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.

$\endgroup$
  • $\begingroup$ 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 $f_C(A)$ ? $\endgroup$ – Sasha Aug 26 '11 at 14:35
  • $\begingroup$ Sorry, I have just corrected. $\endgroup$ – Richard Aug 26 '11 at 14:41
5
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Aug 26 '11 at 15:23
  • 2
    $\begingroup$ 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 $SO(8)$. $\endgroup$ – Jason DeVito Aug 26 '11 at 17:18
  • 1
    $\begingroup$ @Jason Good points! $\endgroup$ – Sasha Aug 26 '11 at 17:31
  • 1
    $\begingroup$ Yet some questions: 1.What is the conjugation by reflections? 2. In case $SO(8)$ are there yet another automorphisms besides $f_C$ ? $\endgroup$ – Richard Aug 26 '11 at 17:34
  • 2
    $\begingroup$ 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 $SO(8)$. $\endgroup$ – Sasha Aug 26 '11 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.