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I had a question which is slightly more general than this one on mathoverflow: I am looking for an explicit description of the isomorphism $\mathbb S_\nu(V\otimes W) \cong \bigoplus C_{\lambda\mu\nu} \mathbb S_\lambda V\otimes \mathbb S_\mu W$ from Exercise 6.11 in Fulton & Harris. In said question, the author asks whether $${\textstyle\bigwedge}^p(V\otimes W) \cong \bigoplus\nolimits_{\substack{\lambda\vdash p\\\\\ell(\lambda)\le n\\\\\lambda_1\le m }} \mathbb{S}_\lambda V \otimes \mathbb{S}_{\bar\lambda}W$$

is given by

$$ (v_1\otimes w_1)\wedge\ldots\wedge(v_p\otimes w_p) \longmapsto \sum\nolimits_{\substack{\lambda\vdash p\\\\\ell(\lambda)\le n\\\\\lambda_1\le m }} c_\lambda(v_1\otimes\ldots\otimes v_p) \otimes c_{\bar\lambda}(w_1\otimes\ldots\otimes w_p). $$

The answer is affirmative and in the comments it is said that this is due to Schur-Weyl duality. I frankly don't understand that argument: I do not see why the map is well-defined. One would have to show that simultaneous permutation of the $v_i$ and the $w_i$ with some $\pi\in S_p$ introduces a factor of $\mathrm{sgn}(\pi)$ on the right, and this is not at all clear to me. Can someone explain?

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Good question; my bet is that the element of $\wedge^p\left(V\otimes W\right)$ has to be preprocessed by antisymmetrization when $p\geq 3$. An answer to this question is also likely to answer mathoverflow.net/questions/77948/… . –  darij grinberg Feb 5 '13 at 17:16
1  
If I understand correctly, you want to define the above map from $\bigotimes^p (V\otimes W)$, mapping $(v_1\otimes w_1)\otimes\cdots\otimes(v_p\otimes w_p)$ to the sum on the right. Then, you want to precompose that map with the inclusion $\bigwedge^p (V\otimes W)\hookrightarrow \bigotimes^p (V\otimes W)$ rather than showing it factors through the corresponding projection $\bigotimes^p (V\otimes W)\twoheadrightarrow \bigwedge^p (V\otimes W)$. Now if we proceed like that, it is no longer obvious that the map is surjective and we can not deduce from the Cauchy formula that it is an isomorphism. –  Garfield Feb 6 '13 at 13:56
    
Yes, which is why I said "bet" and not "answer" ;) . –  darij grinberg Feb 6 '13 at 18:48
    
Fair enough! =D It boils down to the same question, I suppose. –  Garfield Feb 6 '13 at 19:01

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