Three geometric constructions of the irreducible representations of GL

Three geometric constructions of the irreducible representations of GL_n July 3, 2009

Posted by Joel Kamnitzer in Uncategorized.
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The past few weeks there has been a summer school and conference on geometric representation theory and extended affine Lie algebras at University of Ottawa. As part of this event, I gave a week long lecture series entitled “three geometric constructions of the irreducible representations of $GL_n$ “. Specifically I discussed the Borel-Weil theorem, Ginzburg’s construction using Springer fibres, and the geometric Satake correspondence. I focused on $GL_n$ to keep the root system combinatorics and the geometry as elementary as possible.

The typed lecture notes from my talk are now available. If you do read them, please let me know if you have any comments/corrections. (You can also find videos of the talks.)

The other lectures at the summer school were given by Neher, Kang, Wang, Savage, and Chari. I recommend reading their notes/watching their videos if you want to learn more about geometric representation theory, crystals, and affine Lie algebras.

Comments»

1. Anonymous - July 3, 2009: Hi Joel,

This is great! I’ve been wanting to learn about this stuff for a while now. Some comments on the first few pages:

1) Typo on page 2, line 10: “without any using”.

2) Typo on page 2, line -14: “This is equivalent to consider”.

3) Possible typo on page 3, lines 2 and 3: should there be parentheses around the bases?

4) Is it actually true that every group homomorphism $\mathbb{C}^\times \to \mathbb{C}^\times$ is of the form $z \mapsto z^n$ ? Or should we have an extra condition on the homomorphism (such as continuity)?

5) Typo on page 3, line 13: towards the end, you have ^x instead of ^\times.

6) You say that Proposition 2.4 tells us that an arbitrary representation (of $\mathbb{C}^\times$ ) can be written as a direct sum of irreducible representations. Does this mean that the V(n)’s are irreducible? If so, why is this the case?

7) In example 2.10, it would probably be less ambiguous if you replaced “these [...] representations” by “alt(2,C^n) and sym(2, C^n) [...] representations” (I’m too lazy to TeX!).

8) On page 6, paragraphs 2 and 4 are identical.
2. Steven Sam - July 4, 2009: Anonymous: regarding 4), it’s said at the beginning that all representations are assumed to be algebraic (or holomorphic), and in this case, all algebraic homomorphisms $\mathbb{C}^\times \rightarrow \mathbb{C}^\times$ are of the form $z \mapsto z^n$ for some integer n.

Otherwise, you could take the map $z \mapsto |z|$ , and this is certainly a (continuous) group homomorphism, but it’s not algebraic.

Joel: Thanks for these notes!
3. David Speyer - July 6, 2009: OK, I’m trying to work through the Ginzburg construction of the adjoint rep of $\mathfrak{gl}_3$ . I keep winding up with a $\mathfrak{gl}_3$ -crystal rather than a $\mathfrak{gl}_3$ -representation. Can someone explain what I’m doing wrong?

There are seven weights in the adjoint rep. For the six extreme weights, the Springer variety is a point. For the weight in the center, the Springer variety is two $\mathbb{P}^1$ ’s, glued at a single node. I’ll call this variety $C$ . So the dimensions of the weight spaces work.

Now, I try to get the maps. Morally, in each case, we have a subvariety $Z$ of $C \times (\mathrm{pt})$ . This subvariety is isomorphic to $\mathbb{P}^1$ and lies above one of the two components of $C$ ; We are supposed to push-pull through $Z$ . Because you have only defined pull-back in smooth fiber bundles, we are in fact supposed to pull-back to some smooth ambient space, intersect with $Z$ and then push.

I don’t want to write out my attempts here, as they are surely flawed. I keep getting that, for any map to (resp. from) the center weight space, the image (resp. kernel) is spanned by the fundamental class of one or the other component of $C$ . Could someone write out a computation of a case where that doesn’t happen, so I can see where I’m going wrong?

Thanks in advance,
David
4. David Speyer - July 6, 2009: Typo: in the sentence beginning “Now define a map” on page 13, the arrow is backwards.
5. Joel Kamnitzer - July 6, 2009: David-
Thanks for your question. I think that it works like this:

The map going _out_ of the central weight space is the “non-crystal” one. I mean it is non-zero on each of the two fundamental classes. Let me call the two components $A_1, A_2$ . We need to look at the intersection of $Z$ with $\pi_1^{-1}(A_1)$ and $\pi_1^{-1}(A_2)$ . The intersection with one of them will be transverse — they meet in a point. The intersection with the other is not transverse since they meet in a $\mathbb{P}^1$ . So we get a contribution of the Euler characteristic of the normal bundle which turns out to be 2. Hence we get $E([A_1]) = [A_1]$ and $E([A_2]) = 2 [A_2]$ .

Note: there is a typo in the definition of the Ginzburg action. In section 4.3, $\pi_1, \pi_2$ should actually be maps from the product of the cotangent bundles.
6. David Speyer - July 6, 2009: Thanks Joel! On the T home, I figured out how this works for the adjoint rep of GL_2 (that is, why the multipicities aren’t all 1), so I guessed it had to be similar for GL_3. Maybe I’ll write up the GL_2 example for anyone else who is interested.

