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In their paper "New lower bounds for the border rank of matrix multiplication", Landsberg and Ottaviani make use of the fact that

$$\tag{$\dagger$} {\textstyle\bigwedge}^p(V\otimes W) \cong \bigoplus\nolimits_{\substack{\lambda\vdash p\\\\\ell(\lambda)\le n\\\\\lambda_1\le m }} \mathbb{S}_\lambda V \otimes \mathbb{S}_{\bar\lambda}W$$

where $\bar\lambda$ denotes the conjugate partition of $\lambda$. This isomorphism is basically Exercise 6.11 in Fulton & Harris, so there is no doubt about it. However, from what I gather, in Lemma 3.1 of the paper, they use the fact that the above isomorphism is given by the map

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

where $c_\lambda$ denotes the Young symmetrizer corresponding to the partition $\lambda$. I cannot find a proof for this. Can someone explain to me why the above map defines

   a) a morphism of $\mathfrak{S}_p$-modules and
   b) a bijection?

Since all vector spaces involved are of finite dimension and by $(\dagger)$, it would certainly suffice to show that it is either injective or surjective.

Also, if I misunderstood the proof of Lemma 3.1 and the isomorphism is given by another elementary rule, please tell me what it is.

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  • $\begingroup$ In a), you don't mean $\mathfrak S_p$ but you mean $\mathrm{GL}\left(V\right)\times \mathrm{GL}\left(W\right)$, right? $\endgroup$
    – darij grinberg
    Commented May 17, 2012 at 16:57
  • $\begingroup$ I actually think it is both, and the $\mathrm{GL}(V)\times\mathrm{GL}(W)$ part seemed rather clear. Please educate me if I'm wrong. $\endgroup$
    – Jesko Hüttenhain
    Commented May 17, 2012 at 17:35
  • $\begingroup$ I don't see how $\mathbb S_{\lambda} V$ (in general, not just for $\lambda$ being $\left(1,1,...,1\right)$ or $\left(p\right)$) becomes a $\mathfrak S_p$-module. We have $\mathbb S_{\lambda} V = c_{\lambda} V^{\otimes p}$, and $c_{\lambda}$ is not (in general) central in $k\mathfrak S_p$. $\endgroup$
    – darij grinberg
    Commented May 17, 2012 at 23:35
  • $\begingroup$ Hm. That's actually a relief, because most of my problems came from trying to understand that. I suppose it doesn't have to carry $\mathfrak{S}_p$-module structure anyway. $\endgroup$
    – Jesko Hüttenhain
    Commented May 18, 2012 at 5:41

1 Answer 1

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This is well-known in the theory of symmetric functions and is one of two Cauchy identities. You can find this in most books, for example, MacDonald Chapter I, Section 4. Orthogonality equation (4.3').

Knuth's extension of the Robinson-Schensted correspondence gives bijective proofs of both Cauchy identities, for example see Stanley Enumerative Combinatorics 7.14.3 Theorem.

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  • $\begingroup$ What is the title of the book by MacDonald? $\endgroup$
    – Jesko Hüttenhain
    Commented May 17, 2012 at 11:16
  • $\begingroup$ @Jesko Symmetric Functions and Hall polynomials $\endgroup$
    – Bruce Westbury
    Commented May 17, 2012 at 11:34
  • $\begingroup$ The equality $\sum_n e_n(xy) = \sum_\lambda s_{\lambda}(x)s_{\bar\lambda}(y)$ does imply $(\dagger)$, but why does it mean that the isomorphism is of this form? $\endgroup$
    – Jesko Hüttenhain
    Commented May 17, 2012 at 15:26
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    $\begingroup$ This is Schur-Weyl duality. One definition of the Schur functors is that they are the image of the Young idempotent. $\endgroup$
    – Bruce Westbury
    Commented May 17, 2012 at 15:30
  • $\begingroup$ I am thoroughly convinced. Thanks a lot! $\endgroup$
    – Jesko Hüttenhain
    Commented May 17, 2012 at 16:31

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