Representations of $\mathfrak{sl}(3,\mathbb{C})$ and the symmetric group

Question

let $V$ be the fundamental 3 dimensional representation of $\mathfrak{sl}(3,\mathbb{C})$ and consider the product $V^{\otimes N}$. The action of any representation $\rho$ of $\mathfrak{sl}(3,\mathbb{C})$ commutes with the action of the symmetric group $S_N$ that permutes the vector in the tensor product. I read that this implies the Schur Weyl duality

$$V^{\otimes N}=\bigoplus_{\lambda}V_\lambda\otimes S_\lambda $$

where $V_\lambda$ are irreducible representations of $\mathfrak{sl}(3,\mathbb{C})$ and $S_\lambda$ are irreducible representations of $S_N$. I'm confused by how this decomposition works, usually the following example is provided

$$V^{\otimes 2}=S^2V\oplus\Lambda^2V $$

where $S^2V$ is the symmetric part of the tensor product and $\Lambda^2V$ the antisymmetric part. I don't see the promised decomposition in this example, I see a direct sum of two spaces that are irreducible representations of both $\mathfrak{sl}(3,\mathbb{C})$ and $S_2$, not a direct sum of tensor products of representations of $\mathfrak{sl}(3,\mathbb{C})$ and $S_2$. Could somebody help me understand some concrete examples of this decomposition?

Answer 1

You may have been confused by the fact that you have a preconceived idea of how $S_3$ should act on the components. Keep in mind that there is no natural action of $S_N$ on any of the modules $V(\lambda)$. That action comes to being only when we look at subspaces of $V^{\otimes N}$ specifically! So Schur-Weyl seeks to identify subspaces of $V^{\otimes N}$, and identify what they look like as reps for $\mathfrak{sl}_3$ and $S_N$ independently from each other.

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I will try and describe this decomposition in the case $N=3$. I use weight diagrams. Here's the way I draw $V$:

The black dots are the weights. The red arrows are the roots. The blue arrows are the fundamental dominant weights. The green circles give the formal character of the fundamental $\mathfrak{sl}_3$-module $V=V(\lambda_1)$. A single green circle means that the multiplicities of all the weights are $=1$.

The second diagram shows the formal character of $V^{\otimes 3}$. Notice that this time multiplicities $3$ and $6$ occur, and I try to convey that with the appropriate number of concentric green circles.

We immediately spot the highest weight $3\lambda_1$. Indeed, the 10-dimensional $\mathfrak{sl}_3$-module $V(3\lambda_1)$ is a summand of $V^{\otimes3}$. This summand consists of totally symmetric tensors, so it is trivial as an $S_3$-module. What the Schur-Weyl formula is trying to convey is that the 10-dimensional subspace $W_1$ has the structure $V(3\lambda_1)\otimes \mathbf{1}$ when viewed as a mixed module of $\mathfrak{sl}_3\times S_3$. Recall that the formal character of $V(3\lambda_1)$ looks like

In other words, all its weights have multiplicity one.

Next we observe that the second highest weight of $V^{\otimes 3}$ is $\lambda_1+\lambda_2$, appearing with multiplicity $3$. We recall that this is the highest weight of the adjoint representation of $\mathfrak{sl}_3$, of dimension $8$. The conclusion is that the adjoint representation appears as a composition factor of $V^{\otimes3}$ with multiplicity two. So there is a $16$-dimensional subspace $W_2$ of $V^{\otimes3}$ that as an $\mathfrak{sl}_3$ module looks like two copies of $V(\lambda_1+\lambda_2)$. What does it look like as an $S_3$-module? If the weight vectors of $V$ are $x_1$ of weight $\lambda_1$,$x_2$ of weight $-\lambda_1+\lambda_2$, and $x_3$ of weight $-\lambda_2$, then the weight space corresponding to weight $\lambda_1+\lambda_2$ in $V^{\otimes3}$ is the span of $$x_1\otimes x_1\otimes x_2, x_1\otimes x_2\otimes x_1, x_2\otimes x_1\otimes x_1.$$ We see that the group $S_3$ permutes those three vectors according to its natural $3$-dimensional representation. We recall from representation theory of finite groups that this $3$-dimensional rep splits into a direct sum of the trivial representation (here spanned by the averagre of those three vectors belonging to the subspace $W_1$ of symmetric tensors), and a 2-dimensional irreducible representation, call it $M$. Obviously the space $W_2$ must then be a bunch of copies of $M$ as an $S_3$-module. In other words, Schur-Weyl wants us to identify the space $W_2$ as $$W_2=V(\lambda_1+\lambda_2)\otimes M.$$ As a refresher please find the weight diagram of the adjoint representation:

Finally, there is the 1-dimensional subspace $W_3$ of completely antisymmetric tensors. Because the weights of $V$ add up to zero, $W_3\cong V(0)$ as an $\mathfrak{sl}_3$-module. Because the tensors are totally antisymmetric, as a representation of $S_3$ we see that $W_3$ looks like $\mathbf{alt}$, the $1$-dimensional representation affording the sign character.

So the Schur-Weyl decomposition looks like $$ \begin{aligned} V^{\otimes3}&=\left(V(3\lambda_1)\otimes \mathbf{1}\right)\\ &\oplus \left(V(\lambda_1+\lambda_2)\otimes M\right)\\ &\oplus \left(V(0)\otimes\mathbf{alt}\right). \end{aligned} $$

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As a check let's verify differently that $V^{\otimes3}$ splits in the prescribed way as an $S_3$-module. Let $\psi$ be the character of $V^{\otimes3}$ as an $S_3$-module. We see that

• $\psi(1_{S_3})=27$.
• The permutation $(12)$ fixes the nine basic tensors of the form $x_i\otimes x_i\otimes x_j$, $i=1,2,3$, $j=1,2,3$, and permutes the other vectors of the basis. Meaning that $\psi((12))=9$ and the same holds for all the other 2-cycles.
• The 3-cycle $(123)$ fixes the three basis vectors $x_i\otimes x_i\otimes x_i$ so $\psi((123))=3$.
• The multiplicity of the trivial character $\chi_0$ as a component of $\psi$ is thus $$\langle\psi,\chi_0\rangle_{S_3}=\frac16(1\cdot27\cdot1+3\cdot9\cdot1+2\cdot3\cdot1)=10.$$
• Similarly the multiplicity of the $2$-dimensional character $\chi_M:1\mapsto 2, (12)\mapsto0,(123)\mapsto-1$ is $$\langle\psi,\chi_M\rangle_{S_3}=\frac16(1\cdot27\cdot2+3\cdot9\cdot0+2\cdot3\cdot(-1))=8.$$
• And the multiplicity of the sign character $\mathbf{alt}$ is $$\langle\psi,\mathbf{alt}\rangle_{S_3}=\frac16(27\cdot1+3\cdot9\cdot(-1)+2\cdot3\cdot1)=1.$$

All of this matches with the conclusion that $W_1$, $W_2$ and $W_3$ are also the isotypic components of $V^{\otimes 3}$.