Famously $$ \rm Aut(Spin(8))\cong PSO_8(\mathbb{R})\rtimes S_3 $$ which is not equal to $$ \rm Aut(SO_8(\mathbb{R})) \cong PSO_8(\mathbb{R}) \rtimes 2 $$ see https://math.stackexchange.com/a/3105463/758507

I'm interested in other examples of simple Lie groups whose automorphism group is strictly smaller than the automorphism group of the universal cover.

This cannot happen for types $ B_n, C_n,F_4,G_2,E_7,E_8 $ since every automorphism is inner (the Dynkin diagram has no symmetry so no outerautomorphisms) and thus we groups of those types we always have $$ \rm Aut(G)=G/Z(G)=Aut(\tilde{G}) $$

If $ G $ is centerless then we have $$ G \cong \tilde{G}/Z(\tilde{G}) $$ and so, since $ Z(\tilde{G}) $ is a characteristic subgroup of $ \tilde{G} $, we have $$ \rm Aut(G) \cong Aut(\tilde{G}/Z(\tilde{G}))\cong Aut(\tilde{G}) $$ see https://math.stackexchange.com/a/4038199/758507

So that implies for type $ D_n $ $$ \rm Aut(PSO_{2n}(\mathbb{R}))\cong Aut(Spin_{2n}(\mathbb{R})) $$ for all $ n $. The only other form of $ D_n$ is $$ \rm SO_{2n}(\mathbb{R}) $$ For $ n\neq 4 $ then the outer automorphism group is cyclic 2 (comes from swapping short legs of Dynkin diagram) and this outer automorphism can be realized by conjugation by $diag(-1,1,\dots, 1)\in O_{2n}(\mathbb{R}) $. Thus $$ \rm Aut(SO_{2n}(\mathbb{R}))\cong Aut(Spin_{2n}(\mathbb{R})) $$ for $ n \neq 4 $. $ SO_8 $ has exceptionally small automorphism group, as noted above.

What about the $ A_{n-1} $ series? Again we know $$ \rm Aut(PU_{n})\cong Aut(SU_n) $$ what about intermediate groups $$ \rm SU_n/\zeta_d $$ for some $ d $ dividing $ n $ (here $ \zeta_d $ is a primitve $ d $th root of unity). Does $ SU_n/\zeta_d $ ever have strictly smaller automorphism group? Seems probably not since the only outer automorphism of $ SU_n $ is complex conjugation (flipping Dynkin diagram) and complex conjugation preserves the group generated by $ \zeta_{d} $ ( it sends $ \zeta_d \mapsto\zeta_d^{d-1} $).

So I guess my question really is:

Is $ SO_8 $ the only simple Lie group whose automorphism group is not isomorphic to the automorphism group of its universal cover?

  • $\begingroup$ $SO_{2n+1}$ having no outer automorphisms is because the Dynkin diagram has no automorphisms. $\endgroup$
    – Kenta S
    Commented Nov 30, 2022 at 3:09
  • $\begingroup$ @MarianoSuárez-Álvarez I guess every automorphism preserves the center, the center is a characteristic subgroup. Also I've significantly changed the question now with the intent of improving it. $\endgroup$ Commented Nov 30, 2022 at 14:02
  • $\begingroup$ A simple group is centerless if and only if it is not abelian if and only if it is not isomorphic to the cyclic group $\mathbb{Z}/p\mathbb{Z}$ for any prime number $p$. So, the centerlessness assumption could be dropped unless $\mathbb{Z}/p\mathbb{Z}$ could be made into a Lie group for some (or perhaps, all) prime(s) $p$. $\endgroup$ Commented Nov 30, 2022 at 14:32
  • $\begingroup$ @MarianoSuárez-Álvarez They always descend for the centerless case (math.stackexchange.com/a/4038199/758507). For some reason that seemed very unintuitive to me I guess because I know for $ Spin(8) $ passing to $ SO(8) $ shrinks the automorphism group and so my intuition was that shrinking the center would shrink the automorphism group, but clearly that intuition is wrong. Anyway I've changed the question significantly again to be basically "is SO(8) the only simple Lie group with automorphism group smaller than it's universal cover"? $\endgroup$ Commented Nov 30, 2022 at 15:17
  • $\begingroup$ @MarianoSuárez-Álvarez: I disagree that the automorphism group change from $Spin(8)$ to $SO(8)$ is simply that two automorphisms of $Spin(8)$ end up inducing the same automorphism of $SO(8)$. The triality automorphism of $Spin(8)$ (a generator of the non-trivial $3$-cycle in $S_3$) induces a map on the Lie algebra $\mathfrak{spin}(8)\cong \mathfrak{so}(8)$ which is simply not induced by any automorphism of $SO(8)$. (But I do agree that I don't like questions to significantly change once they're posted.) $\endgroup$ Commented Nov 30, 2022 at 18:43

1 Answer 1


Among the (connected) simple Lie groups, the following holds:

Theorem 1: Suppose $G$ is a connected simple Lie group with universal cover $\overline{G}$ which is not of the form $Spin(4k)$. Then $Aut(G) = Aut(\overline{G})$.

You've already indicated the argument except for $A_n$, $D_{2n+1}$, and $E_6$, but I'll give an argument which handles them all simultaneously.

