Given a real $m$-dimensional smooth manifold why do we consider tangent spaces and what not, when we can just use Whiteney's Embedding Theorem and do calculus in $\mathbb{R^{2m}}$?

I assume there is a good reason!

  • $\begingroup$ While I suppose this is more a personal preference, working with things like tangent vector fields in $\mathbb{R}^3$ and lugging dot products etc. everywhere is more tiresome than working with an intrinsic metric. I'd imagine that situation only gets worse in the general case. $\endgroup$ Oct 23 '13 at 13:34

The thing is, there is nothing canonical about such embeddings; a manifold does not come equipped with such an embedding, and of course there are many such. Even if we choose to regard manifolds as certain subsets of Euclidean space, the fact remains that it is possible to embed them in different ways. How do we know which properties are intrinsic to the manifold (i.e., are embedding-independent)?

This consideration goes back to the very beginnings of differential geometry and (what one must call) differential topology, although the latter term wasn't around then. Even though Gauss studied surfaces as special subsets of $\mathbb{R}^3$, a major achievement was the recognition that the Gaussian curvature of a surface (which was defined extrinsically, using e.g. the Gauss normal map) was an intrinsic invariant of the surface, i.e., can be measured directly in terms of the metric restricted to the surface. This is his Theorema Egregium.

I think the advantage of intrinsic definitions (such as the standard definition of manifold) is that they often serve to remove a certain amount of conceptual clutter, so that such considerations are built into the definitional framework from the outset. I think essentially the same question could be raised about finite-dimensional vector spaces (let's say over the real numbers). Of course, they all arise concretely, being isomorphic to $\mathbb{R}^n$, but it is often conceptually clearer not to be tied to a specific such representation, and to develop the theory in a coordinate-free way. Standard manifold theory serves much the same purpose.


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.