Timeline for Equivalent condition to linear operator having trace zero
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2022 at 18:07 | answer | added | Jon Aycock | timeline score: 3 | |
| Aug 19, 2022 at 21:32 | comment | added | Jon Aycock | @R.vanDobbendeBruyn, in fact the ring isn't $\mathbb{C}[t]$, but a group ring for a group with a canonical (topological) generator. Specifically, $V$ and $W$ should be continuous representations of $\widehat{\mathbb{Z}}$, and we're looking at the trace of $1$. (Or really, continuous representations of $D_p/I_p$, and we're looking at the trace of Frobenius.) So maybe I really need some way of using the fact that this is a group element rather than a random linear operator. | |
| Aug 18, 2022 at 21:33 | comment | added | Sam Hopkins | @JonAycock: vague idea, but maybe if you had some analog/generalization of the exponential map, then you can talk about the determinant being one... | |
| Aug 18, 2022 at 21:28 | comment | added | Jon Aycock | @SamHopkins, that was one of my initial thoughts. But it feels like it's hard to make it useful. | |
| Aug 18, 2022 at 20:12 | comment | added | R. van Dobben de Bruyn | Hmm, the trace of $t$ seems like an extra structure that cannot be obtained purely from the underlying rigid monoidal category. Indeed, there are many automorphisms of $\mathbf C[t]$ that do not preserve $t$, so how do you know that you're looking at the trace of $t$ rather than that of $3t-17$? | |
| Aug 18, 2022 at 19:55 | comment | added | Sam Hopkins | Well, the trace zero operators are precisely the ones that belong to the special linear Lie algebra. | |
| Aug 18, 2022 at 19:21 | history | asked | Jon Aycock | CC BY-SA 4.0 |