# Does $Spin(2n+1)$ always contain $Spin(2n) \rtimes Out(Spin(2n))$?

$$D_n = Spin(2n)$$, $$n \geq 3$$, 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 $$O(2n)= SO(2n) \rtimes $$ 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 $$(SU_2 \times SU_2) \rtimes =O(4) \hookrightarrow SO(5)$$ 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.)

• 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 $n\neq 4$. Commented Feb 20, 2023 at 23:15
• @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 $Spin(9)$? Commented Feb 21, 2023 at 16:02

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. $$\square$$.

• 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 $Spin(10)$. Commented Feb 21, 2023 at 19:33
• 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? Commented Feb 21, 2023 at 22:04
• @Ian: $F_4$ only has one form (simple conmected) so this should lift. So, we have narrowed down the magic $n$ to $10\leq n \leq 26$. Commented Feb 21, 2023 at 23:42