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By semisimple I mean the real Lie algebra of $G$ is semisimple. I guess there is not but I can't formulate a rigorous argument.

Ariosto
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  • I just read here:

    http://people.uleth.ca/~dave.morris/talks/NormalSubgroups-Lecture1-Chgo-2-08.pdf

    that if G is connected, then its amenable iff it has a closed normal solvable subgroup, N, such that G/N is compact. But semisimple groups have no nontrivial connected normal solvable subgroups. Not sure if discrete N are an issue or not...

    – Mr. Cooperman Apr 02 '16 at 06:55
  • Since the link provided by @Timkinsella is unfortunately broken now you can use this one: https://web.archive.org/web/20161013155715/http://people.uleth.ca/~dave.morris/talks/NormalSubgroups-Lecture1-Chgo-2-08.pdf – Jannis May 15 '24 at 10:02

1 Answers1

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Let $G$ be a noncompact semisimple Lie group. If $G$ is connected, then it contains an isomorphic copy of $\mathbb{F}_2$, the free group with 2 generators (see Theorem 3.9 in the book by Patterson), thus not amenable.

If $G$ is not connected, then let $G_e$ be the connected component of identity. $G_e$ is an open subgroup, so the factor group $G/G_e$ is discrete. $G$ is amenable iff both $G_e$ and $G/G_e$ are amenable. By definition, $G$ and $G_e$ has the same solvable radical (i.e., the largest solvable connected closed normal subgroup), so $G_e$ is also semisimple. Thus, the connected semisimple Lie group $G_e$ is amenable iff $G_e$ is compact. In summary, if $G$ is not connected, then $G$ is amenable iff $G_e$ is compact and $G/G_e$ is an amenable discrete group.

To find an example of a non-connected non-compact semisimple Lie group, take a compact semisimple Lie group $H$ and an infinite amenable discrete group $D$, and define the cartesian product $G=H\times D$.

Onur Oktay
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