Dimension of the Nilpotent Centralizer of a Nilpotent Matrix – Algebraic Geometry

ag.algebraic-geometrylie-algebrasmatricesreference-request

Fix a natural number $n$ and an algebraically closed field $k$. Let $\mathfrak{g}=\mathfrak{gl}_n(k)$. For any partition of $n$, $\lambda=(\lambda_1,\ldots,\lambda_r)$, let $A_{\lambda}$ be the $n\times n$ nilpotent matrix in Jordan form whose blocks correspond to $\lambda$. I'd like to know the dimension of the nilpotent part of the centralizer of $A_{\lambda}$, i.e., if $\mathcal{N}(\mathfrak{g})$ is the null cone, then I'm interested in the dimension of the variety
$$
X_{\lambda}=\mathcal{N}(\mathfrak{g})\cap C_{\mathfrak{g}}(A_{\lambda})=\{B\in\mathfrak{g}\mid B^n=0,\;[A_{\lambda},B]=0\}.
$$

It seems like this should be known, but I can't seem to find a reference. For the two extreme cases, we have the following:

  1. If $\lambda_i=1$ for all $i$, then $A_{\lambda}=0$, so $C_{\mathfrak{g}}(A_{(1,\ldots,1)})=\mathfrak{g}$, from which it follows that $\dim(X_{(1,\ldots,1)})=\dim(\mathcal{N}(\mathfrak{g}))=n^2-n$.
  2. If $r=1$, then $A_{(n)}$ is regular nilpotent, so its centralizer consists precisely of polynomials in $A_{(n)}$ of degree less than $n$. The nilpotent part is given by those polynomials with constant term $0$, so that $\dim(X_{(n)})=n-1$.

Does anyone know if the answer for arbitrary $\lambda$ is written down anywhere? If not, how can we compute this dimension?

Best Answer

Indeed, this is known.

First, you have some structure on this nilptotent cone. It is the product of an affine space and the nilpotent cone of a reductive Lie algebra.

Namely, embed $A_{\lambda}$ in a $\mathfrak{sl}_2$-triple $(A_{\lambda}, H_{\lambda}, B_{\lambda})$. Then $H_{\lambda}$ yield a characteristic grading $\mathfrak g=\bigoplus_{i\in \mathbb Z} \mathfrak g_i$ where $\mathfrak g_i=\{x\in \mathfrak g| [x,H_{\lambda}]=ix\}.$ Then $$C_{\mathfrak g}(A_{\lambda})=\bigoplus_{i\in \mathbb N} C_{\mathfrak g_i}(A_{\lambda})$$ and, denoting $\mathcal N(\mathfrak g)\cap C_{\mathfrak g_0}(A_{\lambda})$ by $\mathcal N_0$, we get $$X_{\lambda}=\mathcal N_0\oplus\bigoplus_{i\in \mathbb N} C_{\mathfrak g_i}(A_{\lambda}) $$ You can look at A. Premet (Nilpotent commuting varieties of reductive Lie algebras, Invent. Math., 154 (2003), 653-683) for a few more details.

Hence, $X_{\lambda}$ is an irreducible subvariety of $C_{\mathfrak g}(A_{\lambda})$ of codimension $\textrm{codim}_{C_{\mathfrak g_0}(A_{\lambda})} \mathcal N_0=\textrm{rank}(C_{\mathfrak g_0})$. Since the $C_{\mathfrak g}$ and $C_{\mathfrak g_0}$ are known (e.g. see thm 6.1.3 of Collingwood MacGovern for the dimension of the first one), this is computable

We rephrased Premet work in the type A case in http://arxiv.org/abs/1306.4838v1 Section 4 (This may also be heavily based on Basili's works mentioned there). In particular, we mention the fact (Proposition 4.5) that $X_{\lambda}$ is of codimension $d_{\lambda}$, the number of parts of $\lambda$, in $C_{\mathfrak g}(A_{\lambda})$.

Summing this up, a formula for the dimension of $X_{\lambda}$ would be $$(\sum_i \hat\lambda_i^2)-\hat\lambda_1,$$ where $\hat\lambda$ is the transposed partition associated to $\lambda$.

Hope that this helps for my first day on mathoverflow.

Yours,

Michael Bulois

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