This talk is about a theory of "perfectoid spaces", which "compares objects in characteristic p with objects in characteristic 0". What are those spaces, where can one read about them?

Edit: A bit more info can be found in Peter Scholze's seminar description and in Bhargav Bhatt's.

Edit: Peter Scholze posted yesterday this beautiful overview on the arxiv.

Edit: Peter Scholze posted today this new survey on the arxiv.

  • 4
    $\begingroup$ They were defined by the speaker of this talk in his recent work. So I geuss the best place to start is math.uni-bonn.de/people/scholze . $\endgroup$ Commented May 22, 2011 at 20:45
  • 4
    $\begingroup$ I found nothing to read there $\endgroup$ Commented May 22, 2011 at 20:50
  • 6
    $\begingroup$ It might be the case that there is nowhere that one can read about them yet. I learnt about them from a talk that Scholze gave in Nottingham: they are, vaguely speaking, "perfections of affinoid algebras", that is, take an algebra like $\mathbf{Q}_p\langle T\rangle$ and then adjoin all $p^n$th roots of $T$. But it's a bit more delicate than this really. If you really want to know then just email Scholze. Or wait patiently until the author writes something! $\endgroup$ Commented May 22, 2011 at 20:58
  • 3
    $\begingroup$ @Riepe: thanks for asking the question! it elicited Scholze's response! $\endgroup$
    – SGP
    Commented May 31, 2011 at 22:29
  • $\begingroup$ The ending -oid describes that something or somebody is comparable or similar to something. In mathematics comparable is defined: Comparability. So these are algebraic structures that are comparable to perfections of algebras that are comparable to affine ... and now your space comes into play. Mind Space_(mathematics) is simply a set or a universe with structure added. I.e. Tarski-Grothendiek universe. $\endgroup$ Commented Jul 16, 2022 at 17:45

4 Answers 4


Update: The lecture notes of the CAGA lecture series on perfectoid spaces at the IHES can now be found online, cf. http://www.ihes.fr/~abbes/CAGA/scholze.html.

It seems that it's my job to answer this question, so let me just briefly explain everything. A more detailed account will be online soon.

We start with a complete non-archimedean field $K$ of mixed characteristic $(0,p)$ (i.e., $K$ has characteristic $0$, but its residue field has characteristic $p$), equipped with a non-discrete valuation of rank $1$, such that (and this is the crucial condition) Frobenius is surjective on $K^+/p$, where $K^+\subset K$ is the subring of elements of norm $\leq 1$.

Some authors, e.g. Gabber-Ramero in their book on Almost ring theory, call such fields deeply ramified (they do not require that they are complete, anyway).

Just think of $K$ as the completion of the field $\mathbb{Q}_p(p^{1/p^\infty})$, or alternatively as the completion of the field $\mathbb{Q}_p(\mu_{p^\infty})$.

In this situation, one can form the field $K^\prime$, given as the fraction field of $K^{\prime +} = \varprojlim K^+/p$, where the transition maps are given by Frobenius. Concretely, in the first example it is given by the completion of $\mathbb{F} _p((t^{1/p^\infty}))$, where $t$ is the element $(p,p^{\frac 1p},p^{\frac 1{p^2}},\ldots)$ in $K^{\prime +}=\varprojlim K^+/p$.

Now we have the following theorem, due to Fontaine-Wintenberger in the examples I gave, and deduced from the book of Gabber-Ramero in general:

Theorem: There is a canonical isomorphism of absolute Galois group $G_K\cong G_{K^\prime}$.

At this point, it may be instructive to explain this theorem a little, in the example where $K$ is the completion of $\mathbb{Q}_p(p^{1/p^\infty})$ (this assumption will be made whenever examples are discussed). It says that there is a natural equivalence of categories between the category of finite extensions $L$ of $K$ and the category of finite extensions $L^\prime$ of $K^\prime$. Let us give an example, say $L^\prime$ is given by adjoining a root of $X^2 - 7t X + t^5$. Basically, the idea is that one replaces $t$ by $p$, so that one would like to define $L$ as the field given by adjoining a root of $X^2 - 7p X + p^5$. However, this is obviously not well-defined: If $p=3$, then $X^2 - 7t X + t^5=X^2 - t X + t^5$, but $X^2 - 7p X + p^5\neq X^2 - p X + p^5$, and one will not expect that the fields given by adjoining roots of these different polynomials are the same.

