10
$\begingroup$

Is the only continuous outer automorphism of $\operatorname{GL}(n, \mathbb{R})$ the transpose inverse map $g \mapsto (g^\intercal)^{-1}$? If not, what other continuous outer automorphisms are there?

$\endgroup$
7
  • 7
    $\begingroup$ The Dynkin diagram has unique nontrivial automorphism, the one corresponding to the inverse transpose. This implies the result you are after. What book on Lie groups are you reading? Automorphisms of simple Lie algebras (in terms of Dynkin diagrams) are discussed in Bourbaki, but also in other places. $\endgroup$ Jan 27 at 4:26
  • 5
    $\begingroup$ Automorphisms of the group $GL_n(K)$ for an arbitrary (noncommutative) field $K$ have been described for quite a long time. See for example Jean Dieudonne's "On the automorphisms of the classical groups", Memoirs of the AMS, 1951 (Section II). $\endgroup$
    – kabenyuk
    Jan 27 at 9:11
  • 13
    $\begingroup$ I can't believe that people are still voting to close this post. It's been closed once and then reopened, so isn't it time to leave it alone?! $\endgroup$
    – Derek Holt
    Jan 28 at 13:20
  • 4
    $\begingroup$ @DerekHolt It's a PSQ, is it not? Shouldn't all PSQ's be closed? $\endgroup$ Jan 31 at 20:31
  • 11
    $\begingroup$ @MikeEarnest: why? $\endgroup$ Feb 17 at 20:38

1 Answer 1

8
$\begingroup$

All the automorphisms of $\operatorname{GL}(n,\mathbb{R})$ are known, and from this description you will get the answer to your question. There are others besides the transpose inverse map, since inner automorphisms are continuous.

There are three basic ones we can define:

Inner automorphisms: $P_x: \operatorname{GL}(n,\mathbb{R}) \rightarrow \operatorname{GL}(n,\mathbb{R})$, defined by $g \mapsto xgx^{-1}$, where $x \in \operatorname{GL}(n,\mathbb{R})$.

Field automorphisms: $Q_\psi: \operatorname{GL}(n,\mathbb{R}) \rightarrow \operatorname{GL}(n,\mathbb{R})$, defined by $(g_{ij}) \mapsto (\psi(g_{ij}))$, where $\psi: \mathbb{R} \rightarrow \mathbb{R}$ is a field automorphism.

The transpose-inverse automorphism $\varphi: \operatorname{GL}(n,\mathbb{R}) \rightarrow \operatorname{GL}(n,\mathbb{R})$ defined by $g \mapsto g^{-T}$.

For example looking at the book of Dieudonné mentioned in the comments, we see that every automorphism of $\operatorname{GL}(n,\mathbb{R})$ is of the form $P_x \circ Q_{\psi} \circ \varphi^a$ for some $x \in \operatorname{GL}(n,\mathbb{R})$, field automorphism $\psi$, and $a \in \{0,1\}$.

(Note that there is nothing special about $\mathbb{R}$, and this same thing is true for $\operatorname{GL}(n,F)$ with $F$ a field of characteristic zero.)

Anyway, the only continuous field automorphism of $\mathbb{R}$ is the identity, so every continuous automorphism of $\operatorname{GL}(n,\mathbb{R})$ is of the form $P_x \circ \varphi^a$. The outer ones are of course those with $a = 1$.

$\endgroup$
2
  • 3
    $\begingroup$ Remark: Every field automorphism of $\mathbb{R}$ is continuous. So Aut($\mathbb{R})=1$. $\endgroup$ Feb 18 at 5:34
  • 2
    $\begingroup$ @BrauerSuzuki: Very good point. So every automorphism of $GL(n,\mathbb{R})$ should be continuous. $\endgroup$ Feb 18 at 5:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .