I know that the outer automorphisms of compact simply connected simple Lie groups are given by the symmetries of the Dynkin diagrams (DD), so that for example $E_6$ enjoys an outer automorphism which exchanges the long legs of its DD:

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I am interested in the situation where the group $G$ is multiply connected, i.e. $G = \tilde G / H$ where $\tilde G$ is the universal cover of $G$ and $H$ is a subgroup of the center of $\tilde G$. Also, I consider $\tilde G$ to be semi-simple. I am particularly interested in the group $E_6^3/\mathbb{Z}_3$ where the quotient is over the diagonal $\mathbb{Z}_3$, i.e. generated by the product of the generators of the centers of each $E_6$.

The question is then if such a group indeed has an outer automorphism. Intuitively, I would think that a permutation of the three $E_6$ factors is such a transformation. This is the case if $G = E_6^3$, but I'm not sure what happens for $E_6^3/\mathbb{Z}_3$.


1 Answer 1


Given any automorphism $f:G\rightarrow G$, we can always lift $f$ to an automorphism $\tilde{f}:\tilde{G}\rightarrow \tilde{G}$, in the sense that $\pi \circ \tilde{f} = f\circ \pi$ for $\pi:\tilde{G}\rightarrow \tilde{G}/H = G$ the projection map, and $H = \ker \pi$.

If we pick $h\in H$, then $f(\pi(h)) = f(e) = e$, so $\pi(\tilde{f}(h)) = e$ which means $\tilde{f}(h) \in H$. In short, $\tilde{f}(H) = H$.

Conversely, if choose any automorphism $\tilde{f}:\tilde{G}\rightarrow \tilde{G}$ for which $\tilde{f}(H) = H$, then $\tilde{f}$ induces a map $f:G\rightarrow G$, defined by $f(\pi(g')) = \pi\circ \tilde{f}$.

In this way, you have a natural identification of $Aut(G)$ with $\{\tilde{f}\in Aut(\tilde{G}): \tilde{f}(H) = H\}$.

Now, applying this to your specific problem, the automorphism $\tilde{f}:E_6^3\rightarrow E_6^3$ given by, say, $f(a,b,c) = (c,a,b)$ maps $\mathbb{Z}_3 := \{(z,z,z):z\in Z(E_6)\}$ to itself. Thus, $\tilde{f}$ descends to an automorphism of $E_6^3/\mathbb{Z}_3$. An analogous result holds for any other permutation of $(a,b,c)$ as well.

In case you care, if $\rho:E_6\rightarrow E_6$ is any automorphism of $E_6$, then $(a,b,c)\mapsto (\rho(a), \rho(b),\rho(c))$ also descends to an automorphism of $E_6^3/\mathbb{Z}_3$. However, it's not clear to me if, say, the map $(a,b,c)\mapsto (\rho(a), \rho(b), c)$ descends it depends on how the outer automorphism group of $E_6$ acts on $\mathbb{Z}_3$. If the action is trivial, then this latter map descends. Otherwise it doesn't.


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