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Let $M$ be a smooth manifold of dimension $m$ and $N \subseteq M$. Then $N$ is said to be a submanifold of $M$ of dimension $n$ ($\leq m$) if for every point $p \in N$ there is a chart $(U,\phi)$ around $p$ in $M$ such that $\phi(U\cap N)$ is open in $\mathbb{R}^n$ ($\subseteq \mathbb{R}^m$).

Using the above definition how do I prove the following?

The image of the map $\alpha: [-1,1] \to \mathbb{R}^2, t \mapsto (t^3,t^2)$ is not a submanifold of $\mathbb{R}^2$.

In Analysis on Manifolds, Munkres says that the reason for this is the fact that $D\alpha(0)$ does not have rank 1. But how does it follow that all the charts around the origin in $\mathbb{R}^2$ will fail to comply to the above definition of submanifold?

Sayantan
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3 Answers3

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A change of charts is a local diffeomorphism. But then the differential is an isomorphism, so it doesn't change the rank of $D\alpha(0)$, which is $0$. You can see this via the chain rule as well, picking charts $\psi,\phi: \alpha((-\epsilon,\epsilon))\to \Bbb R$ where the first is the neighborhood of $\alpha(0)$ in $\Bbb R^2$ were we can easily assume $\phi(0,0)=\psi(0,0)=0$ by translation, we have that

$\psi\circ \alpha\phi^{-1}:\Bbb R\to \Bbb R$

has the same rank no matter what chart you choose, as

$$D(\psi\circ\alpha\circ\phi^{-1})(0)$$

can easily be multiplied by $\big(D\psi(\alpha\circ\phi^{-1}(0))\big)^{-1}$ on the left and $D\phi(0,0)$ on the right to get $D\alpha(0)=0$ stays the same on both sides.

Adam Hughes
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  • while there is nothing incorrect written here, it should be noted that this doesn’t really answer the question. If we compose on the domain of $\alpha$ by a diffeomorphism, then certainly the resulting map has the same rank as $\alpha$. Note however that the image of a map $\alpha$ is kept the same even if we compose it by an arbitrary bijection $f$ on the domain (i.e $\alpha$ and $\alpha\circ f$ have the same image, even if $f$ is not a diffeo). Case in point the example $t\mapsto (t^3,t^3)$ vs $s\mapsto (s,s)$ below. – peek-a-boo Mar 05 '23 at 09:30
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When $D\alpha(0)$ does not have rank $1$ then something might go wrong at $t=0$, but it doesn't have to. Consider the map $$\beta:\quad\ ]{-1},1[\ \to{\mathbb R}^2,\qquad t\mapsto(t^3,t^3)\ .$$ Here ${\rm rank}\bigl(D\beta(0)\bigr)=0$, but the image set is clearly a submanifold of ${\mathbb R}^2$.

In the case of your $\alpha(\cdot)$ something goes wrong indeed. The image set $S$ is Neil's parabola $$y=|x|^{2\over3}\qquad(-1<x<1)\ .$$ If $S$ were a one-dimensional submanifold of ${\mathbb R}^2$ it would have to be possible to introduce new coordinates $(x',y')$ in a neighborhood $U$ of the origin such that $S\cap U$ has a description of the form $$S\cap U=\bigl\{(x',y')\>\bigm|\>y'=\phi(x')\bigr\}$$ with a $C^1$-function $\phi$, $\>\phi(0)=0$. This is not possible.

Christian Blatter
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It follows from the inverse function theorem (or, if you prefer, the implicit function theorem) that a $k$-dimensional submanifold of $\Bbb R^n$ must, in a neighborhood of each point, be the graph of a smooth function on one of the standard $k$-coordinate planes. (Prove this!)

The curve $y^3=x^2$ clearly does not meet this criterion. (You have only $2$ situations to consider.)

Ted Shifrin
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