12
$\begingroup$

Let $g \geq 2$. Let $S = \langle a_1,b_2,...,a_g,b_g | [a_1,b_1] \cdots [a_g,b_g] \rangle$ be the fundamental group of a genus $g$ surface and let $F_g$ be a free group with $g$ generators. Given two surjections $f_1,f_2 : S \to F_g$ is there a way to determine if there are automophisms $\phi: S \to S$ and $\psi: F_g \to F_g$ so that $f_1 = \phi \circ f_2 \circ \psi$?

Is there an example of two surjections $f_1,f_2$ that are not equivalent in the above way?

I asked the question on MSE before but didn't get much.

$\endgroup$
  • $\begingroup$ A naive question: is it clear such a surjection exists? $\endgroup$ – PseudoNeo Oct 15 '18 at 2:00
  • 3
    $\begingroup$ @PseudoNeo Algebraically, yes: kill all of the $b_i$. Geometrically, yes: the surface is the boundary of a handlebody, equivalent to a wedge of $g$ circles. $\endgroup$ – Mike Miller Oct 15 '18 at 2:02
  • $\begingroup$ Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_{2g}$) and was very confused. $\endgroup$ – PseudoNeo Oct 15 '18 at 2:03
  • 1
    $\begingroup$ For context, there's no homomorphism onto $F_{g+1}$. Out of curiosity, what can be said of the set of surjective homomorphisms $\pi_1(S)\to F_k$ modulo $Aut(\pi_1(S))\times Aut(F_k)$, when $1\le k<g$? is it infinite? $\endgroup$ – YCor Oct 15 '18 at 8:15
8
$\begingroup$

This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly different language, that is they prove that any such epimorphism is induced by choosing a genus $g$ handlebody.

$\endgroup$
  • $\begingroup$ Do you know if the set of surjective homomorphisms onto $F_g$ is a singleton modulo $Aut(\pi_1(S))$ (instead of modding out by $Aut(\pi_1(S))\times Aut(F_g)$)? this is related to the question whether automorphisms of $F_k$ can be lifted to automorphisms of $\pi_1(S)$. $\endgroup$ – YCor Oct 15 '18 at 8:17
  • $\begingroup$ I would say no, because it seems unlikely to me that any automorphism of the free group induces an homeomorphism of the handlebody, and barring that I have no idea about your lifting problem and I see no indication about it in Leininger--Reid's arguments (but I'm not a specialist so I might be missing something obvious). $\endgroup$ – Jean Raimbault Oct 15 '18 at 16:45
  • $\begingroup$ @JeanRaimbault Thanks this is perfect. I am trying to do this algorithmically also - is it clear how lemma 2.2 can be made effective? I.e. given the surface with the labeled generators and a surjection to a free group, how do we find the set of curves to attach to the surface to realize the handlebody that realizes the surjection? $\endgroup$ – user101010 Oct 18 '18 at 2:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.