It is well-known that the Mapping Class Group of a closed surface of genus $g$ surjects onto $Sp(2g, \mathbb{Z})$ (see, for example the Farb-Margalit book). However, I was wondering if there is a simple proof that the set of pseudo-Anosov elements in MCG surjects onto $Sp(2g, \mathbb{Z})$ (and also a reference for where this might have been stated first) -- I can construct a somewhat sophisticated proof, but this should be easier.

  • $\begingroup$ I assume you mean "surjects onto $Sp(2g,\mathbb{Z))$"? $\endgroup$ – Lee Mosher Oct 27 '13 at 20:57
  • $\begingroup$ @LeeMosher Yes, fixed! $\endgroup$ – Igor Rivin Oct 27 '13 at 21:16
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    $\begingroup$ In Lemma 2.5, Darren Long proves that any normal subgroup of the mapping class group contains a pseudo-Anosov. dx.doi.org/10.1017/S0305004100063957 You want to show that a coset of a normal subgroup contains a pseudo-Anosov. One might be able to prove this using a modification of his techniques, which use the action on PML (and might hold in this more general context). $\endgroup$ – Ian Agol Oct 27 '13 at 23:15

Here's an outline of a proof. Consider a pseudo-Anosov mapping class $\phi$ such that the infinite cyclic group $C = \langle \phi \rangle$ is malnormal. Using the methods of Ivanov and/or McCarthy one can show that for any mapping class $\psi \not\in C$, the mapping class $\psi \phi^k$ is pseudo-Anosov for sufficiently large $k$. Now apply this with $\phi$ in the Torelli group and $\psi$ representing any element of $SP(2g,\mathbb{Z})$.

Some details added later:

First part: There is a simple direct construction of a pseudo-Anosov in the Torelli group: use Penner's recipe, which says that if $c,d$ are a filling pair of curves then $\tau_c \tau_d^{-1}$ is pseudo-Anosov. Take $c,d$ to be separating. The malnormality requirement can also be obtained by choosing $c,d$ so that no mapping class fixes both $c,d$ nor interchanges them.

Second part: Since $\psi \not\in C$ then (as in Sam's answer) $\psi$ fixes neither the attracting nor repelling measured geodesic laminations of $\phi$, and neither does it interchange them; thie follows from the assumption that $C$ is malnormal. Now pick a tiny compact neighborhood $K$ of the attracting lamination of $\phi$, so tiny that $\psi(K)$ does not contain the repelling lamination. Pick a high power $\phi^k$ so that $\phi^k(\psi(K))$ is in the interior of $K$ and so that $\phi^k\psi$ is contracting on $K$; this is possible because, in projective train track coordinates, $\phi$ is contracting near its attracting lamination. By this means (and working similarly with the inverse) one shows that $\phi^k\psi$ has north-south dynamics (and hence so does its conjugate $\psi\phi^k$). Finally, no reducible mapping class has north south dynamics, so $\psi\phi^k$ is pseudo-Anosov.

  • $\begingroup$ This approach was my first thought, but this seems pretty heavy duty (why is there such a mapping class in Torelli? What are these methods of McCarthy and Ivanov?) $\endgroup$ – Igor Rivin Oct 27 '13 at 21:15
  • $\begingroup$ To answer your second question, I am referring to the methods of proof of the Tits alternative for mapping class groups. $\endgroup$ – Lee Mosher Oct 29 '13 at 2:24
  • $\begingroup$ As pointed out in my comment, Darren Long has shown that any normal subgroup contains a pseudo-Anosov, so being in Torelli is irrelevant. $\endgroup$ – Ian Agol Oct 29 '13 at 4:54
  • $\begingroup$ I am being dense, I know, but this (expanded) argument seems to show that $\phi^k \psi$ maps any neighborhood of $\lambda^+$ to its. Does this mean that its attracting lamination is the same as that of $\phi?$ This seems implausible. $\endgroup$ – Igor Rivin Oct 29 '13 at 14:10
  • $\begingroup$ No, but it does mean that as $k \to \infty$, the attracting lamination of $\phi^k \psi$ approaches that of $\phi$. $\endgroup$ – Lee Mosher Nov 1 '13 at 1:59

Suppose that $f$ is a pseudo-Anosov and let $\lambda^\pm$ be its stable and unstable laminations. Suppose that $g$ is any mapping class with the following property: $g(\lambda^+) \neq \lambda^-$. Then, for all sufficiently large $n$, the composition $f^n g$ is also pseudo-Anosov. This is proved using the north-south dynamics of $f$.

Now, as in Mosher's answer, choose $f$ in Torelli and let $g$ be a lift of the desired element of $Sp$. If needed, compose $g$ with a bounding pair map to arrange the side-condition. We are done.

Added later: Here are some of the details of the dynamical argument. I've edited this several times to try and make it correct. Full disclosure - I first heard this from Yair Minsky.

Pick $U$, a small neighborhood of $\lambda^+$, chosen so that $\lambda^-$ is not contained in $g(U)$. I'll also want $f|U$ to be strictly contracting. Using north-south dynamics, there is an $m^+ > 0$ so that for all $n > m^+$ we have $f^{m^+} g(U) \subset U$. The contracting property implies that the map $f^n g$ has a unique fixed point in $U$.

On the other hand, pick a small neighborhood $V$ of $g^{-1}(\lambda^-)$, so that $\lambda^+$ is not in $V$. Also, we require $f^{-1}|g(V)$ to be strictly contracting There is a power $m^- > 0$ so that for all $n > m^-$ we have $V \subset f^n g(V)$. Deduce that the map $f^n g$ has a unique fixed point in $V$.

We place one more restriction on $n$. We need $f^n g(V^c)$ to be contained in $U$. I now claim that $h = f^n g$ has no other fixed points, and so is pseudo-Anosov.

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    $\begingroup$ Could you elaborate on the "north-south dynamics"? $\endgroup$ – Igor Rivin Oct 28 '13 at 0:08
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    $\begingroup$ A map $h$ of a topological space $X$ acts with north-south dynamics if $h$ has a pair of distinct fixed points $a$ and $b$ with the following property. For any neighborhoods $U$ and $V$ containing $a$ and $b$, respectively (but not containing $b$ and $a$, respectively) there is a positive integer $n$ so that $f^n(X - V) \subset U$ and $f^{-n}(X - U) \subset V$. The point $a$ is called "attracting" and the point $b$ is called "repelling". The roles of these points are played by $\lambda^{\pm}$ in the case of a pseudo-Anosov map. $\endgroup$ – Sam Nead Oct 28 '13 at 2:26
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    $\begingroup$ I am aware of the definition, but why does it work here? $\endgroup$ – Igor Rivin Oct 28 '13 at 2:49
  • $\begingroup$ This part of my answer will look similar to the second part of Lee's addition. I'll write out my version to see how our side-conditions compare. $\endgroup$ – Sam Nead Oct 30 '13 at 2:52

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