Timeline for How to find all linear transformations commuting with all elements of the image of an isotypic representation of a compact Lie group
Current License: CC BY-SA 4.0
9 events
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| Sep 24, 2023 at 18:52 | comment | added | Vladimir47 | Thanks a lot. My last question - any references??? | |
| Sep 24, 2023 at 17:43 | comment | added | LSpice | Re, I misread your notation, so we can call the simple $W$, and then $V = W^{\oplus n}$. The space $\operatorname{End}(W)$ (I write $\operatorname{End}_G(W)$) consists of linear endomorphisms of $W$ that commute with the action of $G$. These are multiplications by (complex) scalars for $W$ complex. A real $W$ carries a $G$-equivariant structure of a module over $\mathbb R$, $\mathbb C$, or $\mathbb H$, and then the operators are again scalars from the appropriate division algebra. That seems pretty concrete! | |
| Sep 24, 2023 at 9:50 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Math Jaxed + further tagged
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| Sep 24, 2023 at 7:48 | comment | added | Vladimir47 | Thank you for your answers. But I want to have some reference. And what is End(V)? By V you mean a simple G-module (in my question V is a isotypic one)? And I want to have a concrete answer for G=SU(2) (real representations). | |
| Sep 23, 2023 at 20:37 | comment | added | Qiaochu Yuan | With real representations you get $\text{End}(V^n) \cong M_n(\text{End}(V))$ where $\text{End}(V)$ could be $\mathbb{R}, \mathbb{C}$, or $\mathbb{H}$. | |
| Sep 23, 2023 at 19:21 | comment | added | Vladimir47 | No, I work with real representations! I supposed tensor product too. | |
| Sep 23, 2023 at 18:03 | review | Close votes | |||
| Oct 10, 2023 at 3:02 | |||||
| Sep 23, 2023 at 17:45 | comment | added | LSpice | I assume you're working with complex representations. The space of transformations of $V^{\oplus n}$, which is $G$-isomorphic to $V \otimes_{\mathbb C} \mathbb C^n$ (where $\mathbb C^n$ is the $n$-dimensional isotrivial representation of $G$), that commute with $G$ are parameterised by $\operatorname{End}(\mathbb C^n)$, acting purely on the second tensor factor. Proof: for each pair $(i, j)$, the map $V \xrightarrow{\operatorname{in}_i} V^{\oplus n} \to V^{\oplus n} \xrightarrow{\operatorname{out}_j} V$ is $G$-equivariant, hence a scalar. Now assemble the scalars into a matrix. | |
| Sep 23, 2023 at 16:36 | history | asked | Vladimir47 | CC BY-SA 4.0 |