# Reconciling the affine grassmannian and the based loop group

I'm trying to reconcile the differences between the (algebraic) based loop group and the affine grassmannian. I once believed that I understood the relationship, but I just read a paper which has thrown that theory to the wind. There are quite a few spaces I need to introduce, so please forgive the verbosity.

For simplicity, let $K$ be a compact, connected, simply connected Lie group with complexification $G$. Define the following spaces:

• $LK, LG$ the spaces of analytic loops $S^1 \to K$ and $S^1 \to G$ respectively,
• $L_\text{alg} G$ the space of algebraic maps $\mathbb C^\times \to G$, and $L_{\text{alg}}^+ G$ those maps in $L_\text{alg} G$ which extend to maps on $\mathbb C$.
• $\Omega K$ the space of base-point preserving analytic loops; that is, $\gamma \in LK$ such that $\gamma(e_{S^1} ) = e_K$. And $\Omega_\text{alg} K = L_\text{alg} G \cap \Omega K$.
• $\mathcal Gr = \frac{ G((t))}{G[[t]]}, \mathcal Gr' = \frac{ G[t,t^{-1}]}{G[t]}$

Here are some facts:

1. $\mathcal Gr' = \frac{L_\text{alg} G}{L_\text{alg}^+G} = \Omega_\text{alg} K$. There are multiple references for this, one of which is [1, page 124].
2. $\mathcal Gr = \Omega K$, see [2, page 2]
3. $\mathcal Gr$ is ($LG$ -equivariantly) isomorphic to $\mathcal Gr'$, see [3, Prop 1.2.4].

In fact, the equivariance of 3 seems to be an after thought. If I understand the argument correctly, the isomorphism of 3 preserves the filtrations from which $\mathcal Gr$ and $\mathcal Gr'$ are given the direct limit topology, and hence is actually a homeomorphism. Combining these together would thus say that $\Omega K$ is homeomorphic to $\Omega_\text{alg} K$. This is something which I believe is known to be untrue.

So, I figure that I have either greatly confused the categories in which these isomorphisms occur, or that one of the three above facts is dubious. If I had to guess, I cannot see any reason why $\Omega K = \mathcal Gr$, since there is no reason to suspect that $\Omega K$ would have an algebraic structure.

Any suggestions would be greatly appreciated.

Edit: Some additional papers:

[4, Section 1.2] indicates that $\Omega K$ is a topological model of $\mathcal Gr$.

[5, Section 1] indicates that $\mathcal Gr$ is homeomorphic to $\Omega_\text{alg} K$.

[6, Section 6.1] identifies $\mathcal Gr$ with $\Omega K$ (though it is not clear to me whether this is $\Omega K$ or $\Omega_\text{alg} K$).

I'm not sure if it's related, but it seems like a lot of the time we are ignoring the differences between the analytic and formal power series. For example, this write up talks about the Birkhoff decomposition (as does [2]) $$G((t)) = \Omega K \times G[[t]]$$ and the reference is Pressley-Segal, despite the fact that the result in [PS] $LG = \Omega K \times L^+G$, is for analytic loops.

[1] Mitchell, Stephen A. Quillen's theorem on buildings and the loops on a symmetric space. Enseign. Math. (2) 34 (1988), no. 1-2, 123--166.

[2] Magyar, Peter. Schubert classes of a loop group.

[3] Ginzburg, Victor. Perverse sheaves on a loop group and Langlands' duality.

[4] Lam, Schilling, Shimozono. K-Theory Schubert Calculus of the Affine Grassmannian.

[5] Yun, Zhu. Integral homology of loop groups via Langlands dual groups.

[6] Arkhipov, Bezrukavnikov, Ginzburg. Quantum Groups, the loop Grassmannian, and the Springer resolution.

• I'm already confused about what categories each of your spaces is in, let alone what categories those isomorphisms are in. $LK$ and $LG$ are just topological spaces, yes? I'm not sure what category you want $L_{alg}(G)$ to be in. $\Omega K$ is again a topological space. And $\text{Gr}$ and $\text{Gr}'$ are... ind-varieties? – Qiaochu Yuan Oct 30 '14 at 7:27
• $LK$ and $LG$ are topological groups themselves (group operation is pointwise), their central extensions are Kac-Moody groups. $\Omega K$ ends up being a Frechet manifold with a Kahler structure (it might actually be Hilbert, but I can't remember). And yes, $\mathcal Gr$ and $\mathcal Gr'$ are ind-schemes. However, for the purposes of this questions, let's just take everything topologically. – Tyler Holden Oct 30 '14 at 18:32