Proof: Gromov's short basis, volume comparison

1. Sep 21, 2013

Sajet

Hi!

I'm having problems understanding the last step of a derivation for a version of a theorem of Gromov's we had in class:

In short, the proof takes the Riemannian universal covering $(\tilde M, \tilde g, \tilde p)$ of $(M, p)$ and uses a short basis $(\gamma_1, \gamma_2, ...)$ of the deck transformation group. I don't think there is any need to go into much detail about the construction.

Next it is shown that $\alpha_{ij} \geq \pi/3$, where $\alpha_{ij}$ is the angle between the minimal geodesics from the basis point $\tilde p$ to $\gamma_i \tilde p$ and $\gamma_j \tilde p$ respectively.

Now the desired estimate is acquired as follows:

Let $v_i \in T_{\tilde p}^1\tilde M$ be the starting velocity vector of a minimal unit speed geodesic from $\tilde p$ to $\gamma_i\tilde p$.

Because each angle between two such vectors is at least equal to $\pi/3$, we can draw pairwise disjoint open balls $B_{\pi/6}(v_i) \subset T_{\tilde p}^1\tilde M \cong S^n$ around each $v_i$. (Note that these are (n-1)-dimensional neighborhoods within the n-Sphere of unit vectors.)

Now the desired result is acquired by applying Bishop-Gromov's Volume Comparison Theorem:

If $h$ is the number of these unit vectors $v_i$, then:

$h\cdot vol(B_{pi/6}(e_1) \subset S^n) \leq vol(S^n) \Rightarrow h \leq 3^n$.​

I don't understand this last step. It seems to me, the author goes:

$h \leq \frac{vol(B_{pi/2}(e_1) \subset S^n)}{vol(B_{\pi/6}(e_1) \subset S^n)} \leq \frac{(\pi/2)^n}{(\pi/6)^n} = 3^n$​

where the second step follows by applying the volume comparison theorem and comparing to $\mathbb R^n$. But in my opinion $vol(S^n) = vol(B_\pi(e_1)\subset S^n)$ (with a $\pi$ instead of $\pi/2$), which would give an estimate of $6^n$.

Am I making a very elementary mistake or is there an error in the proof?

Last edited: Sep 21, 2013