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Consider $\mathcal{N}=2$ pure SYM theory. If we want to put the theory in a 4-manifold we take its topological twist. The global symmetry group $$G= SU(2)_{+} \times SU(2)_{-} \times SU(2)_I \times U(1)_R $$ goes to $$ G'=SU(2)_{+} \times SU(2)'_{-} \times U(1)_R $$ where $SU(2)'_{-} = \text{diag}( SU(2)_{-} \times SU(2)_I)$

This $SU(2)'_{-}$ goes by the name diagonal embedding of $ SU(2)_{-} \times SU(2)_I$.

  1. What does this actually mean though?
  2. And how can I see how the original busy fields, say the gauge boson $A_{\mu}$ change representation after the twist?
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    $\begingroup$ The diagonal embedding for any mathematical object $X$ is commonly understood to be the map $X\to X\times X,x\mapsto(x,x)$. $\endgroup$
    – ACuriousMind ♦
    Nov 19, 2015 at 11:52
  • $\begingroup$ But what if you have $X \times Y$ as in the case of twisting? $\endgroup$
    – Marion
    Nov 19, 2015 at 14:10
  • $\begingroup$ I don't see what you mean. You're embedding $\mathrm{SU}(2)\to\mathrm{SU}(2)\times\mathrm{SU}(2)$ here, no? The subscripts are just decorations to keep track of which group does what physically, but they're all the same group. $\endgroup$
    – ACuriousMind ♦
    Nov 19, 2015 at 14:21
  • $\begingroup$ Is it possible to give me an example? The diagonal embedding only keeps elements of which parts of $SU(2)_{-} \times SU(2)_R$? $\endgroup$
    – Marion
    Nov 19, 2015 at 18:32
  • $\begingroup$ @Marion, Do you find the explicit construction of the diagonal embedding? I am thankful to you if you write the answer. $\endgroup$
    – Arian
    Jan 25, 2022 at 18:39

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