Proof: Gromov's short basis, volume comparison

[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:

Let [itex](M^n, g)[/itex] be complete with nonnegative sectional curvature. Then [itex]\pi_1(M, p)[/itex] can be generated by at most [itex]3^n[/itex] elements.

In short, the proof takes the Riemannian universal covering [itex](\tilde M, \tilde g, \tilde p)[/itex] of [itex](M, p)[/itex] and uses a short basis [itex](\gamma_1, \gamma_2, ...)[/itex] 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 [itex]\alpha_{ij} \geq \pi/3[/itex], where [itex]\alpha_{ij}[/itex] is the angle between the minimal geodesics from the basis point [itex]\tilde p[/itex] to [itex]\gamma_i \tilde p[/itex] and [itex]\gamma_j \tilde p[/itex] respectively.

Now the desired estimate is acquired as follows:

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

Because each angle between two such vectors is at least equal to [itex]\pi/3[/itex], we can draw pairwise disjoint open balls [itex]B_{\pi/6}(v_i) \subset T_{\tilde p}^1\tilde M \cong S^n[/itex] around each [itex]v_i[/itex]. (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 [itex]h[/itex] is the number of these unit vectors [itex]v_i[/itex], then:

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

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

[itex]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[/itex]​

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

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

 

[jvicens]

Hi there,

Thank you for sharing your question about the last step of the derivation for Gromov's theorem. I can understand why you are confused and I would like to provide some clarification.

From what I can see, there is indeed an error in the proof. The volume comparison theorem states that if a complete Riemannian manifold M has nonnegative sectional curvature, then the volume of any ball in M is less than or equal to the volume of the corresponding ball in Euclidean space with the same radius. In other words, vol(B_r(p) \subset M) \leq vol(B_r(q) \subset \mathbb{R}^n) for any two points p, q \in M and any radius r.

In this case, the author seems to have mistakenly compared the volume of the ball B_{\pi/2}(e_1) \subset S^n (which is the unit sphere in Euclidean space) to the volume of the ball B_{\pi/6}(e_1) \subset S^n (which is a small neighborhood of the unit vectors in the unit sphere). This comparison is incorrect and should instead be vol(B_{\pi/2}(e_1) \subset S^n) \leq vol(B_{\pi/6}(e_1) \subset S^n) since the latter is a smaller ball.

To summarize, the correct final step should be:

h \leq \frac{vol(B_{pi/6}(e_1) \subset S^n)}{vol(B_{pi/6}(e_1) \subset S^n)} \leq \frac{(\pi/6)^n}{(\pi/6)^n} = 1

I hope this helps clarify the confusion. If you have any further questions, please do not hesitate to ask.