# 'Stalk' of vanishing cycles at $k$-point

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.

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.
We didn't use the choice of $\overline{eta}$, so the action is preserved.
• 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 – Alex Youcis Apr 20 '15 at 21:07
• 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 $g^\ast\overline{\mathbb{Q}_\ell}$). – Alex Youcis Apr 20 '15 at 21:10
• 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 – Alex Youcis Apr 20 '15 at 21:12
• 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. – Will Sawin Apr 20 '15 at 21:25
• 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 – Alex Youcis Apr 20 '15 at 21:29