TITLE: The kernel of free group map to surface group
QUESTION [18 upvotes]: $G$ is a surface group of genus $g\geq 2$ (the fundamental group of closed orientable surface of genus g). $F$ is a free group of rank $2g$ with basis $\{x_1,\dots,x_{2g}\}$. $\phi$ is a surjective homomorphism from $F$ to $G$ such that $G$ is generated by $\phi(x_1),\dots,\phi(x_{2g})$. 

What can we say about the kernel of $\phi$? 
Is it always the normal closure of an element $r$ in $F$? 
Can we write $r$ explicitly if it is?

REPLY [10 votes]: As I said in the comments above, this has a positive solution due to a recent preprint of Lars Louder, which can be found here. The paper proves that surface groups have a "single Nielsen equivalence class of generating $2g$-tuples". I'll explain what this means, and then I will explain why this solves the problem. (In the comments below, Lars Louder has pointed out that the quoted result is due to Zieschang for genus $g\neq 3$, and that his own results are much more general.)
Let $(x_1, x_2, \ldots, x_n)$ be a generating tuple for a group $G$. Then clearly all permutations of this tuple are generating tuples for $G$, as are $(x_1^{-1}, x_2, \ldots, x_n)$ and $(x_1x_2, x_2, \ldots, x_n)$. These alterations correspond to Nielsen transformations, or equivalently, to automorphisms of the free group $F(x_1, \ldots, x_n)$. Indeed the Nielsen transformations I described above generate all Nielsen transformations, that is, if you repeat the above alterations then you obtain all of the "obvious" generating tuples for $G$, for example if $n=3$ then $(x_1^{-1}, x_2x_1x_3, x_2x_1)$ will be obtained, and so on. If two generating $n$-tuples can be obtained one from the other in this way then they are said to be Nielsen equivalent. They partition the generating tuples into Nielsen equivalence classes.
Fix $G=\langle y_1, z_1, \ldots, y_g, z_g; [y_1, z_1]\cdots [y_g, z_g]\rangle$. What Lars Louder have proven is that if $\mathcal{L}_1=(x_1, x_2, \ldots, x_{2g})$ generates $G$ (note that each $x_i$ denotes a word over $y_{\ast}$ and $z_{\ast}$) then $\mathcal{L}$ can be obtained from the standard generating tuple $\mathcal{L}_0=(y_1, z_1, \ldots, y_g, z_g)$. It is then an easy exercise to see that when you go from the generating tuple $\mathcal{L}_0$ to $\mathcal{L}_1$ via your Nielsen transformation $\phi$, so $\phi(y_1)=x_1$, $\phi(z_1)=x_2$, etc. you obtain the following presentation of $G$.
$$\langle y_1, z_1, \ldots, y_{g}, z_g; [x_1, x_2]\cdots [x_{2g-1}, x_{2g}]\rangle$$
Recall that each $x_i$ denotes a word over $y_{\ast}$ and $z_{\ast}$, so this presentation makes sense.
There are some old results in a similar vein from the 70s. For example, Steve Pride proved that one-ended two-generator, one-relator groups with torsion have a single Nielsen equivalence class of generating pairs (and this solves the isomorphism problem for such groups), while at around the same time Brunner proved that the Baumslag-Solitar group $BS(2, 3)=\langle a, t; t^{-1}a^3t=a^3\rangle$ has infinitely many Nielsen equivalence classes of generating pairs. In fact, Brunner proves something stronger. Two generating tuples $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_n)$ of a group $G$ are said to lie in the same T-system is there exists an automorphism $\phi$ of $G$ such that $(\phi(x_1), \ldots, \phi(x_n))$ is Nielsen equivalent to $(y_1, \ldots, y_n)$. Brunner proves that $BS(2, 3)$ has infinitely many T-systems. (The "T" stands for transitivity, if I recall correctly.)
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