Setup:
Let $\mathcal{K}=\mathbb{C}((t))$, $\mathcal{O}:= \mathbb{C}[[t]]$ and $G$ be a reductive algebraic group (over $\mathbb{C}$). Let further $\mathcal{K}_n$ denote the $\mathcal{O}$-ideal in $\mathcal{K}$ generated by $t^{-n}$, $\mathcal{O}^l$ the ideal in $\mathcal{O}$ generated by $t^l$. Now $G(\mathcal{O}/\mathcal{O}^l)$ is well known to be a linear algebraic group. Let $G(\mathcal{O}^l)$ denote the kernel of the obvious map $G(\mathcal{O})\to G(\mathcal{O}/\mathcal{O}^l)$
In Nadlers thesis [Na05] $G(\mathcal{K})$ is topologized by $$ G(\mathcal{K}):= \varinjlim \varprojlim G(\mathcal{K_n})/G(\mathcal{O}^l) $$
My question is now: $G(\mathcal{K})$ is a group, but is it also a topological group?
Motivation:
Let $Gr$ denote the affine Grassmannian. In the definition of the convolution product $*:Perv_G(Gr)\times Perv_G(Gr)\to Perv_G(Gr)$ it is used, that the map $G(\mathcal{K})\times_{G(\mathcal{O})}Gr\to Gr$ is continious. I would like to know why this is true
[Na05] Perverse Sheaves on Real Loop Grassmannians, Invent. Math. 159 (2005), no. 1, 1--73.
