# 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 $K(G,1)$?

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

2. $M$ has negative sectional curvature.

3. $M$ has nonpositive sectional curvature.

4. $M$ 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).

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

6. $M$ has negative scalar curvature.

7. $M$ has nonpositive scalar curvature.

8. $M$ 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?

• I'm not sure I've chosen the right tags -- if anybody knows of some better ones to use, please edit them! – Tim Campion Jul 17 '18 at 13:21
• The interesting question is #4, whose answer I feel like I'm supposed to know, but don't off the top of my head. #3 is true, since it implies the exponential map on the universal cover is a diffeomorphism. #5 is false: every manifold has a metric of negative scalar curvature. – John Pardon Jul 17 '18 at 13:40
• 3 (and hence 1 or 2) implies what you want by the Cartan-Hadamard theorem which in fact is true for any CAT(0) space, see en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem. – Igor Belegradek Jul 17 '18 at 13:41
• Regarding 4: The only nontrivial case is when $f$ takes a positive value (else it is a special case of 3), say $f(a)>0$. Then consider the Riemannian product $\epsilon S^2\times g(\epsilon )S^1$ where $S^2$ is the unit sphere and $g$ is a function we choose so that the product has volume $a$. By picking $\epsilon$ large we can arrange the product to have curvature $< f(a)$. Thus 4 does not imply asphericity. – Igor Belegradek Jul 17 '18 at 14:04
• To improve the question consider the following results are in projecteuclid.org/euclid.jdg/1214446030 [K.Fukaya and T.Yamaguchi, J. Differential Geom. 33, Number 1 (1991), 67-90, Almost nonpositively curved manifolds] and numdam.org/item?id=CM_1987__63_2_223_0 [Bavard, C. Courbure presque négative en dimension 3 Compositio Mathematica, Volume 63 (1987) no. 2 , p. 223-236]. – Igor Belegradek Jul 17 '18 at 14:51

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).

• Can the metric of negative curvature be chosen to be complete? – Tim Campion Jul 17 '18 at 19:03
• @TimCampion Actually, the results I mention (for scalar curvature) are for compact manifolds. There may be versions for complete manifolds too, but I am not certain. – Igor Rivin Jul 17 '18 at 19:05
• Lohkamp proved even more. Every manifold has a metric with negative Ricci curvature. Gao and Yau first proved this for 3-manifolds. This is perhaps the most amazing case, since Ricci uniquely determines the full curvature tensor. However, negative Ricci is not equivalent to negative sectional curvature. – Deane Yang Jul 17 '18 at 20:16
• Lohkamp's result is true for manifolds of dimension at least three; it doesn't hold in dimension two thanks to Gauss-Bonnet. – Michael Albanese Jul 18 '18 at 10:01

Let me summarize the information in the comments in a CW post. Feel free to edit.

• 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.

• 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.

• 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.
• Some remarks: a) Closed manifolds of constant negative Ricci curvature need not be aspherical, see e.g. theorem 3 in ncbi.nlm.nih.gov/pmc/articles/PMC431004/?page=1 which is Yau's announcement of his solution of Calabi's conjecture [Calabi's conjecture and some new results in algebraic geometry]. b) The result I linked in comments is due to Fukaya and Yamagichi (there is no Y. Fukaya). Asking whether their result holds for noncompact complete manifolds does not quite make sense because the diameter is infinite and their strategy is to control the diameter. – Igor Belegradek Jul 18 '18 at 18:23