Is the set of singular matrices ever a differentiable manifold?

I can see that invertible matrices are a differentiable manifold nonetheless I do not recognize just how to show that something is not a differentiable manifold so conveniently.

Is it ever before the instance that single matrices create a differentiable manifold?

2022-07-25 20:46:29
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Answers: 1

In the case $n=2$, consider $C = \{(x_{11},x_{12},x_{21},x_{22}) \in {\mathbb R}^4: x_{11} x_{22} - x_{12} x_{21} = 0\}$. With the change of variables $y=x_{11}+x_{22},z=x_{11}-x_{22},v=x_{12}+x_{21},w=x_{12}-x_{21}$, this becomes $\{(y,z,v,w) \in {\mathbb R}^4: y^2 - z^2 - v^2 + w^2 = 0\}$. The intersection of this with the unit sphere is $\{(y,z,v,w) \in {\mathbb R}^4: y^2 + w^2 = z^2 + v^2 = 1/2\}$, which is the Cartesian product of two circles, i.e. a $2$-torus. But in ${\mathbb R}^3$ a region whose boundary is a $2$-torus is not simply connected. So no neighbourhood of the origin in $C$ is homeomorphic to ${\mathbb R}^3$.

2022-07-25 22:38:43