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I know that if one reduces 10 dimensional $\mathcal{N}=1$ SYM theory to 4 dimensions one gets $\mathcal{N}=4$ SYM. There are other examples also. I have two related questions regarding this fact.

  1. How does one know, to obtain a particular theory in lower dimension, which higher dimensional theory to start with?

  2. Doesn't the end product (final theory) depend on the procedure i.e. on which manifold one compactifies it to? In some cases there might be more than one possibility of compactification which preserve SUSY, I guess.

I am not very much familiar with the literature on SUSY gauge theories. So this questions might be very elementary.

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  • $\begingroup$ Hi, if you are interested in this subject there are good and reasonably understandable lecture notes by prof. Samtleben in which some explicit example is given too arxiv.org/abs/0808.4076 $\endgroup$
    – Oscar
    Commented Mar 19, 2019 at 20:12

2 Answers 2

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  1. A priori, it is hard to know without having any experience with dimensional reductions. One has to get a feeling for how certain quantities change under the process, e.g. how components of the higher-dimensional gauge fields may turn into adjoint scalars, how spinors behave. An interesting thing to note is that symmetries of the lower-dimensional theory often have a geometric interpretation in terms of the higher-dimensional theory. For example, the $\mathrm{SL}(2,\mathbb{Z})$ symmetry of $\mathcal{N}=4$ super Yang-Mills in four dimensions can be understood by starting from a six-dimensional theory and compactifying on a torus, which has an $\mathrm{SL}(2,\mathbb{Z})$ modular invariance.
  2. Yes, this is correct. Compactification is by no means unique, and the result definitely depends on the manifold. One can even break supersymmetry by compactifying accordingly.
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  • $\begingroup$ Your answer is very good but is he talking about compactification our dimensional reduction? As I understand dimensional reduction is just replace the covariant derivatives by scalars $\nabla_M\rightarrow \Phi_M$ and split the spinors of the theory, which is physically equivalent to perform a T duality along a D-brane. $\endgroup$
    – Nogueira
    Commented Feb 27, 2020 at 2:08
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A neat way of deciding which higher-dimensional theory to start from is to count the number of spinor components. In ten dimensions, a spinor has $2^{[d/2]} = 32$ components (complex) which are reduced to 16 after imposing the Majorana-Weyl condition consecutively in that order(which can only be done where d mod 8 = 2). Therefore, $\mathcal{N}=1$ SYM in ten dimensions has 16 components (real !)

A Majorana spinor in four dimensions has 4 components. So, suppose you want $\mathcal{N}=4$ SYM in four dimensions (which has 16 [four copies of four components) you want to start with $\mathcal{N}=1$ SYM in ten dimensions.

$\mathcal{N}=1$ SYM in four dimensions similarly has 4 components. So, if I want to construct a supersymmetric theory in two dimensions with four supercharges i.e $\mathcal{N}$ = (2,2) SYM (which has single component after imposing Majorana-Weyl), then one should start from $\mathcal{N}=1$ SYM in four dimensions and not with $\mathcal{N}=4$ SYM.

16 supercharges:
D=10, $\mathcal{N}$ = 1 $~\to~$ D=6, $\mathcal{N}$ = 2 $~\to~$ D=4, $\mathcal{N}$ = 4 $~\to~$ D=3, $\mathcal{N}$ = 8 $~\to~$ D=2, $\mathcal{N} = (8,8)$

8 supercharges:
$\hspace{33mm}$ D=6, $\mathcal{N}$ = 1 $~\to~$ D=4, $\mathcal{N}$ = 2 $~\to~$ D=3, $\mathcal{N}$ = 4

4 supercharges:
$\hspace{64mm}$ D=4, $\mathcal{N}$ = 1 $~\to~$ D=3, $\mathcal{N}$ = 2

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