Automated Proving

Here’s links to Tim Gowers’ posts on his automated theorem prover:

It’s all very long, so you might not want to read all of it, but at least read through the first post, and maybe the first few paragraphs of the second post.

Future talk Schedule

Hey guys,

Sorry this is coming through a bit late, been a crazy week!

The talk schedule which was decided on Wednesday afternoon is as follows:

  1. Reuben, Intro to Lie algebra, nilpotent and solvable (Serre 1, FH 9.1,9.2)
  2. Chris, Semi-simple Lie algebras (Serre 2, FH 9.3,9.4)
  3. Pat, Lie algebras of dim 1,2,3 (FH 10)
  4. Stephen, Rep theory of sl(2,C) (Serre 4, FH 11)
  5. Reuben, Cartan Subalgebras (Serre 3, FH 14 & App. D)
  6. Mitchell, Root Systems I (Serre 5, FH 21)
  7. Jackson, Structure of semi-simple Lie algebras
  8. Alex, Reps of semi-simple Lie algebras

That covers the talks that should take us to the end of the course. Other things that were mentioned on Wednesday were that the midterm would assess questions on reps of finite groups, and then the final project would be about Lie algebras.

(edit 26/3: Swapped Reuben and Alex)

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 dimV \geq 2 any 1-dimensional subspace of V constitutes a strict sub-representation of \rho. This violates irreducibility, so dimV =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:

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!