# 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

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

where

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

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 ${S \mapsto S^{\flat}}$.