# Negative Curvature on Torus

We were looking for an example of a compact non-simply connected riemannian manifold with non positive sectional curvature. We came up with the following idea, which is wrong by Gauss-Bonnet, but we don't know what is wrong. Let $$H^+$$ denote the hyperbolic plane, and consider the metric $$g=\frac{1}{y^2}(dx^2+dy^2)$$ on $$H^+.$$ Consider the diffeomorphism $$\phi:\mathbb{R}^2\to H^+,(x,y) \mapsto (x,\exp(y)),$$ and let $$\tilde{g}$$ denote the metric $$\phi^* g$$ on $$\mathbb{R}^2.$$ Then $$\phi$$ becomes an isometry between $$(\mathbb{R}^2,\tilde{g})$$ and $$(H^+,g).$$ One can induce a metric $$g^\prime$$ on $$\mathbb{T}^2$$ by means of the standard covering map $$\pi:\mathbb{R}^2\to \mathbb{T}^2,$$ in a way such that $$\pi$$ becomes a local isometry. This will imply that $$(\mathbb{T}^2,g^\prime)$$ is a Riemannian manifold with negative curvature, which is not possible. What is wrong with this reasoning?

One can induce a metric $$g'$$ on $$\mathbb T^2$$ by means of the standard covering map... in a way such that $$\pi$$ becomes a local isometry
This is only possible if the metric $$g'$$ is invariant under the action of the deck transformation group of $$\pi$$, which is an action of the group $$\mathbb Z^2$$ on $$\mathbb R^2$$. That group action is generated by $$T_1(x,y) = (x+1,y)$$ and $$T_2(x,y) = (x,y+1)$$. A bit of calculation or other geometric considerations will show that although $$g'$$ is indeed invariant under $$T_1$$, it is not invariant under $$T_2$$. For example, the lengths of the segment $$[0,1] \times \{0\}$$ is equal to $$1$$ whereas the length of the segment $$[0,1] \times \{1\}$$ is equal to $$e \ne 1$$, but the second segment is the image of the first under $$T_2$$.