Is the surface of a cube a 2-dimensional differentiable manifold?

Question

Is the surface of a cube $[0,1]^3 \subset \mathbb{R}^3$ a 2-dimensional differentiable manifold (it is obviously a topological manifold)? Just assume the topology is Euclidean topology please. It seems that the 8 corners of the cube are singularities by intuition. This problem may be easy for you, however I find it hard to prove (or disprove) by the definition of differentiable manifold. I appreciate your help.

Answer 1

No, at least not in the sense that you want.

If it were a manifold, then by the classification theorem for compact 2-manifolds, it'd have to be diffeomorphic to a sphere. So the question then becomes

Is there an embedding $f$ of the sphere in 3-space whose image is exactly the unit cube?

If there were, then (with a little bit of isotopy in the domain sphere) we could assume that $f(0, 1, 0) = (0, 1, 0)$ and $f(1, 0, 0) = (1, 0, 0)$, i.e., that the north pole and the "east pole" of the cube get transformed into the center of the top of the cube and the center of the "east" face of the cube, respectively.

Now consider the arc $c$ defined by $$ t\mapsto (\sin t, \cos t, 0) $$ for $0 \le t \le \pi/2$. That's a smooth path on the sphere (it's part of a line of longitude!), so $b = f \circ c$ must be a smooth path on the cube.

Because the cube with all its edges and vertices removed has six connected components, with the north and east points in different components, but the interval is connected, we see that the path $b$ must meet the set of edges and vertices at some point $P$. Let's say that $$ b(t_0) = P $$ where $t_0$ is neither $0$ nor $\pi/2$, i.e., it's in the interior of the domain of $b$.

What is $b'(t_0)?$

On the one hand, it's $f'(c(t_0)) c'(t_0)$, and must therefore be nonzero, because $f$ is a diffeomorphism onto its image. On the other hand, it has to be a "tangent vector" at $b(t_0)$, a point on the edge of the cube. That pretty much requires that it lie in the direction of the edge. So a smooth curve from the north to the east pole of the sphere could become a smooth curve from the north to the east "pole" of the cube by heading towards an edge, running along the edge, and continuing off the edges into the eastern face.

But now take another unit arc $d$ that meets $c$ orthogonally at $c(t_0)$, and let's say $d(t_0) = c(t_0)$ by a shift of parameter if necessary.

Let $a(t) = f(d(t))$. What's $a'(t_0)$? It's got to be tangent to the cube at $a(t_0) = P$, and it's got to be nonzero and linearly independent from $b'(t_0)$, because $f$ is a diffeomorphism.

So: take any arc on the cube that crosses an edge transversally, and with nonzero speed at the crossing. Can that arc be differentiable at the crossing? No. For just before crossing, the curve lies in (say) the north face, and hence has a tangent orthogonal to $(0,1,0)$; just after the crossing, it lies in the east face, and has a tangent orthogonal to $(1,0,0)$. Hence the tangent at the crossing must be orthogonal to both, and thus be in the direction of $(0,0,1)$. But that makes it NOT linearly independent from $b'(t_0)$. And that's a contradiction.

I've been a little sloppy here, talking about "just before the crossing" and "just after the crossing"...how do I know that for some small time before $t_0$ we have $a(t)$ is on the "north face" and that for some small time after, it's on the "east face"? Answer: a little bit of careful analysis, and the fact that every nonconstant curve is well approximated by its tangent locally.

But the gist here is clear: the tangent plane to $S^2$ at $c(t_0)$ would have to be transformed, by $f'$, isomorphically to a tangent plane to the cube along one of the edges. But that "tangent plane" can really only contain the edge itself, hence is 1-dimensional, and you've got a contradiction.