# Witt vectors, the cotangent complex, and a solid construction of $B_{dR}^+$

In a remarkable lecture delivered on October 29th: New Foundations for functional analysis, Dustin Clausen suggests at the 40 minute mark a remarkable new construction interpretation of Fontaine's ring $$B_{dR}^+$$ making use of solid modules, the condensed analog of non-Archimedean completeness. This relies on a construction of the solid cotangent complex of a perfectoid field, which happens to be as simple as possible, a suspension of the field, and then proceeds to construct the desired period ring by analogy with the Witt vectors of a perfect ring. It's possible that I have misunderstood this story.

With the understanding that mathoverflow may not be the best place to push on unpublished results (feel free to close the question if it crosses the line), I wondered if anyone has a reference for or could offer a fully deformation theoretic construction of the Witt vectors, to at least have the other side of the analogy.

It seems to me that I can get a decent a priori construction of the Witt vectors by leveraging perfection (for cotangent vanishing and multiplicative lifting via p-adic contracting property of the Frobenius), but it might be difficult to get from here to the formulas, the $$F$$ and $$V$$ operators, the fact that when we begin with a field the result is a DVR. Excellent references abound, but I am curious whether there is any which begins and remains with the perspective of the cotangent complex.

The relevant blackboard is:


Edit: In his answer, Z.M explains that the mathematical reason does not apply, and in Peter's answer he explains that the moral reason does not apply! So I was overly pessimistic on both counts, and in spite my ignorance my claims from my talk are perfectly substantiated!

But for now let me just give the take-away:

I should not have implied that we give a new construction of $$\BdR$$. Rather I should have said that we give a universal property for $$\BdR$$: it is the universal pro-nilpotent thickening in solid rings.

The two reasons why I was cheating:

1. I don't think it's actually true that the contangent complex being $$F[1]$$ implies, for purely formal reasons, a full universal one-parameter formal deformation. Certainly it does give the first-order deformation. But that only uses that the $$H_1$$ is $$F$$ and the $$H_0$$ is $$0$$, not that the others vanish. Using the vanishing of the cotangent complex in higher degrees, you do get a good deal of knowledge about higher deformations by iteratively using transitivity triangles, and it feels like it should be saying a lot, but as far as I know it's still incomplete. To get around this problem and actually make a full one-parameter formal deformation, one option is to work relative to an already-existing one. So in the perfect field case, you can use that $$\mathbb Z$$ is already a one-parameter deformation of $$\mathbb F_p$$ and that the cotangent complex of $$k$$ over $$\mathbb F_p$$ vanishes; this is the approach discussed in section six of Bhargav's notes, which is where I learned it. In the perfectoid case, you at least see in the same way that if you construct $$\BdR$$ by hand for a base perfectoid, then you get it for all perfectoids over that one. Not as snazzy as what I claimed, but it's also not nothing: for example, $$\Ainf$$ and $$\BdR$$ for the $$p$$-cyclotomic extension are very explicit.

2. Even if it were true that you get the construction for formal reasons from the cotangent calculation, the way I know how to get the cotangent calculation uses $$\Ainf$$. So it would still be morally, if not mathematically, incorrect to claim this is a new construction of $$\BdR$$.

• I definitely don't mean to catch anyone out. I'm very grateful for the quick clarification and for those notes, which I hadn't seen before. Regardless of the novelty of the construction it sounds like it will be at least illuminating. Viva condensery, this stuff makes me wish I were still a student! Oct 31, 2021 at 21:26
• I believe that what you original said is true. I also thought that an already existing base is necessary as in Bhatt, but when you said this during the conference, I realized that it is redundant: if $B\to R$ is the universal thickening of $R$, and let $J$ be its kernel, then $J^n/J^{n+1}$ is a symmetric product of $J/J^2$, thus if $J/J^2$ is flat, then so are all these associated gradeds. I worked out a variant of this in my thesis.
– Z. M
Nov 2, 2021 at 16:31
• @pupshaw Yes, but after some thoughts, I should not refer to my work in this case, since this case essentially appeared in Arpon Raksit's work, and I write an answer to explain it now.
– Z. M
Nov 3, 2021 at 10:48
• @LSpice $A_{inf}$ is really called $A_{inf}$, I'm afraid. None of us chose it but even very terminologically adventurous people have not seen fit to change it. $A_\infty$ makes me think of operads... Nov 3, 2021 at 22:05
• @pupshaw, fixed. Sorry! Nov 3, 2021 at 22:55

