July 17th: Coxeter group (continued), Maps Between Projective Spaces

Jul 17—2025

Continuing from the previous post, we defined the Coxeter groups B(n) to be the set of actions on $(\mathbb{CP}^1)^n$ (I had thought it was just $\mathbb{CP}^n$ in the previous post, but there is a difference between the two) given by permutation of the factors and antiholomorphic involution with no fixed points on each factor (something to consider: is such antiholomorphic involution on $\tilde{\mathbb{C}}$ unique?).

If we visualize this involution map geometrically, we would see $\tilde{\mathbb{C}}$ as $S^2$ and the map would be the antipodal map. I had asked in the last possible if we could find an explicit expression for such map. We will find such expression below.

Key: we express $\mathbb{CP}^1$ in terms of the space of quaternions, $\mathbb{H}$ .

A point in $\mathbb{H}$ is given by $a + bi + cj + dk = (a + bi) + (c +di)j$ . We can map this to the point $(a + bi, c + di) \in \mathbb{C}^2$ . Thus, $\mathbb{H}/\mathbb{C}* \cong \mathbb{CP}^1$ .

The left multiplication by j in $\mathbb{H}$ gives rise to an antiholomorphic involution with no fixed points in $\mathbb{CP}^1$ .

Suppose $z = a + bi, \omega = c + di$ . Then,

$j( z + \omega j) = j(a + bi) + j(c + di)j = (a - bi)j - (c - di)j = - \bar{\omega} + \bar{z}j$ .

Hence, the left multiplication by j maps $(z, \omega) \Rightarrow (-\bar{\omega}, \bar{z})$ .

In $\mathbb{CP}^1$ , the corresponding action would be $[z, \omega] \Rightarrow [-\bar{\omega}, \bar{z}]$ .

It has no fixed points in $\mathbb{CP}^1$ . We can show this by noticing that if $z = -\bar{\omega}$ and $\omega = \bar{z}$ , then $z = \omega = 0$ .

We can also show this by the following geometric observation. This action maps the dual of $[z, \omega]$ , given by $zx + \omega y = 0$ (a line in the dual space), to $-\bar{\omega}x + \bar{z}y = 0$ , its Hermitian-orthogonal. When this is a real line, it is mapped to its orthogonal line. Hence, no line is mapped to itself. It follows that its corresponding dual will not be mapped to itself.

The explicit formula for this map in $\mathbb{CP}^1$ is thus given by $t \Rightarrow -\frac{1}{\bar{t}}$ .

Goal #2: Describing points in $\mathbb{RP}^2$ with lines in $\mathbb{C}^2$ .

Claim: there exists a bijective map from the set of pairs of Hermitian-orthogonal complex lines passing through the origin in $\mathbb{C}^2$ to points in $\mathbb{RP}^2$ .

From above, a pair of Hermitian-orthogonal complex lines passing through the origin in $\mathbb{C}^2$ , say $zx + \omega y = 0, -\bar{\omega}x + \bar{z}y = 0$ , correspond to $[z, x], [-\bar{\omega}, \bar{z}] \in \mathbb{CP}^1$ .

We construct the following quadratic form from the two linear functionals corresponding to the two lines:

$f(x, y) = (zx + \omega y)(-\bar{\omega}x + \bar{z}y)$

Expanding gives:

$f(x, y) = f_0 x^2 + f_1 xy + f_2 y^2$ ,

where $f_0 = -z\bar{\omega}, f_1 = z\bar{z} - \omega \bar{\omega}, f_2 = \bar{z}\omega$ .

Notice that $f_1 \in \mathbb{R}$ and $f_0 = - \bar{f_2} \in \mathbb{C}$ .

Hence, using the real, imaginary parts of $f_2$ and $f_1$ , we can describe a point in $\mathbb{R}^3$ , it is given by $(Re(\bar{z}\omega, Im(\bar{z}\omega), |z|^2 - |\omega|^2) \in \mathbb{R}^3$ .

Since the equations of the two lines are invariant under scaling, i.e. $\lambda[z, \omega], \lambda[-\bar{\omega}, \bar{z}]$ for any non-zero complex $\lambda$ describe the same two lines, it follows that the two lines get mapped to the set $\{ |\lambda|^2 (Re(\bar{z}\omega, Im(\bar{z}\omega), |z|^2 - |\omega|^2): \lambda \in \mathbb{C}* \}$ , which describes a ray beginning at the origin in $\mathbb{R}^3$ . It is not difficult to show that this map is surjective and that $[z, \omega], [-\bar{\omega}, \bar{z}]$ determines a unique ray centered at the origin in $\mathbb{R}^3$ , and vice versa.