Explicit automorphism map of ${\rm Spin}(8;\mathbb{R})$, ${\rm SO}(8;\mathbb{R})$, ${\rm PSO}(8;\mathbb{R})$ $\DeclareMathOperator{\SO}{\mathrm{SO}}\DeclareMathOperator{\Spin}{\mathrm{Spin}}\DeclareMathOperator{\Inn}{\mathrm{Inn}}\DeclareMathOperator{\Out}{\mathrm{Out}}\DeclareMathOperator{\Aut}{\mathrm{Aut}}$ How do we construct a precise map of inner + outer automorphism of special orthogonal group $\SO(n;\mathbb{R})$?

*

*$d=2$; We can look at $\SO(2;\mathbb{R})=U(1)$ which is abelian, and we know the inner
$$
\Inn(\SO(2;\mathbb{R}))=\SO(2;\mathbb{R})/Z(\SO(2;\mathbb{R}))=1
$$
$$
\Out(\SO(2;\mathbb{R}))=\mathbb{Z}/2
$$
The total $\Aut(\SO(2;\mathbb{R}))=\Inn(\SO(2;\mathbb{R})) \rtimes \Out(\SO(2;\mathbb{R}))=\mathbb{Z}/2$
We have no $\Inn(\SO(2;\mathbb{R}))$ except the identity map.
I believe that we can get the $\Out(\SO(2;\mathbb{R}))=\mathbb{Z}/2$ by flipping $t \to -t$ in
$$
U(1)=\{\exp(i t) | t \in [0, 2 \pi)\}
\to
\{\exp(-i t) | t \in [0, 2 \pi)\}.
$$
I wish to see explicit answer like the above for my following questions ---


*other $n$ but $n\neq 2,8$ is discussed in MSE with answer still pending.


*for $n=8$

Question 1: How do we construct the inner automorphism map explicitly  (if my result is correct?)? Let us consider $\Spin(8;\mathbb{R})$, $\SO(8;\mathbb{R})$, $\SO(8;\mathbb{R})/(\mathbb{Z}/2)$.

for $n=8$
$$
\Inn(\Spin(n;\mathbb{R}))=\Spin(n;\mathbb{R})/Z(\Spin(n;\mathbb{R})) = 
\SO(8;\mathbb{R})/\mathbb{Z}/2 
$$
$$
\Inn(\SO(n;\mathbb{R})/\mathbb{Z}/2)=(\SO(n;\mathbb{R})/\mathbb{Z}/2)/Z(\SO(n;\mathbb{R})/(\mathbb{Z}/2)) = 
\SO(8;\mathbb{R})/\mathbb{Z}/2 
$$
$$
\Inn(\SO(n;\mathbb{R}))=\SO(n;\mathbb{R})/Z(\SO(n;\mathbb{R})) = 
\SO(8;\mathbb{R})/\mathbb{Z}/2 
$$

Question 2: How do we construct the outer automorphism map explicitly
$$
\Out(Spin(8;\mathbb{R}))=S_3
$$
$$
\Out(\SO(8;\mathbb{R}))=\mathbb{Z}/2 
$$
$$
\Out(\SO(8;\mathbb{R})/\mathbb{Z}/2 )=S_3
$$
Given the parametrization of  $\SO(n;\mathbb{R})$ how to map to itself via the $\Out$ map?


