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Let $H$ be a handlebody with $\Sigma = \partial H$. Given an automorphism $f : \Sigma \to \Sigma$ we can glue to obtain a closed 3-manifold $M = H \cup_f H$ and in fact all such 3-manifolds are obtained this way. Since only the isotopy class of $f$ is necessary in order to specify the homeomorphism type of the resulting 3-manifold, we have a map from the mapping class group $\mathop{MCG}(\Sigma)$ to the set of homeomrphism types of closed 3-manifolds. By looking at the action on $H_1(\Sigma;\mathbb{Z})$ which has a symplectic structure via the cup product, we obtain a surjection $$ \psi: \mathop{MCG}(\Sigma) \to \mathop{Sp}(H_1(\Sigma; \mathbb{Z})) $$ What information about $M$ can be obtained from $\psi([f])$?

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In addition to the first homology group, it also determines the Seifert pairing on the torsion in the first homology group. What is more, in an appropriate sense this is all it determines. This is all contained in my paper "Symplectic Heegaard splittings and linked abelian groups" with Joan Birman and Dennis Johnson, which can be downloaded from my webpage here. This paper also shows that the symplectic matrix also determines a certain subtle invariant of the Heegaard splitting (that alas vanishes after a single stabilization).

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Well, from the characteristic polynomial of the matrix, thanks to Casson, one can figure out if the gluing map is pseudo-Anosov (if the characteristic polynomial is irreducible, non-cyclotomic, and does not have the form $P(x) = Q(x^k),$ for some $k>1.$). In principle, there should be a some criterion in terms of the matrix for low-bounding the translation distance in the curve complex, and if that is at least $3,$ and so the manifold will be hyperbolic, but I don't know what the criterion is.

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    $\begingroup$ Since your phrasing is unclear, I want to emphasize (to the OP; I know you know this) that Casson's criterion only goes in one direction; indeed, there are pseudo Anosov mapping classes that act trivially on homology. $\endgroup$ – Andy Putman Mar 17 '17 at 2:39
  • $\begingroup$ @AndyPutman Yes, thanks for clarifying! I think the last question I ask is interesting (I don't know why I had never thought to ask it before :( ), and I would guess the answer is positive (again, in one direction). $\endgroup$ – Igor Rivin Mar 17 '17 at 2:42
  • $\begingroup$ I also expect that there is some criterion, though I am not sure what it would be. It's a good question! $\endgroup$ – Andy Putman Mar 17 '17 at 3:25
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    $\begingroup$ Actually, on further reflection I doubt that such a criterion exists. I bet you could prove this using the results in the paper I cite in my answer. $\endgroup$ – Andy Putman Mar 17 '17 at 12:33
  • $\begingroup$ @AndyPutman A negative result would be interesting too, especially since probabilistically it is true that if your top singular value when acting on homology is large, so is your cc translation distance. $\endgroup$ – Igor Rivin Mar 17 '17 at 14:12

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