How to show that the unit sphere is a topological manifold?
Sorry for this basic question, but I´m not sure of something. I want to see one example. The definition of a n-manifold is a Hausdorff space, such that each point has an open neighborhood homeomorphic to the open n-dimensional disc. How can i prove that $$ S^n = \left\{ {x \in \mathbf{R}^{n + 1} :\left| x \right| = 1} \right\} $$ is an n-manifold
Solution 1:
$S^n$ is a Hausdorff space because it's a subspace of a Hausdorff space ($\mathbb{R}^{n+1}$).
As for an open neighbourhood for every point homeomorphic to some open subset of $\mathbb{R}^n$ (you don't need this open subset to be the open $n$-dimensional disk, but , in this case, you naturally get a disk), you can assume the point is the North pole $(0,\dots, 0, 1) \in S^n$. (Why? -Well... Think about it. :-) ). Then, for instance, consider the open North hemisphere:
$$ S^n_+ = \left\{ (x_1, \dots , x_n , x_{n+1}) \in S^n \ \ \vert \ \ x_{n+1}> 0 \right\} $$
with the natural projection onto the $x_{n+1} = 0$ hyperplane of $\mathbb{R}^{n+1}$, which happens to be homeomorphic to $\mathbb{R}^n$:
$$ p: S^n_+ \longrightarrow \mathbb{R}^n \ , \qquad p(x_1,\dots , x_n, x_{n+1} )= (x_1, \dots , x_n) $$
Exercises:
- Why is $S^n_+$ an open neighbourhood of the North pole inside the sphere $S^n$?
- Who is the image of $p$?
- Find an inverse for $p$, from its image to $S^n_+$.
Hint. As lhf suggests you, try first with $n=1, 2$. Draw some pictures.
Solution 2:
Alternatively, you can use the local form of a submersion to proove tha $S^n$ is a submanifold of $\mathbb{R}^{n+1}$
http://en.wikipedia.org/wiki/Submersion_%28mathematics%29
$f:\mathbb{R}^{n+1}\rightarrow\mathbb{R}$, $f(x_1,...,x_{n+1})=x_1^2+...+x_{n+1}^2$ is a $C^\infty$-submersion on $S^n=f^{-1}(1)$.
Another nice method is to use steregraphic projections
http://en.wikipedia.org/wiki/Stereographic_projection
$S^n=U_N\cup U_S$ where $U_N=S^n\setminus\{(0,...,0,1)\}$ and $U_S=S^n\setminus\{(0,...,0,-1)\}$ are opens for the induced topology of $\mathbb{R}^{n+1}$ on $S^n$.
One can check that : $\varphi_N : U_N\rightarrow\mathbb{R}^n$, $\varphi_N(x_1,...,x_{n+1})=\big(\dfrac{x_1}{1-x_{n+1}},...,\dfrac{x_n}{1-x_{n+1}}\big)$ and $\varphi_S : U_S\rightarrow\mathbb{R}^n$, $\varphi_S(x_1,...,x_{n+1})=\big(\dfrac{x_1}{1+x_{n+1}},...,\dfrac{x_n}{1+x_{n+1}}\big)$ are homomorphisms.