I've been looking at Tate's Algebraic Cycles and Poles of Zeta Functions (hard to find online… Google books outline here) and have a question about his work on (conjecturing!) the Tate conjecture for the Fermat variety $X_m^r$, defined by the equation
$$ X_0^m + X_1^m + \cdots + X_r^m = 0$$
over a field $\mathbb{F}_q$ of characteristic $p$.
Fix some $i$, $0\leq i\leq m$. He starts by saying there is only one non-trivial dimension to consider, namely $r = 2i+1$. This is because, combining the Veronese embedding with the Lefschetz hyperplane theorem, we have that every non-middle cohomology is 0 or 1-dimensional, and so the cycle class map is surjective (i.e., the Tate conjecture holds) trivially. Now, if some power of $p$ is congruent to -1 modulo $m$, then because the map $X_m^r \to X_{q+1}^r$ given by $X_j \to X_j^{(q+1)/m}$ is dominant, we can (without loss of generality) assume $m = q+1$.
The benefit to doing this is because we now have a large group of automorphisms: the maps induced by the maps in the group $U$ of projective transformations
$$ X_j \to \sum a_{ij}X_i $$
where $(a_{ij})$ is a matrix over $\mathbb{F}_{q^2}$ which is unitary with respect to the map $a\to a^q$.
Now he writes that he and John Thompson proved the representation of $U$ on $H^{2i}(\overline{X})$ (where $\overline{X} = X\times_{\mathbb{F}_q} \overline{\mathbb{F}_q}$) decomposes as (the direct sum of) the trivial representation and an irreducible representation, and the desired result follows easily from this.
There is a natural action, applying the automorphisms of $U$ to $\overline{X}$, but I'm not entirely sure how to get a hold of $U$. I'm guessing once I have an idea of how to get my hands on the representation, the rest should follow without too much trouble.
Edit: Some potential progress!
We can decompose the representation of $U$ on $H^{2i}(\overline{X})$ without actually computing it (though I'd still like to hear about how!). Indeed, the action of $U$ on $H^{2i}(\overline{X})$ is transitive, so there are only two conjugacy classes and hence the representation decomposes as the trivial representation and a non-trivial (irreducible) representation.
From here, we may find the eigenvalues of the $q^2$ Frobenius. From the Weil conjectures, we know they are of the form $\zeta q^w$ for some $\zeta$ with aboslute value 1, and integer $w$. Further, Weil computed the Zeta function of the Fermat variety $X_m^r$ (see Number of Solutions of Equations in Finite Fields), and showed the eigenvalues are all Jacobi sums, and combined with the Weil conjectures, we can find (since the characters are nice) that each of these must be $\pm q^i$ (recall, we only need to consider the middle cohomology). Thus, the classes of the (appropriately twisted) middle cohomology are all algebraic, i.e., if there are "enough" cycles, every class in the appropriate cohomology could be the image of a cycle.
On the other hand, to see there are (enough) cycles, one can simply show the cycles don't only map into the trivial part of the representation of $U$. This is done in the answer here .
I would still be happy to have input on approaches that work more closely with $U$. Or corrections to the above, if I've missed something.
Best Answer
It turns out that the Tate conjecture for Fermat varieties follows easily from the fact that the middle cohomology is spanned by algebraic cycles. Indeed, if $X$ is a Fermat variety of dimension $r$ over a field $\mathbb{F}_q$ of characteristic $p$, then the middle cohomology $H^{2r-2}(X)$ is spanned by the cycles of the form
$$ \sum_{i=1}^r a_i X_i^{q^r-1}, $$
where the $a_i$ are elements of $\mathbb{F}_{q^r}$.
To see this, note that the action of $U$ on $\overline{X}$ (where $\overline{X} = X\times_{\mathbb{F}_q}\overline{\mathbb{F}_q}$) decomposes as the direct sum of the trivial representation and an irreducible representation. Thus, the classes of the (appropriately twisted) middle cohomology are all algebraic, i.e., if there are "enough" cycles, every class in the appropriate cohomology could be the image of a cycle.
On the other hand, to see there are (enough) cycles, one can simply show the cycles don't only map into the trivial part of the representation of $U$. This is done in the answer here.
Thus, we conclude that every class in $H^{2r-2}(X)$ is algebraic, and hence the Tate conjecture holds for $X$.