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?
2026-03-25 12:53:07.1774443187
What are the continuous outer automorphisms of the general linear group?
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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 $a = 1$.