I still suspect there is a way to make the excess intersection computation much easier than I am finding it. Here is how I learned to use the excess intersection formula: Suppose that A and B, smooth varieties of dimensions a and b, intersect inside a smooth variety X of dimension a+b, and their intersection, D, has dimension d. Then we have natural maps of normal bundles:

$N_{X/D} \longrightarrow N_{A/D} \oplus N_{B/D}$

Consider the kernel; it is a d-dimensional vector bundle on Z. Its top Chern class is the multiplicity with which A meets B.

In our example, the ambient X is 10-dimensional (namely $(T^* \mathbb{P}^2) \times T^* (\mathcal{F}_3)$ ), the two subvarieties are both 5-dimensional, and $D$ is a $\mathbb{P}^1$ . I’d be a bit intimidated by that normal bundle computation. You clearly aren’t, which makes me suspect that you know something that I don’t.

I notice that, in the few cases I’ve tried, the multiplicity of the intersection seems to be the same as the Euler characteristic of $D$ . That makes me think there is some symplectic magic going on here.

For example, in the GL_2 case, I needed to compute the self intersection of the exceptional locus in the resolution of an $A_1$ singularity. That self intersection is $-2$ , the same (up to sign) as the Euler characteristic of $\mathbb{P}^1$ . If I had needed to compute the self intersection of the exceptional fiber in the blow up of a smooth point, instead, I would have gotten $-1$ . I’m sure this has something to do with the fact that the former is a symplectic resolution and the latter is not. Any clues?
7. David Speyer - July 6, 2009: The $GL_2$ computation: We’ll compute the representation of $GL_2$ with highest weight $(2,0)$ . So we are dealing with $2$ -step flags in $2$ -space.

The nilpotent cone, $\mathcal{N}$ , consists of all matrices of the form $\left( \begin{smallmatrix} a & b \\ c & -a \end{smallmatrix} \right)$ with $a^2+bc=0$ . This is a quadratic singularity in $3$ -space; it looks like the cone on a plane conic.

The Springer variety has three components. They are the cotangent bundles to the spaces of flags of type $(0,0,2)$ , of type $(0,1,2)$ and of type $(0,2,2)$ . Call them $S_0$ , $S_1$ and $S_2$ . The spaces $lates S_0$ and $S_2$ are just points, and the maps to $\mathcal{N}$ send them to the origin. The space $S_1$ is a line bundle over $\mathbb{P}^1$ , specifically the cotangent bundle, and can be thought of as blowing up the origin of $\mathcal{N}$ . We’ll write $E$ for the fiber of $S_1$ over the origin.

The representation we are interested in corresponds to the fiber over the origin in $\mathcal{N}$ . So the Springer fibers are $S_0$ , $E$ and $S_2$ . By definition, $Z$ is contained in $(S_0 \sqcup S_1 \sqcup S_2)^2$ ; in fact, it lives in $S_0 \times S_1$ and $S_1 \times S_2$ . More specifically, $Z = S_0 \times E \ \sqcup E \ \times S_2$ .

Let’s start by computing the maps from the outside to the center. Take the fundamental class of $S_0$ . Pull it back to $S_0 \times S_1$ to get all of $S_0 \times S_1$ . Intersect with $S_0 \times E$ to get $S_0 \times E$ . Pushforward to $S_1$ to get $E$ . The only thing that is hard is not getting frustrated by the trivialities. We conclude that $[S_0]$ and $[S_2]$ are taken to $[E]$ .

Now, let’s go the other way. Start with the fundmental class of $E$ . Pull back to $S_1 \times S_2$ to get $E \times S_2$ . Intersect with $E \times S_2$ — that intersection is not transverse. We get a contribution of the degree of the normal bundle to $E$ , namely $2$ . (Hmmm. Actually, it’s $-2$ . Any idea why the final answer has a positive sign?) Pushing forward doesn’t do anything funny, so our final answer is that $[E]$ is taken to $2*[S_2]$ .

Note that, even though we only cared about the fiber over zero, we still had to know the whole family. I suspect that, had I been smarter, and used some symplectic magic, I could avoid that.
8. David Speyer - July 6, 2009: By the way, thanks for the great notes!
9. Joel Kamnitzer - July 6, 2009: Here is the symplectic trick that you ask for.

Suppose you have two Lagrangians subvarieties $A, B$ of a symplectic variety $X$ . Then the excess intersection bundle of the intersection of A and B is the cotangent bundle of $A \cap B$ .

To see this, let us take everything to be linear. Then $D$ is isotropic, so the symplectic form gives us a surjection $X/D \rightarrow D^\star$ . The kernel of this map is $D^\perp/D$ . However, $D^\perp = A + B$ and so we see that the cokernel of $N_{A/D} \oplus N_{B/D} \rightarrow N_{X/D}$ is $D^\star$ .
10. On the geometry of street signs « Concrete Nonsense - July 13, 2009: [...] On the geometry of street signs I spent three weeks of June and July at a summer school and conference on geometric representation theory at the University of Ottawa. This conference is already mentioned on the mathematical blogosphere (I refuse to talk about the “blathosphere” or the “blathyscape”), at the Secret Blogging Seminar. [...]

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