July 16th: Coxeter groups (an example), an algebraic corollary

Jul 16—2025

In his PDE, Arnold says that the Coxeter group B(n) acts on the factors of $(\mathbb{CP}^1)^n$ by permutation and antiholomorphic involution with no fixed points (we can take the antipodal map).

For an example, let’s take B(2) acting on $\mathbb{CP}^2$ . A point in $\mathbb{CP}^2$ is given by (X, Y), where $X, Y \in \^{C}$ . Then, the orbit of (X, Y) under B(2) is: {(X, Y), (Y, X), (-X, Y), (-Y, X)} (NOTE: I apologize for my abuse of notation, the – sign does not mean the literal -; here, it simply stands for the antipodal map defined on $\^{C}$ . The map that sends z to -z would not work, as 0 and $\infty$ are both fixed points. It would be interesting to see if we can write an expression for such map).

Claim: B(2) is $D_4$ , the symmetries of a square.

Intuitively, when we write out the orbit, it definitely seems like this group would be a subgroup of the symmetries of a square. I give a visual proof below:

An interesting theorem (A.2 of Arnold’s PDE)

A real homogeneous polynomial f of degree n has a unique representation as the sum of two polynomials; one of which is the product of n linear real factors, and the other is divisible by $x^2 + y^2 + z^2$ .

What is more interesting is a remark that Arnold made afterwards: any real algebraic curve of degree n defined in a projective space determines n real “principal axes”. How does this connect to the theorem above? It is clear that the polynomial f contains a product of n linear real factors, which determine n lines. Are these n lines the principal axes that Arnold is referring to?

The proof of the theorem really uses the idea of quotient to the space of algebraic curves; the ideal is the one generated by $x^2 + y^2 + z^2$ . We want to look at f when it is restricted to the surface, S, defined by $x^2 + y^2 + z^2 = 0$ .

S does not have any real solution in $\mathbb{CP}^2$ . Since S is a real curve, it maps to itself under complex conjugation and has no fixed points under this map.

Since S is a degree 2 curve (and is irreducible) and f is degree n, it follows by Bézout’s theorem that either S is a component of (f=0), which means that f is a multiple of $x^2 + y^2 + z^2$ , or (f=0) and S has 2n intersection points.

Since both (f=0) and S are real algebraic curves, the 2n intersection points consist of n conjugate pairs (and no pairs are real).

These n conjugate pairs give n lines; each pair of points define a line. It follows that these n lines are real, i.e. can be defined by ax + by + cz = 0. We can define g to be the product of these n lines. We now show that $f|_{S}$ is proportional to $g|_{S}$ .

Key: we parameterize S by setting $x = 2t, y = t^2 - 1, z = i(t^2 +1)$ . Since the number of intersection points of (f=0) and S is finite, we can choose a suitable coordinate on S so that the point corresponding to $t = \infty$ is not one of the intersection points.

Then, $f|_{S} = f(x(t), y(t), z(t))$ and $g|_{S} = g(x(t), y(t), z(t))$ are 2 polynomials of degree 2n that share 2n roots. It follows that $f|_{S} = c\cdot g|_{S}$ , where c is a constant.