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.



Dimension of orbits in affine Grassmanian
The same way you find the dimension of an orbit in any homogeneous space. The orbit is $G(\mathcal{O})/(G(\mathcal{O})\cap t^{\lambda}G(\mathcal{O})t^{-\lambda})$, so you just have to calculate the corresponding quotient $\mathfrak{g}[[t]]/(\mathfrak{g}[[t]]\cap \operatorname{Ad}\_{t^\lambda}\mathfrak{g}[[t]])$ for Lie algebras.


Since $\operatorname{Ad}\_{t^\lambda}(\mathfrak{g}[[t]])=\sum t^{\langle \alpha,\lambda\rangle}\mathfrak{g}\_{\alpha}[[t]]$, we get that $$\mathfrak{g}[[t]]/(\mathfrak{g}[[t]]\cap \operatorname{Ad}\_{t^\lambda}\mathfrak{g}[[t]])=\sum \mathfrak{g}\_{\alpha}[[t]]/t^{\max(0,\langle \alpha,\lambda\rangle)}\mathfrak{g}\_{\alpha}[[t]].$$ So each negative root contributes nothing, and each positive root contributes $\langle \alpha,\lambda\rangle$; summing over all positive roots, we get $\langle 2\rho,\lambda\rangle$.


