10
$\begingroup$

The Serre-Swan theorem states (at least in one form) that the category of real vector bundles over a compact Hausdorff space $M$ is equivalent to the category of finitely generated projective modules over the ring $C(M)$ of continuous functions on $M$. The equivalence is provided by the functor $\Gamma$ sending a bundle to the totality of all its continuous sections.

Are there any classic applications or uses of this theorem? To me right now it seems like a pristine result to be admired from a distance, as I currently know of no actual use for it. I'd love to remedy that!

$\endgroup$
  • 4
    $\begingroup$ Serre-Swan always struck me as more philosophically important than anything else. $\endgroup$ – Qiaochu Yuan Jun 29 '11 at 3:12
14
$\begingroup$

One application of this is that topological K-theory (i.e., the K-theory of the exact category of vector bundles on the space) is the same thing as the "algebraic" $K_0$ of the ring of continuous functions (i.e., the K-theory of the exact category of finitely generated projective modules over that ring). So topological K-theory is a "special case" of algebraic K-theory (though the tools for proving things like Bott periodicity are very different from those used in proving general results on exact category in algebraic K-theory).

By the way, here is the corresponding result in algebra:

Algebraic vector bundles over an affine scheme $\mathrm{Spec} A$ are the same as finitely generated projective $A$-modules (let's say $A$ is noetherian). So a module is projective if and only if it is locally free, in algebra language.

$\endgroup$
  • $\begingroup$ The result about algebraic vector bundles holds in full generality, not needing that $A$ is Noetherian. See for instance Theorem 7.22 in Pete Clark's notes. $\endgroup$ – Ingo Blechschmidt Mar 10 '15 at 10:00
  • $\begingroup$ Very late comment, but it seems worth mentioning that the higher topological $K$-groups of $X$ do not coincide with the higher algebraic $K$-groups of $C(X)$ (they do coincide with the higher operator algebraic $K$-groups). So "special case" is only true for $K_0$. $\endgroup$ – MaoWao Nov 28 '18 at 12:59
8
$\begingroup$

You could perhaps do worse than consulting $\S 6.4$ of my commutative algebra notes: "Applications of Swan's Theorem." (You could definitely do better: see below.)

The first application I give is to show that the ring of real-valued continuous functions on $[0,1]$ is a connected ring in which each finitely generated projective module is free but for which there is a nonfree infinitely generated projective module. As I admit myself in the notes, it is possible to prove this purely algebraically and I allude to another proof taken from one of Lam's books, but the topological approach is a nice one.

The second application is a big one: it exhibits stably free non-free modules over the ring $\mathbb{R}[x_0,\ldots,x_n]/(x_0^2+\ldots + x_n^2 - 1)$ of polynomial functions on the $n$-sphere when $n \neq 0,1,3,7$. This is done by reducing to the known behavior of tangent bundles to the $n$-sphere in the usual differential topological setting.

By the way, I got this second application directly from Swan's paper. There are other applications given there as well...

$\endgroup$
0
$\begingroup$

One nice application, which took me a while to understand, (and Wikipedia was of great help!) is the beautiful "bridge" between Algebra and Differential Geometry, that the bundle of frames is trivial iff it has a (global, obviously) section. (this means in this case a basis of each tangent space, globally and of course differentiably defined). Serre-Swan theorem measures in this case how far a bunlde is from being isomorphic to a product, and at the same time, how far its space of sections is far from being free (over the ring of differentiable functions on the manifold)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.