Proving the dimension of a manifold is well defined

Question

Show that if $M \subseteq \Bbb R^n$ is a $k$-dimensional manifold, then there is no $x \in M$ and $U_x$ a neighborhood of $x$ such that $M \cap U_x$ has a good parametrization from $V_x \subseteq \Bbb R^m$ where $k \neq m$.

The definition of "good parametrization" of $M$ is as follows: there exists a smooth, regular homeomorphism $r:V_x \to \Bbb R^n $ such that $r(V_x) =M$ where $V_x \subseteq \Bbb R^m$ is an open set.

I'm having a hard time approaching this one, any guidance will be appreciated

Answer 1

If $\phi$, $\psi$ are two good parametrizations (i.e. smooth immersions into $M$ which are homeomorphisms onto their images) with the same non-empty image, then the transition map $\tau=\psi^{-1}\circ\phi$ is a homeomorphism between non-empty open subsets of $\mathbb R^k$ and $\mathbb R^l$ and one can use tools from algebraic topology to show that this implies $k=l$. However in the differentiable setting this is easier:

Assume for a moment that we knew that $\tau$ is in fact a diffeomorphism. Then by the chain rule for all $x$ in the domain of $\tau$ we have

$$id_{\mathbb R^k}=D(id)(x)=D(\tau^{-1}\circ\tau)(x)=D(\tau^{-1})(\tau (x))\circ D(\tau)(x)$$

so $D(\tau)(x)$ is a linear isomorphism between $\mathbb R^k$ and $\mathbb R^l$ which implies $k=l$.

It remains to show that $\tau$ is a diffeomorphism. For this we can use the following lemma:

$\textbf{Lemma:}$
Assume $\psi:\tilde V \to\mathbb R^n$ is a smooth immersion, with $\tilde V\subset\mathbb R^k$ open. Then for all $x\in \tilde V$ there are open neighbourhoods $V\subset\tilde V$ of $x$ and $U$ of $\psi(x)$ with $\psi(V)\subset U$ and a smooth map $\Gamma:U\to V$ such that $\Gamma\circ\psi_{|V}=\text{id}_{|V}$.

So if $z$ is in the domain of $\tau$ we can set $x=\tau(z)$, find $V$, $U$ and $\Gamma$ as above and then $\tau=\Gamma\circ\phi$ on $\tau^{-1}(V)$ which is smooth. In the same way $\tau^{-1}$ will then be smooth.

proof of the Lemma:
Let $T$ be the image of the differential of $\psi$ at $x$ which is a $k$-dimensional subspace of $\mathbb R^n$ and let $L:\mathbb R^n\to\mathbb R^k$ be any linear map which maps $T$ onto $\mathbb R^k$. Then by the chain rule $D(L\circ\psi)(x)=DL(\psi(x))\circ D\psi(x)=L\circ D\psi(x) $ is surjective and linear and hence invertible. By the inverse function theorem there are open neighbourhoods $V\subset\Omega$, $W\subset\mathbb R^k$ of $x$ and $L(\psi(x))$ such that $L\circ\psi:V\to W$ is a diffeomorphism. Now set $U=L^{-1}(W)$ and $\Gamma = (L\circ\psi)^{-1}\circ L$. $\square$