Submanifolds

Historically the theory of differential geometry arose from the study of surfaces in $\mathbb{R}^3$. We want to consider the more general case of submanifolds in $\mathbb{R}^n$. Recall that we have

Definition 4.2   A set $Z \subset \mathbb{R}^n$ is a submanifold of dimension $d$ if there is a smooth map $f \colon \mathbb{R}^n \to \mathbb{R}^{n-d}$ for some $d$ such that $Z = f^{-1}(0)$ and $df(z)$ is onto for all $z \in Z$.

We will show that any submanifold of $\mathbb{R}^n$ is a manifold by constructing co-ordinate charts. The ideas is simple. At any point $z \in Z$ we define the plane tangent to $Z$. This is the kernel $K_z$ of the map $df(z)$. Then we consider orthogonal projection of $K_z$ onto $Z$. This defines co-ordinates on an open neighbourhood of $z$. In fact these are a very special kind of co-ordinate. To prove this we first prove

Proposition 4.2   Let $Z$ be a submanifold of $\mathbb{R}^n$ of dimension $d$. Then at any $z \in Z$ we can find a co-ordinate chart $(U, \psi)$ on $\mathbb{R}^n$ such that if $\psi = (\psi^1,\dots, \psi^n)$ then

$$U \cap Z = \{ x \in U \colon \mid \psi^{d+1}(x)= \dots = \psi^n(x) = 0\}.$$

Proof. Let $K_z$ be the kernel of $df(z)$ and let $\pi \colon \mathbb{R}^n \to K_z$ be the orthogonal projection onto $K_z$. Choose a basis $v^1, \dots,v^d$ and write $\pi_z$ with respect to this basis as

$$\pi(x) = \sum_{i=1}^d \pi^i(v) v^i.$$

Define a map $\chi\colon \mathbb{R}^n \to \mathbb{R}^n$ by

$$\psi(x) = (\pi^1(x),\dots,\pi^d(x),f^1(x), \dots, f^{n-d}(x)).$$

Because $\pi$ is a linear map it is its own derivative so we have

$$d\psi(z) = \pi^1(z),\dots,\pi^d(z), df^1(z), \dots , df^{n-d}(z)).$$

Consider $v$ in the kernel of $d\psi(z)$. Then $d\pi(z)(v) =0$ and $df(z)(v) = 0$. Hence $v$ is both orthogonal to $K_z$ and in $K_z$ so it must be zero. So $d\psi(z)$ is injective and hence a dimension count surjective so a bijection. Now we can apply the inverse function theorem so there is an open set $U$ in $\mathbb{R}^n$ such that $\psi(U)$ is open and

$$\psi_{\vert U}\colon U \to \chi^(U)$$

is a diffeomorphism. But this just means that $(U, \psi_{\vert U})$ is a co-ordinate chart on $\mathbb{R}^n$. Notice that $\psi^{d+1}(x) = \dots = \psi^n(x) = 0$ if and only if $f(x) = 0$ if and only if $x$ is in $U \cap Z$. $\qedsymbol$

Let $(U, \psi)$ be a set of co-ordinates on $\mathbb{R}^n$ such that

$$U \cap Z = \{ x \in U \colon \mid \psi^{d+1}(x)= \dots = \psi^n(x) = 0\}.$$

Then consider $(U \cap Z, \bar \psi)$ where $\bar \psi = (\psi_{\vert U \cap Z}^1, \dots, \psi_{\vert U \cap Z}^{n-d})$. This is a co-ordinate chart. The only thing to check is that that because $\psi(U)$ is open then

$$\bar\psi(U\cap Z) = \{ x \in \mathbb{R}^d \mid (x^1, \dots, x^d, 0, \dots, 0) \in \psi(U) \}$$

is open. This is an elementary fact about the topology of $\mathbb{R}^n$. Consider two such co-ordinate charts $(U, \psi)$ and $(V, \chi)$. We will prove that $(U \cap Z, \bar \psi)$ and $(V \cap Z, \bar \chi)$ are compatible. This follows essentially from the fact that $(U, \psi)$ and $(V, \chi)$ are compatible. First we note that

$$\bar\chi(U\cap V \cap Z) = \{ x \in \mathbb{R}^d \mid (x^1, \dots, x^d, 0, \dots, 0) \in \chi(U \cap V) \}$$

and, again, this is open. For the smoothness of the co-ordinate change map we note that

$$\bar\psi^i\circ{\bar\chi}^{-1}_{\vert\chi(U\cap V \cap Z)}(x) =\psi^i\circ{\chi}^{-1}_{\vert\chi(U\cap V )}(x,0)$$

where $(x,0) = (x^1,\dots, x^d, 0, \dots, 0)$. Hence the result follows.

We have now proved

Theorem 4.1   If $Z$ is a submanifold the set of charts above is an atlas.

This results gives us a lot of examples of submanifolds:

Example 4.3 (Spheres)   Consider the sphere $S^n \subset \mathbb{R}^{n+1}$ defined as the set of points $x$ whose length $\Vert x\Vert$ is equal to one. Here

$$\Vert x\Vert^2 =\sum_{i=1}^n (x^i)^2.$$

We can prove it is a submanifold of $\mathbb{R}^n$ and hence a manifold by considering the map $f\colon \mathbb{R}^{n+1} \to 1$ defined by

$$f(x)= \Vert x\Vert^2 - 1.$$

Clearly this is smooth and has zero set $Z$ equal to the sphere $S^n$. To check that the derivative is smooth note that

$$df(x)(v)= 2\< x, v\>$$

and is a linear map onto a one-dimensional space so to show it is onto we just need to show that it is not equal to the zero linear maps.

Example 4.4 (The orthogonal group)   The orthogonal group is the group of all linear transformations of $\mathbb{R}^n$ that preserve the usual inner product on $\mathbb{R}^n$. We shall think of it as a group of $n$ by $n$ matrices:

$$O(n) = \{ X \mid XX^t = 1\}.$$

We can identify the set of all $n$ by $n$ matrices with $\mathbb{R}^{n^2}$. There are various ways of doing this. So as to be concrete let us assume we have done it by writing down the rows one after the other. With this identification in mind define a smooth map $f \colon \mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$ by $f(X) = X^2 - 1$. It is clear we have $f^{-1}(0) = O(n)$. Define the linear subspace $S \subset \mathbb{R}^{n^2}$ to be the set of all symmetric matrices. This can be identified with $\mathbb{R}^d$ where $d = n(n+1)/2$. It is easy to check that $f$ takes its values in $S$ so we will think of $f$ as a smooth map $f\colon \mathbb{R}^{n^2} \to \mathbb{R}^d$.

We want to calculate $df(X)$ the derivative of $f$ at a matrix $X$. By differentiating the path $t \mapsto X + tY$ we obtain

$$df(X)(Y) = YX^t + XY^t.$$

If $B$ is any symmetric matrix then it is easy to check that $df(X)((1/2)BX) = B$ if we use that fact that $XX^t = 1$ and $B = B^t$. We have therefore shown that $df(X)$ is onto for any $X \in O(n)$ so that $O(n)$ is a submanifold of dimension $n - d = n(n=1)/2$.