Geometric intuition for Fontaine-Wintenberger?

I asked my advisor the question in the title. He told me it was a stupid question and that I should focus on my research. Thus we're asking here.

The statement of Fontaine-Winterberger, per their groundbreaking 1979 paper Extensions algébrique et corps des normes des extensions APF des corps locaux, is as follows.

The (absolute) Galois groups of$$\bigcup_n \mathbb{Q}_p\left(p^{1/p^n}\right)$$and$$\bigcup_n \mathbb{F}_p \left(\left(t^{1/p^n}\right)\right)$$are canonically isomorphic.

Can someone unpack this statement for me, and give their geometric intuition for it?

I do not work in number theory nor arithmetic geometry, but I'm willing to take some stuff on faith and Google around to find what I need to learn to understand whatever answer is provided.

This is a result about Galois theory that has a "classical" flavor. I want my questions to be evaluated with that in mind.

• The answer to that question seems relevant. – Aurel Sep 22 '17 at 7:07
• "canonically isomorphic" : the absolute Galois groups are themselves not canonically defined... – js21 Sep 22 '17 at 7:48
• It's not a stupid question. – anomaly Sep 22 '17 at 16:36
• what an exceptionally strange and offensive response from your advisor :/ – forget this May 12 '20 at 23:28

First, the Fontaine-Winterberger isomorphism can also be recovered from a theorem of Deligne, namely Thm 2.8 here. Deligne showed that if two local fields $K_1$ and $K_2$ (possibly of different characteristic) were such that $O_{K_1}/\mathfrak{m}_{K_1}^N \simeq O_{K_2}/\mathfrak{m}_{K_2}^N$ then the category of finite etale extensions of $K_1$ with ramification bounded by $N$ is isomorphic to the corresponding category for $K_2$. One gets the Fontaine-Winterberger isomorphism by taking $K_1 = \mathbb{Q}_p(p^{\frac{1}{p^N}})$, $K_2 = \mathbb{F}_p((t^{\frac{1}{p^N}}))$ and letting $N \rightarrow +\infty$.
Abbes and Saito gave a geometric description of the ramification filtration of local fields in this paper. They were motivated by the case of imperfect residue fields, but a quick survey of their idea in the classical case can be found in Section $2$ of this article by Shin Hattori.
Shin Hattori used Abbes-Saito's idea in order to reprove and extend Deligne's result. This yields a geometric proof of Deligne's theorem, using rigid geometry and perfectoid spaces. The basic idea is that the ramification of $K_1$, resp. $K_2$, can be read from the connected components of some rigid spaces $X_1^{\leq j}$, $X_2^{\leq j}$ defined by Abbes-Saito, and if $j \leq N$ one can exhibit explicit pro-etale covers $\widetilde{X}_1^{\leq j}$, $\widetilde{X}_2^{\leq j}$ such that $\widetilde{X}_1^{\leq j}$ is the tilt of $\widetilde{X}_2^{\leq j}$. More details can be found in this survey by Shin Hattori, or in the paper here.
Remark: the only thing needed about perfectoid spaces is the fact that tilting preserves connected components. This is even easier than showing that the corresponding adic spaces are homeomorphic, and in any case does not use the equivalence of etale site (so no circular reasoning here). For the sake of completeness: if $A$ is a ring then idempotents of $A$ are in bijection with idempotents of the monoid $A^{\flat} = \lim_{x \mapsto x^p} A$.
• Why do you say "finite étale" extensions of $K_1, K_2$ as opposed to "finite separable" extensions? Is there a general statement that holds for bases other than fields? – dorebell Sep 22 '17 at 8:23