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

• – Q. Zhang Sep 28 at 2:30
• which answer do you suggest to concerns the explicit map of $S_3$? and inner map? thanks! – annie marie heart Sep 28 at 2:32
• An automorphism permits the three extremal simple weights; that's the map from the automorphism group (hence from the outer automorphism group, since it's trivial) to $\operatorname S_3$. An automorphism also carries a so called pinning (Borel subgroup, torus, set of simple root vectors) to another pinning, which is conjugate by a unique inner automorphism to the first; that's the map to the inner automorphism group. What does "Given the parametrization of $\operatorname{SO}(n; \mathbb R)$ mean", and what do you want to map to itself? – LSpice Sep 28 at 3:34
• The assertion that $\mathrm{Out}(\mathrm{SO}(8))$ has order 6 is false. It has order 2, and the full automorphism group is $\mathrm{PO}(8)$. The claim about $S_3$ is correct with $\mathrm{PSO}(8)$ instead. – YCor Sep 28 at 6:30
• I'm not sure that I understand what you are looking for regarding inner automorphisms. The inner automorphisms all come from conjugation via a matrix of ${\rm SO}(8,\mathbb{R})$ (and clearly the central involution acts trivially), so how much more explicit do you want to be? – Geoff Robinson Sep 29 at 8:38

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 $${\frak{so}}(8)$$.