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?