# The Wobbly Table Problem, part 3 •

In part 1 I described the Wobbly Table problem, and explained how a positive solution follows from Dyson’s theorem, and in part 2 I explained how I would prove Dyson’s theorem from the following lemma, in the style of proofs of the Borsuk–Ulam theorem:

There is no $D_4$–equivariant continuous function from $SO(3)$ to $S^2$, where the group acts on $SO(3)$ by right multiplication and on $S^2$ by permuting the vertices of a regular tetrahedron.

The right multiplication of the dihedral group $D_4$ of order 8 is by matrices representing the standard symmetries of a square in the $xy$–plane, with vertices at $(1,0,0)$, $(0,1,0)$, $(-1,0,0)$, and $(0,-1,0)$. The regular tetrahedron in the $2$–sphere comes from the image of the four standard basis vectors of $\R^4$ under projecting to the unit sphere in the $3$–space given by $x+y+z+w=0$, the orthogonal complement of the diagonal $\triangle \R^4$.

For convenience, we’ll give this $3$–space coordinates so that the regular tetrahedron has vertices at $(a,0,b)$, $(0,a,-b)$, $(-a,0,b)$, and $(0,-a,-b)$, with $a=\frac{\sqrt{2}}{\sqrt{3}}$ and $b=\frac{1}{\sqrt{3}}$: the four vertices lie just above and below the vertices of the standard square.

We remark that the action of $D_4$ on the 2–sphere — by permutations of the vertices of the regular tetrahedron — then becomes the product of the standard symmetries of the square with reflection in the $xy$–plane for odd permutations. For example, a counterclockwise turn by 90 degrees in the standard action becomes a counterclockwise turn by 90 degrees followed by a reflection in the $xy$–plane in our permute-the-tetrahedron action.

It will also be to our advantage to use the closed $3$–ball model of $SO(3)$ — that is, the identification of $SO(3)$ with a closed ball in $\R^3$ with radius $\pi / 2$, where a point’s distance from the origin corresponds to amount of rotation, and the ray from the origin to the point determines the axis of rotation. This gives an identification of $SO(3)$ with real projective $3$–space, and we can give this space the round metric it inherits from its double cover the $3$–sphere. See the wikipedia article on the topology of $SO(3)$ for details.

With the identity of $SO(3)$ at the center of the ball, its orbit under the $D_4$ action consists of the 8 points pictured below. The labels, e.g. $(A, B, C, D)$, refer to permutations of the standard square. There appear to be more than 8 points, but recall that points on the boundary are identified with their antipodes. The square is drawn in for convenience, but it’s meant only to remind the viewer of the symmetries; it’s not actually embedded in $SO(3)$.

We’ll make use of the following facts from algebraic topology. First, that maps from the $2$–sphere to itself are classified up to homotopy by their degree. And second, that maps from a closed $2$–disk to the $2$–sphere which restrict on their boundary to a given fixed map, are classified up to homotopy by an integer invariant which is essentially degree. For example, take a disk mapping to the 2-sphere with boundary being sent to the equator. Then one example of such a map is to send the disk to the upper hemisphere. Then we can form other such maps by adding $n$ “bubbles” to the map by gathering material in a neighborhood of a point and wrapping it around the sphere $n$ times, for any integer $n$. Then any map of the disk to the sphere sending the boundary to the equator is homotopic to such a map. This follows from e.g. the long exact sequence for relative homotopy groups. See Allen Hatcher’s book Algebraic Topology [pdf] for a comprehensive reference.

A final remark: This lemma was proved by Hausel, Makai, and Szucs in 1999 by other methods. They actually compute existence/non-existence of equivariant maps from $SO(3)$ to $S^2$ for all subgroups of the symmetric group $S_4$, using characteristic classes and the geometry of ellipsoids in $\R^3$.

Suppose $f: SO(3) \to S^2$ is an equivariant function as described in the hypothesis. The strategy is this: we’ll find a particular embedded $2$–sphere in $SO(3)$ on which the function $f$ restricts a priori to a map with degree 0. Then we’ll use the hypothesis of equivariance to show in fact that the restriction of $f$ actually has odd degree (and therefore non-zero degree), yielding a contradiction.

To find this embedded $2$–sphere, we’ll use the boundary of a fundamental domain of the $D_4$ action on $SO(3)$. In particular take the Voronoi cell around the identity matrix for the diagram formed by the 8 points in the orbit. The metric used for the Voronoi diagram is the standard round metric on $SO(3)$, not the Euclidean metric on a closed $3$–ball. Then our embedded (topological, not smooth) $2$–sphere that we’ll use in this proof is the boundary of this cell.

This $2$–sphere is geometrically an octagonal prism whose faces are pieces of round $2$–spheres. See the picture; the faces are drawn flat although they should bulge outward. They are flat as subspaces in the sense that they’re totally geodesic in $SO(3)$.

The degree of $f$ restricted to the prism is a priori 0 because the prism bounds a $3$–ball. But because of the way we have constructed the prism, we’ll see how various elements of the $D_4$ action swap various pairs of opposite faces of the prism, and we can use this to compute an odd degree for the restriction of $f$.

