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Following Lemma 2.7 from Vogtmann's Rational Homology of Bianchi Groups, I want to define cuspidal cohomology as $$H_{\mathrm{cusp}}(M)=\frac{H_1(M)}{i_*(H_1(\partial M))}$$ where $i:\partial M\to M$ is the inclusion and I take integral coefficients. The motivation is that Abelian covering spaces such that the preimage of a cusp consists of disjoint copies of the same cusp are corresponding to exactly those holonomies $\pi_1(M)\to G$ that factor through the cuspidal cohomology.

I verified for all hyperbolic cusped 3-manifolds in the SnapPea census that if $H_1(M)=\mathbb{Z}^n\oplus T$ where $T$ is torsion, then $H_\mathrm{cusp}(M)=\mathbb{Z}^{n-\mbox{number of cusps}}\oplus T$. This is to be expected rationally by half-lives-half-dies. It seems to also hold integrally for hyperbolic 3-manifolds.

Is there an argument that this holds for all hyperbolic cusped 3-manifolds? (And maybe a counterexample if you take a non-hyperbolic manifold)

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Vogtmann's name is spelled with two n's. – Jim Conant Jun 6 '12 at 23:11

Consider the long exact sequence on homology (coefficients in $\mathbb{Z}$) $$\to H_1(\partial M)\overset{i}{\to} H_1(M)\to H_1(M,\partial M) \to H_0(\partial M) \to $$ You are looking for $H_1(M)/i_\ast(H_1(\partial M)) \cong im\{ H_1(M)\to H_1(M,\partial M)\} \cong ker \{ H_1(M,\partial M)\to H_0(\partial M)\}$ by the exactness of the sequence. If $H_1(M)\cong \mathbb{Z}^n\oplus T$, $T$ torsion, then $H_2(M)\cong \mathbb{Z}^{n-1}$ (the rank $n-1$ follows from Euler characteristic). By universal coefficients, we have a short exact sequence $$0\to Ext(H_1(M),\mathbb{Z})\to H^2(M) \to Hom(H_2(M),\mathbb{Z})\to 0.$$ One computes $Ext(H_1(M),\mathbb{Z})\cong T$, $Hom(H_2(M),\mathbb{Z})\cong \mathbb{Z}^{n-1}$ (see p. 195 of Hatcher), so that $H^2(M)\cong \mathbb{Z}^{n-1} \oplus T$. By Lefschetz duality, $H^2(M)\cong H_1(M,\partial M) \cong \mathbb{Z}^{n-1}\oplus T$. So you are looking for the torsion in $$ker\{ H_1(M,\partial M)\to H_0(\partial M) \} \cong ker\{ \mathbb{Z}^{n-1}\oplus T \to \mathbb{Z}^c\},$$ which is clearly isomorphic to $T$.

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One can make some progress on this by using half-lives--half-dies with coefficients in $\mathbb{F}_p$. Suppose the cuspidal homology is $\mathbb{Z}^{n - c} \oplus S$. If $p$ is coprime to the order of $T$, then the dimension of $H_1(M; \mathbb{F}_p)$ is just $n$ and the mod $p$ cuspidal homology has dimension $n - c$; hence the order of $S$ must also be coprime to $p$. Indeed, this argument shows that if the order of $T$ is square-free then your claim holds. Perhaps some more sophisticated argument with universal coefficients is enough to prove it in general? (If this is true, it's surely not because the manifolds are hyperbolic but rather a general fact about 3-manifolds with torus boundary.)

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I can't vote my own answer down, but Ian's answer (currently below this one) is complete. – Nathan Dunfield Jun 12 '12 at 13:18

It is worth pointing out that $T$ may lie in the image of $i_*$:

There is an $M$ such that $H_1(M) \cong \mathbb{Z}^{n} \oplus T$ and $H_{\mathrm{cusp}}(M) \cong \mathbb{Z}^{n-\mathrm{number\ of\ cusps}} \oplus T$ but $T$ dies under the quotient map $H_1(M) \to H_\mathrm{cusp}(M)$.

Let $N$ be an orientable $I$-bundle over the Klein bottle. Then the "cuspidal homology" of $N$ is $\mathbb{Z}/2\mathbb{Z}$, but the torsion in $H_1(M)$ dies under $H_1(M) \to H_\mathrm {cusp}(M)$.

To find a hyperbolic example, note that a theorem of R. Myers (Excellent 1-manifolds in compact 3-manifolds. Topology Appl., 49(2):115–127, 1993.) produces a null-homotopic hyperbolic knot $K$ in $N$. Thurston's Hyperbolic Dehn Surgery Theorem allows us to perform surgery on $K$ to obtain a hyperbolic $M$ where all the maps on homology are as above.

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@Richard, it seems you are saying A and not A in your answer. – Misha Jun 7 '12 at 16:30
@Misha: The answer is an attempt to emphasize the difference between isomorphism and equality. As the question asks for equality, I think this is relevant. – Richard Kent Jun 7 '12 at 16:41
Edited to clarify. – Richard Kent Jun 7 '12 at 16:54
@Richard, I understand what you are saying now. However, it seems to me that in the context of the question, equality would not make much sense since $H_{cusp}$ is the quotient of $H_1$, not a subgroup. I think the equality sign in question is just the (common) abuse of notation. Nevertheless, you have a very nice example which shows that torsion part of $H_1$ could lie in the image of $i_*$. It also suggests that one should check what happens in the case of other Seifert manifolds. – Misha Jun 7 '12 at 17:15
@Misha: Yeah, my interpretation of equality as meaning that T survives is psychological, I guess. One more edit to further clarify. – Richard Kent Jun 7 '12 at 17:27

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