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martyall
martyall
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My question came up while reading this article by Nicholas Katz, specifically lemma 4.2. I don't think it's necessary to read the article to answer the question, but I'm including it anyways for reference.

Suppose we have a smooth, affine, geometrically integral curve $U_0/\mathbb{F}_{q}$, and let $U/\overline{\mathbb{F}}_{q}$ be it's base change. Then we have the following exact sequence:

$$1 \rightarrow \pi_{1}(U) \rightarrow \pi_{1}(U_{0}) \rightarrow Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \rightarrow 1$$

One has the frobenius $Frob_{q}\in Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q})$, which acts on $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ by transport of structure.

In this proof, we know that $Frob_{q}$ does not have $1$ as an eigenvalue, so that it operates without fixed points. To prove his result he derives a contradiction by constructing a nontrivial continuous additive homomorphism $f:\pi(U_{0})\rightarrow \overline{\mathbb{Q}}_{\ell}$, and claims that by restriction to $\pi_{1}(U)$ we get a fixed point in $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$.

I'm aware of the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$, and as far as I can tell the proof of the claim would go like this. Use the homomorphism $Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \hookrightarrow Out(\pi_{1}(U))/Inn(\pi_{1}(U))$$Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \hookrightarrow Aut(\pi_{1}(U))/Inn(\pi_{1}(U))$ to get an action of $Frob_{q}$ on $\pi_{1}(U)$. For $\phi\in Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})$ define $(Frob_{q}\cdot\phi)(a) = \phi(a^{Frob_{q}})$. Because our $f$ mentioned above is restricted from $\pi_{1}(U_{0})$ where it also factors through $\pi_{1}(U_{0})^{ab}$, $f$ is indeed invariant under this action.

Essentially what I don't see is why the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ commutes with the action of $Frob_{q}$, which seems to be necessary. I only know the above isomorphism through a chain of fairly non-constructive isomorphisms-- something along the lines of what's mentioned in 11.3 of Milne's notes. However, it seems unlikely that something like this wouldn't turn out to be true. Can anyone explain why?

My question came up while reading this article by Nicholas Katz, specifically lemma 4.2. I don't think it's necessary to read the article to answer the question, but I'm including it anyways for reference.

Suppose we have a smooth, affine, geometrically integral curve $U_0/\mathbb{F}_{q}$, and let $U/\overline{\mathbb{F}}_{q}$ be it's base change. Then we have the following exact sequence:

$$1 \rightarrow \pi_{1}(U) \rightarrow \pi_{1}(U_{0}) \rightarrow Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \rightarrow 1$$

One has the frobenius $Frob_{q}\in Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q})$, which acts on $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ by transport of structure.

In this proof, we know that $Frob_{q}$ does not have $1$ as an eigenvalue, so that it operates without fixed points. To prove his result he derives a contradiction by constructing a nontrivial continuous additive homomorphism $f:\pi(U_{0})\rightarrow \overline{\mathbb{Q}}_{\ell}$, and claims that by restriction to $\pi_{1}(U)$ we get a fixed point in $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$.

I'm aware of the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$, and as far as I can tell the proof of the claim would go like this. Use the homomorphism $Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \hookrightarrow Out(\pi_{1}(U))/Inn(\pi_{1}(U))$ to get an action of $Frob_{q}$ on $\pi_{1}(U)$. For $\phi\in Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})$ define $(Frob_{q}\cdot\phi)(a) = \phi(a^{Frob_{q}})$. Because our $f$ mentioned above is restricted from $\pi_{1}(U_{0})$ where it also factors through $\pi_{1}(U_{0})^{ab}$, $f$ is indeed invariant under this action.

Essentially what I don't see is why the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ commutes with the action of $Frob_{q}$, which seems to be necessary. I only know the above isomorphism through a chain of fairly non-constructive isomorphisms-- something along the lines of what's mentioned in 11.3 of Milne's notes. However, it seems unlikely that something like this wouldn't turn out to be true. Can anyone explain why?

