What are "perfectoid spaces"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:54:55Z http://mathoverflow.net/feeds/question/65729 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65729/what-are-perfectoid-spaces What are "perfectoid spaces"? Thomas Riepe 2011-05-22T20:30:25Z 2013-03-26T09:10:43Z <p><a href="http://www.zib.de/de/mathematik-kalender.html?oid=27084144" rel="nofollow" title="link">This</a> 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? </p> <p>Edit: A bit more infos can be found in Peter Scholze's <a href="http://www.math.uni-bonn.de/people/scholze/ARGOS.pdf" rel="nofollow" title="pdf">seminar description</a> and in <a href="http://www-personal.umich.edu/~bhattb/almost_purity_2011/plan.pdf" rel="nofollow" title="pdf">Bhargav Bhatt</a>'s.</p> <p>Edit: Peter Scholze posted yesterday <a href="http://arxiv.org/abs/1111.4914" rel="nofollow" title="arxiv">this beautiful overview</a> on the arxiv.</p> <p>Edit: Peter Scholze posted today <a href="http://de.arxiv.org/abs/1303.5948" rel="nofollow" title="arxiv">this new survey</a> on the arxiv.</p> http://mathoverflow.net/questions/65729/what-are-perfectoid-spaces/65735#65735 Answer by SGP for What are "perfectoid spaces"? SGP 2011-05-22T22:40:02Z 2011-05-22T22:40:02Z <p><a href="http://video.ias.edu/graf/2011/03/22/scholze#attachments" rel="nofollow">video of a talk about it by Scholze at IAS</a></p> http://mathoverflow.net/questions/65729/what-are-perfectoid-spaces/66563#66563 Answer by Peter Scholze for What are "perfectoid spaces"? Peter Scholze 2011-05-31T15:41:03Z 2011-11-04T10:57:55Z <p>Update: The lecture notes of the CAGA lecture series on perfectoid spaces at the IHES can now be found online, cf. <a href="http://www.ihes.fr/~abbes/CAGA/scholze.html" rel="nofollow">http://www.ihes.fr/~abbes/CAGA/scholze.html</a>.</p> <p>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.</p> <p>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 $\leq 1$.</p> <p>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).</p> <p>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 <code>$\mathbb{Q}_p(\mu_{p^\infty})$</code>.</p> <p>In this situation, one can form the field $K^\prime$, given as the fraction field of <code>$K^{\prime +} = \lim_{\leftarrow} K^+/p$</code>, where the transition maps are given by Frobenius. Concretely, in the first example it is given by the completion of <code>$\mathbb{F} _p((t^{1/p^\infty}))$</code>, where $t$ is the element <code>$(p,p^{\frac 1p},p^{\frac 1{p^2}},\ldots)$</code> in <code>$K^{\prime +}=\mathrm{lim}_{\leftarrow} K^+/p$</code>.</p> <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:</p> <blockquote> <p>Theorem: There is a canonical isomorphism of absolute Galois group $G_K\cong G_{K^\prime}$.</p> </blockquote> <p>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.</p> <p>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.</p> <p>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:</p> <p>$K^+$-mod $\rightarrow$ $K^+$-mod / $\mathfrak{m}$-torsion $\rightarrow$ $K^+$-mod / $p$-power torsion.</p> <p>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.</p> <p>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.</p> <p>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) etale maps. In the following, we will often need the notion of an almost finitely presented etale map, which is the almost analogue of a finite etale cover.</p> <blockquote> <p>Theorem (Tate, Gabber-Ramero): If $L/K$ finite extension, then $L^+/K^+$ is almost finitely presented etale. Similarly, if $L^\prime/K^\prime$ finite, then $L^{\prime +}/K^{\prime +}$ is almost finitely presented etale.</p> </blockquote> <p>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 etale, 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 etale over $K_n^+$ up to $p^{1/p^n}$-torsion. Now take the limit as $n\rightarrow \infty$ to see that $L^+$ is almost etale over $K^+$.</p> <p>Now we can prove the theorem above:</p> <p>finite etale covers of $K$ = almost finitely presented etale covers of $K^+$ = almost finitely presented etale covers of $K^+/p$ [because (almost) finite etale covers lift uniquely over nilpotents] = almost finitely presented etale covers of $K^{\prime +}/t$ [because $K^+/p = K^{\prime +}/t$, cf. the example] = almost finitely presented etale covers of $K^{\prime +}$ = finite etale covers of $K^\prime$.</p> <p>After we understand this theory on the base, we want to generalize to the relative situation. Here, let me make the following claim.</p> <blockquote> <p>Claim: $\mathbb{A}^1_{K^\prime}$ "equals" $\lim_{\leftarrow} \mathbb{A}^1_K$, where the transition maps are the $p$-th power map.