# Report from Oberwolfach

Just returned from a workshop on “Arithmetic of Shimura varieties” at Oberwolfach. Some scattered recollections:

• Gabber wasn’t there, but there were some Gabberesque moments anyway. In particular, during Xuhua He’s talk, Goertz observed that a point is an example of a Deligne-Lusztig variety, so any variety is a union of Deligne-Lusztig varieties. Gotta be careful…
• The food was about the same as usual. Worst Prize was tied between two dishes: a depressing vegetable soup which somehow managed to be flavorless and bitter simultaneously, and a dessert which looked like a lovely innocent custard but tasted like balsamic vinegar. The best dishes were all traditional German fare.
• Best Talks (in no particular order): Jean-Stefan Koskivirta, Miaofen Chen, Ben Howard, Timo Richarz.
• Apparently this paper can be boiled down to a page or two.
• There was (not surprisingly) some late-night discussion of the Stanford Mystery. [Redacted] proposed a theory so mind-bogglingly outrageous that it certainly won’t fit in this margin.
• “Fun was never really my goal.” – A representative UChicago alum.
• On Thursday it snowed, and a snowball fight broke out after dinner. This was a lot of fun, but I’m still glad we didn’t follow Pilloni’s suggestion of a match between Team Europe (Pilloni, Stroh, Morel, Anschutz, Richarz, Mihatsch, etc.) and Team USA (me).
• Here’s an innocent problem which turns out to be pretty tricky. Let $X$ be a (separated, smooth) rigid analytic space over $\mathbf{Q}_p$, and let $Y \to X$ be a map from a perfectoid space which is a $\underline{G}$-torsor for some profinite group $G$. In shorthand, you should think that $X =" Y/G$ with G acting freely (this is all literally true in the category of diamonds). It’s easy to cook up examples of this scenario: for instance, you can take $X=\mathrm{Spa}\mathbf{C}_p \left\langle T^{\pm 1} \right\rangle$ and $Y=\mathrm{Spa}\mathbf{C}_p \left\langle T^{\pm 1/p^\infty} \right\rangle$, so then $Y \to X$ is a torsor for the group $\mathbf{Z}_p$. However, there are also much more complicated examples which arise in nature. In particular, if $X$ is a Rapoport-Zink space or abelian-type Shimura variety at some finite level, and $Y$ is the associated infinite level perfectoid guy over it, then we’re in the situation above, with $G$ open in the $\mathbf{Q}_p$-points of some auxiliary reductive group.

Anyway, supposing we’re in the situation above, we can ask the following complementary questions:
Q1. Suppose that $Y$ is affinoid perfectoid. Does this imply that $X$ is an affinoid rigid space?
Q2. Suppose that $X$ is an affinoid rigid space. Does this imply that $Y$ is affinoid perfectoid?

It seems like both of these questions are actually really hard! For Q1, we can (by assumption) write $Y=\mathrm{Spa}(A,A^+)$ for some perfectoid Tate-Huber pair $(A,A^+)$, and then one might guess that $X$ coincides with $X'=\mathrm{Spa}(A^G,A^{+G})$. There is certainly a map $X \to X'$, but now one is faced with the problem of showing that $A^G$ is “big enough” for this map to be an isomorphism. This can be reduced to any one of a handful of auxiliary problems, but they all seem hard (at least to me). For instance, as a warmup one could try to prove either of the following implications:

W1. Under the hypothesis of Q1, $H^1(X,\mathcal{O}_X)$ vanishes.
W2. Under the hypothesis of Q1, $H^1(X,\mathcal{O}_{X}^+)$ is killed by a fixed power of $p.$

Both of these conclusions would certainly hold if we already knew that $X$ was affinoid: the first is just (a consequence of) Tate acyclicity, while the fact that $H^1(X,\mathcal{O}_{X}^+)$ is killed by some power of $p$ for smooth affinoids is a non-trivial theorem of Bartenwerfer. But I have totally failed to prove either W1 or W2.

