I came across this claim: let $M$ be an oriented manifold of $\dim M=n$ and $\mathcal{A}$ an atlas for $M$. Then any $U \in \mathcal{A}$ is diffeomorphic to either $\mathbb{R}^n$ or $\mathbb{H}^n:= \left\lbrace x \in \mathbb{R}^n : x_n \geq 0 \right\rbrace $.

This appears in a proof of Stokes' theorem found on page 33 of Bott and Tu's Differential Forms. They first prove the theorem for cases $\mathbb{R}^n$ and $\mathbb{H}^n$ and use these cases to prove it for a general manifold: one just picks an oriented atlas and a partition of unity associated to it and instead of proving the theorem for a compactly-supported form $\omega$ on $M$, the problem reduces to proving it on forms of the type $f_i \omega$ where $f_i$ is in the partition of unity; $f_i \omega$ will obviously have support in a map in the atlas.

Excerpt from the book: actual excerpt from the book I have two questions.

Firstly, I believe in this proof, there's no need for all the atlases of $M$ to satisfy this property, it suffices to find one that has it, which is easy: just pick charts around each point that are diffeomorphic to a ball in $\mathbb{R}^n$ and hence to the entire $\mathbb{R}^n$. Since the integral is well defined, if we prove the theorem by splitting up $M$ in this atlas using the partition of unity, we will have proven it for any atlas. is this correct? If it were, then why would the authors also prove the theorem for $\mathbb{H}^n$, if it suffices for $\mathbb{R}^n$?

Secondly, independent of this proof, is the claim I mentioned in the beginning true? Intuitively it seems like it can't be but I can't seem to come up with any counterexample. I was thinking for example that an atlas on $\mathbb{R}^2 \setminus \left\lbrace (0,0) \right\rbrace$ can be just the identity map, which makes the (unique) map domain not diffeomorphic to $\mathbb{R}^2$, which I've seen by means of deRham cohomology. I also saw in this question that the upper half plane is not diffeomorphic to the whole plane. But maybe $\mathbb{R}^2 \setminus \left\lbrace (0,0) \right\rbrace$ is diffeomorphic to the upper half plane? This I don't know the answer to.

Thank you.

  • 1
    $\begingroup$ Why do you need both $\mathbb R^n$ and $\mathbb H^n$? They are diffeomorphic. Or do you mean $x_n \ge 0$? Where exactly does the claim occur? Please give more details. $\endgroup$
    – Paul Frost
    May 14, 2020 at 13:48
  • $\begingroup$ Yes, I mean $x_n \geq 0$, thank you. I got confused because one normally defines it with $x_n > 0$ but Bott and Tu work with $x_n \geq 0$. I've added details and the part of the proof itself. $\endgroup$ May 14, 2020 at 15:33

1 Answer 1


Bott and Tu consider manifolds with boundary (otherwise Stokes' theorem wouldn't make much sense).

On p. 30 they define

A manifold $M$ of dimension $n$ with boundary is given by an atlas $\{(U_\alpha, \phi_\alpha) \}$ where $U_\alpha$ is homeomorphic to either $\mathbb R^n$ or the upper half space $\mathbb H^n = \{(x_1, ... , x_n) \mid x_n \ge 0\}$.

In my opinion this is an unusual definition because it imposes serious restrictions on the charts occcurring in an "admissible" atlas. I would require that $U_\alpha$ is homeomorphic to either an open subset of $\mathbb R^n$ or of $\mathbb H^n$. On the other hand, it is no problem to restrict to special charts as you noticed in your question. So let us agree that the definition is a matter of taste.

Bott and Tu's definition provides the answer to your question. It does of course not say that any atlas will do. Your atlas on $M = \mathbb R^2 \setminus \{(0,0\}$ csnistng of the identity map is not adequate because $M$ is not homeomorphic to $\mathbb R^2$ or to $\mathbb H^2$. Both of these spaces are contractible, but $M$ is not.

  • $\begingroup$ Thank you, confusion is now clarified. $\endgroup$ May 14, 2020 at 17:40

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