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Is it true or false that the Inner (Inn), Outer (Out) and Total (Aut) Automorphism of a Lie group $G$ is the same as the covering group of the Lie group, say $\tilde G$ (regardless of how many types of covering groups of $G$ has: $\tilde G_1$, $\tilde G_2$, $\tilde G_3$, ... )?

Note that Inn$(G)=G/Z(G)$, so this seems to be true for the inner automorphism, since we have $$\text{Inn}(G)=G/Z(G)=\tilde G/Z(\tilde G)=\text{Inn}(\tilde G). \tag{1}$$ Also $$ \text{Aut}(G)=\text{Inn}(G) \rtimes \text{Out}(G), $$ so we only need to prove $$ \text{Out}(G)=\text{Out}(\tilde G) (?) \tag{2} $$ to show that also $$\text{Aut}(G) =\text{Aut}(\tilde G) (?)\tag{3} $$

For example,

  1. For the $G=SO(3)=PSU(2)$, there is a double cover $\tilde G =SU(2)$, such that indeed $$\text{Inn}(PSU(2))=\text{Inn}(SU(2))=PSU(2)=SO(3).$$ Also $$\text{Inn}(PSU(2))=\text{Inn}(SU(2))=PSU(2)=SO(3).$$

  2. For the $G=PSU(N)$, there is a double cover $\tilde G =SU(N)$, such that indeed $$\text{Inn}(PSU(N))=\text{Inn}(SU(N))=PSU(N).$$ $$\text{Out}(PSU(N))=\text{Out}(SU(N))=\mathbb{Z}/2.$$ $$\text{Aut}(PSU(N))=\text{Aut}(SU(N))=PSU(N) \rtimes \mathbb{Z}/2.$$

  3. So ar the following statements generally true? eq 1, 2, and 3? $$ \boxed{\text{Out}(G)=\text{Out}(\tilde G) (?), \quad \text{Aut}(G) =\text{Aut}(\tilde G) (?)} $$

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    $\begingroup$ It's not true in general. The easiest example is that of the circle $G=\mathbf{R}/\mathbf{Z}$. Also have a look at the cases of $G=\mathrm{SL}_2(\mathbf{R})\times\mathrm{PSL}_2(\mathbf{R})$ and $G=\mathrm{SO}(4,4)$. $\endgroup$
    – YCor
    Commented Oct 5, 2021 at 18:05
  • $\begingroup$ One can also see that it's not true, because your situation also includes every finite group as $\tilde{G}$ since $\tilde{G} \to 1$ is also a covering. It also shows that $Aut = Inn \rtimes Out$ is false in general. In other words: These kinds of statements should always come with some condition on connectedness. $\endgroup$
    – Johannes Hahn
    Commented Oct 5, 2021 at 18:55
  • $\begingroup$ In addition to @‍JohannesHahn's point about connectedness, is "Lie group" here meant to be "reductive (and connected) Lie group"? Your examples seem all to be of that flavour. (Although, of course, as @YCor points out, it's not true even in that case, basically (my gloss) because of non-trivial automorphisms of the centre; but one could talk about what is true.) $\endgroup$
    – LSpice
    Commented Oct 5, 2021 at 21:27
  • $\begingroup$ My examples are meant to be connected (I should have written $\mathrm{SO}_0(4,4)$, although $\mathrm{SO}(8)$ should work as well). If $G$ is simply connected and $Z$ is a discrete central subgroup, the automorphism group of $G/Z$ is the group of automorphisms of $G$ preserving $Z$. (In terms of Out: if $Z_G$ is the center of $G$, then $\mathrm{Out}(G)$ naturally acts on $Z_G$ and $\mathrm{Out}(G/Z)$ is the stabilizer of $Z$.) $\endgroup$
    – YCor
    Commented Oct 5, 2021 at 21:43
  • $\begingroup$ Am I correct that if $\tilde{G}$ is just (1) a double cover or (2) a universal double cover of $G$, say $1 \to \mathbb{Z}_2 \to \tilde{G} \to G \to 1$, then we can have the Out$(G)$=Out$(\tilde{G})$ always true? Thus since Inn$(G)$=Inn$(\tilde{G})$, so Aut$(G)$=Aut$(\tilde{G})$ also always true? $\endgroup$
    – wonderich
    Commented Oct 6, 2021 at 18:09

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