The Serre-Swan states that given a vector bundle $V$ with base $X$ (compact and Hausdorff), we have that $\Gamma(V)$ is a finitely generated projective $C(X)$-module. Moreover, all finitely generated projective $C(X)$-module are of this form.

If we take an inner-product in each local trivialization of $V$, using a finite partition of unity we can construct a $C(X)$-inner product: $$\langle\quad,\quad\rangle:\Gamma(V)\times \Gamma(V)\rightarrow C(X) $$ This inner product makes $\Gamma(V)$ a $C(X)$-Hilbert module that due to Serre-Swans theorem is finitely generated and projective. Can we conclude that $\Gamma(V)$ is a direct summand of $C(X)^n$ (for some $n$) as a $C(X)$-Hilbert module? This would mean that there exists a Hilbert module $H$ such that $M\oplus H\simeq C(X)^n$ isometrically. I know that if we only consider the module structure this is true. The problem that arises when trying to imitate the proof for the module scenario is that the canonical morphism $f:C(X)^n\rightarrow M$ doesn't need to preserve the Hilbert module structure and therefore we can't assert that $M$ is "projective" in a Hilbert module sense.

The motivation behind this is that I'd like to show that $\Gamma(V)$ has a finite Parseval frame (also know as a normalized tight frame).

  • $\begingroup$ I'm not completely sure of what you're asking, but I'll give it a go. Since $\Gamma(V)$ is finitely generated and projective, there exists $n \in \mathbb{N}$ and a projection $p \in M_n(C(X))$ so that $\Gamma(V)$ is isomorphic to $pC(X)^n$ as right Hilbert $C(X)$-modules. We then have $pC(X)^n \oplus (1_n - p)C(X)^n \cong C(X)^n$. $\endgroup$ – Zorngo Jun 9 '18 at 2:03

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