In fact there are five pairs of opposite faces — four pairs of opposite squares and the one pair of octagons on top and bottom — each of which get swapped by an element of $D_4$.

The line from the identity matrix to each of the four group elements representing reflections of the square pierces one of the four pairs of opposite squares, and right multiplying by one of these group elements swaps the associated pair of faces, along with a counterclockwise twist of that square by 90 degrees. For example:

The line which passes through the identity matrix and the group elements corresponding to rotations of the square pierces the two octagonal faces. Right multiplication by the group element corresponding to a 90 degrees counterclockwise rotation of the square moves the bottom octagon to the top, with a counterclockwise rotation (viewed from below) of an eight of a full turn. (Likewise, the group element corresponding to 90 degrees clockwise turn of the square is the inverse of this operation, and hence moves the top face to the bottom face in the reverse direction.)

To see that these group elements permute various faces in the way that I’ve indicated is left as an exercise to the reader. A hint: the whole point of choosing a fundamental domain given by Voronoi cells is that the group acts on $SO(3)$ by isometries, and a Voronoi wall is equidistant to (at least) two points in the group orbit.

Because these pairs of faces are related by group elements in $SO(3)$, their images in the $2$–sphere under $f$ are also related by the symmetries of the tetrahedron. To make use of this now, we will take the restriction of $f$ to the octagonal prism, and equivariantly deform it to a position where we can easily compute its degree.

This will proceed in three stages: first, we’ll move the images of the vertices of the prism to the north and south poles of the $2$–sphere, second, we’ll move the images of the edges to four specific arcs on the sphere, and finally, we’ll deform the faces.

Note that although the prism has 16 vertices, 24 edges, and 10 faces, we don’t have complete freedom in deforming their images. Because we want to keep equivariance, we actually only have freedom in moving the images of 4 vertices, 8 edges, and 5 faces: the opposite-face-identification means that each vertex, edge, and face has 3, 2, and 1 partners respectively.

We can specify the image of any of these vertices, and then the image on the other three is determined:

We can specify the image of any of these edges, and then the image on the other two is determined:

The first two stages can be carried out because the $2$–sphere is simply connected, so there is no obstacle to moving vertices and edges wherever we please, although of course the faces must be adjusted near their boundaries for this to be done continuously.

Consider the following special points and edges in the $2$–sphere: we’ll take the north and south poles, and look at four arcs which run from south to north. These are useful to consider because the symmetries of the tetrahedron also permute these arcs:

We’ll deform our map equivariantly and send four vertices to the south pole. By equivariance that determines our map on all the other vertices to go to either the south or north pole. We’ll also deform our map equivariantly on eight edges, sending four of them to our special arcs from south to north, and the other four to the constant arc at the south pole. See the image below:

We specify the destinations using colors and arrows; the color of each vertex indicates where it goes, and likewise for the edges.

Note that we’ve specified the image of some edges where we only know where one of its endpoint goes. That’s okay, because if we move the four indicated vertices first, it turns out that all the other endpoints of the indicated edges wind up in the appropriate spots. Here is where all the vertices and edges get sent, using equivariance to compute their destinations:

Now finally we deform the faces. Note that each face of the prism can be deformed to cover exactly one quadrant of the $2$–sphere (carved up by our four special arcs from bottom to top), and additionally cover the entire sphere an additional $n$ times (by “bubbles”) for some integer $n$. We show that the degree of this map on the prism is odd.

In fact note that the additional extra $n$ “bubbles” all cancel out (at least as far as parity is concerned). Each face and its opposite (whether square or octagonal) have exactly the same number of bubbles, or possibly negative, depending on whether the group element swapping them is orientation preserving or reversing. Thus the sum from opposite faces is even. And so we’re left with considering only the contribution on the quadrants.

Note that the front four faces (indicated in the picture below) each cover one of the four quadrants:

The back four faces, and the top and bottom face, each cover a single quadrant with alternately positive and negative orientation. Thus the total degree is 1, plus the even contribution from the bubbles, and is therefore odd.

That’s our contradiction — the map has degree zero a priori on the boundary of the octagonal prism, and yet has odd degree from equivariance. Thus there is no such equivariant map.

## Some concluding remarks. •

We’ve proved that a square table can always be placed stably on a continuous floor without wobbling, using Dyson’s theorem and proving that via this lemma about equivariant functions.

It’s sort of lucky that this lemma is sufficient to prove Dyson’s theorem, because actually the function from $SO(3)$ to the $2$–sphere (which we get by looking at the value of a function on squares) has more symmetry than just $D_4$–equivariance: after all, fixing two diagonal points of the square and rotating the other pair preserves the value of the function on the first two points along the whole circle of rotation, but this is not captured by equivariance.

In fact we’re unlucky if we try to prove Livesay’s generalization of Dyson’s theorem by the same method. For rectangles, the symmetry group is the dihedral group $D_2$ of order 4, and again this fails to capture all the symmetry of the map, and for this group there are equivariant functions from $SO(3)$ to the $2$–sphere.

Dyson’s theorem and Livesay’s theorem generalize to higher dimensions and it would be very annoying to try to use this method to prove the generalizations. So this is sort of a one-shot deal.