I liked this TED talk by the Fields Medallist Cedric Villani.

# Author: steveylynch

# Progress on irreps of abelian groups

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?

# Norms come cheap

We had a brief discussion today about when exactly the norm on a vector space is induced by an inner product, that was left unresolved. The following Wikipedia article treats the question well enough, for those who care:

http://en.wikipedia.org/wiki/Parallelogram_law

Tl;dr the answer is almost never – most norms you can think of are not born of an inner product.

This is quite an important fact I think, particularly in infinite dimensions where Hilbert spaces have a richer algebraic structure than their lesser counterparts, Banach spaces. For example this natural isomorphism we discussed, between a vector space and it’s double dual still ‘works’ for a Hilbert space (sort of – we distinguish between the *algebraic *dual and the *continuous* or *topological* dual of an infinite dimensional topological vector space, and I’m talking here about the latter), but one can only say in general that a Banach space is *contained* in it’s double dual.

For those wanting to learn more about homeopathy, the ‘doctor’ has you covered:

Also, would it be possible for someone to upload our speaking schedule for the coming weeks? Thank you!

# An easy exercise

This is a simple question to test your understanding of tensor products, if you wish. If and are vector spaces over some field of dimension and respectively, then has dimension . Vector spaces of equal dimension are automatically isomorphic, so why don’t we just define to be the ‘unique’ vector space of dimension ?

If the contents of my talk made sense to you, the answer should be obvious.

Thanks Jackson, for setting up this blog.