Christian's course on Lie algebras advanced and so did the problem sheets. Here is another exercise that nearly by accident ended up in the problem set: You are supposed to show that all unitary irreducible representations of SU(2) are finite dimensional.

OK, we physicists know how to do this: In terms of a Cartan basis (H,E,F) we take an eigenvector of H and than act on it a couple of times with E to show that if this does not terminate we end up with negative norm vectors. Easy.

But this was too easy. If we have to consider infinite dimensional representations one should be careful with functional analysis. It could be that H is only essentially self-adjoint (like the momentum operator) and thus does not have eigenvectors because it only has a continious spectrum. Can you repair the above prove in this case or show that it cannot happen or find a counter example?

## Thursday, November 18, 2004

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## 3 comments:

Contrary to the brainwashing we've all undergone, not all irreducible representations of the group SU(2) are finite-dimensional.

Counterexample: Let V be the space of all functions on SU(2) that vanish at all but finitely many points. SU(2) acts on V by left translation.

(Take real or complex-valued functions, depending on whether you want a real or complex representation.)

V is not irreducible: for example, it has a subrepresentation consisting of functions f such that when you sum f(x) over all x in SU(2), you get zero.

However, I think it's pretty easy to see that V has no finite-dimensional subrepresentations.

That's not a unitary representation, though.

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