The Unapologetic Mathematician

Coordinate Patches

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 open regions of the plane, we did so in a “compatible” way, which I’ll explain soon. This notion of compatible coordinates is key to making a lot of differential topology and geometry work out right.

For the moment, though, let’s introduce a useful term. A “coordinate patch” on a manifold $M$ is an open subset $U\subseteq M$ together with a map $\phi:U\to\mathbb{R}^n$ that is a homeomorphism between $U$ and its image $\phi(U)$. Armed with this definition, we might say that a manifold is a topological space where every point can be contained in some coordinate patch. The only subtle point here is that this definition would put too much emphasis on the patches rather than on the local topology of the manifold itself.

The useful thing about a coordinate patch is that it lets us pull back coordinates from $\mathbb{R}^n$ to our manifold, or at least to the open set $U$. Let’s say $p\in U$ is sent to the point $\phi(p)\in\mathbb{R}^n$. We can now use the coordinate functions $x^i:\mathbb{R}^n\to\mathbb{R}$ to read off coordinates. In fact, when working in a particular coordinate patch, we will often abuse the notation and simply write

$\displaystyle x^i(p)=x^i(\phi(p))$

Of course, when we write $x^i(p)$ the actual number we get out for the $i$th component depends immensely on our coordinate homeomorphism $\phi$, and yet we’ve made no mention of it in our notation! This is one of the most confusing things about doing differential geometry and topology calculations involving coordinates, and it’s important to keep it in mind.

February 23, 2011

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