Computing the trace of the following automorphism of the elliptic curve $y^2 = x^3+x$ Consider the elliptic curve $E$ defined by $y^2z= x^3 +xz^2$ over an algebraic closure $\overline{\mathbf{Q}}$ of $\mathbf{Q}$. Consider the endomorphism $f:E\to E$ given by $(x:y:z)\mapsto (-x:iy:z)$. Note that $f$ has precisely two fixed points: $(0:0:1)$ and $(0:1:0)$. 
How do I compute the trace of $f$ on the (etale) of $E$? More precisely, there are three cohomology groups: $H^0$, $H^1$ and $H^2$. 
If I'm right, $H^0$ and $H^2$ are $1$-dimensional. So in each case the action is given by the multiplication of some number. What are these numbers? (Note that $f$ is an automorphism. Maybe this helps?)
The $H^1$ is two-dimensional. How do I find the action of $f$ on $H^1$. (It's given by a two by two matrix.)
My motivation for this question is the trace formula. I want to see examples! I plan on working out more examples with CM elliptic curves. Then maybe some higher genus curves and finally some higher-dimensional varieties.
 A: An endomorphism always acts trivially on $H^0$, and on $H^2$ it acts as multiplication by the degree, hence again trivially in your case.
To compute the action on $H^1$, of course you can use the trace formula,
together with your fixed point computation, to see that trace is zero.
Since $H^2 = \wedge^2 H^1$, the determinant on $H^1$ is the previously computed
action on $H^2$.  Thus $f$ acts on $H^1$ via a matrix of order dividing $4$,
trace zero, and determinant $1$.  This pins it down pretty well.
Of course, you would like to compute the action on $H^1$ in an a priori way,
so as to verify the trace formula.
To this end,
note that $f^2 = [-1]$, and so your elliptic curve $E$ is endowed with an action of $\mathbb Z[i]$.  The $\ell$-adic $H^1$ is then a free module of rank one over $\mathbb Z_{\ell}[i]$.  (See the discussion of the endomorphism action on
$\ell$-adic Tate modules in Silverman for a proof of the analogous fact
for the $\ell$-adic Tate-module, and note that $H^1$ is dual to the $\ell$-adic Tate module.)
Thus the action of $f$ on $H^1$ has the same matrix as the action of $i$ on
$\mathbb Z_{\ell}[i]$, which is $\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}.$  Thus we get a direct proof that it acts via a matrix of determinant $1$ and trace $0$.

More generally, if $E$ has CM by $\mathcal O$, then the $\ell$-adic $H^1$
is free over $\mathbb Z_{\ell}\otimes_{\mathbb Z} \mathcal O$.  Thus the characteristic polynmomial of an element of $\alpha$ acting on $H^1$ is the same
as the degree $2$ monic polynomial over $\mathbb Z$ that it satisfies, and its trace on $H^1$ is the same as its trace as an algebraic integer in $\mathcal O$.  This is discussed in the chapter on endomorphisms in Silverman, and is a basic ingredient in the theory of CM elliptic curves (as discussed e.g. in Silverman 2).
