Timeline for Peter–Weyl decomposition of a group representation rather than group algebra
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 20 at 2:06 | comment | added | Conifold | I found a chapter from Goodman-Wallach's book (pp. 186-189) and Sengupta's book (pp. 249, 257) that seem to use double commutants to decompose representations. The notation and terminology are over my head and I am not sure if they are decomposing $T$ or $\mathbb{C}[T]$, but is this what you have in mind? Maybe I can decipher them. | |
| Jun 20 at 1:21 | comment | added | Qiaochu Yuan | Artin-Wedderburn is itself a pretty straightforward result, it doesn't use anything complicated, just Schur's lemma and "Cayley's theorem for rings," see e.g. Wikipedia: en.wikipedia.org/wiki/Wedderburn%E2%80%93Artin_theorem It doesn't have to feel like a black box, you can just dig into the details of the proof. | |
| Jun 20 at 1:19 | comment | added | Qiaochu Yuan | @Conifold: the version of the double commutant theorem I gave in that old blog post uses Artin-Wedderburn so it is not independent of the Artin-Wedderburn decomposition. The von Neumann version for $^{\ast}$-algebras may give an independent proof but I'm not familiar with how the details go; it should still use the fact that we can take orthogonal complements to invariant subspaces so basically still semisimplicity is the key. $M(T)$ is naturally a subspace of the dual space of $\mathbb{C}[G]$ but with the obvious self-duality, yes, we should have $\mathbb{C}[T] \cong M(T)$. | |
| Jun 20 at 1:10 | vote | accept | Conifold | ||
| Jun 20 at 1:10 | comment | added | Conifold | Thank you for the elaboration. It is just that I am more familiar with the harmonic analysis side of the representation theory that uses $T(g)$ and equivariant maps explicitly, and not so much with modules and ideals, but I'll try to work through your proof. Would the double commutant theorem give a proof that does not require using the decomposition of $\mathbb{C}[G]$? $M(T)$ is the subspace of $L^2(G)$ spanned by the matrix elements $T_{ij}(g)$ in some basis, quoted here. | |
| Jun 19 at 23:41 | comment | added | Qiaochu Yuan | @Conifold: I don't really understand what you're unsatisfied with (and I also have no idea what $M(T)$ means) but I added some details. | |
| Jun 19 at 23:41 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jun 19 at 4:27 | comment | added | Conifold | Finite $G$ is all good, but I am having hard time translating between the language of semisimple algebras and modules and group representations and matrix elements. What does "modules break up", etc., mean in the latter? Any reference that does the translating? Is it true that $\mathbb{C}[T]\simeq M(T)$ or, perhaps, $M(T^*)$, and if so how does one 'see' it? Can this be proved without decomposing $\mathbb{C}[G]$ first so that it follows as a special case? There was a question about double commutant for representations, but alas, no answer. | |
| Jun 18 at 18:24 | history | answered | Qiaochu Yuan | CC BY-SA 4.0 |