Sphere as an embedded submanifold

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In summary, the conversation discusses different ways of proving that a sphere is an embedded submanifold, using various theorems and examples. It mentions the importance of specifying which manifold the submanifold is embedded in and the concept of proper submanifolds. Ultimately, it is determined that the simplest way to prove this is by using the inclusion map, which is always an embedding.
  • #1
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hello,
i already have 2 ways using 2 theorems to prove that a sphere is an embedded submanifold.
i just want to check if i can prove it using this specific theorem:
let M and N be manifolds
if \phi : M --> N is an embedding, then \phi (M) is an embedded submanifold.
because i am having trouble finding this \phi.
thank you.
 
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  • #2
Submanifold of what? It makes no sense to speak of a submanifold without referring to what manifold it's a submanifold of. For instance in your theorem you assert that [itex]\phi(M)[/itex] is a submanifold of N.

If you don't care what manifold it's a submanifold of you can just use the rather trivial example of letting [itex]\phi : \mathbb{S}^n \to \mathbb{S}^n[/itex] be the identity, and then getting that [itex]\mathbb{S}^n[/itex] is a submanifold of itself (or do you perhaps only want to deal with proper submanifolds?).
 
  • #3
yeah sorry, i forgot to mention in the theorem, a submanifold of N, as you said.
and i do want it a proper submanifold,
i guessed N might be R^3, for instance.
 
  • #4
Well in that case you know the n-sphere is a subset of the (n+1)-dimensional Euclidean space [itex]\mathbb{S}^n \subseteq \mathbb{R}^{n+1}[/itex]. [itex]\mathbb{R}^{n+1}[/itex] is of course a manifold so we just define [itex]\phi : \mathbb{S}^n \to \mathbb{R}^{n+1}[/itex] to be the inclusion [itex]\phi(x)=x[/itex]. This is an embedding.
 
  • #5
" Well in that case you know the n-sphere is a subset of the (n+1)-dimensional Euclidean space .S^n is of course a manifold so we just define Phi: S^n -->R^(n+1) to be the inclusion Phi(x)=x . This is an embedding." (Sorry,I don't know well how to use
quoting function)

I think this is somewhat tautological, i.e., that if A is a subset of X, A given
the subspace topology, then the inclusion map of A in X is an embedding
into X, but I may be wrong:

Given a space X, and a subspace A of X, the inclusion map i:A-->X is always an
embedding:

i) Let U be open in X. Then i^-1(U)=U/\A , open in subspace of A in X.

ii) Let V open in (A, subspace) . Then V=W/\A ; W open in X
then i(V) is open in the subspace topology of X.
 
  • #6
Bacle said:
I think this is somewhat tautological, i.e., that if A is a subset of X, A given
the subspace topology, then the inclusion map of A in X is an embedding
into X, but I may be wrong:
This is correct. I'm sorry if I somehow conveyed with my post that it was a non-trivial fact. jem05 was asking for an embedding from the n-sphere into some larger topological space, so I just provided the simplest one I could think of.
 
  • #7
ramshop wrote, in part:

" This is correct. I'm sorry if I somehow conveyed with my post that it was a non-trivial fact. jem05 was asking for an embedding from the n-sphere into some larger topological space, so I just provided the simplest one I could think of. "

No problem. It is the only one I can think of myself, other than maybe the cube, or
minor variations of it.
 

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