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?

@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...

oooh. I see

2 hours later…
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

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

right, for example DAG VII 8.9
so on 7.5.4.5 one can remove the connectivity hypothesis...?

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

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

IIRC For connective rings étale=TAQ-étale=THH-étale
In general you only have TAQ-étale ⇐ THH-étale

11:55 AM
doesn't HA 7.5.4.6 say that arbitrary étale maps are THH-étale?

Isn't there a "connective" hypothesis there?

I don't see it

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

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

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

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)

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

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

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

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

Right, Ok, this makes sense

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?

Yeah, that is standard: thh-étale means simply 0-cotruncated
No, it's not used

4 hours later…
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?

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. $e_2$)