Kazhdan-Lusztig theory gives a correspondence between irreducibles of the affine Hecke algebra for a simply connected linear algebraic group $G$ and certain homological data extracted from the Steinberg variety (for $q$ not a root of unity). I want to see this correspondence explicitly for $SL_2$ by (1) looking at generators and relations for the Hecke algebra $H$ and (2) looking at the geometric data. My problem is that for the 1-dimensional representations I can't get them to match up. I expect it's a somewhat silly error but I can't find it, in any case.

Edit: The sources I'm relying on Criss/Ginzburg (paper version here: http://arxiv.org/abs/math/9802004). The affine Hecke algebra in this case I think is some deformation of the group algebra of the affine Weyl group. On the geometric side it's $G \times \mathbb{C}^*$-equivariant K-theory on the Steinberg variety.

Method 1: The affine Hecke algebra for $SL_2$ is an algebra $H$ with generators $T, X, X^{-1}$ and relations $$(T + 1)(T - q) = 0$$ $$TX^{-1} - X^1 T = (1-q)X^1$$ $$X X^{-1} = X^{-1}X = 1$$ It's not too hard to compute the one-dimensional representations. For central character $\sqrt{q} + 1/\sqrt{q}$ (the scalar which $X + X^{-1}$, the generator for $\mathcal{O}(T//W)$, acts by), one has the representations $$(T, X) \mapsto (-1, 1/\sqrt{q}), (q, \sqrt{q})$$ and for central character $-\sqrt{q} - 1/\sqrt{q}$ $$(T, X) \mapsto (-1, -1/\sqrt{q}), (q, -\sqrt{q})$$

In total, giving four representations, two for each central character.

Method 2: We should be able to read off the irreducibles by Kazhdan-Lusztig theory. Namely, for generic $q$, fix $g$ with desired central character, and take the $g$-fixed locus of the Springer resolution $\mu^g: \tilde{\mathcal{N}}^g \rightarrow \mathcal{N}^g$. The total Borel-Moore homology of $\mu^{g, -1}(x)$, choosing representatives $x$ in each $C(g)$-orbit of $\mathcal{N}^g$, is a representation of $H$ specialized at this central character.

For $g = \pm \text{diag}(\sqrt{q}, 1/\sqrt{q})$ one has $\mathcal{N}^g = \mathbb{A}^1$ the nilpotent radical of the usual Borel. It has two $C(g)$ orbits: $x = 0$ and $x \in \mathbb{A}^1 - 0$.

(1) For $x = 0$, the fiber has two-dimensional homology. The double centralizer $C(g, 0) = T$, which is connected, so we don't have to worry about isotypical components of $A(g, 0) = C(g, 0)/C(g, 0)^\circ$.

(2) For $x = \left(\begin{array}{cc}0&1\\0&0\end{array}\right)$, the fiber is a point, so we have a one-dimensional homology. The component group isn't relevant here.

So here, we only find two 1-dim representations, one for each central character. I must have misunderstood something.

  • $\begingroup$ It's essential to distinguish carefully between the "affine Weyl group" (or Hecke algebra) and the "extended" version, both of which enter into the work of Kazhdan-Lusztig and others influenced by them. (In this direction, it would also be very helpful to indicate which sources you are primarily relying on.) $\endgroup$ Jul 25, 2015 at 14:12
  • $\begingroup$ I'm relying on Criss/Ginzburg (paper version here: arxiv.org/abs/math/9802004). The affine Hecke algebra in this case I think is some deformation of the group algebra of the affine Weyl group. On the geometric side it's $G \times \mathbb{C}^*$-equivariant K-theory on the Steinberg variety. I'll add this to the main body too. $\endgroup$ Jul 25, 2015 at 14:51
  • $\begingroup$ Specifically, pp40 second definition 11.3 is the first method I am using, and theorem 11.8 is the second method $\endgroup$ Jul 25, 2015 at 15:04
  • $\begingroup$ In the geometric statements on the Kazhdan-Lusztig side, there is a parameter $t \in \mathbb{C}^\times$ that shows up, and we consider fixed point sets of $a = (g,t)$. Why do we fix $t = 1$ (the $q$) in this setup (I assume this is what is meant by the notation $\mathcal N^g$ in the place of $\mathcal N^a$)? $\endgroup$
    – mi.f.zh
    May 31, 2021 at 6:04
  • $\begingroup$ @mi.f.zh I think above I want to take q to be not a root of unity, i.e. generic. My notation is just sloppy, but you're right that one wants q-commuting elements (not commuting elements). $\endgroup$ Jun 1, 2021 at 7:18

2 Answers 2


It turns out I was confused about the statement on the geometric side. I'm using Ginzburg's 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 $\mathbb{A}^1$.

Anyway, thanks for your help, everyone! The comments and answers were very informative.

  • $\begingroup$ Thank you for sharing the resolution of the apparent paradox! Although I'd gone over this part of CG before, it wasn't easy to see what was amiss. $\endgroup$ Jul 30, 2015 at 6:44
  • $\begingroup$ @Harrison Chen I know it's been 8 years, but is there any chance you, or someone else who sees this knows how to determine which module corresponds to which orbit? $\endgroup$ 2 days ago

The foundations seem to come from the book Representation Theory and Complex Geometry (Birkhauser, 1997) by Victor Ginzburg (with his former student Chriss), though in the question here you rely mainly on the more limited lectures given by Ginzburg in Montreal (1997) and written up by Baranovsky. I'm not familiar with all details of the book or lecture notes, but it looks to me as if the confusion I cautioned about in my comment is the source of your problem. Ginzburg's formulation in Section 11 of the notes involves both the affine Weyl group and the extended version, which differ in your case because the "fundamental group" (weight lattice modulo root lattice) has order 2. Probably this factor of 2 leads to your finding 4 rather than 2 representations of degree 1.

Though it may complicate things further from your viewpoint, I'd be more comfortable if you referred also to the original results of Kazhdan-Lusztig. In any case the notation in this subject can be treacherous.

ADDED DETAILS: Looking further at how both versions (lecture notes and book) are structured, I'm more convinced that the source of your problem is the shift in the literature toward the convention that the term "affine Hecke algebra" refers mainly to the algebra coming from the extended affine Weyl group (denoted $\widetilde{W}$ in the lecture notes). Indeed, in Section 7.1 of the Chriss-Ginzburg book the "affine Weyl group" (there denoted $W_{aff}$) refers to the extended version which is not usually a Coxeter group, as explained in Remark 7.1.3. Similarly, the algebra denoted $\textbf{H}$ is the main object of interest (and is called the "affine Hecke algebra" in Definition 7.1.9 of the book): this plays the leading role in the crucial transition to a more geometric viewpoint, worked out more explicitly for $\mathrm{SL}_2$ in Section 7.5.

The moral of the story is that you have to distinguish more carefully between the two versions of the affine Hecke algebra when discussing finite dimensional representations.


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