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!


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 $\chi(X)$.

  • 1
    $\begingroup$ 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 $\endgroup$
    – marlu
    Jun 11 '20 at 1:57
  • $\begingroup$ @marlu It's a special case of Example 4.6 of arxiv.org/pdf/1405.6381.pdf (let $f=\phi$ and $C=\Gamma_{\phi}$). $\endgroup$
    – user796022
    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!

  • $\begingroup$ Thanks!!!!!!!!!! $\endgroup$
    – user796022
    Jun 12 '20 at 1:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.