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. 



A question on algebraic loop groops
The answer is yes. See for example


Pressley, Andrew; Segal, Graeme (1986), Loop groups, Oxford Mathematical Monographs. Oxford Science Publication


