Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 24H (Differential Geometry)

The determinant is continuous and $GL(n)={\det }^{-1}(\mathbb{R}\setminus \{0\})$ is open in $M(n)\simeq {\mathbb{R}}^{n^2}$. Every open subset inherits the ambient smooth charts, so it is a submanifold and

For an invertible $A$, factor $\det (A+tB)=\det A\det (I+t{A}^{-1}B)$. The determinant's permutation expansion gives $\det (I+tC)=1+t\mathrm{tr}C+O(t^2)$. Differentiating at zero yieldsAt a determinant-one matrix choose $B=A/n$; the derivative is one. Thus one is a regular value of the smooth determinant map on $GL(n)$. The regular level set theorem makes $SL(n)$ a submanifold of codimension of a submanifold one, withThis includes $n=1$, when $SL(1)=\{1\}$ has dimension zero.

For $n=1$, the singular set is the point $\{0\}$, a zero-dimensional submanifold. For $n\ge 2$, every elementary matrix $E_{ij}$ is singular and the smooth curve $t\mapsto t{E}_{ij}$ lies in $X$ through zero. If $X$ were a submanifold, all $E_{ij}$ would belong to $T_{0}X$ and span $M(n)$. Its dimension at zero would therefore be $n^2$, making $X$ locally open there. But arbitrarily small matrices $tI$ are invertible, a contradiction. ThusThe failure is at the zero matrix; fixed-rank pieces of the singular set can separately be smooth submanifolds.