$M =\{(t, \vert t \vert) \text{ }\vert t \in \mathbb{R} \} $ isn't a Smooth Submanifold

Question

I want to show formally that

$$M =\{(t, \vert t \vert) \text{ }\vert t \in \mathbb{R} \} $$

is not a smooth $C^{\infty}$-submanifold of $\mathbb{R}^2$.

My attempts: Intuitively it's clear that the problem is the origin point $(0,0)$. Indeed, $M$ is the graph of the absolute value function which is not differentiable in the origin.

But I have some problems to show that $M$ isn't smooth manifold in a rigorous formal way.

In the lecture we are working with following definition: $M$ is a $n$-dimensional smooth (so $C^{\infty}$) submanifold of $\mathbb{R}^{n+k}$ iff for every $p \in M$ there exist an open subset $U \subset \mathbb{R}^{n+k}$ with $p \in U$, an open $V \subset \mathbb{R}^n$ and a smooth function $\gamma \in C^{\infty}(V,U)$ with following properties

$\gamma(V) = U \cap M$

$rank(D\gamma \vert _v) = n$ at every $v \in V$ where $D\gamma \vert _v$ is the differential of $\gamma$ at $v$

$\gamma$ is a homeomorphism from $V$ to $M \cap U$

I know that there are some other equivalent definitions of smooth manifolds but I want to know how to get a contradiction using this criterion.

The problem is that there exist no such function with properties as above so if I try to find some $\gamma$ which maps onto $(0,0)$ how to show that there exist some $v_0 \in V$ with $rank(D\gamma \vert _{v_0}) = 0$.

Another idea would be to get a contradiction showing that $\gamma'$ along some can't be continuous, right? But here also I don't find a way how to construct the contradiction formally using the submanifold criterion above.

Answer 1

Reduce it to calculus: $\gamma$ is a smooth map from an open subset of $\mathbb{R}^1$ into $\mathbb{R}^2$; i.e. it's a smooth curve. Write $\gamma(s) = (x(s), y(s))$ and suppose $\gamma(s_0) = (0,0)$. Note that since $(x(s), y(s))$ is a point of $M$, we have $y(s) = |x(s)|$.

If $y'(s_0) \ne 0$ then by definition of the derivative, there would exist $s$ near $s_0$ with $y(s) < y(s_0) = 0$, which is impossible. So $y'(s_0) = 0$. Now note that $$|x'(s_0)| = \lim_{s \to s_0} \frac{|x(s)-x(s_0)|}{|s-s_0|} = \lim_{s \to s_0} \frac{|x(s)|}{|s-s_0|} = \lim_{s \to s_0} \frac{|y(s)|}{|s-s_0|} = |y'(s_0)| = 0.$$

Since $x'(s_0) = y'(s_0) = 0$, we have $D\gamma_{s_0} = 0$ whose rank is 0, not 1.

Answer 2

It's not always easy to show that a subset isn't a smooth submanifold using the definition. May I suggest another approach?

Maybe you have seen a version of the Implicit Function Theorem like this:

Let $S\subset\mathbb{R}^2$ be a submanifold (curve) and $p\in S$. Then there is a neighborhood $V$ of $p$ in $S$ such that $V$ is the graph of a differentiable function of the form $y=f(x)$ or $x=g(y)$.

Suppose $M$ is a smooth submanifold (curve). Then around $(0,0)$ it locally is the graph of a function. This function cannot be of the form $x=g(y)$, since projection of the curve on the $y$-axis is not surjective. So locally the curve must be of the form $(t,f(t))$. But this implies that $f(t)=|t|$. Since this $f$ is not differentiable at $0$, we arrive at a contradiction and must conclude that $M$ is not a smooth submanifold.