Group Theory query based on Green Schwarz Witten volume 2

1. Nov 26, 2014

maverick280857

Hi,

In chapter 12 of GSW volume 2, the authors remark, "spinors form a representation of SO(n) that does not arise from a representation of GL(2,R)."

What do they mean by this?

More generally, since SO(n) is a subgroup of GL(2,R) won't every representation of GL(2,R) be a representation of SO(n) as well?

I know the Dirac matrices of the spinor representation of SO(n) will have different matrix dimension depending on whether n is even or odd. Is that related?

Thanks!

2. Nov 27, 2014

Eredir

Check out this answer from MathOverflow:
http://mathoverflow.net/questions/121620/why-does-gln-have-no-spinor-representations

3. Dec 9, 2014

lpetrich

Here's what I think is going on.

GL(2,R) ~ GL(1,R+) * SL(2,R)
SL(2,R) ~ SO(2,1)
SO(2,1) is an analytic continuation of SO(3)
For SO(3),
U(2) ~ U(1) * SU(2)
SU(2) ~ SO(3)

Only SO(2) is. SO(n) is a subgroup of GL(n,R) ~ GL(1,R+) * SL(n,R), and of SL(n,R) also. Reps of GL(n,R) can indeed be decomposed into reps of SO(n), but an irreducible rep of GL(n,R) is not in general irreducible in SO(n).

That's a separate issue. The Dirac matrices are matrices in a "Clifford algebra", and one indeed uses Clifford algebras to get spinor-rep generators. For algebra elements Xi:
Xi.Xj + Xj.Xi = 2 gij

for algebra metric g. One finds algebra generator Lij from a multiple of the commutator of Xi and Xj.

For SO(2n) and SO(2n+1), one constructs 2n+1 Clifford-algebra matrices using outer products of n Pauli matrices, giving their spinor reps dimension 2n. One needs all 2n+1 for the generators of the SO(2n+1) spinor rep, and it is thus irreducible. But one needs only 2n of them for the generators of the SO(2n) spinor rep, with the remaining one splitting that rep into two equal irreducible halves, both with dimension 2n-1.