What is the diagonal $U(1) \subset SU(2)$ and the diagonal $u(1) \subset su(2)$? What is meant by the diagonal $U(1) \subset SU(2)$ and the diagonal $u(1)\subset su(2)$? I have read it above eqn. (10) in this paper http://arxiv.org/abs/0812.3572 but have also heard it mentioned in physics talks before.
We can choose the Pauli matrices $\sigma^1,\sigma^2,\sigma^3$ as a basis for the Lie algebra $su(2)$. Does this have something to do with $\sigma^3$ being the only one of these that is a diagonal matrix?
 A: Yes, of course, the "diagonal $U(1)$ symmetry inside $SU(2)$" just refers to the group of matrices that are diagonal. The $2\times 2$ matrices  that are diagonal are ${\rm diag}(a,b)$. Their belonging to $SU(2)$ means that $|a|^2=|b|^2=1$ – from the unitarity – and $ab=1$, from the special condition (unit determinant).
So the diagonal $SU(2)$ matrices are simply matrices of the form ${\rm diag}(\exp(i\phi),\exp(-i\phi))$, which form a group isomorphic to a $U(1)$ as announced in that paper.
Note that, as AcuriousMind pointed out, the term "diagonal subgroup" is usually used differently, as the subgroup isomorphic to $G$ of elements of the form $(g,g)$ inside a group $G\times G$.
A: Your terminology is hard to comprehend. My understanding regarding your question goes as follow.
What we know is group $U(1)$ isomorphic to $SO(2)$-rotation in a 2D plane. On the other hand the Lie algebra of $SU(2)$ is same as of $SO(3)$. Which means $SU(2)$ is isomorphic (locally they have same Lie algebra) to $SO(3)$. 
One can write
\begin{equation}
SO(3) = 
\begin{bmatrix}
SO(2) & 0 \\
0 &  1
\end{bmatrix}
\end{equation}
This implies $SO(2)$ is a subgroup of $SO(3)$. Since $SU(2)$ is homomorphic to $SO(3)$. Consequently $SO(2)$ is diagonal to $SU(2)$, as shown.
