2

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.

NateZ
  • 63
  • 5

2 Answers2

3

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).

Navid Nabijou
  • 61
  • 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
1

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.

John Hughes
  • 93,729