Can the surface of a 3-dimension cube be diffeomorphic to a 3-dimension sphere?

Question

We all know the surface of a cube homeomorphic to a sphere $S^3$ by retraction, but I'm confused that whether the surface of a cube can be diffeomorphic to $S^3$.

Answer 1

Let me answer this in lower dimensions:

First, consider the bi-unit square $S$ (the boundary of $[-1, 1] \times [-1, 1]$ in $\Bbb R^2$) and the unit circle $C$ in $\Bbb R^2$.

There's a nice map ("radial projection") from the first to the second defined as follows:

$$ f(x, y) = \begin{cases} (\frac{x}{\sqrt{y^2+1}}, \frac{y}{\sqrt{y^2 + 1}})& x = \pm 1 \\ (\frac{x}{\sqrt{x^2+1}}, \frac{y}{\sqrt{x^2 + 1}})& y = \pm 1 \\ \end{cases} $$ which cane be written, more generally, as $f(x, y) = \frac{1}{\sqrt{x^2 + y^2}} (x, y)$, but I wanted to emphasize that it's defined on the square, and that at the four corners, the expressions involved shift from a dependence on $x$ to a dependence on $y$ or vice versa.

The inverse of $f$ is $$ g(x, y) = \begin{cases} (\frac{x}{|x|}, \frac{y}{|x|}) & |x| \ge |y|\\ (\frac{x}{|y|}, \frac{y}{|y|}) & |x| \le |y| \end{cases} $$

Now one smoothness structure on $S$ can be described as follows: we say that a function $$ u: S \to \Bbb R $$ is smooth if and only iff $$ u \circ g : C \to \Bbb R $$ is smooth.

With this definition, the square $S$ is a smooth manifold, and it's diffeomorphic to the circle $C$, with $g$ being the diffeomorphism.

There's a small caveat here, though: the inclusion map $$ i : S \to \Bbb R^2: (x, y) \mapsto (x, y) $$ that embeds the square in the plane is not a smooth embedding.

What about higher dimensions? Well, an exact analog to $f$ can be written using more variables (and the "general" formula that I gave after the "cases" formula); the analog of $g$ involves 3 cases for a cube in 3-space (look at $|x|, |y|, |z|$: each case handles the situations where one of these is the largest).

When we get to 4-space, it's the same deal, but with another variable, and the formula for $g$ has four cases.

But everything else I said (including the fact that with the resulting smoothness structure, the standard embedding of the square/cube/hypercube is not a smooth embedding) remains true.