Video

Something Interesting

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

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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 $G$ is an abelian group, $V$ is a finite dimensional vector space over $\mathbb{C}$ and $\rho: G \to GL(V)$ is an irreducible representation. Fix $g$ in $G$ and let $\lambda$ be an eigenvalue of $\rho(g)$ whose corresponding eigenspace is $E_\lambda \subset V$. Then if $v$ is an element of $E_\lambda$, for any $h$ in $G$ we have: $\rho(g)\rho(h)v = \rho(gh)v = \rho(hg)v = \rho(h)\rho(g)v = \lambda \rho(h)v$, so $\rho(h)v$ is also an element of $E_\lambda$. Therefore $E_\lambda$ is closed under the action of $G$ and forms a sub-representation of $\rho$. By irreducibility it must hold that $E_\lambda = V$, so $\rho(g) = \lambda I$, where $I$ is the identity on $V$. Since $g$ was arbitrary we may conclude that $\rho$ sends every group element to some scalar multiple of the identity and it’s clear then that if dim$V \geq 2$ any 1-dimensional subspace of $V$ constitutes a strict sub-representation of $\rho$. This violates irreducibility, so dim$V =1$.

The other nice thing about this argument is that it seems to work for infinite $G$. 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 $U$ and $V$ are vector spaces over some field of dimension $n$ and $m$ respectively, then $U \otimes V$ has dimension $nm$. Vector spaces of equal dimension are automatically isomorphic, so why don’t we just define $U \otimes V$ to be the ‘unique’ vector space of dimension $nm$?

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

Thanks Jackson, for setting up this blog.