In any case, the essential point with Q1 seems to be the following. If $H$ is some open subgroup, then $(A^+/p^n)^H$ will always have plenty of elements, and indeed taking the direct limit as $H$ shrinks recovers $A^+/p^n$. But the obstruction to lifting an element of $(A^+/p^n)^H$ to an element of $(A^+)^H$ is a torsion class in $H^1(H,A^+)$, and the latter group seems hard to control.

For Q2, there is maybe a slightly clearer path through the forest: it would follow from the following conjecture, which I explained during my talk in the workshop.

To set things up, let $(A,A^+)$ be any uniform Tate-Huber pair over $(\mathbf{Q}_p,\mathbf{Z}_p)$, and let $X=\mathrm{Spa}(A,A^+)$ be the associated pre-adic space. Let $X_v$ denote the site given by perfectoid spaces over $X$ with covers given by v-covers, and let $\mathcal{O}$ and $\mathcal{O}^+$ be the obvious structure sheaves on $X_v$. Set $\breve{A}^+ = H^0(X_v,\mathcal{O}^+)$ and $\breve{A} = \breve{A}^+ [1/p] = H^0(X_v,\mathcal{O})$, so the association $(A,A^+) \mapsto (\breve{A}, \breve{A}^+)$ is an endofunctor on the category of uniform Tate-Huber pairs over $\mathbf{Q}_p$. One can check that breve’ing twice is the same as breve’ing once, and that the natural map $(A,A^+) \to (\breve{A},\breve{A}^+)$ induces an isomorphism of diamonds. If $A$ is a smooth (or just seminormal) affinoid $K$-algebra for some $K/\mathbf{Q}_p$, or if $A$ is perfectoid, then breve’ing doesn’t change $A$.

Conjecture. Let $(A,A^+)$ be a uniform Tate-Huber pair over $\mathbf{Q}_p$ such that every completed residue field of $\mathrm{Spa}(A,A^+)$ is a perfectoid field. Then $\breve{A}$ is a perfectoid Tate ring.

Aside from disposing of Q2, this conjecture would settle another notorious problem: it would imply that if $A$ is a uniform sheafy Huber ring and $\mathrm{Spa}(A,A^+)$ is a perfectoid space, then $A$ is actually perfectoid.

It may be instructive to see an example of a non-perfectoid (uniform) Tate ring which satisfies the hypothesis of this conjecture. To make an example (with $p>2$), set $A=\mathbf{C}_p \left\langle T^{1/p^\infty} \right\rangle$, and let $B=A[\sqrt{T}]$ with the obvious topology. Set $C=\mathbf{C}_p \left\langle T^{1/2p^\infty} \right\rangle$, so there are natural maps $A \to B \to C$. Then $A$ and $C$ are perfectoid, but $B$ isn’t: the requisite $p$-power roots of $\sqrt{T}$ mod $p$ don’t exist. Nevertheless, every completed residue field of $B$ is perfectoid (exercise!), and the map $B \to C$ induces an isomorphism $\breve B \cong C$.

OK, this bullet point turned out pretty long, but these things have been in my head for the last couple months and it feels good to let them out. Besides, Yoichi Mieda asked me about Q1 during the workshop, so despite the technical nature of these questions, I might not be the only one who cares.

• Oberwolfach continues to be one of the best places in the world to do mathematics.

Thanks to the organizers for putting together such an excellent week!

## 9 thoughts on “Report from Oberwolfach”

1. pupshaw says:

The Stanford mystery is so frustrating. Does anyone claim to have seen anything of substance? Killer titles, for sure.

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1. No one I’ve talked to has seen any substance at all. It seems like everyone is equally in the dark.

Dig your pseudonym – I’m a big Frank fan!

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1. pupshaw says:

Always glad to meet a unifactor denizen. I certainly hope they can find a way to carve off something substantial and resolve this situation. Seems to me that any fraction of what’s claimed in those titles (or the abstract that appeared and disappeared) would go a long way. Wouldn’t it be nice if it worked?

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2. AP says:

By the “Stanford mystery”, do you mean a certain PhD student with a prominent advisor making rather strong claims? I’ve heard all kinds of rumors regarding that, none of them positive.

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1. curious41 says:

The Stanford mystery is not officially listed anymore on their website. Any news one way or another ?

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