Uniqueness of the dimension of a manifold

Question

Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m \ne k$. Meaning, there is no $x \in M$ and $x \in U_x$ a neighborhood of $x$, such that $M \bigcap U_x$ has a good parametrization from $V_x \subset R^m$ (where $m \ne k$).

I am not really sure how to prove it. I thought that I could use the fact that for every $x \in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.

However, I got stuck. Any help would be appreciated.

Answer 1

If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $\mathbb R^k$. So if $M$ is also $m$-dimensional then $\mathbb R^m\simeq\mathbb R^k$ (as vector spaces) which implies $m=k$.

Answer 2

I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?

It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.

It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.

Answer 3

For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $x\in X$ has Euclidean nbhd $U\ni x$ with homeomorphism $f\colon U\rightarrow \mathbb R^n$, then $$ H_i(X,X\setminus x) \xleftarrow\sim H_i (U,U\setminus x) \xrightarrow\sim H_i(\mathbb R^n,\mathbb R^n\setminus f(x)) $$ where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (\mathbb R^n,\mathbb R^n\setminus f(x))\cong \tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $\mathbb R^n\setminus f(x)\subset\mathbb R^n$ and htpy equivalence $\mathbb R^n\setminus f(x) \simeq S^{n-1}$. In conclusion, for each $x\in X$ and $i > 0$ we find that $$ H_i (X,X\setminus x)\cong\tilde H_{i-1} (S^{n-1})\cong\begin{cases}\mathbb Z\quad \text{if}\, i=n,\\ 0\quad\text{else.}\end{cases} $$ This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $x\in X$. (Uniqueness of dimension is obviously false for the manifold $X = \emptyset$. Note that this is a manifold.)

Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $\mathbb R^m\approx \mathbb R^n\Rightarrow m = n$ without algebraic topology.