Overview: the mapping class group maps into $\text{Aut}(F_{2g})$ and its image stabilizes the surface relation. I am trying to check this for a specific example and am doing something wrong.
The mapping class group of the genus 3 surface with one boundary component, $\text{Mod}_3^1$, acts on the fundamental group of the surface $\langle \alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3 \rangle \cong F_6$.
This action gives a map $\text{Mod}_3^1 \hookrightarrow \text{Aut}(F_6)$. Since $\text{Mod}_3^1$ fixes the boundary pointwise, its corresponding map in $\text{Aut}(F_6)$ must fix the boundary curve $[\alpha_1, \beta_1][\alpha_2, \beta_2][\alpha_3, \beta_3]$.
Consider the bounding pair map $T_d T_e^{-1}$. I want to check that $T_d T_e^{-1}$ stabilizes the boundary curve.
I believe $T_d T_e^{-1}$ acts on $\pi_1(\Sigma_3^1, *)$ in the following way (so corresponds to the obvious element of $\text{Aut}(F_6)$):
- $\alpha_1 \mapsto \beta_2 \alpha_1 \beta_2^{-1}$
- $\beta_1 \mapsto \beta_2 \beta_1 \beta_2^{-1}$
- $\alpha_2 \mapsto [\alpha_1,\beta_1]\alpha_2$
- $\alpha_3$, $\beta_3$ are fixed
The Primer on Mapping Class Groups walks through this starting on page 195.
Now, I would expect this element of $\text{Aut}(F_6)$ to stabilize $[\alpha_1, \beta_1][\alpha_2, \beta_2][\alpha_3, \beta_3]$. It sends $$[\alpha_1, \beta_1][\alpha_2, \beta_2][\alpha_3,\beta_3] \rightarrow [\beta_2 \alpha_1 \beta_2^{-1},\beta_2 \beta_1 \beta_2^{-1}][[\alpha_1,\beta_1]\alpha_2, \beta_2][\alpha_3, \beta_3]$$
It is easy to check that $[\beta_2 \alpha_1 \beta_2^{-1},\beta_2 \beta_1 \beta_2^{-1}] = \beta_2[\alpha_1, \beta_1]\beta_2^{-1}$. This gives us $$[\alpha_1, \beta_1][\alpha_2, \beta_2][\alpha_3,\beta_3] \rightarrow \beta_2 [\alpha_1, \beta_1] \beta_2^{-1} [[\alpha_1,\beta_1]\alpha_2,\beta_2][\alpha_3, \beta_3]$$
which I don't think simplifies to $[\alpha_1, \beta_1][\alpha_2, \beta_2][\alpha_3, \beta_3]$, so at least one of my steps is wrong. Some places I may be going wrong:
- My map into $\text{Aut}(F_6)$ could be incorrect.
- I could be using the wrong element of $F_6$ for the boundary curve. This paper by Dennis Johnson says $\Pi_{i=1}^g [\alpha_i, \beta_i]$ is represented by $\partial M$, but with orientation opposite to that acquired in the natural way by $M$.
- Calculation or simplification error
- I'm fundamentally misunderstanding something
Edit on 1/8: simpler question
How does $T_d$ act on $F_6$? I would guess $\alpha_1 \rightarrow \beta_2 \alpha_1 \beta_2^{-1}$, $\beta_1 \rightarrow \beta_2 \beta_1 \beta_2^{-1}$, and $\alpha_2 \rightarrow \beta_2 \alpha_1$. This would send the boundary curve to $\beta_2 [\alpha_1, \beta_1][\alpha_2, \beta_2] \beta_2^{-1} [\alpha_3, \beta_3]$?