I can’t sleep and I’ve had an idea, so here’s another post. We’ve discussed a few times the possibility of proving that every irreducible representation of an abelian group is 1-dimensional, by elementary means.

Alex proved this statement in his talk today (as a corollary of some wider considerations), using the character machinery that he and Pat outlined, however I think I’ve got a proof that circumvents the need for this. I claim very little originality – the key idea is stolen right from the proof of ‘Schur’s lemma’ in the form that it is presented in MATH3103. Here goes:

Suppose is an abelian group, is a finite dimensional vector space over and is an irreducible representation. Fix in and let be an eigenvalue of whose corresponding eigenspace is . Then if is an element of , for any in we have: , so is also an element of . Therefore is closed under the action of and forms a sub-representation of . By irreducibility it must hold that , so , where is the identity on . Since was arbitrary we may conclude that sends every group element to some scalar multiple of the identity and it’s clear then that if dim any 1-dimensional subspace of constitutes a strict sub-representation of . This violates irreducibility, so dim.

The other nice thing about this argument is that it seems to work for infinite . Thoughts?

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Is this fact true if one replaces the complex numbers by, say, real numbers? That is, should we necessarily have that the underlying field is algebraically closed?