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

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?

• All the irreps of $S_2$ are 1-dimensional by virtue of it being abelian. That is why $N=2$ is a simpler case. Aug 12, 2019 at 9:42
• @JyrkiLahtonen then I imagine that one of the missing tensor product is the trivial representation, the only other is the alternating one. Should I interpret that as $S^2 V\otimes \mathbf{1} \oplus \Lambda^2 V \otimes \mathbf{alt}$? That looks weird to me as $\Lambda^2 V$ is already the alternating representation of $S_2$ Aug 12, 2019 at 9:48
• I have some old Mathematica snippets lying around. I will try and draw a picture related to $N=3$. Aug 12, 2019 at 9:56
• The fundamental dominant weights. $\lambda_1$ is the highest weight of the fundamental 3-dimensional rep $V(\lambda_1)$. $\lambda_2$ is the highest weight of the dual 3-dimensional rep $V(\lambda_2)$. Physicists denote these reps $3$ and $\overline{3}$ respectively. Aug 12, 2019 at 10:02
• In that case we have opposite conventions as to which is $\lambda_1$ and which is $\lambda_2$. The choice does not really matter :-) Aug 12, 2019 at 10:09

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 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}

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 $$V^{\otimes 3}$$.

• Thank you very much. It will take me a bit to read the answer in detail, I'll report back! Aug 12, 2019 at 11:06
• I failed to explain that $V^{\otimes 3}$ splits as an $S_3$-module accordingly: The trivial rep appears with multiplicity ten, $M$ with multiplicity eight, and $\mathbf{alt}$ with multiplicity one. This can be verified by calculating the character of $V^{\otimes3}$ and the appropriate inner products. Aug 12, 2019 at 11:29
• Sorry, I'm a bit confused by the blue arrows in the drawings: if I understand correctly the green circles are the weights of the fundamental rep of $\mathfrak{sl}(3,\mathbb{C})$ which are $\lambda_1,\lambda_2,\lambda_3$. This would make of the two blue arrows $\lambda_2$ and $\lambda_1+\lambda_2$, yet you call one of the weights with multiplicity $3$ in the three-fold tensor product $\lambda_1+\lambda_2$, but that looks to me like $\lambda_1+2\lambda_2$ Aug 12, 2019 at 11:39
• No, @user438666. The weights of the fundamental rep $V$ are $\lambda_1$, $-\lambda_1+\lambda_2$ and $-\lambda_2$. The root system has rank two, so there are only two fundamental dominant weights. The difference between any two weights in the same irreducible representation must differ by an element of the root lattice (= the free abelian group generated by the red arrows). Aug 12, 2019 at 11:46
• ah, I think we are using different conventions. The reference I'm following uses weights $\lambda_i$ such that $\lambda_1+\lambda_2+\lambda_3=0$ and are exactly the green circles in your picture, I know how to "translate" it now I think Aug 12, 2019 at 11:53