Shortest path connecting two opposite points on a cube

Consider the sphere with equator 4. Divide it into spherical cubes, the central projections from an inscribed cube.

Note that the exponential map from tangent plane to the sphere is short. Note also that if one maps a unit cube centered at the origin by the exponential map it will cover the spherical cube. It is easy to modify the map to get a short symmetric map from the unit cube to each spherical cube.

Joining all these maps, we get a short map fro the surface of the cube to our sphere, where the statement is evident.

[This is an attempt to explain the details in Anton Petrunin's answer to this question, since the comments suggest that a number of people have found it hard to understand as it was written.]

Let $C$ be the surface of the unit cube in $\mathbb{R}^d$ (for $d\geq 2$); when considered as a metric space, it is endowed with the distance-on-the-surface: so the goal is to show that the smallest possible distance between two antipodal points is $2$ (clearly it is at most $2$). Let $S$ be the ($(d-1)$-dimensional) sphere of radius $2/\pi$, also endowed with the intrinsic distance: certainly the distance between two antipodal points is $2$. So we are done if we can find a $1$-Lipschitz (="short", =which doesn't increase distances) map $h\colon C\to S$ mapping antipodal points to antipodal points. Note that this map does not have to be injective.

Let $Q_i$, for $1\leq i\leq 2d$ be the facets of $C$: each can be considered as a $(d-1)$-dimensional Euclidean unit cube with its usual distance; let $o_i$ be the center of $Q_i$. Let $K_i$ be the image of $Q_i$ by the central (=gnomonic) projection map $g\colon C\to S$ (i.e., if we imagine both $C$ and $S$ as centered at the origin of $\mathbb{R}^d$, then $g$ takes an element of $C \subseteq \mathbb{R}^d$ to the unique point of $S$ having that direction; note that $g$ is not the sought-after map $h$). So $K_i$ is a "spherical cube"; let $o'_i = g(o_i)$ be its center (the $o'_i$ are the $2d$ vertices of a regular cross-polytope on $S$, and the $K_i$ are the Voronoi cells of the $o'_i$, if we want).

We will define $h\colon Q\to K$ merely on one of the $Q_i$ (henceafter simply written "$Q$", and "$K$" is the corresponding $K_i$), commuting with all the symmetries of $(Q,K)$, so that we can use the same map on each $Q_i$ to get the sought-after $h\colon C\to S$.

Let $e\colon T_{o'} S\to S$ be the exponential map centered at the center $o'$ of $K$ (which can also be called the azimuthal equidistant projection): it takes a tangent vector to $S$ at $o'$ to the point of $S$ which is "in that direction, at that distance" (the direction and distance being given by the tangent vector). The map $e$ is $1$-Lipschitz (a fact which, for the sphere, is completely elementary, as the distance between two points can be expressed in polar coordinates around $o'$). Identify $T_{o'} S$ with the Euclidean space in which $Q$ lives, so that the origin is $o$ (center of $Q$): so now we have a map $e\colon Q\to S$ (taking $o$ to $o'$, and clearly commuting with all symmetries).

Note that this $e$ (restricted to $Q$) is a homeomorphism taking $Q$ to a set $e(Q)$ containing $K$: in other words, the distance in a given direction from $o$ to the boundary point of $Q$ in that direction is always at least the distance from $o'$ to the boundary point of $K$ in the same direction. (This is again an elementary fact that can be checked by comparing explicit formulæ for the gnomonic projection and the azimuthal equidistant projection; it is here that the choice of $2/\pi$ for the radius of $S$ is crucial.) So we can restrict $e$ to the inverse image $P := e^{-1}(K) \subseteq Q$ of $K$ (the azimuthal equidistant projection of a spherical cube: it looks like a kind of "puffy cube", tangent to $Q$ in the centers of its facets). Note that $P$ is convex. (I don't have an argument here, but it's surely not too difficult.)

What follows is an illustration of $P$ for $d=2$ (the boundary of $P$ is in red, the boundary of $Q$ is, of course, the black square):

This map still needs to be modified a little: let $h\colon Q\to K$ be obtained by composing the nearest point projection $Q\to P$ (which makes sense since $P$ is convex compact) with the exponential map $P\to K$ (restricted to $P$, as explained above, where it is a homeomorphism). So $h$ is $1$-Lipschitz (and not injective on all of $Q$, although it is a homeomorphism on $P$). Also, $h$ commutes with all the symmetries of $(Q,K)$ because the construction is completely symmetric. In particular, the various copies of $h\colon Q_i\to K_i$ match at the boundaries, so we get a map $h\colon C\to S$ as announced.

(Alternatively, instead of using the nearest point projection $Q\to P$, I think we can "truncate" the azimuthal projection: given a point in $x$ in $Q$, map it to the point in $P$ or $K$ which is in the same direction and at the same distance except if the distance exceeds the distance to the boundary, in which case map it to the boundary point in that direction. This avoids the argument that $P$ is convex but then makes it less clear that the map is $1$-Lipschitz.)

Take a path that joins the antipodes and concatenate it with its symmetric image. Get a centrally symmetric closed path on the boundary of the cube. If this path avoids one of the facets of the cube (and hence the antipodal facet as well), then we can project it to the boundary of the cube one dimension less and get a centrally symmetric path of at most the same length. If the path goes through all the faces, then its length is at least $2\sqrt{n}$ (apply the quadratic-arithmetic mean inequality to the components of the velocity vector). So, if $n \ge 4$, a path is either long or can be projected.

One has to deal with the $n=3$ case by hand, for example by using the development of the cube.