We primarily see 2 types of notation \(\mathcal{N}=(2,2)\) and \(\mathcal{N} = 2\). The ambiguity is resolved by considering the spacetime dimensions we are working in. David Tong’s notes on supersymmetry has a chapter “Counting Supersymmetries” which explains this in greater detail. I aim to summarise.

“Spinor supercharges” vs “Supersymmetry Generators/Supercharge”

\(Q^I_\alpha\) (collecting the \(\alpha\) components into a column vector) is said to be a spinor supercharge. Index \(I\) goes from \(1,...,(\mathcal{N} \text{ or } L \text{ or } R)\) (more on this ambiguity later), and index \(\alpha\) goes from \(1,...,\text{dim of Weyl spinor representation in dim }D\). Each component of the spinor supercharge is a SUSY generator. SUSY generators are also called supercharges confusingly. Therefore I find it important to state whether a quantity is a “spinor supercharge” or a “supercharge”. A spinor supercharge consists of a few component supercharges. Actually, I prefer to use supercharge to exclusively mean “spinor supercharge” and use “SUSY generator” to mean the individual components.

We use \(N\) (non-curly N) to denote the number of SUSY generators. And \(\mathcal{N}\) (mathcal N) to denote the number of spinor supercharges.

\(\mathcal{N} = (L,R)\) vs \(\mathcal{N} = n\)

And then we see things like \(\mathcal{N} = (2,2)\) vs \(\mathcal{N} = 1\). What’s going on? Again Tong’s notes explains it well.

In dimensions \(D \equiv 2 \text{ mod } 4\), left and right handed Weyl spinors are NOT related by complex conjugation, meaning that we can have a left handed spinor supercharge without corresponding right handed spinor supercharge. That is why we use 2 numbers to describe the number of spinor supercharges, each one being possible left or right handed.

On the contrary, in \(D\equiv 0 \text{ mod } 4\), left and right handed spinors are related by complex conjugation. They come in pairs. Existence of left implies existence of right and vice versa. Hence one number describes the number of spinor supercharges. If we wanted to express \(\mathcal{N} = n\) in terms of \((L,R)\) notation, it would be \((n,n)\).

So if you see something like \((L,R), \text{dim }D\), the number of SUSY generators are \(N=(L+R)\times (\text{dim of Weyl spinor representation in dim }D)\).

And if you see something like \(\mathcal{N}, \text{dim }D\), the number of SUSY generators are \(N=\mathcal{N} \times 2 \times (\text{dim of Weyl spinor representation in dim }D)\).

I have a question: What happens in odd dimensions?

Some Examples

The following examples are taken from David Tong “Counting Supersymmetries” Chapter.

\(N=32\):

  • \(\mathcal{N} = 1\), \(D=11\)
    The spinor representation in \(11\) dimensions consists of \(16\) component generators. So \(N=32\) is obtained by \(\mathcal{N} \times 2 \times 16\).
  • \(\mathcal{N} = (1,1)\), \(D=10\)
    The spinor representation in \(10\) dimensions consists of \(16\) component generators. So \(N=32\) is obtained by \((L+R) \times 16\).
  • \(\mathcal{N} = (2,2)\), \(D=6\)
    The spinor representation in \(6\) dimensions consists of \(8\) component generators. So \(N=32\) is obtained by \((L+R) \times 8\).
  • \(\mathcal{N} = 8\), \(D=4\)
    The spinor representation in \(4\) dimensions consists of \(2\) component generators. So \(N=32\) is obtained by \(\mathcal{N} \times 2 \times 2\).

\(N=16\):

  • \(\mathcal{N} = (1,0)\), \(D=10\) (check yourself)
  • \(\mathcal{N} = (1,1)\), \(D=6\) (check yourself)
  • \(\mathcal{N} = 4\), \(D=4\) (check yourself)
  • \(\mathcal{N} = 8\), \(D=3\)
    The spinor representation in \(3\) dimensions consists of \(1\) component generators. So \(N=16\) is obtained by \(\mathcal{N} \times 2 \times 1\).
  • \(\mathcal{N} = (8,8)\), \(D=2\)
    The spinor representation in \(2\) dimensions consists of \(1\) component generators. So \(N=16\) is obtained by \((L+R) \times 1\).

\(N=8\)

  • \(\mathcal{N} = (1,0)\), \(D=6\) (check yourself)
  • \(\mathcal{N} = 2\), \(D=4\) (check yourself)
  • \(\mathcal{N} = 4\), \(D=3\) (check yourself)
  • \(\mathcal{N} = (4,4)\), \(D=2\) (check yourself)

\(N=4\)

  • \(\mathcal{N} = 1\), \(D=4\) (check yourself)
  • \(\mathcal{N} = 2\), \(D=3\) (check yourself)
  • \(\mathcal{N} = (2,2)\), \(D=2\) (check yourself)

\(N=2\)

  • \(\mathcal{N} = 1\), \(D=3\) (check yourself)
  • \(\mathcal{N} = (2,0)\), \(D=3\) (check yourself) This arises from twisting, and we usually say \((0,2)\) to avoid confusion with the \((2,0), D=6\) theory
  • \(\mathcal{N} = (1,1)\), \(D=2\) (check yourself)

Weyl Spinor Representation

This can be understood by reading Clifford Algebra.