# Do manifolds with no Ricci lower bounds for any metric exist?

Is there a smooth (noncompact) manifold $M$ such that for any Riemannian metric $g$ on $M$ there are $p_i$ and unit tangent vectors $v_i \in T_{p_i}M$ such that $Ricc(g)|_{p_i}(v_i,v_i) \leq -i$? This seems unlikely, but I'm not sure how to prove it.

Alternatively, what is the simplest example of a fixed $(M,g)$ with no lower Ricci bounds in the sense above? It seems that some conformal change of the standard Euclidean metric could accomplish this, but I dont see a simple way to do this.

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## 1 Answer

Any smooth manifold $M$ admits a metric with quadratic curvature decay. In fact there is a complete Riemannian metric $g$ on $M$ such that $$|K_x|=o(|x_0x|^{-2}),$$ here $|K_x|$ is absolute bound for sectional curvature of $g$ at point $x$, $x_0$ is a fixed point and and $|x_0x|$ denotes distance induced by $g$.

In particular there is no examples which you are looking for.

See p. 96 in Gromov M., Volume and bounded cohomology, Publ. Math. IHES 56 (1982) 5–99.

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Perfect! Thank you! –  Otis Chodosh Apr 5 '11 at 16:12