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A point of a manifold with corners is a boundary point by definition if one of its coordinates is $0$ by some (hence in all) chart with corners (see here).

In the same page one can read:

The boundary of a smooth manifold with corners, however, is in general not a smooth manifold with corners (e.g., think of the boundary of a cube). ... It is, however, a union of finitely many such

But how is the boundary of such an object defined? What is the definition of the boundary of the unions of manifolds with corners? In topological sense it is itself (or at least a subset of itself), but I think we should expect here a definition so, that the boundary of the boundary of a manifold with corner is empty.

Second attempt to formulate my question without error

Is there a definition of the boundary of the boundary of a manifold with corners so, that we can state, that the boundary of the boundary of a manifold with corners is empty? (Similarly as we can say that the boundary of the boundary of a manifold with boundary is empty)

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There's no need to define the "boundary of a union of manifolds with corners" in general. If you look at the full quotation that you excerpted from my book, it says

The boundary of a smooth manifold with corners, however, is in general not a smooth manifold with corners (e.g., think of the boundary of a cube). In fact, even the boundary of ${\overline {\mathbb R}{}^n_+}$ itself is not a smooth manifold with corners. It is, however, a union of finitely many such: $\partial {\overline {\mathbb R}{}^n_+}=H_1\cup \dotsb\cup H_n,$ where $H_i = \big\{ (x^1,\dots, x^n)\in {\overline {\mathbb R}{}^n_+}: x^i = 0\big\}$ is an $(n-1)$-dimensional smooth manifold with corners contained in the subspace defined by $x^i=0$.

I wasn't claiming that the boundary of every manifold with corners is a union of finitely many manifolds with corners, only the "model" manifold with corners $\overline {\mathbb R}{}^n_+$ (which is the subset of $\mathbb R^n$ where all coordinates are nonnegative). Each of the subsets $H_i$ is itself a manifold with corners, whose boundary and corner points are defined in the same way as for any manifold with corners.

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  • $\begingroup$ Dear Jack, thank you for the clarification. I clearly misunderstood this statement. However, I feel that this has only a little to do with my real intended question. Unfortunately I formulated my question totally wrong, so I try to formulate it better: Is there a definition for the boundary of the boundary of a manifold with corners, that we can state, that the boundary of the boundary of a manifold with corners is empty? $\endgroup$ – mma May 9 '15 at 6:31
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    $\begingroup$ This paper by Dominic Joyce gives a more involved definition of boundary, under which the boundary of a manifold with corners is again a manifold with corners. It's not generally true that the boundary of a boundary is empty, however. Another approach is to triangulate your manifold with corners and regard it as a smooth chain (in the sense of algebraic topology). Then the boundary is again a smooth chain, and $\partial(\partial M)=0$. $\endgroup$ – Jack Lee May 9 '15 at 18:23

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