# Does a small-area sphere in a 3-manifold bound a small ball?

Let $M$ be a closed riemannian 3-manifold. I think that the following fact should be true and should have a relatively simple proof, but I cannot figure it out.

For every $\varepsilon>0$ there is a $\delta>0$ such that every smooth 2-sphere in $M$ of area smaller than $\delta$ bounds a ball of volume smaller than $\varepsilon$.

Roughly, small-area spheres must bound small-volume balls.

Note that:

• If $M\neq S^3$ then $M$ contains spheres that bound regions of arbitrarily small volume that are not balls (just take a spine of $M$ and small regular neighborhoods of it).
• It suffices to prove that the 2-sphere is contained in a small-volume ball and invoke Alexander theorem.
• The same fact stated for 3-spheres in $S^4$ would imply the (open) Schoenflies problem (every 3-sphere bounds a 4-ball), since every 3-sphere in $S^4$ can be shrinked to have arbitrarily small area.
• It is not true in general that a torus of small area is contained in a ball (pick neighborhoods of a homotopically non-trivial knot).
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How about arguing this way: if the 2-sphere is null homotopic, it bounds a ball (by the Poincare conjecture). So apply an isoperimetric inequality. This reduces your question to showing the infimum of non-null homotopic $S^2$'s in a 3-manifold is not zero. – Ryan Budney Feb 27 '14 at 20:43
AFAIK, the isoperimetric inequality tells you that a null-homotopic small sphere bounds a small region, but that small region might not be a ball: the ball could be on the other side. This confuses me a bit. – Bruno Martelli Feb 27 '14 at 20:46

In their paper "The classical Plateau problem and the topology of 3-manifolds", Meeks and Yau claim that for any fixed closed Riemannian 3-manifold $M$ there is a lower bound for the area of non-trivial two-spheres. Furthermore, the least area such is embedded (or double covers an essential $RP^2$.)

There are many ways to use this to answer the question. Here is one possibility:

We first induct on $s(M)$: the number of essential spheres in a maximal system in $M$. If $s(M) = 0$, then $M$ has universal cover $R^3$ or $S^3$, and we are done, using the proof of Alexander's theorem, say. (This step is not obvious, but let's move along.)

Suppose that $S$ is a minimal area essential sphere in $M$, and so is embedded. Let $T$ be the given sphere with small area, which is necessarily inessential. If $T \cap S$ is not generic, then move $T$ slightly to make it so. We now induct on $|T \cap S|$. If $T \cap S = \emptyset$ then we can cut along $S$ and cap off with a ball to get a manifold $M'$. Note that $s(M') < s(M)$. (We must also check that the area of a smallest essential sphere in $M'$ is greater than the area of $S$.)

If $|T \cap S| \neq \emptyset$, then let $\alpha$ be an innermost curve of intersection. Thus $\alpha$ bounds a disk in $S$ that has area less than either disk it bounds in $T$. (This uses the fact that a surgery of an essential sphere yields at least one essential sphere, and the minimality of $S$.) So we may surger $T$ to get a pair of inessential spheres. (This is because both of them have area less than that of $T$.) This reduces $|T \cap S|$ and we are done.

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thanks! The adaptation of Alexander's proof does not seem totally obvious to me: you should use that R^3 has a metric which is periodic over a group acting co-complactly, otherwise the theorem is clearly false (since R^3 is not compact). – Bruno Martelli Feb 28 '14 at 12:16
now you may try to use the isoperimetric inequality for such periodic metrics. – user126154 Feb 28 '14 at 15:35

This is a direct corollary of Federer -Flemming deformation Lemma saying that vary small area sphere can be homotoped to 1-skeleton of the fixed triangulation of the manifold. The dimension assumption is irrelevant. Proof of this Lemma can be found somewhere in Federer's book (I will chase down the precise reference when I can).

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the Federer-Fleming deformation gives you a cobordism between the given ball and a PL chain, right? Is that enough to conclude? (also in dimension 4?). – user126154 Feb 28 '14 at 14:49