However, there is the following way out: $L^\prime$ can be defined as the splitting field of $X^2 - 7t^{1/p^n} X + t^{5/p^n}$ for all $n\geq 0$, and if we choose $n$ very large, then one can see that the fields $L_n$ given as the splitting field of $X^2 - 7p^{1/p^n} X + p^{5/p^n}$ will stabilize as $n\rightarrow \infty$; this is the desired field $L$. Basically, the point is that the discriminant of the polynomials considered becomes very small, and the difference between any two different choices one might make when replacing $t$ by $p$ become comparably small.

This argument can be made precise by using Faltings's almost mathematics, as developed systematically by Gabber-Ramero. Consider $K\supset K^+\supset \mathfrak{m}$, where $\mathfrak{m}$ is the maximal ideal; in the example, it is the one generated by all $p^{1/p^n}$, and it satisfies $\mathfrak{m}^2 = \mathfrak{m}$, because the valuation on $K$ is non-discrete. We have a sequence of localization functors:

$K^+$-mod $\rightarrow$ $K^+$-mod / $\mathfrak{m}$-torsion $\rightarrow$ $K^+ $-mod / $p$-power torsion.

The last category is equivalent to $K$-mod, and the composition of the two functors is like taking the generic fibre of an object with an integral structure.

In this sense, the category in the middle can be seen as a slightly generic fibre, sitting strictly between an integral structure and an object over the generic fibre. Moreover, an object like $K^+/p$ is nonzero in this middle category, so one can talk about torsion objects, neglecting only very small objects. The official name for this middle category is $K^{+a}$-mod: almost $K^+$-modules.

This category is an abelian tensor category, and hence one can define in the usual way the notion of a $K^{+a}$-algebra (= almost $K^+$-algebra), etc. . With some work, one also has notions of almost finitely presented modules and (almost) étale maps. In the following, we will often need the notion of an almost finitely presented étale map, which is the almost analogue of a finite étale cover.

Theorem (Tate, Gabber-Ramero): If $L/K$ finite extension, then $L^+/K^+$ is almost finitely presented étale. Similarly, if $L^\prime/K^\prime$ finite, then $L^{\prime +}/K^{\prime +}$ is almost finitely presented étale.

Here, $L^+$ is the valuation subring of $L$. As an example, assume $p\neq 2$ and $L=K(p^\frac 12)$. For convenience, we look at the situation at a finite level, so let $K_n=\mathbb{Q}_p(p^{1/p^n})$ and $L_n=K_n(p^\frac 12)$. Then $L_n^+ = K_n^+[X] / (X^2 - p^{1/p^n})$. To check whether this is étale, look at $f(X)= X^2 - p^{1/p^n}$ and look at the ideal generated by $f$ and its derivative $f^\prime$. This contains $p^{1/p^n}$, so in some sense $L_n^+$ is étale over $K_n^+$ up to $p^{1/p^n}$-torsion. Now take the limit as $n\rightarrow \infty$ to see that $L^+$ is almost étale over $K^+$.

Now we can prove the theorem above:

Finite étale covers of $K$ = almost finitely presented étale covers of $K^+$ = almost finitely presented étale covers of $K^+/p$ [because (almost) finite étale covers lift uniquely over nilpotents] = almost finitely presented étale covers of $K^{\prime +}/t$ [because $K^+/p = K^{\prime +}/t$, cf. the example] = almost finitely presented étale covers of $K^{\prime +}$ = finite étale covers of $K^\prime$.

After we understand this theory on the base, we want to generalize to the relative situation. Here, let me make the following claim.

Claim: $\mathbb{A}^1_{K^\prime}$ "equals" $\varprojlim \mathbb{A}^1_K$, where the transition maps are the $p$-th power map.

