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?

  • $\begingroup$ Could you change the link to the abstract page, please? $\endgroup$ – S. Carnahan Dec 19 '11 at 2:03
  • $\begingroup$ I've changed it now. $\endgroup$ – Vinoth Dec 19 '11 at 3:03
  • $\begingroup$ This arXiv preprint, filling in details of the 2003 Comptes Rendus note by the authors, hasn't been formally published (in part perhaps because Vybornov left academia). It might be simplest to send an email inquiry to Mirkovic. Though he is busy with many projects these days, he could best answer your question. $\endgroup$ – Jim Humphreys Dec 19 '11 at 15:35

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