodge-de-Rham spectral sequence for X degenerates at $E_1$ (see [13]). This is in fact true for every smooth, proper curve X over k, see Example (2) in Section (1.5) of [13].

Proposition 3.1

The following sequence is exact:

$$\begin{aligned} 0 \rightarrow H^0(X,\Omega _X) \xrightarrow {i} H^1_\mathrm {dR}(X/k)\xrightarrow {p} H^1(X, {\mathcal {O}}_X) \rightarrow 0. \end{aligned}$$

(11)

We will call the sequence (11) the Hodge-de-Rham short exact sequence.

An elementary proof of Proposition 3.1 (that works for every smooth projective curve) can be found in [12, Proposition 4.1.2]; the main ingredient there is just the fact that the residue of differentials of the form df vanishes at every point of X and that hence the obvious composition $H^1(X, {\mathcal {O}}_X) \rightarrow H^1(X, \Omega _X) \overset{\sim }{\rightarrow } k$ is the zero map. For a hyperelliptic curve X, the surjectivity of p will also be verified in the proof of Theorem 3.2 below.

In order to state a basis of $H^1_\mathrm {dR}(X/k)$, we now define certain polynomials. To this end, we introduce the notations $f^{\le m}(x):= a_0 + \cdots +a_mx^m$ and $f^{> m}(x):=a_{m+1}x^{m+1}+ \cdots + a_nx^n$ for any polynomial $f(x):= a_0 + \cdots + a_n x^n \in k[x]$ and any $m \ge 0$. Let $1 \le i \le g$.

When $p\ne 2$ we define

$$\begin{aligned} s_i(x) := xf'(x) - 2if(x) \in k[x] \end{aligned}$$

and put $\psi _i(x):= s_i^{\le i}(x)$ and $\phi _i(x):= s_i^{>i}(x)$ so that $s_i(x) = \psi _i(x) + \phi _i(x)$.

When $p = 2$ we define

$$\begin{aligned} s_i(x,y) := xf'(x) + (xh'(x) + ih(x))y\in k[x]\oplus k[x]y \subseteq k(x,y) \end{aligned}$$

(where $k[x]\oplus k[x]y$ denotes the k[x]-module generated by 1 and y) and put $\psi _i(x,y) := s_i^{\le i}(x,y)$ and $\phi _i(x,y) := s_i^{> i}(x,y)$ where now the operations $\le i$ and $> i$ are applied to both the coefficients $xf'(x)$ and $xh'(x) + ih(x)$. Again we have $s_i(x,y) = \psi _i(x,y)+ \phi _i(x,y)$.

We now give a basis of $H^1_\mathrm {dR}(X/k)$ in terms of the polynomials just introduced and using the presentation of $H^1_\mathrm {dR}(X/k)$ developed above.

Theorem 3.2

If $p \ne 2$, the residue classes

$$\begin{aligned} \gamma _i:= \left[ \left( \frac{\psi _i(x)}{2x^{i+1}y} dx, \frac{-\phi _i(x)}{2x^{i+1}y} dx, \frac{y}{x^i} \right) \right] , \quad i=1, \ldots ,g, \end{aligned}$$

(12)

along with the residue classes

$$\begin{aligned} \lambda _i:= \left[ \left( \frac{x^{i}}{y} dx , \frac{x^{i}}{y} dx, 0 \right) \right] , \quad i = 0,\ldots ,g-1, \end{aligned}$$

(13)

form a k-basis of $H^1_\mathrm {dR}(X/k)$.

On the other hand, if $p=2$, the residue classes

$$\begin{aligned} \gamma _i := \left[ \left( \frac{\psi _i(x,y)}{x^{i+1}h(x)} dx, \frac{\phi _i(x,y)}{x^{i+1}h(x)} dx, \frac{y}{x^i} \right) \right] , \quad i =1, \ldots , g, \end{aligned}$$

(14)

together with the residue classes

$$\begin{aligned} \lambda _i := \left[ \left( \frac{x^{i}}{h(x)} dx, \frac{x^{i}}{h(x)} dx, 0 \right) \right] , \quad i=0, \ldots , g-1, \end{aligned}$$

(15)

form a k-basis of $H^1_\mathrm {dR}(X/k)$.

Remark 3.3

• (a) If $p \not =2$ and $f(x) = x^p -x$, an easy calculation shows that the basis elements given above are the same as those given in Theorem 3.1 of [4].

• (b) If $p=2$, another basis of $H^1_\mathrm {dR}(X/k)$ is given in [2, Section 4].

Proof

The elements in (13) and (15) are the images under the map i of the differentials $\frac{x^i}{y}dx, i=0, \ldots , g-1$, and $\frac{x^i}{h(x)}dx, i=0, \ldots , g-1$, respectively, which form a basis of $H^0(X, \Omega _X)$ by Lemma 2.2. Furthermore, provided the elements in (12) and (14) are well-defined elements of $H^1_\mathrm {dR}(X/k)$, these elements are mapped to the elements $[\frac{y}{x^i}], i=1, \ldots , g$, under p, which form a basis of $H^1(X, {\mathcal {O}}_X)$ by Proposition 2.5. By Proposition 3.1, it therefore suffices to check that the elements in (12) and (14) are well-defined elements of $H^1_\mathrm {dR}(X/k)$.