Let me point out that, the original statement in the video of Dustin Clausen's talk is essentially correct (although it might have been inadvertent, according to his answer), thanks to Arpon Raksit's paper Hochschild homology and the derived de Rham cohomology revisited (arXiv link).

For sake of simplicity, we fix a base ring $$R$$. Roughly speaking, there is a nonconnective version of animated $$R$$-algebras due to Bhatt–Mathew, which is named after derived (commutative) $$R$$-algebras in Raksit's paper (§4). The point is that, animated $$R$$-algebras are precisely modules over the monad $$\mathbb{L}{\operatorname{Sym}} : D (R)_{\geq 0} \rightarrow D (R)_{\geq 0}$$ on the $$\infty$$-category of connective $$R$$-module spectra (or connective $$R$$-modules, by abuse of terminology). Using Goodwillie calculus, one can extend to the whole $$\infty$$-category $$D (R)$$ of $$R$$-modules, obtaining a monad $$\mathbb{L}{\operatorname{Sym}} : D (R) \rightarrow D (R)$$, and derived $$R$$-algebras are modules over this monad, whose $$\infty$$-category will be denoted by $${\operatorname{DAlg}}_R$$.

This formalism applies more generally to derived algebraic contexts (4.2.1) in place of $$D (R)$$, and in particular, to the $$\infty$$-category $$\widehat{{\operatorname{DF}}}^{\geq 0} (R)$$ of completely (nonnegatively decreasingly) filtered $$R$$-modules, obtaining a monoid, whose modules are completely filtered derived $$R$$-algebras, or equivalently, nonnegative $$h_-$$-differential graded derived $$R$$-algebras $${\operatorname{DG}}_-^{\geq 0} {\operatorname{DAlg}}_R$$ due to 5.1.5.

Similar to 5.3.2 (and even easier), the functor $$\operatorname{gr}^0\colon{\operatorname{DG}}_-^{\geq 0} {\operatorname{DAlg}}_R \rightarrow {\operatorname{DAlg}}_R$$ admits a left adjoint $${\operatorname{Cpl}} : {\operatorname{DAlg}}_R \rightarrow {\operatorname{DG}}_-^{\geq 0} {\operatorname{DAlg}}_R$$. There are two ways to understand this functor:

1. By adjunction, $$\operatorname{Cpl}(A)$$ is the initial completely filtered derived $$R$$-algebra $$B$$ equipped with a map $$A\to\operatorname{gr}^0(B)$$ of derived $$R$$-algebras. In some sense, this is closely related to the final object in the "pro-infinitesimal site" of $$A/R$$ (reminder: the site is opposite to the category of rings);
2. If we start with a quotient $$R / I$$ where $$I \subseteq R$$ is an ideal generated by a Koszul-regular sequence, then the image under this left adjoint is the $$I$$-completion $$R_I^{\wedge}$$ along with the $$I$$-adic filtration (the readers could convince themselves that this object is somehow initial), so this functor should be understood as a variant of adic completion, which justifies the notation.

