$SO(6) \to SU(2) \times SU(2) \times U(1)$ branching rules What do these branching rules mean?
\begin{eqnarray*} SO(6)_E &\to& SU(2)_\ell \times SU(2)_r \times U(1)_\Sigma 
\end{eqnarray*}
I am taking these examples from a paper of Gukov (on p.51) but more examples all over this paper.  And this discussion of Lie groups and this notation could be in any number of hep-th papers.
This could be the branching rules of a group representation, however this is likely the transformations of a sections of a vector bundle over a 6-dimensional space.  


*

*$SO(6)$ acts on a 6-manifold (such as $\mathbb{R}^6$ or $M_4 \times \Sigma$, where $M_4$ is a 4-manifold and $\Sigma$ is a Riemann surface.

*$SO(6)$ acts on sections of a vector bundle (e.g. the tangent bundle).  In fact the paper mentions various particles (such as a "Weyl fermion") - what kind of bundle is that?


Here are a few of the branching rules that he mentions:
\begin{eqnarray*}
 \mathbf{4}_+ &\to& (\mathbf{2}, \mathbf{1})^{+1}\oplus  (\mathbf{1}, \mathbf{2})^{-1} \\
 \mathbf{4}_- &\to& (\mathbf{2}, \mathbf{1})^{-1}\oplus  (\mathbf{1}, \mathbf{2})^{+1} \\
\mathbf{6} &\to & (\mathbf{2}, \mathbf{2})^{0}\oplus  (\mathbf{1}, \mathbf{1})^{+2}\oplus  (\mathbf{1}, \mathbf{1})^{-2}
\end{eqnarray*}
Topologically twisted 6d (0,2)-theory is a bit out of my reach, but we know they will be solutions to a differential equation involving sections of a reasonable-looking bundle such as:
$$ \Lambda_+^2M_4 \times T^\ast \Sigma  $$
with the wedge on the 4-manifold $M$ and the co-tangent space on the surface $\Sigma$.
 A: $SO(6) = SU(4)/Z_2$ (i.e. the $Alt^2$ rep of $SU(4)$ preserves an $\mathbb R^6$ inside that $\mathbb C^6$), by the way. 
Your subgroup is of the same rank as the whole, so by Borel-de Siebenthal theory it must be obtained by iterating the following two operations: erase a vertex of the Dynkin diagram, or affinize-a-component-then-erase-a-vertex-of-that. Since your ambient group $A_3$ is a product of type $A$ groups, the affinize-then-erase step does nothing. So all we get to do is erase a vertex. Since you want to get $A_1 \times A_1$ when you're done, the one to erase is the middle vertex of the $A_3$.
One way to think about your groups is to go from $SO(6)$ to $SO(4)\times SO(2)$, and then identify $SO(4)$ as $(SU(2)\times SU(2))/Z_2$ (which I think of as the left-right multiplication of $U(1,\mathbb H)$ on $\mathbb H$).
Anyway this branching is easy to compute positively since it's to a Levi. For example, you could use SSYT to describe your $A_3$-representation, then look at those tableaux that are unraisable w.r.t. the first and third tableau crystal operators.
