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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 $(T\times T^\vee)/W$?

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  • $\begingroup$ Actually, Bezrukavnikov-Finkelberg-Mirković showed $K^{G(\mathcal{O})}(Gr_G)$ is the coordinate ring $\mathfrak{B}^{G^\vee}_{G}$ of Toda phase space (Theorem 2.15). I wonder if somebody could tell me what $K^{I}(Fl_G)$ is, indeed. $\endgroup$ Jul 25, 2017 at 15:21

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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 $\mathcal{\tilde{Z}}$.

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  • $\begingroup$ I guess it's a good time to ask one more stupid question. Can I consider $Fl_G$ as a $(G/B)$-bundle over $Gr_G$? Or is $Fl_G$ a direct product $(G/B) \times Gr_G $? $\endgroup$ Jul 25, 2017 at 13:19
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    $\begingroup$ It's a bundle, but definitely not a direct product (there are line bundles on $Fl_G$ which would have to pullback to line bundles on $Gr_G$ that couldn't exist). However, for purposes of cohomology or K-theory it behaves like one, basically because of Schubert decomposition. $\endgroup$
    – Ben Webster
    Jul 27, 2017 at 15:37
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Sorry, this question itself is wrong!

Bezrukavnikov-Finkelberg-Mirković actually showed that the $G(\mathcal{O})$-equivariant $K$-theory of affine Grassmannian Steinberg variety $\mathcal{R} = \{ (x,[g])\in \mathfrak{g}(\mathcal{O})\times Gr_G \mid \operatorname{Ad}_{g^{-1}}(x) \in \mathfrak{g}(\mathcal{O})\}$ is the coordinate ring of the following phase space $$ \textrm{Spec}\; K^{G(\mathcal{O})}(\mathcal{R})=(T\times T^\vee)/W $$ where $T$ is the maximal torus and W is the Weyl group.

I wanted to ask the case when we replace $\mathcal{R}$ by affine flag Steinberg variety $\mathcal{Z} = \{ (x,[g])\in \textrm{Lie}(I)\times Fl_G \mid \operatorname{Ad}_{g^{-1}}(x) \in \textrm{Lie}(I)\}$. Indeed, Varagnolo-Vasserot has already showed $$ K^{I}(\mathcal{Z})=\mathbb{C}[T\times T^\vee]\rtimes \mathbb{C}[W]~. $$ Thank you very much to Hiraku Nakajima for pointing out my mistake and telling me the reference. Sorry to MO reader for this stupid question!

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    $\begingroup$ I edited your second displayed equation. The LHS as written before wasn't well-defined, since it doesn't have an action of the full $G(\mathcal{O})$. $\endgroup$
    – Ben Webster
    Jul 24, 2017 at 15:45

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