Can anyone give me a reference (or proof sketch) for the fact that there are psuedo-Anosov diffeomorphisms of closed hyperbolic surfaces which do not extend over any handlebody? Thanks.
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$\begingroup$ There is even a pA map $f$ so that $f^n$ extends over a handlebody if and only if $n = 0$. $\endgroup$– Sam NeadCommented Jun 2, 2012 at 20:31
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5 Answers
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If you take a mapping class $\psi:\Sigma\to \Sigma$ whose characteristic polynomial of its action on homology $\psi_\ast: H_1(\Sigma;\mathbb{Z}) \to H_1(\Sigma;\mathbb{Z})$ is irreducible, then it cannot extend over a handlebody. The point is that if $\psi= \Psi_{| \Sigma}$, where $\Psi: H\to H, \partial{H}=\Sigma$, $H$ a handlebody, then $\psi$ preserves $K=ker\{H_1(\Sigma)\to H_1(H)\}$, which is an integral Lagrangian subspace of $H_1(\Sigma)$. So the characteristic polynomial of $\psi_{\ast | K}$ divides the characteristic polynomial of $\psi_\ast$. In fact, this argument shows that $\psi$ does not extend over any manifold $M$ with $\partial M=\Sigma$.
Since $Mod(\Sigma)\to Sp(H_1(\Sigma))$ is surjective, one also has a pseudo-Anosov element giving any symplectic matrix (see Sam Nead's comment for one possible argument, or one may restrict to a matrix satisfying Casson's criterion for irreducibility as in Rivin's answer), so there is a pA element satisfying the irreducibility criterion.
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$\begingroup$ Why does one have a pseudo-ANosov giving any symplectic matrix? $\endgroup$ Commented Jun 2, 2012 at 15:57
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1$\begingroup$ Igor - take any pA element $f$ that acts trivially on homology. Take any $g$ with the desired action on homology. Then for some $n$ the element $f^n \circ g$ will be pA. $\endgroup$– Sam NeadCommented Jun 2, 2012 at 17:18
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2$\begingroup$ Arrange matters so that $g$ doesn't send $\lambda^+(f)$ to $\lambda^-(f)$; post-compose $g$ with a separating Dehn twist if necessary. Now use the Brouwer fixed point theorem and the north-south dynamics of $f$ to prove (for $n$ sufficiently large) that the action of $f^n \circ g$ has exactly two fixed points: one close to $\lambda^+(f)$ and one close to $g^{-1}(\lambda^-(f))$. $\endgroup$– Sam NeadCommented Jun 2, 2012 at 20:22
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You can find a proof of this in the paper
MR1885215 (2002m:57019) Leininger, Christopher J.(1-TX); Reid, Alan W.(1-TX) The co-rank conjecture for 3-manifold groups. (English summary) Algebr. Geom. Topol. 2 (2002), 37–50 (electronic).
It works for any surface of genus at least $2$.
It was originally proven by Casson and by Johnson-Johannson, though they did not publish their proofs (they don't mention the pseudo-Anosov condition, only that the mapping classes do not extend over any handlebody; however, I'm pretty sure that you can get a pseudo-Anosov mapping class by following their proofs). I have a photocopy of the preprint of Johnson-Johannson; if you want it, let me know and I can scan it.
You might also be interested in the paper "Relative Weight Filtrations on Completions of Mapping Class Groups" by Hain (available here) and the thesis of Jamie Jorgensen (available here). They prove that there exist mapping classes that don't extend to any handlebody arbitrarly deep in the "Johnson filtration" of the mapping class group (so these are very algebraically complicated). I'm pretty sure you can follow their ideas to get ones that are pseudo-Anoson.
answered Jun 2, 2012 at 1:26
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In the section 3 of the article "Extending Pseudo-Anosov maps into compression bodies" (by Biringer, Johnson, and Minsky), you will find examples of pseudo-Anosov maps on the boundary of a genus 3 handlebody that do not extend to a handlebody automorphism.
Reference for the article: arXiv:1011.0021v1
answered Jun 2, 2012 at 0:14
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Really a long comment on @Agol's answer:
One does not actually need the fact (which I was not aware of) to show what the OP wants from what @Agol says, since by Casson-Bleiler, any mapping class whose characteristic polynomial is irreducible, non-cyclotomic, and does not have the form $f(x^k),$ for $k>1$ is pseudo-Anosov, and by my results, the characteristic polynomial of a generic element of the symplectic group has all three properties (see Walks on groups, counting reducible matrices, polynomials, and surface and free group automorphisms, Igor Rivin Duke Mathematical Journal 142, 2, pp353-379), and hence, the generic element of the mapping class group, by Ian's remark, has the property the OP wants.
answered Jun 2, 2012 at 16:02
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I accidentally found this Rice University thesis (never published, it seems) which studies the question in some depth:
Author Jamie Bradley Jorgensen Title Surface homeomorphisms that do not extend to any handlebody and the Johnson Filtration.
answered Jun 12, 2012 at 18:53
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$\begingroup$ @Igor Rivin : If you look at my answer, I gave a link to a version of his thesis that he posted on the arXiv. Jamie left mathematics after completing his PhD and wasn't interested in publishing his thesis. $\endgroup$ Commented Jun 12, 2012 at 20:19
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$\begingroup$ (though I see now that I mangled the link. I justed edited my answer to give a correct link). $\endgroup$ Commented Jun 12, 2012 at 20:21
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$\begingroup$ @Andy: oops, I did not notice.... $\endgroup$ Commented Jun 12, 2012 at 21:50