Homotopy Theory

3:01 AM

I have a basic question, but hopefully not too basic for this chatroom (apologies if it is).

If G is a topological group, then it follows from Brown's representability theorem that there is a classifying space BG (unique up to homotopy equivalence) such that isomorphism classes of principal bundles on X correspond to [X, BG]. which are basepoint-preserving homotopy classes of pointed maps. However, it seems like one should instead have [X, BG] which is free homotopy classes of maps. If G is connected, then BG is simply connected so [X, BG] and [X, BG]. coincide.

But what if G is not connected? Shouldn't we still have [X, BG] instead of [X, BG]. ? Is there a way to make this work? One example where the two differ is when G is the fundamental group of a genus g > 1 surface (so that the surface is a model for BG).

Arun Debray

3:17 AM

If Σ is a closed, connected surface with basepoint, is it not true that [X, Σ]. = [X, Σ] ?

Mike Miller

@MichaelAlbanese If $G$ is discrete, principal $G$-bundles up to isomorphism correspond to representations $\pi_1 X \to G$ up to conjugacy, which is indeed $[X, BG]$

Arun Debray

I thought it was, but could be mistaken

Mike Miller

@ArunDebray I am pretty sure the difference is precisely the difference between conjugacy classes of homomorphisms and actual homomorphisms

Michael Albanese

@ArunDebray: I don't think so, take X = S^1. Then the left hand side is the fundamental group and the right hand side is conjugacy classes of that group.

@MikeMiller: Is there a way to see that via Brown's representability theorem?

Mike Miller

I'm not sure if there is a version of Brown's without basepoints, but I am an ignoramus.

Arun Debray

3:20 AM

oh yes, you're right. thank you! @MikeMiller and @MichaelAlbanese

Marc Hoyois

@MichaelAlbanese As far as I know there is no Brown representability without base points.

Mike Miller

Here is a relevant MO question: mathoverflow.net/questions/104866/…

Marc Hoyois

Brown representability actually gives you a pointed space BG that classifies pointed principal G-bundles, i.e., principal G-bundles that are trivialized over the base point.

Michael Albanese

@MarcHoyois: Do you know why pointed principal G-bundles and principal G-bundles coincide for G connected but not for G disconnected?

Marc Hoyois

The set of iso classes principal G-bundles is the quotient of that of pointed principal G-bundles by the action of π_1(G).

There is probably a version of Brown representability for groupoid-valued functors that would apply directly to arbitrary CW complexes.

Michael Albanese

4:10 AM

So how would one show that for any topological group G, there is a space BG such that [X, BG] classifies principal G bundles?

Mike Miller

4:36 AM

@MichaelAlbanese Well, you can explicitly make a space EG that G acts freely and properly on via the bar / infinite join construction, and then you check that EG/G classifies principal G-bundles. But I think the proof you want is a pointed space that represents pointed principal bundles, and then to examine what happens when you forget about the based part of the homotopy.

Or maybe I'm being too silly. I guess you can just apply that result to X plus a disjoint basepoint

Marc Hoyois

@MikeMiller That doesn't work because Brown representability is only for pointed connected spaces...

Mike Miller

Ah yes, thanks. Something felt off

Urs Schreiber

8:02 AM

regarding super-algebra in stable homotopy theory: I would think that the dominance of even periodic phenomena is evident rather for than against the relevance of super-algebra.

Notice that supercommutative algebras may be identified with Z-graded-commutative algebras that happen to be even periodic.

But general algebra in stable homotopy theory is Z-graded-commutative, (in fact is S-graded commutative). And so it is natural to think of even graded ring spectra as spectral stand-ins for supercommutative superalgebras.

This point was made by Charles Rezk in 2009

After some discussion with him, I had expanded a little on this perspective

here: ncatlab.org/nlab/show/spectral+super-scheme

Denis Nardin

9:34 AM

@MichaelAlbanese This old answer of mine might tell you what you are looking for

Michael Albanese

11:25 AM

@DenisNardin: That seems to be what I was looking for. So Brown's representability gives you a pointed space BG, and pointed homotopy classes of maps gives you isomorphism classes of principal G-bundles with a choice of point in the fiber over the basepoint (which gives an identification of the fiber with G). But if you regard BG as an unpointed space and take free homotopy classes, you get isomorphism classes of principal G-bundles. Right?

Denis Nardin

Yes, although you need to work a little to prove the last statement

It's not 100% trivial

Michael Albanese

Right.

Thanks for the answer.

Abstractly defining BG by Brown's representability theorem has always bugged me.

I'm glad it can be made to work (with some effort).

