The Unapologetic Mathematician

Mathematics for the interested outsider

Spheres as Submanifolds

With our extension of the implicit function theorem in hand, we have another way of getting at the sphere, this time as a submanifold.

Start with the Euclidean space \mathbb{R}^{n+1} and take the smooth function f:\mathbb{R}^{n+1}\to\mathbb{R} defined by f(x)=\langle x,x\rangle. In components, this is \sum_i\left(x^i\right)^2, where the x^i are the canonical coordinates on \mathbb{R}^{n+1}. We can easily calculate the derivative in these coordinates: f_{*x}(v)=2\langle x,v\rangle. This is the zero function if and only if x=0, and so f_{*x} has rank 1 at any nonzero point x. The point x=0 is a critical point, and every other point is regular.

On the image side, we see that f(0)=0, so the only critical value is 0. Every other value is regular, though f^{-1}(y) is empty for y<0. For f^{-1}(a^2) we have a nonempty preimage, which by our result is a manifold of dimension (n+1)-1=n. This is the n-dimensional sphere of radius a, though we aren’t going to care so much about the radius for now.

Anyway, is this really the same sphere as before? Remember, when we first saw the two-dimensional sphere as an example, we picked coordinate patches by hand. Now we have the same set of points — those with a fixed squared-distance from the origin — but we might have a different differentiable manifold structure. But if we can show that the inclusion mapping that takes each of our handcrafted coordinate patches into \mathbb{R}^3 is an immersion, then they must be compatible with the submanifold structure.

We only really need to check this for a single patch, since all six are very similar. We take the local coordinates from our patch and the canonical coordinates on \mathbb{R}^3 to write out the inclusion map:

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

Then we use these coordinates to calculate the derivative

\displaystyle\begin{aligned}g_{*(x,y)}(u,v)&=\left(1,0,\frac{-x}{\sqrt{1-x^2-y^2}}\right)u+\left(0,1,\frac{-y}{\sqrt{1-x^2-y^2}}\right)v\\&=\left(u,v,\frac{-xu-yv}{\sqrt{1-x^2-y^2}}\right)\end{aligned}

This clearly always has rank 2 for x^2+y^2<1, and so the inclusion of our original sphere into \mathbb{R}^3 is an immersion, which must then be equivalent to the inclusion of the submanifold f^{-1}(1), since they give the same subspace of \mathbb{R}^3.

About these ads

April 25, 2011 - Posted by | Differential Topology, Topology

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 389 other followers

%d bloggers like this: