Why does not a closed 3-manifold modelled on SL(2,R) admit a metric of nonpositive curvature?

If you read our paper a bit further, you will find that on page 348 we mention that this result is due to Eberlein and give a reference to his 1982 paper. More precisely, he proves a more general theorem that a nonpositively curved compact Riemannian manifold whose fundamental group has nontrivial center has a finite-sheeted covering space which splits smoothly as a product of a torus with another factor. For Seifert manifolds, this is equivalent to the types $E^3$ and $H^2\times R$.

I am not sure that this is what the OP is looking for, but here is a justification of the fact that the unit tangent bundle of a hyperbolic surface cannot be endowed with a nonpositively curved metric.

Let $\Sigma$ be a closed surface of genus $\geq 2$. The unit tangent bundle $\mathrm{UT}(\Sigma)$ naturally projects onto $\Sigma$, each fiber being a circle $\mathbb{S}^1$: $$\mathbb{S}^1 \to \mathrm{UT}(\Sigma) \to \Sigma \hspace{1cm} (1)$$ This observation yields a short exact sequence $$1 \to \mathbb{Z} \to \pi_1(\mathrm{UT}(\Sigma)) \to \pi_1(\Sigma) \to 1.$$ In particular, the fundamental group of the $3$-manifold $\mathrm{UT}(\Sigma)$ contains a normal infinite cyclic subgroup $\langle a \rangle$. Assume for a moment that $\mathrm{UT}(\Sigma)$ can be endowed with a nonpositively curved metric. So $\pi_1(\mathrm{UT}(\Sigma))$ is a CAT(0) group, and it must contain a finite-index subgroup admitting $\langle a \rangle$ as a direct factor, say $\langle a \rangle \times K$. Let $\Xi \to \mathrm{UT}(\Sigma)$ denote the corresponding finite cover. Then, by construction, $\Xi$ is homeomorphic to some product $\mathbb{S}^1 \times M$, whose decomposition is compatible with (1). In other words, the fiber bundle (1) is trivial up to a finite cover, which is clearly false.

The arguments are essentially extracted from the proof of Theorem II.7.27 from Bridson and Haefliger's book Metric spaces of non-positive curvature.