Is the cube with boundary and corners a manifold with boundary?

Question

The definition of a n-manifold with boundary as I understand it, is that the manifold without boundary is an n-manifold, and the boundary is an (n-1)-manifold. Thus because the boundary of cube has edges and corners, which are of a lower dimension than 2, would the boundary not be a 2-manifold? This seems right, but I wanted to verify that my logic is correct.

edges and corners don't do anything topologically, so the boundary of a cube is a topological sphere. — Igor Rivin, Dec 16 '13 at 21:43
If you are referring to a differentiable manifold, then this is correct. — Alex Becker, Dec 16 '13 at 21:43

Navid Nabijou · Accepted Answer · 2013-12-18T09:06:29.027

4

I assume you're talking about differentiable manifolds here. In that case (as has been noted in comments) the boundary of a cube is not a 2-manifold with boundary.

It's worth looking up the definition of "manifold with corners" (or trying to guess what it should be yourself). ~~You'll see that the boundary of cube is a 2-dimensional manifold with corners.~~

Edit: Sorry, I made a mistake before: the solid cube is a 3-dimensional manifold with corners, but I don't believe the boundary is (though I guess it can be thought of as a piecewise-linear manifold).

edited Dec 18 '13 at 09:06

answered Dec 16 '13 at 21:57

Navid Nabijou

71

Thank you! Yes, I was talking about differentiable manifolds. My class covered them very loosely in the last few days of class, and this was a book question. – NateZ Dec 16 '13 at 22:11
Thank you for your answer. I've a dumb question: how exactly do we wee that boundary of a solid 3-cube is not a 2-dimensional manifold with corners? One has to come up with an argument of the transition maps between charts not being smooth diffeomorphism, right? So, how do we show that if we equip it with a charts $(U,\phi), U:=$ open in $[0,\infty)^k\times \mathbb{R}^{3-k}, 1\le k \le 3, \phi(U)$ open in $X,$ then the transition maps $\psi\circ \phi^{-1}$ cannot be smooth? – Mathguest Jan 14 '25 at 09:16
In this regards, I'm looking at https://arxiv.org/pdf/0910.3518, where the author defined manifold with corners and proved in Proposition 2.7 that the boundary of an $n$-manifold with corners is an $(n-1)$-manifold with corners; but I understand that his definitions are perhaps different from the previous ones. – Mathguest Jan 14 '25 at 09:19

score 1 · Answer 2 · answered Dec 16 '13 at 22:11

The boundary of a solid cube in 3-space seems to be the object you're talking about. Topologically, it's the same as the 2-sphere (the surface of the earth, for example). In the smooth category, it also happens to be the image of smooth map from the sphere to 3-space, but this smooth map is singular at each vertex and along each edge.

Is the cube with boundary and corners a manifold with boundary?

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