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

I'm currently learning some stuffs about systolic inequalities. While reading the relevant sections (p329 to 340) in Berger's Panoramic View of Riemannian Geometry, I noticed a gap in one of the proofs (starting at page 331). The goal is to prove :

For any non simply connected compact surface $(M,g)$, $Area(M,g)\geq Sys(M,g)^2/2$.

Berger considers a periodic geodesic $c$ realizing the systole $L=Sys(M,g)$, pick a point $m$ on $c$ and claims that $Vol B(m,L/2)\geq L^2/2$, which is enough to show the result.

However in the proof of the claim, he invokes the following fact : for $r<L/2$, $B(m,r)$ is a topological disk. He says that this comes from the "very definition of the systole". But this claim is false, it is enough to consider a torus with a long thin finger glued at some point to see it (fig 7.26 on the page where the claim is made shows exactly this). In the paper "Systolic and inter-systolic inequalities", Gromov, facing the same kind of situation, just says "chop the fingers", while this is intuitively convincing I don't see a way to make it rigorous.

My question is the following : is this fixable ?

Thinking a little bit about it, it seems enough for the rest of the argument to show that all but one of the connected component of $M\backslash B(m,r)$ are disks, and that the systole $c$ doesn't meet the components which are disks. But I am not able to prove this at the moment.

I'm aware that another proof is available, through estimating the homological systole. But I like the proof on the next page of Berger that the systolic ration grows at least like the square root of the genus, which uses the same argument.

I should also say that I don't have access the article of Hebda to which Berger refers.

share|improve this question
    
The constant 2 in your claimed systolic inequality seems very small. If i'm not totally delusional, then the argument Berger presents would at least establish (via a Minkowski convex body type argument) the systolic inequality with constant $2^2=4$. Rigorously fingers' are detectable as points of large sectional curvature where the area of an embedded disk is less than we expect (like with long fingers) and chopping fingers is forgivable if we are just looking to establish a systolic inequality of the type $sys_1^2 \leq const. area$. I'm confused about what is this' in your question. –  J. Martel Apr 9 '13 at 15:06
    
You can fill in the chopped-off fingers by disks, because generically each of the additional boundary components is contractible, hence bounds a disk. –  katz Apr 9 '13 at 15:06
    
I mean, as you say, the Berger's area estimate is wrong unless one chops fingers. So i'm uncertain what you would like to try and fix. –  J. Martel Apr 9 '13 at 15:07
    
I'm sorry, in my first comment i was referring to the 2-torus. –  J. Martel Apr 9 '13 at 15:08
1  
The constant "2" in Berger's correct inequality can be improved to "4/3" for all surfaces other than the sphere and the projective plane. This was done by Gromov in 1983. –  katz Apr 9 '13 at 15:15

1 Answer 1

up vote 3 down vote accepted

You don't really need to reattach chopped off fingers to get the estimate. The point is that for almost all (in the sense of Sard) $r$, the points $\gamma(-r)$ and $\gamma(r)$ are in the same connected component of the $r$-level curve of the distance function. Applying the coarea formula and the triangle inequality then gives the estimate.

share|improve this answer
    
Thanks, but I'm not sure I understand. How do you show that $\gamma(-r)$ and $\gamma(r)$ are in the same connected component of $S_r=\{d(m,p)=r\}$ (call it $S_r'$) ? Also, the next step in Berger is to notice that each of the components of $S_r'$ linking $\gamma(-r)$ to $\gamma(r)$ has length greater than $2r$. But to do this, you need to show that the curve $\gamma$ with $\gamma([-r,r])$ replaced by one of the two component of $S_r'$ is not contractible, I am not sure how to do this without knowing in advance that $S_r'$ bounds a disk containing $m$. –  Thomas Richard Apr 9 '13 at 11:24
    
If $\gamma(\pm r)$ were in separate components $S_{\pm}$, then choosing say $S_+$ note that $S_+$ has intersection $\pm 1$ with $\gamma$ and is therefore nontrivial, contradicting the assumption that $2r$ is less than the systole. –  katz Apr 9 '13 at 11:55
    
Sorry I'm a bit slow... The fact that $S_+$ is not contractible implies that it has length greater than the systole, but I don't see how to derive a contradiction from this. –  Thomas Richard Apr 9 '13 at 14:12
    
Since $S_+$ is in a ball of radius smaller than half the systole, it has to be contractible. To see this, partition $S_+$ into infinitesimal line segments, and then decompose it into a product of many loops based at the central point $\gamma(0)$ by joining each partition point to $\gamma(0)$. –  katz Apr 9 '13 at 14:38
    
I like this one, thanks a lot ! –  Thomas Richard Apr 9 '13 at 16:01

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.