# Tate conjecture for Fermat varieties

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.

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Have you looked at Shioda and Katsura's paper projecteuclid.org/download/pdf_1/euclid.tmj/1178229881 and Ulmer's paper arxiv.org/pdf/math/0109163.pdf ? – lee Dec 25 '15 at 18:55
Also, I know very little about Tate's conjecture about cycles of arbitrary codimension. But isn't it true that if X dominant Y of same dimension and T(r) holds for X would imply T(r) holds for Y? – lee Dec 25 '15 at 18:57
@lee I have looked Shioda-Katsura before (long time ago) when I was first coming to terms with everything, so it was actually quite nice to look at it again! I have not seen Ulmer's paper before, but a quick gloss makes it look like they do not address the residual representation, which is what I was specifically interested in with this question. – Alex Dec 28 '15 at 18:50
It is possible Tate's claim actually goes through some of the methods used in the papers, but I was curious as to whether there was something about the representation, not coming from the geometry, that was particularly simple (since it only got a comment and I did not find it in any other paper's by Tate). – Alex Dec 28 '15 at 18:52
I thought in the situation that X dominant Y, then you can express chow group of Y as a subspace of chow group of X, the same for etale cohomology. So then you will get that T(r) for X imply T(r) for Y. And we know that Fermat Varieties are dominated by product of Fermat curves, and Tate's conjecture holds for product of curves. Then we'd conclude that T holds for Fermat Varieties. But maybe I was wrong... – lee Dec 30 '15 at 15:34