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Is the following true? I cannot see a counterexample and it seems very intuitively clear, at least in the embedded case.

Claim: Consider the set $S$ of closed immersed Riemann surfaces $\Sigma \subset (X,g)$ (I am particularly interested in spheres), with the magnitude of the mean curvature bounded from above by some $C>0$, where $X$ is also closed. Let $Vol(\Sigma)$ denote the area, and $Diam(\Sigma)$ the diameter. Then for every $\epsilon >0$ there is a $\delta>0$ s.t. if $$Vol (\Sigma) < \delta, \text{ then } Diam (\Sigma)< \epsilon $$ where $\Sigma \in S$.

I really just want to say here that diameter of $\Sigma$ can be assured to be arbitrarily small by requiring that its volume is small.

edit 1: For more clarity let me add that $\epsilon, \delta$ above are assumed to depend on $\Sigma, X, g$.

edit 2: $C$ should be upper bound for magnitude of the mean curvature.

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  • $\begingroup$ You really need to sharpen your question: For example, are you only considering compact surfaces $\Sigma$ without boundary? Are you making any hypotheses on the ambient Riemannian manifold $(X,g)$? Without some hypotheses such as these, it is hopeless to prove any such estimate, since counterexamples are easily constructed. $\endgroup$
    – Robert Bryant
    Commented Dec 3, 2015 at 15:37
  • $\begingroup$ Yes sorry I meant everything to be compact without boundary of course, I edited to say closed. $\endgroup$
    – Yasha
    Commented Dec 3, 2015 at 15:50
  • $\begingroup$ And what about $(X,g)$? If you take $X = \Sigma\times S^1$ and let $g$ be a product metric (for some metric on $\Sigma$), then $\Sigma\times\{1\}\subset X$ is totally geodesic (and embedded to boot), so the mean curvature is identically $0$, and yet there obviously is no relationship between the diameter of $\Sigma$ (and hence it's diameter in $X$) and its volume. $\endgroup$
    – Robert Bryant
    Commented Dec 3, 2015 at 16:10
  • $\begingroup$ Edited again, does this help? $\endgroup$
    – Yasha
    Commented Dec 3, 2015 at 16:21
  • $\begingroup$ Sadly, no. See my comment directly above. $\endgroup$
    – Robert Bryant
    Commented Dec 3, 2015 at 16:24

2 Answers 2

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Assuming I am interpreting your question correctly, this is true when the ambient space is Euclidean and the submanifold is closed (i.e. compact and without boundary). To see this one may invoke a result of Topping that bounds the (intrinsic) diameter of a closed, immersed submanifold of dimension $m$ by the $L^{m-1}$ norm of mean curvature over the immersed submanifold, e.g. for a surface take the $L^1$ norm (this scales correctly).

For a general ambient Riemannian manifold, by isometricly embedding the ambient metric in Euclidean space, one should be able to get a related bound as long as the embedding (of the ambient space) has bounded mean curvature.

See this answer to a related question for the reference.

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  • $\begingroup$ Thanks I will have a look, I have just slightly clarified the question I hope. $\endgroup$
    – Yasha
    Commented Dec 3, 2015 at 16:44
  • $\begingroup$ I think Topping indeed answers the question positively. Thanks! $\endgroup$
    – Yasha
    Commented Dec 3, 2015 at 18:07
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Consider the helicoid, it has vanishing mean curvature. Rescaling it, you can get arbitrary large volume inside the unit ball. That is, diameter can not be bounded in terms of area.

A big piece from the helicoid can be closed by a surface with arbitrary small mean curvature. So you may assume that the surface is closed, moreover you can assume it is a sphere.

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  • $\begingroup$ Can you please remove your answer it has nothing to do with the question. $\endgroup$
    – Yasha
    Commented Oct 11, 2017 at 17:10

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