dimension of a connected Manifold

Question

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.

Answer 1

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.

Answer 2

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.

Let us start showing the dimension of a connected topological manifold is defined without ambiguity.

First note the following lemma.

Lemma 1: If $U$ is an open set of $\Bbb R^n$, $V$ is an open set of $\Bbb R^k$ and there is a homeomorphism $h : U \to V$ then $n=k$.

This lemma is a consequence of the Invariance of Domain Theorem (which is a theorem in algebraic topology).

Now let $\{(U_\alpha,\phi_\alpha)\}_\alpha$ be a topological atlas of $M$. For each $\alpha$, $\phi_\alpha : U_\alpha \subset M \to \phi(U_\alpha) \subset \mathbb{R}^{n_\alpha}$ is a homeomorphism.

Let us define a function $f: M \to \Bbb N$, in the following way: given any $p\in M$ there is $\alpha$ such that $p \in U_\alpha$, define $f(p) = n_\alpha$.

If there is $\beta$ such that $p \in U_\beta$, then $\phi_\beta \circ \phi_\alpha^{-1}$ is a homeomorphism from $\phi_\alpha(U_\alpha \cap U_\beta) \subset \Bbb R^{n_\alpha}$ onto $\phi_\beta(U_\alpha \cap U_\beta) \subset \Bbb R^{n_\beta}$, so, by the lemma above, $n_\alpha=n_\beta$. So $f$ is well defined.

Now, for all $n \in \Bbb N$, if $p \in f^{-1}(\{n\})$ then there is $\alpha$ such that $p \in U_\alpha$, $n = n_\alpha$ and $U_\alpha \subseteq f^{-1}(\{n\})$. So, for all $n \in \Bbb N$, $f^{-1}(\{n\})$ is open.

Suppose there are $n, m \in \Bbb N$ such that $n\neq m$ and $n, m \in \text{Image}(f)$. Then $f^{-1}(\{n\})$ and $f^{-1}(\Bbb N \setminus \{n\})$ are non-empty disjoint open sets covering $M$. But this contradicts the connectness of $M$. So $f$ must be constant.

The only value of $f$ is the dimension of $M$. This completes the proof of the first part.

Now, since a $C^r$ connected manifold is a connected topological manifold, it follows immediately that the dimension of a $C^r$ connected manifold is defined without ambiguity.

The second part consists in showing that in the case of a $C^r$ connected manifold, we can prove that the dimension of the manifold is defined without ambiguity, without using the lemma 1 above (and so without using the Invariance of Domain Theorem and algebraic topology).

The proof is very similar to the proof above. We are going to use the following lemma.

Lemma 2: If $U$ is an open set of $\Bbb R^n$, $V$ is an open set of $\Bbb R^k$ and there is a $C^r$ diffeomorphism $h : U \to V$ then $n=k$.

Lemma 2 is a consequence of the Inverse Function Theorem. In fact, by such theorem, we know that at each point of $p\in U$, $h'(p)$ is an invertible linear transformation from $\Bbb R^n$ onto $\Bbb R^k$. So, $n=k$.

Now let $\{(U_\alpha,\phi_\alpha)\}_\alpha$ be a $C^r$ atlas of $M$. For each $\alpha$, $\phi_\alpha : U_\alpha \subset M \to \phi(U_\alpha) \subset \mathbb{R}^{n_\alpha}$ is a $C^r$ diffeomorphism.

Let us define a function $f: M \to \Bbb N$, in the following way: given any $p\in M$ there is $\alpha$ such that $p \in U_\alpha$, define $f(p) = n_\alpha$.

If there is $\beta$ such that $p \in U_\beta$, then $\phi_\beta \circ \phi_\alpha^{-1}$ is a $C^r$ diffeomorphism from $\phi_\alpha(U_\alpha \cap U_\beta) \subset \Bbb R^{n_\alpha}$ onto $\phi_\beta(U_\alpha \cap U_\beta) \subset \Bbb R^{n_\beta}$, so, by the lemma 2 above, $n_\alpha=n_\beta$. So $f$ is well defined.

Now, exactly as above, we prove that, for all $n \in \Bbb N$, $f^{-1}(\{n\})$ is open and then we prove that $f$ is be constant.

The only value of $f$ is the dimension of $M$. This completes the proof of the second part.

Answer 3

If $M$ is a smooth manifold of dimensions $m$ and $n$ then for each $p\in M$, we can find charts about $p$ with images open in $\mathbb{R}^n$ and $\mathbb{R}^m$. Hence $T_pM$ is isomorphic to $\mathbb{R}^n$ and $\mathbb{R}^m$