# 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 $$g: S^2 \to {\mathbb{R}}$$ 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 $$g: S^2 \to {\mathbb{R}}$$ 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 $$h: SO(3) \to {\mathbb{R}}^4$$ 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 $${\mathbb{R}}^4$$ by permuting coordinate factors. Think of the four coordinate factors of $${\mathbb{R}}^4$$ 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 $${\mathbb{R}}^4$$, $$\triangle {\mathbb{R}}^4 = \{ (x,x,x,x) : x \in {\mathbb{R}}\}$$. 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 $$\triangle {\mathbb{R}}^4$$. In the picture below, we show the diagonal in $${\mathbb{R}}^4$$ 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 $${\mathbb{R}}^3$$, 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 $${\mathbb{R}}^4$$.

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.