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Suppose $G$ is a compact Lie group, we know the center of $G$ is a compact Abelian subgroup, so it must be isomorphic to a direct product of finite abelian subgroup and a torus.

Now suppose the center is not finite (i.e the torus part of the center is not trivial), then can we split the torus out? For example, if the torus part is $S^1$, can we claim $G$ is a direct product of $S^1$ and some subgroup $H$ of $G$. I think this is true, but I am not sure.

If this is not true, can anyone provide me with an counterexample? Thank you!

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  • $\begingroup$ you must assume that $G$ is connected; otherwise, this is obviously false. AS Bryant has said, it is false even for $U(2)$. $\endgroup$
    – Venkataramana
    Commented Nov 7, 2014 at 1:27
  • $\begingroup$ which is connected, and its center is connected, so I don't see why you bring up connectedness. $\endgroup$
    – Allen Knutson
    Commented Nov 7, 2014 at 3:01
  • $\begingroup$ Right. I should have formulated my statement carefully. The original motivation comes from a trick in Knapp's book, when we use some kind of Unitarian trick to simplify the argument for Harish Chandra's $\Xi$-function. But there he assumes G is linear connected semisimple group, I just think if his argument may be useful for general real reductive group. Of course we need to assume connectedness. $\endgroup$
    – Fangyang Tian
    Commented Nov 7, 2014 at 4:49
  • $\begingroup$ @Allen; I meant to say, even f $G$ is connected, it is false. But for disconnected $G$, it is also false, by "easier" considerations $\endgroup$
    – Venkataramana
    Commented Nov 7, 2014 at 7:11

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This is not true. Take $\mathrm{U}(2)$, for example. It is not the direct product of its center (the matrices of the form $\mathrm{e}^{i\theta} \mathrm{I}_2$) with the simple part $\mathrm{SU}(2)$. The point is that the center, which is an $S^1$, intersects the simple part $\mathrm{SU}(2)$ in two distinct points, $\{\pm \mathrm{I}_2\}$.

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  • $\begingroup$ Thanks you! I think the fallacy may come from the outer automorphism in general, am I right? $\endgroup$
    – Fangyang Tian
    Commented Nov 6, 2014 at 17:16
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    $\begingroup$ Nevertheless $G$ and $Z(G)\times H$ (where $H=G/Z(G)$) have the same Lie algebra, so their universal covers coincide. $\endgroup$
    – Xin Nie
    Commented Nov 6, 2014 at 17:53
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    $\begingroup$ There are no outer automorphisms for SU(2). The phenomenon of not splitting off central tori has nothing to do with outer automorphisms. $\endgroup$
    – user27920
    Commented Nov 6, 2014 at 18:14
  • $\begingroup$ Yep, but for $SU(n)$, (n>3), the outer automorphism group comes from Dynkin Diagram. So it is not trivial, right? When I try to justity my claim, I use an exact sequence. But it is not split in general. My concern is that when we construct an homomorhism from $Z(G)$ to $Aut(G/Z(G))$, it is not necessarily an inner automorphism. Am I right? $\endgroup$
    – Fangyang Tian
    Commented Nov 7, 2014 at 0:05

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