Let $(M, g)$ be a (complete) Kähler manifold with Ricci curvature $\geq c$.

Is it true that the volume ratio of geodesic balls in $M$ with respect to balls in the corresponding (simply connected) complex space form with Ricci $\equiv c$ is a decreasing function?

I suspect the answer is no. Examples would be nice, also showing why the Riemannian proof breaks would help.

EDIT: A friend made me notice Gang Liu's paper (https://arxiv.org/pdf/1108.4231v1.pdf), where it is shown that, for real analytic metrics, the volume ratio is decreasing for small values of the radius.


1 Answer 1


It looks like the answer is indeed no with the quadric $\mathbb CP^1\times \mathbb CP^1$ a counterexample. Here the corresponding complex model space is $\mathbb CP^2$.

Recall that to get an Einstein metric with coefficient $\lambda=1$ we should choose it is as curvature of the anti-canonical bundle $-K$. Now, for $\mathbb CP^1\times \mathbb CP^1$ we have $-K\cong O(2)\times O(2)$. The diagonal $\mathbb CP^1\subset \mathbb CP^1\times \mathbb CP^1$ is geodesic and $-K$ restricts as $O(4)$ to it. It follows that the diameter of $\mathbb CP^1\times \mathbb CP^1$ is larger than that of $\mathbb CP^2$ (where $-K\cong O(3)$). Inddeed the diameter of the former space is equal to the diameter of its diagonal and the diameter of the latter space is equal to the diameter of any line in it.

As for where the usual proof breaks down, I don't know.

  • $\begingroup$ Right, I did not think about the diameter. Thanks. $\endgroup$ Aug 21, 2018 at 18:28
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    $\begingroup$ So the volume ratio starts equal to 1 at r=0, who knows what it does until r= diameter of the projective plane, then increases to 8/9 at r=diameter of the product. There is a recent result of Fujita (arxiv.org/abs/1508.04578) which asserts that the volume of any Kahler-Einstein Fano is less or equal than that of the Fubini-Study. The proof is purely algebraic, I guess there is no differential geometric interpretation then. $\endgroup$ Aug 21, 2018 at 18:40

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