The question is a special case of a previous question.

Let $M$ be a compact smooth manifold, then it is clear that $C^{\infty}(M)$ is a Frechet algebra with pointwise multiplication and a collection of semi-norm defined by $p_{\alpha}(f):=\sup_{\beta\leq\alpha}||\partial^{\beta}(f)||$.

Now let $E\to M$ be a $C^{\infty}$-fiber bundle with compact base and compact fiber. Then it is clear that $C^{\infty}(E)$, the space of smooth functions on the total space of the fiber bundle, is a Frechet $C^{\infty}(M)$-module.

My question is: is $C^{\infty}(E)$ always a projective Frechet $C^{\infty}(M)$-module?

I think this question is trivial for experts. Please let me know if there is any references or it the question is not suitable for mathoverflow.

  • $\begingroup$ what is the module structure while the fibers are not vector space? $\endgroup$ – Ali Taghavi Jul 1 '18 at 11:34
  • $\begingroup$ I am sorry I just realize the module structure(by composition) $\endgroup$ – Ali Taghavi Jul 1 '18 at 11:35
  • 1
    $\begingroup$ Spaces of sections of smooth VECTOR bundles over $M$ are exactly the finitely generated projective $C^\infty(M)$-modules: Namely, choose a a second VECTOR bundle $F$ such that $E\oplus F$ is trivial, isomorphic to the space of smooth sections of $M\times \mathbb R^n \to M$, which is the free module $C^\infty(M)^n$. Projection with kernel $F$ shows that $\Gamma(E)$ is a direct summand in there. This argument can be read in the other direction also (Theorem of Serre-Swan). $\endgroup$ – Peter Michor Jul 1 '18 at 16:26

The answer is Yes. Basically, one can realize $C^\infty(E)$ as a direct summand in a larger projective $C^\infty(M)$-module. Perhaps this really is trivial to experts, but to me the argument occurred only after staring sufficiently long at related arguments in this article, which came up in your other question:

Ogneva, O. S., Coincidence of the homological dimensions of the Fréchet algebra of smooth functions on a manifold with the dimension of the manifold, Funct. Anal. Appl. 20, 248-250 (1986); translation from Funkts. Anal. Prilozh. 20, No. 3, 92-93 (1986). ZBL0626.46057.

Namely, start by recalling the isomorphism $C^\infty(U \times F) = C^\infty(U) \hat{\otimes} C^\infty(F)$, where $\hat{\otimes}$ is the projective tensor product of Fréchet spaces. This means that $C^\infty(U\times F)$ is a free $C^\infty(U)$-module (freely generated by $C^\infty(F)$, in the category of Fréchet $C^\infty(U)$-modules), and hence projective. By the result of Ogneva, for any open $U\subset M$, $C^\infty(U)$ is projective over $C^\infty(M)$. Hence, $C^\infty(U\times F) \cong C^\infty(U) \otimes_{C^\infty(M)} C^\infty(M\times F)$ is also projective over $C^\infty(M)$, being the tensor product of projective modules.

Next, consider a countable, locally finite open cover of $(U_i)$ of $M$, trivializing the fiber bundle $E\to M$ as $E|_{U_i} \cong U_i\times F \to U_i$, where $F$ is the typical fiber, and let $(\chi_i)$ be a partition of unity subordinate to this cover. Such a cover and corresponding partition of unity certainly exist if the base $M$ is compact, but also more generally even if $M$ is not compact, but satisfies a suitable countability condition (second countable, paracompact). Now, the countable direct product $\prod_i C^\infty(U_i \times F)$ is still Fréchet and projective over $C^\infty(M)$, and it has $C^\infty(E)$ as a direct summand, as evinced by the inclusion/projection maps \begin{align*} C^\infty(E) \to \prod_i C^\infty(U_i \times F) &\colon f \mapsto (f|_{E_{U_i}}) , \\ \prod_i C^\infty(U_i \times F) \to C^\infty(E) &\colon (f_i) \mapsto \sum_i \chi_i f_i . \end{align*}


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