I found out that there are outer automorphisms of compact Lie group $E_6$. For example complex conjugation $\tau$ as defined in this Yokota paper is outer automorphism of $E_6$ which fix $F_4$ subgroup. This means that automorphism group of $E_6$ is bigger than $E_6$ (embedded into $Aut(E_6)$ as inner automorphisms divided by 3-element center $\mathbb Z_3$). Is it known what is the full automorphism group of $E_6$ ?
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For a compact simply-connected semisimple Lie group $G$ the quotient group $Out(G)=Aut(G)/Inn(G)$ is isomorphic to the automorphism group of the Dynkin diagram, which is $\mathbb{Z}/2$ in case of $E_6$, see here. Now we have the exact sequence $$ 1\rightarrow Inn(G) \rightarrow Aut(G) \rightarrow Out(G)\rightarrow 1, $$ which is an extension problem. For the automorphism groups of the corresponding semisimple Lie algebras, the exact sequence splits, so that $Aut(\mathfrak{g})$ is a semidirect product of $Inn(\mathfrak{g})$ and $Out(\mathfrak{g})$.
answered Sep 8, 2016 at 9:06
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$\begingroup$ Thank you ! Semidirect product by cyclic group $\mathbb Z_2$ means that in $K=Aut(G)$ there are two components of $E_6/\mathbb Z_3$. Let $K_0$, $K_1$ be these components, $1\in K_0$. Let's consider set $M$=$g\tau g^{-1}$ for $g$ in $E_6$. It is subset of $K_1$. According to Robert Bryant answer to this mathoverflow.net/questions/249091/… question set $M$ is equal to symmetric space $E_{IV}$. Topologically both components $K_0$ and $K_1$ are the same. Can we find $E_{IV}$ in $K_0$ ? For example $M_0$=$\{\tau g:g\in M\}$ ? $\endgroup$– kobiSep 8, 2016 at 9:30
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$\begingroup$ I see. There is a long discussion in the comments on your MO question. Perhaps your question about $E_{IV}$ belongs then there, if it is reopened, or you pose a new question, which focuses more on this point. $\endgroup$ Sep 8, 2016 at 9:36