Here's Cisinski's talk this morning on his new proof of straightening/unstraightening with Kim Nguyen: youtube.com/watch?v=OnMUka9bLAw

@AlexanderCampbell Amazing! This answers my question from a while back I believe.

@AlexanderCampbell Do you know whether this should appear in a paper soon or not?

Notation question: I am writing a little piece on sections of bundles of smooth manifolds. I care about the regularity of these bundles: I need to switch between smooth, L^2, H^1 (Sobolev) etc regularity. How do you denote the space of sections of these bundles? I have seen C^\infty(TM), L^2(TM), for example, but this is ambiguous. Any suggestions?

@ThomasRot You could always use C^{\infty}(M;TM), L^2(M;TM) etc... maybe you want something shorter though.

Yeah, maybe that is what I will settle on. I also need all functions M\rightarrow TM, which I would denote by $C^\infty(M,TM)$. This seems to be a good option

@ThomasRot I like the convension in which the sheaf of sections is denoted, e.g., L^2_E(-) with the bundle as a subscript.

@S.carmeli: That is also nice So for $L^2$ vector fields you would write $L^2_{TM}(M)$?

Possibly dumb question: it's common wisdom that for (non-connective) dg algebras A over k characteristic 0, there's a computation of the derived fiber product along the diagonal map: $$A \times_{A \times A} A \simeq A \oplus A[-1]$$. In the cdga setting I think one can show this via a Dold-Kan argument, or by choosing a resolution.

In characteristic 0 the same statement should be true for $E_\infty$ algebras. How does one prove this directly in the $E_\infty$-ring setting?

@Bbb It's a general fact that $A×_{A×A} A\cong \lim_{S^1}A$ in every ∞-category with finite limits (because $S^1\cong \ast\amalg_{S^0}\ast$). Then you just use the fact that there is a split fiber sequence $ΩA→\lim_{S^1}A→A

So this is just true, in every stable category

(you don't really use the algebra structure on either side of the equality)

Unless you meant to give the rhs the square zero multiplication, in which case it is true only in a $\mathbb{Z}$-linear setting

I did mean square-0 multiplication: does it make sense to talk about square-0 extensions in an general Z-linear setting?

Yes, it makes sense to talk about square-zero extensions in general. But in general the cohomology of $S^1$ is not square zero (it is, however, if you consider integral cohomology)

For example if you take $A=\mathbb{S}$ the sphere spectrum, the rhs has the multiplication induced by $\eta:\mathbb{S}[-1]\otimes\mathbb{S}[-1]\cong\mathbb{S}[-2]\to\mathbb{S}[-1]$ (multiplication by the Hopf map)

In fact this is true in general. I think (but I haven't verified the details) that a necessary and sufficient condition for the rhs to be a square 0 extension is that the image of the Hopf map is zero under the unit $\mathbb{S}\to A$

Nice, gonna meditate on this and maybe come back with followups, thank you!

Hmm.. wait the last statement cannot literally be true

Regardless, that's pretty much the intuition

what's a reference for talking about square-0 extensions in an Z-linear setting?

what do you mean? Square 0 extensions of E_∞-algebras? (Or E_1-algebras)?

(to be clear: when I say $\mathbb{Z}$-linear, I mean what people sometimes call "dg settings" as opposed to more homotopical setups)

Oh I thought you meant something along the lines of it makes sense in stable infnity categories tensored over Z-modules or something

Yeah, that's the same thing

But square-0 extensions can be defined in an arbitrary stable category

See Higher Algebra, 7.4.1

Ahh right, forgot about this, makes sense now, thanks for clarifying.

@ThomasRot exactly.