Currently, I am studying smooth manifolds and I want to solve some exercises. There is a question that says:

Show that the dimension of a connected topological manifold is defined without ambiguity! Meaning that if $\dim M=n$, with the change of charts, the dimension still is $n$. Then show that this is true for a $C^r$ connected manifold using Inverse Function Theorem.

To prove the first part, let $(U,\varphi)$ and $(V,\psi)$ be two charts with $\varphi(U) \subseteq \mathbb{R}^m$ and $\psi(V) \subseteq \mathbb{R}^n$ and suppose that $U \cap V \neq \emptyset$. Then, since $\psi \circ \varphi : \varphi (U \cap V) \rightarrow \psi (U \cap V) $ is a homeomorphism, $m=n$. Now, using this, if $M$ is a connected topological manifold with atlas $\mathcal{A}$, for all charts $(U_{\alpha}, \varphi_{\alpha})\in \mathcal{A}$, $\varphi_{\alpha}(U_{\alpha}) \subseteq \mathbb{R}^n$ and $n$ is constant for all the charts. Am I right?

But I have a problem showing the second part! I think that since every $C^r$ manifold is also a topological manifold, the answer to this question would be trivial! Why do we need "Inverse Function theorem" to answer it??

Any help is appreciated.

  • $\begingroup$ Why do we need "Inverse Function theorem" to answer it?? We don't. That doesn't make much sense to be honest. Also and n is constant for all the charts. requires a proof. This is typically done by considering $f:M\to\mathbb{R}$, $f(x)=\text{local dimension around }x$ and proving that $f$ is continuous (and so constant if $M$ is connected). What you've proven is that $f$ is well defined (i.e. independent on the choice of atlas). $\endgroup$
    – freakish
    May 31 '19 at 7:59
  • $\begingroup$ What seems strange to me about the exercise is that proving the statement for topological manifolds seems much harder than the case of differentiable manifolds. I think you need algebraic topology for the topological case, while only linear algebra is needed for the differentiable case. $\endgroup$
    – JWL
    May 31 '19 at 8:02
  • $\begingroup$ @JWL Algebraic topology to prove that the dimension is well defined on connected topological manifolds? What are you talking about? (see my previous comment) $\endgroup$
    – freakish
    May 31 '19 at 8:07
  • $\begingroup$ @freakish Yeah, you're right. I'll write it down in detail. But do you have any ideas about the second part? $\endgroup$ May 31 '19 at 8:08
  • $\begingroup$ @JWL No, we don't!!!! The first part is as I've written. $\endgroup$ May 31 '19 at 8:09

Your argument about overlapping chart sets uses invariance of domain: the fact that a homeomorphism between open sets of $\mathbb{R}^m$ and $\mathbb{R}^n$ exists implies $n=m$ is quite non-trivial. It's actually not quite invariance of domain, but the related fact that topological dimension (there are several equivalent definitions for separable metric spaces) is a topological invariant and $\dim(\mathbb{R}^n)=n$ (where $\dim$ is one of those equivalent topological dimension functions; topologically speaking the main fact we need for that is Brouwer's fixed point theorem.) This can be applied for simple topological manifolds, as all we need is homeomorphisms in the charts.

Now the chart open sets of the manifold form an open cover and so if $M$ is connected we can apply the chain characterisation of connectedness to conclude that the local dimension is constant: take any points $x$ of local dimension $m$ and $y$ of local dimension $n$. So there are finitely many chart open sets $U_1, \ldots, U_k$ such that $x \in U_1$, $y \in U_k$ and $U_i \cap U_{i+1} \neq \emptyset$ for all $i \in \{1,\ldots,k-1\}$. By the "overlapping chart sets implies equal dimension" argument you also gave and which is, as said, justified by non-trivial topological results on Euclidean spaces, we conclude that the local dimension of all $U_i$ is $m$: $U_1$ because of $x$, $U_2$ because it intersects $U_1$, $U_3$ because it intersects $U_2$ up to $U_k$ because it intersects $U_{k-1}$ (or make it a finite induction). So $m=n$ because local dimension is well-defined.

For $C^r$ manifolds we don't need the inverse function theorem, we know all charts maps are homeomorphisms by definition (or maybe you need the inverse function theorem to see that if it's not part of the definition?). You just fall back to the fact on topological manifolds.

  • $\begingroup$ The first part is right. But if we need to prove the second part without the topological manifolds, we need IFT. I will post my answer ASP! Thanks any way... $\endgroup$ May 31 '19 at 13:39
  • $\begingroup$ @Yasinowski What is the definition of $C^r$ manifolds that it is not a topological manifold? $\endgroup$ May 31 '19 at 14:39
  • $\begingroup$ $C^r$ manifolds are definitely topological manifolds. But I need proof using IFT. $\endgroup$ May 31 '19 at 14:55
  • $\begingroup$ @Yasinowski don't get fixated on that. A simpler proof is often a better proof. $\endgroup$ May 31 '19 at 15:02
  • 1
    $\begingroup$ In the differentiable case, one can construct the 'tangent space' and maps between manifolds induce vector space maps between tangent spaces. This easily implies the well-defined-ness of the dimension since dimensions of vector spaces can be defined in an elementary way. This argument might use IFT in some implicit way, as many arguments in basic differentiable manifold theory does, but I don't think so. $\endgroup$
    – JWL
    May 31 '19 at 15:29

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