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I am trying to understand Cauchy formula for exterior powers of tensor products.

Let $k$ be a characteristic 0 field. Let $E$ and $F$ be two vector spaces over $k$. Then we have $$ \wedge^d(E\otimes F) \cong \bigoplus_{|\lambda|=d} L_\lambda E \otimes L_{\lambda'}F, \ \text{where $\lambda'$ is dual of $\lambda$}.$$ Here $L_{\lambda}$ is the Schur functor and $L_\lambda E$ is the Schur module. I am following the notations of Weyman's book( Cohomology of vector bundles and syzygies).

I am curious in the map of this natural isomorphism. There is a proof given in the Weyman's book but I am unable to understand it. If someone can explain the maps even in $d = 2 \ or \ 3$ case that will work for me. Any ideas\hints are welcome.

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  • $\begingroup$ What is $L_\lambda$? $\endgroup$
    – KReiser
    Jan 5, 2021 at 6:57
  • $\begingroup$ Sorry, I should have mentioned that. I have edited the question. $\endgroup$
    – user270331
    Jan 5, 2021 at 7:09

1 Answer 1

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This fact can be deduced from Schur-Weyl duality.

Consider the vector space $(E \otimes F)^{\otimes d}$, which carries commuting actions of the three groups $GL(E)$, $GL(F)$, and $S_d$. We can rearrange things by writing $(E \otimes F)^{\otimes d} = E^{\otimes d} \otimes F^{\otimes d}$, so now $GL(E)$ acts on the first $d$ factors, $GL(F)$ on the last $d$ factors, and this time the symmetric group acts along the diagonal: $\sigma$ acts by permuting the first $d$ and last $d$ factors simultaneously. It is helpful to think of this $S_d$ action as being diagonal: $S_d \hookrightarrow S_d \times S_d$.

By Schur-Weyl duality, considering the pairs $GL(E) \times S_d$ and $GL(F) \times S_d$ indepdently, we get $$ \begin{aligned} E^{\otimes d} \otimes F^{\otimes d} &\cong \left( \bigoplus_{|\lambda| = d} L_\lambda E \otimes S^\lambda \right) \otimes \left( \bigoplus_{|\mu| = d} L_\mu F \otimes S^\mu \right) \\ & = \bigoplus_{|\lambda| = |\mu| = d} L_{\lambda} E \otimes L_{\mu} F \otimes S^\lambda \otimes S^\mu. \end{aligned}$$

Now take the sign-isotypic component of the $S_d$ action on both sides. On the left, we get the sign-isotypic component of $(E \otimes F)^{\otimes d}$, which is isomorphic to the exterior power $\bigwedge^d(E \otimes F)$ because we are in characteristic zero. On the right, the tensor product $S^\lambda \otimes S^\mu$ of Specht modules contains the sign irrep once if $\mu = \lambda'$, and does not contain the sign irrep otherwise. (This is hom-tensor-dual to the fact that tensoring with the sign representation sends $S^\lambda$ to $S^{\lambda'}$). We get $$ \bigwedge^d(E \otimes F) \cong \bigoplus_{|\lambda| = d} L_\lambda E \otimes L_{\lambda'} F.$$

This is essentially the argument given in 4.1 of Howe's "Perspectives on invariant theory", and he calls this theorem "skew $(GL_n, GL_m)$-duality" (others sometimes call it "skew Howe duality").

As for what the isomorphism looks like, Howe also offers some explanation. Fix bases $e_1, \ldots, e_n$ and $f_1, \ldots, f_m$ of $E$ and $F$ respectively, so that $E \otimes F$ has a basis $\{v_{ij} = e_i \otimes f_j\}$. The highest-weight vector corresponding to some partition $\lambda$ on the right can be written down by imagining the $v_{ij}$ in a rectangular matrix, overlaying the Young diagram of $\lambda$, and wedging everything together. For instance, for the partition $(4, 2, 1)$ would have highest weight vector $$ v_{(4, 2, 1)} := (v_{11} \wedge v_{12} \wedge v_{13} \wedge v_{14}) \wedge (v_{21} \wedge v_{22}) \wedge (v_{31}),$$ which we can see has $GL(E)$-weight $(4, 2, 1)$ and $GL(F)$-weight $(3, 2, 1, 1)$.

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