The proof of 5.3.6 essentially leads to the following:

Proposition. Given a derived $$R$$-algebra $$A$$, the $$n$$-th associated graded piece $${\operatorname{gr}}^n ({\operatorname{Cpl}} (A))$$ is given by the symmetric product $$\mathbb{L}{\operatorname{Sym}}_A^n ({\operatorname{gr}}^1 ({\operatorname{Cpl}} (A)))$$ (and in particular, $${\operatorname{gr}}^0 ({\operatorname{Cpl}} (A)) = A$$), and we have an equivalence $$\mathbb{L}_{A / R} \simeq \Sigma {\operatorname{gr}}^1 ({\operatorname{Cpl}} (A))$$ of $$A$$-modules where $$\mathbb{L}_{A / R}$$ is the (algebraic) cotangent complex of $$R$$.

In particular, consider the case that $$R =\mathbb{Z}$$. If $$A =\mathbb{F}_p$$, or more generally, $$A$$ is given by a perfect $$\mathbb{F}_p$$-algebra, then the cotangent complex $$\mathbb{L}_{A / R}$$ has $${\operatorname{Tor}}$$-amplitude in $$[1, 1]$$, therefore the $$A$$-module $${\operatorname{gr}}^1 ({\operatorname{Cpl}} (A))$$, and consequently the $$A$$-modules $${\operatorname{gr}}^{\ast} ({\operatorname{Cpl}} (A))$$, are flat. It follows that the underlying derived ring $${\operatorname{Fil}}^0 ({\operatorname{Cpl}} (A))$$ of $${\operatorname{Cpl}} (A)$$ is concentrated in degree $$0$$, and there is a surjective map $${\operatorname{Fil}}^0 ({\operatorname{Cpl}} (A)) \rightarrow {\operatorname{gr}}^0 ({\operatorname{Cpl}} (A)) = A$$ of rings whose kernel $$I$$ is quasiregular, and by Quillen's result, $${\operatorname{Fil}}^n ({\operatorname{Cpl}} (A))$$ are simply given by $$I^n$$.

On the other hand, if we consider a reasonable ($$\infty$$-)category of deformations of the $$R$$-algebra $$A$$, say the category of surjective maps $$B \twoheadrightarrow A$$ of $$R$$-algebras whose kernel is finitely generated (the completeness with respect to a non-finitely generated ideal is usually pathologic), then this ($$\infty$$-)category should be a full subcategory $${\operatorname{Def}}_R$$ of $$\{ S \in {\operatorname{DG}}_-^{\geq 0} {\operatorname{DAlg}}_R | {\operatorname{gr}}^0 (S) \simeq A \}$$. Having this in mind, when the cotangent complex $$\mathbb{L}_{A / R}$$ is, as an $$A$$-module, equivalent to a shift $$P [1]$$ where $$P$$ is a finite projective $$A$$-module, then $${\operatorname{Cpl}} (A)$$ lies in this subcategory $${\operatorname{Def}}_R$$ and the universal property follows.

• I'm probably missing something, but paragraphs 4 and 5 seem incompatible outside characteristic $0$ to me. For instance, if $A = R/I$ with $I$ regular (say), then $L_{A/R} = I/I^2[1]$, so $Sym^n(L_{A/R}[1]) = \Gamma^n(I/I^2)[1]$, which suggests that the left adjoint produces the completed PD-envelope rather than the completion. (This affects the perfect ring assertion, but is of course not an issue in characteristic 0 where Dustin wanted to apply it.) Nov 3, 2021 at 12:20
• @Anonymous $\operatorname{gr}^n$ is the symmetric product of $\operatorname{gr}^1$, not the cotangent complex (by the way, a typo in your formula, the divided power of the cotangent complex should be shifted by $n$, not $1$), and the cotangent complex is the suspension of $\operatorname{gr}^1$. This is precisely the difference between $h_+$ and $h_-$ differential graded algebras — the $h_+$ ones have a PD-structure (roughly speaking) while the $h_-$-ones do not.
– Z. M
Nov 3, 2021 at 13:18
• Ah, thanks. But it seems you wrote (in the answer) that $gr^1$ is the suspension of the cotangent complex instead of the other way around ? Nov 3, 2021 at 13:24
• @Anonymous Thanks, that was a typo.
– Z. M
Nov 3, 2021 at 13:25
• @pupshaw For the acyclicity of LSym, it follows from Lazard's theorem by realizing flat modules as filtered colimits of free modules. See Lurie's Spectral Algebraic Geometry, Cor 25.2.3.3. In fact, I mistakenly referred to my thesis (the chapters of which are available at my homepage and on arXiv — click my profile if you are interested), which addresses a special version of a PD-variant of this which is sufficient to study the crystalline cohomology. Since the rest of that comment is informative, I did not remove it.
– Z. M
Nov 3, 2021 at 15:02