We first check the equality in  (7). When $p \not =2$, this is verified as follows:

$$\begin{aligned}&\left( \frac{\psi _i(x)}{2x^{i+1}y} - \frac{-\phi _i(x)}{2x^{i+1}y} \right) dx = \frac{s_i(x)}{2x^{i+1}y} dx \\& \quad = \frac{xf'(x) - {2if(x)}}{2x^{i+1}y} dx = \frac{x^i}{2y} \left( \frac{f'(x)}{x^{2i}} -\frac{2if(x)}{x^{2i+1}} \right) dx\\& \quad = \frac{x^i}{2y}d\left( \frac{f(x)}{x^{2i}}\right) = \frac{x^i}{2y} d\left( \left( \frac{y}{x^{i}}\right) ^2\right) = d\left( \frac{y}{x^{i}}\right) . \end{aligned}$$

When $p=2$, we obtain

$$\begin{aligned} h'(x)ydx+h(x)dy=f'(x)dx \end{aligned}$$

by differentiating equation (2) and then verify the equality in (7) as follows (note that we replace all minus signs with plus signs):

$$\begin{aligned}&\left( \frac{ \psi _i(x,y)}{x^{i+1}h(x)} + \frac{\phi _i(x,y)}{x^{i+1}h(x)} \right) dx = \frac{s_i(x,y)}{x^{i+1}h(x)}dx \\& \quad = \left( \frac{f'(x)}{x^ih(x)} + \frac{h'(x)y}{x^ih(x)} + \frac{iy}{x^{i+1}} \right) dx\\& \quad = \frac{dy}{x^{i}} + \frac{iy}{x^{i+1}} dx = d\left( \frac{y}{x^{i}}\right) . \end{aligned}$$

It remains to prove that the first two entries of the triples in (12) and (14) are regular differentials on $U_0$ and $U_\infty$, respectively.

We first consider the case $p \not = 2$. As $\frac{dx}{y}$ is a regular differential on $X= U_0 \cup U_\infty$ by Lemma 2.2, it suffices to observe that each of the functions $\frac{\psi _i(x)}{x^{i+1}}$, $i=1, \ldots , g$, is regular on $U_0$ (in fact has a zero at $\infty$) and that each of the functions $\frac{\phi _i(x)}{x^{i+1}}$, $i=1, \ldots , g$, is regular on $U_\infty$.

We now turn to the case $p=2$. As above, we know from Lemma 2.2 that $\frac{dx}{h(x)}$ is regular on $X=U_0 \cup U_\infty$. Furthermore, for every $i \in \{1, \ldots , g\}$, the function $\frac{\phi _i(x,y)}{x^{i+1}}$ is regular on $U_\infty$ since y is regular on $U_\infty$ and since, by definition of $\phi _i(x,y)$, the k[x]-coefficients of 1 and y in $\phi _i(x,y)$ are divisible by $x^{i+1}$. Hence $\frac{\phi _i(x,y)}{x^{i+1}h(x)} dx$ is regular on $U_\infty$, as was to be shown. It remains to show that $\frac{\psi _i(x,y)}{x^{i+1}h(x)} dx$ is regular on $U_0$. As $\frac{\psi _i(x,y)}{x^{i+1}}$ and $\frac{dx}{h(x)}$ are regular on $U_0 \cap U_\infty$, this amounts to showing that $\frac{\psi _i(x,y)}{x^{i+1}h(x)} dx$ is regular above $\infty$.

We first consider the case when $\infty$ is not a branch point of $\pi$. By Lemma 2.4, the differential dx has a pole of order 2 at each of the two points $P_\infty$, $P'_\infty \in X$ above $\infty$. Furthermore, the k[x]-coefficient of 1 in $\psi _i(x,y)$ has a pole at $P_\infty$ and $P'_\infty$ of order at most i and the k[x]-coefficient of y has a pole at $P_\infty$ and $P'_\infty$ of order at most $i-1$ since the coefficient of $x^i$ in $xh'(x) + ih(x)$ is zero (remember $\text {char}(k)=2$). Moreover, y has a pole at $P_\infty$ and $P'_\infty$ of order at most $g+1$ by Lemma 2.1. Finally, $\frac{1}{h(x)}$ has a zero at $P_\infty$ and $P'_\infty$ of order $d=\text {deg}(h(x)) = g+1$. Putting all this together we obtain

$$\begin{aligned} \mathrm {ord}&_{P} \left( \frac{\psi _i(x,y)}{x^{i+1} h(x)}dx \right) \\&= \mathrm {ord}_{P}\left( \psi _i(x,y)\right) + \mathrm {ord}_{P}\left( \frac{1}{x^{i+1}} \right) + \mathrm {ord}_{P}\left( \frac{1}{h(x)}\right) + \mathrm {ord}_{P}(dx)\\&\ge \min \{-i, -(i-1)-(g+1)\} + (i+1) + (g+1) -2 =0 \end{aligned}$$

for $P \in \{P_\infty , P'_\infty \}$, which shows that $\frac{\psi _i(x,y)}{x^{i+1}h(x)} dx$ is regular at $P_\infty$ and $P'_\infty$.

We finally assume that $\infty$ is a branch point of $\pi$ and prove that $\frac{\psi _i(x,y)}{x^{i+1}h(x)} dx$ is regular at the unique point $P_\infty \in X$ above $\infty$. By Lemma 2.4, the order of the differential dx at $P_\infty$ is $2(g-1-d)$ where $d= \mathrm {deg}(h(x))$. For similar reasons as above, the k[x]-coefficients of 1 and y in $\psi _i(x,y)$ have a pole at $P_\infty$ of order at most 2i and $2(i-1)$, respectively, and $\frac{1}{h(x)}$ has a zero at $P_\infty$ of order 2d. Finally, y has a pole at $P_\infty$ of order at most $2(g+1)$ by Lemma 2.1. Putting all this together we obtain

$$\begin{aligned} \mathrm {ord}&_{P_\infty } \left( \frac{\psi _i(x,y)}{x^{i+1} h(x)}dx \right) \\&= \mathrm {ord}_{P_\infty }\left( \psi _i(x,y)\right) + \mathrm {ord}_{P_\infty }\left( \frac{1}{x^{i+1}} \right) + \mathrm {ord}_{P_\infty }\left( \frac{1}{h(x)}\right) + \mathrm {ord}_{P_\infty }(dx)\\&\ge \min \{-2i, -2(i-1)-2(g+1)\} + 2(i+1) + 2d + 2(g-1-d) =0, \end{aligned}$$

which shows that $\frac{\psi _i(x,y)}{x^{i+1}h(x)} dx$ is regular at $P_\infty$. $\square$

