Degrees of freedom in $(A,B)$ representation of the Lorentz group

Question

Matthew D. Schwartz in his QFT and the Standard Model subsection 10.1.2, shows that $SO(1,3) \cong SU(2) \times SU(2)$ which according to this PhysicsSE post is actually untrue. I was confused about this because of the usage of direct product to actually mean tensor product in texts such as, for eg., Wu Ki Tung's Group Theory in Physics. But here in this context, it is now actually clear that what is meant is a direct product so the following makes sense.

Schwartz writes:

\begin{equation} \mathfrak{so}(1,3) \cong \mathfrak{su}(2)\oplus \mathfrak{su}(2).\tag{10.27} \end{equation}

So that would mean that if we have an spin $A$ and $B$ representations of $\mathfrak{su}(2)$, we can construct(?) a representation of $\mathfrak{so}(1,3)$ by taking the direct sum of these two representations, Schwartz calls this (and I think this is standard) $(A,B)$. I'll be speaking terms of matrix representations to avoid being too abstract. Now since this is a direct sum of a $2A+1$ dimensional representation with a $2B+1$ dimensional representation, the matrix representation of $(A,B)$ is $2A+2B+2$ dimensional, which is what I would expect.

The point is that Schwartz says below eq. (10.27) that this representation has $(2A+1)(2B+1)$ degrees of freedom, which is not clear to me, since I only see $(2A+1) + (2B+1)$. Can someone clarify?

Answer 1

Now since this is a direct sum of a $2A+1$ dimensional representation with a $2B+1$ dimensional representation, the matrix representation of $(A,B)$ is $2A+2B+2$ dimensional, which is what I would expect.

No, absolutely not! You have made a hash between the Lie algebra consisting of the 3+3=6 generators of the Lie group, and the Lie group itself, the infinity of exponentials of linear combinations of these generators, or rather their representations!

Before you get into nonlocal effects remote from the identity and the somewhat subtle but not technically decisive diference between Cartesian and tensor products, it is important you appreciate the big picture of the corresponding matrices acting on vectors.

You have three linearly independent (2+1) × (2+1) matrices acting on 2+1 dimensional vectors, and leaving 2+1 dimensional vectors alone, and three (2+1)×(2+1) matrices acting on 2+1 dimensional vectors and leaving 2+1 vectors alone; so these matrices "don't talk to each other", and are not involved in nontrivial commutators; and act on (2+1)×(2+1) dimensional vectors living in the tensor product of the two algebra and group representations $(A,B)$.

Consider a generic element of the group, with $\mathfrak{a}$ from the left $\mathfrak{su}(2)$, and $\mathfrak{b}$ from the right one, $$ e^{ i\theta ~\mathfrak{a} \otimes \mathbb{I}_B + i \phi~\mathbb{I}_A\otimes\mathfrak{b} }= e^{i\theta \mathfrak{a}}\otimes e^{i \phi \mathfrak{b}}, $$ obviously acting on vectors $v_A\otimes v_B$, of dimension (2+1)×(2+1) as stated. Your error is in thinking of this exponent as a block matrix with $\mathfrak{a} $ and $\mathfrak{b}$ blocks!

Infinitesimally, it expands to $$ (\mathbb{I}_A + i\theta \mathfrak{a}+...) \otimes (\mathbb{I}_B + i \phi \mathfrak{b}+...)= \mathbb{I}_A \otimes \mathbb{I}_B + i\theta ~\mathfrak{a} \otimes \mathbb{I}_B + i \phi~\mathbb{I}_A\otimes\mathfrak{b}+... $$

Work out a simple example with A=1/2 and B=1, and write the 6×6 tensor product matrices involved for two representative generators from each factor group of your choice. The exponentials for these two irreps are known in closed from, so... (Again, you do not have 5×5 matrices, instead!)