# Explicit expressions of inner / outer automorphism of special orthogonal group SO(N)

I am aware the general statement on inner / outer automorphism of special orthogonal group SO($N$). Here I am trying to show them very explicitly.

Consider SO(3) Lie algebra generators: $$[X_i,X_j]=i \epsilon^{ijk} X_k\\ X_1=i {\begin{pmatrix} 0& 0&0\\ 0& 0&-1\\0&1&0 \end{pmatrix}}, X_2=i {\begin{pmatrix} 0& 0&1\\ 0& 0&0\\-1&0&0 \end{pmatrix}}, X_3=i {\begin{pmatrix} 0& -1&0\\ 1& 0&0\\0&0&0 \end{pmatrix}}$$

Note $X_j^\dagger=X_j$.

Write the SO(3) Lie group elements as $$g=e^{i \theta \alpha_j X_j}$$ where $g^{-1}=g^\dagger$. And $g^{-1}g=g^\dagger g=1$. $g=e^{i \theta \alpha_j X_j}$ is the unitary Rep of SO(3) Lie group. We can choose $\sum_{j=1}^3|\alpha_j|^2=1$ so let to $\vec \alpha_j$ to be a unit vector on $S^2$.

1. For SO(3) Lie group, we have an order 2 inner automorphism $\mathbb{Z}_2$, but no outer automorphisms. (Look at the symmetries of Dynkin diagram $B_n$.) How do we find the explicit $k$ such that $$k g_{\text{SO(3)}} k^{-1} = g_{\text{SO(3)}}^{'}?$$ which is the inner automorphism arisen by the conjugation of $k$?

Similarly, I believe that we can write SO(N) Lie algebra generators: $$[X_i,X_j]=i \epsilon^{ijk} X_k,$$ with $i,j,k \in \{1,2,..., N(N-1)/2 \}$, such that all the SO(n) Lie group elements as $$g_{\text{SO(N)}}=e^{i \theta \alpha_j X_j},$$ I suppose the $\theta$ has a 2$\pi$ periodicty? Then

1. For SO$(2n+1)$=SO$(2n+1, \mathbb{R})$ Lie group, when $2n+1$ is odd and $n \geq 1$, we have an order 2 inner automorphism $\mathbb{Z}_2$, but no outer automorphisms. (Look at the symmetries of Dynkin diagram $B_n$.) How do we find the explicit $k$ such that $$k g_{\text{SO(2n+1)}} k^{-1} = g_{\text{SO(2n+1)}}^{'}?$$ which is the inner automorphism arisen by the conjugation of $k$?

2. For SO$(2n)$=SO$(2n, \mathbb{R})$ Lie group, when $2n$ is even and $n > 1$ (but not $n=1$?), we have an order 2 outer automorphism $\mathbb{Z}_2$, but no outer automorphisms. (Look at the symmetries of Dynkin diagram $D_n$.) How do we find the explicit map $$g_{\text{SO(2n+1)}} \to g_{\text{SO(2n+1)}}^{'}?$$ which explicit map is the out automorphism?

• You write $k g_{\text{SO(3)}} k^{-1} = g_{\text{SO(3)}}^{-1}$ etc. Does this mean you are seeking $k\in \text{SO}(3)$ with $kgk^{-1}=g^{-1}$ for all $g\in \text{ SO}(3)$? There is no such $k$, as SO$(3)$ is non-Abelian so $g\mapsto g^{-1}$ is not an automorphism. – Lord Shark the Unknown Aug 17 '18 at 16:07
• $\text{SO}(3)$ is the special orthogonal group, not the special unitary group. – Lord Shark the Unknown Aug 17 '18 at 17:07
• thanks, so many typos in my questions..., I correct it to $$g_{\text{SO(2n+1)}} \to g_{\text{SO(2n+1)}}^{'}.$$ – wonderich Aug 17 '18 at 17:32