As a first step towards understanding this, let us check this on points. Here it says that $K^\prime = \varprojlim K$. In particular, there should be map $K^\prime\rightarrow K$ by projection to the last coordinate, which I usually denote $x^\prime\mapsto [x^\prime]$ (because it is a related to Teichmüller representatives) and again this can be explained in an example:

Say $x^\prime = t^{-1} + 5 + t^3$. Basically, we want to replace $t$ by $p$, but this is not well-defined. But we have just learned that this problem becomes less serious as we take $p$-power roots. So we look at $t^{-1/p^n} + 5 + t^{3/p^n}$, replace $t$ by $p$, get $p^{-1/p^n} + 5 + p^{3/p^n}$, and then we take the $p^n$-th power again, so that the expression has the chance of being independent of $n$. Now, it is in fact not difficult to see that

$\lim_{n\rightarrow \infty} (p^{-1/p^n} + 5 + p^{3/p^n})^{p^n}$

exists, and this defined $[x^\prime]\in K$. Now the map $K^\prime\rightarrow \varprojlim K$ is given by $x^\prime\mapsto ([x^\prime],[x^{\prime 1/p}],[x^{\prime 1/p^2}],\ldots)$.

In order to prove that this is a bijection, just note that

$K^{\prime +} = \varprojlim K^{\prime +}/t^{p^n} = \varprojlim K^{\prime +}/t = \varprojlim K^+/p \leftarrow \varprojlim K^+$.

Here, the last map is the obvious projection, and in fact is a bijection, which amounts to the same verification as that the limit above exists. Afterwards, just invert $t$ to get the desired identification.

In fact, the good way of approaching this stuff in general is to use some framework of rigid geometry. In the papers of Kedlaya and Liu, where they are doing extremely related stuff, they choose to work with Berkovich spaces; I favor the language of Huber's adic spaces, as this language is capable of expressing more (e.g., Berkovich only considers rank-$1$-valuations, whereas Huber considers also the valuations of higher rank). In the language of adic spaces, the spaces are actually locally ringed topological spaces (equipped with valuations) (and affinoids are open, in contrast to Berkovich's theory, making it easier to glue), and there is an analytification functor $X\mapsto X^{\mathrm{ad}}$ from schemes of finite type over $K$ to adic spaces over $K$ (similar to the functor associating to a scheme of finite type over $C$ a complex-analytic space). Then we have the following theorem:

Theorem: We have a homeomorphism of underlying topological spaces $|(\mathbb{A}^1_{K^\prime})^{\mathrm{ad}}|\cong \varprojlim |(\mathbb{A}^1_K)^{\mathrm{ad}}|$.

At this point, the following question naturally arises: Both sides of this homeomorphism are locally ringed topological spaces: So is it possible to compare the structure sheaves? There is the obvious problem that on the left-hand side, we have characteristic $p$-rings, whereas on the right-hand side, we have characteristic $0$-rings. How can one possibly pass from one to the other side?

Definition: A perfectoid $K$-algebra is a complete Banach $K$-algebra $R$ such that the set of power-bounded elements $R^\circ\subset R$ is open and bounded and Frobenius induces an isomorphism $R^\circ/p^{\frac 1p}\cong R^\circ/p$.

Similarly, one defines perfectoid $K^\prime$-algebras $R^\prime$, putting a prime everywhere, and replacing $p$ by $t$. The last condition is then equivalent to requiring $R^\prime$ perfect, whence the name. Examples are $K$, any finite extension $L$ of $K$, and $K\langle T^{1/p^\infty}\rangle$, by which I mean: Take the $p$-adic completion of $K^+[T^{1/p^\infty}]$, and then invert $p$.

Recall that in classical rigid geometry, one considers rings like $K\langle T\rangle$, which is interpreted as the ring of convergent power series on the closed annulus $|x|\leq 1$. Now in the example of the $\mathbb{A}^1$ above, we take $p$-power roots of the coordinate, so after completion the rings on the inverse limit are in fact perfectoid.

