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?



Dimension of the nilpotent centralizer of a nilpotent matrix
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


