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I will be so thankful for any comments or help about the following question.

Is it possible to obtain all maximal centralizers in the General linear group $GL_n(F)$, for an arbitrary field $F$ and positive integer $n$ ? Here, by maximal centralizer I mean: $C=C_G(x)$ is maximal if $C⊆C_G(y)$ , then $C=C_G(y)$ or $y∈Z(G)$.

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  • $\begingroup$ For centralizers of elements $x$, whose order is coprime to the characteristic of the field, you should be able to find the maximal centralizers (you may assume $x$ to have prime order). The unipotent case is probably hard. $\endgroup$
    – j.p.
    Feb 5, 2014 at 18:57
  • $\begingroup$ If $x$ is unipotent and $x-1$ is nilpotent of degree $d$, then the centralizer of $x$ is contained in the centralizer of $1+(x-1)^{d-1}$. So the unipotent case reduces to centralizers of non-identity elements $x$ such that $(x-1)^2 = 0$. See also mathoverflow.net/questions/148704/…. $\endgroup$
    – Mark Wildon
    Feb 5, 2014 at 21:55
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    $\begingroup$ If $x$ is semisimple and $F$ is algebraically closed or a finite field then one can appeal to the theory of reductive algebraic groups to obtain the centralisers. In particular, $C_G(x)$ will be a Levi subgroup. So one simply needs to compute maximal Levi subgroups in $G$ (which can be determined by the root system). You may find Bonnaf\'e's paper "Quasi-Isolated Elements in Reductive Groups" helpful. $\endgroup$
    – Jay Taylor
    Feb 6, 2014 at 6:40

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As Mark Wildon points out, there is a closely related question already on MO here. Both questions are answered in a relatively elementary style by the old method of Frobenius, which is presented in modern form in textbooks such as Springer GTM 136 by Adkins and Weintraub (see 5.5). There is also a version for algebraic groups in section 1.2 of my 1995 AMS book Conjugacy Classes in Semisimple Algebraic Groups, though the proof works the same over an arbitrarty field and just involves matrix manipulations in the case of general or special linear groups.

There is no need here to deal separately with semisimple and unipotent elements. Indeed, the selling point of the Frobenius method is that it works straightforwardly over any field $F$ (even finite). Given a linear operator $T$ on a finite dimensional vector space $V$, one works with the rational canonical form over $F$, which splits $V$ into a direct sum of subspaces of nondecreasing dimension with each being a cyclic $F[T]$-module and the invariant factor polynomials dividing as you go along (the last of these being the minimal polynomial of $T$). Then it's easy to cmpute the dimension over $F$ of the space of operators commuting with $T$. (In case $T$ is nonsingular, you get the centralizer group size from this.)

From this type of calculation, I think it's clear that some maximal proper centralizers occur when the degrees of the invariant factor polynomials are 1, 1, ..., 1, 2. (For instance, you get the unique nontrivial minimal unipotent conjugacy class this way in the algebraic group setting.)

The advantage of working with the rational canonical form (rather than the Jordan form) is that it applies to any field $F$ and gives a uniform comptuation. (Naturally, it's much more complicated to study arbitrary semisimple or reductive algebraic groups relative to a field of definition.)

ADDED: I don't want to claim too much here. The Frobenius method only leads readily to a uniform family of maximal proper centralizers having largest possible dimension over $F$ (measured in the full matrix algebra). Beyond this I'm not sure how to locate other possible maximal centralizers, which would presumably depend more on the nature of $F$. The ones I've singled out at the end all have centralizers of the same dimension, and moreover each matrix is either unipotent or is semisimple with two distinct eigenvalues in $F$, one repeated $n-1$ times.

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  • $\begingroup$ Is it possible to recognize explicitly all maximal centralizers? Mark Wildon pointed out one type. $\endgroup$
    – maryam
    Feb 6, 2014 at 20:11
  • $\begingroup$ In addition on MO author had used Jordan form which works on perfect fields. Is it possible to use that discussion in arbitrary field? $\endgroup$
    – maryam
    Feb 6, 2014 at 20:13

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