Explicit complete metric on $\mathbb{C}\backslash \{ 0,1\} $ with Gaussian curvature $K \leqslant - 1$. I wonder how to give a complete metric on $\mathbb{C}\backslash \{ 0,1\} $ with Gaussian curvature $K \leqslant  - 1$ explicitly. 
I tried to modify the Poincaré disk model since it has constant curvature $-1$ under the hyperbolic metric. However, I still could not see how to make it right. 
Is it possible to use cutoff functions to give an explicit expression?
Any help will be appreciated.
 A: You can get an explicit formula with $K=-1$ on the nose, if you wish. Here's an outline.
There exists a holomorphic universal covering map $f : \mathbb{H} \to \mathbb{C} \setminus \{0,1\}$ which is invariant under a certain index 6 subgroup $\Gamma < \text{SL}_2(\mathbb{R})$ acting by isometries of the hyperbolic metric $\frac{\sqrt{dx^2+dy^2}}{y}$ on the upper half plane $\mathbb{H}$. Pushing forward the hyperbolic metric under $f$ will give you the Riemannian metric you desire.
The index 6 subgroup is generated by the two parabolic transformation 
$$z \mapsto z+2
$$ 
$$z \mapsto \frac{1}{\frac{1}{z} + 2} = \frac{z}{z+2}
$$
It has a fundamental domain which is the ideal quadrilateral in $\mathbb{H}$ with ideal vertices $-1$, $0$, $+1$, $\infty$. That quadrilateral is a union of 6 translates of the standard fundamental domain for $\text{SL}(2,\mathbb{Z})$, namely the triangle with vertices $i$, $\frac{1}{2} + \frac{\sqrt{3}}{2} i$, $\infty$ (which is one way to see that the index equals $6$).
The universal covering map $f$, when restricted to the ideal triangle with infinite vertices $0,1,\infty$, maps the interior of that triangle to the upper half plane fixing the three points $0,1,\infty$. I think that information is enough to let one construct an explicit formula for $f(z)$. It'll have some complex logarithms in it, in order to convert the angle $0$ corners into angle $\pi$ corners at each of $0,1,\infty$. There might be a complex analyst around here who knows the explicit formula for $f(z)$, or who knows a reference for it; unfortunately I don't know either of those.
