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The Soul Theorem states that in every complete, connected riemannian manifold $M$ with $\mathrm{sec}(M)\geq 0$, there exists compact, totally convex, totally geodesic submanifold $S$ such that $M$ is diffeomorphic to the normal bundle of $S$.

I don't know if I'm being too formal on my approach but I don't totally get how a manifold could be diffeomorphic to a bundle given that they're in different classes of objects. What is the precise meaning of diffeomorphic here? In general the normal bundle of $S$ (any submanifold) will be of dimension les than $\mathrm{dim}(M)$ right? So how is it possible that they're diffeomorphic?

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    $\begingroup$ You're missing 2 hypotheses: $M$ must be noncompact and the Riemannian metric you've chosen must be nonnegatively curved. $\endgroup$ – Jason DeVito May 23 '12 at 1:36
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The total space of any vector bundle over a manifold is itself a manifold; if it has rank $r$ and the base space has dimension $n$, it has dimension $r+n$ (since it locally looks like a product). Since the rank of the normal bundle $NS$ is equal to the codimension of $S$ in $M$, the dimension of $NS$ as a manifold is equal to the dimension of $M$.

The general version of the picture you want to be visualizing (which is more elementary than the Soul Theorem) is the tubular neighborhood theorem; one phrasing of it states that if $A$ is a submanifold of $M$ then $A$ has a neighborhood in $M$ which is diffeomorphic to $NA$.

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  • $\begingroup$ ok. so I should think of the bundle as a union of the fibers right? $\endgroup$ – Sak May 23 '12 at 1:37
  • $\begingroup$ Yeah, with the topology and smooth structure that come from being a local product. $\endgroup$ – Micah May 23 '12 at 1:40

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