In the proofs in the next section, we will need a refined description of the basis elements given in (12) when another open subset is added to our standard cover ${\mathcal {U}}= \{U_0, U_\infty \}$. To this end, we now fix $a \in {\mathbb P}^1_k \backslash \{0, \infty \}$ and define the covers ${\mathcal {U}}':=\{U_a, U_\infty \}$ and ${\mathcal {U}}'' := \{U_0, U_a, U_\infty \}$ of X. Similarly to (7) and (8), the first de-Rham cohmology group $H^1_\mathrm {dR}(X/k)$ is then isomorphic to the k-vector space

$$\begin{aligned}&\left\{ (\omega _0, \omega _a, \omega _\infty , f_{0a}, f_{0 \infty },f_{a \infty }) \in \right. \nonumber \\&\Omega _X(U_0) \times \Omega _X(U_a)\times \Omega _X(U_\infty ) \times {\mathcal {O}}_X(U_0 \cap U_a)\nonumber \\&\times {\mathcal {O}}_X(U_0 \cap U_\infty ) \times {\mathcal {O}}_X(U_a \cap U_\infty ) \mid \nonumber \\&\left. f_{0a} - f_{0\infty } + f_{a \infty } = 0,\; df_{0a} = \omega _0 - \omega _a, \; df_{0\infty } = \omega _0 - \omega _\infty , \; df_{a\infty } = \omega _a - \omega _\infty \right\} \end{aligned}$$

(16)

quotiented by the subspace

$$\begin{aligned}&\left\{ ( df_0, df_a, df_\infty , f_0- f_a, f_0 - f_\infty , f_a - f_\infty ) | \right. \nonumber \\&\quad \left. f_0 \in {\mathcal {O}}_X(U_0), f_a \in {\mathcal {O}}_X(U_a ), f_\infty \in {\mathcal {O}}_X(U_\infty )\right\} . \end{aligned}$$

(17)

We use the notations $\check{H}^1_\mathrm {dR}({\mathcal {U}})$ and $\check{H}^1_\mathrm {dR}({\mathcal {U}}'')$ for the representations of $H^1_\mathrm {dR}(X/k)$ introduced in (7), (8) and (16), (17), respectively. The canonical isomorphism $\rho :\check{H}^1_\mathrm {dR}({\mathcal {U}}'') \rightarrow \check{H}^1_\mathrm {dR}({\mathcal {U}})$, is then induced by the projection

$$\begin{aligned} \rho :(\omega _0, \omega _a, \omega _\infty , f_{0a}, f_{0 \infty },f_{a \infty }) \mapsto (\omega _0, \omega _\infty , f_{0 \infty }). \end{aligned}$$

(18)

When $p \not =2$, the next proposition explicitly describes the pre-image of the basis elements $\gamma _i = \left[ \left( \frac{\psi _i(x)}{2x^{i+1}y}dx, \frac{-\phi _i(x)}{2x^{i+1}y}dx, \frac{y}{x^i}\right) \right] , i=1,\ldots ,g$, of $H^1_\mathrm {dR}(X/k)$ under $\rho$. To this end, we define the polynomials

$$\begin{aligned} g(x) := (x-a)^g, \quad r_i(x) := g^{\le i -1}(x) \quad \text {and} \quad t_i(x) := g^{> i-1}(x) \end{aligned}$$

in k[x] for $1 \le i \le g$ so that $r_i(x) + t_i(x) = (x-a)^g$.

Proposition 3.4

Let $p \not = 2$. For $i \in \{1, \ldots , g\}$, let

$$\begin{aligned} \omega _{0i}:= & {} \frac{\psi _i(x)}{2x^{i+1}y}dx, \quad \omega _{\infty i}:= \frac{-\phi _i(x)}{2x^{i+1}y}dx,\\ \omega _{ai}:= & {} \frac{(\psi _i(x)t_i(x) - \phi _i(x)r_i(x))(x-a) - 2if(x)(-1)^{g-i}\left( {\begin{array}{c}g\\ i\end{array}}\right) a^{g-i+1}x^i}{2x^{i+1}(x-a)^{g+1}y}dx \end{aligned}$$

and

$$\begin{aligned} f_{0ai}:= \frac{r_i(x)y}{x^i(x-a)^g}, \quad f_{0 \infty i}:= \frac{y}{x^i}, \quad f_{a \infty i}:= \frac{t_i(x)y}{x^i(x-a)^g}. \end{aligned}$$

Then we have:

$$\begin{aligned} \rho ^{-1}(\gamma _i) =[( \omega _{0 i}, \omega _{a i}, \omega _{\infty i}, f_{0 a i}, f_{0 \infty i}, f_{a \infty i} )]. \end{aligned}$$

(19)

Remark 3.5

This description of $\rho ^{-1}(\gamma _i)$ does not generalise the description given in Lemma 3.3 of [4] in case of the hyperelliptic curve $y^2 = x^p - x$. In fact, the proof of that lemma seems to contain various mistakes.

Proof

We fix $i \in \{1, \ldots , g\}$. We obviously only need to show that the sextuple on the right-hand side of (19) is a well-defined element of the space (16).

From the proof of Theorem 3.2 we already know that ${d(f_{0 \infty i}) = \omega _{0 i} - \omega _{\infty i}}$ and that ${f_{0 \infty i}, \omega _{0 i}}$ and ${\omega _{\infty i}}$ are regular on the appropriate open sets.

Since $r_i(x)+t_i(x)=(x-a)^g$, we have

$$\begin{aligned} {f_{0 a i} - f_{0 \infty i}+ f_{a \infty i} = }\frac{r_i(x)y}{x^i(x-a)^g} - \frac{y}{x^i} + \frac{t_i(x)y}{x^i(x-a)^g} = {0}, \end{aligned}$$

as desired.

The function $f_{0ai}$ is obviously regular above ${\mathbb P}^1_k \backslash \{0, a, \infty \}$. We furthermore observe that $\mathrm {ord}_\infty \left( \frac{r_i(x)}{x^i (x-a)^g}\right) \ge -(i-1) +i +g = g+1$ and that, by Lemma 2.1, the order of y above $\infty$ is at least $-2(g+1)$ or at least $-(g+1)$ depending on whether $\infty$ is a branch point of $\pi$ or not. Thus, ${f_{0ai}}$ is regular above $\infty$ and hence on ${U_0 \cap U_a}$.

