# Mean curvature upper bounds and area, or geodesic curvature upper bounds and length

Let $M$ be a closed manifold with non-torsion $\pi_2$, and $A$ a non-trivial free homotopy class of a map $f: S^2 \to M$. Let $S$ be the set of (immersed) class $A$ surfaces in $(M,g)$ with mean curvature bounded from above by a fixed constant $C$. Is there a $C=C(A)>0$ sufficiently small, so that the $g$-area function is bounded on $S=S(A,C)$?

A preliminary version of the question would be to ask if this holds for loops in $M$, with mean curvature replaced by geodesic curvature, again assuming $\pi_1(M)$ is non-torsion. But it is surfaces I need.

• What do you mean by "universal constant"? What does it depend on, exactly? Or do you just mean "fixed constant"? – Sebastian Goette Oct 13 '17 at 21:35
• I was trying to emphasize that it is the same constant for all the surfaces. – Yasha Oct 13 '17 at 23:31
• I guess I should use the word uniform or nothing instead? – Yasha Oct 13 '17 at 23:33

I suspect the answer is that there is no bound on the area -- at least not without restrictions on the metric $g$. My reasoning comes from the following paper which shows that on any three-manifold there is an open set of metrics which admit a sequence of embedded minimal $S^2$s with unbounded area.

In particular, if your manifold $M$ is three-dimensional, then you can modify the metric $g$ in a geodesically convex ball so that the above construction holds inside the ball and so find null homotopic minimal spheres with arbitrarily large area (inside the ball). One can then do a connect sum with a representative of your class.

The intuition for the construction in the paper I linked to (which I admit I haven't read carefully) is, I believe, as follows: think about a surface with two pegs" (i.e. two distinct regions that look like large spheres connect summed to the original space by thin necks. One should be able to make long closed geodesics by wrapping back and forth around the pegs.

• Why can you do connect sum? Suppose I have a pair of geodesics in $R^2$ intersecting at a right angle, how do I get an approximate geodesic out of that? Do you mean for a very special metric if things happen just right you can do a connect sum? – Yasha Oct 14 '17 at 0:49
• You could just have used the tubular neighborhood of some curve connecting the spheres to the image of $f$ and maintained a uniform bound on the mean curvature, but now I see you have edited the question so now the constant $C$ needs to be small. I suspect you could still make this construction work with a catenoidal like neck, but it is more subtle. – Rbega Oct 14 '17 at 2:54
• You could probably also look at the construction in the linked paper and see if you could use it to build a counterexample more directly (for instance, if you do the construction in $S^3$ and can show that there are two balls that are always separated by the sequence of spheres, you could modify the topology in each ball of so that the minimal spheres are all in some non-trivial homotopy class (which will necessarily be the same). – Rbega Oct 14 '17 at 2:56

The answer is No. Indeed it is no for spheres of all dimension assuming one can find an immersed class $A$ $S^n$ with mean curvature less than a given $C(A)$. The counterexamples can be generalized from the following example for $n=1$.

Suppose we have a closed surface $(M,g)$, and a free homotopy class $A$ geodesic $\gamma$ in $M$. Deform $g$ to $g'$ keeping the geodesic $\gamma$ fixed, so that there is a half spherical bump near $\gamma$. That is so that we have an isometric embedding of $$\phi: H =\{ (x,y,z) \in S^2 \subset R^2 | z \geq 0 \} \to (M,g').$$ So that $\phi (1,0,0)$ is close to $\gamma (0)$ and $\phi_* (0,1,0)$ is approximately parallel to $\dot {\gamma} (0)$, with respect to the $g'$ Levi-Civita connection. Given this we may "tie" $\gamma$ to multiple covers of the loop $c$, given by the restriction of $\phi$ to the $z=0$ equator. And this will produce our counterexample. It is somewhat more tricky to do this with $C(A)=0$ but I believe it can be done.