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I am searching for examples of connected locally compact group $G = N \rtimes H$, where $N$ is a simply connected nilpotent non-abelian Lie group, $H$ is linear reductive and $H$ operates on $N$ without non-trivial fixed points. Please enlighten me.

P.S. I added the ergodic theory tag because I believe such groups are seen there.

This is also posted in stackexchange. https://math.stackexchange.com/questions/3037426/examples-of-non-abelian-simply-connected-nilpotent-lie-groups

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  • $\begingroup$ I would try to stick $H$ into the title somehow, because without it the question seems really trivial (all groups of upper triangular matrices are contractible) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 13, 2018 at 8:18

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Consider the nilpotent group $N={\mathbb R} \rtimes {\mathbb R}^2$ (the group of $3\times 3$ upper triangular unipotent matrices with real coefficients. If $v,w \in {\mathbb R}^2$, then their commutator $[v,w]$ in $N$ is simply the wedge $v\wedge w \in \wedge ^2 {\mathbb R}^2\simeq {\mathbb R}$.

The group $H=SL(2,{\mathbb R})$ operates on $N$ since it preserves the symplectic form $v\wedge w$ on ${\mathbb R}^2$.

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  • $\begingroup$ Is it obvious that $H$ acts without fixed points? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 13, 2018 at 8:20
  • $\begingroup$ If the action of $H$ on $N$ is as follows: for $h \in H, (t,v) \in N, h.(t,v) = (t,hv)$ then $H$ fixes $\mathbb{R} \rtimes \{(0,0)\}$ ! What exactly is your action? $\endgroup$
    – Mambo
    Commented Dec 13, 2018 at 8:37
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    $\begingroup$ OK. You replace $SL(2,{\mathbb R})$ by $H=GL(2,{\mathbb R})$ except that on $\mathbb R$ the group $H$ acts via determinant, and hence does not fix anything except identity. It acts on ${\mathbb R}^2$ preserving the symplectic form up to sclars, and hence still acts on $N$. $\endgroup$
    – Venkataramana
    Commented Dec 13, 2018 at 11:54
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    $\begingroup$ $N$ is not a semidirect product $\mathbf{R}\rtimes\mathbf{R}^2$; I don't know if you mean to think of it as a central (not split) extension with kernel $\mathbf{R}$ and quotient $\mathbf{R}^2$, or as a semidirect product with kernel $\mathbf{R}^2$ and quotient $\mathbf{R}$. Anyway, it works. $\endgroup$
    – YCor
    Commented Dec 13, 2018 at 19:33
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    $\begingroup$ @Mambo: yes. The simple connectedness follows from the homotopy exact sequence since $N \rightarrow {\mathbb R}^2$ is a locally trivial fibration with fibre $\mathbb R$. $\endgroup$
    – Venkataramana
    Commented Dec 15, 2018 at 8:09
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You can take $H=T$ to be the maximal torus in a (quasi)split reductive group $\Gamma$, $N$ to be the unipotent radical of a Borel subgroup containing $T$, and the action is by conjugation. If some element $x\in N$ is fixed by $T$ then it centralizes $T$ and so must be in $T$, which is impossible.

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    $\begingroup$ You need to exclude the edge case that $N$ is abelian (e.g., $SL_n$ works for $n \geq 3$). I think more generally $H$ can be taken to be a Levi factor of a parabolic subgroup $P=L \ltimes U$ with the same caveat (your example is the case $P=B = T\ltimes N$). $\endgroup$
    – Guntram
    Commented Dec 14, 2018 at 17:45

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