As above, the function $f_{a \infty i}$ is regular above ${\mathbb P}^1_k \backslash \{0, a, \infty \}$. Furthermore, the functions $\frac{t_i(x)}{x^i}$, y and $\frac{1}{(x-a)^g}$ are obviously regular above 0. Therefore, ${f_{a \infty i}}$ is regular above 0 as well and hence on ${U_a \cap U_\infty }$.

We next show that ${df_{0 a i} = \omega _{0 i} - \omega _{a i}}$. Using the product rule and the chain rule we obtain

$$\begin{aligned} df_{0 a i}&= d \left( \frac{r_i(x)y}{x^i(x-a)^g} \right) \\&= \frac{r_i(x)}{x^i(x-a)^g}dy + d\left( \frac{r_i(x)}{x^i(x-a)^g} \right) y\\&= \frac{f'(x)r_i(x)}{2x^i(x-a)^gy}dx + \left( \frac{r_i'(x)}{x^i(x-a)^g} -\frac{i r_i(x)}{x^{i+1}(x-a)^g} - \frac{gr_i(x)}{x^i(x-a)^{g+1}}\right) ydx \\&= \frac{xf'(x)r_i(x)(x-a) + 2f(x)\left( xr_i'(x)(x-a) - ir_i(x)(x-a) - gxr_i(x)\right) }{2x^{i+1}(x-a)^{g+1}y} dx. \end{aligned}$$

We now recall that

$$\begin{aligned} xf'(x) - 2if(x) =\psi _i(x) + \phi _i(x). \end{aligned}$$

We furthermore recall that $r_i(x) = g^{\le i-1}(x)$ where $g(x) = (x-a)^g$. Therefore

$$\begin{aligned} r'_i(x)&\cdot (x-a) - g \cdot r_i(x)\\&\quad = \left[ g'(x)\right] ^{\le i-2}\cdot (x-a) - g \cdot g^{\le i-1}(x)\\&\quad = \left( \left[ g'(x)\cdot (x-a)\right] ^{\le i-1} + a\cdot b_{i-1}\cdot x^{i-1} \right) - g \cdot g^{\le i-1}(x)\\&\quad = a\cdot b_{i-1}\cdot x^{i-1} \end{aligned}$$

where $b_{i-1} = (-1)^{g-i} i \left( {\begin{array}{c}g\\ i\end{array}}\right) a^{g-i}$ denotes the coefficient of $x^{i-1}$ in $g'(x)$.

Thus we obtain

$$\begin{aligned} df_{0 a i}&= \frac{(\psi _i(x) + \phi _i(x)) r_i(x)(x-a) +2 f(x) (ab_{i-1} x^i)}{2x^{i+1}(x-a)^{g+1}y} dx\\&= \frac{\psi _i(x) \left( (x-a)^{g+1} - t_i(x)(x-a)\right) +\phi _i(x) r_i(x)(x-a) + 2 f(x) (ab_{i-1} ) x^i}{2x^{i+1}(x-a)^{g+1}y}dx\\&= \frac{\psi _i(x)}{2x^{i+1}y}dx - \frac{(\psi _i(x)t_i(x) - \phi _i(x)r_i(x))(x-a) - 2 f(x) (ab_{i-1} ) x^i}{2x^{i+1}(x-a)^{g+1}y}dx\\&= \omega _{0i} -\omega _{ai}, \end{aligned}$$

as claimed.

From the above we moreover obtain that

$$\begin{aligned} {df_{a \infty i} =} \, df_{0 \infty i} - df_{0 a i} = (\omega _{0 i} - \omega _{\infty i }) - (\omega _{0 i} - \omega _{a i}) = {\omega _{a i} - \omega _{\infty i}}. \end{aligned}$$

Finally, $\omega _{ai}$ is regular on $U_a$ because $\omega _{ai}=\omega _{0i}-df_{0ai}$ and $\omega _{ai}$ is hence regular on $U_0 \cap U_a$ and because $\omega _{ai}= \omega _{\infty i} + df_{a \infty i}$ and $\omega _{ai}$ is hence regular on $U_a \cap U_\infty$. $\square$

4 Actions on $H^1_\mathrm {dR}(X/k)$

In this section we study the action of certain automorphisms on $H^1_\mathrm {dR}(X/k)$. We first prove that the hyperelliptic involution acts by multiplication by $-1$ on $H^1_\mathrm {dR}(X/k)$ when $p\not = 2$ and as the identity when $p=2$. We then give a family of hyperelliptic curves for which the Hodge-de-Rham short exact sequence (11) does not split equivariantly.

Theorem 4.1

The hyperelliptic involution acts on $H^1_\mathrm {dR}(X/k)$ by multiplication by $-1$.

Proof

Recall that the hyperelliptic involution is the unique non-trivial automorphism $\sigma$ of X such that $\pi \circ \sigma = \pi$.

If $p\not = 2$, the involution $\sigma$ acts on K(X) by $(x,y) \mapsto (x,-y)$. Hence, $\sigma$ maps each entry of the triples in (12) and (13) to its negative. Thus, the $p \not = 2$ part of Theorem 3.2 implies Theorem 4.1.