Let me add to Z. M's answer, and note that Dustin has no reason to apologize at all: What he said is literally correct.

Namely, one can directly show that $$L_{F/\mathbb Z}^\blacksquare$$ is isomorphic to $$F[1]$$, for any perfectoid field (or even ring) $$F$$. Here are the steps:

1. One has $$L_{F/\mathbb Z}^\blacksquare = L_{\mathcal O_F/\mathbb Z}^\blacksquare\otimes_{\mathcal O_F} F$$. This is in fact true already before solidification, and basically formal.

2. The solidification $$L_{\mathcal O_F/\mathbb Z}^\blacksquare$$ is derived $$p$$-complete. This holds true more generally for any derived $$p$$-complete solid ring $$A$$ in place of $$\mathcal O_F$$. The key input is that solid tensor products preserve derived $$p$$-complete (connective) complexes of solid modules.

3. Thus, to show that $$L_{\mathcal O_F/\mathbb Z}^\blacksquare\cong \mathcal O_F[1]$$, it suffices to show that $$L_{(\mathcal O_F/p)/\mathbb F_p}^\blacksquare\cong \mathcal O_F/p[1]$$. In fact, the latter holds even before solidification.

4. Now $$\mathcal O_F/p = \mathcal O_F^\flat/t$$ for some perfect ring $$\mathcal O_F^\flat$$ and nonzerodivisor $$t$$, from which the computation follows easily (using the vanishing of the cotangent complex of perfect rings).

(These types of computations of cotangent complexes, and the avoidance of $$A_{\mathrm{inf}}$$, are very much as in my thesis. The new thing is really that we can directly characterize $$B_{\mathrm{dR}}^+(F)$$ as the universal solid pro-infinitesimal thickening of $$F$$: In classical language the required "completeness" of the thickening was extremely hard to formulate.)

• Thanks, Peter! I didn't know one could avoid A_inf. This is much nicer than what I was thinking of. Nov 4, 2021 at 10:47
• @pupshaw Another way to rephrase this map is that this is essentially the first Breuil–Kisin twist. See the first line on page 9 of arxiv.org/abs/2111.01830
– Z. M
Nov 4, 2021 at 17:47
• @Z.M Well, I guess I just meant it's nicer from the perspective of making this discussion independent of the usual definition of B_dR. I'm a an of A_inf too. Nov 4, 2021 at 18:28
• Another question: I am not sure about the meaning of the solidifcation of $L_{A/\mathbb Z}$ when $A$ is a $p$-complete ring. For example, when $A$ is $p$-completion of a discrete ring, seemingly the "natural" one to consider is the $(A,A)_\blacksquare$-solidification rather than $(A,\mathbb Z)_\blacksquare$.
– Z. M
Nov 4, 2021 at 18:33
• @Z.M Which solidification to take doesn't actually matter: It's enough to take the one for $\mathbb Z$, the result will already be solid over $A$. It's part of the magic :-). And at least in my thesis, I found it psychologically quite useful to avoid $A_{\mathrm{inf}}$; I think it's no coincidence that the tilting equivalence was first proved without the intermediary of $A_{\mathrm{inf}}$, even if with it it's quite simple. These big rings are just a bit confusing... Nov 4, 2021 at 18:47