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Jun 12, 2010 at 7:27 answer added Victor Protsak timeline score: 4
Jun 12, 2010 at 4:46 comment added S. Carnahan Oops, the continuous parameter comes from the noncompact part of the torus, which doesn't really intersect SU(2) much. Sorry about the mistake.
Jun 11, 2010 at 21:59 vote accept Daniel
Jun 11, 2010 at 20:48 answer added B R timeline score: 11
Jun 11, 2010 at 19:57 comment added Daniel I've just checked now and actually if the principal series of $SL_2(C)$ is labelled by $(\nu,\rho)$ where $\nu$ is half-integer, $\rho$ is continuous, then if it is restricted to $SU(2)$ it will decompose into an infinite sum of representations with $j = \nu, \nu+1, \nu+2, ....$: $(\nu,\rho) = \sum_{j=\nu}^{\infty} j$ So we do not get infinite dimensional unitary irreducible representations of $SU(2)$ this way.
Jun 11, 2010 at 19:26 comment added S. Carnahan You can restrict principal series representations of $SL_2(\mathbb{C})$ to $SU(2)$. They have a continuous parameter, and they are unitary.
Jun 11, 2010 at 19:14 history asked Daniel CC BY-SA 2.5