Find the curvature of the metric $ds=\frac{|dz|}{(1+|z|^2)}$ on $\mathbb{C}$

Question

The curvature of the metric $g$ is defined as $$k(z)=-\bigg(\frac{2}{\alpha(z)}\bigg)^2 \partial \bar\partial log \alpha(z)$$ where $\alpha$ is positive and real valued.

Also $\partial=\frac{(\partial_x -i \partial_y)}{2}$ and $\bar\partial\frac{(\partial_x + i \partial_y)}{2}$

I am trying to find the curvature of the metric $ds=\frac{|dz|}{(1+|z|^2)}$ on $\mathbb{C}$

How do I do this?

Answer 1

The expression $ds=\frac{|dz|}{(1+|z|^2)}$ is an instruction for how to measure the length of a smooth (actually rectifiable) curve $\gamma: [a,b]\rightarrow \mathbb{C}$. When $\gamma$ is smooth, this length is just $$L(\gamma)=\int_\gamma \ ds =\int_a^b \gamma^*ds=\int_a^b\frac{|\gamma'(t)|}{1+|\gamma(t)|^2}dt. $$

A conformal metric on a domain $U\subset \mathbb{C}$ is simply a function $\alpha: U\rightarrow \mathbb{R}^{>0}$; one uses the conformal metric to scale the Euclidean metric $|dz|$ in order to measure the $\alpha$-length of an arc. So, defining $\alpha(z)=\frac{1}{1+|z|^2}$, we have $ds=\alpha(z)|dz|$.

Deigression: If $\beta: \mathbb{C}\rightarrow \mathbb{R}^{>0}$ is smooth, then $\beta(z)|dz|$ defines a Riemannian metric $\langle , \rangle_\beta$ on $\mathbb{C}$ by $$\langle v,w\rangle_{\beta(z)}:=\beta(z)\langle v,w\rangle, \text{ where }v,w\in T_z \mathbb{C}\cong \mathbb{C}$$ and $\langle, \rangle$ is the ordinary dot product in $\mathbb{R}^2$. It is a theorem that the Gaussian curvature of the Riemannian surface $(\mathbb{C},\langle,\rangle_\beta)$ is equal to $k(z)$ as defined above. End Digression.

The computation of $k(z)$ is now a matter of computing ordinary derivatives, observing that some general rules like $\bar\partial z=0$, $\partial\bar z=0$, $\partial |z|^2=\partial z\bar z = \bar z$, etc. all hold. We therefore have that $$\bar\partial\log(\alpha(z))= -\bar\partial \log(1+z\bar z)=-\frac{z}{1+z\bar z} $$ applying the $\partial$ operator and applying the quotient rule yields $$\partial\bar\partial \log(\alpha(z))=\frac{-1}{(1+z\bar z)^2}.$$ Multiplying by $-(\frac{2}{\alpha(z)})^2=-4(1+|z|^2)^2$ gives $k(z)=4$. The metric $ds$ gives the plane the geometry of a sphere of radius 1/2 (minus a single point), as is suggested in the comments.

Alfors makes extensive use of conformal metrics in his book `Lectures on Quasiconformal mappings,' which is an excellent resource for anyone interested in learning Teichmüller theory.