In characteristic $p$, one can pass from usual affinoid algebras to perfectoid algebras by taking the completed perfection; the difference between the two is small, at least as regards topological information on associated spaces: Frobenius is a homeomorphism on topological spaces, and even on étale topoi. [This is why we don't have to take $\varprojlim \mathbb{A}^1_{K^\prime}$: It does not change the topological spaces. In order to compare structure sheaves, one should however take this inverse limit.]

The really exciting theorem is the following, which I call the tilting equivalence:

Theorem: The category of perfectoid $K$-algebras and the category of perfectoid $K^\prime$-algebras are equivalent.

The functor is given by $R^\prime = (\varprojlim R^\circ/p)[t^{-1}]$. Again, one also has $R^\prime = \varprojlim R$, where the transition maps are the $p$-th power map, giving also the map $R^\prime\rightarrow R$, $f^\prime\mapsto [f^\prime]$.

There are two different proofs for this. One is to write down the inverse functor, given by $R^\prime\mapsto W(R^{\prime \circ})\otimes_{W(K^{\prime +})} K$, using the map $\theta: W(K^{\prime +})\rightarrow K$ known from $p$-adic Hodge theory. The other proof is similar to what we did above for finite étale covers:

perfectoid $K$-algebras = almost $K^{+}$-algebras $A$ s.t. $A$ is flat, $p$-adically complete and Frobenius induces isom $A/p^{1/p}\cong A/p$ = almost $K^+/p$-algebras $\overline{A}$ s.t. $\overline{A}$ is flat and Frobenius induces isom $\overline{A}/p^{\frac 1p}\cong \overline{A}$,

and then going over to the other side. Here, the first identification is not difficult; the second relies on the astonishing fact (already in the book by Gabber-Ramero) that the cotangent complex $\mathbb{L}_{\overline{A}/(K^+/p)}$ vanishes, and hence one gets unique deformations of objects and morphisms. At least on differentials $\Omega^1$, one can believe this: Every element $x$ has the form $y^p$ because Frobenius is surjective; but then $dx = dy^p = pdy = 0$ because $p=0$ in $\overline{A}$.

Now let me just briefly summarize the main theorems on the basic nature of perfectoid spaces. First off, an affinoid perfectoid space is associated to an affinoid perfectoid $K$-algebra, which is a pair $(R,R^+)$ consisting of a perfectoid $K$-algebra $R$ and an open and integrally closed subring $R^+\subset R^\circ$ (it follows that $\mathfrak{m} R^\circ\subset R^+$, so $R^+$ is almost equal to $R^\circ$; in most cases, one will just take $R^+=R^\circ$). Then also the categories of affinoid perfectoid $K$-algebras and of affinoid perfectoid $K^\prime$-algebras are equivalent. Huber associates to such pairs $(R,R^+)$ a topological spaces $X=\mathrm{Spa}(R,R^+)$ consisting of continuous valuations on $R$ that are $\leq 1$ on $R^+$, with the topology generated by the rational subsets $\{x\in X\mid \forall i: |f_i(x)|\leq |g(x)|\}$, where $f_1,\ldots,f_n,g\in R$ generate the unit ideal. Moreover, he defines a structure presheaf $\mathcal{O}_X$, and the subpresheaf $\mathcal{O}_X^+$, consisting of functions which have absolute value $\leq 1$ everywhere.

Theorem: Let $(R,R^+)$ be an affinoid perfectoid $K$-algebra, with tilt $(R^\prime,R^{\prime +})$. Let $X=\mathrm{Spa}(R,R^+)$, with $\mathcal{O}_X$ etc., and $X^\prime = \mathrm{Spa}(R^\prime,R^{\prime +})$, etc. . i) We have a canonical homeomorphism $X\cong X^\prime$, given by mapping $x$ to $x^\prime$ defined via $|f^\prime(x^\prime)| = |[f^\prime] (x)|$. Rational subsets are identified under this homeomorphism. ii) For any rational subset $U\subset X$, the pair $(\mathcal{O}_X(U),\mathcal{O}_X^+(U))$ is affinoid perfectoid with tilt $(\mathcal{O}_{X^\prime}(U),\mathcal{O}_{X^\prime}^+(U))$. iii) The presheaves $\mathcal{O}_X$, $\mathcal{O}_X^+$ are sheaves. iv) For all $i>0$, the cohomology group $H^i(X,\mathcal{O}_X)=0$; even better, the cohomology group $H^i(X,\mathcal{O}_X^+)$ is almost zero, i.e. $\mathfrak{m}$-torsion.

