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How do we deduce the following statement from the Chebotarev density theorem? The statement is from Ngo's Fundamental Lemma paper.

Let $U$ be a scheme of finite type over $\mathbb{F}_q$. Let $\mathcal{L}_1$, $\mathcal{L}_2$ be two local systems of pure weight $i$. Suppose for any $u \in U$ and $k \in \mathbb{N}$, we have $Tr(\sigma_u^k,\mathcal{L}_1)=Tr(\sigma_u^k,\mathcal{L}_2)$, where $\sigma_u$ is the Frobenius conjugacy class of $u$ in $\pi_1(U)$. Then $\mathcal{L}_1$ and $\mathcal{L}_2$ are isomorphic up to semisimplification.

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    $\begingroup$ I guess $U$ has to be normal for this to hold? Otherwise you can take the constant local system of rank $r$ on $\mathbf{P}^1$ and identify the fibers over $0$ and $\infty$ using any matrix in $\mathrm{GL}_r(\overline{\mathbf{Q}}_\ell)$, obtaining a local system over a nodal curve $C$. For $r=1$ the resulting local systems will be semisimple and will violate your statement. $\endgroup$
    – Piotr Achinger
    Commented Mar 23, 2022 at 21:36

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As Piotr says, we must assume $U$ normal. The purity assumption is not needed.

There are two steps to this proof

(1) Suppose for any $u \in U$ and $k \in \mathbb{N}$, we have $Tr(\sigma_u^k,\mathcal{L}_1)=Tr(\sigma_u^k,\mathcal{L}_2)$, where $\sigma_u$ is the Frobenius conjugacy class of $u$ in $\pi_1(U)$. Then for any $\sigma \in \pi_1(U)$, we have $Tr(\sigma^k,\mathcal{L}_1)=Tr(\sigma^k,\mathcal{L}_2)$

(2) Suppose that for any $\sigma \in \pi_1(U)$, we have $Tr(\sigma^k,\mathcal{L}_1)=Tr(\sigma^k,\mathcal{L}_2)$ Then $\mathcal{L}_1$ and $\mathcal{L}_2$ are isomorphic up to semisimplification.

The first one follows from the Chebotarev density theorem. Because the traces are $\ell$-adic numbers, it suffices to prove the identity mod $\ell^n$ for each $n$. Whether this is true or not for a given $\sigma$ depends only on the image of $\sigma$ in some finite group (a subgroup of $GL_m (\mathbb Z/\ell^n\mathbb Z)^2$ where $m$ is the rank of these representations).

But in a finite quotient of a Galois group (here is where we use normalcy, to get that the fundamental group is a quotient of a Galois group), every element arises from a Frobenius conjugacy class. Since we have the identity on every Frobenius element, we have it on every element.

The second one is a general fact about representations of groups over fields of characteristic zero. In particular, we do not use the $\ell$-adic topology, so we may assume the base field is $\mathbb C$. Let $G$ be the Zariski closure of the image of $\pi_1$ in $GL_m \times GL_m$ acting on the two (semisimplified) representations. Because $G$ has a faithful semisimple representation, it is reductive. Take a maximal compact subgroup. If two representations are isomorphic on restriction to this compact subgroup they are isomorphic (Weyl). Now use character theory on this compact subgroup.

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  • $\begingroup$ Could you elaborate on the use of being normal? More specifically, what is the Galois group you are referring to in "the fundamental group is a quotient of a Galois group". $\endgroup$
    – userabc
    Commented Mar 25, 2022 at 3:52
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    $\begingroup$ @userabc For $X$ a variety, there is a natural map from the Galois group of $k(X)$ to the etale fundamental group of $X$ (the induced map on etale fundamental groups from the inclusion of the generic point). If $X$ is normal, this map is a surjection. (A map of fundamental groups is surjective iff the pullback map on covers sends connected covers to connected covers. For $X$ normal, a cover of $X$ is normal, so if it is connected, it is necessarily irreducible, and thus remains connected on pullback to the generic point.) $\endgroup$
    – Will Sawin
    Commented Mar 25, 2022 at 13:35

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