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

by user438666   Last Updated August 12, 2019 11:20 AM

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?

Answers 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.

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

enter image description here

The black dots are 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.

enter image description here

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

enter image description here

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 $x_1,x_2,x_3$, 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 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), 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 $$W_2=V(\lambda_1+\lambda_2)\otimes M.$$

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 $$ \begin{aligned} V^{\otimes3}&=V(3\lambda_1)\otimes \mathbf{1}\\ &\oplus V(\lambda_1+\lambda_2)\otimes M\\ &\oplus V(0)\otimes\mathbf{alt}. \end{aligned} $$

Jyrki Lahtonen
Jyrki Lahtonen
August 12, 2019 11:02 AM

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