TITLE: Grothendieck's Vanishing Cycles
QUESTION [7 upvotes]: Suppose $S$ is the spectrum of a strict henselian ring $R$ which is also a discrete valuation ring (DVR), then $S$ consists of a closed point $s$ and a generic point $\eta$. We have a henselian trait,
\begin{equation}
\eta \rightarrow S \leftarrow s
\end{equation} 
If $f:X \rightarrow S$ is a (flat) morphism, then Grothendieck studied the nearby cycle functor $R \Psi_f$ and vanishing cycle functor $R \Phi_f$. Suppose we have a fibration over 
\begin{equation}
\pi: X \rightarrow C
\end{equation}
where $X$ is a variety over $k$ and $C$ is a curve over $k$. For a point $x$ in $C$, I have seen papers talking about the punctured henselian neighbourhood $S$ of $x$, which requires a local parameter. Is $S$ the strict henselization of the local ring $\mathcal{O}_{C,x}$? People thinks $S$ is like a small neighbourhood of $x$ in the analytic topology if there is an embedding $k \rightarrow \mathbb{C}$, could someone explain why more carefully?
Another question is does the definition of this punctured henselian neighbourhood of $x$ depends on the choice of a local parameter around $x$? If so, why?

REPLY [6 votes]: The analogy with the classical picture is the following : 

A small disk around $x$ is the spectrum $\tilde{S}$ of a strict henselization $\mathcal{O}_{C,x}^{sh}$ of $\mathcal{O}_{C,x}$
The point $x$ is the closed point of $\tilde{S}=\operatorname{Spec}\mathcal{O}_{C,x}^{sh}$ (hence a geometric point lying above $x$).
The punctured disk around $x$ is the spectrum $\tilde{\eta}$ of the fraction field of $\mathcal{O}_{C,x}^{sh}$ (in other words, this is $\tilde{\eta}=\tilde{S}\setminus x$ : the disk without the point).

None of these depends of a local parameter, but it depends on the choice of a geometric point lying above $x$ (in other words, it depends on the choice of an algebraic closure of $\kappa(x)$). Because you are speaking of nearby cycles, these even depend on the choice of an algebraic closure $\overline{\eta}$ of $\tilde{\eta}$.

Now I try to explain why this analogy is valid.
First, étale neighborhood are finer than Zariski one. Hence they are "closer" to the classical neighborhood. For example, the curve $y^2=x^3+x^2$ is irreducible and has a node at $x=(0,0)$. The local ring at $x$ for the Zariski topology remain integral (as a localization of an integral ring). However, the strict henselization (in other words, the local ring for the étale topology) is isomorphic the strict henselization of the local ring at $(0,0)$ of $k[u,v]/(uv)$. In other words, the étale topology sees the two branches at $(0,0)$. This is good because this is also the case for the classical topology.
Second, their categories of locally constant sheaves are (almost) the same :

On a small disk, every locally constant sheaf is constant since the disk is contractible. The same is true for $\tilde{S}$ : every locally constant étale sheaf is constant.
Obviously, the same is true for the center of the disk and the geometric point.
On the punctured disk, the category of locally constant sheaf is the category of representation of $\pi_1(D^*)=\mathbb{Z}$. On $\tilde{\eta}$, the category of locally constant étale sheaf is the category of representation of $\operatorname{Gal}(\overline{\eta}/\tilde{\eta})$. Now this group may be very different from $\mathbb{Z}$. However, if the residue field has characteristic 0, this group is (isomorphic to) $\hat{\mathbb{Z}}$ the profinite completion of $\mathbb{Z}$. This is because every étale cover of $\tilde{\eta}$ is obtained by adding a $n$-th root the uniformizer (after picking one). 


Note however that in the classical picture, one need to choose a generator of $\pi_1(D^*)$. In the étale picture in characteristic 0, this is a bit more complicated. The isomorphism $\operatorname{Gal}(\overline{\eta}/\tilde{\eta})$ depends on the choice of the local parameter $\pi$ and the choice of primitive $n$-th root of $\pi$ for every $n$ (this is in fact $\hat{\mathbb{Z}}(1)$).