« first day (2768 days earlier)      last day (635 days later) » 

Alexander Campbell
Alexander Campbell
2:33 AM
@JonathanBeardsley Yes (if I've understood your question). One way to see this is that the $p$-cartesian morphisms form the right class of a factorisation system on the total category $E$.
 
Jonathan Beardsley
Jonathan Beardsley
@AlexanderCampbell oh yes that's a perfect argument! So they're closed under limits in the arrow category. Thanks!
I just knew that had to be true
 
Tyler Lawson
Tyler Lawson
3:37 AM
the odd-primary signs are tough because you have to be very very careful with casual identification. (eg: have you ever had to worry about what the sign is on the connecting homomorphism in the long exact sequence on homology, or the Mayer-Vietoris sequence?) there's an ugly normalization in Steenrod-Epstein's definition of operations to make them satisfy (a) stability and (b) the p'th power identity.
those two properties are kind of what you have to trace everything back to.
(A senior mathematician once told me that they spent two months reconciling signs in the Nishida relations.)
if somebody can write something that nails it down in a systematic way, that would be fantastic.
 
Tyler Lawson
Tyler Lawson
3:54 AM
but i mean, let's be serious here for a second.

at p=3, β Q^1 goes from degree k to degree k+3.

so if I have an HF_3-algebra A, with a map S^k -> A representing my element ... what is is β Q^1?

a map from S^{k+3}?

a map from S^k ⊗ S^3?

a map from S^3 ⊗ S^k?

...which of these are the same?
it seems to require great patience.
2
 
 
4 hours later…
Edoardo Lanari
Edoardo Lanari
7:38 AM
@JonathanBeardsley Another way to see this is that cartesian arrows can be characterized as those inducing a certain pullback square (see Prop.2.7. of ncatlab.org/nlab/show/Cartesian+morphism ), and these are stable under limits.
 
 
2 hours later…
Bryan Shih
Bryan Shih
9:43 AM
Hi all, I wonder if there is a reference on the remark in 7.1.3 of Bhatt, Lurie's and Mathew's paper https://arxiv.org/abs/1805.05501 :

The statement is:
The chain complexes of free Z-modules whose homotopy category is identified with derived category of Z modules is a consequence of Z having finite projective dimension.


I'd be glad if some there is soome commentary on this statement too.
(I'm not clear what is meant there. Don't all rings R not have finite projective dimensiono in D(R)? )
 
Dustin Clausen
Dustin Clausen
10:04 AM
Hi Bryan, R as an R-modules always has projective dimension 0, right. But what they mean when they say "R has finite projective dimension" is that there is a d such that every R-module has projective dimension \leq d
If R has finite projective dimension, you can represent every unbounded complex by a complex of free R-modules. But otherwise you get stuck. You might want to check out the Spaltenstein reference cited by Bhatt Lurie Mathew
If we were dealing with homological complexes bounded below there would be no issue, so this is really an unbounded complex thing
 
Bryan Shih
Bryan Shih
Hi Dustin, ok, will check, and thanks again!
 
Dustin Clausen
Dustin Clausen
10:22 AM
Sure thing :)
 
Bryan Shih
Bryan Shih
11:13 AM
That's great, so I see one has to go through the proof of Lemma 3.3 of your referenced paper!

Though I am still unclear on 7.1.1. (2). Do we have any good controll of the mapping set of objects in Ch(Z)^{tf} chain complexes of p-torsion free abelian groups. It seems to be implied that in 7.1.16 these are given by homtopy classes of chain maps.
 
 
4 hours later…
Jonathan Beardsley
Jonathan Beardsley
3:22 PM
@EdoardoLanari thankyou! That seems like a very useful fact in general
 
 
1 hour later…
Rune Haugseng
Rune Haugseng
4:23 PM
@JonathanBeardsley For F : C -> D left adjoint to G, which is lax monoidal, the oplax structure map F(x \otimes y) -> Fx \otimes Fy is always the adjoint of the composite x \otimes y -> GFx \otimes GFy -> G(Fx \otimes Fy) where the first map is tensoring two copies of the unit and the second is the lax structure map for G
 
