# Hyperbolic metric on the torus?

Here is a silly mistake I am making: where exactly is the mistake?

I know that torus cannot hold a metric of constant curvature -1 ( hyperbolic metric ).

But what if I do this:

The upper half-plane and $\mathbb{C}$ are diffeomorphic by a diffeomorphism, say $\phi$; pull the hyperbolic metric from the upper half plane to the complex plane $\mathbb{C}$, and then quotient it by $Z\oplus Z$ to get a hyperbolic metric on the torus. Impossible! But what am I missing?

• If I'm not misunderstanding, the problem is the that $\mathbb{Z}\oplus\mathbb{Z}$ doesn't act by isometries in the pulled back metric. – Jason DeVito Mar 13 '11 at 4:23
• $\mathbb Z^2$ does not act by isometries, so the quotient does not inherit a hyperbolic metric. – Ryan Budney Mar 13 '11 at 4:23
• @Ryan or @Jason, do write that as an answer, which it is :) – Mariano Suárez-Álvarez Mar 13 '11 at 7:21

You have some action of $\mathbb{Z}^2$ on the plane. Conjugate that action by $\phi$ to find an action of $\mathbb{Z}^2$ on the upper-half plane. The question now is: how can $\mathbb{Z}^2$ act on the upper-half plane via isometries?
Let $\alpha$ and $\beta$ be the two generators of $\mathbb{Z}^2$ and suppose that neither acts via the identity. The classification of isometries tells us that $\alpha$ and $\beta$ are either elliptic, parabolic, or hyperbolic. Since they commute with each other an analysis of fixed points shows that they must have the same type. Thus their action on the plane can be reduced to an isometric action on a circle (elliptic case) or on a line. After some $1$-dimensional work we find that the reduced, and hence the original, action is either unfaithful or not discrete.