Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 25I / Solution

A point $x$ is critical for a smooth map $f:X\to Y$ when $D{f}_x:T_{x}X\to T_{f(x)}Y$ is not surjective. A critical value is the image of a critical point; a regular value has no critical point in its fiber, including the case of an empty fiber.

To prove the regular level set theorem, take $x_0\in f^{-1}(y)$ and write $n=\dim X$, $m=\dim Y$. In local coordinates, surjectivity supplies an invertible $m\times m$ derivative minor; relabel source coordinates so it uses the first $m$ columns. The maphas an invertible derivative. The inverse function theorem makes it a local diffeomorphism. In these coordinates the fiber is obtained by fixing its first $m$ coordinates, so it is locally an open subset of ${\mathbb{R}}^{n-m}$. These smooth charts prove that a nonempty regular fiber is a smooth embedded manifold of dimension $n-m$, with tangent space $\ker D{f}_x$.

For the special orthogonal group, use $F:M_n(\mathbb{R})\to {\mathrm{Sym}}_n(\mathbb{R})$ defined by $F(A)=A{A}^T$. Its derivative is $D{F}_A(H)=H{A}^T+A{H}^T$. At an orthogonal $A$, any symmetric matrix $B$ is obtained by taking $H=BA/2$, so $I$ is a regular value. Therefore $O(n)=F^{-1}(I)$ has dimension $n^2-n(n+1)/2=n(n-1)/2$. The determinant takes only the values $\pm 1$ there, so $SO(n)$ is an open and closed part of $O(n)$ and has the same manifold dimension:It is closed in matrix space and bounded because every orthogonal matrix column has norm one. The Heine-Borel theorem proves compactness. Finally the tangent space is the derivative kernel,Thus its translated tangent space is the vector space of skew-symmetric matrices.