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Proof: Gromov's short basis, volume comparison

  1. Sep 21, 2013 #1

    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 [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?
    Last edited: Sep 21, 2013
  2. jcsd
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