# The Unapologetic Mathematician

## An Example of a Manifold

Let’s be a little more explicit about our example from last time. The two-dimensional sphere consists of all the points in $\mathbb{R}^3$ of unit length. If we pick an orthonormal basis for $\mathbb{R}^3$ and write the coordinates with respect to this basis as $x$, $y$, and $z$, then we’re considering all triples $(x,y,z)$ with $x^2+y^2+z^2=1$. We want to show that this set is a manifold.

We know that we can’t hope to map the whole sphere into a plane, so we have to take some points out. Specifically, let’s remove those points with $z\leq0$, just leaving one open hemisphere. We will map this hemisphere homeomorphically to an open region in $\mathbb{R}^2$.

But this is easy: just forget the $z$-component! Sending the point $(x,y,z)$ down to the point $(x,y)$ is clearly a continuous map from the open hemisphere to the open disk with $x^2+y^2<1$. Further, for any point $(x,y)$ in the open disk, there is a unique $z\geq0$ with $x^2+y^2+z^2=1$. Indeed, we can write down

$\displaystyle(x,y)\mapsto\left(x,y,\sqrt{1-x^2-y^2}\right)$

This inverse is also continuous, and so our map is indeed a homeomorphism.

Similarly we can handle all the points in the lower hemisphere $z<0$. Again we send $(x,y,z)$ to $(x,y)$, but this time for any $(x,y)$ in the open unit disk — satisfying $x^2+y^2<0$ we can write

$\displaystyle(x,y)\mapsto\left(x,y,-\sqrt{1-x^2-y^2}\right)$

which is also continuous, so this map is again a homeomorphism.

Are we done? no, since we haven’t taken care of the points with $z=0$. But in these cases we can treat the other coordinates similarly: if $y>0$ we have our inverse pair

$\displaystyle(x,y,z)\mapsto(x,z)\qquad(x,z)\mapsto\left(x,\sqrt{1-x^2-z^2},z\right)$

while if $y<0$ we have

$\displaystyle(x,y,z)\mapsto(x,z)\qquad(x,z)\mapsto\left(x,-\sqrt{1-x^2-z^2},y,z\right)$

Similarly if $x>0$ we have

$\displaystyle(x,y,z)\mapsto(y,z)\qquad(y,z)\mapsto\left(\sqrt{1-y^2-z^2},y,z\right)$

while if $x<0$ we have

$\displaystyle(x,y,z)\mapsto(y,z)\qquad(y,z)\mapsto\left(-\sqrt{1-y^2-z^2},y,z\right)$

Now are we done? Yes, since every point on the sphere must have at least one coordinate different from zero, every point must fall into one of these six cases. Thus every point has some neighborhood which is homeomorphic to an open region in $\mathbb{R}^2$.

This same approach can be generalized to any number of dimensions. The $n$-dimensional sphere consists of those points in $\mathbb{R}^{n+1}$ with unit length. It can be covered by $2(n+1)$ open hemispheres, each with a projection just like the ones above.

February 23, 2011 - Posted by | Differential Topology, Topology

## 4 Comments »

1. is there a requirement to make sure the maps are compatible? ie if a point’s domain is in more than one map that the composite is continuous (in both directions)?

Comment by scot | February 23, 2011 | Reply

2. We’re getting to that scot, stay tuned!

Comment by John Armstrong | February 23, 2011 | Reply

3. […] Let’s look back at yesterday’s example of a manifold. Not only did we cover the entire sphere by open neighborhoods with homeomorphisms to […]

Pingback by Coordinate Patches « The Unapologetic Mathematician | February 23, 2011 | Reply

4. […] our example of a manifold, we covered the two-dimensional sphere with coordinate patches, and so we have an […]

Pingback by Atlases and Structures « The Unapologetic Mathematician | March 2, 2011 | Reply