# How do I show that a tensor product representation of $L(SU(2))\equiv su(2)$ is reducible?

So I have been reading about the irreducible representations of the Lie algebra $L(SU(2))$ and came across the Cartan-Weyl basis:

$$H = \sigma_3$$ $$E_+ = \frac{1}{2}(\sigma_1+i \sigma_2)$$ $$E_- = \frac{1}{2}(\sigma_1-i\sigma_2)$$

where $\sigma_1,\sigma_2,\sigma_3$ are the Pauli matrices.

## My understanding so far

If I choose to represent $L(SU(2))$ using the Cartan-Weyl basis with a representation $R$, I find that the eigenvectors of $R(H)$ form an $n$ dimensional basis of my representation space $V$, and the $R(E_{\pm})$ transition between these eigenvectors. There are no invariant subspaces of my representation space because each eigenvector can be transformed to another via repeated action of $R(E_{\pm})$ and therefore I have an $n$-dimensional irreducible representation of $L(SU(2))$. The eigenvectors $\{ v_\lambda \}$ have weights (eigenvalues) $\lambda \in \{ -\Lambda, -\Lambda + 2, ... , \Lambda \}$, where the dimension of $V$ is $n = \Lambda + 1$.

## My confusion with tensor product representations

If I took the tensor product of two different representations $R_\Lambda$ and $R_{\Lambda '}$ of the Lie algebra with highest weights $\Lambda$ and $\Lambda'$ respectively, I will get a represnetation space $V_\Lambda \otimes V_{\Lambda'}$ which is spanned by $\{ v_\lambda \otimes v'_{\lambda'} \}$. The representatives of $L(SU(2))$ are given by

$$R_{\Lambda \otimes \Lambda'}(X) = R_{\Lambda}(X) \otimes I_{\Lambda'}+ I_\Lambda \otimes R_{\Lambda '}(X).$$

for $X \in L(SU(2))$. Therefore, we find that $\{ v_\lambda \otimes v'_{\lambda'} \}$ are eigenvectors of $R_{\Lambda \otimes \Lambda'}(H)$ with eigenvalues $\lambda + \lambda'$. This is all fine, but at this point I have read that you can decompose our representation into irreps. I cannot wrap my head around this. My understanding is that a representation is reducible if all the representation matrices can be written in a block diagonal form, or equivalently that the representation space has invariant subspaces w.r.t. the representation elements $R(X)$. How can I show this?

I have seen that the highest weight of $R_{\Lambda \otimes \Lambda'}$ is $\Lambda + \Lambda '$, and it has multiplicity 1, and we can write:

$$R_{\Lambda \otimes \Lambda'} = R_{\Lambda + \Lambda'} \oplus \widetilde{R}_{\Lambda,\Lambda' }$$

for some remainder $\widetilde{R}_{\Lambda,\Lambda' }$. This is the part I do not understand at all. Why can I just do that? If I have one eigenvector $v_\Lambda \otimes v'_{\Lambda '}$ with weight $\Lambda + \Lambda'$, why can I just split the representation space up and say we have a representation $R_{\Lambda + \Lambda'}$ on an invariant subspace? Do I need to show that there are multiple invariant subspace w.r.t. the ladder operators $R(E_\pm)$?

I think this question is suitable for Maths stack exchange however I am a physicist studying a particle physics symmetry module so I have asked it here as I sometimes struggle to understand rigorous mathematical proofs.

• Particle physics? Have you "added" two simple spins or angular momenta in your elementary quantum mechanics? Add spin 1/2 to a spin 1/2? – Cosmas Zachos Dec 27 '17 at 0:49

Let's work this out explicitly for the case of adding two spins of $1/2$. In your notation, the weights we’re adding are $\{-1, 1\}$ and $\{-1, 1\}$. Upon taking the tensor product, we get a four-dimensional vector space with a vector of weight $2$, a vector of weight $-2$, and a two-dimensional plane of vectors with weight $0$.

You want to know how we can generate an irreducible subspace by starting with the vector of weight $2$. The irreducible subspace containing a vector is the set of vectors we can reach from that vector by applying raising and lowering operators. So we start with the vector of weight $2$, lower to get a vector of weight $0$, and lower again to get a vector of weight $-2$. Further lowering can't get us anything, because there's nothing of weight $-4$.

You might wonder why we can't get the other vector of weight $0$. Well, we already know all the irreducible representations of $\mathfrak{su}(2)$, and all of them have nondegenerate weights, so $0$ can't show up twice. But more explicitly, the only way to get a different $0$ is to start at $-2$ and apply the raising operator, and you can use the commutation relations to show you end up with the exact same $0$ weight vector as before.

Thus three vectors with weights $\{-2, 0, 2\}$ form an irreducible subrepresentation. We can then forget about them and proceed to the remaining weight $0$ which also forms an irreducible subrepresentation; the reasoning is similar for arbitrary spins.

• Thank you for your answer. So essentially, from our knowledge of the irreducible Lie Algebras of $SU(2)$, we know that all the weights are unique, so if we have a representation that has root degeneracy, we must have some decomposition! This makes sense. One question: are the invariant subspaces invariant to the action of the Cartan Weyl basis or the angular momenta operators, or both? – Matt0410 Dec 27 '17 at 0:55
• @Matt0410 Both. After all, they're just linear combinations of each other. – knzhou Dec 27 '17 at 0:57
• And of course once you go to more general Lie algebras like $\mathfrak{su}(3)$ the weights are no longer nondegenerate for irreps, so you need more machinery. (You never get extra vectors by 'going forwards and then backwards' but you can get them by 'going in a triangle'.) – knzhou Dec 27 '17 at 0:58

I) What OP calls weights $$\lambda ~\in~ \{ -\Lambda, -\Lambda + 2, \ldots ,\Lambda-2, \Lambda \},\qquad \Lambda~\in~\mathbb{N}_0,\tag{1}$$ of an irreducible $sl(2,\mathbb{C})$ representation (irreps) are twice as big as physicists' traditional half-integer notation $$m ~\in~ \{ -j, -j + 1, \ldots ,j-1, j \},\qquad j~\in~\frac{1}{2}\mathbb{N}_0. \tag{2}$$

A tensor product of $sl(2,\mathbb{C})$ irreps decomposes into a direct sum of irreps via the Clebsch-Gordan fusion rule

$$\underline{2j_1+1} \otimes \underline{2j_2+1} ~=~ \oplus_{j=|j_1-j_2|}^{j_1+j_2} \underline{2j+1}.\tag{3}$$

Here $\underline{1}$, $\underline{2}$, $\underline{3}$, $\ldots$, refer to the singlet irrep, doublet irrep, triplet irrep, $\ldots$, respectively. It is a nice exercise to check that the dimensions on the right- and left- hand sides of (3) match. See also this & this Phys.SE questions.