This allows one to define general perfectoid spaces by gluing affinoid perfectoid spaces. Further, one can define étale morphisms of perfectoid spaces, and then étale topoi. This leads to an improvement on Faltings's almost purity theorem:

Theorem: Let $R$ be a perfectoid $K$-algebra, and let $S/R$ be finite étale. Then $S$ is perfectoid and $S^\circ$ is almost finitely presented étale over $R^\circ$.

In particular, no sort of semistable reduction hypothesis is required anymore. Also, the proof is much easier, cf. the book project by Gabber-Ramero.

Tilting also identifies the étale topoi of a perfectoid space and its tilt, and as an application, one gets the following theorem.

Theorem: We have an equivalence of étale topoi of adic spaces: $(\mathbb{P}^n_{K^\prime})^{\mathrm{ad}}_{\mathrm{et}}\cong \varprojlim (\mathbb{P}^n_K)^{\mathrm{ad}}_{\mathrm{et}}$. Here the transition maps are again the $p$-th power map on coordinates.

Let me end this discussion by mentioning one application. Let $X\subset \mathbb{P}^n_K$ be a smooth hypersurface. By a theorem of Huber, we can find a small open neighborhood $\tilde{X}$ of $X$ with the same étale cohomology. Moreover, we have the projection $\pi: \mathbb{P}^n_{K^\prime}\rightarrow \mathbb{P}^n_K$, at least on topological spaces or étale topoi. Within the preimage $\pi^{-1}(\tilde{X})$, it is possible to find a smooth hypersurface (of possibly much larger degree) $X^\prime$. This gives a map from the cohomology of $X$ to the cohomology of $X^\prime$, thereby comparing the étale cohomology of a variety in characteristic $0$ with the étale cohomology of characteristic $p$. Using this, it is easy to verify the weight-monodromy conjecture for $X$.

  • 106
    $\begingroup$ Yeah... And so we watch math history in the making. $\endgroup$
    – Olivier
    Commented May 31, 2011 at 18:17
  • 8
    $\begingroup$ @Scholze: Many thanks for this! @Oliver: well put indeed! $\endgroup$
    – SGP
    Commented May 31, 2011 at 21:57
  • 8
    $\begingroup$ The polynomial $X^2−7p^{1/p^n}X+p^{5/p^n}$ actually splits over $\mathbb{Q}_p(p^{1/p^n})$ (Newton), and similarly for $x^2-7tX+t^5$. :-) For $p > 2$ one could take e.g. $X^2−7p^{1/p^n}X+p^{1/p^n}$ instead (which doesn't split because the roots have valuation outside the value group of the extensions considered here). However, it seems quadratic extensions are too simple to fully illustrate what's going on, because the quadratic extension defined is the same for any $n$. $\endgroup$
    – fherzig
    Commented Feb 13, 2014 at 20:00
  • 2
    $\begingroup$ you said 'At least on differentials Ω1Ω1, one can believe this'... I was wondering given perfectoid space over char 0 perfectoid field, is there a notion of sheaf of (relative) differential? I tried to google around yet got nothing... $\endgroup$
    – S. Li
    Commented Apr 19, 2016 at 3:46
  • $\begingroup$ I made the effort in calculating what is the limit n to infinity following "In fact it is not difficult to see that. My result is that this limit is $1$ if $ln[p]<0$. This is divergent for $p>1$. I used for reference Mathematica. $\endgroup$ Commented Oct 1, 2022 at 20:01

Here is a completely different kind of answer to this question.

A perfectoid space is a term of type PerfectoidSpace in the Lean theorem prover.

