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Let $V$ be the defining representation of $GL(4,\mathbb C)\to GL(V)$ with $V=\mathbb{C}^4$. Let $Ext\;V$ be the exterior square of $V$ which is a 6-dim repsentation.

My question: How does $$V\otimes Ext\; V$$ decompose into irreducibles? I am especially interested in the multiplicity of $V$ and its contragredient. Root system: $A_3$ I fail to visualize like $A_2$.

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1 Answer 1

This is an example in Fulton and Harris's Representation Theory: A First Course (Springer GTM 129 (1991)), except that they work only with $SL(V)$, not $GL(V)$. The representation you call $Ext\;V$ is denoted $\wedge^2 V$. Its tensor product with $V$ has a natural map to $\wedge^3 V$, which is isomorphic with the contragredient $V^*$ as a representation of $SL(V)$, but not quite for $GL(V)$ (where $\wedge^3 V$ is the tensor product of $V^*$ with the 1-dimensional determinant representation). For the kernel of this map, Fulton and Harris say on page 220 that it contains "the irreducible representation $\Gamma_{1,1,0}$ with highest weight $2L_1+L_2$", and then Exercise 15.10 asserts that the kernel is in fact this irreducible representation.

So in summary Fulton and Harris say that $V \otimes \wedge^2 V$ is the direct sum of irreducible representations of dimensions $4$ and $20$, the former of which is $\wedge^3 V = V^* \otimes \det$.

P.S. The Fulton-Harris text also shows pictures of the root system etc. for $A_3$.

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