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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.

Gabi G
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  • If a manifold is a topological space where for every $x \in M$ there is an open set $U_x \ni x$ and a chart making it homeomorphic to an open set of $\Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ? – reuns Mar 22 '19 at 01:56
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    The statement is false, since $M = \emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $M\neq\emptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology. – o.h. Mar 22 '19 at 07:11

3 Answers3

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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.

o.h.
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    Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do – Gabi G Mar 22 '19 at 09:21
  • @Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer – o.h. Mar 22 '19 at 15:01
  • @GabiG But you need topology for the proof. – Filippo May 28 '21 at 06:49
  • @o.h. Has trii deleted his answer? – Filippo May 28 '21 at 06:50
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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 $m=k$.

Claire
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  • I think your answer makes no sense. If $M$ is a topological space, an $m$-dimensional differentiable atlas defines an $m$-dimensional tangent space in each point - by the definition of the tangent space. You are right that differentiable structures (atlases) with different dimensions induce tangent spaces with different dimensions in each point, but that's not a contradiction and therefore doesn't prove that differentiable structures always have the same dimension. – Filippo May 28 '21 at 06:47
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    @Filippo I assumed that $M$ is a smooth manifold to begin with ("...$M$ is a graph of a smooth function"), so that a differential structure is part of the data. Now the definition of "of dimension m" here seems to be, that $M$ is locally diffeomorphic to an open subset of $\mathbb R^m$ ("...good parametrization from $V_x\subset \mathbb R^m$"). I also assumed $M\neq\emptyset$. The (short) argument here uses, that tangent spaces of $M$ are allready well defined and that the derivative of a (local) diffeomorphism consists of linear isomorphisms. – Claire May 30 '21 at 16:25
  • Thank you for remaining constructive in spite of my harsh words (I should have formulated the first sentence differently, sorry). I think that I understand what you mean: Your argument seems to be based on the Diffeomorphism Invariance of the dimension: There cannot be a diffeomorphism $F\colon(M,T,A)\to(M,T,A′)$ if $A$ and $A′$ are differentiable structures with different dimensions. And that's because there cannot be two charts $x\colon U\to\mathbf{R}^m$ and $x′\colon U\to\mathbf{R}^n$ defined on an open subset $U\subset M$ such that $m\neq n$ and the transition charts are differentiable – Filippo May 30 '21 at 20:48
  • But this doesn't exclude the possibility that there are two non-compatible atlases with different dimensions (even if we require that both atlases are smooth). – Filippo May 30 '21 at 20:54
  • However, rereading the question (in particular the first part with "good parametrization"), the diffeomorphism invariance might actually be exactly what the OP was looking for...That's not what I would call the proof of the "uniqueness of the dimension of a manifold", but that's not your fault. – Filippo May 30 '21 at 21:06
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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.

jmerry
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  • Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds – Gabi G Mar 22 '19 at 09:19
  • @GabiG a topological manifold does not need to have a differentiable structure. – Zest Mar 23 '19 at 01:49