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?

  • $\begingroup$ I may be wrong but I believe the formula you want is written in Springer-Steinberg, Cor. 1.8 (ii) page 251, also reproduced in Carter, page 398. $\endgroup$ – Francois Ziegler Mar 20 '14 at 1:00
  • 1
    $\begingroup$ @FrancoisZiegler: Thank you for those references, but I don't think that's what I'm looking for. Carter gives the dimension of the unipotent radical of the connected centralizer of a unipotent element, which need not be the set of all unipotent elements in the centralizer. In fact, it is almost always smaller. For instance, if $u=I_n$, the identity, then the centralizer is $\mathrm{GL}_n(k)$ which has no unipotent radical. However, there are plenty of unipotent elements in $\mathrm{GL}_n(k)$. Is there an easy way to get what I want from your references that I'm not seeing? $\endgroup$ – Jared Mar 20 '14 at 2:12
  • $\begingroup$ Ah, I see. Sorry -- and no, I didn't mean to say that your problem easily reduces to theirs. $\endgroup$ – Francois Ziegler Mar 20 '14 at 2:18

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.


Michael Bulois

  • $\begingroup$ Welcome to MO, Michael! $\endgroup$ – Filippo Alberto Edoardo Mar 20 '14 at 10:08
  • $\begingroup$ Wonderful. Thank you for this very helpful response! $\endgroup$ – Jared Mar 20 '14 at 16:35

As Michael indicates in his first sentence, the answer is basically straightforward (though not usually written down in the very explicit way you request). It may be helpful to add some perspective, since this type of question comes up in different levels of generality. The recent research literature provides some fine-tuning, especially in prime characteristic, but is not at all essential for the concrete case you are looking at.

Classically, study of the centralizer of a typical $n \times n$ matrix starts with Jordan canonical form and arrives quickly at an overall dimension formula. This usually requires an algebraically closed field, but in the case of a nilpotent (or unipotent) matrix that isn't essential. In the language of partitions, you have to pass to the transpose partition as Michael shows. It makes no real difference whether you study nilpotent matrices or unipotent matrices at this point.

More generally, this kind of question arises for a reductive or semisimple Lie algebra in characteristic 0, or for a corresponding algebraic group. The simple case is the essential one, for which the Dynkin/Kostant theory provides a systematic method: here you arrive at a direct sum decomposition of the centralizer involving its nilradical and a reductive complement. Here the respective dimensions are readily computable for the general or special linear case. Moreover, the dimension of the nilpotent variety in a reductive Lie algebra (or unipotent variety in the group) is easy to compute here, since the Lie algebra is just a direct sum of certain general linear Lie algebras determined by the partition. In a reductive Lie algebra the nilpotent cone has dimension equal to the total number of roots, which together with the rank adds up to the total dimension of the Lie algebra.

For arbitrary Lie types (now working over an algebraically closed field of characteristic 0), the story is similar but less straightforward combinatorially. In good prime characteristic the answers are pretty much the same, but for bad primes the results diverge a bit for the groups and Lie algebras (except in the special linear case).

By now there are reasonable textbook references, some mentioned by Francois and by Michael, for the basic theory and dimensions involved (which make it very easy to answer the question here). Here are more explicit details:

Springer-Steinberg lectures, section IV.1, in Lecture Notes in Mathematics 131 (Springer, 1970).

Carter, Finite Groups of Lie Type (Wiley Interscience, 1985), 13.1.

Collingwood-McGovern, Nilpotent Orbits in Semisimple Lie Algebras (Van Nostrand Reinhold, 1993), Chapter 3.

Humphreys, Conjugacy Classes in Semisimple Algebraic Groups (AMS, 1995), 1.3 and 7.10 (with minor corrections on the AMS bookpage or my homepage).


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