Timeline for Is the ability to define Haar measure the main (or only) reason to consider locally compact topological groups?
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| Mar 22, 2019 at 6:24 | comment | added | KConrad | Also, every locally compact (Hausdorff) group $G$ has a $\sigma$-compact open subgroup $H$ (proof: pick symmetric compact neighborhood $K$ of the identity and such a subgroup is $H = \bigcup_{n \geq 1} K^n$ where $K^n = KK\cdots K$ ($n$ times)), so $H$ is $\sigma$-finite wrt the restriction to $H$ of a Haar measure on $G$. Haar measure on a coset $gH$ looks like Haar measure on $H$, so the "only" reason Haar measure on a locally compact group $G$ wouldn't be $\sigma$-finite is due to cardinality issues for the discrete coset space $G/H$. | |
| Mar 22, 2019 at 5:52 | comment | added | Marty | Thanks Keith! That is a more satisfying converse theorem to me too. | |
| Mar 21, 2019 at 14:38 | comment | added | KConrad | Very briefly, the idea of the proof is to show that $G$ contains a compact subset $K$ such that $KK^{-1}$ contains a neighborhood of the identity. Since $KK^{-1}$ is compact, it follows that $G$ is locally compact. | |
| Mar 21, 2019 at 14:34 | comment | added | KConrad | where $K$'s are compact then $G$ must be locally compact. This is a theorem of Chandra Gowrisankaran. It is from her PhD thesis at McGill (see digitool.library.mcgill.ca/webclient/…) and appeared in Proc. AMS 25 (1970) 381–384. The inner regularity hypothesis for all Borel sets is stronger than what is always true about Haar measures, but for $\sigma$-finite $G$ a Haar measure must be inner regular on all Borel sets, so I think this is a pretty satisfying converse theorem to the existence of Haar measure, and should be cited instead of Weil. | |
| Mar 21, 2019 at 14:32 | comment | added | KConrad | Marty, I learned of a much more satisfying converse result than Weil's theorem, which always felt awkward since it does not say that $G$ itself is locally compact. If you assume the measure on $G$ is not just left invariant, but also has other properties resembling those of Haar measure, then $G$ is in fact locally compact. Theorem: If $G$ is a Hausdorff topological group and it has a left-invariant Borel measure $\mu \not\equiv 0$ that is locally finite (i.e., each $g \in G$ has a neighborhood $U_g$ such that $\mu(U_g) < \infty$) and for Borel $A$, $\mu(A) = \sup_{K \subset A} \mu(K)$ (contd) | |
| May 31, 2010 at 18:58 | history | answered | Marty | CC BY-SA 2.5 |