# Exterior powers of tensor products

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.

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

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 $$(3, 2, 1, 1)$$.