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? 



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


Note that Shioda et al have studied cycles on general Fermat hypersurfaces and
verified the Tate conjecture in many (all?) cases.


**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.


