# Why is $\partial\partial M=\varnothing$?

Why is the border of the border of an oriented differentiable $n$-dimensional Manifold $M$ empty, that is $$\partial\partial M = \emptyset?$$

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Well, what is the boundary of the boundary of a closed half-space in $\mathbb R^n$? Since manifolds are locally like closed half-spaces of $\mathbb R^n$ and the boundary is computed locally, this gives you the answer. –  Mariano Suárez-Alvarez Apr 14 '13 at 14:48

This simply says that the boundary of the manifold $M$ is a manifold without boundary. Indeed, by definition, it is locally homeomorphic to $\mathbb{R}^{n-1}$ by restriction of the atlas of $M$ to $\partial M$. Note this has nothing to do with the differential structure.

Note: I think it is worth noting that the boundary $\partial^\mathrm{man}$ of a manifold is a notion which is different from that of boundary in a general topological space $\partial^\mathrm{top}$. For instance, $\partial^\mathrm{man}(0,1)=\emptyset$ while $\partial^\mathrm{top}(0,1)=\{0,1\}$. To avoid confusion, some people call $\partial^\mathrm{top}$ the fronteer.

It is not true that $\partial^\mathrm{top}\partial^\mathrm{top} S=\emptyset$ in general. Precisely, in the case of a closed half-space, $\partial^\mathrm{top}\partial^\mathrm{top} S=\partial ^\mathrm{top}S\neq \emptyset$. For instance, $$\partial^\mathrm{top} \{(x,y)\in\mathbb{R}^2\;;\;y\geq 0\}=\partial^\mathrm{man} \{(x,y)\in\mathbb{R}^2\;;\;y\geq 0\}=\{(x,y)\in\mathbb{R}^2\;;\;y=0\}=B$$ and $$\partial^\mathrm{top}B=B\qquad\text{while}\qquad \partial^\mathrm{man}B=\emptyset$$ as $B$ is homeomorphic to $\mathbb{R}$.

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Thanks a lot to whoever took the time to do the $\rm$ edit. –  1015 Apr 14 '13 at 16:28
Can you help me with this please ? math.stackexchange.com/questions/361342/mass-of-a-rectangle/… –  user43418 Apr 14 '13 at 16:35

Let M be a manifold and $\omega>0$ be a function. By twice use of Stokes Theorem:

$$\int_M d^2 \omega=\int_M d(d(\omega))=\int_{\partial M}d\omega=\int_{\partial{\partial M}} \omega$$

Since $d^2 \omega=0$ always (property of exterior derivative) then $\int_{\partial{\partial M}} \omega=0$ and because $\omega>0$ then $\partial\partial M=0$.

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You can only integrate top-dimensional forms on a manifold. So if $\dim(M) = n$, $\omega$ needs to be a $(n - 2)$-form, not a function. –  Henry T. Horton Apr 14 '13 at 15:35