VOLUMES OF HYPERBOLIC 3-MANIFOLDS OF BETTI NUMBER AT LEAST 3

ANDREW PRZEWORSKI

Abstract. We provide a new, simpler proof of the fact that the smallest volume hyperbolic 3-manifold has betti number at least 3. In the process, we improve the best known lower bound on the volume of such a manifold.

Since it was shown that there is a hyperbolic 3-manifold of smallest volume, there has been great interest in determining this manifold. Conjecturally, it is the Weeks manifold W, obtained by (-5/1,-5/2) Dehn surgery on the Whitehead link. The volume of the Weeks manifold is 0.9427 . . .. As of now, the best known lower bound on volume for an arbitrary orientable hyperbolic 3-manifold is 0.28, established in [10]. Much better results can be achieved by placing some restrictions on the types of manifolds under consideration [2][6][7][8]. One such result, due to Culler, Hersonsky, and Shalen [4], is that if an orientable hyperbolic 3-manifold has betti number at least 3, then its volume is at least 0.94689 and hence it is not the smallest hyperbolic 3-manifold. We will provide a slight improvement of this result, as well as a much simpler proof.

First, we cite the necessary background results. Rather than repeat the preliminary arguments from [4] we condense these arguments into the following result, which draws upon [3] and [5]:

Theorem 1. If M is a closed orientable hyperbolic 3-manifold of betti number at least 3 whose shortest geodesic has length at most λ then

Vol(M) ≥ V (λ) = \frac{πλ}{e^{λ} − 1} (\frac{e^{2λ} + e^{λ} + 5}{2(cosh \frac{λ}{2})(e^{λ} + 3)}) − \frac{πλ}{2}

Our simplification of [4] will be facillitated by a result of Gabai, Meyerhoff, and Thurston [9]

Theorem 2. [9] Every closed orientable hyperbolic 3-manifold except Vol3 contains an embedded tube of radius at least 0.52959 … about its shortest geodesic. Unless the shortest geodesic has length greater than 1.0595 …, there is a tube of radius \frac{log 3}{2} about this geodesic.

2000 Mathematics Subject Classification. Primary 57M50; Secondary 57N10.

Key words and phrases. betti, volume, hyperbolic 3-manifold.

These two results are all that we need.

Theorem 3. If the first betti number of a closed orientable hyperbolic 3-manifold M is at least 3 then the manifold has volume at least 1.015.

Proof. Since M has betti number at least 3, M is not Vol3. Let l be the length of the shortest geodesic in M. If l ≥ 1.0595 then by Theorem 2, there is a tube of radius at least 0.52959 about the shortest geodesic. The volume of this tube is then at least π1.0595 sinh^{2} 0.52959 = 1.024 . . .. If 0.9698 ≤ l ≤ 1.0595, then there is a tube of radius at least \frac{log 3}{2} about the shortest geodesic. The volume of this tube is at least π0.9698 sinh^{2} \frac{log 3}{2} = 1.0155 . . .. If l ≤ 0.9698, then by Theorem 1, the volume of the maximal tube about the shortest geodesic is at least V (0.9698) = 1.0158 . . .. □

A simple corollary of this is:

Corollary 4. The smallest volume orientable hyperbolic 3-manifold has betti number at most 2.

Proof. In [1], it is shown that the smallest volume noncompact hyperbolic 3-manifold has volume 1.014 … > Vol(W). Hence, the smallest volume orientable hyperbolic 3-manifold must be closed. If it has betti number at least 3, then its volume would be at least 1.015 > Vol(W), a contradiction. □

References

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Department of Mathematics, University of Texas, Austin, Texas 78712

 E-mail address: prez@math.utexas.edu