Grunsky matrix

Remarks

The Grunsky matrix is symmetric: $c_{mn} = c_{nm}$, because $Q(z, \zeta) = Q(\zeta, z)$. The coefficients are polynomials in $b_1, b_2, \ldots$; for example $c_{11} = b_1$. The Grunsky inequalities say that for every finite sequence $\lambda_1, \ldots, \lambda_N \in \C$, $$\Bigl| \sum_{m, n = 1}^{N} \sqrt{mn}\, c_{mn} \lambda_m \lambda_n \Bigr| \le \sum_{n=1}^{N} \abs{\lambda_n}^2.$$ Example: $g(z) = z + 1/z$ is injective on $\Delta$, since $g(z) = g(w)$ with $z \ne w$ forces $zw = 1$. Here $Q(z, \zeta) = 1 - 1/(z\zeta)$, so $\log Q = -\sum_{n \ge 1} \frac{1}{n} (z\zeta)^{-n}$, and $c_{nn} = 1/n$ while $c_{mn} = 0$ for $m \ne n$. The weighted matrix $(\sqrt{mn}\, c_{mn})$ is the identity, so the Grunsky inequalities reduce to $\abs{\sum_n \lambda_n^2} \le \sum_n \abs{\lambda_n}^2$, with equality for real $\lambda_n$. Non-example: $g(z) = z + 2/z$ is not injective on $\Delta$, since $g(i\sqrt{2}) = g(-i\sqrt{2}) = 0$. Then $Q(z, \zeta) = 1 - 2/(z\zeta)$ vanishes at points of $\Delta \times \Delta$, and no holomorphic logarithm exists there. Sign and normalisation conventions differ between authors; some use $\log\frac{g(z) - g(\zeta)}{z - \zeta}$ with a plus sign or with the factors $\sqrt{mn}$ built in.