Background My question is about the paper http://arxiv.org/PS_cache/arxiv/pdf/0712/0712.4160v2.pdf; specifically about the isomorphism in Theorem $1.2$ (in the Introduction), $T_{\lambda} \cap \overline{\mathcal{O}_{\mu}} \simeq L^{<0}G \cdot \lambda \cap \overline{L^{\geq 0} \cdot \mu}$. To explain the notation, $G=GL_n(\mathbb{C})$, $\mu, \lambda$ are partitions of $n$ with $\lambda \leq \mu$ in the dominance ordering. $T_{\lambda}$ is a transverse slice to a nilpotent of type $\lambda$, slightly different from the Slodowy slice; see pg 11, equations (15) and (16) for a definition and example.

Let $L^{\geq 0}G = G(\mathbb{C}[[t]]), L^{<0}G = \text{ker}(G(\mathbb{C}[t^{-1}]) \rightarrow G)$, $\text{Gr}_G$ be the affine Grassmanian for $G$, and let $\lambda \in \text{Gr}_G$ be the element corresponding to $g L^{\geq 0}G$, where $g$ is the diagonal matrix with entries $t^{-\lambda_1}, t^{-\lambda_2}, \cdots , t^{-\lambda_n}$.

Question I was trying to understand the above isomorphism when $n=2, \lambda = (1,1), \mu=(2,0)$. On the left side, we get $\mathcal{N}$ (all $2 \times 2$ nilpotent matrices), since $T_{\lambda} = \mathfrak{g}$. On pg 19, specifically equation (33), it is stated that the isomorphism $T_{\lambda} \cap \overline{\mathcal{O}_{\mu}} \simeq L^{<0}G \cdot \lambda \cap \overline{L^{\geq 0} \cdot \mu}$ is given by the following map $\psi$ (here I choose $L_b$, defined in 4.4.1, to be the lattice denoted above by $\lambda$):

$\psi(x+f_1) = (1 + \sum_{k=1}^{\infty} t^{-k} f_1 (t+f_1)^{k-1}) \lambda$

Here $x=0, f_1$ is arbitrary. However the above sum doesn't make sense to me; e.g. when I take $f_1$ to be the nilpotent with first row $(0, 1)$ and second row $(0,0)$, since $f_1^2=0$,I get:$t^{-k} f_1 (t+f_1)^{k-1}=t^{-1}f_1$, so that the above sum is an infinite sum with each term being $t^{-1}f_1$, which doesn't make sense. For similar reasons I can't get it to work with other matrices. What am I doing wrong?

Background My question is about the paper http://arxiv.org/abs/0712.4160; specifically about the isomorphism in Theorem $1.2$ (in the Introduction), $T_{\lambda} \cap \overline{\mathcal{O}_{\mu}} \simeq L^{<0}G \cdot \lambda \cap \overline{L^{\geq 0} \cdot \mu}$. To explain the notation, $G=GL_n(\mathbb{C})$, $\mu, \lambda$ are partitions of $n$ with $\lambda \leq \mu$ in the dominance ordering. $T_{\lambda}$ is a transverse slice to a nilpotent of type $\lambda$, slightly different from the Slodowy slice; see pg 11, equations (15) and (16) for a definition and example.

Let $L^{\geq 0}G = G(\mathbb{C}[[t]]), L^{<0}G = \text{ker}(G(\mathbb{C}[t^{-1}]) \rightarrow G)$, $\text{Gr}_G$ be the affine Grassmanian for $G$, and let $\lambda \in \text{Gr}_G$ be the element corresponding to $g L^{\geq 0}G$, where $g$ is the diagonal matrix with entries $t^{-\lambda_1}, t^{-\lambda_2}, \cdots , t^{-\lambda_n}$.

Question I was trying to understand the above isomorphism when $n=2, \lambda = (1,1), \mu=(2,0)$. On the left side, we get $\mathcal{N}$ (all $2 \times 2$ nilpotent matrices), since $T_{\lambda} = \mathfrak{g}$. On pg 19, specifically equation (33), it is stated that the isomorphism $T_{\lambda} \cap \overline{\mathcal{O}_{\mu}} \simeq L^{<0}G \cdot \lambda \cap \overline{L^{\geq 0} \cdot \mu}$ is given by the following map $\psi$ (here I choose $L_b$, defined in 4.4.1, to be the lattice denoted above by $\lambda$):

$\psi(x+f_1) = (1 + \sum_{k=1}^{\infty} t^{-k} f_1 (t+f_1)^{k-1}) \lambda$

Here $x=0, f_1$ is arbitrary. However the above sum doesn't make sense to me; e.g. when I take $f_1$ to be the nilpotent with first row $(0, 1)$ and second row $(0,0)$, since $f_1^2=0$,I get:$t^{-k} f_1 (t+f_1)^{k-1}=t^{-1}f_1$, so that the above sum is an infinite sum with each term being $t^{-1}f_1$, which doesn't make sense. For similar reasons I can't get it to work with other matrices. What am I doing wrong?