Examples

Wild topological manifolds

Wild topological manifolds

Topological spaces can be stranger than one imagines on the basis of simple examples. Here is Alexander’s Horned Sphere (https://en.wikipedia.org/wiki/Alexander_horned_sphere; explanatory essay):

The horned sphere $S$ is a subset of ${\mathbb{R}}^3$ homeomorphic to ${\mathbb{S}}^2$. The outside ${\mathbb{R}}^3-S$ contains non-collapsible loops (so it is not simply connected), which is unlike the standard inclusion of ${\mathbb{S}}^2$ in ${\mathbb{R}}^3$.

The map $f:{\mathbb{S}}^2\hookrightarrow {\mathbb{R}}^3$ with $f({\mathbb{S}}^2)=S$ is an embedding of topological spaces and thus topological manifolds. (It is continuous, injective, and induces a homeomorphism onto the image which is endowed with the subspace topology; so the image is homeomorphic to ${\mathbb{S}}^2$ as a topological manifold.)

The map $f$ is not an embedding of differentiable, smooth, or piecewise linear manifolds.

Parametrized paraboloid

Parametrized paraboloid

Let $\phi :{\mathbb{R}}^2\to {\mathbb{R}}^3$ by $\phi :(x,y)\mapsto \left(x,y,x^2+y^2\right)$.

Then $[d\phi ]$ is given by:

$$[d_{(x,y)}\phi ]=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 2x& 2y\end{array}\right)$$

Notice that the determinant of the upper $2\times 2$ square is nonzero.

Singular curve

Singular curve

Consider $\phi :\mathbb{R}\to {\mathbb{R}}^2$ given by $\phi (t)=(t^2,t^3)$:

Then: $[d_t\phi ]=\left[\begin{array}{c}2t\\ 3t^2\end{array}\right]$ and $[d_0\phi ]=\left[\begin{array}{c}0\\ 0\end{array}\right]$.

Parametrized 2-sphere

Parametrized 2-sphere

Spherical coordinates define a parametrization of the 2-sphere:

$$f(\theta ,\phi )=\left(\sin \phi \cos \theta ,\sin \phi \sin \theta ,\cos \phi \right)$$

The differential:

$$[d_{(x,y)}\phi ]=\left(\begin{array}{cc}\cos \phi \cos \theta & -\sin \phi \sin \theta \\ \cos \phi \sin \theta & \sin \phi \cos \theta \\ -\sin \phi & 0\end{array}\right)$$

Notice that poles of the sphere correspond to points with $\phi =n\pi$, and there the right column is all zeros. So poles are not regular points of $\phi$. One can check that every other point is regular by computing the cross product of the column vectors.

Also, at the poles the function is not injective.

Therefore one must restrict the domain, for example to $\phi \in (-\pi ,\pi )$ and $\theta \in (0,2\pi )$, to get a true parametrization. Unfortunately this does not cover the entire sphere; indeed it is impossible to cover the sphere with a single parametrization.

Parametrized 2-torus in R4

Parametrized 2-torus in ${\mathbb{R}}^4$

The 2-torus can be parametrized in ${\mathbb{R}}^4$ very easily:

$$f(\phi ,\theta )=\left(\cos \phi ,\sin \phi ,\cos \theta ,\sin \theta \right)\phi ,\theta \in (0,2\pi )$$

This parametrization is regular everywhere, but the domain cannot be extended to cover the entire torus without becoming non-injective (and no longer a homeomorphism onto its image), for topological reasons. The torus cannot be covered by a single parametrization.

Smooth injection phi but inverse of phi not smooth

Smooth injection $\phi$ but ${\phi }^{-1}$ not smooth

Let $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=x^3$. Clearly $f$ is infinitely differentiable and injective.

Notice $f^{\prime }(0)=0$. So the differential does not have maximal rank at $x=0$.

The inverse function $\sqrt[3]{x}$ is not smooth, indeed it is not differentiable at $x=0$.

Circle requires two charts

Circle requires two charts

Show that it is not possible to cover the unit circle ${\mathbb{S}}^1\subset {\mathbb{R}}^2$ with a single coordinate chart.

Solution

Suppose the contrary, that $\phi :U\to {\mathbb{R}}^2$ for $U\subset \mathbb{R}$ is a coordinate chart covering ${\mathbb{S}}^1$.

(We assume this means $U$ is connected.)

---

Let $p\in {\mathbb{S}}^1$ be some point. Set $A={\mathbb{S}}^1-\{p\}$.

• Then $A$ is connected.
• Therefore ${\phi }^{-1}(A)$ is connected since ${\phi }^{-1}$ is a homeomorphism.

---

Set $q={\phi }^{-1}(p)$.

• Then ${\phi }^{-1}(A)=U-\{q\}$.
• $U-\{q\}$ is disconnected: the open sets $U\cap \{x<q\}$ and $U\cap \{x>q\}$ separate $U-\{q\}$.
• So ${\phi }^{-1}(A)$ is disconnected and connected, a contradiction.