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Let $X$ be a compact Hausdorff space.

  • The Serre-Swan theorem allows us to identify complex vector bundles with projective finitely generated modules over the ring $C(X;\mathbb{C})$ of complex-valued functions on $X$.
  • The Gelfand-Naimark theorem tells us that $X$ is determined up to homeomorphism by the C*-algebra $C(X;\mathbb{C})$.
  • The Gelfand-Kolmogorov theorem says that $X$ is determined up to homeomorphism by the $\mathbb{R}$-algebra $C(X;\mathbb{R})$.

There are several things that I'm bothered with.

  • Considering just how many functions a given space admits, and considering that any function is determined by its local behaviour at every point, I find it difficult to believe that compactness is really necessary.
  • The asymmetry between $\mathbb{R}$ and $\mathbb{C}$ is surprising. Why is there no Serre-Swan theorem for real vector bundles? And why does $C(X;\mathbb{C})$ need more algebraic structure to recover $X$ than $C(X;\mathbb{R})$ does? A search for a real Serre-Swan theorem or a complex Gelfand-Kolmogorov theorem gave me nothing.

The above considerations bring me to the following conjectures. Let $X$ be a locally compact Hausdorff space.

  • Conjecture 1. Complex vector bundles correspond to projective modules over the ring $C(X;\mathbb{C})$, and real vector bundles correspond to projective modules over the ring $C(X;\mathbb{R})$.
  • Conjecture 2. The space $X$ is determined up to homeomorphism by the $\mathbb{C}$-algebra $C(X;\mathbb{C})$, as well as by the $\mathbb{R}$-algebra $C(X;\mathbb{R})$. Perhaps even the ring structure suffices.

My question is of course whether the conjectures are true or not.

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    $\begingroup$ $C^\ast(X)$ and $C(\beta X)$ are isomorphic rings. Same for $C(X)$ and $C(\nu X)$. Probably as $\Bbb R$-algebras they're isomorphic too. $\endgroup$ – Henno Brandsma Sep 10 at 8:47
  • $\begingroup$ What's your topology on $C(X;\mathbb{C})$? You can't lift the uniform norm. However, you could consider $C_0(X;\mathbb{C})$ (the space of functions vanishing at $\infty$), but the natural generalisation of the above theorems seems to work mostly for, say, $\sigma$-compact spaces. $\endgroup$ – WoolierThanThou Sep 10 at 8:48
  • $\begingroup$ @WoolierThanThou He only looks at the algebraic structure.. $\endgroup$ – Henno Brandsma Sep 10 at 8:53
  • $\begingroup$ Oh, right... well, see below. $\endgroup$ – WoolierThanThou Sep 10 at 8:55
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    $\begingroup$ There is a duality for locally compact spaces, but it's not as nice as one might hope, see math.stackexchange.com/questions/170984/… $\endgroup$ – Alessandro Codenotti Sep 10 at 9:17
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Take $X=\omega_1$, in the order topology. Any real or complex valued function on it is bounded so $C(X)$ and $C(\beta X)$ (here $\beta X \simeq \omega_1 +1$, of course) are isomorphic as rings (and as (real) algebras as well, an isomorphism of rings $C(X)$ is always an $\Bbb R$-algebra isomorphism too; I haven't studied the complex case well enough, but the real case is classical.) So in general locally compact Hausdorff spaces you cannot distinguish a space from its (C-S) compactification on ring or algebra structure alone...

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  • $\begingroup$ I did not expect that! It's clear that I have to retune the intuition that I thought I had... $\endgroup$ – user554397 Sep 10 at 11:09
  • $\begingroup$ Why is $beta X \cong \omega_1 + 1$ "of course"? Presumably that's the one-point compactification but I'd expect the Stone-Cech compactification to be much larger a priori. $\endgroup$ – Qiaochu Yuan Sep 12 at 3:06
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    $\begingroup$ @QiaochuYuan this is a rare but classical case where they’re the same. Because we can extend all real functions on $\omega_1$ to $\omega_[+1$. $\endgroup$ – Henno Brandsma Sep 12 at 4:32
  • $\begingroup$ @HennoBrandsma What if we require our spaces to be CW complexes? $\endgroup$ – user554397 Oct 9 at 16:25
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Conjecture 1 is true.

The main technical content of the proof is that over a compact Hausdorff space, every real or complex vector bundle is a direct summand of a trivial real or complex vector bundle; see this math.SE question (which doesn't have a proof but does have a comment with a link to a proof).

When $X$ is compact Hausdorff, the C*-algebra structure of $C(X, \mathbb{C})$ can be recovered from its $\mathbb{C}$-algebra structure, so the $\mathbb{C}$-algebra structure determines $X$.

To see this, first observe that the range of a function $f : X \to \mathbb{C}$ can be recovered by considering its spectrum $\sigma(f)$ (the set of $\lambda \in \mathbb{C}$ such that $f - \lambda$ fails to be invertible), which only depends on the $\mathbb{C}$-algebra structure. From the spectrum we can recover the norm as the spectral radius

$$\| f \| = \sup_{\lambda \in \sigma(f)} \| \lambda \|$$

so the norm can be recovered from the $\mathbb{C}$-algebra structure. Next, we can also get the *-structure by considering the subspaces of $C(X, \mathbb{C})$ consisting of elements with purely real resp. purely imaginary spectrum; $C(X, \mathbb{C})$ is always the direct sum of these, and the *-structure acts by the identity on the first bit and by $-1$ on the second bit.

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  • $\begingroup$ Thanks for the answer. But is the vector bundle being a direct summand of a trivial bundle also true for non-compact (but still locally compact) spaces? $\endgroup$ – user554397 Sep 17 at 12:58
  • $\begingroup$ @guest: I don't know, but I don't think so. The proof crucially uses compactness to find a finite open cover trivializing the bundle. $\endgroup$ – Qiaochu Yuan Sep 17 at 19:07

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