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?

  • $\begingroup$ 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. $\endgroup$ – Gabe K Nov 6 '18 at 4:48
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    $\begingroup$ 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. $\endgroup$ – Yiyue Zhang Nov 7 '18 at 0:41
  • $\begingroup$ 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/… $\endgroup$ – Gabe K Nov 7 '18 at 14:52

A connected sum of two spheres can be done with arbitrary thin neck while keeping the lower bound on scalar curvature.

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.


In general, the scalar curvature does not give enough control to get these sorts of volume estimates. However, if you use the Ricci curvature instead, you can get these kinds of estimates using the volume comparison theorem or something similar.

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 $\mathbb{S}^4$.


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