The key feature in common among all these examples is:

Proposition 2: Suppose $\overline{G}$ is a simply connected simple Lie group which is not isomorphic to $Spin(4k)$ for some $k$. Then the center $Z(\overline{G})$ is cyclic.

Proof: The centers are listed this website. (If someone wants to fill in a more official reference, please do!) $\square$

We will also need the following

Lemma 3. Suppose $Z(\overline{G})$ is cyclic and let $H\subseteq Z(\overline{G})$ be any subgroup. If $\overline{f}:\overline{G}\rightarrow \overline{G}$ is any automorphism, then $\overline{f}(H) = H$.

Proof: As you already mentioned, the center of a group is always characteristic, so $\overline{f}$ restricts to a map $\overline{f}:Z(\overline{G})\rightarrow Z(\overline{G})$. But in a cyclic group, a subgroup is uniquely characterized by its order, so $H\subseteq Z(\overline{G})$ is characteristic. $\square$

Now we can prove Theorem 1. I'll write $\mathfrak{g}$ for the Lie algebra of $G$.

Proof: (of Theorem 1) Since $Aut(\overline{G})\cong Aut(\mathfrak{\overline{g}})\cong Aut(\mathfrak{g})$, it is enough to show that every automorphism of $\phi:\mathfrak{g}\rightarrow \mathfrak{g}$ comes from an automorphism $f:G\rightarrow G$. But this follows easily from the following observations:

  1. Since $\overline{G}$ is simply connected, $\phi$ is induced by an automorphism $\overline{f}:\overline{G}\rightarrow \overline{G}$.

  2. $G$ is isomorphic to $\overline{G}/H$ for some $H\subseteq Z(\overline{G})$.

  3. The map $\overline{f}$ maps $H$ to $H$ (Proposition 2 and Lemma 3) $\square$

I now claim that Theorem 1 is, in some sense, as good as possible.

Theorem 4 For each $k\geq 2$ (so that $Spin(4k)$ is actually simple), there is a Lie group $G$ with universal cover $\overline{G} = Spin(4k)$ for which $Aut(G)$ is smaller than $Aut(\overline{g})$.

Proof: Set $\overline{G} = Spin(4k)$ and note that $Z(\overline{G})\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}.$ When $k=2$, you've already noted Theorem 2 is true (coming from the triality automorphism of $Spin(8)$). So, let's assume $k\geq 3$.

Then the outer automorphism group $Out(\overline{G})$, which is isomorphic to the symmetries of the Dynkin diagram, is $\mathbb{Z}/2\mathbb{Z}$. Let $\overline{f}:\overline{G}\rightarrow \overline{G}$ be such an outer automorphism.

As seen on page 50 of

Siebenthal, Jean de. "Sur les groupes de Lie compacts non connexes.." Commentarii mathematici Helvetici 31 (1956/57): 41-89. http://eudml.org/doc/139134,

the map $\overline{f}:Z(\overline{G})\rightarrow Z(\overline{G})$ is non-trivial. One the other hand $\overline{f}^2$ is inner, so fixes $Z(\overline{G})$ points wise. It follows that $Z(\overline{G})\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ has a generating set $\{a,b\}$ for which $\overline{f}$ simply swaps $a$ and $b$.

Consider the quotient of $\overline{G}$, $\pi:\overline{G}\rightarrow G:= Spin(4k)/\langle a\rangle$.

Proposition 5: There is no automorphism $f:G\rightarrow G$ which induces the same map as $\overline{f}$ on the Lie algebra level. That is, the automorphism $\overline{f}_\ast:\mathfrak{spin}(4k)\rightarrow \mathfrak{spin}(4k)$ is not induced by a map $f:G\rightarrow G$.

Proof: Suppose for a contradiction that $f$ is such a map, then it must lift to $\overline{f}$ (since the two maps to the same thing on the Lie algebra level), so we obtain the commutative diagram $$\begin{matrix} \overline{G} & \xrightarrow{\overline{f}} & \overline{G}\\ \downarrow{\pi} & & \downarrow{\pi}\\ G & \xrightarrow{f} & G \end{matrix}.$$

But now start with $b\in \overline{G}$ and follow it. We have $(\pi\circ \overline{f})(b) = \pi(\overline{f}(b)) = \pi(a) = e$, where $e$ denotes the identity element. On the other hand, $b\notin\ker \pi$, so $\pi(b)\neq e$ and $f$ is an automorphism, so $f(\pi(b))\neq e = \pi(\overline{f}(b))$. Thus, we have reached a contradiction. $\square$

  • 1
    $\begingroup$ When $k=1$, the group $Spin(4)/\langle a\rangle $ is $SO(3)\times SU(2)$, which is obviously not isomorphic to $SO(4)$ (though it is diffeomorphic!) When $k=2$, the groups $Spin(8)/\langle a\rangle$ and $SO(8)$ are accidentally isomorphic, as a consequence of triality. For larger $k$, $Spin(4k)/\langle a\rangle$ is not isomorphic to $SO(4k)$. Indeed, Proposition 5 essentially establishes that $Spin(4k)/\langle a\rangle$ has trivial outer automorphism group, so it can't be isomorphic to $SO(4k)$. $\endgroup$ Commented Nov 30, 2022 at 22:42

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