Prove that the $n-sphere$ is a manifold

Question

Prove that the $n$-sphere :

$$S^n=\{(x_1,...,x_{n+1})\in\mathbb{R}^{n+1}:x^2+...+X^2_{n+1}=1\}$$

is a manifold by showing that $0$ is a regular value for the function:

$$f:\mathbb{R}^{n+1}\rightarrow\mathbb{R}$$ defined by: $(x_1,...,x_{n+1})\in\mathbb{R}^{n+1}\mapsto x_1^2+...+x^2_{n+1}$

My attempt:

A regular value: of $f:U \rightarrow \mathbb{R}^m$ is a point in $\mathbb{R}^m$\ $f(C_f)$, the complement of the image of all critical points of $f$. (not sure what this is saying)...

Don't know how to do the problem..

Answer 1

So the definition I'm operating with for a regular value is that if you have $G: U\to \mathbb{R}^k$ where $U$ is an open, nonempty subset of $\mathbb{R}^n$ and $k < n$, then $y$ is a regular value if $DG(x)$ has rank $k$ for every $x\in G^{-1}(y)$. This is identical to the preimage of a regular value not having any critical points, since the rank of the derivative is not maximal at those points.

Now, in the case of $f(x) = \sum_i x_i^2$, we have that $Df(x) = 2x$. This has rank 1, since $Df(x) = 0 \implies x = 0$, and $0 \notin f^{-1}(1)$. Thus, $1$ is a regular value, and the preimage of $1$ (the unit sphere) is a manifold.

Answer 2

Given a function $f:\mathbb{R}^n \to \mathbb{R}^m$, a regular value is an element $x \in \mathbb{R}^m$ such that the derivative of $f$ on $f^{-1}(x)$ is surjective.

In this case, the derivative of $f$ is its gradient $\nabla f=(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\dots,\frac{\partial f}{\partial x_{n+1}})=(2x_1,2x_2,\dots,2x_{n+1})$, which is surjective $\forall x \in \mathbb{R}^{n+1}, x \neq 0$.

Since $x=0 \notin f^{-1}(1)$, one can conclude that $n$-spheres are manifolds.

Answer 3

Don't you mean that 1 is a regular value?

Here is the intuition behind the theorem that you are confused by. For the sake of being concrete, I'll describe it in $\mathbb{R}^3$. Take any point $x$ in the sphere. Draw the plane tangent to the sphere at that point. Draw 2 vectors in this plane that put a coordinate system on it. Next draw the line at right angles for a third vector. Those 3 vectors make a basis for the tangent space in $\mathbb{R}^3$ around $x$. And the image of the third vector makes a basis for the tangent space in $\mathbb{R}$ of $f(x) = 1$.

That mental picture should look something like http://standards.sedris.org/18026/text/ISOIEC_18026E_SRF/image022.jpg.

Now the fun thing is that the coordinate system for the tangent space can be projected back to the sphere to wind up with a coordinate space in $\mathbb{R}^3$ for a neighborhood around the point. If we can do that for every point, then the sphere is a manifold because we can cover it in little coordinate systems.

Got that image clearly in your head?

The question is, when can we make this image/construction work out in general? In higher dimensions? With arbitrary manifolds?

Well, suppose that we have a smooth function $f$ from a manifold $N$ of dimension $n$ to a manifold $M$ of dimension $m$. Suppose that we can find $m$ vectors in the tangent space around $x$ in $N$ that are mapped to a basis of the tangent space of $f(x)$ in $M$. What then?

Well, we can always take those $m$ vectors and complete them to a basis for the tangent space of $x$. However the $n - m$ new vectors will generally not be in the space tangent to the surface we are trying to describe. But each of those additional vectors will, after we subtract the right linear combination of the first $m$, map to the $0$ vector in the tangent space of $f(x)$. And this gives us a picture that looks like the one we had for the sphere. And it works out the same way.

We don't have to do the construction. We just have to know that it can be completed.

And that is the intuition behind saying that if the derivative of $f$ is surjective for the tangent space of $f(x)$, then $x$ is regular and at least locally looks like a manifold. If every point is regular, then the preimage of $f$ is a manifold.

So in your case all you have to do is note that for each $x$ in the hypersphere, the gradient vector of $f$ is a basis for the tangent space of 1 in $\mathbb{R}$. And now we know that the construction could be carried out, and the hypersphere is a manifold.