Timeline for Local homology of a space of unitary matrices
Current License: CC BY-SA 3.0
17 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
May 11, 2017 at 23:00 | vote | accept | John Klein | ||
S May 11, 2017 at 22:06 | history | bounty ended | John Klein | ||
S May 11, 2017 at 22:06 | history | notice removed | John Klein | ||
May 7, 2017 at 1:42 | vote | accept | John Klein | ||
May 11, 2017 at 22:06 | |||||
May 6, 2017 at 8:23 | comment | added | Mark Grant | @JohnKlein: I don't know yet. Two unitary matrices are unitarily similar iff they have the same eigenvalues, so that map is actually the orbit map of the conjugation action, and $\mathcal{D}$ is the preimage of the inclusion $SP^{n-1}(S^1)\subset SP^n(S^1)$ obtained by adding a 1. Surely its not a bundle though. Greg's answer seems to be more a propos. | |
May 5, 2017 at 21:59 | answer | added | Gregory Arone | timeline score: 11 | |
May 5, 2017 at 16:05 | comment | added | John Klein | @MarkGrant I am aware of that map. It can be used to equip $U(n)$ with the structure of a Whitney stratified space (I learned this from David Ayala). But how can you use the map to compute the local homology of $\cal D$? | |
May 5, 2017 at 15:30 | comment | added | Mark Grant | This is pure speculation, but can you use the map $U(n)\to SP^n(S^1)$ which takes a unitary matrix to its spectrum of eigenvalues? The symmetric power is not a manifold, but maybe its local homology is easier to relate to multiplicities of eigenvalues than that of $\mathcal{D}$. | |
S May 4, 2017 at 22:33 | history | bounty started | John Klein | ||
S May 4, 2017 at 22:33 | history | notice added | John Klein | Authoritative reference needed | |
May 3, 2017 at 15:15 | comment | added | Neil Strickland | You can filter $U(n)$ by the dimension of the relevant eigenspace. The $k$'th filtration quotient is (essentially by the Cayley transform) the Thom space of the bundle of hermitian endomorphisms of the tautological bundle over the Grassmannian $G_k(\mathbb{C}^n)$. This circle of ideas also shows that $\mathcal{D}^c$ is just $\mathbb{R}^{n^2}$. The filtration is stably split by a theorem of Haynes Miller. This does not obviously answer your question, but it seems like relevant context. | |
May 3, 2017 at 0:23 | history | edited | John Klein | CC BY-SA 3.0 |
added 1 character in body
|
S May 2, 2017 at 23:03 | history | edited | John Klein | CC BY-SA 3.0 |
Three small typos: deleted one parenthesis, and added another. Added period.
|
S May 2, 2017 at 23:03 | history | suggested | jeq | CC BY-SA 3.0 |
Three small typos: deleted one parenthesis, and added another. Added period.
|
May 2, 2017 at 22:48 | review | Suggested edits | |||
S May 2, 2017 at 23:03 | |||||
May 2, 2017 at 22:48 | history | edited | John Klein | CC BY-SA 3.0 |
added 1 character in body
|
May 2, 2017 at 22:25 | history | asked | John Klein | CC BY-SA 3.0 |