If $p =2$, the involution $\sigma$ acts on K(X) by $(x,y) \mapsto (x,y+h(x))$. In particular, it fixes the basis elements (15) of $H^1_\mathrm {dR}(X/k)$. According to the $p=2$ part of Theorem 3.2, it remains to show that $\sigma$ also fixes the residue classes $[(\omega _{0i}, \omega _{\infty i}, f_{0 \infty i})], i=1, \ldots , g$, in (14). For $i \in \{1, \ldots , g\}$, this follows from the description of $H^1_\mathrm {dR}(X/k)$ given in (7) and (8) and from the equation

$$\begin{aligned} \sigma ((\omega _{0i}, \omega _{\infty i}, f_{0 \infty i}))&- (\omega _{0i}, \omega _{\infty i}, f_{0 \infty i}) \\&= \left( d \left( \frac{h^{\le i}(x)}{x^i}\right) , d\left( \frac{h^{> i}(x)}{x^i}\right) , \frac{h^{\le i}(x)}{x^i} -\frac{ h^{> i}(x)}{x^i}\right) \end{aligned}$$

which in turn is verified in the following three lines (where we replace all minus signs with plus signs):

$$\begin{aligned}&\sigma \left( \frac{\psi _i(x,y)}{x^{i+1}h(x)} dx\right) + \frac{\psi _i(x,y)}{x^{i+1}h(x)}dx = \frac{[xh'(x) + ih(x)]^{\le i}h(x)}{x^{i+1}h(x)}dx = d\left( \frac{h^{\le i}(x)}{x^i} \right) \\&\sigma \left( \frac{\phi _i(x,y)}{x^{i+1}h(x)} dx\right) + \frac{\phi _i(x,y)}{x^{i+1}h(x)}dx = \frac{[xh'(x) + ih(x)]^{> i}h(x)}{x^{i+1}h(x)}dx = d\left( \frac{h^{> i}(x)}{x^i} \right) \\&\quad \sigma \left( \frac{y}{x^i}\right) + \frac{y}{x^i} = \frac{h(x)}{x^i}. \end{aligned}$$

$\square$

Remark 4.2

If $p\not =2$, Theorem 4.1 can also be proved as follows. By Lemma 2.2, the involution $\sigma$ acts by multiplication by $-1$ on $H^0(X,\Omega _X)$. By Serre duality, it then acts by multiplication by $-1$ also on $H^1(X, {\mathcal {O}}_X)$. Finally, by Maschke’s Theorem (for the cyclic group of order 2) applied to the Hodge-de-Rham short exact sequence (11), it acts by multiplication by $-1$ also on $H^1_\mathrm {dR}(X/k)$.

Before we state the main result of this paper, we recall that any automorphism $\tau$ of X induces a map $\bar{\tau } :{\mathbb {P}}_k^1 \rightarrow {\mathbb {P}}_k^1$ since $\mathbb {P}_k^1$ is the quotient of X by the hyperelliptic involution and since the hyperelliptic involution $\sigma$ belongs to the centre of $\mathrm {Aut}(X)$ (see [7, Corollary 7.4.31]). The following commutative diagram visualises this situation:

Theorem 4.3

Let $p \ge 3$. We assume that the degree of the polynomial f(x) defining the hyperelliptic curve X is odd. We furthermore assume that there exists $\tau \in \mathrm {Aut}(X)$ such that the induced map $\bar{\tau }:{\mathbb {P}}_k^1 \rightarrow {\mathbb {P}}_k^1$ is given by $x \mapsto x+a$ for some $a \not = 0$. Let G denote the subgroup of $\mathrm {Aut}(X)$ generated by $\tau$. Then the Hodge-de-Rham short exact sequence (11) does not split as a sequence of k[G]-modules.

The following example explicitly describes hyperelliptic curves that allow an automorphism $\tau$ as assumed in the previous theorem.

Example 4.4

Let $p \ge 3$, let $a \in k^\times$ and let $q(z) \in k[z]$ be any monic polynomial without repeated roots. Then $f(x) := q(x^p - a^{p-1}x) \in k[x]$ obviously has no repeated roots either and thus $y^2=f(x)$ defines a hyperelliptic curve X. Moreover, $(x,y) \mapsto (x+a,y)$ defines an automorphism $\tau$ of X and the induced automorphism $\bar{\tau }$ is given by $x \mapsto x+a$.

Remark 4.5

• (a) When applied to $a=1$ and to the hyperelliptic curve X given by $q(z) = z$ in Example 4.4, Theorem 4.3 becomes the main theorem of [4].

• (b) Theorem 4.3 and Example 4.4 imply that, for every algebraically closed field k of characteristic $p\ge 3$, there exist infinitely many $g \ge 2$ and hyperelliptic curves over k of genus g for which the Hodge-de-Rham spectral sequence does not split equivariantly.

• (c) Suppose $g \ge 2$ is given. If p is a prime divisor of $2g+1$ then, according to Theorem 4.3 and Example 4.4, every monic polynomial $q(z) \in k[z]$ of degree $(2g+1)/p$ without repeated roots defines a hyperelliptic curve X of genus g for which the Hodge-de-Rham sequence does not split equivariantly.

The following proposition shows that any hyperelliptic curve satisfying the assumptions of Theorem 4.3 is in fact of the form as given in Example 4.4.

Proposition 4.6

Let $p \ge 3$ and let $\tau \in \mathrm {Aut}(X)$. If the induced isomorphism $\bar{\tau } :{\mathbb {P}}_k^1 \rightarrow {\mathbb {P}}_k^1$ is given by $x \mapsto x+ a$ for some $a \not = 0$, then the action of $\tau ^*$ on y is given by $\tau ^*(y) = y$ or $\tau ^*(y) = -y$ and f(x) is of the form $f(x) = q(x^p-a^{p-1} x)$ for some polynomial $q \in k[z]$ without repeated roots.

Proof

We first show that $\tau ^*(y) = \pm y$. There exist $g_1(x)$ and $g_2(x)\not = 0$ in k(x) such that

$$\begin{aligned} \tau ^*(y) = g_1(x) + g_2(x)y \in k(x,y). \end{aligned}$$

Hence

$$\begin{aligned} f(x+a) = \tau ^*(y^2) = (\tau ^*(y))^2 = g_1(x)^2 + 2g_1(x)g_2(x)y + g_2(x)^2f(x). \end{aligned}$$

(20)

This implies that $g_1(x) =0$ because otherwise

$$\begin{aligned} y = \frac{f(x+a) - g_1(x)^2 - g_2(x)^2f(x)}{2g_2(x)g_1(x)} \end{aligned}$$

would belong to k(x). By comparing the degrees in (20) we see that $g_2(x)$ is a constant, and then by comparing coefficients in the same equation we see that $g_2(x)^2 = 1$. Hence $\tau ^*(y) = \pm y$, as claimed.

