# Why does in string theory the amount of supersymmetry have to be $\cal{N} \leq 2$?

Why is this, that in string theory the maximum amount of supersymmetry is $\cal{N} = 2$, whereas in supergravity one can have up to $\cal{N} = 8$ ?

-
I think its important to know which dimension you are in, knowing only say $\mathcal N=2$ is not enough to determine how much supersymmetry there is and therefore hard to compare to $\mathcal N=8$. –  Heidar Apr 16 '13 at 22:39
My point was that those two statements are not necessarily inconsistent with each other, one has to specify the space-time dimension. The fact that $\mathcal N\leq 2$ (understood appropriately) comes from analyzing the spectrum of superstring theory (see here for the results en.wikipedia.org/wiki/…). However, this is a statement in 10 dimensions. If $\mathcal N=1$ that means you have a spinor $Q_\alpha$, but the number of components $\alpha$ is given by representation theory of Clifford algebra. In higher dimensions, spinors have (cont.) –  Heidar Apr 17 '13 at 0:22
more components. As you see in the above link, type I string theory has 16 supercharges ($\mathcal N=1$) and type II has 32 ($\mathcal N=2$), this is related to spinors in 10 dimensions (which I think have 16 components/spinor dimension). In lower dimensions you can still have, say, 32 supercharges but you can't do it with only one spinor since the spinors have fewer components there, thus you must have $\mathcal N\geq 2$. When people talk about $\mathcal N=8$ SUGRA, it's in $D=4$ I think. In four dimensions Dirac spinors have 4 components, so $8\times 4 = 32$ super charges. (cont.) –  Heidar Apr 17 '13 at 0:24
The same as $\mathcal N=2$ in D=10. All I'm saying here is correct morally, but there might be some details which aren't completely correct. However, more details can be found in most SUSY notes. –  Heidar Apr 17 '13 at 0:27
Correction: "can't do it with only one spinor" should say "can't do it with only two spinors". –  Heidar Apr 17 '13 at 0:32