Let $G$ be a real semi-simple Lie group. Let $\Gamma$ be a non-uniform lattice in $G$. Then it is known that $\Gamma$ contains a non-trivial unipotent element. When $\mathbb{R}$-rank of $G$ is 1, it is known that a unipotent element will be contained in a unique maximal unipotent subgroup of $G$, which then will be contained in a proper parabolic subgroup of $G$.

Question is , in higher rank case, can one find a proper parabolic subgroup of $G$ containing a unipotent element of $\Gamma$?


1 Answer 1


The question can be formulated in a better way: given a non-trivial unipotent element $u$ in $G$,how to associate a proper parabolic subgroup containing the unipotent element? This is always possible: over $\mathbb C$, this is just the Borel fixed point theorem for the unipotent group acting on $G/P$ for any parabolic subgroup. Over $\mathbb R$ a more "canonical way" is to take the unipotent one parameter group $U_1$ in which your unip lies, then take its normaliser $N_1$. Let $U_2$ be the unipotent radical of $N_1$ (it contains $U_1$). Let $N_2$ be the normaliser of $U_2$. Keep repeating this process, and at some time you get a unipotent group $U=U_n$ which is the unipotent radical of the normaliser $P=N_n$: $U_n=U_{n+1}$. Then a result of Borel-Tits says that $P$ is a parabolic proper subgroup containing $U\supset U_1\supset\{u\}$.

[Edit]I will post the precise reference to Borel Tits tomorrow since I do not have access to it at home (this is not necessary, thanks to grghxy's link). If we assume (for the sake of simplicity) that $G$ is a real simple algebraic group of real rank at least two, then a lattice is necessarily arithmetic, and since it is non-uniform (contains a unipotent element) the lattice is contained in the $\mathbb Q$ points of a $\mathbb Q$ -structure on $G$. The above procedure actually gives us a parabolic subgroup defined over $\mathbb Q$ since all the groups $N_k$ and $U_k$ are defined over $\mathbb Q$.

  • $\begingroup$ This is a useful approach, but it's a good idea to add an explicit reference/link to the Borel-Tits result. (The question asked could also use more background.) Note too that the "unipotent" concept here requires identification of the complex Lie group with an algebraic group where the Jordan decomposition makes sense. Then the real group is the group of real points, etc. $\endgroup$ Jul 30, 2015 at 12:51
  • $\begingroup$ mathoverflow.net/questions/104201/… $\endgroup$
    – grghxy
    Jul 30, 2015 at 14:00
  • $\begingroup$ Since the 1972 Borel-Tits paper is referred to here and in the link just posted by grghxy, maybe it's a good idea to add a link to it (though it's in French): gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002088894 $\endgroup$ Jul 30, 2015 at 14:18

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