# The Unapologetic Mathematician

## 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 $\mathbb{R}^3$.

April 25, 2011