TITLE: Relation between affine flag and Grassmannian Steinberg variety
QUESTION [10 upvotes]: Let $\mathcal{K}=\mathbb{C}((t))$ be the field of formal Laurent series over $\mathbb{C}$, and by $\mathcal{O}=\mathbb{C}[[t]]$ the ring of formal power series over $\mathbb{C}$. 
Given a semi-simple Lie group $G$, affine Grassmannian $Gr_G$ is defined by the coset space $G(\mathcal K)/G(\mathcal O)$ wheareas the affine flag variety is defined by $Fl_G=G(\mathcal K)/I$ where $I$ is the Iwahori subgroup which is the preimage of a Borel subgroup $B$ under the map $G(\mathcal O)\to G$. 
Bezrukavnikov-Finkelberg-Mirković showed that the $G(\mathcal{O})$-equivariant $K$-theory of affine Grassmannian $Gr_G$ is the coordinate ring of the following phase space
$$
\textrm{Spec}\; K^{G(\mathcal{O})}(Gr_G)=(T\times T^\vee)/W
$$
where $T$ is the maximal torus and W is the Weyl group.
My question is as follows: if you replace affine Grassmannian $Gr_G$ by affine flag variety $Fl_G$, then is the corresponding space
$$
\textrm{Spec}\; K^{G(\mathcal{O})}(Fl_G)
$$
just the direct product with the contangent bundle $T^*(G/B)$ of the flag variety (or nil-cone $\mathcal N$) 
$$
((T\times T^\vee)/W)\times (T^*(G/B))~~~?
$$
Or is $\textrm{Spec}\; K^{G(\mathcal{O})}(Fl_G)$ a non-trivial $T^*(G/B)$-bundle (or some another bundle) over $(T\times T^\vee)/W$?

REPLY [3 votes]: To clarify the issue you're having: you can look at the spaces 
$$\mathcal{\tilde{R}}=\{(x,g)\in \mathfrak{g}(\mathcal{O})\times G((t)) \mid \mathrm{Ad}_{g^{-1}}(x)\in \mathfrak{g}(\mathcal{O})\}$$
$$\mathcal{\tilde{Z}}=\{(x,g)\in \mathrm{Lie}(I)\times G((t)) \mid \mathrm{Ad}_{g^{-1}}(x)\in \mathrm{Lie}(I)\}$$
For former has left and right $G(\mathcal O)$ actions via $$g'\cdot (x,g) \cdot g''=(\mathrm{Ad}_{g'}(x),  g'gg'')$$ and the same formulae define left and right actions of $I$ on the latter.  To get an algebra, you should mod out by the same group on both sides, so you get $$K\big(G(\mathcal{O})\backslash \mathcal{\tilde{R}}/G(\mathcal{O})\big) \cong \mathbb{C}(T\times T^{\vee})^W$$
If instead, you consider $K\big(I\backslash \mathcal{\tilde{R}}/I\big)$, then you get the above algebra tensored with matrices on the vector space $H^*(G/B)$, and this is always hold true; the bimodules $K\big(G(\mathcal{O})\backslash \mathcal{\tilde{R}}/I\big)$ and $K\big(I\backslash \mathcal{\tilde{R}}/G(\mathcal{O})\big)$ induce the Morita equivalence.  So, in order to get something more interesting, you need to change the underlying space to $\mathcal{\tilde{Z}}$.