$ D_n = Spin(2n) $, $ n \geq 3 $, is a simply connected compact simple group of dimension $ \frac{2n(2n-1)}{2} $. For $ n \neq 4 $ the outer automorphism group of $ Spin(2n) $ is cyclic 2. For $ n=4 $ the outer automorphism group of $ Spin(8) $ is $ S_3 $. This exceptionally large automorphism group is known as triality.

Is it true that $$ Spin(2n) \rtimes Out(Spin(2n)) $$ is always a subgroup of $ Spin(2n+1) $?

Update: Jason points out that there is a natural $ O(2n) $ subgroup of $ SO(2n+1) $. Recall that $ O(2n+1)= SO(2n+1) \times \{ \pm I \} \times $ is a direct product while $ O(2n)= SO(2n) \rtimes <diag(-1,1,1,1,1,\dots,1)> $ is a semi direct product with $ diag(-1,1,1,1,1,\dots,1) $ (a hyperplane reflection with $ 2n-1 $ dimensional fixed space and one $ -1 $ eigenvalue) inducing the unique nonidentity automorphism of $ SO(2n) $. So we can take this $$ SO(2n) \rtimes Aut(SO(2n)) =O(2n) \hookrightarrow SO(2n+1) $$ And lift it to get $$ Spin(2n) \rtimes Aut(Spin(2n)) \hookrightarrow Spin(2n+1) $$ The only exception is for $ n=4 $ when $ Aut(SO(8))=C_2 \neq S_3 = Aut(Spin(8)) $. In that case, the procedure above yields $$ Spin(8) \rtimes C_2 \hookrightarrow Spin(9) $$ In other words, a semi direct product of $ Spin(8) $ with a cyclic $ 2 $ subgroup of the full automorphism group of $ Spin(8) $. So this answer all cases of the title question except for:

Is $ Spin(8) \rtimes Aut(Spin(8)) $ a subgroup of $ Spin(9) $?

Note that for small $ n $ this story still holds for $ n=2 $. We have $$ SO(4) \rtimes Out(SO(4))=O(4) \hookrightarrow SO(5) $$ lifts to $$ (SU_2 \times SU_2) \rtimes <SWAP>=O(4) \hookrightarrow SO(5) $$ Famously $ SU_2 $ has no outer automorphisms since complex conjugation corresponds to the inner automorphism of conjugation by $ \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} $. So the only outer automorphism of $ Spin(4)=SU_2 \times SU_2 $ is the order 2 automorphism swapping to the two simple factors.

For $ n=1 $ I'm not really sure what to say. We have $$ O(2) \hookrightarrow SO(3) $$ which lifts to $$ O^*(2) \hookrightarrow Spin(3)=SU_2 $$ but at this point there is no more relationship with outer automorphism groups. (Also common misconception that $ O_2^* $ is isomorphic to $ O_2 $. This is not the case. You can see the difference by looking at elements of order 2.)

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    $\begingroup$ What do you get by lifting (a cover of) the usual $O(n)$ in $SO(2n+1)$ to $Spin(2n+1)$? That may give the answer for $n\neq 4$. $\endgroup$ Commented Feb 20, 2023 at 23:15
  • $\begingroup$ @JasonDeVito Ok that covers the other cases. I updated my question. What do you think about the $ n=4 $ case? Is $ Spin(8) \rtimes Aut(Spin(8)) $ a subgroup of $ Spin(9) $? $\endgroup$ Commented Feb 21, 2023 at 16:02

1 Answer 1


The $n=4$ works differently: $Spin(8) \rtimes Out(Spin(8))$ does not embed into $Spin(9)$.

To see this, first let's classify embeddings of $Spin(8)$ into $Spin(9)$, and then we'll worry about extending them to the other components.

Proposition 1: Up to conjugacy, the only subgroup of $Spin(9)$ which is isomorphic to $Spin(8)$ is the usual one, obtained from lifting the standard $SO(8)\subseteq SO(9)$.

Proof: By projection to $SO(9)$, a subgroup $H\subseteq Spin(9)$ can be thought of as a $9$-dimensional real representation of $H$. A result of Mal'Cev (I will dig up the reference later) implies that if two real representations $H,H'\subseteq SO(9)$ are equivalent (i.e., they are conjugate in $Gl_9(\mathbb{R})$) then they are conjugate in $SO(9)$. Lifting to $Spin(9)$, it follows that two subgroups $H,H'\subseteq Spin(9)$ are conjugate if and only if the corresponding $9$-dim real reps of $H$ and $H'$ are equivalent.

