Timeline for Chern classes of PU(n)-bundles
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2017 at 17:06 | vote | accept | Ulrich Pennig | ||
| Oct 12, 2017 at 17:06 | |||||
| Apr 4, 2016 at 13:50 | comment | added | Ulrich Pennig | @DylanWilson Can you modify your answer accordingly? Then I can accept it. Thanks! | |
| Apr 2, 2016 at 18:50 | comment | added | Dylan Wilson | Ah- apologies, I guess I spoke too quickly. So then I guess something like first $n^2-n$ will be nonzero and the rest will be zero? | |
| Apr 2, 2016 at 12:33 | comment | added | Ulrich Pennig | @DylanWilson Isn't the second to last one (i.e. the $(n^2 - 1)$th one) the sum over all the monomials, which are products of all $x_{i,j}$, where I leave out one of them (e.g. $x_{11}x_{12}x_{21} + x_{11}x_{21}x_{22} + x_{11}x_{12}x_{22} + x_{12}x_{21}x_{22}$ for $n = 2$)? If this is true, then every such monomial will contain a diagonal element. | |
| Mar 30, 2016 at 14:39 | comment | added | Dylan Wilson | Indeed, every symmetric polynomial except the last one has at least some monomial that doesn't contain one of these diagonal elements! (Which is essentially the content of the last paragraph). | |
| Mar 30, 2016 at 14:38 | comment | added | Dylan Wilson | It's not the case that each monomial has such an element. For example, the first symmetric polynomial is just the sum of all the $x_{i,j}$. Most of those will map to something nonzero. | |
| Mar 30, 2016 at 9:50 | comment | added | Ulrich Pennig | There is one thing I don't understand though. What is wrong with the following reasoning: The $(n^2-1)$th Chern class in $H^*(BU(n^2), \mathbb{Z})$ is mapped to the $(n^2-1)$th elementary symmetric polynomial. This is a sum of monomials with each one containing at least one "diagonal" element $x_{ii}$. Since the induced map $H^*(BU(n^2)) \to H^*(BPU(n))$ is a ring homomorphism, this should be mapped to $0$ by your argument. | |
| Mar 25, 2016 at 12:45 | comment | added | Dylan Wilson | What do you mean? Ad* takes x_ij to t_i-t_j and Chern classes go to the corresponding symmetric polynomials. | |
| Mar 25, 2016 at 12:11 | comment | added | Sebastian Goette | How do these classes behave under $\mathrm{Ad}^*$? | |
| Mar 23, 2016 at 23:33 | comment | added | Dylan Wilson | (To compute the map on cohomology you need to know that inversion on the circle gives multiplication by -1 on group cohomology and multiplication sends a generator to the sum of the two natural generators, since the Chern class of a tensor product is the sum.) | |
| Mar 23, 2016 at 23:28 | history | answered | Dylan Wilson | CC BY-SA 3.0 |