Proving [imath]\{(x,y,z)\in \mathbb{R}^3 \vert x^2-y^2-2z=0\}[/imath] is a submanifold of [imath]\mathbb[R}^2[/imath]

[MathNugget]

I have some ideas, but I need a bit of help.

Firstly, I think the path to get this done is proving that set is a manifold, then proving it is an embedding (I use the definition of submanifold through an embedding).

I think I have to first rearrange things a bit:
[math]M=\{(x,y,z)\in \mathbb{R}^3 \vert x^2-y^2-2z=0\}=\{(x,y,\frac{x^2-y^2}{2})\mid x,y,z\in \mathbb{R}\}[/math]
Then I define [imath]f:M\rightarrow \mathbb{R}^2[/imath], [imath]f(x,y,\frac{x^2-y^2}{2})=(x,y)[/imath]. M is Hausdorff and second-countable as part of [imath]\mathbb{R}^3[/imath], and this function f would be the homeomorphism. The atlas also would have a single map...

That function has injective differential, is injective and is a diffeomorphism...does this suffice to prove it?

 

[blamocur]

I have some ideas, but I need a bit of help.

Firstly, I think the path to get this done is proving that set is a manifold, then proving it is an embedding (I use the definition of submanifold through an embedding).

I think I have to first rearrange things a bit:
[math]M=\{(x,y,z)\in \mathbb{R}^3 \vert x^2-y^2-2z=0\}=\{(x,y,\frac{x^2-y^2}{2})\mid x,y,z\in \mathbb{R}\}[/math]
Then I define [imath]f:M\rightarrow \mathbb{R}^2[/imath], [imath]f(x,y,\frac{x^2-y^2}{2})=(x,y)[/imath]. M is Hausdorff and second-countable as part of [imath]\mathbb{R}^3[/imath], and this function f would be the homeomorphism. The atlas also would have a single map...

That function has injective differential, is injective and is a diffeomorphism...does this suffice to prove it?

Would not you rather consider [imath]g : \mathbb R^2 \rightarrow M[/imath] or [imath]\mathbb R^2 \rightarrow \mathbb R^3[/imath] ?

 

[MathNugget]

Would not you rather consider [imath]g : \mathbb R^2 \rightarrow M[/imath] or [imath]\mathbb R^2 \rightarrow \mathbb R^3[/imath] ?

I guess I would consider [imath]g: \mathbb{R}^2\rightarrow M[/imath], [imath]g(x, y)=(x, y, \frac{x^2-y^2}{2})[/imath], since it is the inverse of f, and it would help me prove f (and g) are homeomorphisms. I don't think I understand what you mean.


On a side note: I realize now, when I want to prove something is a submanifold of [imath]\mathbb{R}^n[/imath], it is enough to provide the diffeomorphism providing the embedding, as all diffeomorphisms are homeomorphisms, so the supposed submanifold is immediately locally euclidean. Right?

 

[blamocur]

I guess I would consider [imath]g: \mathbb{R}^2\rightarrow M[/imath], [imath]g(x, y)=(x, y, \frac{x^2-y^2}{2})[/imath], since it is the inverse of f, and it would help me prove f (and g) are homeomorphisms. I don't think I understand what you mean.


On a side note: I realize now, when I want to prove something is a submanifold of [imath]\mathbb{R}^n[/imath], it is enough to provide the diffeomorphism providing the embedding, as all diffeomorphisms are homeomorphisms, so the supposed submanifold is immediately locally euclidean. Right?

I meant that [imath]g:\mathbb R^2 \rightarrow \mathbb R^3[/imath] can be shown to be an embedding whose image is [imath]M[/imath], which would prove that [imath]M[/imath] is a submanifold -- wouldn't it?

 

[MathNugget]

I meant that [imath]g:\mathbb R^2 \rightarrow \mathbb R^3[/imath] can be shown to be an embedding whose image is [imath]M[/imath], which would prove that [imath]M[/imath] is a submanifold -- wouldn't it?

I guess? I thought for [imath]M \subset N[/imath], the function should be [imath]f: M \rightarrow N[/imath] to prove it is an embedding.
I suppose it is the same thing if we look for [imath]f^{-1}: N \rightarrow M[/imath], but all the properties would look different...