Denis Nardin

It's more subtle than people usually realize. Even if it is the first application of the theorem that Brown presents in his paper (although that paper is rather concise in its exposition)

Michael Albanese

Is there a similar way to deal with cohomology being represented by Eilenberg-MacLane spaces? That is, the difference between [X, K(G, 1)]. and [X, K(G, 1)] ?

Denis Nardin

Yes, but there it's simpler

I assume you are referring to the case G abelian? (otherwise it is literally the same thing)

The pointed homotopy classes represent reduced cohomology, while the unpointed ones represent unreduced cohomology. But it is quite easy to see that $\tilde H^*(X_+;A)\cong H^*(X;A)$ and that $[X_+,K(A,n)]_*\cong [X,K(A,n)]$, so it suffices to prove that reduced cohomology is representable

Michael Albanese

11:35 AM

Isn't it the same thing for G abelian and different for G non-abelian? For example X = S^1, G the fundamental group of a genus g > 1 surface.

Denis Nardin

I'm sorry I must be confused. Isn't BG=K(G,1) for all finite groups G?

Michael Albanese

Right.

Denis Nardin

And if you want to verify that it is the case with the above definition of BG, well we have seen that π_nBG=Bun_{G,0}(S^n) and then you can show that the latter is G for n=1 and 0 otherwise

This is just a classical argument with clutching functions

Michael Albanese

My comment about G abelian vs. non-abelian was intending to point out that $[S^1, \Sigma_g]. = \pi_1(\Sigma_g)$ while $[S^1, \Sigma_g]$ is the collection of conjugacy classes in $\pi_1(\Sigma_g)$ (which is not the same thing). Note that $\Sigma_g$ is a model for $B\pi_1(\Sigma_g) = K(\pi_1(\Sigma_g), 1)$.

Denis Nardin

I probably have completely misunderstood what you were saying

Can you rephrase your original question?

Michael Albanese

11:43 AM

Sorry. My question was as follows: again by Brown representability, there is a pointed space $K(G, 1)$ such that $H^1(X; G) = [X, K(G, 1)].$, but we should have $[X, K(G, 1)]$ instead. But I think your earlier comment about representing reduced cohomology should fix that. In particular, Brown representability actually applies to $\tilde{H}^1$ not $H^1$.

Denis Nardin

Yes, exactly

Remember if you want to apply (unstable) Brown representability, you must work in pointed connected spaces

Otherwise the theorem is just plain false

Michael Albanese

Right.

Thanks again.

Saal Hardali

3:40 PM

Why is it that cycles in $KU$-homology can be expressed as maps from $Spin^c$ manifolds equipped with complex vector bundles? I'm tempted to say that it follows from the conner-floyd isomorphism:

$$MSpin^c_*(X) \otimes_{MSpin^c_*} KU_*$$

But isn't the same true with $MU$ replacing $MSpin^c$? (ny landwever exavtness of $K$-theory for instance?) Yet the same statement is not true for $MU$ i suspect. Also the fact that a homoloy theory has geometric interpretation for cycles must be a nontrivial assertion i'd be very happy to understand what makes it possible in this case?

Maybe every cycle in $K$ theory can also be expressed as a map from a (stably almost) complex manifold equipped with a (virtual) complex vector bundle and then the answer is that $Spin^c$ is not the interesting thing here. I suspect this is not true though...

Eric Peterson

7:09 PM

@SaalHardali a half-remembered guess: there's a (non-multiplicative) splitting of MSpin into HF2s, HZs, and kOs; i think something similar is true for MSpin^c with kOs judiciously replaced by kUs. in the MSpin case, the unit class is supported on a kO, so there's a (non-mult) splitting kO --> MSpin --> kO. the MSpin homology of X has a description just in terms of maps from Spin-manifolds to X, and so the task is to pick out what extra data says "i'm living on the summand sitting on the unit class"

Eric Peterson

7:26 PM

(and you're right that MU does not enjoy such a splitting)

Saal Hardali

8:05 PM

@EricPeterson Interesting! Do you know a reference for this stuff?

Arun Debray

9:41 PM

@SaalHardali The splitting of MSpin is Anderson, Brown, Peterson, "The Structure of the Spin Cobordism Ring"

I have seen the result for spinc attributed to them, but they don't explicitly discuss it. In any case, it's explicitly given in Bahri-Gilkey, "Pinc and equivariant spinc cobordism of cyclic 2-groups."

Eric Peterson

i imagine you'll also find the hopkins-hovey paper spin cobordism determines real k-theory (and its bibliography) helpful

but i don't know a reference for the literal result you asked about, only for the ancillary results i named

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