I.8: Tilting

Tilting is the key construction for perfectoid rings which allows to naturally associate a perfectoid (perfect) ring in characteristic {p} to a perfectoid ring of any characteristic (in particular, {0}). Thus, some of the geometrical problems in characteristic {0} might be translated to characteristic {p}, solved there and the result can be translated again to characteristic {0}. This was done with the weight-monodromy conjecture by Peter Scholze. The following blog post (as was the previous) is based mostly on notes of Jared Weinstein from Scholze’s lectures at Berkeley.

Let {R} be a perfectoid Tate ring. The tilt of {R} is

\displaystyle R^{\flat} = \varprojlim _{x \mapsto x^p} R

with the inverse limit topology. A priori this is only a topological multiplicative monoid, but we can endow it with a ring structure, by defining the addition

\displaystyle (x^{(0)},x^{(1)},...)+(y^{(0)},y^{(1)},...) = (z^{(0)}, z^{(1)},...)


\displaystyle z^{(i)} = \lim _{n \rightarrow \infty}(x^{(i+n)}+y^{(i+n)})^{p^n} \in R

We have to check that it is well-defined.

Lemma 33 The limit {z^{(i)}} exists and defines a ring structure, which makes {R^{\flat}} into a topological {\mathbb{F}_p}-algebra that is a perfect uniform Tate ring. The subset {R^{\flat \circ}} of power-bounded elements is given by the topological ring isomorphism

\displaystyle R^{\flat \circ} = \varprojlim _{\Phi} R^{\circ} \simeq \varprojlim _{\Phi} R^{\circ} /p \simeq \varprojlim _{\Phi} R^{\circ} / \varpi

where {\varpi \in R} is a pseudo-uniformizer which divides {p} in {R^{\circ}}. Moreover there exists a pseudo-uniformizer {\varpi \in R} with {\varpi ^p | p} in {R^{\circ}} which admits a sequence of {p}-th power roots {\varpi ^{1/p^{n}}}, giving rise to an element

\displaystyle \varpi ^{\flat} = (\varpi, \varpi ^{1/p},...) \in R^{\flat \circ}

which is a pseudo-uniformizer of {R^{\flat}}. We have {R^{\flat} = R^{\flat \circ}[1/\varpi ^{\flat}]}.

Proof: By definition {R^{\flat}} is perfect. Let {\varpi _0} be a pseudo-uniformizer of {R}. Let us check that the maps

\displaystyle \varprojlim _{\Phi} R^{\circ} \rightarrow \varprojlim _{\Phi} R^{\circ} /p \rightarrow \varprojlim _{\Phi} R^{\circ} / \varpi _0

are isomorphisms. The key is that any sequence {(\bar{x}_0,\bar{x}_1,...) \in \varprojlim _{\Phi} R^{\circ} / \varpi _0} lifts uniquely to a sequence {(x_0,x_1,...) \in \varprojlim _{\Phi} R^{\circ} }. This is standard: {x_i = \lim _{n \rightarrow \infty} \tilde{x} _{n+i} ^{p^n}}, where {\tilde{x}_j \in R^{\circ}} is any lift of {\bar{x}_j}. Hence {R^{\flat \circ}} is a well-defined ring.

Now assume {\varpi _0 ^p |p} in {R^{\circ}}. We construct {\varpi ^{\flat}}. In fact the preimage of {\varpi _0} under {R^{\flat \circ} = \varprojlim _{\Phi} R^{\circ} / \varpi _0 ^p \rightarrow R^{\circ} / \varpi _0 ^p} is an element {\varpi ^{\flat}} with the right properties. It is congruent to {\varpi _0} modulo {\varpi _0 ^p} and therefore it is also a pseudo-uniformizer. Then the projection {\varpi} of {\varpi ^{\flat}} to the {0}-th coordinate is the desired pseudo-uniformizer of {R^{\circ}}. \Box

As in the last line of the proof above, we have a continuous multiplicative (but not additive) map

\displaystyle R^{\flat} = \varprojlim _{\Phi} R \rightarrow R

which projects onto the {0}-th coordinate. We denote it by {f \mapsto f^{\sharp}}. This projection defines a ring isomorphism {R^{\flat \circ} / \varpi ^{\flat} \simeq R^{\circ} / \varpi}. Let us note the following fact which follows from this isomorphism, the nilpotence conditions and a fact that an open subring is integrally closed in {R^{\circ}} (resp. {R^{\circ \flat}}) if it its image in the quotient {R^{\circ} / \varpi} (resp. {R^{\flat \circ} / \varpi ^{\flat}}) is integrally closed.

Lemma 34 The set of rings of integral elements {R ^+ \subset R^{\circ}} is in bijection with the set of rings of integral elements {R^{\flat +} \subset R^{\flat \circ}} by {R^{\flat +} = \varprojlim _{x \mapsto x^p} R^+}. Also {R ^{\flat +} / \varpi ^{\flat} = R^{+} / \varpi}.

We say that a Huber pair {(R,R^+)} is perfectoid if {R} is perfectoid. We have the following theorem due to Scholze (see also Kedlaya-Liu’s work)

Proposition 35 Let {(R,R^+)} be a perfectoid Huber pair. Then for all rational subsets {U \subset X = \mathrm{Spa} (R,R^+)}, the {\mathcal{O}_X(U)} is again perfectoid. In particular {(R,R^+)} is stably uniform, hence sheafy (by Buzzard-Verberkmoes, as explained earlier).

The following two fundamental theorems are due to Scholze. We leave the without proof. A reader should consult Scholze’s “`Perfectoid spaces”‘.

Theorem 36 Let {(R,R^+)} be a perfectoid Huber pair with tilt {(R^{\flat}, R^{\flat +})}. There is a homeomorphism

\displaystyle \mathrm{Spa} (R,R^+) \simeq \mathrm{Spa} (R^{\flat}, R^{\flat +})

which sends {x} to {x ^{\flat}} such that that {|f(x^{\flat})| = |f^{\sharp}(x)|}. This homeomorphism preserves rational subsets.

Theorem 37 Let {R} be a perfectoid ring with tilt {R^{\flat}}. Then there is an equivalence of categories between perfectoid {R} algebras and perfectoid {R^{\flat}}-algebras via {S \mapsto S^{\flat}}.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s