So, we need to understand real $9$-dim reps of $Spin(8)$. Representation theory tells us that $Spin(8)$ has precisely $3$ non-trivial $9$-dimensional real representations. They are the standard rep or either of the half spin reps, summed with a $1$-dim trivial rep.

The half-spin reps are defined by non-trivial maps $Spin(8)\rightarrow SO(8)$. (We don't need it, but these maps are the compositions $Spin(8)\rightarrow Spin(8)\rightarrow SO(8)$ where the first map is a traility automorphism and the second map is the usual projection.) Since $Spin(8)$ is simple, such a map must be a covering map. In particular, the image is all of $SO(8)$. It now follows that regardless of rep we use, the image of $Spin(8)$ in $Spin(9)$ is, up to conjugacy, the usual one. $\square$

Let $N$ denote the normalizer of (the usual embedding of) $Spin(2n)$ in $Spin(2n+1)$ and let $N_0$ denote the identity component.

Proposition 2: We have $N_0 = Spin(2n)$.

Proof: Consider the chain of subgroups $Spin(2n)\subseteq N_0\subseteq Spin(2n+1)$. This gives a homogeneous fibration $$N_0/Spin(2n)\rightarrow Spin(2n+1)/Spin(2n)\rightarrow Spin(2n+1)/N.$$ The total space is $Spin(2n+1)/Spin(2n) = SO(2n+1)/SO(2n) = S^{2n}$. Moreover, the base and fiber are orientable since all the Lie groups involved are connected. But the formula $2 = \chi(S^{2n}) = \chi(Spin(2n+1)/N)\chi(N/Spin(2n))$ implies that either the base or fiber has Euler characteristic $1$. For homogeneous spaces, this implies that the base or fiber is a single point. Thus, $N_0 = Spin(2n)$ or $N_0 = Spin(2n+1)$. But $Spin(2n+1)$ is simple, so it has no positive dimensional normal subgroups, so $N_0 = Spin(2n)$. $\square$

Proposition 3: Suppose $Spin(2n)\subseteq Spin(2n+1)$ is the "usual" embedding. Then $N$ consists of precisely two components.

Proof: We have $Spin(2n)\subseteq N\subseteq Spin(2n+1)$, which gives a homogeneous fibration $N/Spin(2n)\rightarrow Spin(2n+1)/Spin(2n)\rightarrow Spin(2n+1)/N$. From Proposition 2, we already know that $Spin(2n)$ is the identity component of $N$, so $N/Spin(2n)$ is a finite group. In particular, this is a covering.

Since $\chi(S^{2n}) = 2$, $S^{2n}$ can only double cover, so $N$ must have at most two components. The argument you gave in the original post establishes that $N$ has at least two components, so $N$ has precisely two components. $\square$

Proposition: $Spin(8) \rtimes Out(Spin(8))$ does not embed into $Spin(9)$.

Proof: Notice that $Spin(8)$ is normal in $Spin(8)\rtimes Out(Spin(8))$. If this embedded into $Spin(9)$, by Proposition 1, we may assume the identity component $Spin(8)$ embeds in the usual fashion. This implies that $Spin(8)\rtimes Out(Spin(8))\subseteq N$, which implies $N$ has at least $6$ components. This contradictions Proposition 3. $\square$.

  • $\begingroup$ Assuming I didn't make a mistake above, the leads to an obvious question: what is the smallest $n$ for which $Spin(8)\rtimes Out(Spin(8))$ embeds into $Spin(n)$? I don't know a single $n$ for which it embeds. I also want to note that the Mal'cev result works differently for $2n$-dimensional reps. Conjugacy in $Gl_{2n}$ implies conjugacy in $O(2n)$, but not necessarily in $SO(2n)$. So, it's not clear to me whether my argument also rules out $Spin(10)$. $\endgroup$ Commented Feb 21, 2023 at 19:33
  • $\begingroup$ This answer mathoverflow.net/a/435474/387190. seems to imply that $ N(T)G_{long}=Spin_8 \rtimes Out(Spin_8) $ is a subgroup of $ F_4 $. Since $ F_4 $ is a subgroup of $ SO_{26} $ my guess is that $ Spin_8 \rtimes Out(Spin_8) $ embeds in $ Spin_{26} $? But maybe the lift through the double cover messes that up? $\endgroup$ Commented Feb 21, 2023 at 22:04
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    $\begingroup$ @Ian: $F_4$ only has one form (simple conmected) so this should lift. So, we have narrowed down the magic $n$ to $10\leq n \leq 26$. $\endgroup$ Commented Feb 21, 2023 at 23:42

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