# Representation theory course

Well, like David, I am teaching a course this semester and writing up notes.

My course is on representation theory. More specifically, I hope to cover the basics of the representation theory of complex reductive groups, including the Borel-Weil theorem. In my class, I have started from the theory of compact groups, for two reasons. First, that is the way, I learned the subject from my advisor Allen during a couple of great courses. Second, I am following up on a course last semester taught by Eckhard Meinrenken on compact groups.

Feel free to take a look at the notes on the course webpage and give me any feedback.

Very soon, I will reach the difficult task of explaining complexification of compact groups. As I complained about in my previous post, I don’t feel that this topic is covered properly in any source, so I am bit struggling with it. Anyway, the answers to that post did help me out, so we will see what happens.

## 9 thoughts on “Representation theory course”

1. I did the compact groups approach to representations of SU(2) at the end of my representations of finite groups class last semester (following Teleman’s notes), and it’s really quite nice. I’d heard the propaganda from all of you who had Allen’s course that the groups approach was great, and I kept meaning to learn it, but somehow kept putting it off until last semester.

2. David Speyer says:

Looks good! Out of curiosity, what proof are you giving for Theorem 4.1?

3. Joel Kamnitzer says:

David – Theorem 4.1 states that every group element is conjugate into the maximal torus. I actually didn’t give a proof for this theorem because it was covered in Eckhard’s course last semester. Allen’s proof was to consider the flag variety. Every element of G defines a homeomorphism of the flag variety which has a fixed point if it is conjugate into the maximal torus. To see that there must be a fixed point, you can use that the cohomology is even dimensional, so the Lefschetz theorem shows that there must be a fixed point.

4. Joel Kamnitzer says:

Noah – I feel that there are two advantages to the compact groups method. First, you immediately reach the complete reducibility theorem. Second, you see that the restriction to the maximal torus determines the representation because it determines the character. On a related note, you also see that the generating function for the dimensions of the weight spaces is the character (defined by trace).

5. David Speyer says:

That’s the proof I know. I was just wondering if there is a clever way to do this without actually developing the geometry of flag manifolds. (Looking for clever tricks in anticipation that I will wind up teaching a similar course myself one day.)

6. I’ll note that there is a slicker way of proving this than Joel’s method (though it is fundamentally the same); every element of a compact group lies in a torus of rank >0 (just take the closure of a 1 parameter subgroup going through it). For any torus action on any non-terrible topological space, the Euler characteristic of the space and its fixed point set coincide (look at the complement, and pull back a triangulation on the quotient). Since there is a torus action on the flag manifold with finitely many fixed points, there can be no free torus action on it, and we are done.

7. I should have taken a course like this (long ago). Instead I picked up bits and pieces of the subject on my own. For me the transition from compact Lie groups to complex semisimple Lie algebras is conceptually the hardest step. Maybe it’s due to my one-sided algebraic upbringing and general discomfort around analytic methods.

8. I just read through chapter 5, and I think you might be a bit confused in section 5.1.3.

Given $X$ a subset of $\mathbb{R}^n$, the ideal $I(X) \otimes \mathbb{C}$ is always radical. Proof: Consider $a+bi$, with $a$ and $b \in \mathbb{R}[x_1, \ldots, x_n]$. Suppose that $(a+bi)^k$ vanishes on $X$. Then $(a+bi)^k (a-bi)^k = (a^2+b^2)^k$ vanishes on $X$. So $a$ and $b$ vanish on $X$ and we have $a$ and $b \in I(X)$. So $a+bi \in I(X) \otimes \mathbb{C}$. This seems like it is what you need for your current purposes.

Now, there is a harder statement, which is about arbitrary ideals in $\mathbb{R}[x_1, \ldots, x_n]$: If $L/K$ is a separable extension of fields, and $I$ is an ideal in $K[x_1, \ldots, x_n]$, then $I \otimes L$ is radical if and only if $I$ is. I couldn’t find any way to simplify this proof in the case of $\mathbb{C}/\mathbb{R}$.

Also, you seem to be suggesting that, if $I$ is an ideal in $\mathbb{R}[x]$, and $\mathbb{R}[x]/I$ is a domain, then $\mathbb{C}[x]/(I \otimes C)$. This is NOT true. In one variable, let $I = (x^2+1)$.

9. Joel Kamnitzer says:

Thanks David, again!