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Let $S$ be a fixed hyperbolic surface with genus $g$ and $n$ punctures. Given any pseudo-Anosov map $f$ on $S$ (with stretch factor $\lambda$) with stable and unstable measured foliations $\mu^s$ and $\mu^u$ respectively. Given any simple closed curve $a$ let $I(\mu^s(a))=min\{\mu^s(\alpha): \alpha \text{ is in the homotopy class of }a \}$ where $\mu^s(a)$ is the length of $a$ with respect to the measure $\mu^s$.

My question is: does there exists $\epsilon>0$ (may depend on $g,n$) such that given any pseudo-Anosov map $f$ on $S$, there exists a simple closed curve $a$ such that $I(\mu^s(a))>\epsilon$.

PS: I have used the notations of the book "A primer on mapping class group" by Farb and Margalit. See Lemma 14.22 and Theorem 14.23.

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    $\begingroup$ This is true for every $\epsilon,g,n,S,f$, because if one starts with any $a$ at all then $I(\mu^s(f^n(a))) \to \infty$ as $n \to -\infty$, so one can replace $a$ with $f^n(a)$ for $n$ sufficiently close to $-\infty$. $\endgroup$
    – Lee Mosher
    Commented Dec 13, 2017 at 17:27
  • $\begingroup$ Thanks for the answer. I have another question: If $x=\underset{\text{set of all simple closed curves}}{inf}I(\mu^s(a))$, then is there an $\epsilon$ (depending on $g,n$) such that $x>\epsilon$? $\endgroup$
    – Cusp
    Commented Dec 13, 2017 at 18:26
  • $\begingroup$ No there does not, because $I(\mu^s(f^n(a))) \to 0$ as $n \to +\infty$. $\endgroup$
    – Lee Mosher
    Commented Dec 13, 2017 at 19:19
  • $\begingroup$ @LeeMosher, I hope you'll post your comment as an answer! $\endgroup$
    – HJRW
    Commented Dec 14, 2017 at 10:39

1 Answer 1

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To answer the main question as well as the question in the comments, for every simply closed curve $a$ we have $I(\mu^s(a)) \in (0,\infty)$, and every $n$ we have $I(\mu^s(f^n(a))) = \lambda^{-n} I(\mu^s(a))$. It follows that $I(\mu^s(f^n(a)))$ limits to $0$ as $n \to+\infty$ and to $+\infty$ as $n \to -\infty$. So there are no positive upper or lower bounds on $I(\mu^s(a))$ as $a$ varies over all simple closed curves.

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