# Show that there is no metric on $\mathbb{C}^{\ast}$ with $K\leq -1$ everywhere.

I try to solve this problem:

Show that there is no metric on $\mathbb{C}^{\ast}$ with $K\leq -1$ everywhere.

Here, the metric on a domain $U$ is a $C^2$ function $\rho: U\rightarrow \mathbb{R}^{\geq 0}$ which vanishes only at isolated points.

And a curvature of $\rho$ is defined as follows $$K=-\frac{1}{\rho^2}\Delta\log \rho$$ where $\Delta$ is the Laplace operator.

My try:

I want to apply Ahlfor's lemma and lead some contradiction. According to Ahlfor's lemma, let $\rho$ be a metric in a domain $U$ with $K_{\rho}\leq -1$ when defined. Then $$\rho\leq \lambda_{\mathbb{D}},$$ where $\lambda_{\mathbb{D}}$ is the hyperbolic metric on $\mathbb{D}$ is given by $$\lambda_{\mathbb{D}}=\frac{2}{1-|z|^2}.$$ Then I got stuck... I don't know if I could get some contradiction in this way since $\rho\leq \lambda_{\mathbb{D}}$ is not like restriction that all metrics on $\mathbb{C}^{\ast}$ cannot satisfy.

• To invoke Ahlfors' Lemma, you need a mapping $f\colon \Bbb D\to \Bbb C^*$. Otherwise your inequality "$\rho\le\lambda_{\Bbb D}$" doesn't make any sense. – Ted Shifrin Dec 7 '17 at 18:11
• The domain $\mathbb{C}^*$ can be smoothly embedded in the unit disc, which admits a hyperbolic metric. Hence, so does $\mathbb{C}^*$. – Amitai Yuval Dec 8 '17 at 9:21