Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(M,g)$ be a compact riemannian manifold with sectional curvature $|K_g| \leq 1$. A lemma due to Klingenberg asserts that then either the injectivity radius $i_g \geq \pi$ or $(M,g)$ contains a geodesic loop $\gamma$ of length $L(\gamma) = 2i_g$.

Let $n = 2$. Suppose for a moment that $i_g < \pi$ and $M$ is orientable. Note that $\gamma$ is automatically simple. (I am note sure how often this occurs. An example, however, is a thin long cylinder with two balloon-like 2-spheres attached.) Then $\gamma$ cuts $M$ into two pieces.

Intuition says that these two pieces should be rather large, as the manifold cannot curve together very quickly due to the curvature condition. By another Klingenberg lemma, known as the "Long Homotopy Lemma", any null-homotopy $\gamma_t$ of $\gamma_0 = \gamma$ must sweep through some curve $\gamma_{t_0}$ of length $L(\gamma_{t_0}) \geq \pi$. (This lemma holds for all $n$ and non-orientable manifolds. It is Exercise 1 in Chapter 10 of do Carmo's text book.)

To some extend, this is the kind of theorem I am looking for. It asserts that away from $\gamma$, $(M,g)$ must be slightly larger (on the scale of the curvature bound) than near $\gamma$. However, I'm looking for some result describes the geometry of $M$ away from $\gamma$ in a "more global" fashion. For instance: Is there any estimate of the diameter or the volume of the pieces that $M$ is cut into?

In view of the examples given by horse with no name, one should assume that $M$ has $g(M) \neq 0$ (or that $M$ has some other non-vashing characteristic class in the case that $n > 2$).

As pointed out in the comments and Anton Petrunin's answer, $M$ should either be a sphere or $\gamma$ must also be assumed to cut $M$ in half.

share|improve this question
2  
If your on an oriented surface then doesn't Gauss-Bonnet together with your curvature bound tell you that $Area(\Omega_0)\geq 2\pi$? Here $\Omega_0$ is a component of $M\backslash \gamma$. I guess you could do better if the genus of $\Omega_0$ was known to be bigger than 1. –  Rbega Apr 26 '13 at 12:24
    
Thank you, Rbega. This, of course, doesn't take the length of $\gamma$ into account. (One should expect the volume of $\Omega_i$ to increase as $i_g \rightarrow 0$. –  Malte Apr 26 '13 at 12:45
    
I guess you could rescale the metric so you should be able to prove that $Area(\Omega_0)\geq 2\pi i_g^{-2}$. Perhaps I'm missing something here. –  Rbega Apr 26 '13 at 13:29
    
For negative curvature, there is the thin-thick decomposition by Margulis' lemma... –  Asaf Apr 26 '13 at 15:26
3  
Why should $\gamma$ disconnect your surface? think of a torus, or a hyperbolic surface with small non-separating systole. –  Benoît Kloeckner Apr 26 '13 at 17:24
show 1 more comment

2 Answers

Cheeger's examples:

Arbitrarily small volume (but bounded diameter and $i_g\rightarrow 0$ ): Flat torus $S^1(1)\times S^1(\frac{1}{n})$ where the first circle has diameter 1 and the second diameter $1/n$.

Arbitrarily large diameter (bounded volume and $i_g\rightarrow 0$): Flat torus $S^1(n)\times S^1(\frac{1}{n})$ where the first circle has diamter $n$ and the second diamter $1/n$.

share|improve this answer
add comment

It seems that you want $M$ to be a sphere (although you do not mention it).

The natural example is a tube with curvature $-1$ closed by two spherical cups with curvature $1$. (This is a surface of revolution.) This example should have the minimal area and minimal diameter among the spheres with given length of closed geodesic. But at the moment I do not see a proof.

Some rough estimates will follow from the Klingenberg's result. Take the neighborhood $U$ of your closed geodesic of size near $e^{1/\ell}$ where $\ell$ is the length of the geodesic. Note that $U$ has to be a cylinder otherwise you end in singular point. It implies that diameter has to have order $e^{1/\ell}$.

Rough bounds for the area follow Rbega comment. They are not far from the area of the surface of revolution, somewhere in the range between $4{\cdot}\pi$ to $(4+2{\cdot}\sqrt{2}){\cdot}\pi$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.