# Reducibility of tensor products of Lorentz group representations

Consider the statement: (34.29 in Srednicki's QFT text) $$\tag{34.29} (2,1)\otimes(1,2)\otimes(2,2)~=~(1,1)\oplus\ldots$$

Where of course, $(a,b)$ label representations of Lorentz group in the usual way.

In particular, it says this tensor product of the group representions is completely reducible, into the direct sum of $(1,1)$ with some other representations.

Why is this so?

It would be helpful if someone could point me to good literature or lecture notes on this- my understanding of representation theory and Lie groups & algebras is at the level of arXiv:math-ph/0005032.

• Relevant: physics.stackexchange.com/q/6108 – DJBunk May 21 '14 at 21:19
• I would recommend checking out Georgi's text in particular, he gives a nice presentation how to decompose reducible representations. – DJBunk May 21 '14 at 21:20
• Unfortunately I don't have Georgi handy right now. First of all, did you read chapter 33 in Scrednicki? He at least states the result you are asking about in the last couple of paragraphs of that chapter. Additionally, Ramond covers this in chapter 5 of his group theory textbook- section 5.2 – DJBunk May 21 '14 at 21:42
• If you know and understand the answer now, I suggest you post it so others can benefit. – DJBunk May 22 '14 at 0:39
• I've posted my own solution, please check for correctness. – zzz May 23 '14 at 16:05

SU(2) representations

For representations of $SU(2)$, we know: $$(2m+1) \otimes(2n+1) = (2(m+n)+1)\oplus (2(m+n-1)+1)\oplus...\oplus(2(m-n)+1)$$ We get this simply from angular momenta addition. For example: $$2\otimes2 = 1 \oplus 3$$ Where the numbers label the dimension of (target space of) the representation (or the components of the representation).

This says the space that is invariant under the $2 \otimes 2$ representation, let's call is $S_2$ ( the target space of that representation), is isomorphic to the direct sum of 2 other spaces $S_1 \oplus S_3$, such that $S_1$ is invariant under the $1$ representation, and $S_3$ is invariant under $3$. In an explicit derivation for angular momenta addition in Hilbert space, $1$ would correspond to the 1-dimensional subspace of spin-0 states, $3$ corresponds to the 3-dimensional subspace of spin-1 states. The 1 spin-0 and 3 spin-1 spin eigenstates span the space, hence indeed the direct sum decomposition above holds.

Lorentz group representations

Lie algebra of the lorentz group is generated by some operators $L_i$, a useful decomposition is $L_i= N_i + N_i^{\dagger}$, where $N_i$, $N_i^{\dagger}$ are elements in an $SU(2)$ representation. Since $N_i$ and $N_i^{\dagger}$ commute, we can obtain the eigenspace of $L_i$ through angular momenta addition. Hence we can view the spaces spanned by the eigenvalues of each of the $N_k$ and $N_k^{\dagger}$ (for an arbitrary choice of k) as a Hilbert space, then $L_i$ is an operator in the tensor product of these Hilbert spaces. Namely, $L_i$ act the linear combinations of tensor product states $|... \rangle | ...\rangle$. This means we can view the representation $L_i$ as a tensor product of the 2 $SU(2)$ representations $N_i$ and $N_i^{\dagger}$. Hence the precise meaning of the notation $(1,2)$ is: $$(a,b) \equiv a \otimes b$$

From here we can use the above result for tensor products of $SU(2)$ representations. For example:

$$(2,1)\otimes(2,1) = (2 \otimes 1)\otimes (2 \otimes 1) = (2 \otimes (1\otimes 2) \otimes 1) = (2 \otimes 2 \otimes 1) \\=((1 \oplus 3)\otimes 1)= (1 \otimes 1) \oplus (3 \otimes 1) = (1,1) \oplus (3,1)$$

Where we've use the various properties of the tensor and direct products (associativity, distributivity). In the third equality we reduced the tensor product using the formula at the very top. Repeating the procedure for:

$$(2,1)\otimes(1,2)\otimes(2,2)$$

We clearly see $(1,1)$ is one of the terms in the direct sum decomposition.