Let $\mathcal{K} = \mathbb{C}((t)), \mathcal{O}=\mathbb{C}[[t]]$, $G$ be a reductive group, and $\text{Gr}_G=G(\mathcal{K})/G(\mathcal{O})$; there is a left action of $G(\mathcal{O})$ on $\text{Gr}_G$. Let $X_*(T)=\text{Hom}(\mathbb{C}^{\times},T)$ (note that there is a natural embedding of $X_*(T)$ inside $G(\mathcal{K})$). Let $B$ be a Borel subgroup. Let $X^*(T) = \text{Hom}(T, \mathbb{C}^{\times})$.

Choose $\lambda \in X_*(T)$ to be dominant, and (abusing notation), let $\lambda$ also denote the image of $\lambda \in X_*(T) \subset G(\mathcal{K})$ in the quotient $\text{Gr}_G$. Define $\text{Gr}^{\lambda} = G(\mathcal{O}) \cdot L_{\lambda}$.

My question is how to show that $\text{dim}(\text{Gr}^{\lambda}) = 2 \rho (\lambda)$. Here $\rho \in X^*(T)$ is half of the sum of the positive roots with respect to $B$.

This is stated in the third paragraph on pg $5$ of Mirkovic-Vilonen (http://arxiv.org/pdf/math/0401222v4.pdf).

I know how to do this for $G=SL_2(\mathbb{C})$, but I'm not sure how to generalize it to arbitrary groups.