Is the square pyramid a manifold with corners?

Solution 1:

If there were a diffeomorphism taking (a neighborhood of) the top vertex of the pyramid to a (neighborhood of) the vertex in $[0,\infty)^3$, taking its derivative it would in particular preserve the tangent cone (the set of tangent vectors $v \in T_p M$ such that $v = \gamma'(0)$ for some curve $\gamma: [0,1] \to M$ with $\gamma(0)$ the vertex). If, for convenience, we set the top point of the pyramid to be $0 \in \mathbb R^3$, the tangent cone of the top vertex of the pyramid is the set of scalar multiples of points in the pyramid; the tangent cone of the vertex of the cube is $[0,\infty)^3$. So we've reduced this to a linear algebra problem: showing a linear isomorphism cannot take the first cone to the second. But, as you predicted, a linear isomorphism preserves the number of faces of the tangent cone. (I do not want to write down the details, so I leave this to you.)

As for why we don't want to include the pyramid: I haven't ever used manifolds with corners much myself, so I can only speculate, but I suspect it's because much of the interest is in a) developing a theory that works with simplices, so that we can integrate along chains; and b) developing a theory that will allow us to define cobordisms between manifolds with boundaries. Such cobordisms would, near the 'boundary', look like $M \times I$, where $M$ is a manifold with boundary; this gives us the local model $[0,\infty)^2 \times \mathbb R^{n-1}$. Now we're going to want cobordisms between these, etc, inductively saying we should want to include all local models $[0,\infty)^k \times \mathbb R^{n-k}$. In doing so we never really needed to have corners like the top of a square pyramid.