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quasar987
Oct3-08, 03:49 PM
My book says that the cusp y=x^2/3 is not an embedded submanifold of Rē. Why is that?

jostpuur
Oct4-08, 10:18 AM
According to my notes the embedding is defined like this:

Let M and N be differentiable manifolds, and f:M\to N a smooth (C^{\infty}) mapping. If for all p\in M the tangent space mapping f_{*p}:T_p(M)\to T_{f(p)}(N) is injective, and f:M\to f(M) is a homeomorphism when f(M) has the induced topology from N, then f is an embedding of M in N.

If we set the natural differentiable structures on \mathbb{R} and \mathbb{R}^2, then a mapping

f:\mathbb{R}\to\mathbb{R}^2,\quad\quad f(x)=(x, (x^2)^{1/3})

is not an embedding, because it is not smooth at origo.

quasar987
Oct4-08, 12:15 PM
True, but an embedded submanifold is by definition (or characterisation) the image of a smooth embedding. Couldn't there be a pair (M, f) other than M=R and f given in your post such that f(M) = the cusp, and such that f is an smooth embedding?

jostpuur
Oct4-08, 04:01 PM
So is the real question this: We give \mathbb{R}^2 the natural differentiable structure, define f:\mathbb{R}\to\mathbb{R}^2 like in my post, and ask that does this mapping somehow induce a differentiable structure on \mathbb{R} so that f becomes smooth?

Or perhaps the question is, that why cannot this f induce differentiable structure in such way, assuming that the book's claim is right?

quasar987
Oct4-08, 08:50 PM
The real question is this:
Consider Rē with the natural differentiable structure, and C the subset of Rē defined by the equation y=x^2/3. Is there a smooth manifold M and an embedding f:M-->Rē whose image is C.

mathwonk
Oct5-08, 12:52 AM
every line through the origin of that set has intersection number ≥ 2 with the set. for a manifold, the generic intersection number will be one.

jostpuur
Oct5-08, 03:46 AM
Isn't the content of the posts #4 and #5 the same?

quasar987
Oct5-08, 10:21 AM
every line through the origin of that set has intersection number ≥ 2 with the set. for a manifold, the generic intersection number will be one.

Well, the cusp is a smooth manifold, with smooth atlas consisting of the unique chart "projection onto the x coordinate".

Isn't the content of the posts #4 and #5 the same?

Well, it seems to me that the question in post #5 is more general than any of the 2 questions of post #4. Or perhaps they are equivalent?