representations of the Lorentz group in 4 dimensions - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:54:19Z http://mathoverflow.net/feeds/question/102537 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102537/representations-of-the-lorentz-group-in-4-dimensions representations of the Lorentz group in 4 dimensions elelias2 2012-07-18T13:05:29Z 2012-07-18T15:37:08Z <p>Hi,</p> <p>First of all I should say I am quite uneducated in group theory, so my question can be very naive. Sorry about that.</p> <p>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.</p> <p>In particular, with the definitions:</p> <p>$J_i = \frac{1}{2} \epsilon_{ijk} M^{jk}$ ; $K_i = M^{i0}$</p> <p>where $M$ are the generators of the Lorentz group, one can build:</p> <p>$N_i = \frac{1}{2}(J_i - iK_{i})$</p> <p>$N^{\dagger}_i = \frac{1}{2}(J_i + iK_{i})$</p> <p>so that, in terms of the Ns:</p> <p>$[N_i,N_j] = i\epsilon_{ijk}N_k$</p> <p>$[N_i^{\dagger},N_j^{\dagger}] = i\epsilon_{ijk}N_k^{\dagger}$</p> <p>$[N_i,N_j^{\dagger}] = 0$</p> <p>now there are two su(2) representations that do not mix with each other. </p> <p>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:</p> <p>$J_i = N_i + N_i^{\dagger}$, </p> <p>which seems to imply that both Ns have to be matrices of the same dimension. </p> <p>Thanks a lot.</p>