Let $E$ be locally convex topological vector space. Let $c^\infty E$ denote the same vector space equipped with the $c^\infty$-topology (i.e. the finest topology on it, s.t. all smooth curves $\mathbb{R} \rightarrow E$ are continous, see "The convenient setting of global analysis" by A.Kriegl and P.Michor, I.2.12). If now $E,F$ are two such spaces, it may happen that the identity $c^\infty(E \times F) \rightarrow c^\infty(E) \times c^\infty(F)$ is continous but not a homeomorphism (see I.4.16 of the same reference).

I am wondering whether it is a homeomorphism in the following situation: Let $M$ be a smooth, finite-dimensional (but not necessarily compact) manifold and $V = V_1 \oplus V_2$ the Whitney sum of two finite-rank vector bundles over $M$. Then we have $\Gamma_c(V) \cong \Gamma_c(V_1) \times \Gamma_c(V_2)$ as vector spaces and I think this is also an isomorphism of locally convex spaces if all these spaces are equipped with the usual strict inductive limit topology. But are the spaces $c^\infty \Gamma_c(V_1) \times c^\infty\Gamma_c(V_2)$ and $c^\infty\Gamma_c(V)$ homeomorphic ? As far as I can see, they are linearly diffeomorphic (since the isomorphism maps smooth curves to smooth curves) but I don't know, under which (additional) conditions this map is also a homeomorphism.

  • $\begingroup$ @Semon Choi: Thanks for correcting the tag, "topological-vector-spaces" is certainly the right place, I was not aware of it. Actually, it reminds me, that I could/should have added another piece of information: In the case of $E = \Gamma_c(V)$, it is actually known, that $c^\infty E$ is not a topological vector space if $M$ is not compact. (Kriegl-Michor, I.4.26 (ii)). So in particular, the $c^\infty$-topology does not coincide with the locally convex one. $\endgroup$ – fhanisch Aug 16 '13 at 15:49

$\def\Gmc#1{\Gamma_{\rm c}(#1)}\def\ci{{\rm c}^\infty} $Generally the linear isomorphism $\ci\Gmc{V_1\oplus V_2}\to\ci\Gmc{V_1}\times\ci\Gmc{V_2}$ is not a homeomorphism. For example, when taking $E=\mathcal D(\mathbb R)$, the identity map $\iota:c^\infty E\times c^\infty E\to c^\infty(E\times E)$ is not continuous. This follows from Proposition 4.26(ii) on page 45 in Kriegl and Michor's book. To see this, just observe that there it is shown that the vector addition $a:c^\infty E\times c^\infty E\to c^\infty E$ is not continuous. If $\iota$ were continuous, then $a$ would also be such since it is the composition $a\circ\iota:c^\infty E\times c^\infty E\to c^\infty(E\times E)\to c^\infty E$, and $a$ as a linear map is continuous $c^\infty(E\times E)\to c^\infty E$ as it is $E\times E\to E$.

Added. (in response to the comment, 23.8.2013) Applying the above argument to $E=s^{\kern.5mm(\mathbb N)}$ instead of $\mathcal D(\mathbb R)$ together with the existence of linear homeomorphisms $\Gmc{V_i}\simeq s^{\kern.5mm(\mathbb N)}$ provided in the proof of Theorem 6.14 on page 73 in Kriegl and Michor's book, when $V_1$ and $V_2$ are non-zero-finite-rank (smooth) vector bundles over a non-compact but second countable non-zero-finite-dimensional manifold, from the diagram $$\require{AMScd} \begin{CD} \ci\Gmc{V_1\oplus V_2} @>\simeq>> \ci(\Gmc{V_1}\times\Gmc{V_2}) @>\simeq>> \ci(E\times E) \\ @. @| @| \\ {} @. \ci\Gmc{V_1}\times\ci\Gmc{V_2} @>>\simeq> \ci E\times\ci E \end{CD} $$ one directly obtains the general result required.

Remark 1. There is a misprint in Kriegl and Michor's book at the end of the proof of Lemma 4.23 on page 44. Namely, there should be

... $b_{N,K}\in U$ by (b''). Thus, ...

in place of

... $b_{N,K}\in U$ by (b'). Thus, ...

The proofs are taken almost verbatim from Frölicher and Kriegl's book Linear Spaces and Differentiation Theory, and the same misprint occurs also there on page 194.

Remark 2. In Kriegl and Michor's Theorem 6.14, it is stated that the base manifold be separable. However, in view of the knowledge of existence of the two-dimensional smooth (or even analytic) Gauld manifold which is separable but not paracompact nor second countable, $\sigma$-compact, etc., I strongly suspect that this be sufficient. Rather, one should require the base to be e.g. second countable.

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  • $\begingroup$ Thanks a lot ! That's a really nice and simple way to show that the identity is not a homeomorphism. Since the spaces of compactly supported sections of different vector bundles are isomorphic as lcs (see e.g. proof of Theorem I.6.14 in Kriegl's and Michor's book and the refs in there), it is probably not even necessary to modify the proof for the more general case, assuming that the base manifold is non-compact and ranks of bundles are nonzero. $\endgroup$ – fhanisch Aug 22 '13 at 11:36

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