6
$\begingroup$

Hi everyone.

Let $S$ be a closed surface with genus at least 3, $\alpha, \beta$ be the two vertices of curve complex of $S$ such that $d_{\mathcal {C}(S)}(\alpha, \beta)\geq 3$.

My question is

Is there a non-trivial finite ordered element $f$ of $MCG(S)$ such that $f(\alpha)=\alpha$ and $f(\beta)=\beta$ in $\mathcal {C}(S)$?

Thanks!

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Are you asking whether $f$ exists given $\alpha, \beta$, or whether there exists $\alpha,\beta,f$ with that property? $\endgroup$
    – Ian Agol
    Dec 14, 2012 at 4:21
  • $\begingroup$ Either way, the answer is yes, with $f=Id.$ $\endgroup$
    – Igor Rivin
    Dec 14, 2012 at 4:56
  • 4
    $\begingroup$ @Igor Rivin : Of course, that's not what they OP is looking for. And it's not a silly question (at least under Agol's second interpretation) -- there definitely exist nontrivial finite order mapping classes which fix a filling pair of curves (they permute the discs in the complement). $\endgroup$
    – Andy Putman
    Dec 14, 2012 at 5:17
  • $\begingroup$ @Agol, the last one. $\endgroup$
    – yanqing
    Dec 14, 2012 at 6:11
  • $\begingroup$ Actually it's not silly under Agol's 1st interpretation either. Existence of $\alpha,\beta$ for which there is no such $f$ is interesting to ponder. $\endgroup$
    – Lee Mosher
    Dec 14, 2012 at 13:58

2 Answers 2

Reset to default
7
$\begingroup$

Here is a way to find lots of examples. Suppose that $\Sigma$ is a surface and suppose that $f$ is a periodic mapping class. Let $S$ be the quotient orbifold $\Sigma/f$. Then taking full preimages gives a quasi-isometric embedding of the curve complex of $S$ into the curve complex of $\Sigma$. See

arXiv:1104.3492 and arXiv:math/0701719

for two different proofs. When $S$ has an infinite diameter curve complex we can pick $a$ and $b$, curves in $S$, that are as well separated as you want. Then the preimages $\alpha$ and $\beta$ are also well separated, and fixed as simplices by $f$. A bit more work (which I haven't done) should give examples where $\alpha$ and $\beta$ are single vertices. There are also examples where $S$ is honestly a surface, not just an orbifold.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ Here is a way to get a pair of curves rather than pair of multi-curves: Let $S$ be a surface of genus $\ge 2$, $c\in H_1(S,Z_n)$ a nontrivial primitive class, $a, b$ simple curves representing this class. By applying powers of pseudo-Anosov in the Torelli subgroup to $b$, we can assume that $dist(a,b)$ is large. Now, take a homomorphism $f:\pi_1(S)\to Z_n$ sending $c$ to the generator and let $\Sigma$ be the cyclic cover of $S$ corresponding to the kernel of $f$. Then lifts of $a, b$ to $\Sigma$ are connected. Now, proceed as in Sam's answer. $\endgroup$
    – Misha
    Dec 14, 2012 at 17:27
1
$\begingroup$

Here is an example which can be easily generalized for all $S_g$, $g\geq 3$. With similar approaches, one can construct many examples, not just starting with the building block $S_{1,2}$. Notice the Dehn twist about one boundary component in $S_{1,2}$.

http://s11.postimage.org/ajzyt2xsz/finiteorderfill.png

Given a finite group $G$ of mapping classes, can one find $S_g$, and curves $a,b$ on $S_g$ that fill such that $a,b$ are stabilized by each element of $G$? (off the top of my head, the example I have given does each finite cyclic group of order at least 3)

Improve this answer
$\endgroup$
3
  • $\begingroup$ If you look closely at Sam's construction and my comments, you will see that it gives examples for all genera $\ge 3$ since every such surface is a cyclic cover of genus 2 surface. $\endgroup$
    – Misha
    Dec 19, 2012 at 2:31
  • 1
    $\begingroup$ You are correct, thank you for your comment. In case yanqing wanted an explicit example, yanqing now has one. Having explicit examples can be beneficial. $\endgroup$
    – user30022
    Dec 19, 2012 at 11:59
  • $\begingroup$ great, thank you for your explicit example. $\endgroup$
    – yanqing
    Dec 19, 2012 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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