Extension-field subgroups of $\operatorname{GL}(n, K)$

Question

$\newcommand{\op}[1]{\operatorname{#1}}\def\GL{\op{GL}}$I am interested in the subgroups of $\GL(n,K)$ of the form $\GL(n/s, E)$ for $E$ some extension field of $K$ of degree $s$. These arise in the following manner: If $V = K^n$ then $V \cong E^{n/s}$ as $K$-vector spaces, so $\GL(n/s, E)$ acts on $V$. Since an $E$-linear action is in particular $K$-linear, this amounts to an embedding of $\GL(n/s, E)$ in $\GL(n, K)$.

Said a bit less clumsily, the assertion is that if $E$ is a subfield of $\op{End}(V)$ of degree $s$ over $K$ then $\op{Aut}_E(V)$ (the invertible elements of the centralizer $C_E(V)$) is isomorphic to $\GL(n/s, E)$, and $E$ can be recovered as the center of $\op{Aut}_E(V)$. Thus embedded copies of $\GL(n/s, E)$ in $\GL(n, K)$ are parameterized by homomorphisms $E \to \op{End}(V)$.

(There may be other subgroups of $\GL(n, K)$ abstractly isomorphic to general linear groups, but we ignore those: we want only the naturally embedded ones.)

My question is about this lattice of subgroups. Specifically: $\def\Aut{\op{Aut}}$

What can we say about $\langle\op{Aut}_E(V), \op{Aut}_F(V)\rangle$ inside $\op{Aut}_{E \cap F}(V)$?

For example, there are many copies of $\GL(1, \mathbf{C})$ embedded in $\GL(2,\mathbf{R})$: the only restriction is that the matrix representing $i$ should have square $-I$. Let $E$ and $F$ be the embedded copies of $\mathbf{C}$ in $\op{End}(\mathbf{R}^2)$ corresponding to $i \mapsto (0,-1;1,0)$ and $i\mapsto(-1,-2;1,1)$, respectively. Then one can check that $\Aut_E(V)$ and $\Aut_F(V)$ generate $\GL^+(2,\mathbf{R})$, the group of all matrices with positive determinant, which has index 2 in $\Aut_{E\cap F}(V) = \GL(2,\mathbf{R})$.

On the other hand if we replace $\mathbf{R}$ with $\mathbf{Q}$ and $\mathbf{C}$ with $\mathbf{Q}(i)$, then $\Aut_E(V)$ and $\Aut_F(V)$ are both contained in the subgroup of matrices having determinant of the form $a^2+b^2$. I suspect that they generate this subgroup.

Optimistically, are these determinant-type obstructions the only obstructions?

Assume $E \cap F = K$. Is it true that $\langle \Aut_E(V), \Aut_F(V)\rangle$ contains $\op{SL}(n,K)$?

I am actually most interested in the case of finite fields, and in this case I actually know a proof that $\langle \Aut_E(V), \Aut_F(V)\rangle = \Aut_{E\cap F}(V)$. But the proof is fairly non-elementary and it doesn't feel very natural to me, so I'm hoping to find a better one.

In the case of finite fields, is there a simple proof that $\langle \Aut_E(V), \Aut_F(V) \rangle = \Aut_{E\cap F}(V)$?

Any comments (even if it's just a different way of looking at the problem, an interesting special case, a relevant reference, etc.) are much appreciated!

Answer 1

I should record what I've learned since asking this question, though there are still gaps in my understanding.

First, the question is equivalent to asking for a classification of overgroups of $\operatorname{GL}(1, E)$ in $\operatorname{GL}(n, K)$ for $E$ an extension field of $K$. A natural conjecture is that every such group lies between $\operatorname{SL}(d, F)$ and $\operatorname{\Gamma L}(d, F)$ for some field $F$ intermediate between $K$ and $E$, where $d = \dim_F E$. I don't know whether this is true in general, but it was proved by Kantor for finite fields: see Kantor [2].

