For simplicity, suppose $X$ is a smooth $n$-dimensional variety defined over $\mathbb{Q}$. Then the etale cohomology of $X$, denoted by $H^i_{\text{et}}(X,\mathbb{Q}_\ell)$, gives a representation of the Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, which is only ramified at finitely many prime numbers. Suppose $p\neq \ell$ is an unramified prime, then there is a well-defined (geometric) Frobenius action on $H^i_{\text{et}}(X,\mathbb{Q}_\ell)$. \begin{equation} \text{Fr}_p:H^i_{\text{et}}(X,\mathbb{Q}_\ell) \rightarrow H^i_{\text{et}}(X,\mathbb{Q}_\ell). \end{equation} Is there a reasonable way to define the limit $$\lim_{p \rightarrow \infty}\text{Fr}_p,$$ which acts on a cohomology space?

The thing that comes to my mind is that the completion of $\mathbb{Q}$ at the infinite prime is $\mathbb{R}$, while the Galois group $\text{Gal}(\mathbb{C}/\mathbb{R})$ is generated by the conjugation $c$, and it has an action \begin{equation} c:H^i(X(\mathbb{C}),\mathbb{R}) \rightarrow H^i(X(\mathbb{C}),\mathbb{R}). \end{equation} I don't know whether it is reasonable to say $$\lim_{p \rightarrow \infty}\text{Fr}_p=c?$$ One thing that confuses me is that the Frobenius contains lots of arithmetic information of $X$, while $c$ does not. Another thing that confuse me is that, when define the full $L$-function of $X$, at the (unramified) prime number $p$, the local $L$-factor is defined by the characteristic polynomial of the Frobenius $\text{Fr}_p$. While at the infinite prime of $\mathbb{Q}$, the local $L$-factor is given by the Gamma function, which (to me) does not look like the "characteristic polynomial" of the complex conjugation. (Even though complex conjugation plays an important part in the definition of the $L$-factor at infinite prime.)

  • 5
    $\begingroup$ To whatever extent there is a Frobenius at the infinite prime, it doesn't make sense to consider it a limit of the finite Frobeniuses: In the function field setting, there is an Frobenius at the infinite prime, but it can't be expressed as a limit like this in a reasonable way. $\endgroup$ – Will Sawin May 9 at 2:46
  • 2
    $\begingroup$ The analogue of the Galois group at the infinite prime should be the Weil group $\mathbb C^\times \rtimes \mathbb Z/2$, which acts on the cohomology and describes its Hodge numbers and complex conjugation structure. There may be different ways to decide what part of that counts as inertia and what part counts as Frobenius, but regardless the action of this group determines the Gamma factors. $\endgroup$ – Will Sawin May 9 at 2:50
  • 2
    $\begingroup$ @WillSawin, I think that the extension $1 \to \mathbb C^\times \to W_{\mathbb R} \to \mathbb Z/2\mathbb Z \to 1$ is non-split (there's a lift of the non-trivial element of $\mathbb Z/2\mathbb Z$ to an element of $W_{\mathbb R}$ whose square is $-1 \in \mathbb C^\times$, but not one whose square is $1$), so that one shouldn't describe the Weil group as a semi-direct product like that. $\endgroup$ – LSpice May 9 at 2:54
  • 2
    $\begingroup$ I remember Tasho Kaletha talking about how some of the computations in his regular supercuspidals paper indicate whether the extension $\mathbb C/\mathbb R$ should be considered as ramified or unramified … but I don't remember which way the verdict fell. $\endgroup$ – LSpice May 9 at 2:55
  • 2
    $\begingroup$ @LSpice Thanks, good point. $\endgroup$ – Will Sawin May 9 at 11:45

Your Answer

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

Browse other questions tagged or ask your own question.