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Hi,

First of all I should say I am quite uneducated in group theory, so my question can be very naive. Sorry about that.

I'm reading Srednicki's "Quantum Field Theory" and I have a bit of trouble understanding how one can label Lorentz representations as two different su(2) algebras.

In particular, with the definitions:

$J_i = \frac{1}{2} \epsilon_{ijk} M^{jk}$ ; $K_i = M^{i0}$

where $M$ are the generators of the Lorentz group, one can build:

$N_i = \frac{1}{2}(J_i - iK_{i})$

$N^{\dagger}_i = \frac{1}{2}(J_i + iK_{i})$

so that, in terms of the Ns:

$[N_i,N_j] = i\epsilon_{ijk}N_k$

$[N_i^{\dagger},N_j^{\dagger}] = i\epsilon_{ijk}N_k^{\dagger}$

$[N_i,N_j^{\dagger}] = 0$

now there are two su(2) representations that do not mix with each other.

However, I cannot see how one can have, simultaneously, different dimensions for the two representations since they are obtained from the very same matrices. For example:

$J_i = N_i + N_i^{\dagger}$,

which seems to imply that both Ns have to be matrices of the same dimension.

Thanks a lot.

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Notation used here is somewhat confusing. You can rather understand it like this : Take two copies of su(2) algebra. Denote their generators as $A_1,A_2,A_3$ and $B_1,B_2,B_3$ respectively, which thus satisfy relations : $[A_i,A_j]=i\epsilon_{ijk} A_k$ (sum over k), and similarly for $B_i$'s. Now let $V$ be a representation space for the first copy of su(2), and $W$ be a representation space for the second copy of su(2). Form the tensor product $V\otimes W$, and define action of lorentz algebra on it as : $J_i=A_i\otimes I+I\otimes B_i$, and $K_i=i(A_i\otimes I-I\otimes B_i)$. –  dushya Jul 18 '12 at 19:47
    
Incidentally, the Lorentz group $\text{Spin}(1,3)$ is not isomorphic to $SU(2) \times SU(2)$ (note that the latter is compact while the former is not), but their complexifications are isomorphic. –  S. Carnahan Jul 19 '12 at 2:34

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