My question was motivated by looking for a more direct proof than Kantor's, hopefully something that would generalize more easily, going something like as follows. Start with a group $G$ and find the smallest field $F$ (equivalently: largest $d$) such that $\operatorname{SL}(d, F) \subset G$. If $G$ normalizes $\operatorname{SL}(d, F)$ then we're done. Otherwise by conjugation $G$ contains $\operatorname{SL}(d, F)$ and $\operatorname{SL}(d, F')$ for some other field $F'$, so it contains the group they generate, which is...? The details are quite unclear.

However, Stefan Virchow has drawn my attention to a spectacular paper of Shangzhi Li [4], which carries out something like this programme (though many further ideas are needed). Li's theorem is the following.

Let $d \geq 2$ and suppose $E/K$ a field extension of degree $n/d$. Assume $\operatorname{SL}(d,E) \leq X \leq G = \operatorname{GL}(n, K)$. Then one of the following holds:

1. $X$ lies between $\operatorname{SL}(d', F)$ and $\operatorname{\Gamma L}(d', F)$ for some division ring $F$ between $K$ and $E$ (where $d' = d \dim_F E$);
2. $d=2$, $E$ is commutative, and $X$ lies between $\operatorname{Sp}(d',F)$ and its normalizer for some intermediate field $F$ (where again $d' = d \dim_F E$); or
3. $d=2$, $E/K = \mathbf{F}_4/\mathbf{F}_2$, and $X \cong A_6$ or $X \cong A_7$.

The theorem gives both more and less. On the one hand, it classifies overgroups of $\operatorname{SL}(d, E)$. If you assume further that $\operatorname{GL}(d, E) \leq X$, then the latter two cases go away (as far as I can tell) and you get the classification I wanted. On the other hand, because of the restriction $d \geq 2$, it says nothing about overgroups of $\operatorname{GL}(1, E)$.

Glancing at the introduction of Koibaev [3], it seems the $d=1$ case is much more complicated indeed, and I understand lamentably little of it.

---

Since I mentioned the real case explicitly, let me complete the story in that case. Since $\mathbf{C}$ is the only finite extension of $\mathbf{R}$ it suffices to classify overgroups $G$ of $\operatorname{GL}(1,\mathbf{C})$ in $\operatorname{GL}(2,\mathbf{R})$. The Lie algebra $\mathfrak{g}$ of $G$ lies between $\mathfrak{gl}_1(\mathbf{C})$ and $\mathfrak{gl}_2(\mathbf{R})$, and it is not hard to check that therefore $\mathfrak{g} = \mathfrak{gl}_1(\mathbf{C})$ or $\mathfrak{g}=\mathfrak{gl}_2(\mathbf{R})$. Let $G_0$ be the identity component of $G$. In the former case we have $G_0 = \operatorname{GL}(1,\mathbf{C})$, so $G$ normalizes $\operatorname{GL}(1,\mathbf{C})$, so $G$ is contained in $\operatorname{\Gamma L}(1,\mathbf{C}) \cong \operatorname{GL}(1, \mathbf{C}) \ltimes C_2$. In the latter case $G_0 = \operatorname{GL}^+(2,\mathbf{R})$, so $G \geq \operatorname{GL}^+(2,\mathbf{R})$.

This result is contained in Hou--Li [1] (though the above proof seems simpler). They also look at the quaternion case.

[1] Hou, Xin; Li, Shangzhi, On the overgroups of $\mathrm{SL}(1,K)$ in $\mathrm{GL}(r,F)$, Ital. J. Pure Appl. Math. 36, 871-878 (2016). ZBL1369.20045.

[2] Kantor, William M., Linear groups containing a Singer cycle, J. Algebra 62, 232-234 (1980). ZBL0429.20004.

[3] Koĭbaev, V. A.; Shilov, A. V., Transvections in the overgroups of a nonsplit torus., Vladikavkaz. Mat. Zh. 11, No. 4, 22-31 (2009). ZBL1324.20030.

[4] Li, Shangzhi, Overgroups in GL(nr,F) of certain subgroups of SL(n,K). I, J. Algebra 125, No. 1, 215-235 (1989). ZBL0676.20030.