Rune Haugseng
Rune Haugseng
4:51 PM
In other words, it's the composite F(x \otimes y) -> F(GFx \otimes GFy) -> FG(Fx \otimes Fy) -> Fx \otimes Fy where the last map is the counit of the adjunction. Now in your case if you separate out the two copies of the unit map, you have p_!(f \otimes 1_X) -> p_!(p^*p_! F \otimes 1_X) -> p_!(p^*p_! F \otimes p^*p_! 1_X) -> p_!p^*(p_!F \otimes p_! 1_X) -> p_! F \otimes p_! 1_X.
Here the first map is precisely p_! applied to the unit map of F, while the composite of the remaining maps is exactly the "projection formula" map p_!(p^*p_!F \otimes 1_X) -> p_!F \otimes p_! 1_X, which in your case is an equivalence. (I'm writing 1_X for the unit, for some reason.) So under this equivalence the oplax structure map exactly corresponds to the colimit of the unit map, as you suggest.
 
Jonathan Beardsley
Jonathan Beardsley
5:25 PM
@RuneHaugseng oh wow! that's really fantastic. Thank you so much. That says then that "the coaction induced by oplax-monoidality" is equivalent to the Thom diagonal described in ABGHR
 
Rune Haugseng
Rune Haugseng
That was a lot more fun to work out than the grant application I'm supposed to be writing... :-)
 
Jonathan Beardsley
Jonathan Beardsley
Haha... yeah.... luckily I'm done with my grant apps for the year. Sadly though I've made immense progress on an interesting math project the past few weeks, but teaching starts tomorrow... so it may be a while before I actually write it all down.
 
Rune Haugseng
Rune Haugseng
6:28 PM
Speaking of Thom spectra, is there some nice way to describe the fundamental class of a manifold using the Thom isomorphism for the tangent bundle (ideally just thinking of the Thom spectrum non-geometrically as a colimit, as in ABGHR)
 
 
1 hour later…
Denis Nardin
Denis Nardin
7:55 PM
@RuneHaugseng As luck would have it, I'm teaching a class about this right now :). For a compact manifold M the trick is going to be constructing a map S→M^{-TM} where TM is the tangent stable spherical fibration sending a point x to M/(M\x). I think you can characterize this by the fact that for every x∊M the composition S→M^{-TM}→ S^{n-T_xM} is an equivalence (where the latter is the Pontryagin-Thom collapse map for the embedding x→M).
Once you have this the fundamental class for an E-orientable manifold is just the class in $π_0(E⊗M^{-TM})\cong π_0(E⊗Σ^{-n}Σ^∞M_+)=E_n(M)$ induced by functoriality and the Thom isomorphism
For noncompact manifolds you have to be a bit more careful, but I believe you can do something similar
In general I believe spectral Verdier duality is the right setting to ask these questions though, working only with local systems gets quickly quite limiting
 
S. carmeli
S. carmeli
@DenisNardin @RuneHaugseng just a tiny comment/wonder on Dennis's beautiful explanation. Actually, the Thom collapse map canonically takes the form M--> S^{T_xM} so there's seem to be an absolutely canonical identification of S^0 with the "S^{n-T_xM}" in Dennis's answer, isn't it?. I think you want it to be that specific identification (e.g. this takes into account orientation issues nicely).
*Denis. sorry!
 
Denis Nardin
Denis Nardin
8:32 PM
Hmm.. I hadn't realized that, but of course it's clear (the composition comes from a PT collapse map associated with an embedding x→S^N for N>>0). On the other hand I don't think we can be too picky in fixing the identification...
Also this story should really be independent of orientation issues: the whole point is to offload all of them to the Thom isomorphism :)
 
S. carmeli
S. carmeli
Sure, I believe that taking this extra TxM into account exactly explain this independence on orientations right? Because if you swap the local orientation it gets in twice right?
 
Denis Nardin
Denis Nardin
Yes, but it seems weird to me to ask that a map is homotopic to the identity, rather than it is an equivalence... Oh well
Said it differently: for a map being an equivalence it's a condition, while a homotopy to the identity is structure. I'd really like what I wrote to be a condition, not structure
 

« first day (2768 days earlier)      last day (635 days later) »