Timeline for Are Chern classes well defined up to contractible choice?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2022 at 2:26 | comment | added | Ben Wieland | It seems to me that this splits into two pieces. I think producing a canonical first Chern cycle is not trivial. Having fixed that, I'm more optimistic about producing higher Chern classes from the first as symmetric polynomials. People have suggested several approaches, but I suggest that they should exist over $BNT$, of which $BG$ is a stable summand by the Becker-Gottlieb transfer. | |
| Oct 13, 2022 at 12:30 | comment | added | D.-C. Cisinski | The choice of an orientation gives a map of ring spectra $MU\to H\mathbb Z$, so that it suffices to produce Chern classes in complex cobordism. Maybe we could try to charakterize Chern classes from Thom classes. Then $MU$ has canonical Thom classes... | |
| Oct 12, 2022 at 8:56 | answer | added | Johannes Ebert | timeline score: 5 | |
| Oct 12, 2022 at 8:11 | comment | added | Brian Shin | To refine Chris Schommer-Pries's and Dmitri Pavlov's suggestions to cohomology with integer coefficients, perhaps we could use the fact that characteristic classes produced by Chern--Weil theory lift to Cheeger--Simons differential cohomology. (I know very little about this subject, so maybe this is not a helpful comment. Also, I see that @DmitriPavlov mentions $\mathrm{K}(\mathbb{Z},2k)$, so perhaps I am just rehashing their suggestion.) | |
| Oct 12, 2022 at 5:32 | comment | added | Z. M | I guess that the concept of cocyles depends on the choice of your cohomology theory (singular, etc.). | |
| Oct 12, 2022 at 2:49 | comment | added | John Wiltshire-Gordon | Is there a "best" way to pick a cocycle representing a given orientation class of $S^1$? I'd guess no, and moreover that the required choice is not contractible. | |
| Oct 12, 2022 at 0:32 | history | became hot network question | |||
| Oct 11, 2022 at 20:46 | history | edited | André Henriques | CC BY-SA 4.0 |
fixed a small typo and changed the capitalisation in the title.
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| Oct 11, 2022 at 20:38 | answer | added | Dmitri Pavlov | timeline score: 2 | |
| Oct 11, 2022 at 20:24 | comment | added | André Henriques | @ChrisSchommer-Pries. Your answer is good when working with real coefficients. But it doesn't settle the questions when working with $\mathbb Z$ coefficients. | |
| Oct 11, 2022 at 20:19 | answer | added | Neil Strickland | timeline score: 11 | |
| Oct 11, 2022 at 19:05 | comment | added | Aleksandar Milivojević | To talk about connections on the universal vector bundle we do have to pick a geometric representative of the underlying base homotopy type BU (maybe that’s ok though); if we are allowed to pick a representative of BU we can also choose the one with only even dimensional cells, where the lift from cohomology to cocycles is unique. | |
| Oct 11, 2022 at 16:47 | comment | added | Chris Schommer-Pries | Hmm. Can you choose a connection on the universal vector bundle and then define cocycle representatives via the curvature of the connection? So then these classes would be parametrized by the space of connections. | |
| Oct 11, 2022 at 16:45 | comment | added | Denis Nardin | I'm not sure if this can be turned into a complete answer, but $c_1$ is completely determined as the map $BU\to BU_1=K(\mathbb{Z},2)$ given by the determinant homomorphism $U\to U_1$ (and the identification $BU_1=K(\mathbb{Z},2)$ is the same as the choice of a generator of $\pi_1U_1$, say pick a counterclockwise rotation :)). Then one can hope to describe $c_n$ in a similar way using symmetric polynomials in $c_1$... | |
| Oct 11, 2022 at 16:26 | history | asked | André Henriques | CC BY-SA 4.0 |