I was looking for a simple example of a covariant derivative on a bundle, where the bundle is not projective. If necessary, the example could be from complex or noncommutative geometry, but I would hope for a simple example with the usual calculus on a manifold.

In noncommutative geometry, there is a theorem by Cuntz and Quillen saying that for the universal calculus, a module on an algebra has a covariant derivative if and only if the module is projective. The universal calculus should be the most restrictive in this context, there should be other calculi with non-projective modules having covariant derivatives. I suspect that algebraic geometry has lots of these, but I would like an easy one to explain!

Added: I guess that an example might be based around a skyscraper sheaf, but I don't see how to give such a thing a connection! (I am likely being stupid here...)


The space of smooth sections $\Gamma(E)$ of vector bundle $E\to M$ over a manifold $M$ is always a finitely generated projective module over the algebra $C^\infty(M)$ of smooth functions: it is a direct summand in a finitely generated free module (Choose a second vector bundle $F$ such that the fiber sum $E\oplus F = M\times \mathbb R^n$ is a trivial bundle.You can do this with the help of a finite trivializing atlas for the bundle $E$ which exists by the use of covering dimension. So in differential geometry there is no chance for this.

  • $\begingroup$ Thanks - there is no chance for a locally trivial bundle to do this. But is there any sheaf-like structure (expressed as a module with connection over the algebra of functions) for which it can be done? $\endgroup$ – Edwin Beggs May 8 '15 at 12:25
  • $\begingroup$ Don't we need compactness of $M$ for this? $\endgroup$ – Thomas Rot Jun 7 '15 at 15:48
  • $\begingroup$ @Thomas Rot: Since $M$ is second countable, there is a finite trivializing atlas (consisting of not-connected open sets in general) for the bundle $E$. This follows from dimension theory. $\endgroup$ – Peter Michor Jun 8 '15 at 6:34

OK, on more thought (apologies, I didn't think about things the right way before...): Let $A=C^\infty(\mathbb{R}/\mathbb{Z})$ (just to keep to a compact manifold), and $B$ be the algebra of complex valued power series with coefficients decaying faster than any power. The Taylor expansion at the point zero gives an algebra map $T_0:A\to B$, and we use this map to make $B$ into a left $A$-module. There is a connection $\nabla:B\to\Omega^1 A\otimes_A B$ (with the usual calculus on $A$) defined by $$ \nabla \big(\sum_{n\ge 0} b_n\,x^n\big)=dx\otimes \sum_{n\ge 0} n\,b_n\,x^{n-1}\ . $$ Now suppose that $\theta:B\to A$ is a left $A$-module map. Then for all $b\in B$ and $f\in A$, $\theta(T_0(f)\,b)=f\,\theta(b)$. If there is a $b\in B$ so that $\theta(b)\neq 0$, then there is an $y\in\mathbb{R}$ so that $y\neq 0$ and $\theta(b)(y)\neq 0$ (this as $\theta(b)$ must be continuous). Now choose a smooth function $f$ so that $f(y)\neq 0$, but $f$ vanishes on a neighbourhood of zero. We get a contradiction. Thus the only left $A$-module maps $\theta:B\to A$ must be zero, and in particular $B$ cannot be a projective module, as then it would be a submodule of a free module, and free modules have lots of such maps.


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