Because I haven't seen locally compact topological groups used for anything except Fourier analysis.
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$\begingroup$ A lot of examples of topological groups in the "real world" are locally compact, for instance Lie groups, groups of ideles and adeles etc. And Fourier analysis is a powerful technique for these; see Tate's thesis. Non-commutative analogues lead to the wonders of the Langlands programme etc. $\endgroup$– Robin ChapmanMay 31, 2010 at 10:29
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1$\begingroup$ Closed. Try phrasing your questions more tactfully, and with more context. See the "how to ask" page for tips. $\endgroup$– S. Carnahan ♦Jun 2, 2010 at 2:42
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3 Answers
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I can think of two ways to answer this question.
First, regarding your comment "I haven't seen locally compact topological groups used for anything except Fourier analysis," I think the appropriate answer is given by Charles Matthews and Robin Chapman. Locally compact topological groups "arise in nature". Here's a general theorem explaining how -- a more general statement and proof can be found in Brian Conrad's notes (PDF file).
Let $A$ be a local topological ring, such that $A^\times$ is open in $A$ and has continuous inversion. If $A$ is locally compact and Hausdorff, and $G$ is a algebraic group (separated group scheme locally of finite type suffices) over $A$, then $G(A)$ is a locally compact Hausdorff topological group.
This implies that $GL_n(A)$ is locally compact, for a wide class of important rings $A$. Conrad's notes also explain how this can be extended, under some hypotheses, to the adelic points of algebraic groups (following Weil).
A second way to answer your question is the following: if one cares about harmonic analysis on groups, then the setting of locally compact topological groups is precisely the right setting to work in. Besides the existence of Haar measure, an important converse is given in an appendix to Weil's "L'intégration dans les groupes topologiques et ses applications." I refer to Weil's result that -- given a group $G$ with a left-invariant measure $\mu$, satisfying an additional very general condition or two (one condition is that if $f$ is a measurable function on $G$, then $(x,y) \mapsto f(x^{-1} y)$ is a measurable function on $G \times G$), there is a unique topology on $G$ for which its completion $\hat G$ is locally compact and has Haar measure equal to the natural extension of $\mu$. In other words, not only is the "locally compact" condition sufficient for existence of Haar measure, but it is a necessary condition for anything resembling Haar measure to exist.
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$\begingroup$ 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) $\endgroup$– KConradMar 21, 2019 at 14:32
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$\begingroup$ 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. $\endgroup$– KConradMar 21, 2019 at 14:34
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$\begingroup$ 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. $\endgroup$– KConradMar 21, 2019 at 14:38
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$\begingroup$ Thanks Keith! That is a more satisfying converse theorem to me too. $\endgroup$– MartyMar 22, 2019 at 5:52
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$\begingroup$ 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 $G/H$. $\endgroup$– KConradMar 22, 2019 at 6:24
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"Locally compact", for example, forces a Banach space to have a finite dimension. Restricting to locally compact groups excludes many "big" examples, and keeps many useful ones, such as adele groups in number theory. It includes Lie groups. It is a sensible class in which to work out representation theory, including non-commutative analogues of Pontryagin duality (Fourier theory).
answered May 31, 2010 at 10:32
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An example of relevance of the local compactness is given by the Chabauty topology: it is a natural topology defined on the set $C(G)$ of closed subgroups of a topological group $G$. When $G$ is locally compact, then $C(G)$ is compact (and can be used for example to construct compactifications). To learn more one this, Pierre de la Harpe has written a nice survey (on arxiv and his web page, I guess).
answered Jun 1, 2010 at 9:57