Timeline for Trace and exterior product
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 23, 2019 at 5:09 | comment | added | Per Alexandersson | @BugsBunny Ah, i see. Yes, i have already done the enumerative part - I was hoping for a way to avoid that in the proof. I am trying to prove a cyclic sieving phenomenon by using representation-theory, see background here: math.upenn.edu/~peal/polynomials/… | |
| Aug 22, 2019 at 18:12 | comment | added | Bugs Bunny | @Per Alexandersson It is doable as well. It is equivalent to figuring out all eigenvalues. It is clear how to do it for each $n$ but I am too drunk to think of a general formula. The basis elements correspond to 2-ary length $n$ necklaces (see en.wikipedia.org/wiki/Necklace_(combinatorics) ). Thus, you need to know the sizes of all the necklaces. Each necklace of size $m$ contributes all roots of $z^m-(-1)^m$ to the eigenvalues. Now it is up to your enumerative combinatorics skills (mine are $-\infty$) to figure out the general answer. | |
| Aug 22, 2019 at 17:36 | comment | added | Per Alexandersson | @BugsBunny How about powers of T then? | |
| Aug 22, 2019 at 17:31 | comment | added | darij grinberg | And of course, the OP's guess $\prod_{j=1}^n \left(1+\xi^j\right) = 0$ as well, due to the $1 + \xi^n = 1 + \left(-1\right) = 0$ factor :) | |
| Aug 22, 2019 at 15:55 | history | answered | Bugs Bunny | CC BY-SA 4.0 |