Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 24H (Differential Geometry)

A point $x\in X$ is a critical point of a smooth map if $d{f}_x:T_{x}X\to T_{f(x)}Y$ is not surjective. A critical value is the image of at least one critical point. A regular value is a point $y\in Y$ for which every point of $f^{-1}(y)$ has surjective differential. A point outside the image is therefore a regular value vacuously. These definitions use rank relative to $\dim Y$, not merely whether a derivative vanishes in chosen coordinates.

Let $m=\dim X$, $n=\dim Y$, and $x\in f^{-1}(y)$. In coordinate charts centered at $x$ and $y$, write the map as $F:{\mathbb{R}}^m\to {\mathbb{R}}^n$. Surjectivity of its derivative gives an invertible $n\times n$ minor; reorder the domain coordinates so it is the first one. The maphas invertible derivative at the point. By the inverse function theorem, it is a local coordinate change. In these coordinates the fiber is simply the set whose first $n$ coordinates are zero, so it is locally a coordinate plane of dimension $m-n$. These charts are inherited from $X$, provingIf the fiber is empty, it is the empty submanifold under the usual convention. The no-boundary assumptions avoid any extra boundary conditions in this local model.

Regard the space of real $n\times n$ matrices as ${\mathbb{R}}^{n^2}$, and use the smooth determinant map. For invertible $A$, factor $A+tH=A(I+t{A}^{-1}H)$. Expanding the determinant to first order givessince the linear coefficient of $\det (I+tK)$ is the sum of the diagonal entries of $K$. At every $A$ with determinant one, taking $H=A$ makes this derivative $n\ne 0$. Thus one is a regular value, and part (ii) makes $\mathrm{SL}(n,\mathbb{R})$ a submanifold of dimension $n^2-1$.

Its tangent space is the kernel of this differential. At the identity,Indeed tangent derivatives of curves with constant determinant lie in that kernel, and the local regular-value coordinates show every kernel vector is tangent to the level set.