What is the weakest negative curvature condition ensuring a manifold is a $K(G,1)$? The only statement I'm sure of is that any hyperbolic or Euclidean manifold is a $K(G,1)$ (i.e. its higher homotopy groups vanish), since its universal cover must be $\mathbb H^n$ or $\mathbb E^n$. But for example, if a complete Riemannian manifold $M$ satisfies one of the following, can I conclude that $M$ is a $K(G,1)$?


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*$M$ has sectional curvature bounded above by some negative number.

*$M$ has negative sectional curvature.

*$M$ has nonpositive sectional curvature.

*$M$ has sectional curvature bounded above by $f(\operatorname{vol}(M))$ (where $f: \mathbb R \to \mathbb R$ is some function depending only on the dimension of $M$ that I don't know).

*$M$ has scalar curvature bounded above by some negative number.

*$M$ has negative scalar curvature.

*$M$ has nonpositive scalar curvature.

*$M$ has scalar curvature bounded above by $f(\operatorname{vol}(M))$.
Do the answers change if I assume that $M$ is compact? Have I left out a relevant condition of some kind?
 A: Let me summarize the information in the comments in a CW post. Feel free to edit.


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*For "weaker" notions of curvature, negative curvature seems to not imply that a manifold is a $K(G,1)$. As Deane Yang pointed out, Lohkamp showed that for each $d \geq 3$, there are numbers $a(d) > b(d) > 0$ such that every manifold $M$ of dimension $d$ admits a complete metric $g$ with $-a(d) < \operatorname{Ric}(M,g) < -b(d)$. I believe this implies an analogous result for scalar curvature.


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*The only loophole I can see is that there might be a smaller interval $a(d) \geq a' \geq b' \geq b(d) > 0$ such that if $M$ admits a metric $g$ with $-a'\leq \mathrm{Ric}(M,g) \leq -b'$, then $M$ is a $K(G,1)$. (Possibly $a',b'$ might depend on further parameters such as $\operatorname{diam}(M,g)$ or $\operatorname{vol}(M,g)$).
For instance, if $M$ admits a metric of constant negative Ricci curvature, does this imply that $M$ is a $K(G,1)$? Igor Belegradek points out below that the answer is no in this case as shown by Yau.


*For sectional curvature, the story is different. As several people pointed out, the Cartan-Hadamard theorem says that any manifold admitting a complete metric of nonpositive sectional curvature is a $K(G,1)$.
We may ask if this can be improved to allow a small amount of positive curvature. As Igor Belegradek pointed out, "small amount" can't be specified in terms of volume, since $R S^2 \times g(R) S^1$ has constant volume $a$ for appropriate $g(R)$, but by choosing $R$ sufficiently large, it has arbitrarily small positive curvature. But as Igor Belegradek also pointed out, Fukaya and Yamaguchi showed that there is a positive number $\epsilon(d,D)$ dependent only on the dimension $d$ and diameter $D$, such that any compact Riemannian manifold $M$ with $-1 \leq \operatorname{sec}(M) < \epsilon(\operatorname{dim}(M), \operatorname{diam}(M))$ is a $K(G,1)$. The lower bound on the curvature is necessary; Fukaya says that Gromov constructed metrics on $S^3$ with fixed diameter and arbitrarily small sectional curvature.


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*I don't know if Yamaguchi - Fukaya's result holds for complete Riemannian manifolds (as Igor Belegradek points out the question doesn't even make sense in this case).

*Another direction which might be interesting would be to control "small amounts of positive curvature" in some other way. For instance, rather than controlling the $L^\infty$ norm of the sectional curvature, one might ask for control over some averaged version of it -- this might allow the curvature to become very positive at a point so long as it's not very positive in a large region. Somehow the necessity of the lower curvature bound in Fukaya's result suggests to me that something like this might be a good idea.


A: For 1-3, yes, by the Cartan-Hadamard Theorem.5-... No. For example, every 3-manifold admits a metric of negative scalar curvature (I think this is actually true for any manifold, due to Lohkamp).
