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I want to count the number of (isomoprhism clases) of one-dimensional representations of the affine Hecke algebra for $G = \text{SL}_2$. I'm doing it in two ways:
(1) by explicitly looking at generators and relations, and
(2) by looking at the cohomology of flags fixed by two $q$-commuting elements.
The problem is by (1) I count four, and by (2) I only count two. I'm not entirely sure where I am going wrong.

Method 1: The affine Hecke algebra for $\text{SL}_2$ is an algebra $H$ which is a free module over $k[q, q^{-1}]$ with basis $\{e^n, e^n T\}$ with multiplicative relations $$(T + 1)(T - q) = 0\,,$$ $$Te^{-1} - e^1 T = (1-q)e^1\,,$$ $$e^{-1} e^1 = 1\,.$$ Then, the one-dimensional dimensional irreducibles are as follows. For one-dimensional representations, one has $$T \mapsto -1\,,$$ $$e^1 \mapsto \pm q^{-1/2}\,,$$ and $$T \mapsto q\,,$$ $$e^1 \mapsto \pm q^{1/2}\,.$$ i.e. there are four $1$-dim representations.

Method 2: We should be able to read off the irreducibles by Kazhdan-Lusztig theory. Namely, for generic $q$, for $(g, x) \in G \times \mathcal{N}$ and $g$ semisimple, such that $gxg^{-1} = qx$, the cohomology of "Springer" fiber $\mathcal{B}_{g, x}$ (i.e. fixed by both $g, x$) is a big direct sum whose summands are the tensor product of an irreducible $H$-module with an irreducible representation of the component group of the double centralizer $C(g, x)/C(g, x)^\circ$. Further, any given irreducible can only occur once in this list.

So, I have for conjugacy classes of $(g, x)$:

(1) $x = 0$, and $g$ is parameterized by $\mathbb{A}^1 = T//W$. When $g$ is regular semisimple, the fiber is two points, and when it is the identity, the fiber is $\mathbb{P}^1$. In either case the total cohomology is two dimensional. Now, the centralizer when $g$ is regular is the torus $T$, which is connected, so the $H$-representation is really two dimensional (and not, say, the direct sum of two one-dimensional representations tensored with representations of the component group). When $g = 1$, then the centralizer is $G$, which is still connected.

(2) $x = \left(\begin{array}{cc}0&1\\0&0\end{array}\right)$, and $g = \pm \left(\begin{array}{cc} \sqrt{q}&0\\0&1/\sqrt{q}\end{array}\right)$. The fiber here is a point. So we have a one-dimensional cohomology. The centralizer is two points, so not connected, but twisting with an irreducible representation of $\mathbb{Z}/2$ won't change our dimension count in any case.

So here, we only find two $1$-dim representations. Where is the discrepancy?

Batominovski
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user148177
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2 Answers2

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The discrepancy comes from the fact that there are two different affine Hecke algebras, one for the weight lattice and one for the root lattice. The root lattice version is a subalgebra of the weight lattice version. Concretely, the root lattice version is the one appearing in your method 1. It may be realised as the quotient of the group algebra of the free group on two generators $T_0$ and $T_1$ modulo the quadratic relations $$(T_i+1)(T_i-q)=0 \ \text{for} \ i=0,1,$$ explaining the appearance of the four one-dimensional representations: each generator has two possible eigenvalues, which may be chosen independently. The weight lattice version is obtained from this by adjoining the outer automorphism interchanging the two generators. It has only two $1$-dimensional representations because $T_0$ and $T_1$ must now have the same eigenvalues.

Geometrically, the difference is that one of these comes from $\text{SL}_2$ and one from $\text{PSL}_2$.

