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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 (

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

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up vote 6 down vote accepted

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 $\langle 2\rho,\lambda\rangle$.

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