# Can scalar curvature and diameter control volume?

Scalar curvature can control the volume of geodesic ball locally, however, it can not bound the diameter. As far as I know, the example for a manifold with a large scalar curvature and volume has large diameter.

Comparing with the n sphere with the standard metric, if a smooth manifold has scalar curvature larger than this sphere, and the diameter of the manifold is smaller, should the volume of the manifold be smaller than the sphere?

• As a follow up, to build a counter-example I ended up needing there to be some negative Ricci curvature. I experimented a little bit and couldn't figure out how to build any counter-examples with non-negative Ricci curvature. I wonder if it might be possible to get some volume control with non-negative Ricci, positive scalar and bounded diameter that's better than what the volume comparison theorem states. – Gabe K Nov 6 '18 at 4:48
• Thanks. For scalar and Ricci curvature, H. Bray proved a theorem in dimension 3, assuming the scalar is larger than n(n-1), Ricci $\ge\epsilon$, he gave a sharp upper bound of the volume. For Ricci and diameter, I only find a paper in 1993: Ricci curvature, diameter and optimal volume bound. Let $M_k=\{M|Ric(M)\ge k(n-1), diam(M)=\pi\}$. $F_k(D)$ is the volume of D-ball in the simple connected space form with the constant sectional curvature k. It was still open that whether $F_k(\pi)$ is the optimal volume bound of $M_k$ in this paper. – Yiyue Zhang Nov 7 '18 at 0:41
• Thanks for the references and for the interesting question. After thinking about it some more, my thoughts got too long for a comment so I posed a follow-up question. mathoverflow.net/questions/314701/… – Gabe K Nov 7 '18 at 14:52

Take a sphere of radius $$\tfrac1{10}$$ and attacch to it 1000000 spheres of the same radius by very thin necks. You get this way a space of small diameter, large scalar curvature and huge volume.
For an example of why scalar curvature bounds and diameter bounds are not enough to compare the volume to a sphere, consider the manifold $$M_1$$ which is the metric product of a ball of radius r in hyperbolic space $$\mathbb{H}^2$$ (with sectional curvature 1) and the sphere $$\mathbb{S}^2(r)$$. For the second manifold, consider the $$4$$-sphere $$\mathbb{S}^4(s)$$.
The volume of $$M_1$$ is $$\sinh(t) \times 4 \pi r^2$$ while its diameter is $$\sqrt{\pi^2 r^2+ 4t^2} < \pi r +2 t$$. The scalar curvature is $$2(\frac{1}{r^2}-1)$$. Whenever $$r<1$$, the scalar curvature of $$M_1$$ is positive. Meanwhile, the volume of the $$4$$-sphere is $$\frac{8}{3}\pi^2 s^4$$, its diameter is $$\pi s$$, and its scalar curvature is $$\frac{12}{s^2}$$.
To force our first manifold to have much larger volume then the second, we set $$s$$ very large so that the scalar curvature is less than $$1$$. We also set $$r=1/2$$ and $$t=s-1/2$$. In this case, the scalar curvature of $$M_1$$ is 6 and the diameter of $$M_1$$ is strictly smaller than that of $$\mathbb{S}^4(s)$$. However, since hyperbolic sine grows exponentially in $$t$$ whereas the volume of the $$4$$-sphere grows as $$s^4$$, for sufficiently large $$s$$ the volume of $$M_1$$ is larger than that of the sphere. By setting $$s$$ arbitrarily large, we can force $$M_1$$ to have much larger volume than $$\mathbb{S}^4$$.