My question came up while reading this article by Nicholas Katz, specifically lemma 4.2. I don't think it's necessary to read the article to answer the question, but I'm including it anyways for reference.

Suppose we have a smooth, affine, geometrically integral curve $U_0/\mathbb{F}_{q}$, and let $U/\overline{\mathbb{F}}_{q}$ be it's base change. Then we have the following exact sequence:

$$1 \rightarrow \pi_{1}(U) \rightarrow \pi_{1}(U_{0}) \rightarrow Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \rightarrow 1$$

One has the frobenius $Frob_{q}\in Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q})$, which acts on $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ by transport of structure.

In this proof, we know that $Frob_{q}$ does not have $1$ as an eigenvalue, so that it operates without fixed points. To prove his result he derives a contradiction by constructing a nontrivial continuous additive homomorphism $f:\pi(U_{0})\rightarrow \overline{\mathbb{Q}}_{\ell}$, and claims that by restriction to $\pi_{1}(U)$ we get a fixed point in $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$.

I'm aware of the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$, and as far as I can tell the proof of the claim would go like this. Use the homomorphism $Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \hookrightarrow Aut(\pi_{1}(U))/Inn(\pi_{1}(U))$ to get an action of $Frob_{q}$ on $\pi_{1}(U)$. For $\phi\in Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})$ define $(Frob_{q}\cdot\phi)(a) = \phi(a^{Frob_{q}})$. Because our $f$ mentioned above is restricted from $\pi_{1}(U_{0})$ where it also factors through $\pi_{1}(U_{0})^{ab}$, $f$ is indeed invariant under this action.

Essentially what I don't see is why the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ commutes with the action of $Frob_{q}$, which seems to be necessary. I only know the above isomorphism through a chain of fairly non-constructive isomorphisms-- something along the lines of what's mentioned in 11.3 of Milne's notes. However, it seems unlikely that something like this wouldn't turn out to be true. Can anyone explain why?

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martyall
martyall
  • 21
  • 2

Why is this frobenius acting like that frobenius?

My question came up while reading this article by Nicholas Katz, specifically lemma 4.2. I don't think it's necessary to read the article to answer the question, but I'm including it anyways for reference.

Suppose we have a smooth, affine, geometrically integral curve $U_0/\mathbb{F}_{q}$, and let $U/\overline{\mathbb{F}}_{q}$ be it's base change. Then we have the following exact sequence:

$$1 \rightarrow \pi_{1}(U) \rightarrow \pi_{1}(U_{0}) \rightarrow Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \rightarrow 1$$

One has the frobenius $Frob_{q}\in Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q})$, which acts on $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ by transport of structure.

In this proof, we know that $Frob_{q}$ does not have $1$ as an eigenvalue, so that it operates without fixed points. To prove his result he derives a contradiction by constructing a nontrivial continuous additive homomorphism $f:\pi(U_{0})\rightarrow \overline{\mathbb{Q}}_{\ell}$, and claims that by restriction to $\pi_{1}(U)$ we get a fixed point in $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$.

I'm aware of the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$, and as far as I can tell the proof of the claim would go like this. Use the homomorphism $Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \hookrightarrow Out(\pi_{1}(U))/Inn(\pi_{1}(U))$ to get an action of $Frob_{q}$ on $\pi_{1}(U)$. For $\phi\in Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})$ define $(Frob_{q}\cdot\phi)(a) = \phi(a^{Frob_{q}})$. Because our $f$ mentioned above is restricted from $\pi_{1}(U_{0})$ where it also factors through $\pi_{1}(U_{0})^{ab}$, $f$ is indeed invariant under this action.

Essentially what I don't see is why the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ commutes with the action of $Frob_{q}$, which seems to be necessary. I only know the above isomorphism through a chain of fairly non-constructive isomorphisms-- something along the lines of what's mentioned in 11.3 of Milne's notes. However, it seems unlikely that something like this wouldn't turn out to be true. Can anyone explain why?