We now show that f(x) is of the form $q(x^p-a^{p-1}x)$. The extension $k(x)= k(z,x)$ of the rational function field k(z) obtained by adjoining an element x satisfying the equation $x^p-a^{p-1}x-z=0$ is a Galois extension with cyclic Galois group generated by the automorphism $x \mapsto x+a$. We derived above that $f(x) = f(x+a)$. Hence f(x) belongs to k(z). Furthermore, x and hence f(x) is integral over k[z]. Therefore, $f(x)\in k(z)$ belongs to k[z], i.e., $f(x) = q(x^p-a^{p-1}x)$ for some $q \in k[z]$ without repeated roots, as was to be shown. $\square$

Proof of Theorem 4.3

We suppose that the sequence (11) does split and that

$$\begin{aligned} s :H^1(X, {\mathcal {O}}_X)\rightarrow H^1_\mathrm {dR}(X/k)\end{aligned}$$

is a k[G]-linear splitting map. Then we have

$$\begin{aligned} s(\tau ^*(\alpha )) = \tau ^*(s(\alpha )) \in H^1_\mathrm {dR}(X/k)\end{aligned}$$

(21)

and

$$\begin{aligned} p(s(\alpha )) = \alpha \end{aligned}$$

(22)

for all $\alpha \in H^1(X, {\mathcal {O}}_X)$. We will show that these equalities give rise to a contradiction when $\alpha$ is the residue class $\left[ \frac{y}{x^g}\right]$ in $H^1(X, {\mathcal {O}}_X)$ (see Proposition 2.5).

We first show that $\left[ \frac{y}{x^g}\right] \in H^1(X, {\mathcal {O}}_X)$ is fixed by $\tau ^*$. To this end, we recall that ${\mathcal {U}}=\{U_0, U_\infty \}$, ${\mathcal {U}}'= \{U_a, U_\infty \}$ and ${\mathcal {U}}''=\{U_0, U_a, U_\infty \}$ and consider the following obviously commutative diagram of isomorphisms where $\rho$ and $\rho '$ are defined as in (18) and the equalities denote the identification (4):

By (the proof of) Proposition 3.4, the triple $\left( \frac{r_g(x)y}{x^g(x-a)^g}, \frac{y}{x^g}, \frac{t_g(x)y}{x^g(x-a)^g} \right)$ defines a well-defined element of $\check{H}^1({\mathcal {U}}'', {\mathcal {O}}_X)$. Hence we have

$$\begin{aligned} \rho ^{-1} \left( \left[ \frac{y}{x^g}\right] \right)&= \left[ \left( \frac{r_g(x)y}{x^g(x-a)^g}, \frac{y}{x^g}, \frac{t_g(x)y}{x^g(x-a)^g} \right) \right] \\&= \left[ \left( \frac{((x-a)^g-x^g)y}{x^g(x-a)^g}, \frac{y}{x^g}, \frac{y}{(x-a)^g} \right) \right] \text { in } \check{H}^1({\mathcal {U}}'', {\mathcal {O}}_X). \end{aligned}$$

We therefore obtain

$$\begin{aligned} \tau ^* \left( \left[ \frac{y}{x^g}\right] \right)&= \tau ^* \left( \rho ' \left( \rho ^{-1}\left( \left[ \frac{y}{x^g}\right] \right) \right) \right) \\&= \tau ^*\left( \rho ' \left( \left[ \left( \frac{((x-a)^g-x^g)y}{x^g(x-a)^g}, \frac{y}{x^g}, \frac{y}{(x-a)^g} \right) \right] \right) \right) \\&= \tau ^*\left( \left[ \frac{y}{(x-a)^g} \right] \right) = \left[ \frac{y}{x^g} \right] , \end{aligned}$$

as claimed.

By Theorem 3.2, the elements $\lambda _i$, $i=0, \ldots , g-1$, defined in (13) together with the elements $\gamma _i$, $i=1, \ldots , g$, defined in (12) form a basis of $H^1_\mathrm {dR}(X/k)$. Since the canonical projection $p: H^1_\mathrm {dR}(X/k)\rightarrow H^1(X, {\mathcal {O}}_X)$ is k[G]-linear and maps $\gamma _g$ to the residue class $\left[ \frac{y}{x^g} \right]$, it follows that

$$\begin{aligned} \tau ^*(\gamma _g) = \gamma _g + \sum _{i=0}^{g-1} c_i\lambda _i \end{aligned}$$

(23)

for some $c_0, \ldots , c_{g-1} \in k$. On the other hand, we have

$$\begin{aligned} s\left( \left[ \frac{y}{x^g} \right] \right) = \gamma _g + \sum _{i=0}^{g-1}d_i \lambda _i \end{aligned}$$

for some $d_0, \ldots , d_{g-1} \in k$. Now the action of $\tau ^*$ on $\lambda _i$ for $0 \le i \le g-1$ is easily seen to be given by

$$\begin{aligned} \begin{aligned} \tau ^*(\lambda _i)&= \tau ^*\left( \left[ \left( \frac{x^i}{y}dx, \frac{x^i}{y}dx, 0\right) \right] \right) \\&= \left[ \left( \frac{(x+a)^i}{y}dx, \frac{(x+a)^i}{y}dx, 0 \right) \right] = \sum _{k=0}^i \left( {\begin{array}{c}i\\ k\end{array}}\right) a^{i-k} \lambda _k. \end{aligned} \end{aligned}$$

(24)

Plugging the equations obtained so far into Eq. (21) we obtain

$$\begin{aligned} \gamma _g + \sum _{i=0}^{g-1} d_i\lambda _i&= s \left( \left[ \frac{y}{x^g} \right] \right) = s \left( \tau ^*\left( \left[ \frac{y}{x^g} \right] \right) \right) \\&= \tau ^* \left( s \left( \left[ \frac{y}{x^g} \right] \right) \right) = \tau ^* \left( \gamma _g + \sum _{i=0}^{g-1} d_i \lambda _i \right) \\&= \left( \gamma _g + \sum _{i=0}^{g-1} c_i \lambda _i \right) + \sum _{i=0}^{g-1} d_i \left( \sum _{k = 0}^i \left( {\begin{array}{c}i\\ k\end{array}}\right) a^{i-k} \lambda _k \right) . \end{aligned}$$

By comparing coefficients of the basis element $\lambda _{g-1}$, we see that $c_{g-1} = 0$. On the other hand, we will below derive the equation $c_{g-1} = a/4$ from the defining Eq. (23). Since we assumed that $a \ne 0$, this gives us the desired contradiction.

