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I have a simple question on notation.

Let $S$ be a Henselian trait with closed point $s$ (with finite residue field $k$) and generic point $\eta$. Let $X/S$ be a variety. Then, we have the functor

$$R\Psi:D^b_c((X_\eta)_\mathrm{\acute{e}t},\overline{\mathbb{Q}_\ell})\to D^b_c(X_s\times_s \eta,\overline{\mathbb{Q}_\ell})$$

where $D^b_c(X_s\times_s\eta,\overline{\mathbb{Q}_\ell})$ denotes the category of constructible $\overline{\mathbb{Q}_\ell}$-sheaves on $X_{\overline{s}}$ with "an action of $\mathrm{Gal}(\overline{\eta}/\eta)$ compatible with the action of $\mathrm{Gal}(\overline{s}/s)$." This means (a la "Le Formalisme de Cycles Evanescents" in SGA 7) if, for example, we're dealing with a constructible $\overline{\mathbb{Q}_\ell}$-sheaf $\mathcal{F}$ on $X_{\overline{s}}$, that for all $g\in \mathrm{Gal}(\overline{\eta}/\eta)$ we have isomorphisms:

$$\sigma(g):\overline{g}_\ast \mathcal{F}\to\mathcal{F}$$ (where $\overline{g}\in\mathrm{Gal}(\overline{s}/s)$) such that $\sigma(gh)=\sigma(g)\sigma(h)$.

Something which I commonly see is the following. People say that for $x\in X_s(k)$ that considering "$(R\Psi\overline{\mathbb{Q}_\ell})_x$" one gets an element of the derived category of finite-dimensional $\mathrm{Gal}(\overline{\eta}/\eta)$-representations.

Questions:

1) What does $(R\Psi\overline{\mathbb{Q}_\ell})_x$ even mean? This doesn't make any literal sense to me. Here are two possibilities I've considered:

a) If I forget the extra structure of the $\mathrm{Gal}(\overline{\eta}/\eta)$ action, it doesn't make sense (unless I am being silly) to take the stalk an $\overline{\mathbb{Q}_\ell}$-sheaf at a point $x\in X_s(k)$. One could interpret it at the choice of $\overline{\eta}$ gives you a canonical $\overline{x}\in X_{\overline{s}}(\overline{k})$ and so $(R\Psi\overline{\mathbb{Q}_\ell})_x$ might be shorthand for $(R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$. If that's the case, I don't see why it's $\mathrm{Gal}(\overline{\eta}/\eta)$-stable, so that one actually gets an action of $\mathrm{Gal}(\overline{\eta}/\eta)$.

b) Similar to a), but instead of taking the stalk $R\Psi\overline{\mathbb{Q}_\ell}$ thought about as an element of $D^b_c((X_s)_\mathrm{\acute{e}t},\overline{\mathbb{Q}_\ell})$ take the canonical pair $(\overline{x},\overline{\eta})$ (a point of the topos $X_s\times_s \eta$) and consider the stalk of this point. This gives the cohomology of the 'Milnor fiber', which also seems wrong. (EDIT: Ignore b) People think I'm claiming that the stalk at the point $(\overline{x},\overline{\eta})$ is not the cohomology of the Milnor fiber--I know this to be true. What I meant to say, even though I highly doubted it, that the stalk at $(\overline{x},\overline{\eta})$ was another interpretation of $(R\Psi\overline{\mathbb{Q}_\ell})_x$ but I very much don't think that now.)

2) Once I figure out what $(R\Psi\overline{\mathbb{Q}_\ell})_x$ means, how is it a finite dimensional continuous $\mathrm{Gal}(\overline{\eta}/\eta)$-representation (if not obvious from the definition).

Thanks so much!

EDIT: As examples of this notation see the second to last paragraph on page 13 of this article or Theorem 7.10 of this article.

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Every point in $X_s(k)$ extends uniquely to an element of $X_{\overline{s}}(\overline k)$ when we take its $\overline{k}$-points as an $\overline{k}$-scheme. This is the same as saying, if I have a variety defined over $\mathbb Q$, and I have a rational point, there is a canonical complex point associated to it. You don't need a choice of $\overline{eta}$ at all.

Applying this definition, you can see that it is the same as the cohomology of the Milnor fiber. Why do you think this is wrong?

We didn't use the choice of $\overline{eta}$, so the action is preserved.

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  • $\begingroup$ I think your first paragraph is what I said. The choice of $\overline{\eta}$ gives me the choice of $\overline{s}$ (by normalizing $S$ in $\overline{\eta}$ and taking the residue field) which is why there is a canonical $\bar{x}\in X_{\bar{s}}(\overline{k})$ associated to $x$. But, what you didn't answer (unless I am misunderstanding you) is why this actually gives me a well-defined action of $\mathrm{Gal}(\overline{\eta}/\eta)$ on $(R\Psi\overline{\mathbb{Q}}_\ell)_{\overline{x}}$. My worry is that we should get the action just because the compatible action $\endgroup$
    – Alex Youcis
    Apr 20, 2015 at 21:07
  • $\begingroup$ of $\mathrm{Gal}(\overline{\eta}/\eta)$ preserves the geometric point. Of course, it doesn't though. Given $g\in\mathrm{Gal}(\overline{\eta}/\eta)$ we actually have a map between the $\overline{g}(\overline{x})$ and the $\overline{x}$ stalk of $R\Psi\overline{\mathbb{Q}_\ell}$. I think the confusion might be in some canonical identification between these stalks, but I'm not sure (e.g. how when we define the $G_K$ action on $H^i(X_{\overline{K}},\mathbb{Q}_\ell)$ we make the 'canonical' identification between $\overline{Q}_\ell$ and $g^\ast\overline{\mathbb{Q}_\ell}$). $\endgroup$
    – Alex Youcis
    Apr 20, 2015 at 21:10
  • $\begingroup$ Just to be explicit. Literally taking stalks gives me an isomorphism $\sigma(g)_{\overline{x}}:(\overline{g}^\ast (R\Psi\overline{\mathbb{Q}_\ell}))_{\overline{x}}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$. But, this is just an isomorphism $(R\Psi\overline{\mathbb{Q}_\ell})_{\overline{g}(\overline{x})}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$. NOT an isomorphism $(R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$ which is what I would want to have an action. Like I said, maybe this is just some 'canonical identification $\endgroup$
    – Alex Youcis
    Apr 20, 2015 at 21:12
  • $\begingroup$ But we have a canonical isomorphism $\overline{g} (\overline{x} ) = \overline{x}$, because $\overline{x}$ comes from lifting $x$. Similarly the $G_K$-action on $H^i(X_{\overline{K}}, \mathcal F)$ is defined when $\mathcal F$ arises from a sheaf on $X_K$, which gives the required canonical isomorphisms. $\endgroup$
    – Will Sawin
    Apr 20, 2015 at 21:25
  • $\begingroup$ I apologize for being annoying. As you can see, this is a purely silly notational question. So, we want a canonical isomorphism $\overline{g}(\overline{x}}=\overline{x}$. I guess this is because $\overline{g}\circ\overline{x}=\overline{x}\circ\overline{g}$ since $\overline{x}$ lies over a $k$-point (where on the right $\overline{g}$ acts on $\mathrm{Spec}(\overline{k})$). And then we just identify this RHS with $\overline{x}$ via $\overline{g}^{-1}$? Is that right? Also, is there a way of phrasing this without having to make such an identification? For example, we can ignore the $\endgroup$
    – Alex Youcis
    Apr 20, 2015 at 21:29

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