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.

4more comments