From Mitchell’s talk on -modules:
- If is a division algebra, then check that every -module is of the form , and as a corollary, all -modules are of the form .
- Prove the following lemma: If is a field, then is a (finitely generated) -module, then is a (finite dimensional) vector space over .
- Is the ring semisimple? (we did this one in class)
- Prove the following theorem of Wedderburn: If is algebraically closed, then any semisimple algebra over deomposes as . As a corollary, this holds for . What does represent? What does this mean in the context of representations?
From my talk on representations of group products, and induced representations:
- Let be representations. Consider the representation defined appropriately. Is it interesting?
- Show that the character of is .
- We have shown that if are irreducible, is irreducible. Show then that in fact every irreducible representation of is of this form.
- Verify and experiment with the examples in my notes (to be posted soon-ish).
From Reuben’s talk on Lie algebras:
- Show there is a bijection between representations of , and left -modules.
- Show that under the cross product can be realised (by an isomorphism) as , the Lie subalgebra of containing matrices satisfying .
- Suppose there exist such that and are groups, and . What can we say about ? Does it necessarily hold that ?
- Find an example of an indecomposable -module which is not irreducible (i.e. contains nontrivial proper submodules).
- Prove the following lemmas: For a Lie algebra , and , ; and for , . (It follows directly that nilpotent Lie algebras are solvable.)
- Show that Lie algebras consisting of upper triangular matrices are solvable, and strictly upper triangular matrices are nilpotent.
- Show that every irreducible representation of a solvable Lie algebra is one-dimensional.
And finally, from Chris’ talk on semisimple Lie algebras:
- For , show that the radical is given by the centre, the subalgebra of nonzero scalar matrices, of the form . Show also that .
- Prove the following direction of the Cartan-Killing criterion: If is a semisimple Lie algebra, then its Killing form is nondegenerate.
- Prove that , , are simple Lie algebras.
Be sure to let me know what mistakes or gaps still remain.
* Not really “complete”.
Here you go.
The notes contain more detail on the standard representation of S3 and inter-relatedness (via characters) that arise between the different representations, their tensor products, direct sums and so on.
Sorry this is coming through a bit late, been a crazy week!
The talk schedule which was decided on Wednesday afternoon is as follows:
- Reuben, Intro to Lie algebra, nilpotent and solvable (Serre 1, FH 9.1,9.2)
- Chris, Semi-simple Lie algebras (Serre 2, FH 9.3,9.4)
- Pat, Lie algebras of dim 1,2,3 (FH 10)
- Stephen, Rep theory of sl(2,C) (Serre 4, FH 11)
- Reuben, Cartan Subalgebras (Serre 3, FH 14 & App. D)
- Mitchell, Root Systems I (Serre 5, FH 21)
- Jackson, Structure of semi-simple Lie algebras
- 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)
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?
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!