Let $f: \mathbb{R}^2\rightarrow \mathbb{R}$ with $f(x,y) = xy$ and $M=f^{-1}({0})$. Show that: The set $M$ is not a submanifold.

Question

Assignment:

Let $f: \mathbb{R}^2\rightarrow \mathbb{R}$ with $f(x,y) = xy$ and $M=f^{-1}({0})$. Show that: The set $M$ is not a submanifold.

I've been able to show that sets are submanifolds before by parameterising the set and finding the atlas as having an atlas is equivalent to being a submanifold. For showing the opposite, frankly, I could use a tip or something to start with.

Thanks in advance!

Answer 1

So we want to look at the set $\{xy=0\}$, which consists of the $x$- and $y$-axis in the plane. This set is obviously connected.

So you would need a chart around every point in $\{xy=0\}$, in order to prove it a submanifold. But if you would have such a chart, taking out a point would correspond to taking out a point in $\mathbb R^n$ which is then still connected (or at most two components if $n=1$), hence your manifold should be still connected (or has two components if $n=1$). But $\{xy=0\} - 0$ is not connected and has $4>2$ components.

Answer 2

Hint: in a manifold, any point has a neighbourhood homeomorphic to a disk in an Euclidean space. Does $(0,0)$ have such a neighbourhood in $M$?