The Stacks project

Theorem 63.16.1. Let $k$ be a finite field and $X$ a finite type, separated scheme of dimension at most 1 over $k$. Let $\Lambda $ be a finite ring whose cardinality is prime to that of $k$, and $K\in D_{ctf}(X, \Lambda )$. Then

63.16.1.1
\begin{equation} \label{trace-equation-trace-formula-again} \text{Tr}(\pi _ X^* |_{R\Gamma _ c(X_{\bar k}, K)}) = \sum \nolimits _{x\in X(k)} \text{Tr}(\pi _ x |_{K_{\bar x}}) \end{equation}

in $\Lambda ^{\natural }$.

Proof of Theorem 63.16.1. The proof proceeds in a number of steps.

Step 1. Let $j : \mathcal{U}\hookrightarrow X$ be an open immersion with complement $Y = X - \mathcal{U}$ and $i : Y \hookrightarrow X$. Then $T''(X, K) = T''(\mathcal{U}, j^{-1} K)+ T''(Y, i^{-1}K)$ and $T'(X, K) = T'(\mathcal{U}, j^{-1} K)+ T'(Y, i^{-1}K)$.

This is clear for $T'$. For $T''$ use the exact sequence

\[ 0\to j_!j^{-1} K \to K \to i_* i^{-1} K \to 0 \]

to get a filtration on $K$. This gives rise to an object $\widetilde K \in DF(X, \Lambda )$ whose graded pieces are $j_!j^{-1}K$ and $i_*i^{-1}K$, both of which lie in $D_{ctf}(X, \Lambda )$. Then, by filtered derived abstract nonsense (INSERT REFERENCE), $R\Gamma _ c(X_{\bar k}, K)\in DF_{perf}(\Lambda )$, and it comes equipped with $\pi _ x^*$ in $DF_{perf}(\Lambda )$. By the discussion of traces on filtered complexes (INSERT REFERENCE) we get

\begin{eqnarray*} \text{Tr}(\pi _ X^* |_{R\Gamma _ c(X_{\bar k}, K)}) & = & \text{Tr}(\pi _ X^* |_{R\Gamma _ c(X_{\bar k}, j_!j^{-1}K)}) + \text{Tr}(\pi _ X^* |_{R\Gamma _ c(X_{\bar k}, i_*i^{-1}K)}) \\ & = & T”(U, i^{-1}K) + T”(Y, i^{-1}K). \end{eqnarray*}

Step 2. The theorem holds if $\dim X\leq 0$.

Indeed, in that case

\[ R\Gamma _ c(X_{\bar k}, K) = R\Gamma (X_{\bar k}, K) = \Gamma (X_{\bar k}, K) = \bigoplus \nolimits _{\bar x\in X_{\bar k}} K_{\bar x} \leftarrow \pi _ X*. \]

Since the fixed points of $\pi _ X: X_{\bar k}\to X_{\bar k}$ are exactly the points $\bar x\in X_{\bar k}$ which lie over a $k$-rational point $x\in X(k)$ we get

\[ \text{Tr}\big (\pi _ X^*|_{R\Gamma _ c(X_{\bar k}, K)}\big ) = \sum \nolimits _{x\in X(k)}\text{Tr}(\pi _{\bar x}|_{K_{\bar x}}). \]

Step 3. It suffices to prove the equality $T'(\mathcal{U}, \mathcal{F}) = T''(\mathcal{U}, \mathcal{F})$ in the case where

  • $\mathcal{U}$ is a smooth irreducible affine curve over $k$,

  • $\mathcal{U}(k) = \emptyset $,

  • $K=\mathcal{F}$ is a finite locally constant sheaf of $\Lambda $-modules on $\mathcal{U}$ whose stalk(s) are finite projective $\Lambda $-modules, and

  • $\Lambda $ is killed by a power of a prime $\ell $ and $\ell \in k^*$.

Indeed, because of Step 2, we can throw out any finite set of points. But we have only finitely many rational points, so we may assume there are none1. We may assume that $\mathcal{U}$ is smooth irreducible and affine by passing to irreducible components and throwing away the bad points if necessary. The assumptions of $\mathcal{F}$ come from unwinding the definition of $D_{ctf}(X, \Lambda )$ and those on $\Lambda $ from considering its primary decomposition.

For the remainder of the proof, we consider the situation

\[ \xymatrix{ \mathcal{V} \ar[d]_ f \ar[r] & Y \ar[d]^{\bar f} \\ \mathcal{U} \ar[r] & X } \]

where $\mathcal{U}$ is as above, $f$ is a finite étale Galois covering, $\mathcal{V}$ is connected and the horizontal arrows are projective completions. Denoting $G=\text{Aut}(\mathcal{V}|\mathcal{U})$, we also assume (as we may) that $f^{-1}\mathcal{F} =\underline M$ is constant, where the module $M = \Gamma (\mathcal{V}, f^{-1}\mathcal{F})$ is a $\Lambda [G]$-module which is finite and projective over $\Lambda $. This corresponds to the trivial monoid extension

\[ 1\to G\to \Gamma = G \times \mathbf{N}\to \mathbf{N}\to 1. \]

In that context, using the reductions above, we need to show that $T''(\mathcal{U}, \mathcal{F}) = 0$.

