The question turns out quite long and maybe a bit vague, I apologize in advance for that.

I am currently trying to understand Cheeger and Colding proof of the almost splitting theorem. Currently I am having trouble understanding how they use the so-called segment inequality. Recall that the segment inequality is the following estimate :

Theorem (Segment inequality, Cheeger and Colding)

Let $(M^n,g)$ be a Riemannian Manifold with $Ric\geq -(n-1)g$. Let $B_x$ and $B_y$ be two open sets in $M$. Let $f$ be a nonnegative function on $M$, for almost every pair $(x,y)$ in $M^2$, there is a unique unit speed minimizing geodesic $\gamma$ from $x$ to $y$. Set $\mathcal{F}_f(x,y)=\int_0^Lf\circ\gamma(s)ds$. Set $D=\sup_{x\in B_x,y\in B_y}d(x,y)$. Then:

$$\int_{B_x\times B_y}\mathcal{F}_f(x,y)dx\,dy\leq CD(|B_x|+|B_y|)\int_M f(w)dw $$ where $|A|$ denotes the riemannian volume of a set $A$, $C$ is an explicit constant depending only the dimension and the integrals are computed with respect to the riemannian volume element.


Then fix $\varepsilon>0$ set $B_x=B(x_0,\varepsilon)$ and $B_y=B(y_0,\varepsilon)$. The previous estimate shows that one can find $x^*\in B(x_0,\varepsilon)$ and $y^*\in B(y_0,\varepsilon)$ such that:

$$ \mathcal{F}_f(x^*,y^*)\leq CD\frac{|B(x_0,\varepsilon)|+|B(x_0,\varepsilon)|}{|B(x_0,\varepsilon)||B(y_0,\varepsilon)|}\int_M f(w)dw $$

When proving the almost splitting theorem, one has an upper on $\int_M f$ for some geometrically relevant function $f$ which can be made has small as wanted using some geometric freedom we have in the context, and one needs for given points $x$ and $y$ to find $x^*\in B(x_0,\varepsilon)$ and $y^*\in B(y_0,\varepsilon)$ such that $ \mathcal{F}_f(x^*,y^*) $ is as small as wanted, in a way depending only on the geometric parameters at hand.

The problem I have is that the upper bound I have doesn't seem to do the job as the lower bound on the Ricci curvature will not give me a universal upper bound on $\frac{|B(x_0,\varepsilon)|+|B(x_0,\varepsilon)|}{|B(x_0,\varepsilon)||B(y_0,\varepsilon)|}$. I would like to understand why one can bound this.

It would be easy to remedy this by adding an a priori lower bound on the volume of balls (non collapsing assumption), but it is not part of the hypothesis.


My question here is really "what's going on here ? did I miss something obvious ?", but to phrase it a bit more formally :

How can one bound $\mathcal{F}_f(x^*,y^*)$ without a non collapsing assumption ?

References :

These interrogations came from reading Degeneration of Riemannian Metrics under Ricci curvature bounds by J. Cheeger, specifically page 50 of these notes. I looked at the Cheeger Colding paper on the almost Rigidity of warped products but couldn't find an answer.


1 Answer 1


I think that the point is that you need to use the Bishop-Gromov inequality (assuming $\mathrm{ricci}_g\ge -(n-1)\kappa^2g$ : $$\frac{\mathrm{vol}_g B(x,R)}{\mathrm{vol}_g B(x,\epsilon)}\le \frac{v_n(\kappa, R)}{v_n(\kappa,\epsilon)}$$ Hence you will get some points $x^*\in B(x_0,\epsilon), y^*\in B(y_0,\epsilon)$ with $$\mathcal{F}_f(x^*,y^*)\le C(n,\kappa,R,\epsilon)\frac{1}{\mathrm{vol} B(o,10000R)} \int_{B(o,10000R)}f $$

You will find some hints in Gallot's Bourbaki seminar (see the remarque 2.8 :http://archive.numdam.org/ARCHIVE/SB/SB_1997-1998__40_/SB_1997-1998__40__7_0/SB_1997-1998__40__7_0.pdf )

  • $\begingroup$ I actually had come to the same conclusion, and forgot I asked the question. Thanks anyway. $\endgroup$ Commented Mar 30, 2016 at 8:25
  • $\begingroup$ Anyway, I asked another question which came out while reading the same set of notes : here. If you have any ideas to share, feel free to do it ! $\endgroup$ Commented Mar 30, 2016 at 8:29

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