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I am reading Witten's paper 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.

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    $\begingroup$ I have replaced $\otimes$ with $\times$ in this question. Occasionally physicists speak of tensor products like $SU(2) \otimes SU(2)$, but they mean cartesian products like $SU(2) \times SU(2)$, and more mathematicians will understand the question if we use standard notation. $\endgroup$
    – John Baez
    Commented Sep 12, 2015 at 6:40
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    $\begingroup$ I'm not sure what you mean by "as it is in the case of global symmetries", but 1) there is no standard mathematical definition of a tensor product of nonabelian groups, and 2) the Standard Model gauge group $SU(3) \times SU(2) \times U(1)$ is a cartesian product. $\endgroup$
    – John Baez
    Commented Sep 15, 2015 at 7:25
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    $\begingroup$ Hi, yes this is what I wanted to clear out. You see, in most of my courses, they would use the tensor product for the groups. Thanks for clearing out for me this " there is no standard mathematical definition of a tensor product of nonabelian groups" $\endgroup$
    – Marion
    Commented Sep 15, 2015 at 9:11
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    $\begingroup$ Some physicists use the tensor product notation here, but mathematicians never do, because it's really just the cartesian product. $\endgroup$
    – John Baez
    Commented Sep 15, 2015 at 13:08
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    $\begingroup$ Sorry, I can't read that stuff. But I know I'm right. You can take the tensor product of vector spaces, or Hilbert spaces, or abelian groups, or bimodules, but not groups. Check out Wikipedia: en.wikipedia.org/wiki/Tensor_product I also know that physicists sometimes. use the tensor product symbol for groups when they mean Cartesian product. It used to annoy and confuse me. $\endgroup$
    – John Baez
    Commented Sep 15, 2015 at 14:07

2 Answers 2

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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.

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  • $\begingroup$ This is a ver very good answer. I am wondering if you could show the same for the $\mathcal{N}=4$ twist though. There things are harder since the global rotation group is $SU(2)_L \times SU(2)_R \times SU(4)_R$ which can be written as $SU(2)_L \times SU(2)_R \times SU(2)_A \times SU(2)_B \times U(1)$. Now, I do not see how exactly the spinor $(2,1,\bar{4}) \oplus (1,2,4)$ representation changes (that is how the supercharges change under the twist). $\endgroup$
    – Gorbz
    Commented Jun 14, 2017 at 15:00
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I am confused by the notation and terminology, but my guess would be that you just need to calculate the branching from $SU(2)_L \times SU(2)_R \times SU(2)_I$ to the subgroup $SU(2)_L \times SU(2)$, where $SU(2)$ is the diagonal subgroup of $SU(2)_R \times SU(2)_I$, i.e. the subgroup $\{ (g,g) \in SU(2)_R \times SU(2)_I \, | \, g\in SU(2) \}$.

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  • $\begingroup$ Hi, it is physics notation. Thanks for your answer but I think I will require something more closely related to the way the twist works and how the fields change representation. $\endgroup$
    – Marion
    Commented Sep 10, 2015 at 13:37
  • $\begingroup$ My (rather poor) understanding of physics (jargon) is that the fields doesn't change representation. Merely, you just forget the action of $SU(2)_R \times SU(2)_I$ and remember from it only the action of the diagonal subgroup. $\endgroup$
    – Vít Tuček
    Commented Sep 10, 2015 at 13:45
  • $\begingroup$ Well, the fields change transform differently under the twisting. This is what I meant. I want to understand how this happens. $\endgroup$
    – Marion
    Commented Sep 10, 2015 at 14:28
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    $\begingroup$ I still think that the group action is the same. What differs is the weight coordinates by which you denote this action and that is because you have a 2-rank group instead of 3-rank one. $\endgroup$
    – Vít Tuček
    Commented Sep 10, 2015 at 14:38

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