Step 4. There is a natural action of $G$ on $f_*f^{-1}\mathcal{F}$ and the trace map $f_*f^{-1}\mathcal{F}\to \mathcal{F}$ defines an isomorphism

\[ (f_*f^{-1}\mathcal{F})\otimes _{\Lambda [G]} \Lambda = (f_*f^{-1}\mathcal{F})_ G \cong \mathcal{F}. \]

To prove this, simply unwind everything at a geometric point.

Step 5. Let $A = \mathbf{Z}/\ell ^ n \mathbf{Z}$ with $n\gg 0$. Then $f_*f^{-1}\mathcal{F} \cong (f_*\underline A) \otimes _{\underline A} \underline M$ with diagonal $G$-action.

Step 6. There is a canonical isomorphism $(f_*\underline A \otimes _{\underline A} \underline M) \otimes _{\Lambda [G]} \underline\Lambda \cong \mathcal{F}$.

In fact, this is a derived tensor product, because of the projectivity assumption on $\mathcal{F}$.

Step 7. There is a canonical isomorphism

\[ R\Gamma _ c(\mathcal{U}_{\bar k}, \mathcal{F}) = (R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)\otimes _ A^\mathbf {L} M)\otimes _{\Lambda [G]}^\mathbf {L} \Lambda , \]
compatible with the action of $\pi ^*_\mathcal {U}$.

This comes from the universal coefficient theorem, i.e., the fact that $R\Gamma _ c$ commutes with $\otimes ^\mathbf {L}$, and the flatness of $\mathcal{F}$ as a $\Lambda $-module.

We have

\begin{eqnarray*} \text{Tr}( \pi _\mathcal {U}^* |_{R\Gamma _ c(\mathcal{U}_{\bar k}, \mathcal{F})}) & = & {\sum _{g \in G}}' \text{Tr}_{\Lambda }^{Z_ g} \left( (g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)\otimes _ A^\mathbf {L} M} \right) \\ & = & {\sum _{g\in G}}' \text{Tr}_ A^{Z_ g} ( (g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)} ) \cdot \text{Tr}_\Lambda (g|_ M) \end{eqnarray*}

where $\Gamma $ acts on $R\Gamma _ c(\mathcal{U}_{\bar k}, \mathcal{F})$ by $G$ and $(e, 1)$ acts via $\pi _\mathcal {U}^*$. So the monoidal extension is given by $\Gamma = G \times \mathbf{N} \to \mathbf{N}$, $\gamma \mapsto 1$. The first equality follows from Lemma 63.15.9 and the second from Lemma 63.15.8.

Step 8. It suffices to show that $\text{Tr}_ A^{Z_ g}((g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)}) \in A$ maps to zero in $\Lambda $.

Recall that

\begin{eqnarray*} \# Z_ g \cdot \text{Tr}_ A^{Z_ g}((g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)}) & = & \text{Tr}_ A((g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)})\\ & = & \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(\mathcal{V}_{\bar k}, A)}). \end{eqnarray*}

The first equality is Lemma 63.15.7, the second is the Leray spectral sequence, using the finiteness of $f$ and the fact that we are only taking traces over $A$. Now since $A=\mathbf{Z}/\ell ^ n\mathbf{Z}$ with $n \gg 0$ and $\# Z_ g = \ell ^ a$ for some (fixed) $a$, it suffices to show the following result.

Step 9. We have $\text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(\mathcal{V}, A)}) = 0$ in $A$.

By additivity again, we have

\begin{eqnarray*} & \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(\mathcal{V}_{\bar k} A)}) + \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(Y-\mathcal{V})_{\bar k}, A)}) \\ & = \text{Tr}_ A((g^{-1}\pi _ Y)^* |_{R\Gamma (Y_{\bar k}, A)}) \end{eqnarray*}

The latter trace is the number of fixed points of $g^{-1}\pi _ Y$ on $Y$, by Weil's trace formula Theorem 63.14.4. Moreover, by the 0-dimensional case already proven in step 2,

\[ \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^*|_{R\Gamma _ c(Y-\mathcal{V})_{\bar k}, A)}) \]

is the number of fixed points of $g^{-1}\pi _ Y$ on $(Y-\mathcal{V})_{\bar k}$. Therefore,

\[ \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(\mathcal{V}_{\bar k}, A)}) \]

is the number of fixed points of $g^{-1}\pi _ Y$ on $\mathcal{V}_{\bar k}$. But there are no such points: if $\bar y\in Y_{\bar k}$ is fixed under $g^{-1}\pi _ Y$, then $\bar f(\bar y) \in X_{\bar k}$ is fixed under $\pi _ X$. But $\mathcal{U}$ has no $k$-rational point, so we must have $\bar f(\bar y)\in (X-\mathcal{U})_{\bar k}$ and so $\bar y\notin \mathcal{V}_{\bar k}$, a contradiction. This finishes the proof. $\square$

[1] At this point, there should be an evil laugh in the background.

Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03UG. Beware of the difference between the letter 'O' and the digit '0'.