</p> </blockquote> <p>As a first step towards understanding this, let us check this on points. Here it says that $K^\prime = \lim_{\leftarrow} K$. 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 Teichm"uller representatives) and again this can be explained in an example:</p> <p>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</p> <p>$\lim_{n\rightarrow \infty} (p^{-1/p^n} + 5 + p^{3/p^n})^{p^n}$</p> <p>exists, and this defined $[x^\prime]\in K$. Now the map $K^\prime\rightarrow \lim_{\leftarrow} K$ is given by $x^\prime\mapsto ([x^\prime],[x^{\prime 1/p}],[x^{\prime 1/p^2}],\ldots)$.</p> <p>In order to prove that this is a bijection, just note that</p> <p>$K^{\prime +} = \lim_{\leftarrow} K^{\prime +}/t^{p^n} = \lim_{\leftarrow} K^{\prime +}/t = \lim_{\leftarrow} K^+/p \leftarrow \lim_{\leftarrow} K^+$.</p> <p>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.</p> <p>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:</p> <blockquote> <p>Theorem: We have a homeomorphism of underlying topological spaces $|(\mathbb{A}^1_{K^\prime})^{\mathrm{ad}}|\cong \lim_{\leftarrow} |(\mathbb{A}^1_K)^{\mathrm{ad}}|$.</p> </blockquote> <p>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?</p> <blockquote> <p>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 $R^\circ/p^{\frac 1p}\cong R^\circ/p$.</p> </blockquote> <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 $p$.</p> <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.</p> <p>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 etale topoi. [This is why we dont have to take $\lim_\leftarrow \mathbb{A}^1_{K^\prime}$: It does not change the topological spaces. In order to compare structure sheaves, one should however take this inverse limit.]</p> <p>The really exciting theorem is the following, which I call the tilting equivalence:</p> <blockquote> <p>Theorem: The category of perfectoid $K$-algebras and the category of perfectoid $K^\prime$-algebras are equivalent.</p> </blockquote> <p>The functor is given by $R^\prime = (\lim_{\leftarrow} R^\circ/p)[t^{-1}]$. Again, one also has $R^\prime = \lim_{\leftarrow} R$, where the transition maps are the $p$-th power map, giving also the map $R^\prime\rightarrow R$, $f^\prime\mapsto [f^\prime]$.</p> <p>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 etale covers:</p> <p>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 $\overline{A}/p^{\frac 1p}\cong \overline{A}$,</p> <p>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 $\overline{A}$.</p> <p>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 *pre*sheaf $\mathcal{O}_X$, and the sub*pre*sheaf $\mathcal{O}_X^+$, consisting of functions which have absolute value $\leq 1$ everywhere.</p> <blockquote> <p>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 <code>$|f^\prime(x^\prime)| = |[f^\prime] (x)|$</code>. Rational subsets are identified under this homeomorphism. ii) For any rational subset $U\subset X$, the pair <code>$(\mathcal{O}_X(U),\mathcal{O}_X^+(U))$</code> is affinoid perfectoid with tilt <code>$(\mathcal{O}_{X^\prime}(U),\mathcal{O}_{X^\prime}^+(U))$</code>. 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.</p> </blockquote> <p>This allows one to define general perfectoid spaces by gluing affinoid perfectoid spaces. Further, one can define etale morphisms of perfectoid spaces, and then etale topoi. This leads to an improvement on Faltings's almost purity theorem:</p> <blockquote> <p>Theorem: Let $R$ be a perfectoid $K$-algebra, and let $S/R$ be finite etale. Then $S$ is perfectoid and $S^\circ$ is almost finitely presented etale over $R^\circ$.</p> </blockquote> <p>In particular, no sort of semistable reduction hypothesis is required anymore. Also, the proof is much easier, cf. the book project by Gabber-Ramero.</p> <p>Tilting also identifies the etale topoi of a perfectoid space and its tilt, and as an application, one gets the following theorem.</p> <blockquote> <p>Theorem: We have an equivalence of etale topoi of adic spaces: <code>$(\mathbb{P}^n_{K^\prime})^{\mathrm{ad}}_{\mathrm{et}}\cong \lim_{\leftarrow} (\mathbb{P}^n_K)^{\mathrm{ad}}_{\mathrm{et}}$</code>. Here the transition maps are again the $p$-th power map on coordinates.</p> </blockquote> <p>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 etale cohomology. Moreover, we have the projection $\pi: \mathbb{P}^n_{K^\prime}\rightarrow \mathbb{P}^n_K$, at least on topological spaces or etale 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 etale cohomology of a variety in characteristic $0$ with the etale cohomology of characteristic $p$. Using this, it is easy to verify the weight-monodromy conjecture for $X$.</p>