Example 12.6.5327A22321E298

Proof. The extreme sets are the points $[1:0:0]$ and $[0:0:1]$. In affine coordinates, $a$ and $1-a$ vanish there to exactly second order, so both real codimensions are four and both orders are two. To compute the law of $a$ under $\sigma $, polar integration in the chart $Z_0=1$ gives $2\, \mathrm{d}s_1\mathrm{d}s_2/(1+s_1+s_2)^3$, where $s_ j=|z_ j|^2$. The transformation $u_ j=s_ j/(1+s_1+s_2)$ gives the uniform density two on the triangle $u_1,u_2\geq 0$, $u_1+u_2\leq 1$, and $a=u_1/2+u_2$. For $0\leq s\leq 1/2$, the triangle $u_1/2+u_2\leq s$ has area $s^2$, so the law of $a$ has density $4s$ on $[0,1/2]$. With $u_0=1-u_1-u_2$, exchanging $u_0$ and $u_2$ preserves the uniform measure and replaces $a$ by $1-a$; hence the law has density $4(1-s)$ on $[1/2,1]$. Integrating against these two densities gives $H_ a=H_-=4\log 2$ and the formula for $Q$, and the integral definition of $Q$ shows that $Q(0)=1$ is its continuous extension. Theorem 12.6.41 gives the boundary lines and the finite width, with Euclidean coordinates $Y=\sqrt2\log b$ and $R(v)$. $\square$