What is the isospin of the photon? As already mentioned in this Phys.SE post, the PDG booklet says that the isospin of a photon is $0,1$. Is it referring to the strong or to the weak isospin of the photon? What would be the 3rd component of the isospin? And finally, what would be the total $I$ and $I_3$ quantum numbers on the right side of the equation in the case of the following decay:
$$\eta \rightarrow \gamma + \gamma$$
and what would be the type of interaction?
 A: The decays involving photons are, of course, electromagnetic. You may forget about weak intermediaries.
The e.m. current coupling to the photon in the lagrangian in terms of the lightest quarks is 
$$eA_μ(2/3\bar{u}γ^μu−1/3\bar{d}γ^μ d)=eA_\mu \bar{q}(\tfrac{1}{6} \mathbb {1}+I_3)\gamma^\mu q,$$ 
since diag(2/3,-1/3)=diag(1/6,1/6) + diag( 1/2,-1/2). Each quark couples to the photon through its (different) charge. So the photon couples to both isoscalars (I=0) and isovectors (I=1), and electromagnetism violates strong isospin. However, if you imagine this coupling term to preserve strong isospin, then you must represent the photon as a linear combination of (strong) isoscalar and isovector.
As a consequence, if you had to preserve strong isospin, as a way to track its violation, both vertices $\eta \gamma \gamma$ and $\pi^0 \gamma \gamma$ are allowed (they both have isospin zero) in this accounting, and, indeed, both particles decay to two γs as specified by the vertices.
Considerations of weak isospin, spontaneously broken anyway, are irrelevant in these decays.
