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 $H^1$.
.
 A: 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.
A: 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.
A: 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.
