Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a hyperbolic isometry $g$ of the hyperbolic plane, denote by $ax(g)$ the oriented geodesic running from the repelling fixed point $p_g$ and the attracting one $q_g$. Let $G$ be the fundamental group of a hyperbolic once-punctured torus and suppose that $G$ is generated by two hyperbolic elements $a$ and $b$. In particular, the projections of their axes $ax(a)$ and $ax(b)$ are simple closed geodesics on the torus with intersection number equal to $1$. Assume also that the commutator $aba^{-1}b^{-1}$ is a parabolic element, hence it has a unique fixed point. The boundary of the disc model $\partial D$ is subdivided into two arcs by the points $q_a$ and $q_b$. Denote by $J$ the arc which does not contain the point $p_a$. If the pair $(a,b)$ is positively oriented (namely $ax(b)$ crosses $ax(a)$ from the right to the left), then the fixed point of $[a,b]$ lies in $J$. How can we prove this?

share|cite|improve this question

I think you can just prove this by computation for one explicit example, and then it will hold for all other such punctured tori by connectedness of Teichmuller space.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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