Because I haven't seen locally compact topological groups used for anything except Fourier analysis.
closed as not a real question by Harry Gindi, François G. Dorais♦, Robin Chapman, Gjergji Zaimi, S. Carnahan♦ Jun 2 '10 at 2:39It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. 


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 leftinvariant 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. 


"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 noncommutative analogues of Pontryagin duality (Fourier theory). 


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). 

