# Are these strengthenings of Serre-Swan and Gelfand-Naimark true?

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.

• $C^\ast(X)$ and $C(\beta X)$ are isomorphic rings. Same for $C(X)$ and $C(\nu X)$. Probably as $\Bbb R$-algebras they're isomorphic too. – Henno Brandsma Sep 10 at 8:47
• 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. – WoolierThanThou Sep 10 at 8:48
• @WoolierThanThou He only looks at the algebraic structure.. – Henno Brandsma Sep 10 at 8:53
• Oh, right... well, see below. – WoolierThanThou Sep 10 at 8:55
• There is a duality for locally compact spaces, but it's not as nice as one might hope, see math.stackexchange.com/questions/170984/… – Alessandro Codenotti Sep 10 at 9:17

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...

• I did not expect that! It's clear that I have to retune the intuition that I thought I had... – user554397 Sep 10 at 11:09
• 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. – Qiaochu Yuan Sep 12 at 3:06
• @QiaochuYuan this is a rare but classical case where they’re the same. Because we can extend all real functions on $\omega_1$ to $\omega_[+1$. – Henno Brandsma Sep 12 at 4:32
• @HennoBrandsma What if we require our spaces to be CW complexes? – user554397 Oct 9 at 16:25

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.

• 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? – user554397 Sep 17 at 12:58
• @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. – Qiaochu Yuan Sep 17 at 19:07