This is somehow connected to this question.

I can think of at least four topologies to put on $C_c(M)$:

  1. Topologize $C^{\infty}_c(M)\subseteq C^{\infty}(M)$ as a subspace with the weak Whitney $C^\infty$ topology on $C^{\infty}(M)$.
  2. Topologize $C^{\infty}_c(M)\subseteq C^{\infty}(M)$ as a subspace with the strong Whitney $C^\infty$ topology on $C^{\infty}(M)$.
  3. Topologize $C^{\infty}_c(M)=colim_{K\subseteq M} C^{\infty}_K(M)$ as direct limit over compact subsets $K\subseteq M$ where $C^{\infty}_K(M)$ are the functions, which have support in $K$ with the Whitney topology. (The strong and the weak one should coincide in that case.)
  4. Take $h\in C^{\infty}_c(M)$ a smooth function $\epsilon\colon M\rightarrow (0,\infty)$, vector fields $X_1,...,X_k$ and declare the subsets of the shape $\{g\in C^{\infty}_c(M)\colon\text{ }|X_1..X_k(h-g)(x)|<\epsilon(x)\forall x\in M\}$ as a subbase varying over $h,k\in\mathbb{N}_0, X_1,...,X_k$ and $\epsilon$.

How are these topologies related?

My vague guesses are:

  • Topology 2.) is finer than topology 1.), but those are in general not equal.
  • Topology 2.) is finer than topology 3.), but in general not the same.

I have no idea how to relate topology 4.) to the others.

  • $\begingroup$ This answer obviously gives 2=3 but probably one needs there to have some additional assumption like $M$ being second countable. Otherwise probably 2$<$3. $\endgroup$
    – TaQ
    Mar 31, 2015 at 10:15
  • $\begingroup$ I have just asked a new question closely related to this one. $\endgroup$ Aug 25, 2020 at 7:24

2 Answers 2


First, I will describe three topologies on $C^\infty(M,N)$ for smooth manifolds $M$ and $N$, and then relate them to the topologies 1, 2, and 4 from the question. For $N=\mathbb{R}$, all of the below statements relating the topologies remain valid if we pass to the subspaces of compactly supported functions.

It seems that terminology is not completely standardized, I will call these three topologies the weak $C^\infty$-topology, the Whitney $C^\infty$-topology, and the strong $C^\infty$-topology. Assume we are given

  • a locally finite family $(\phi,U)$ of charts of $M$,
  • a family of compact subsets $K_i\subseteq U_i$,
  • a smooth map $h:M\to N$ such that $h(K_i) \subseteq V_i$,
  • a family $(\psi,V)$ of charts of $N$,
  • a family $\varepsilon$ of positive reals,
  • a family $k$ of non-negative integers, and

all with the same indexing set. Consider the set of smooth $g:M\to N$ such that for all indices $i$, we have $g(K_i)\subseteq V_i$ and

$$\|D^\alpha(\psi_i g\phi^{-1}_i - \psi_i h\phi^{-1}_i)\|(m) < \varepsilon_i$$

for all $m\in K_i$ and multiindices $\alpha$ of order at most $k_i$. Sets of this form are a basis for the strong $C^\infty$-topology on $C^\infty(M,N)$. If we additionally require that the $k_i$ are bounded, then we get the Whitney $C^\infty$-topology, that is, we would not have needed a family of $k_i$, but just one non-negative integer $k$. If we require that the indexing set is finite, we get the weak $C^\infty$-topology.

If $M$ is compact, all three topologies agree, and if $M$ is not compact, they all differ. The strong topology is stronger (finer) than the Whitney topology, and the Whitney topology is stronger than the weak topology. Note that for $l< \infty$, the $C^l$-versions of the strong topology and the Whitney topology are equal, which may be why the words "strong" and "Whitney" are sometimes used interchangably. I took this way of differentiating between them from Chris Schommer-Pries's lecture notes, which contains a lot of information about these and other topologies. These two topologies are much more similar to each other than to the weak topology. For example, the Whitney and strong $C^\infty$-topologies have the same weak homotopy type, because they even have the same convergent sequences.

Using these definitions, I do not know whether you meant the Whitney or the strong $C^\infty$-topology in 2. In any case, your first guess is correct, as can be seen from the above description. For $N=\mathbb{R}$, the Whitney $C^\infty$-topology is equal to topology 4. To see this, express the vector fields as linear combinations of partial derivatives in coordinates. Also, it is possible to find a smooth $\varepsilon$ which takes values at most $\varepsilon_i$ on $K_i$ for all $i$ by using bump functions and that $U$ was assumed to be locally finite.

In addition to the above link, see Hirsch's "Differential Topology" and Golubitsky's and Guillemin's "Stable Mappings and Their Singularities" for more information about the weak and the Whitney $C^\infty$-topologies, which use the more intrinsic language of jet bundles in addition to something like the above.

  • 1
    $\begingroup$ I think there's a small typo in the condition for $h$. Shouldn't it be that $h(K_i) \subseteq V_i$? $\endgroup$ Mar 31, 2015 at 19:36
  • $\begingroup$ Let $M=\mathbb{R}$. In topology 4, all functions with compact support whose second derivative is smaller than $1$ would be open. In the Whitney $C^{\infty}$ topology they would'nt since their values and first derivatives must be close too, if one requires something for the second derivative. Am I missing something or does this show topology 4 differs from the Whitney topology? $\endgroup$
    – Tom
    Apr 1, 2015 at 11:54
  • 1
    $\begingroup$ Let $O$ denote this set. For every $f$ in $O$, it is possible to find a Whitney $C^\infty$ basis set which is a subset of $O$ and contains $f$. This basis set is always a proper subset of $O$, because, as you noted, the condition defining it involves the function values and the first derivative, too. $\endgroup$ Apr 1, 2015 at 14:34
  • $\begingroup$ It's easy to see that the Whitney topology is finer than topology 4, but I don't see how to proof the converse. Could you please elaborate on that? $\endgroup$
    – Tom
    May 19, 2015 at 13:19

A lot of detailed information is also contained in sections 3 and 4 of

  • Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158. (pdf)

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