Differential Geometry – Reconstructing Base Manifold from Smooth Vector Bundles

ct.category-theorydg.differential-geometryvector-bundles

It's probably a really naive question, but I didn't find any references. Given a category $V$ and we know that it is equivalent to the category $\mathbf{Vect}(X)$ of smooth vector bundles over a smooth manifold $X$, under what assumptions could we determine $X$? Or is this generally hopeless?

Best Answer

Yes, it is possible to recover the manifold through the following steps:

  • Smooth Serre-Swan theorem: $\mathbf{Vect}(X)$ is equivalent to the category of finitely generated projective modules over $C^\infty(X)$.
  • If $R$ is any commutative ring, then $R$ is canonically isomorphic to the ring of natural endomorphisms of the identity functor on the category of finitely generated projective modules. Thus, we obtain $C^\infty(X)$ as the ring of natural endomorphisms of $\mathrm{id}_{\mathbf{Vect}(X)}$.
  • Finally, it is well-known that a manifold can be recovered from its ring of smooth functions. More precisely, the functor $C^\infty$ from manifolds to the opposite category of commutative rings is fully faithful.
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