Automorphisms of $\pi_1$ induced by pseudo-Anosov maps Suppose $X$ is an orientable surface with non-empty boundary and $f:X\to X$ is a pseudo-Anosov automorphism that acts identically on $H_1(X,\mathbf{Z})$. Let $x$ be a fixed point of $f$.
For any $\gamma\in\pi_1(X,x)$ we have $\gamma^{-1}f(\gamma)\in [\pi_1(X,x),\pi_1(X,x)]$, the commutant of $\pi_1(X,x)$. More generally, we have $\gamma\cdot g^{-1}f g(\gamma)\in [\pi_1(X,x),\pi_1(X,x)]$ where $g$ is an automorphism of $X$ that fixes $x$.
I would like to ask what one can say about the normal closure in $\pi_1(X,x)$ of the set of all elements $\gamma\cdot g^{-1}f g(\gamma)$ where $\gamma$ runs through $\pi_1(X,x)$ and $g$ runs through the set of all diffeomorphisms $X\to X$ that fix $x$. In particular, does this closure  coincide with the commutant of $\pi_1(X,x)$?
 A: No. Let $\Gamma_i$ be the lower central series defined by $\Gamma_1=\pi_1(X,x)$, $\Gamma_{i+1}=[\Gamma_1,\Gamma_i]$. The Johnson filtration $\text{Mod}_g(k)$ is the descending filtration of the mapping class group relative to $x$ defined by:
$f\in \text{Mod}_g(k)\iff f$ acts trivially on $\Gamma_1/\Gamma_k$
The first term $\text{Mod}_g(2)$ is the Torelli group, consisting of diffeomorphisms acting trivially on homology. The next term $\text{Mod}_g(3)$ is the Johnson kernel. By a beautiful theorem of Johnson, this is the subgroup generated by Dehn twists around separating curves.
By residual nilpotence of surface groups, we have $\bigcap \text{Mod}_g(k)=\{1\}$, but every individual term in the filtration is nontrivial. It is not hard to see that every term of the Johnson filtration contains pseudo-Anosovs. Indeed every normal subgroup of the mapping class group contains pseudo-Anosovs (see Lemma 2.5 of Long, "A note on the normal subgroups of mapping class groups") from which Long concluded that any two normal subgroups intersect nontrivially!
Thus since $\text{Mod}_g(k)$ is normal, if we take $f\in\text{Mod}_g(k)$ we have $\gamma^{-1}\cdot g^{-1}fg(\gamma)\in \Gamma_k$ for all $g$ and all $\gamma$.
