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: Is $G(\mathcal{K})$ a topological group?


[Na05]  Perverse Sheaves on Real Loop Grassmannians, Invent. Math. 159 (2005), no. 1, 1--73.