Here's a quote from the source code:

structure perfectoid_ring (R : Type) [Huber_ring R] extends Tate_ring R : Prop :=
(complete  : is_complete_hausdorff R)
(uniform   : is_uniform R)
(ramified  : ∃ ϖ : pseudo_uniformizer R, ϖ^p ∣ p in Rᵒ)
(Frobenius : surjective (Frob Rᵒ∕p))

CLVRS ("complete locally valued ringed space") is a category
whose objects are topological spaces with a sheaf of complete topological rings
and an equivalence class of valuation on each stalk, whose support is the unique
maximal ideal of the stalk; in Wedhorn's notes this category is called 𝒱.
A perfectoid space is an object of CLVRS which is locally isomorphic to Spa(A) with
A a perfectoid ring. Note however that CLVRS is a full subcategory of the category
`PreValuedRingedSpace` of topological spaces equipped with a presheaf of topological
rings and a valuation on each stalk, so the isomorphism can be checked in
PreValuedRingedSpace instead, which is what we do.

/-- Condition for an object of CLVRS to be perfectoid: every point should have an open
neighbourhood isomorphic to Spa(A) for some perfectoid ring A.-/
def is_perfectoid (X : CLVRS) : Prop :=
∀ x : X, ∃ (U : opens X) (A : Huber_pair) [perfectoid_ring A],
  (x ∈ U) ∧ (Spa A ≊ U)

/-- The category of perfectoid spaces.-/
def PerfectoidSpace := {X : CLVRS // is_perfectoid X}

The perfectoid space project home page is here, and the source code is on github. Information for mathematicians on how to read the code above is here, and an informal write-up is here. Johan Commelin, Patrick Massot and I are in the process of writing up a more formal document explaining what we had to do to construct this type.

This definition of a perfectoid space is completely mathematically unambiguous. All terms are defined precisely, and if you don't know what something means then you can right click on it and look at its definition directly, if you compile the software using the Lean theorem prover. Lean is a computer program which checks that definitions and theorems are formally complete and correct. The definitions and theorems have to be written in Lean's language of course. The definition above says that a perfectoid space is a topological space equipped with the appropriate extra structure (sheaf of complete rings, valuation on stalks etc) and which is covered by spectra of perfectoid rings, and a perfectoid ring is a complete uniform Tate ring with the usual perfectoid property. The main work is developing enough of the theory of completions of topological rings and valuations to put the appropriate structure on the adic spectrum of a Huber pair; in total this was over 10,000 lines of code.

Computer scientists have been developing and using formal proof verification tools for decades. I want to make the slightly contentious statement that 99.9% of the time they are working on statements about the kind of mathematical objects which we teach to undergraduates (groups, graphs, 2-spheres etc). There have been some huge successes in this area (computer-checked proofs of Kepler conjecture, odd order theorem, four colour theorem etc). My impression is that on the whole mathematicians are either unaware of, or not remotely interested in, this work, which for the most part concentrates on proving tricky theorems about "low-level" objects such as finite groups. I have now seen what these systems are capable of and in particular I believe that if mathematicians, rather than computer scientists, start working with formal theorem provers then we can build a database of definitions and statements of theorems which would be far far more interesting to the mathematical community than what the computer scientists have managed to do so far. Tom Hales is leading the Formal Abstracts project which aims to do precisely this. Furthermore, I now believe that theorem provers will inevitably play some role in the future of mathematical research. I am not entirely clear about what this role is yet, but one thing I am 100 percent convinced of is that AI will not be proving hard theorems that humans can't do, any time soon. However I am equally convinced that computers will soon be helping us to do research -- perhaps by giving us powerful targeted search through databases of theorems which humans have claimed to prove, and perhaps in the future pointing out gaps in the mathematical literature, as human mathematicians begin to formalise computer proofs more and more.

