# Reference request: Intrinsic definition of the strong Whitney topology on $\mathcal{C}^{\infty}(M,\mathbb{R})$ without using charts or jets

Let $M$ and $N$ be smooth manifolds. There is a description of the strong Whitney topology on $\mathcal{C}^{\infty}(M,N)$ in terms of partial derivative in charts (using locally finite sets of charts in the domain and the codomain) and there is a description coming from the strong topology on $\mathcal{C}(M, J^\infty(M,N))$.

However, if $N=\mathbb{R}$ I think there must be an easier description of the topology of $\mathcal{C}^{\infty}(M,\mathbb{R})$ without using charts or jets, but just vector fields instead.

Does anyone know if such a description appears anywhere in the literature?

• Any reason to prefer vector fields to jets? – Igor Khavkine Jun 20 '15 at 16:56
• @Kathrin Is "Differential Topology" by M. Hirsch useful for your question? – Ali Taghavi Jun 20 '15 at 17:51
• @IgorKhavkine I'm just more used to them. – Kathrin L. Jun 21 '15 at 12:31
• @AliTaghavi I know Hirsch's book, but it did not answer me this question. – Kathrin L. Jun 21 '15 at 12:31

A basis of open neighborhoods of $f$ is given as follows:

• For any choice of finitely many vector fields $X_1,\dots,X_k \in \mathfrak X(M)$ and any function $h\in C^\infty(M)$ let $U_{X_1,\dots,X_k}(f) = \{g\in c^\infty(M,\mathbb R): |X_1X_2\dots X_k(g-f)|<1, |h(f-g)|<1\}$.

This describes the Whitney topology: If you go out to infinity, you can control finitely many derivatives better and better: this is the projective limit of all the Whitney $C^k$-topologies.

There is the finer Whitney topology: If you go out to infinity on $M$, you can control more and more derivatives better and better. To describe this you need families of countably many vector fields such that on each compact set only finitely many are nonzero, and repeat the above description.

• Thank you. Does this this appear anywhere in the literature, ideally with a proof of the equivalence to one of the common descriptions in terms of charts or jets? – Kathrin L. Jun 20 '15 at 12:06
• Not that I know. But the proof is straightforward: Use appropriately nested compact exhaustions of $M$ and partitions of unity to connect with the description using charts. – Peter Michor Jun 20 '15 at 13:07
• I'm a bit confused. Your neighborhoods seem not to control the $C^0$-behaviour. Consider the set of functions with values $\epsilon$-close to $f$. This should be a neighborhood of $f$ in the Whitney topology, which is not open in the topology you suggest. – Kathrin L. Jun 20 '15 at 15:18
• You are right. Use also any function to remedy this. – Peter Michor Jun 20 '15 at 15:37
• Could you please elaborate a bit on the proof of the equivalence? I managed to show that the Whitney topology is finer than the one you suggested in your answer, but was not able to prove the converse. I try to use the definition with charts. – Kathrin L. Jun 25 '15 at 10:53