Week 2 Exercises

Some questions we left open last time:

Let $\rho: \mathbb{Z}/n \to \mathbb{C}^*$ be given by $1 \mapsto \zeta$ for some $n$ -th root of unity $\zeta$ . Show that $\rho^*$ is $1 \mapsto \bar{\zeta}$ . Conclude that $\rho, \rho^*$ are not isomorphic.
Let $A$ be an $n \times n$ matrix of finite order ( $A^m = I$ for some $m \in \mathbb{Z}$ ). Show that $tr(A^{-1}) = \overline{tr(A)}$ , without using eigenvalues.
In the proof of Prop. 1 of Serre (see Section 2.1), he states that “… $\rho_s$ can be defined to be a unitary matrix”. Show that every representation of a finite group is unitary. (As part of this exercise, give an appropriate definition of a unitary representation.)
Prove that the character of $V_1 \otimes V_2$ is given by $\chi_1 \cdot \chi_2$ . If $\rho$ is a representation, prove that the character of a dual representation is $\chi_{\rho^*} = \overline{\chi_\rho}$ . Find the characters of $\textrm{Sym}^2(\rho)$ and $\rho \wedge \rho = \textrm{Alt}^2(\rho)$ .
If $f: (\rho_1, V_1) \to (\rho_2, V_2)$ is a map of representations, show that the kernel and image of $f$ are subrepresentations of $(\rho_1, V_1), (\rho_2, V_2)$ respectively.
Midterm question: Give your own proof of Serre Thm. 3 (section 2.3), that characters of irreducible representations form an orthonormal system.
Prove Prop. 6 of Serre (section 2.5): If $(\rho, V)$ is an irreducible representation on $G$ of degree $n$ and character $\chi$ , and $f$ is a class function on $G$ , then $\rho_f := \sum_{t \in G} f(t)\rho(t)$ is a homothety of ratio $\frac{1}{n} \sum_{t \in G} f(t) \chi(t)$ .
Prove that the number of degree 1 representations on a finite group $G$ is $|G^{ab}|$ .

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:

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)

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!

Link to my talk

If anyone would like to download the pdf of my written materials, the dropbox link

Click to access BasicDefinitionsTalk.pdf

Should work for that. I’ve rewritten it slightly from what I gave, namely removed the discussion of the proof of Theorem 1, given the mid term question. I’ve also fleshed out some of the material a bit more, but its still a bit bare.

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.

Exercises from Week 1

Here’s a list of some questions which had been left unanswered from the week 1 meeting:

Consider representations on vector spaces where the base field is not $\mathbb{C}$ (e.g. vector spaces over $\mathbb{F}_2$ ). What can go wrong?
If you choose two different characters (degree 1 representations) of a group, why aren’t they isomorphic?
Let $G$ be a finite group. How many different degree 1 representations does $G$ have?
Consider a group $G$ which acts on a set $X$ , and show that this induces a corresponding action on $Fun(X)$ (the set of all functions mapping from $X$ into some other set, say $\mathbb{C}$ ). Show that the regular representation of $G$ is a particular case of this.
Theorem 1 of Serre states: Let $\rho: G\to GL(V)$ be a representation, and let $W$ be a subspace of $V$ which is stable under $\rho$ ; then there is a complement space $W^c$ which is also stable.
Write a proof of this theorem in your own words. Highlight where we use the following facts: (1) $G$ is finite; (2) $V$ is finite-dimensional; (3) $V$ is over the field $\mathbb{C}$ .
For each of the three conditions, find a counter-example to show that the theorem doesn’t hold if the condition is dropped.
(This will also be a midterm exam question.)
If $G$ is a group, why is $F(G)$ (the free vector space on $G$ ) the same as the regular representation of $G$ ?
For vector spaces $U,V,W,X$ , does it hold that $\textrm{Hom}(U,W)\otimes\textrm{Hom}(V,X) \simeq \textrm{Hom}(U\otimes V,W\otimes X)$ ?
If $U$ and $V$ are finite dimensional vector spaces, prove that $(U \otimes V)^* = U^* \otimes V^*$ . Show that this does not hold in general when $U, V$ are infinite-dimensional.
For vector spaces $U, V, W$ , prove that $\textrm{Hom}(U\otimes V,W)=\textrm{Hom}(U,\textrm{Hom}(V,W))$ .
Let $V$ be a finite dimensional vector space, and $V^*$ denote its dual and $V^{**}=(V^*)^*$ denote its double dual. One knows that $V$ is isomorphic to $V^*$ and to $V^{**}$ . However, the isomorphism $V\simeq V^{**}$ is much better than the isomorphism $V\simeq V^*$ !! Here is one reason:

Suppose $G$ acts on $V$ ; i.e., we have a representation $\rho:G\to GL(V)$ . Show that $G$ acts on $V^*$ and $V^{**}$ . Let us denote these representations by $\rho^*$ and $\rho^{**}$ . Show that the representations $(\rho, V)$ and $(\rho^{**}, V^{**})$ are always isomorphic, where as, it could happen that $(\rho,V)$ and $(\rho, V^*)$ are not isomorphic!!
One says that the isomorphism $V \simeq V^{**}$ is canonical or functorial, where as the isomorphism $V\simeq V^*$ is not (it depends on a choice of basis).