If $M$ is a smooth paracompact manifold, then what is the usual topology of $C^\infty_c(M) $, i.e., the smooth function with compact support?
$\begingroup$
$\endgroup$
4
-
4$\begingroup$ Crossposted to math.SE: math.stackexchange.com/q/137701/264 $\endgroup$– Zev ChonolesCommented Apr 27, 2012 at 16:29
-
1$\begingroup$ I don't know if this is usual, but it should be possible to define a metric by $$d(f,g) = \sum_n \frac{1}{2^{n+A(n)}}\sum_{|\alpha|=n}\frac{\left|\sup_K\frac\partial{\partial x^\alpha}(f-g)\right|}{1 + \left|\sup_K\frac\partial{\partial x^\alpha}(f-g)\right|}$$ where $A(n)$ is the number of $\alpha$ s.t. $|\alpha|=n$. The space should be complete in the induced topology. $\endgroup$– Todd LeasonCommented Apr 27, 2012 at 17:46
-
$\begingroup$ Added: $K$ has to be taken to include the support of $f,g$. $\endgroup$– Todd LeasonCommented Apr 27, 2012 at 17:50
-
3$\begingroup$ Todd: smoothly truncating $e^{-x^2}$ on $\mathbb R$ so as to obtain a sequence of compactly supported functions appropriately should give a Cauchy sequence in that metric which does not converge, no? $\endgroup$– Mariano Suárez-ÁlvarezCommented Apr 27, 2012 at 18:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
Topologizing $C_c^\infty(M)\subseteq C^\infty(M)$ with the subspace topology (where $C^\infty(M)$ has the Whitney topology, generated by the seminorms $\left|\sup_K\frac\partial{\partial x^\alpha}f\right|$), makes it a dense subspace; in particular it is not itself complete. So I wouldn't really call this the "usual topology" on $C_c^\infty(M)$. (it would be sort of like saying the usual topology on $C(M)$ is given by the $L^2$ norm).
To me the usual topology is the inductive limit topology $C_c^\infty(M)=\lim_{K\subseteq M}C_c^\infty(K)$ (which Mariano calls the colimit topology). This topology is not metrizable when $M$ is noncompact (since it's not even first-countable), but is "nicer" in the sense that it gives a well-understood dual space, namely the space of distributions on $M$.
In comparison, the dual space of $C^\infty(M)$ with the Whitney topology is the space of compactly supported distributions on $M$.
answered Apr 27, 2012 at 17:28
-
$\begingroup$ I agree with this answer, of course. I only want to remark that, in the book "Differential Topology" (by Morris W. Hirsch), the term Whitney topology is used for the strong $C^\infty$ topology, and you are using the weak $C^\infty$ topology. $\endgroup$ Commented Dec 20, 2017 at 9:20
-
$\begingroup$ The (strict) colimit topology also has the benefit of making $C^\infty_c(M)$ quasi-complete (a.k.a., locally complete). $\endgroup$ Commented Dec 20, 2017 at 20:24
-