Timeline for Reference for a dual version of the Cauchy decomposition.
Current License: CC BY-SA 3.0
7 events
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| May 2, 2012 at 9:38 | comment | added | Igor Makhlin | Yes, the second identity can certainly be derived from the first in one way or another. It can also be proven in almost any fashion the first one can. I was just hoping to avoid any kind of proof by giving a reference. Well, no such luck. | |
| May 1, 2012 at 18:09 | comment | added | darij grinberg | I admit that the above argument was not very canonical, though (in the sense that my above automorphism $w$ was defined using the standard basis of $V$, so I do not know how to define something like this naturally for $\mathrm{GL}\left(V\right)$ instead of $\mathrm{GL}_n$), so I do not think it is optimal. | |
| May 1, 2012 at 18:07 | comment | added | darij grinberg | Remark: To prove $\left(V^{\lambda}\right)^{\ast} \cong \left(V^{\ast}\right)^{\lambda}$, it is enough to recall that representations of $S_n$ are self-dual (since all irreducible representations are real) and use the definition of Schur functors as Hom's from irreducible representations of $S_n$. | |
| May 1, 2012 at 17:47 | comment | added | darij grinberg | ... it is enough to show that every partition $\lambda$ of $p$ satisfies $\left(V^{\lambda}\right)^{\ast} \cong \left(V^{\lambda}\right)^w$. But this follows from $V^{\ast} \cong V^w$ using $\left(V^{\lambda}\right)^{\ast} \cong \left(V^{\ast}\right)^{\lambda}$ (this should be pretty easy to check) and $\left(V^{\lambda}\right)^w \cong \left(V^{w}\right)^{\lambda}$ (this follows from functoriality of Schur functors). | |
| May 1, 2012 at 17:45 | comment | added | darij grinberg | I might be totally wrong, but can't the analogous identity be derived from the first one? Define an automorphism $w$ of $\mathrm{GL}_n\left(\mathbb C\right)$ by $w\left(A\right)=A^{T-1}$ for all matrices $A\in \mathrm{GL}_n\left(\mathbb C\right)$. Then, it is easy to see that $V^{\ast}\cong V^w$ as representations of $\mathrm{GL}_n\left(\mathbb C\right)$. Hence, $V^{\ast}\otimes W \cong V^w \otimes W \cong \left(V\otimes W\right)^{w\times \mathrm{id}}$. So much for the left hand side. For the right hand side, ... | |
| Apr 26, 2012 at 19:50 | history | edited | Igor Makhlin | CC BY-SA 3.0 |
added 71 characters in body; deleted 1 characters in body
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| Apr 26, 2012 at 19:44 | history | asked | Igor Makhlin | CC BY-SA 3.0 |