Some problems in Cheeger's "A lower bound for the smallest eigenvalue of the Laplacian"

Question

Pictures below is from Cheeger's "A lower bound for the smallest eigenvalue of the Laplacian"

$f$ is the eigenfunction of smallest eigenvalue of Laplacian on $M$, where $M$ is Riemannian manifold with $\partial M =\varnothing $.

1, I don't know why $V(M_1)\le V(M_2)$ means that $h_1 \ge h$. From the first below picture, the definition of $h_1$ and $h$ are different. I don't know how to get $h_1 \ge h$ ?

2, $M_1$ is a part of $M$, in my view, $\dim M_1 = \dim M$. There should be not any submanifold $A$ of $M$ such that $\partial A = M_1$.

3, What is the critical levels ? Is it the set of critical points ? Why the regions of $M_1$ lying between the critical levels of $f^2$ have a natural product strcture ? I can't see it. In fact, I don't know the image of level surface and orthogonal trajectories.

PS: I feel my problem is too much to it is hard to answer. If so, is there any book contain the Cheeger's paper and not hard to read ?

Answer 1

Let me try to answer your questions in order:

1. Any $S$ in the set over which infimum is taken in the definition of $h_1$ is going to lie in the set over which infimum is taken in the definition of $h$. Let $M^*$ denote what Cheeger denoted as $M_1$ in the definition of $h_1$ (Since $M_1$ is already taken by $f^{-1}([0, \infty))$). If $V(M_1) \leq V(M_2)$,

$\frac{A(S)}{V(M^*)} \geq \frac{A(S)}{V(M_1)}$ (as $M^* \subset M_1$) $\geq \frac{A(S)}{V(M_2)}$ (as $V(M_1) \leq V(M_2)$), so

$\frac{A(S)}{V(M^*)} \geq \frac{A(S)}{min(V(M_i))}$.

2. What Cheeger presumably means is that $M_1$ is a manifold-with-boundary.

3. I would guess critical level is meant to be the preimage of critical value. The fact that between critical levels piece of the manifold has product structure is in fact one of the bases of Morse theory; I would refer to the beginning of Milnor's "Morse Theory" for the proof. Short outline is that you put any Riemmanian metric on $M$, and consider a gradient flow of $f$ with respect to this metric; and the fact that there are no critical points allows you to integrate the system of ODE and get the coordinate system.