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I would like to know about the literature concerning the group of outer automorphisms of the Lie algebra $\mathfrak{sl}(2,R)$. This question is addressed in different places in a contradictory way. In certain works, e.g.

M.A. Farinati and A.P Jancsa, Three dimensional real Lie bialgebras, Revista de la union matematica argentina Vol. 56, No. 1, 2015, Pages 27–62, 2015,

it is implicitly claimed that the aforesaid group is trivial and all automorphisms of $\mathfrak{sl}(2,R)$ are inner. By googling the question, I also found several sources claiming that the group of inner automorphisms of $\mathfrak{sl}(2,R)$ is PSL(2,R) (see https://groupprops.subwiki.org/wiki/Special_linear_group:SL(2,R)), and the outer automorphisms are given by PGL(2,R) (see Outer automorphisms of simple Lie Algebras) which are obviously different.

Similarly, I would like to know about the structure of the group of outer automorphisms of the Lie algebra $\mathfrak{su}(2)$

Thank you in advance for your comments.

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    $\begingroup$ Maybe you first want to define what an inner automorphism is. It can be defined as an automorphism in the group generated by the $\exp(\mathrm{ad}(x))$, where $x$ ranges over ad-nilpotent elements. But in the case of $SU(2)$ it makes a problem: only $x$ is such an element. The question you link also does not bother with giving a definition. $\endgroup$
    – YCor
    Commented Aug 24, 2017 at 15:10
  • $\begingroup$ I'm sorry to introduce this mistake! we had in mind the complex case, where outer automorphisms are controled by automorphisms of the Dynkin Diagram, that is trivial in the sl2 case, but over the complex.. $\endgroup$
    – Marco Farinati
    Commented Mar 1, 2018 at 18:34

1 Answer 1

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It is clear that $PGL(2,\mathbb{R})$ acts as automorphisms. It is easy to check that the reflections act in a manner unlike any positive determinant matrices. Hence the automorphism group is larger than $PSL(2,\mathbb{R})$. Since we know the answer over $\mathbb{C}$ (as in Fulton and Harris, Representation Theory, p. 498), by complexification, we know that all automorphisms arise from conjugation by some matrices. We can easily check that complex matrices give us real automorphisms only when they are real up to a constant scaling. So we see that the automorphism group of $\mathfrak{sl}(2,\mathbb{R})$ is $PGL(2,\mathbb{R})$.

For $\mathfrak{su}(2)$, any automorphism complexifies to an automorphism of $\mathfrak{sl}(2,\mathbb{C})$, so arises by conjugation from a complex matrix $g$, again from Fulton and Harris. To get conjugation by $g$ to preserve the real subspace $\mathfrak{su}(2)$, we need $gAg^{-1}$ to be special unitary for any special unitary $A$. Plug this in and check that this forces $g^*g$ to commute with all such $A$, so by Schur's lemma, $g^* g=\lambda I$ for some complex number $\lambda$. Take determinant to find that $\lambda=\pm 1$. Replace $g$ by $ig$ if needed to get $\lambda=1$, so $g \in SU(2)$. Check that $g$ acts trivially if and only if $g=-I$ to see that the automorphism group of $\mathfrak{su}(2)$ is $SO(3)=SU(2)/\pm 1$.

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    $\begingroup$ I think it might be fair to say that the lesson you are imparting is that it's easier to understand automorphisms of complex reductive Lie algebras that preserve real structures than to understand automorphisms of real Lie algebras directly. $\endgroup$
    – LSpice
    Commented Aug 24, 2017 at 14:09
  • $\begingroup$ @Ben McKay: Does this also imply that the automorphisms of the Lie group $\text{SL}(2,\mathbb{R})$ are precisely the ones given by conjugation with elements of $\text{GL}(2,\mathbb{R})$? $\endgroup$
    – Abenthy
    Commented Feb 1, 2019 at 9:55
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    $\begingroup$ @Abenthy: yes, because $SL(2,\mathbb{R})$ is connected. Its automorphisms clearly give automorphisms of its Lie algebra, and these arise from $\mathbb{P}GL(2,\mathbb{R})$. Each automorphism of $SL(2,\mathbb{R})$ commutes with the exponential map, so is determined near the identity element by the automorphism of the Lie algebra. By analyticity, and connectedness of $SL(2,\mathbb{R})$, the automorphism is determined uniquely, and so must be that element of $\mathbb{P}GL(2,\mathbb{R})$. $\endgroup$
    – Ben McKay
    Commented Feb 1, 2019 at 11:42

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