# 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?

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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 $$V$$:

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.

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 $$x_1,x_2,x_3$$, 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
August 12, 2019 11:02 AM