Let $M$ be orientable 3-manifold admitting a Heegaard splitting $V\cup_{S}W$.

Let $X$ be a carrier graph of $M$ such that rank($X$)=rank($\pi_{1} M$).

Note: A connected graph is called a carrier graph of $M$ if there is a map $f: X\rightarrow M$ such that $f: \pi_{1} X\rightarrow \pi_{1}M$ is surjective. And we call $f$ a carrier map of $X$.

Thank Agol for comments. I have editted my questions again.

Now I want to know

If we fix the carrier graph$X$, is it possible that there is a carrier map $f$ of $X$, $f(X)\subset S$? Can we ask $f(X)$ to be embedded into $S$?

Or more weakerly, Is there a pair of $(X,f)$ such that $X$ is a carrier graph of $M$ with rank(X)=rank($\pi_{1} M$) and $f(X)$ can be embedded into $S$?

  • $\begingroup$ I'm not sure I understand the question: any map of a graph may be homotoped to have image in the Heegaard surface by general position with respect to the cores of the two handlebodies of the Heegaard splitting. Do you want it to embed in S? $\endgroup$ – Ian Agol Dec 23 '12 at 6:43
  • $\begingroup$ @Agol. Yes, I hope it can be embedded. $\endgroup$ – yanqing Dec 23 '12 at 7:52

When the rank of the fundamental group equals the Heegaard genus then there always a carrier graph that embeds in the Heegaard surface $S$. I don't know the answer when the rank is less than the Heegaard genus, but I strongly suspect that the answer is "no". The place to start is the paper of Boileau and Zieschang, titled "Heegaard genus of closed orientable Seifert 3-manifolds". Hyperbolic examples where rank is less than Heegaard genus are given by Tao Li, in his paper "Rank and genus of 3-manifolds".


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.