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I have been trying to calculate the explicit images of the Schur functors for the action of $\mathfrak{S}_3$ on $V^{\otimes 3}$ where $V$ is some vector space, for the sake of concreteness of dimension 2.

$\mathfrak{S}_3$ has 3 irreducible representations, namely the trivial, alternating and standard. Using their characters and the character of the representation on $V^{\otimes 3}$, it is easy to see that their multiplicities in the decomposition of $V^{\otimes 3}$ into irreducibles are 4, 0 and 2 respectively.

Now, the Schur functor associated to the partition $(3,0,0)$ is precisely the symmetrisation functor, and its image is $\mathrm{Sym}^3 V$, which has dimension 4, as is to be expected. On the other hand, the image of the projector associated to the partition $(1,1,1)$ is the antisymmetrisation functor, and its image is 0, which makes sense since $V$ has no third exterior power.

What puzzles me is the functor corresponding to the standard representation. The Young symmetriser in this case is $1 + (12) - (13) - (123)$. While doing the calculation I was expecting its image to be 4 dimensional, which would then further decompose into the copies of the standard representation predicted by the characters. But, unless my calculation is wrong, the subspace that I get is only of dimension 2. If $a$ and $b$ are a basis for $V$, then the images of $a\otimes a \otimes a$ and $b\otimes b \otimes b$ both vanish. All three products of two $a$s and one $b$ get sent to multiples of $b\otimes a \otimes a - a \otimes a \otimes b$, and exchanging $a$ and $b$, the image of the products of two $b$s and one $a$ are multiples of $a\otimes b \otimes b - b \otimes b \otimes a$. This space is not even a subrepresentation.

I would have hoped to get instead the span of $b\otimes a \otimes a - a \otimes a \otimes b$ and $a\otimes a \otimes b - a \otimes b \otimes a$, which is isomorphic to the standard representation, and then the corresponding subspace under the exchange of $a$ and $b$. Is this correct? Is the image of a Young symmetriser not a representation or am I missing something?

Arnau Mas
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1 Answers1

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The discussion in Schur-Weyl duality for qubits might help answer this question!

Your $a \otimes a \otimes a$ vanishes under this Young Symmetrizer (which corresponds to the space $\mathbb{S}_{2,1} V$, see https://en.wikipedia.org/wiki/Schur_functor) since it belongs to $Sym^3V$. It won't vanish under the symmetrizer for the partition $(3, 0, 0)$, which is just the sum $\sum_{\sigma \in S_3} \sigma$.

Finally, the space you obtained ($Span \{ b \otimes a \otimes a - a \otimes a \otimes b, \ a \otimes b \otimes b - b \otimes b \otimes a \} $) Is a representation of $GL(2, \mathbb{C})$ (you can check that this subspace is invariant under the operation $U\otimes U\otimes U $). On the other hand, $Span \{ b \otimes a \otimes a - a \otimes a \otimes b, \ a \otimes a \otimes b - a \otimes b \otimes a \} $ is a representation of $S_3$.

Jed Burkat
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