# The Wobbly Table Problem, part 2

In part 1 I described the Wobbly Table problem, and explained how a positive solution follows from Dyson’s theorem:

Let $g: S^2 \to \R$ be a continuous function from the 2-sphere to the real numbers. Then there are two orthogonal diameters of the sphere whose endpoints $\pm x$ and $\pm y$ all take the same value of $g$.

In this post I’ll give a proof of Dyson’s theorem. Whereas Dyson’s original proof uses a geometric technique called $T$–sets, my proof uses techniques that might be found in a first course in algebraic topology and follows the style of standard proofs of the Borsuk–Ulam theorem.

Let $g: S^2 \to \R$ be a continuous function from the 2-sphere to the real numbers. We’re interested in the value of $g$ on 4-tuples of points which are the endpoints of two orthogonal diameters, so we’ll define a function $h$, associated with $g$, which records the value of $g$ on four points of a square at once.

Let $h: SO(3) \to \R^4$ take the set of $3\times 3$ orthogonal matrices to Euclidean 4-space, defined as follows:

$h([v_1, v_2, v_3]) = (g(v_1), g(v_2), g(-v_1), g(-v_2))$

Here $[v_1, v_2, v_3]$ is a matrix expressed in terms of its column vectors. The vectors $v_1, v_2, -v_1, -v_2$ form the vertices of a square, and $h$ records the values of $g$ on this square (see the picture).

$\overset{h}{\longrightarrow} (g({\color{red} A}), g({\color{blue} B}), g({\color{darkgreen} C}), g(D))$

Now we make two observations. First, the function $h$ has symmetry: it’s keeping track of labels on the vertices of the squares, but if we rotate the square, the value of $h$ on that different square is related to the value of $h$ on the original square. We express this symmetry as equivariance with respect to a group action, which we now explain.

The group of symmetries of a square is the dihedral group $D_4$ (of order 8). This group has a representation $\rho: D_4 \to SO(3)$ as standard rotation matrices (as the symmetries of a square with vertices at the first two standard basis vectors and their negatives).

The group $D_4$ also acts on $\R^4$ by permuting coordinate factors. Think of the four coordinate factors of $\R^4$ as being labeled by the vertices $A, B, C, D$ of a square; then, applying a symmetry of a square permutes the labels.

Therefore we can express the equivariance of $h$ by:

$h([v_1, v_2, v_3] \cdot \rho(a)) = h([v_1, v_2, v_3]) \cdot a, \forall a \in D_4.$

Second, phrased in terms of $h$, Dyson’s theorem is equivalent to saying that the image of $h$ intersects the diagonal of $\R^4$, $\triangle \R^4 = \{ (x,x,x,x) : x \in \R \}$. If Dyson’s theorem failed to hold for a given function $g$, then the function $h$ could be deformed so that its image lies in the unit 2-sphere in the 3-space orthogonal to $\triangle \R^4$. In the picture below, we show the diagonal in $\R^4$ as well as the 2-sphere which lies orthogonal to it. Note that we show a 3-dimensional analogue: we have the diagonal pictured in $\R^3$, with a 1-sphere (a circle) in the complementary 2-plane to the diagonal. In 4 dimensions, there is actually a 2-sphere lying in the complementary 3-space to the diagonal.

Furthermore, this deformation remains equivariant: now $D_4$ permutes the vertices of a regular tetrahedron in $S^2$ rather than permuting the coordinates in $\R^4$.

After these two observations, Dyson’s theorem will follow from:

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.

I give the proof of this lemma in part 3.