# Is it easy to compute the action of an endomorphism on $H^1$

Let $X_0$ be a smooth projective geometrically connected curve over $\mathbf{F}_p$ and let $X = X_0 \otimes_{\mathbf{F}_p} \overline{\mathbf{F}_p}$.

Assume that the genus $g\geq 2$.

Let $f:X\to X$ be an endomorphism. Then $f$ is (EDIT) bijective if (EDIT) $f$ is non-constant.

How can one determine the action of $f$ on $H^1(X_{et},\mathbf{Q}_\ell)$ computationally?

Explicitly:

Consider the curve $X$ given by the equation $x^{100}+y^{100} = z^{100}$. Let $f:X\to X$ be defined as $(x:y:z)\mapsto (\zeta_{100} x: y: \zeta_{100} z)$, where $\zeta_{100}^{100} = 1$.

It is easy to determine the number of fixed points with the trace formula, but I need more. I need a $(2g\times 2g)$-matrix over $\overline{\mathbf{Q}_\ell}$ giving the action of $f$ on $H^1$.

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Can't you use comparison theorems to reduce the problem to the lift of your curve over the Witt vectors, and then in turn reduce the problem to the associated complex curve? For a complex curve, you can easily work out the associated action on the two Hodge groups (on the "complexification" of the cohomology). – Jason Starr Mar 8 '12 at 13:07
Ahh that's a good idea. So we reduce to the case of complex manifolds basically. Then we simply need to determine the action on $H^{0,1}$ and $H^{1,0}$. What is the general strategy for working out this action? (I'm not asking you to work out a concrete example. Just to give the idea if possible...) – Harized Mar 8 '12 at 13:20
It is not true that endomorphisms are always automorphisms in positive characteristic. E.g. the Frobenius. – Felipe Voloch Mar 8 '12 at 15:14
you're right. they're all bijective though... – Harized Mar 8 '12 at 16:00
The composition of the structure map $X \to Spec k$ with any $k$-point $\Spec k \to X$ is an example of a non-bijective endomorphism. – Sasha Mar 9 '12 at 6:57

edit: this assumes that $X$ and its endomorphism can be lifted to characteristic zero.

A natural way would be to compute the action of $G$ on $H^0(X,\Omega)$. If $X$ is given in coordinates as in your example it seems natural to do this by trying to find a basis of differentials also in coordinates, which makes everything really explicit. Serre duality then gives you all of $H^1$. For hyperelliptic curves, exactly this computation appears in a paper of Getzler (but it is surely classical), http://arxiv.org/abs/math/9909171, section 3.

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$H^1(X,O_X)$ is a vector space in char. p and of dimension half of $H^1(X,Q_{\ell})$. It does provide some info but not all of it. One can do the computation on the obvious lift of the curve to char 0 (as Jason suggested) but you still only get half of it. – Felipe Voloch Mar 8 '12 at 17:03

Do the computation in the deRham cohomology (differentials of second kind modulo exact differentials) in the obvious characteristic zero lift of your curve, as suggested by Jason. A good reference is Lang's book "introduction to algebraic and abelian function". There he does the general theory (see ch I sec. 8, in particular) and then he does the Fermat curves in great detail.

EDIT since apparently people seem to be making a big deal out of this, using the basis $\omega_{rs}$ given in Lang ch II, it immediately follows that the matrix is diagonal with the 99 non-trivial 100th-roots of unity occurring as eigenvalues with multiplicity 98.

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Here's a method for computing the action on the integral homology of the associated complex curve $x^{100} + y^{100} = z^{100}$; from this you can derive computations with other coefficients as suggested above.

The composite $[x:y:z] \mapsto [x^{100}:y^{100}:z^{100}] \mapsto [x^{100}:z^{100}]$ exhibits your curve as a cover of $\mathbb P^1$ ramified only over $0$, $1$, and $\infty$. Such covers are classified by the fundamental group of $\mathbb P^1 \setminus \{0,1,\infty\}$, and such (finite) covers are in bijective correspondence with finite sets with a right action of the free group on two generators $F$ (an application of the Riemann existence theorem). I'll write $F = \langle a,b,c | abc = 1 \rangle$ where $a$ is a curve giving monodromy around $0$, $b$ around $1$, and $c$ around $\infty$. Your particular cover comes from $H \backslash F$ where $H$ is the kernel of the homomorphism $F \to \mathbb{Z}/100 \times \mathbb{Z}/100$ sending $a$ to $(1,0)$ and $b$ to $(0,1)$, and the group you're interesting in acting by is a group of deck transformations via monodromy around $\infty$ (so by $(-1,-1)$).

So let's suppose you have a curve classified by $H \backslash F$ and let's compute its homology in an $NH/H$-equivariant fashion.

Take the preimages of $0,1,\infty$; of the real intervals $(-\infty,0),(1,\infty)$; of the open set which is left, which is homeomorphic to an open disc. Taking preimages of these gives you a cell structure on your curve, with three $F$-orbits of points, two $F$-orbits of edges, and one $F$-orbit of 2-cells. Some computation with how edges are glued together allows you to describe the cell structure on your curve as follows:

• Points $F \cdot p$, $F \cdot q$, $F \cdot r$ where $p, q, r$ have stabilizers generated by $a$, $b$, and $c$ respectively

• Edges $F \cdot \ell$, $F \cdot \ell'$ where $\ell$ is an edge from $p$ to $br$, $m$ is an edge from $q$ to $r$

• Two-cells $F \cdot u$, where the boundary of $u$ is attached by the path $m ({}^a \ell)^{-1} \ell ({}^b m)^{-1}$

So the homology is computed $NH/H$-equivariantly by a chain complex \begin{align*} \mathbb Z [H \backslash F] \cdot u &\to \mathbb Z [H \backslash F] \cdot \ell \times \mathbb Z [H \backslash F] \cdot m\\ &\to \mathbb Z [H \backslash F / \langle a \rangle] \cdot p \times \mathbb Z [H \backslash F / \langle b \rangle] \cdot q \times \mathbb Z [H \backslash F / \langle c \rangle] \cdot r \end{align*} Here the boundary of $u$ is $(1-b)m + (1-a)\ell$, the boundary of $\ell$ is $br-p$, and the boundary of $m$ is $r-q$.

If you want cohomology, take Hom out.

This leaves the difficult - but mechanical - process of computing the homology groups with the action of the specific generator that you've listed.

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