The left-hand side of Eq. (23) is $\tau ^*(\gamma _g)$. To compute $\tau ^*(\gamma _g)$ we consider the following commutative diagram of isomorphisms where $\rho$ is the canonical projection (18), $\rho '$ is given by $(\omega _0, \omega _a, \omega _\infty , f_{0 a}, f_{0 \infty }, f_{a \infty }) \mapsto (\omega _a, \omega _\infty , f_{a \infty })$ and the equalities denote the identification given by Eqs. (7) and (8):

(25)

Then, by Proposition 3.4, we have:

$$\begin{aligned} \begin{aligned} \tau ^*(\gamma _g)&= \tau ^*(\rho '(\rho ^{-1}(\gamma _g))) \\&= \tau ^*\left( \left[ \omega _{a g}, \frac{- \phi _g(x)}{2x^{g+1}y} dx, \frac{y}{(x-a)^g} \right) \right] \\&= \left[ \left( \tau ^*(\omega _{a g}) , \frac{-\phi _g(x+a)}{2(x+a)^{g+1}y}dx, \frac{y}{x^g} \right) \right] . \end{aligned} \end{aligned}$$

(26)

On the other hand, the right hand side of Eq. (23) is equal to

$$\begin{aligned} \left[ \left( \frac{\psi _g(x)}{2x^{g+1}y}dx, \frac{-\phi _g(x)}{2x^{g+1}y}dx, \frac{y}{x^g} \right) \right] + \sum _{i=0}^{g-1} c_i \left[ \left( \frac{x^i}{y}dx, \frac{x^i}{y}dx, 0 \right) \right] . \end{aligned}$$

(27)

Note that the third entry in both (26) and (27) is $\frac{y}{x^g}$. Now, if the third entry of a triple $(df_0, df_\infty , f_0|_{U_0 \cap U_\infty }- f_\infty |_{U_0\cap U_\infty })$ in the subspace (8) of the space (7) vanishes, then $f_0$ and $f_\infty$ glue to a global and hence constant function and the whole triple vanishes. Hence, the triples in (26) and (27) are equal already before taking residue classes. By comparing the second entries of (26) and (27) we therefore obtain the equation

$$\begin{aligned} - \frac{\phi _g(x+a)}{2(x+a)^{g+1}y} dx = -\frac{\phi _g(x)}{2x^{g+1}y}dx + \sum _{i = 0}^{g-1} c_i \frac{x^i}{y}dx \quad \text { in } \quad \Omega _{K(X)}. \end{aligned}$$

Since dx is a basis of $\Omega _{K(X)}$ considered as a K(X)-vector space, the equation above is equivalent to the equation

$$\begin{aligned} \frac{\phi _g(x+a)}{2(x+a)^{g+1}} = \frac{\phi _g(x)}{2x^{g+1}} - \sum _{i=0}^{g-1} c_i x^i \end{aligned}$$

(in the rational function field k(x)) which in turn is equivalent to the equation

$$\begin{aligned} \phi _g(x+a)x^{g+1} = \phi _g(x)(x+a)^{g+1} - 2(x+a)^{g+1}x^{g+1}\sum _{i=0}^{g-1}c_i x^i \quad \text { in } \quad k[x]. \end{aligned}$$

Now, the assumption that the degree of f(x) is odd means that the degree of f(x) is precisely $2g+1$. By definition, the terms of highest degree in $\phi _g(x)$ are the same as the terms of highest degree in

$$\begin{aligned} s_g(x) = xf'(x) - 2gf(x) = x^{2g+1} + 0\cdot x^{2g} + \cdots . \end{aligned}$$

We therefore have

$$\begin{aligned}&\left( (x+a)^{2g+1} + 0 \cdot (x+a)^{2g} + \cdots \right) x^{g+1}\\&\quad = (x^{2g+1} + 0 \cdot x^{2g} + \cdots )(x+a)^{g+1} - 2(x+a)^{g+1}x^{g+1}(c_{g-1}x^{g-1} + \cdots ). \end{aligned}$$

Hence, by comparing the coefficients of $x^{3g+1}$, we obtain

$$\begin{aligned} (2g+1)a = (g+1)a - 2c_{g-1}. \end{aligned}$$

and hence

$$\begin{aligned} c_{g-1} = \frac{((g+1) - (2g+1))a}{2} = - \frac{g}{2} a. \end{aligned}$$

Finally, we have $2g + 1= \deg (f(x)) \equiv 0 {\;\mathrm mod}\; p$ by Proposition 4.6 and hence

$$\begin{aligned} c_{g-1} = \frac{a}{4}, \end{aligned}$$

as claimed above. This concludes the proof of Theorem 4.3. $\square$

Remark 4.7

While the method of calculating $\tau ^*\left( \left[ \frac{y}{x^g}\right] \right)$ and $\tau ^*(\gamma _g)$ in the proof above is the same as in [4], the actual computations do not generalise those in [4], not only due to Remark 3.5 but also because of further mistakes in [4]. Finally, the argument in the proof above for obtaining the desired contradiction is different from the one in [4] the very end of which has unfortunately not been carried out anyway.

We conclude with an example which demonstrates that the requirement in Theorem 4.3 of f(x) to be of odd degree is a necessary condition.