  • 11
    $\begingroup$ I wish I could upvote this answer many times. The news that people are tackling the formalisation of the Stacks Project (not to mention cutting-edge mathematics, like perfectoid spaces) is very exciting. As a point of tribal pride, it's nice to see modern, dependent-type-based proof assistants used for this task, rather than eg Metamath or a ZFC-style system. $\endgroup$
    – David Roberts
    Commented Aug 8, 2018 at 6:47
  • 3
    $\begingroup$ So they are now defined in Lean? Congratulations! $\endgroup$
    – David Roberts
    Commented May 12, 2019 at 10:02
  • 4
    $\begingroup$ They are now done! In fact here's an observation which I find sociologically interesting: if I think I have proved a theorem, there comes a point where I begin to start telling other people "I can prove this theorem" -- probably before I have written anything, and perhaps even before I have checked all the final lemmas. But with this perfectoid spaces work, there was no ambiguity. For months it compiled with "a couple of warnings" indicating that, strictly speaking, work still needed to be done. But now the warnings are gone: it was a very clear-cut experience. It now compiles. $\endgroup$ Commented May 12, 2019 at 10:43
  • 1
    $\begingroup$ Hi Kevin, you mention the computer-checked proof of the Kepler conjecture, but I am sure you are aware that this proof was only for dimension 3 or less, whereas the mathematician Viazovska proved the sphere-packing theorem in dimensions 8 and 24 using simple arguments and no long computer calculations, surely this is evidence that this is kind of proof does not necessarily have an essential place in the business of proving things, as Viazovska used clever arguments to prove what took arduous computer calculations to prove only in the Euclidean case. $\endgroup$ Commented Sep 25, 2019 at 11:44
  • 4
    $\begingroup$ " surely this is evidence that this kind of proof does not necessarily have an essential place in the business of proving things" -- I dunno. Is that a logical deduction? The five colour theorem doesn't need a computer proof, but I don't think that is evidence that the four colour theorem doesn't need a computer proof. $\endgroup$ Commented Sep 26, 2019 at 20:50

Here's a much more elementary (and thus incomplete) answer.

A rough definition is given in:

What Is ... a Perfectoid Space?, by Bhargav Bhatt, Notices of the AMS Volume 61, Number 9, October 2014, pp. 1082-1084

Below I give some motivation.

To start, there is an obvious analogy between the ring of $p$-adic integers, $\mathbf{Z}_p$ and the ring of formal Taylor series over the field with $p$ elements, $\mathbf{F}_p[[t]]$ (a ring is an algebraic object like the integers: you can add, subtract, multiply, but not divide). As sets, these are both naturally identified with infinite sequences of integers mod $p$, aka $\mathbf{F}_p \cong \mathbf{Z}/(p)$. For example, if $p = 2$, so $\mathbf{Z}/(2) \cong \{0, 1\}$ (evens and odds), the sequence $(1, 1, 1, \ldots)$ corresponds to the 2-adic number $\ldots 111$, and the formal series $1x^0 + 1x^1 + 1x^2 + \cdots$ (aka, $1 + x + x^2 + \cdots$). In general, $p$-adic integers are enumerated by $a_0p^0 + a_1p^1 + a_2p^2 + \cdots$ and formal Taylor series mod $p$ by $a_0t^0 + a_1t^1 + a_2t^2 + \cdots$, so these correspond (as sets!) by replacing $p$ by $t$ (or conversely).

Just as obviously, these are different algebraically: $\mathbf{Z}_p$ has characteristic 0 (if you keep adding 1 to itself finitely many times, you don't get back to 0), but $\mathbf{F}_p[[t]]$ has characteristic $p$ (add 1 to itself $p$ times and you get 0). Both are obtained as an inverse limit, respectively of $\mathbf{Z}/(p^n)$ (integers mod $p^n$) and $\mathbf{F}_p[t]/(t^n)$ (represented by polynomials with coefficients mod $p$ and degree below $n$).

