pseudo-Anosovs with given action on homology

Question

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.

Answer 1

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.

Answer 2

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.