# Monday, November 22

:::{.remark}
Considering the infinite dimensional case, $\tilde A_2$.
Here $W = W(\tilde A_2) = \gens{s_1,s-2,s_3 \st s_i^2=1, (s_i s_j)^3=1}$, and we can form $X_w \subseteq G/B$ for any $w\in W$.
This will be a finite dimensional projective variety with a Torus action, and there are BSDH resolutions for reduced words given by $T\dash$equivariant maps
\[
P_{i_1}\mix{B} \cdots P_{i_n}/B \mapsvia{\mu} X_w
.\]
These are resolutions of singularities, and in particular birational.
Note that $W$ is infinite here.
:::

:::{.remark}
Article by Graham-Li: say $w\in W$ is **spiral** iff $w = (s_j s_j s_k)^\ell$ for $i,j,k \in \ts{1,2,3}$.
This produces a nice family of Schubert varieties.
For $\rank A = 2$, we have $\dim \lieh = 3+1=4$.
Up to a change of coordinates, we can use $\alpha_1\dual = \tv{1,0}^t$ and $\alpha_2\dual = \tv{1,0}^t$ and let $V \da R \tensor_\ZZ \ts{\alpha_1\dual, \alpha_2\dual}$ be the ambient Euclidean space and set $L\da \ZZ\gens{ \alpha_1, \alpha_2}$.
Then use the action of $W$ to define $W_{\text{aff}} \da L \semidirect W_f$ where $W_f = W(A\dual)$, and $s_i(\chi) = \chi - \inner{ \alpha, \lambda} \alpha\dual$.
Here we think of $L$ as translations.
The dual roots are \( \alpha_1 = \tv{2, -1}^t. \alpha_2 = \tv{-1, 2} \) and so $\tilde \alpha = \alpha_1 + \alpha_2 = \tv{1,1}^t$
Define hyperplanes $H_{\alpha, n} \da \ts{v\in V \st \inner{\alpha}{v} = n \in \ZZ}$.
There is a *fundamental alcove* enclosed by the positive sides of the various hyperplanes and within distance 1 of $H_{\tilde \alpha, 0}$.
If you draw the picture and now act on the fundamental alcove by simple reflections, the image "spirals" out away from the origin.
:::

:::{.remark}
The article doesn't use BSDH resolutions, maybe compare and contrast with what we've done.
:::

:::{.remark}
Back to $\mu$.
For $\ell = 1$, we have
\[
P_1 \mix{B} P_2 \mix{B} P_3/B \mapsvia{\mu} X_w
.\]

The Bruhat order yields

\begin{tikzcd}
	&& {s_1s_2s_3} \\
	{s_1s_2} && {s_1s_3} && {s_2s_3} \\
	\\
	{s_1} && {s_2} && {s_3} \\
	\\
	&& 1
	\arrow[from=4-1, to=6-3]
	\arrow[from=4-3, to=6-3]
	\arrow[from=4-5, to=6-3]
	\arrow[from=2-1, to=4-1]
	\arrow[from=2-3, to=4-3]
	\arrow[from=2-5, to=4-5]
	\arrow[from=2-1, to=4-3]
	\arrow[from=2-3, to=4-1]
	\arrow[from=2-3, to=4-5]
	\arrow[from=2-5, to=4-3]
	\arrow[from=1-3, to=2-1]
	\arrow[from=1-3, to=2-3]
	\arrow[from=1-3, to=2-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMiwwLCJzXzFzXzJzXzMiXSxbMCwxLCJzXzFzXzIiXSxbMiwxLCJzXzFzXzMiXSxbNCwxLCJzXzJzXzMiXSxbMCwzLCJzXzEiXSxbMiwzLCJzXzIiXSxbNCwzLCJzXzMiXSxbMiw1LCIxIl0sWzQsN10sWzUsN10sWzYsN10sWzEsNF0sWzIsNV0sWzMsNl0sWzEsNV0sWzIsNF0sWzIsNl0sWzMsNV0sWzAsMV0sWzAsMl0sWzAsM11d)

Note that there are no braid relations.
We can consider the $T\dash$equivariant multiplicity $\mce_x^T X_w = \sum_{z\in \mu\inv(x)^T} \mce_z^T(z)$ given by summing over the $T\dash$equivariant fixed points in the fiber.
Here this just equals $\mce_z^T(z)$ where $\mu(z) = x$, since there is a unique $T\dash$fixed point in the fiber.
A basic AG argument shows that the resolution is an isomorphism and thus $X_w$ is smooth, so there is no singular locus.
The paper gives a nice formula for $\ell \geq 6$.
:::

:::{.remark}
Starting the calculation: 

1. Consider $e_x X_w\in \ff(S(\lieh\dual))$ the equivariant multiplicity, then $x\in X_w$ is smooth iff a certain change of basis $c_{w, x}$ corresponds to the equivariant multiplicity.
2. In the rationally smooth locus, they show smoothness iff there is a single $T\dash$fixed point in the fiber.

:::