Prove that sphere $\mathbb{S}\subset\mathbb{R}^{n+1}$ is a dimension $n$ surface.

Question

A set $M\subset\mathbb{R}^n$ is called a $m$-dimension surface when, for every point $x\in M$, exists $A$ open on $M$ which contains $x$, $A_0$ open in $\mathbb{R}^m$ and a homeomorphism $\varphi:A\to A_0$. Prove the sphere $\mathbb{S}^{n}=\{x\in\mathbb{R}^{n+1};\langle x,x\rangle=1\}$ is a $n$-dimension surface.

It looks like I may have to play with some function like stereographic projection but no progress was made by me.

Answer 1

Consider the hyperplane $H$ defined by $x_n=1$, which is tangent to $\Bbb S$ at the point $A=(0,\ldots, 0, 1)$. Note that it doesn't really matter that it is tangent, so you don't have to worry about what this precisely means or how to prove it. The point is that there is a correspondence between points in $\Bbb S\setminus (0,\ldots, 0, -1)$ and points in the hyperplane $H$, given as you guessed by a stereographic projection (set $B=-A=(0,\ldots, 0, -1)$).

For $X=(x_1,\ldots, x_{n}, 1)\in H$, find the intersection between the line $BX$ and the sphere, and call it $\varphi(B)$. Then prove that $\varphi$ is a homeomorphism.

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Another option is to use a map easier to define, but on a reduced domain, and then argue that the sphere has the same structure around every point due to its symmetry. In a small neighbourhood of the north pole $A$ in $H$, map $X=(x_1,\ldots, x_n, 1)$ to the point of the sphere that is close to $A$ and has the same first $n$ coordinates: $$\psi(X)=\left(x_1,\ldots, x_n, \sqrt{1-x_1^2-\ldots - x_n^2}\right)$$ It is easier to define $\psi$ and see that it is a homeomorphism, but you have this additional step of arguing that this can be done around every point in $\Bbb S$ (technically, all you need is to conjugate this $\psi$ by a rotation).

Answer 2

Both $S^n\setminus\{(0,\ldots,0,1)\}$ and $S^n\setminus\{(0,\ldots,0,-1)\}$ are homeomorphic to $\mathbb R^n$ and the union of these two sets is $S^n$. This is enough to prove that $S^n$ is a $n$-dimensional surface.

Answer 3

Pick $x^* \in S^n \subset \mathbb{R}^{n+1}$. Let $L = \{ y | \langle x^*, y \rangle = 0 \}$ and note that $\dim L = n$. Let $b_1,...,b_n$ be an orthonormal basis for $L$, let $B = \begin{bmatrix} b_1 & \cdots b_n\end{bmatrix}$ and let $A_0 = \mathbb{R}^n$.

Note that $B^TB = I$ and if $x \in L$ then $BB^T x = x$.

Let $A= \{ x \in S^n | \langle x^* , x \rangle >0\}$, $A$ is clearly open.

Pick $x \in A$ and choose $t$ such that $tx-x^* \bot x^*$, that is $t(x) = {1 \over \langle x^* , x \rangle}$. $t$ is clearly smooth on $A$.

Define $\phi(x) = B^T(t(x)x-x^*)$ and note that $\phi(x) \in A_0$.

For $z \in A_0$, let $\beta(z) = { Bz +x^* \over \|Bz+x^* \| }$, note that $\beta$ is smooth.

Note that $B\phi(x)+x^* = BB^T(t(x)x-x^*)+x^* = t(x)x$, and so we see that $\beta(\phi(x)) = x$.

Note that $\phi(\beta(z))= B^T( t(\beta(z))\beta(z)-x^*) = t(\beta(z))B^T\beta(z) = z$.

Hence $\phi$ is a diffeomorphism.