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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?

Joe
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Craig
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    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. – Moishe Kohan Jan 27 '24 at 04:26
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    Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Shaun Jan 27 '24 at 08:21
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    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). – kabenyuk Jan 27 '24 at 09:11
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    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?! – Derek Holt Jan 28 '24 at 13:20
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    @DerekHolt It's a PSQ, is it not? Shouldn't all PSQ's be closed? – Mike Earnest Jan 31 '24 at 20:31
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    @MikeEarnest: why? – Martin Argerami Feb 17 '24 at 20:38
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    @MartinArgerami Actually, I cannot think of a reason. I was just echoing the zeitgeist of MSE (or so I thought). I will vote to leave this open, if it pops up in the queue later. – Mike Earnest Feb 18 '24 at 19:34

1 Answers1

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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 four 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.

Multiplication by a linear character: $R_{\chi}: \operatorname{GL}(n,\mathbb{R}) \rightarrow \operatorname{GL}(n,\mathbb{R})$ defined by $g \mapsto \chi(g)g$, where $\chi: \operatorname{GL}(n,\mathbb{R}) \rightarrow \mathbb{R}^*$ is a homomorphism into the multiplicative group.

The transpose-inverse automorphism $\varphi: \operatorname{GL}(n,\mathbb{R}) \rightarrow \operatorname{GL}(n,\mathbb{R})$ defined by $g \mapsto (g^{-1})^T$.

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 $R_{\chi} \circ P_x \circ Q_{\psi} \circ \varphi^a$ for some $x \in \operatorname{GL}(n,\mathbb{R})$, field automorphism $\psi$, homomorphism $\chi: \operatorname{GL}(n,\mathbb{R}) \rightarrow \mathbb{R}^*$, 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 $R_{\chi} \circ P_x \circ \varphi^a$. The outer ones are those which have $a = 1$, or $\chi$ nontrivial.

testaccount
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    Remark: Every field automorphism of $\mathbb{R}$ is continuous. So Aut($\mathbb{R})=1$. – Brauer Suzuki Feb 18 '24 at 05:34
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    @BrauerSuzuki: Very good point. So every automorphism of $GL(n,\mathbb{R})$ should be continuous. – testaccount Feb 18 '24 at 05:54
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    Actually, this answer is not correct. For example, there is the continuous outer automorphism $\phi(g) = |!\det(g)|, g$, which is not of the claimed form. – Robert Bryant Sep 05 '24 at 20:46
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    @RobertBryant: Thanks, I think you are right, what is missing is are the automorphisms $\phi(g) = \chi(g) g$, where $\chi$ is a homomorphism $\chi: GL(n,\mathbb{R}) \rightarrow \mathbb{R}^*$. To be clear this is just my mistake and the result in stated correctly in the book by Dieudonné. – testaccount Sep 06 '24 at 05:35
  • @testaccount: You should also amend the last sentence of your answer since the $R_\chi$ are also outer automorphisms (when $\chi$ is not trivial). There's also a misprint in your definition of $\varphi$; it's not $g\mapsto g^{-T}$, but $g\mapsto (g^{-1})^T$. – Robert Bryant Sep 06 '24 at 08:43
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    @RobertBryant: Thanks again - but isn't $g^{-T}$ a common notation for inverse transpose? – testaccount Sep 06 '24 at 08:55
  • @testaccount: It could be. I've never seen it before, though. – Robert Bryant Sep 06 '24 at 09:07
  • @testaccount: Unfortunately, the answer is still not quite complete because not all of the $R_\chi$ (which are homomorphisms) are automorphisms; they may not be injective. For example, when $n$ is odd, the homomorphisms $g\mapsto \det(g),g$ and $g\mapsto (\det(g))^{-1/n},g$ are not injective. (I forgot about this, but, in fact, I had proposed $g\mapsto |\det(g)|,g$ in my earlier comment instead of $g\mapsto \det(g),g$ precisely because I wanted the map to be bijective even when $n$ is odd.) – Robert Bryant Sep 09 '24 at 08:42
  • @RobertBryant Is the problem in this case that, say for $O(n)$ with odd $n$, this homomorphism maps every element to the subgroup $SO(n)$? – Craig Sep 10 '24 at 16:06
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    @Craig: In particular, yes; it's not an automorphism if it's not injective and surjective. – Robert Bryant Sep 10 '24 at 16:37