# How does Tate verify his own conjecture for the Fermat hypersurface?

This question is about Tate's 1963 paper "Algebraic Cycles and Poles of Zeta Functions". Here he announces a conjecture (now known as "the Tate conjecture") which states that certain classes in the cohomology of a projective variety are always explained by the existence of algebraic cycles. In the case of a variety $X/\mathbf{F}_q$ of dimension $n$, the conjecture predicts that the subspace of classes in $H^{2d}(X\otimes\overline{\mathbf{F}}_q,\mathbf{Q}_\ell)(d)$ which are Frobenius-invariant is spanned by the image of the space of algebraic cycles in $X$ of codimension $d$.

As an example, Tate gives the projective hypersurface $X$ defined by $$X_0^{q+1}+\dots X_r^{q+1} =0,$$ where $r=2i+1$ is odd. Here $X$ admits a large group of automorphisms $U$, namely those projective transformations in the $X_i$ which are unitary with respect to the semilinear form $\sum_i X_iY_i^q$. Using the Lefschetz theorem, it isn't at all hard to compute $H^{2i}(X)$ as a $U$-module: it decomposes as a trivial $U$-module and an irreducible $U$-module of dimension $q(q^r+1)/(q+1)$. And then when you attempt to compute the $q^2$-power Frobenius eigenvalues on the middle cohomology, you find (once again by Lefschetz) that each one is a Gauss sum which in this case is nothing but $\pm q$. (If there is enough demand, I can supply all these calculations here.) Thus, miraculously, all classes in $H^{2i}(X\otimes\overline{\mathbf{F}}_q,\mathbf{Q}_\ell)(i)$ are fixed by some power of Frobenius.

My question is: How did Tate confirm the existence of the necessary cycles? Surely the hyperplane section of codimension $i$ lands in the part of $H^{2i}$ which has trivial $U$-action (for $U$ translates hyperplanes to other hyperplanes, and these all cohomologically equivalent). In order to verify Tate's conjecture, all you need to do is produce a cycle in $X$ whose projection into the big $U$-irreducible part of $H^{2i}$ is nonzero. How did Tate produce this cycle? Did he lift to characteristic zero and appeal to the Hodge conjecture, or what?

I don't know how Tate did it but here is one way. Let $\zeta$ be such that $\zeta^{q+1}=-1$ and put $a_j=(0\colon\cdots\colon1\colon\zeta:\cdots\colon0)$, $j=0,\ldots,i$ with the $1$ in coordinate $2j$. Then the linear span $L$ of these points is contained in the Fermat hypersurfaces and gives a subvariety of middle dimension. I claim that its class is not a multiple of the $i+1$'st power of the hyperplane section. Indeed, transform $L$ by the automorphism permuting the two first coordinates to get $L'$. Then (assuming that $q$ is odd say) $L$ and $L'$ are disjoint so that $[L]\cdot [L']=0$ but if the class of $L$ would be a multiple of the linear subspace section this cannot be. Hence, $[L]$ projects non-trivially to the non-trivial irreducible representation.
Addendum: It is tricky to get dimensions and stuff correct so let me expand upon this in the simple case when $i=1$ and $q=2$ (this is a classical example appearing for instance on pp. 176-177 of Mumford: Algebraic Geometry I Complex projective varieties -- everything works in characteristic different from $3$). Then we get the line $(a\colon a\zeta\colon b\colon b\zeta)$ and by letting the monomal automorphisms of the surface act we get all $27$ lines on the Fermat cubic. It is of course true that the line is the intersection of the cubic and a linear subspace but the intersection is not proper so the class of $L$ is not a power of the hyper plane section.
Note incidentally that in comments (p. 180) Mumford points the special nature of characteristic $2$ for this example essentially saying that there an index $2$ subgroup of the automorphism group of the Néron-Severi lattice preserving the canonical class can be realised by automorphisms of the surface.
• Hi Torsten, I'm a bit confused. That linear span will contain points such as $(a:a\zeta+b:b\zeta:0:\cdots)$ and these do not lie on $X$ in general. I agree that Shioda et al should have all the answers; it seems that they work primarily in characteristic 0. May 13, 2011 at 6:23
• The coordinate 1 is in position $2j$ so we will have points $(a:a\zeta:b:b\zeta:\cdots)$ (the supports of the points are disjoint). May 13, 2011 at 7:04
• Ah! I should read more carefully. I retract my comment. I like how your construction requires that $r$ be odd, which is what we expect. I also like that it requires passing to the degree 2 extension, which is also what we expect! You're probably correct about this, but I'm still having trouble understanding your argument. Isn't $L$ the intersection of $X$ with the linear subspace defined by $x_{2j+1}=\zeta x_{2j}$ $(j=0,\dots,i)$, and therefore isn't $L$ necessarily a power of the hyperplane section? May 13, 2011 at 7:20