Example 4.8

Let $p = 3$ and X be the hyperelliptic curve of genus 2 defined by the equation

$$\begin{aligned} y^2 = f(x) = x^6 + x^4 + x^2 + 2. \end{aligned}$$

As in Theorem 4.3 let $\tau$ denote the automorphism of X given by $(x,y) \mapsto (x+1,y)$ and let $G:= \langle \tau \rangle$.

By Theorem 3.2, a basis of $\check{H}^1_\mathrm {dR}({\mathcal {U}})$ is given by

$$\begin{aligned} \lambda _0= & {} \left[ \left( \frac{1}{y}dx, \frac{1}{y}dx, 0 \right) \right] , \quad \lambda _1 =\left[ \left( \frac{x}{y}dx, \frac{x}{y}dx, 0\right) \right] ,\\ \gamma _1= & {} \left[ \left( \frac{1}{x^2y}dx, \frac{x^4+2x^2}{y}dx, \frac{y}{x} \right) \right] , \quad \gamma _2 =\left[ \left( \frac{x^2 +1}{2x^3y}dx, \frac{2x^3}{y}dx, \frac{y}{x^2}\right) \right] . \end{aligned}$$

By Proposition 2.5, the residue classes $\bar{\gamma }_1:= \left[ \frac{y}{x}\right]$ and $\bar{\gamma }_2:= \left[ \frac{y}{x^2}\right]$ form a basis of $H^1(X, {\mathcal {O}}_X)$. We define a map

$$\begin{aligned} s :H^1(X, {\mathcal {O}}_X)\rightarrow H^1_\mathrm {dR}(X/k)\end{aligned}$$

of vector spaces over k by

$$\begin{aligned} \bar{\gamma }_1 \mapsto \gamma _1 \quad \text {and} \quad \bar{\gamma }_2 \mapsto \gamma _2 + \lambda _1. \end{aligned}$$

Clearly $p \circ s$ is the identity map on $H^1(X, {\mathcal {O}}_X)$, and hence, if s is k[G]-linear, the sequence in Proposition 3.1 does split as a sequence of k[G]-modules.

We now show that s is k[G]-linear. By Proposition 3.4, the pre-images of $\gamma _1$ and $\gamma _2$ under $\rho$ in $\check{H}^1_\mathrm {dR}({\mathcal {U}}'')$ are the residue classes of

$$\begin{aligned} \nu _1 = \left( \frac{1}{x^2y}dx, \frac{x^4+2x^3+2x^2}{2(x-1)^3y}dx, \frac{x^4+2x^2}{y}dx, \frac{y}{x(x-1)^2}, \frac{y}{x}, \frac{(x+1)y}{(x-1)^2} \right) \end{aligned}$$

and

$$\begin{aligned} \nu _2 = \left( \frac{x^2+1}{2x^3y}dx, \frac{x^3+x^2+x+1}{2(x-1)^3y}dx, \frac{2x^3}{y}dx, \frac{(x+1)y}{x^2(x-1)^2}, \frac{y}{x^2}, \frac{y}{(x-1)^2} \right) , \end{aligned}$$

respectively. Using a computation similar to (26), it is easy to verify that

$$\begin{aligned} \tau ^*(\gamma _1)&= \tau ^* (\rho '(\nu _1)) \\&=\left[ \left( \frac{x^4+2x^2 +2x+2}{2x^3y}dx, \frac{x^4+x^3+2x^2+2x}{y}dx, \frac{(x+2)y}{x^2}\right) \right] \\&= \gamma _1 + 2\gamma _2 + 2\lambda _1 \end{aligned}$$

and that

$$\begin{aligned} \tau ^*(\gamma _2)&= \tau ^*(\rho '(\nu _2)) \\&=\left[ \left( \frac{x^3+x^2+1}{2x^3y}dx, \frac{2x^3+2}{y}dx, \frac{y}{x^2}\right) \right] \\&= \gamma _2 + 2 \lambda _0. \end{aligned}$$

Furthermore, we have seen in (24) that

$$\begin{aligned} \tau ^*(\lambda _0) = \lambda _0 \quad \text {and} \quad \tau ^*(\lambda _1) = \lambda _1 + \lambda _0. \end{aligned}$$

We finally conclude that

$$\begin{aligned} s(\tau ^*(\bar{\gamma }_1)) = s(\bar{\gamma }_1 + 2\bar{\gamma }_2) = \gamma _1 + 2\gamma _2 + 2\lambda _1 = \tau ^*(\gamma _1) = \tau ^*(s(\bar{\gamma }_1)) \end{aligned}$$

and

$$\begin{aligned} s(\tau ^*(\bar{\gamma }_2)) = s(\bar{\gamma }_2) = \gamma _2 + \lambda _1 = \tau ^*(\gamma _2 + \lambda _1) = \tau ^*(s(\bar{\gamma }_2)). \end{aligned}$$

Hence s is k[G]-linear, and the Hodge-de-Rham short exact sequence (11) splits.

Remark 4.9

The curve X defined in the previous example is isomorphic to the modular curve $X_0(22)$.

To see this, we first note that, by [6, Table  2], the modular curve $X_0(22)$ is the hyperelliptic curve of genus 2 defined by

$$\begin{aligned} y^2 = f(x) = x^6 + 2x^4 +x^3 +2x^2 +1. \end{aligned}$$

Now, $x \mapsto x-1$, $y \mapsto y$ defines an isomorphism between $X_0(22)$ and the curve defined by $y^2 = x^6 + 2x^4 + 2x^2 + 2$. We finally apply the isomorphism described in the following general procedure.

If $g(x) = a_sx^s + \cdots + a_0$ with $a_0 \ne 0 \ne a_s$, we define $g^*(x) := a_0^{-1}x^sg\left( \frac{1}{x}\right)$. It is stated after Lemma 2.6 in [6] that, if $y^2 = g(x)$ defines a hyperelliptic curve and s is even, then the curves defined by $y^2=g(x)$ and $y^2 = g^*(x)$ are isomorphic.