TITLE: 1-dim representations of the affine Hecke algebra for $G = \text{SL}_2$
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: 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!