I am reading Witten's [paper](https://projecteuclid.org/euclid.cmp/1104161738) on topological field theories, in specific the topological twist in page 359. In order to perform the twist he takes the diagonal subgroup of $K = SU(2)\_{\text{Right}} \times SU(2)\_{\text{Isospin}}$ where the first comes from the $SU(2)\_{\text{Left}} \times SU(2)\_{\text{Right}} \simeq SO(4) $ while the latter is part of the $\mathcal{R}$-symmetry group $SU(2)\_{\text{Isospin}} \times U(1)\_{\mathcal{R}}$.


What does it mean to take the diagonal of $K$ and how can I understand how the representations of the sections/fields change after the twist? I mean, he says it is "easy to see" how the fields transform under the new global group but I cannot see it.



What does it mean to take the diagonal of the group $SU(2) \times SU(2) $?
As noted in Vit's post, the diagonal group $G\_D \subset G \times G$ consists of the elements $\{(g,g) : g \in G \}$ with multiplication defined to be $(g,g) \cdot (h,h) = (gh,gh)$. A representation $(R\_1,R\_2)$ of $G \times G$ thus transforms as the representation $R\_1 \otimes R\_2$ under $G\_D$, where $\otimes$ denotes the tensor product.


Let me illustrate through an example in the paper you are reading. In equation (2.16), it is stated that the fermions are in the representation
\begin{equation}
(1/2,0,1/2)^1 \oplus (0,1/2,1/2)^{-1}
\end{equation}
under $SU(2)\_L\times SU(2)\_R \times SU(2)\_I \times U(1)$. Now under
\begin{equation}
SU(2)\_L\times SU(2)\_{R'} \times U(1) \subset SU(2)\_L\times SU(2)\_R \times SU(2)\_I \times U(1) \,,
\end{equation}
where $SU(2)\_{R'}$ is the diagonal subgroup of $SU(2)\_R \times SU(2)\_I$, this representation transforms as
\begin{equation}
(1/2,0 \otimes 1/2)^1 \oplus (0,1/2 \otimes 1/2)^{-1} \,.
\end{equation}
We know the tensor product decomposition rule for $SU(2)$ pretty well:
\begin{equation}
0 \otimes 1/2 = 1/2 \,, \quad
1/2 \otimes 1/2 = 0 \oplus 1 \,.
\end{equation}
Recall that the numbers here denote the "spin" of the representation. A spin $j$ representation has dimension $(2j+1)$. Note that the representation ``0" is the trivial (one-dimensional) representation. Thus we recover equation (2.18):
\begin{equation}
(1/2,1/2)^1 \oplus (0,0)^{-1} \oplus (0,1)^{-1} \,.
\end{equation}


I hope this makes everything clear to you now.


