Timeline for Is the linear span of irrep matrices a complete matrix basis?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2018 at 6:46 | vote | accept | Māris Ozols | ||
| May 11, 2018 at 20:21 | answer | added | Benjamin Steinberg | timeline score: 5 | |
| May 11, 2018 at 20:18 | comment | added | Benjamin Steinberg | If you would like. | |
| May 11, 2018 at 20:11 | comment | added | Māris Ozols | @BenjaminSteinberg, thanks, this is what I was looking for! Would you like to submit this as an answer? And thanks Vincent for the link to the short proof. | |
| May 11, 2018 at 16:13 | comment | added | Benjamin Steinberg | @LSpice, in this case it is the theorem that any subalgebras of matrices over an algebraically closed field that acts irreducibly is the whole algebra of matrices. | |
| May 11, 2018 at 15:56 | comment | added | LSpice | @BenjaminSteinberg, as a representation theorist, I'm deeply embarrassed to have to ask: what is Burnside's theorem? The one that springs to mind is the solubility of groups with only two distinct prime factors, which surely isn't the relevant one here. | |
| May 11, 2018 at 14:09 | comment | added | Benjamin Steinberg | I think Burnside's theorem is pretty well known. It works basically subalgebras of matrices over an algebraically closed field acting irreduciblly. | |
| May 11, 2018 at 13:52 | comment | added | Vincent | I cannot upvote Benjamin Steinbergs comment enough. I spent many weeks trying to prove a very similar result that also follows from Burnside's theorem. Burnside's theorem is absolutely amazing and deserves to be better known! (I also do not agree with David Handelman above that it is fully obvious that if $\mathbb{C}^d$ is a $G$-irrep so is $\mathbb{C}^{d \times d}$ even if it is true.) A good link to a (modern) proof is this: ac.els-cdn.com/S0024379503007225/… | |
| May 11, 2018 at 13:40 | comment | added | David Handelman | Or simply that the span is a subalgebra that is invariant under the action of $G$. | |
| May 11, 2018 at 13:32 | comment | added | Benjamin Steinberg | Burnside's theorem | |
| May 11, 2018 at 12:20 | history | asked | Māris Ozols | CC BY-SA 4.0 |