These finite steps are very simple objects that require only elementary math to work with. For example $\mathbf{Z}/(2^2) \cong \{0, 1, 10, 11\}$ (base 2, so $\{0, 1, 2, 3\}$ base 10) and $\mathbf{F}_2[t]/(t^2) \cong \{0, 1, 0 + 1t, 1 + 1t\} = \{0, 1, t, 1 + t\}$. You can easily play with the arithmetic, for example $11 \cdot 11 = 1001 \equiv 1 \pmod{100}$ (base 10: $3^2 = 9 \equiv 1 \pmod 4$, and corresponds to $3 \equiv -1$). Similarly, $(1 + t) \cdot (1 + t) = 1 + 2t + t^2 \equiv 1 \pmod{2, t^2}$.

You can think of $\mathbf{Z}_p$ as a "twisted" version of $\mathbf{F}_p[[t]]$: at each step, you need to "go around" a factor of $p$ more times to get back to 0: the characteristic is $p^n$, so in the limit the characteristic is 0, since you'd need infinitely many steps to get back to 0; while for $\mathbf{F}_p[[t]]$ the $t$ means you're growing independently from the characteristic (in a different direction): the characteristic is always $p$. (Forgetting the multiplication, the additive groups are respectively $\mathbf{Z}/(p^n)$ and $(\mathbf{Z}/(p))^n$: the former are a non-trivial group extension, while the latter is just a direct product.)

This suggestive analogy was formalized by Fontaine and Wintenberger in 1979, who showed that if you expand these algebraic objects (allow division (by $p$ or $t$), so you get a field, and $p$th roots), you get objects whose symmetries are naturally identified (the absolute Galois groups are naturally isomorphic). To make the geometry better, you also "fill in the holes" (topologically complete the space, like going from the rationals $\mathbf{Q}$ to the real numbers $\mathbf{R}$). Thus it's analogous steps as going from the ring of integers $\mathbf{Z}$ to the field of rationals $\mathbf{Q}$ to the (topologically) complete field of real numbers $\mathbf{R}$, and to the (algebraically) closed field of complex numbers $\mathbf{C}$ by including the square root of $-1$ (or fourth root of $1$ if you prefer). (The steps aren't quite the same: you add roots of unity first, and don't take the algebraic closure, but same idea.) For a field in characteristic $p$ having $p$th roots is equivalent to the $p$th power ($x \mapsto x^p$, aka Frobenius endomorphism) being 1-to-1 (thus an automorphism, meaning a self-symmetry preserving the algebraic structure), which is equivalent to being a perfect field, hence the name.

Turning to geometry, these two rings (algebraic structures), or rather the fields, correspond geometrically to points (0-dimensional spaces, geometric structures): this is why it's "algebraic geometry". You then define a perfectoid $K$-algebra as a higher-dimensional analogy. In basic algebraic geometry a field $K$ corresponds to a point, polynomials $K[x]$ in 1 variable to a line, polynomials $K[x,y]$ in 2 variables to a plane, etc.; same idea here but more technical.

Finally, a perfectoid $K$-space is what you get by gluing together perfectoid $K$-algebras. For example, in basic algebraic geometry you can glue together two lines to get a circle by gluing $x \mapsto 1/x$ (notice that we used a rational function); here "gluing" means "rigid analytic geometry", and is very technical.

Scholze's key result is that these spaces (starting either from $p$-adics or from formal Taylor series) also correspond to each other naturally (the categories are identified and preserve the topology) ...and that this clarifies many existing results and proves new ones.


video of a talk about it by Scholze at IAS

  • 6
    $\begingroup$ Yes, for the moment I suspect watching the video of the IAS talk is your best bet (though if you're in the neighborhood of Bonn, I think Scholze is giving a seminar this coming Tuesday, the 24th). $\endgroup$
    – D. Savitt
    Commented May 23, 2011 at 0:50
  • 2
    $\begingroup$ Thanks, but I'd prefer reading to watching videos. $\endgroup$ Commented May 23, 2011 at 10:17
  • 5
    $\begingroup$ @Riepe: in addition to the items you have linked to (in the main question), there is also the work of Kedlaya-Liu (see www-math.mit.edu/~kedlaya/math/relative-padic-Hodge1.pdf) as well as the work of Gabber-Ramero and Andreatta-Iovita referred to by Scholze in his talk. $\endgroup$
    – SGP
    Commented May 23, 2011 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.