# Automorphisms of $SO(n,\mathbb R)$

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.

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

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