Batominovski
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Stephen
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  • Thanks for the answer. I'm a bit confused though; I think the root lattice and weight lattice Hecke algebras should specialize to the same thing when $q = 1$, because the affine Weyl group we get by taking the semidirect product, and there's only one nontrivial way for $\mathbb{Z}/2$ to act on a lattice $\mathbb{Z}$. I'm probably not understanding correctly, but I think the above doesn't satisfy this? – user148177 Jul 26 '15 at 03:40
  • I'm also not sure why the affine Weyl group is the group algebra on $T_0, T_1$ modulo that relation -- the group algebra of the affine Weyl algebra should be a subalgebra, so there should be some kind of relation which captures the action of $W$ on the lattice? (Here I'm thinking of $T_i$ as corresponding to $w \in W$) – user148177 Jul 26 '15 at 03:40
  • If it helps, I'm using http://arxiv.org/pdf/math/9802004v3.pdf as a reference, pp39, definition 2. The relation I get is by taking $\lambda = -1$. – user148177 Jul 26 '15 at 03:59
  • @user148177 "I think the root lattice and weight lattice Hecke algebras should specialize to the same thing when q=1". No, the root lattice version specialises to the group algebra of $Q \rtimes W$ and the weight lattice version to the group algebra of $P \rtimes W$, where $Q$ is the root lattice and $P$ is the weight lattice of the finite Weyl group $W$. – Stephen Jul 26 '15 at 21:12
  • But for $SL_2$, aren't those two groups isomorphic? They aren't for $SL_3$ and I imagine others. For example, if we identify $Q$ with $2\mathbb{Z}$ and $P$ with $\mathbb{Z}$, with $W$ the obvious action, the semidirect product seems to give isomorphic groups. Namely, it's the group freely generated by $T, e^n$ for $n \mathbb{Z}$ with the relations $e^{n} e^{m} = e^{n+m}$ and $T e^n = e^{-n} T$. If we use the root lattice we use only even $n$ and if we take the weight lattice we use all $n$. But there's an isomorphism of these groups sending $e^n \mapsto e^{2n}$. – user148177 Jul 26 '15 at 21:21
  • That aside, I'm not sure why we are taking two free generators $T_0, T_1$. In my understanding, the generators of the affine Hecke algebra are $T_w$ for $w \in W$ and $e^\lambda$ for $\lambda$ in either the root or weight lattice. Of course, the relations end up implying $T_e = 1$ (where $e$ is the group identity $e \in W$). – user148177 Jul 26 '15 at 21:25
  • @user148177 When I have time, I will edit my answer to describe $T_0$ an $T_1$ in terms of your presentation. You might have a look at the last chapter of Macdonald's book "Affine Hecke algebras and orthogonal polynomials" in the meantime. – Stephen Jul 26 '15 at 22:04
  • Thanks for the reference. I'll read it and try to understand the above presentation. – user148177 Jul 26 '15 at 22:10
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I think I know what's wrong. It turns out I was confused about the statement on the geometric side. In the Ginzburg/Chriss text, the irreducible representations are not exactly the Borel-Moore homology of the fibers but the image of Borel-Moore homology of some tubular neighborhood. If one takes such a neighborhood the two-dimensional representation I referenced above should be the Borel-Moore homology of a point union a small disk in $\mathbb{A}^1$. However, Borel-Moore homology of open disks vanishes, so the representation I claimed was two-dimensional is really one-dimensional.

Perhaps more convincingly, one can refer to paper as a reference. One doesn't just take the (co)homology of the fibers for each orbit representative. The statement (page 44) is really that the irreducibles (for a given central character) are the summands we get by applying the decomposition to the map $\mu$: $$\mu_* \mathbb{C}_{\tilde{\mathcal{N}}^a} = \bigoplus_{\phi} L_\phi(k) \otimes IC(k)$$

In our case, the map is $\mu: \mathbb{A}^1 \cup \text{pt} \rightarrow \mathbb{A}^1$. The cohomology upstairs is $\mathbb{C}^2$, and decomposes into two one-dimensional representations. So for each central character has two one-dimensional representations as expected; one for each $C(g) = \mathbb{G}_m$ orbit in $\mathbb{A}^1$.

The following was useful for me, just to verify computations if anything: https://www.math.hmc.edu/~davis/ThesisFinal.pdf Also, thanks to the comments and answers, they were useful and informative nonetheless!

user148177
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