I wonder, whatever they are, if their geometric theory is more conventional than spectral AG

@PedroTamaroff there are examples of derived equivalent dgas which are not quasi-isomorphic, see e.g. homepages.math.uic.edu/~bshipley/slides3.pdf

@ArasErgus Thanks. :)

Suppose I have a vector bundle over a manifold M and I take its kth Pontrjagin class w, living in $H^4k(M, R)$ (de Rham cohomology). Then it is known that w is an integral differential form, i.e. that if I integrate w over any compact oriented submanifold of dimension 4k, I get an integer.

Is there a nice reference for this, or a somewhat quick proof?

@Dedalus What's your definition of Pontryagin class? Because with mine it is tautological

The standard reference for these facts is Milnor-Stasheff though

I guess I use the one associated to curvature. Namely, take a connection on your vector bundle and continue from there

Of course, the POV from classifying spaces is very nice and clean, but I would be interested in something which works through curvature.

I guess one proof would be to show that these approaches yield the same result. However, this seems rather involved...

I guess Corollary 1 at page 308 of Milnor-Stasheff?

Let me see

@Denis Nardin That seems to be it. But is it clear why, if I integrate it along a compact oriented submanifold of dimension 4k, I get an integer? Is it coming from something like Poincaré duality? In my mind the argument should go something like this: The class w lives in H^4k(M,Z ) \subset H^4k(M,R). By Poincaré duality, this gives a class in the dual of H^(n-4k)_c(M,R).

Yeah, you need to know that integrating on a submanifold realizes the pairing of Poincaré duality

I don't remember off the top of my head a reference for that. Maybe Bott-Tu?

Or maybe it's in Milnor-Stasheff too, in the same appendix

The fundamental class of a submanifold is, pretty much by definition, living in $H_n^{BM}(M;Z)$

Ugh sorry, confused homology and cohomology again :)

What is H_n^BM?

Bordism?

Borel-Moore homology

The homology version of compactly supported cohomology

ah

The thing that is dual to ordinary cohomology in Poincaré duality

Thanks

And why do we conclude from this that it is always an integer? I suppose that it would be that integrating on submanifolds represents a class in H_*(M,Z) (I drop indexing so I won't make any mistakes), so that the pairing must land in Z by some functoriality arguments or something

But it is a bit murky

Sorry, what is always an integer?

Sorry, let me get the story right before I write it down :)

Are your submanifolds compact? And if not, how are you integrating differential forms on them?

It is compact!

Ok, great. Then if M is a compact oriented manifold there is a perfect pairing $H_i(M;A)×H^{n-i}(M;A)→A$ for every ring of coefficients A. The map is in fact functorial in A

It just so happens that when $A=\mathbb{R}$ you can describe the map via integration

Right

That is what I wanted to say

Actually, why am I even using Poincaré duality?

All that I need is that the isomorphism $H^n_c(M;\mathbb{R})→\mathbb{R}$ dual to the fundamental class of $M$ is given by integration. But this is just an immediate consequence of the deRham theorem...

In particular if you have a class coming from $H^n(M;\mathbb{Z})$ its integral must be an integer

But you want to do this for all submanifolds as well, right?

Well, I can pull back to the submanifold and work there

Right

And the statement I wrote can be checked locally on open subsets, so you can reduce to check it on $\mathbb{R}^n$ where it's just an easy computation

Does HoTT have any flavor that doesn't include the principle of explosion? Contradictions are members of the initial type 0, which has an arrow 0 -> T for any type T by assumption...

@JustAskin I feel like you'd get a better answer on the nForum

or math.SE

nforum because that's where the hott people are, or m.SE because the logic people are there

and I don't think it's at a high enough level for MO