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$\DeclareMathOperator{\SO}{\mathrm{SO}}\DeclareMathOperator{\Spin}{\mathrm{Spin}}\DeclareMathOperator{\Inn}{\mathrm{Inn}}\DeclareMathOperator{\Out}{\mathrm{Out}}\DeclareMathOperator{\Aut}{\mathrm{Aut}}$ 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

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  • $\begingroup$ See mathoverflow.net/questions/116666/triality-of-spin8?rq=1 $\endgroup$ – Q. Zhang Sep 28 at 2:30
  • $\begingroup$ which answer do you suggest to concerns the explicit map of $S_3$? and inner map? thanks! $\endgroup$ – annie marie heart Sep 28 at 2:32
  • $\begingroup$ 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? $\endgroup$ – LSpice Sep 28 at 3:34
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    $\begingroup$ 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. $\endgroup$ – YCor Sep 28 at 6:30
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    $\begingroup$ 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? $\endgroup$ – Geoff Robinson Sep 29 at 8:38
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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 ${\frak{so}}(8)$.

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