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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?

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You write:

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 $T_2$.

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