Question 3: How do we construct the total automorphism map explicitly

$$
\Aut(\Spin(8;\mathbb{R}))=\Inn(\Spin(8;\mathbb{R})) \rtimes \Out(\Spin(8;\mathbb{R}))
=(\SO(8;\mathbb{R})/\mathbb{Z}/2 ) \rtimes  S_3 ?
$$
$$
\Aut(\SO(8;\mathbb{R}))=\Inn(\SO(8;\mathbb{R})) \rtimes \Out(\SO(8;\mathbb{R}))
=(\SO(8;\mathbb{R})/\mathbb{Z}/2 ) \rtimes  \mathbb{Z}/2 ?
$$
$$
\Aut(\SO(8;\mathbb{R})/\mathbb{Z}/2)=\Inn(\SO(8;\mathbb{R})/\mathbb{Z}/2) \rtimes \Out(\SO(8;\mathbb{R})/\mathbb{Z}/2)
=(\SO(8;\mathbb{R})/\mathbb{Z}/2 ) \rtimes  S_3 ?
$$
Given the parametrization of   $\Spin(8;\mathbb{R})$, $\SO(8;\mathbb{R})$, $\SO(8;\mathbb{R})/(\mathbb{Z}/2)$, how to map to itself via the $\Aut$ map?
P.S. Possible useful link but with not explicit (not enough) constructions in Automorphism group of real orthogonal Lie groups
 A: If you just want an explicit realization the of outer automorphisms of $\mathrm{Spin}(8)$, here is one, assuming that you know about the algebra of octonions $\mathbb{O}$, the unique $8$-dimensional (and hence nonassociative) inner product algebra over $\mathbb{R}$ with positive definite inner product.
The subgroup $\mathrm{Spin}(8)\subset{\mathrm{SO}(8)}^3$ is defined as the set of triples $g = (g_1,g_2,g_3)$ that satisfy
$$
\mathrm{Re}\bigl(g_1(a_1)g_2(a_2)g_3(a_3)\bigr) = \mathrm{Re}(a_1a_2a_3)
$$
for all $a_i\in\mathbb{O}$.  (Here, $\mathrm{Re}(a_1a_2a_3) = (a_1a_2a_3)\cdot\mathbf{1}$, where $\mathbf{1}\in\mathbb{O}$ is the multiplicative unit.)  The group of outer automorphisms of $\mathrm{Spin}(8)$ is generated the element $\beta:\mathrm{Spin}(8)\to \mathrm{Spin}(8)$ of order $3$ defined by
$$
\beta(g_1,g_2,g_3) = (g_2,g_3,g_1)
$$
and the element $\alpha:\mathrm{Spin}(8)\to \mathrm{Spin}(8)$ of order $2$ defined by
$$
\alpha(g_1,g_2,g_3) = \bigl(\ cg_2c,\ cg_1c,\ cg_3c\ \bigr)
$$
where $c:\mathbb{O}\to\mathbb{O}$ is octonionic conjugation, i.e., $c(x) = 2(x{\cdot}\mathbf{1})\,\mathbf{1} - x$.  (Note that $c$ belongs to $\mathrm{O}(8)$ but not $\mathrm{SO}(8)$.)
The facts that $\mathrm{Spin}(8)$, as defined as above is a subgroup of $\mathrm{SO}(8)^3$ and that each of the projections $\pi_i:\mathrm{Spin}(8)\to\mathrm{SO}(8)$ defined by $\pi_i(g_1,g_2,g_3) = g_i$ is a nontrivial double cover of $\mathrm{SO}(8)$ and that $\alpha$ and $\beta$ are outer automorphisms of $\mathrm{Spin}(8)$ follow from basic facts about the algebra $\mathbb{O}$.
If the above description is not explicit enough, or a description that does not mention the octonions is preferred, here is Cartan's description at the level of the Lie algebra ${\frak{so}}(8)$, drawn from his paper Le principe de dualité et la théorie des groupes simples et semi-simples (Bull. Sc. Math. 49 (1925), 361–374:
Let indices run from $0$ to $7$ with the understanding that, if a formula gives an index greater than $7$, one subtracts $7$.  (Thus, $8=1$, but $7\not=0$.)  Then an element $a\in {\frak{so}}(8)$ is a skew-symmetric matrix with entries $a = (a_{i,j})$ where $a_{i,j}=-a_{j,i}$.  There are essentially 28 distinct entries, and these break up into $7$ groups of $4$:
$$
b_i = \begin{pmatrix}a_{0,i}\\a_{i+1,i+5}\\a_{i+4,i+6}\\ a_{i+2,i+3}\\\end{pmatrix},\qquad i=1,\ldots,7
$$
Let
$$
H=\frac12\,\begin{pmatrix} -1&-1&-1&-1\\ \phantom{-}1&\phantom{-}1&-1&-1\\ \phantom{-}1&-1&\phantom{-}1&-1\\ \phantom{-}1&-1&-1&\phantom{-}1\\\end{pmatrix}\quad\text{and}\quad
K = \begin{pmatrix}-1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix}.
$$
Note that $H^3=K^2=I_4$ and $KHK = H^2$.
Consider the linear mappings $h:{\frak{so}}(8)\to {\frak{so}}(8)$ and $k:{\frak{so}}(8)\to {\frak{so}}(8)$ induced by the transformations
$$
b_i\mapsto Hb_i\,\quad\text{and}\quad b_i\mapsto Kb_i,
\quad i = 1,\ldots,7.
$$
Then $h$ and $k$ are automorphisms of ${\frak{so}}(8)$ that generate a group of order $6$ that maps isomorphically onto the group of outer automorphisms of ${\frak{so}}(8)$.
