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Let $\mathbb{H}_\mathbb{C}^n$ be n-dimensional complex hyperbolic space. This space is a complex analog of hyperbolic space. It is isometric to the quotient of hyperboloid $$|z_0|^2-|z_1|^2-\dots-|z_n|^2=1$$ in $\mathbb{C}^{n+1}$ by $S^1$.

Question 1. Is it known that round balls in $\mathbb{H}_\mathbb{C}^n$ minimize the surface area among all bodies of given volume?

(I am almost sure that the answer is not known.)

Question 2. Was it conjectured somewhere?

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I am pretty sure it is a somewhat reputed conjecture, but I do not have a clear reference where it is stated.

Edit: It is stated in Gromov's "Metric structures for Riemannian and non-Riemannian spaces", 6.28 (1/2+) (in fact, it is stated for all rank one symmetric spaces).

It might be evoked in a paper of Hsiang and Hsiang in Inventiones, where they prove that the isoperimetric domains in products of hyperbolic and euclidean spaces are invariant under the group of all isometries fixing the center of gravity. It seems a reasonable conjecture that this is true in all symmetric spaces of non-positive curvature. That conjecture might be stated in the Hsiang and Hsiang paper, and is a broad generalization of the conjecture you are interested in.

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    $\begingroup$ I heard it from Marcel Berger in 1994 or 1995 when he was lecturing in Stony Brook. I'm pretty sure I saw the problem (also for rank one symmetric spaces) in one of his books. $\endgroup$
    – alvarezpaiva
    Commented Jul 21, 2013 at 18:43
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My answer might be ten years late, but a great advance toward a proof of this conjecture, considering a lower bound on the Hermitian mean curvature, has been made (at least when the boundary is sufficiently smooth) by Xiaodong Wang in An integral formula in Kahler geometry with applications, Comm. Contemporary Math. 19 (2017), no. 5, 1650063. See Theorem D therein.

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  • $\begingroup$ While your reference is very interesting I don't think it shows the isoperimetric inequality rather it shows that if the least area set containing a given volume had sufficiently large Hermitian mean curvature, then it would have to be a ball -- it can't rule out exotic examples that are more efficient for area but have smaller Hermitian mean curvature. Theorem E shows the result if the minimizer is Hopf. $\endgroup$
    – RBega2
    Commented Oct 25, 2022 at 17:26
  • $\begingroup$ @RBega2 You are aright. I edited my answer. $\endgroup$
    – Didier
    Commented Oct 25, 2022 at 19:32

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