Homotopy Theory

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9:58 AM

In HA, remark 7.4.1.12, Lurie writes that the cotangent complex of $A\to B$ is the fiber of the multiplication map $B\otimes_A B\to B$. The more classical definition of TAQ is given by the indecomposables of the augmentation ideal of $B\otimes_A B$. Lurie's definition seems to bypass the step of taking indecomposables. What am I missing?

Denis Nardin

@BrunoStonek Isn't he talking about the E_1-cotangent complex? (which is very different from the E_∞ one, what's usually called TAQ)

From what I understand A and B are not even commutative algebras here

It could be useful to have a notation that indicates the operad wrt we are taking the cotangent complex though...

Bruno Stonek

oooh. I see

2 hours later…

Bruno Stonek

11:42 AM

here's another confusion. HA 7.5.4.5 says that an étale map of connective E_infty algebras is TAQ-étale. On the other hand, 7.5.4.6 says that étale maps are THH-étale, i.e. the unit of THH is an equivalence (no connectivity hypothesis). But THH-étale implies TAQ-étale, so this would say that an étale map is TAQ-étale, with no connectivity hypothesis... which is not true

Tyler Lawson

i think etale does imply taq-etale in general. if you have an etale map it's base-extended from the connective covers, and TAQ is also base-extended

i think that there's a reverse implication (TAQ-etale => formally etale) that does require connectivity

Bruno Stonek

right, for example DAG VII 8.9

so on 7.5.4.5 one can remove the connectivity hypothesis...?

Tyler Lawson

uff. i may have assumed commutativity rather than E_k for my base-extension formula

i missed that we were talking about E_k-TAQ

Bruno Stonek

hmm, I don't know what you have in mind, but I'm fine with E_infty

Denis Nardin

IIRC For connective rings étale=TAQ-étale=THH-étale

In general you only have TAQ-étale ⇐ THH-étale

Bruno Stonek

11:55 AM

doesn't HA 7.5.4.6 say that arbitrary étale maps are THH-étale?

Denis Nardin

Isn't there a "connective" hypothesis there?

Bruno Stonek

I don't see it

Denis Nardin

Oh wait I thought you were talking about 7.5.4.5

And I am dumb because the implications are obviously in the other direction

Ok, now magic has happened and my statement is correct

Bruno Stonek

ok, you write étale implies TAQ-étale. In HA this is 7.5.4.5 but he puts a connectivity hypothesis

Denis Nardin

Hmm... I am a little bit unsure about it to be honest. I am unconfortable with étale maps between nonconnective rings

Bruno Stonek

12:01 PM

wait, you wrote in general TAQ-étale implies THH-étale but I think it's the other way around...

(Rognes, "Galois extensions of structured ring spectra", 9.4.4)

Denis Nardin

Ok, I am making a mess of things

What is true is that THH-étale is a stronger condition than TAQ-étale

Let me retract what I wrote before and say that every THH-étale morphism is TAQ-étale (this is proved by Rognes in that paper) and for connective rings the two things coincide and coincide with the "strong" notion of étale

But in general neither THH-étale nor TAQ-étale morphisms need to be flat

Bruno Stonek

right. the connectivity hypothesis in 7.5.4.5 still puzzles me, though...

Denis Nardin

I thought Toën and Vezzosi proved it without that hypothesis, but I cannot seem to find it

Bruno Stonek

well, 7.5.4.6 proves that étale implies THH-étale in general. then Rognes proved that THH-étale implies TAQ-étale in general

that seems to give a proof

Denis Nardin

Right, Ok, this makes sense

Bruno Stonek

12:10 PM

(that the conclusion in 7.5.4.6 is equivalent to THH-étaleness is remarked by Mathew in "THH and base-change for Galois...", section 5)

wonder why Lurie wrote "connective" in 7.5.4.5. is it even used in the proof?

Denis Nardin

Yeah, that is standard: thh-étale means simply 0-cotruncated

No, it's not used

4 hours later…

user131753

4:12 PM

I was recently going through this notes. However I didn't understand the proof of Proposition 1.11 (page 8). Specifically the following, "But then using a compatibility of $e_1$ and $ε_1$ we get that the composite natural transformation is $e_2$ as required." I don't understand how we can get $e_2$. Since I haven't understood this part I didn't go further. Can anyone help me here?

Denis Nardin

4:31 PM

@user170039 They're just using that the composition $\epsilon_1F\circ Fe_1$ is the identity, by the definition of adjoint equivalence. If you apply $G_2$ to that you see that the composition of the last two arrows is the identity, and so the whole composition coincides with the first arrow (i.e. $e_2$)

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