5
$\begingroup$

Sorry, I'm not a specialist, I want to ask about automorphisms of the group $SO(n,\mathbb{R})$ ($\mathbb{R}$ - field of reals). It is easy that a function of the form $f_C(A)=CAC^{-1}$ for $A \in SO(n, \mathbb{R})$, where $C\in O(n,\mathbb{R})$, is an automorphism.

But, is it true that each automorphism of $SO(n,\mathbb{R})$ is of the form $f_C$ with $C \in O(n,\mathbb{R})$ or maybe with $C \in SO(n,\mathbb{R})$?

Thanks.

  • 0
    I improved formatting, but it seems you mean to say $C \in O(n, \mathbb{R})$. If so please correct. Also, your question does not read right. Are you asking if every automorphism of $SO(n, \mathbb{R})$ is of the form of $f_C(A)$ ?2011-08-26
  • 0
    Sorry, I have just corrected.2011-08-26

1 Answers 1

5

See Outer automorphism group wiki page, the section on real Lie groups. It says that outer automorphism groups are symmetries of Dynkin diagram.

From this it follows that for $SO(2n-1, \mathbb{R})$, i.e. series $B_n$, all automorphisms are inner. For $SO(2n, \mathbb{R})$ there is order 2 outer automorphism which indeed coincides with conjugation by reflections.

So it follows that the answer to your question is in affirmative, and $C \in SO(n, \mathbb{R})$ for odd $n$, and in $O(n, \mathbb{R})$ for even.

  • 4
    This sounds a bit magical, but it is not too complicated, in fact. An automorphism $f$ of the group $G$ maps a maximal torus to a maximal torus; since all maximal tori are conjugate, up to composing the automorphism with a conjugation, we can assume it in fact *fixes* a maximal torus. Now one can look at what the action of $f$ on the Lie algebra: by our little adjustment, it fixes a Cartan subalgebra, and then it has to preserve the whole structure one constructs from it---in particular, $f$ induces an automorphism of the Dynkin diagram.2011-08-26
  • 0
    @Mariano Why don't you write it up as an answer?2011-08-26
  • 2
    2 Points. Symmetries of the Dynkin Diagram give rise to outer automorphisms of the Lie *algebra*. In the case where the Lie group is simply connected, such automorphisms are in 1-1 correspondance with group automorphisms. But when the Lie group is *not* simply connected all bets are off. Since $\pi_1(SO(n))$ is of of order 2 (for $n>2$), some care must be taken. The second point is that the Dynkin Diagram of $SO(8)$ actually has more symmetry, leading to the so called Triality automorphism. This automorphism, in particular, maps $Spin(8)$ to itself and does *not* descend to $SO(8)$.2011-08-26
  • 1
    @Jason Good points!2011-08-26
  • 1
    Yet some questions: 1.What is the conjugation by reflections? 2. In case $SO(8)$ are there yet another automorphisms besides $f_C$ ?2011-08-26
  • 2
    Conjugation by reflection is $f_C$ , where $C \in O(n, \mathbb{R})$ such that $\det C = -1$. In case of $SO(8)$ if I read Jason's answer correctly, the additional symmetry would be manifest on a universal covering of $SO(8)$, i.e. $Spin(8)$, but is not manifest on $SO(8)$.2011-08-26
  • 1
    To continue what @Sashha wrote, SO(8) does not admit an automorphism corresponding to the "triality" automorphism of the D_4 Dynkin diagram. The group Spin(8) does, but it is quite magical. In particular, the usual construction of Spin(8) embeds it as a subgroup of the Clifford algebra Cliff(8). The triality automorphism *does not* extend to a linear automorphism of Cliff(8). Googling for "triality" will find descriptions.2016-03-15