# Outer automorphisms of multiply connected compact semisimple Lie groups

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:

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 $$E_6^3/\mathbb{Z}_3$$.

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.