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Natural Science Forum / Physics / Research / July 2008Supersymmetry generator(s) ?
Hi, I'm confused about the following three questinos. Suppose we have local N=(1,0) supersymmetry in 10-dimensional minkowski spacetime. In 10-dimensions we have two real irreducible unique 16-dimensional majorana-weyl representations of Spin(9,1) denoted by S^+ and S^- respectively. For N=(1,0) local supersymmetry, the super poincare algebra is now given by (V \oplus o(V)) \oplus \Pi (S^+)*. Then by choosing a basis {P_\mu} for V and {Q_a} for S^* the odd part of the supersymmetry algebra is given by {Q_a,Q_a} = -2 \Gamma^\mu_{ab} P_\mu. Now it is clear that the supersymmetry generators are given by the 16 basis vectors Q_a. So the a-index denotes different vectors. In many physics books however they write down the same supersymmetry algebra bracket and claim that the Q_a is just a single 16-dimensional spinor, where a denotes the component- index. So my question is, what is going on ? Then, another question that is bugging me is this. This Q_a is a lie algebra basis vector of the super Poincare group, which is the isometry group of the super Minkwowski spacetime. The problem is that I have read at some places that the Q_a is a section of a spinor bundle over Minkowski spacetime, but I'm not entirely sure that I understand how a basis vector of a lie algebra of the Poincare group can be interpreted as a section of a spinor bundle over Minkowski spacetime. Could anyone perhaps clarify this for me? I hope anyone can help me out on the above questions, Thanks in advance! Ygor On Jun 21, 4:59 am, ygor.geu...@gmail.com wrote: > The problem is that > I have read at some places that the Q_a is a section of a spinor > bundle over Minkowski spacetime, but I'm not entirely sure that I > understand how a basis vector of a lie algebra of the Poincare group > can be interpreted as a section of a spinor bundle over Minkowski > spacetime. The momentum generator P_{mu} is directly related to the partial differential operator, which I'll denote here @_{mu}. This factors. For Minkowski space, take l = (@_t + @_z)/root(2), m = (@_x + i @_y)/ root(2), m* = (@_x - i @_y)/root(2), n = (@_t - @_z)/root(2), in units where c = 1. Factor the basis 4 = (l,m,m*,n) into 4 = 2_R x 2_L where 2_R: (o, i) and 2_L: (o*, i*) are the right-helicity and left-helicity spinor bases. The factoring is as l = o x o*, m = o x i*, m* = i x o*, n = i x i*, where x denotes tensor product. The elements of the respective bases are "dyads". Impose the convention l = o* x o, m = i* x o, m* = o* x i, n = i* x i (this is essentially equivalent to requiring that the "sigma" coefficients form Hermitian matrices). The dyads (i,o) = (Q_0, Q_1) and (i*, o*) = (Q_0', Q_1') are, theselves, the generators. They're spinor fields, so they're sections over the left and right helicity spinor bundles. They're generators, with the relation P_l = l = 1/2 (o x o* + o* x o) = 1/2 (Q_0 Q_0* + Q_0* Q_0) = 1/2 {Q_0, Q_0*}. By convention, a different scaling is adopted for the Q's so that the 1/2 factor is normalized to 1. The spinors, themselves, in each subspace 2_R and 2_L are treated as Grassmann numbers, so one also has {Q_a, Q_b} = 0 = {Q_a', Q_b'}. The coefficients with respect to the bases 4, 2_R and 2_L are sigma^l_{00'} = sigma^m_{01'} = sigma^m*_{10'} = sigma^n_{11'} = 1, all other sigma^{mu}_{AB'} being 0. I don't think you'll see anywhere in the literature the explicit connection of the spinor dyads and Q operators. But it's implied and is probably what is indirectly being referred to by the passage you're asking the question of. |
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