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12:48 AM
Everyone at Talbot is going insane.
Here is the reason:
Claim 1. (Basterra, Richter, ...) TAQ_* of Z[x] augmented over Z is HZ_*HZ.
Claim 2. Let U: SCR \to CAlg be the forgetful map (where the RHS denotes E_infty rings over A, where A is viewed as an E_infty ring by forgetting). Then U(L_{A/B}) \cong L_{UA/UB}.
Okay, so Claim 2 is certainly false, and Claim 1 is certainly true.
Here is a proof of Claim 2. Find the mistake:
Step 1. Sp(SCR_{/A}) \cong Mod_{UA}. Proof sketch: Square zero extensions come from taking the free things and killing everything above arity 1. Sym^n(M) and M^n_{h\Sigma_n} are usually different, but they happen to be the same for n=0,1.
Step 2. The functor SCR \to CAlg^{cn}_{HZ/} preserves all homotopy limits and colimits. Proof sketch: we're okay for sifted things, and coproducts is a computation.
Step 3. It follows that stabilization for over-categories in both worlds are the same, since this is computed via hocolim of \Omega^n\Sigma^n.
Step 4. The claim follows.
Find the flaw!!
Love,
talbot
(Claim 1 should read "TAQ with coefficients in Z...". And the reason it's mentioned is that AQ, computed in the simplicial world, produces just Z again, which is a lot less interesting than HZ_*HZ)
(Step 4 should maybe read "L_{A/B} for SCRs computed a la Lurie agrees with Andre-Quillen homology classically. Or so he claims in his thesis, because it has the right answer for free algebras L_{(-)/B} is nice wrt sifted colimits. I suppose our confusion is exactly this claim about "being correct for free algebras"....)
 
 
1 hour later…
2:04 AM
Hey Talbot,
I think that the problem is that spectra & square-zero extensions / ab group objects are different
An object in Sp(SCR_{/A}) gives you a representable functor from SCR_{/A} to spectra
Whereas if I have an HA-module M and I form the associated family of square-zero extensions(A semidirect Sigma^n M) [yadda yadda connective covers] making a spectrum, I think that gives me a representable functor from SCR_{/A} to chain complexes / HZ-modules
So I don't think "abelianization = stabilization" holds
Could be wrong
oops. @Dylan
 
3:05 AM
Oh, but maybe what's specifically wrong with your program is Step 1. If you take a free algebra and kill the E_infty operations above arity 1, then you take Sym^2 (M) and mod out by M^2_{hSigma_2} (and plus lots and lots of other stuff). So that leaves you with some typically nontrivial cofiber of that map
 
 
1 hour later…
4:25 AM
You are the best
 
Who or what is "Talbot"?
 
4:44 AM
@JonathanBeardsley that's a good thing :-)
 
 
4 hours later…
8:47 AM
:37632868, I was wondering, do you have any kind of reference for this Thom isomorphism with twisted coefficients? I try to look for anything but I'm unable to find anything. with a reference (i hope) I'll be able to understand what kind of local coefficient I need to use and so I can formulate more precise questions about it
 
 
2 hours later…
10:28 AM
@Dylan I am not sure you need it anymore, but Ab(SCR_/A) "only" admits a fully faithful embedding to Sp(SCR_/A).
 
 
6 hours later…
4:51 PM
@LuigiM The second Stiefel-Whitney class of an orientable vector bundle over $X$ can be calculated as follows from a cellular decomposition. Let's assume there's one 0-cell for convenience. Each 2-cell is attached by some element of the fundamental group; tracing it out determines whether it's orientation preserving or reversing. Then consider the 2-cochain that sends a 2-cell to $\pm 1$ depending on whether its boundary is an orientation preserving or reversing loop.
It is not hard to show that this gives a cocycle, and the corresponding cohomology class is the second Stiefel-Whitney class. So you could probably carry this out for your question.
I don't know a reference for the Thom isomorphism with twisted coefficients but if you go through the proof in the case that the vector bundle is non-orientable the modifications are pretty natural.
 
5:06 PM
@AaronMazel-Gee thanks, I have to spend some time trying to understand that model structure then. My situation is actually pretty nice in terms of B, as it is just sset (so everything is cofibrant). Unfortunately my A is not even a model category (much less simplicial!). I may have to do something else in the end
 

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