An intersection calculation over a finite field

Question:

I made a calculation that must be wrong, but am having trouble spotting the error. Which steps below are invalid? Thank you in advance for your attention!

Setup:

Let $$p$$ and $$\ell \neq p$$ be prime numbers. Let $$X_0$$ be a geometrically connected, proper, smooth variety over $$k=\mathbb{F}_p$$. Let $$X$$ be the base change of $$X_0$$ to $$\overline{k}$$.

Let $$[Z]\in H^{2i}(X,\mathbb{Q}_{\ell}(i))$$ be the cycle class of a subvariety $$Z \subset X$$ of codimension $$i$$. Write $$H^i(-)$$ (and suppress Tate twists) for $$\ell$$-adic cohomology.

Let $$\phi_0:X_0 \rightarrow X_0$$ be the geometric frobenius endomorphism; let $$\phi:X \rightarrow X$$ be the base change of $$\phi_0$$ to $$\overline{k}$$. which is a morphism of $$\overline{k}$$-schemes that is finite of degree $$p$$.

From now on, suppose for concreteness that $$X_0$$ is a curve.

Let $$F=\text{id} \times \phi:X \times X \rightarrow X \times X$$. Then $$F$$ is finite of degree $$p$$. For $$k \geq 1$$, let $$\Gamma \subset X \times X$$ be the graph of the morphism $$F$$. Let $$\Delta$$ be the diagonal subvariety inside $$X \times X$$. We have isomorphisms

$$F^*, F_*:H^2(X \times X) \rightarrow H^2(X \times X)$$

which satisfy $$F_* \circ F^* =p\text{Id}$$.

Some equalities inside $$H^2(X \times X)$$:

1. $$F_*([\Delta])=[\Gamma]$$, because $$F$$ restricts to an isomorphism $$\Delta \xrightarrow{\sim} \Gamma$$.
2. $$F^*([\Gamma])=p[\Delta]$$. This can be checked after applying $$F_*$$; use equality 1 and the formula $$F_* \circ F^* =p\text{Id}$$.

An intersection calculation:

Let $$\cdot$$ denote the intersection pairing on $$H^2(X \times X)$$ (with values in $$\mathbb{Q}_{\ell}$$).

We have that $$[\Gamma] \cdot [\Gamma]=\sum_i (-1)^i\text{Tr}((\phi \circ \phi)^*|H^i(X,\mathbb{Q}_l))$$ (see reference in comments), and by a similar formula, $$[\Delta] \cdot [\Delta]=\chi(X)$$ (euler characteristic). Using the formulas above and the projection formula, we calculate that

$$\chi(X)=F_*([\Delta] \cdot [\Delta]) = F_*(\frac{1}{p}[\Delta] \cdot F^*([\Gamma]) = \frac{1}{p}F_*([\Delta]) \cdot [\Gamma] = \frac{1}{p} [\Gamma] \cdot [\Gamma]$$,

So $$\sum_i (-1)^i\text{Tr}((\phi \circ \phi)^*|H^i(X,\mathbb{Q}_l))=[\Gamma] \cdot [\Gamma]=p\chi(X)$$

which is not true: the left side is the cardinality of $$X_0(\mathbb{F}_{p^2})$$ and certainly does not only depend on $$\chi(X)$$.

• How do you get $[\Gamma] \cdot [\Gamma]=\sum_i (-1)^i\text{Tr}((\phi \circ \phi)^*|H^i(X,\mathbb{Q}_l))$? The right hand side equals $[\Gamma_{\phi \circ \phi}]\cdot [\Delta]$ by the trace formula Jun 11 '20 at 1:57
• @marlu It's a special case of Example 4.6 of arxiv.org/pdf/1405.6381.pdf (let $f=\phi$ and $C=\Gamma_{\phi}$). Jun 11 '20 at 2:48

The calculation of $$[\Gamma] \cdot [\Gamma] = p \chi(X)$$ is correct, the mistake is in the formula $$[\Gamma] \cdot [\Gamma]=\sum_i (-1)^i\text{Tr}((\phi \circ \phi)^*|H^i(X,\mathbb{Q}_l)).$$ You derive that by choosing $$f = \phi$$ and $$C = \Gamma$$ in Example 4.6 of the reference in the comments. Confusingly, they view the graph of a function $$f\colon X_2 \to X_1$$ as a subscheme of $$X_1 \times X_2$$, with the order of the factors switched compared to the usual definition of the graph. Let us denote that by $$\tilde \Gamma_f$$. Then Example 4.6 reads $$[\Gamma]\cdot [\tilde\Gamma_\phi] = \mathrm{Tr}(\phi^* \circ H^*([\Gamma])).$$ Here, $$H^i([\Gamma])$$ is the endomorphism $$(p_1)_* \circ (p_2)^*$$ of $$H^i(X, \mathbb{Q}_\ell)$$ with $$p_1$$ and $$p_2$$ the two projections $$X \overset{p_2}{\leftarrow} \Gamma \overset{p_1}{\to} X.$$ In 4.5(c) they show that $$H^i([\tilde \Gamma_f]) = f^*$$. Your mistake comes from identifying $$\Gamma$$ with $$\tilde \Gamma_\phi$$ and thereby $$H^i([\Gamma])$$ with $$\phi^*$$. However, $$\tilde \Gamma_\phi$$ is $$\Gamma = \Gamma_\phi$$ with the factors switched!