Timeline for Are there locally compact groups which have no compact open subgroups and no discrete infinite cyclic subgroups?
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4 events
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| Aug 3, 2010 at 6:29 | comment | added | Keivan Karai | Thanks. That's probably a better way of finishing the argument. | |
| Aug 2, 2010 at 21:38 | comment | added | Pierre-Yves Gaillard | Say that an endomorphism $x$ of a real finite dimension vector space $V$ satisfies condition (C) if it is semisimple with imaginary eigenvalues, or, equivalently, if the subgroup $\exp(\mathbb Z x)$ of $GL(V)$ is not infinite discrete. Let $\mathfrak g$ be the Lie algebra of $G$, and $x$ be in $\mathfrak g$. If $x$ satisfies (C), so does $ad(x)$, implying $Killing(x,x)\le 0$. If it were so for all $x$ in $\mathfrak g$, then $\mathfrak g$ would be compact. | |
| Aug 2, 2010 at 21:38 | comment | added | Pierre-Yves Gaillard | I agree with the reduction to the case where our noncompact connected semisimple Lie group $G$ is in $GL_n(\mathbb R)$, but I'd express the rest of the argument as follows. [See next comment.] | |
| Aug 2, 2010 at 10:49 | history | answered | Keivan Karai | CC BY-SA 2.5 |