# If $i(x,z)\neq 0$ and if $y$ is conjugate of $x$, then what can we say about $i(x*y,z)$?

Let $S_g$ denote the closed oriented surface of genus $g\geq 2$. Let $x,y$ be two different (upto fixed base point homotopy) but freely homotopic curves, i.e. $y$ is a non-trivial element from a fixed conjugacy class of $x$. If there exist a simple closed loop $z$ (non-trivial) with $i(x,z)\neq 0$ , where $i(\ , \ )$ denotes the geometric intersection number. Then is it true that $i(x*y,z)\neq 0$?

One can prove that if $y=x$ (upto fixed base point homotopy) , then $i(x,z)\neq 0$ implies $i(x^2,z)\neq 0$ , because the hyperbolic axis correspond to the translation $[x],[x^2]\in \pi_1(S_g)$ in $\mathbb H^2$ are same. Also the reason I've chosen $z$ to be a simple closed loop, becasue I think $x*y$ cannot be a simple closed loop here.

I'm unable to find some proof/counter-example. Although my intuition says that it should be true but I would be happy if some one give me a counter-example with a good (general) explanation.

• I could not able to think about a suitable title. So please feel free to edit the title. – Anubhav Mukherjee Jan 3 '17 at 18:14
• What does $x\ast y$ mean? – skd Jan 3 '17 at 19:22
• – Anubhav Mukherjee Jan 3 '17 at 19:26
• Oh, ok. I thought you meant something else. – skd Jan 3 '17 at 19:29
• This question is closely related to this one: mathoverflow.net/q/258189/1345 – Ian Agol Jan 5 '17 at 22:31

If $x$ is simple, then this is true. Here is an outline of the proof.

1) As $x$ and $y$ are freely homotopic, $y=x^g$ for some $g$. As $x$ is simple, $A_x$ and $A_y$ are disjoint.

2) Consider the Theorem 7.38.3 of Beardon's "Geometry of discrete group." Observe that $g$ is the common perpendicular, therefore $\epsilon =+1$.

3) This implies if length of $x$ increases, so does length of $x*y$ (in Teichnuller space).

4) Take right twists along the curve $z$. This will increase the length of $x$ (as $i(x,z)\neq 0$) and hence must increase the length of $x*y$ which implies $i(x*y,z)\neq=0$.

As long as $A_x$ and $A_y$ are disjoint, this proof will work. The other case is when $A_x\cap A_y\neq \emptyset.$ Then $x$ is not simple and their product formula is given by Theorem 7.38.3 of Beardon's "Geometry of discrete group," where $v_2$ will be a lift of the self-intersection point. Now there are two cases:

1) If $cos\theta$ is positive at $v_2$ then the above arguments hold true.

2) If $cos\theta$ is negative, it is the same question as thismentioned by Ian Agol for the following reason. Take the self intersection point and consider the two branches of the curves starting and ending at this intersection point. Name them $x$ and $y$. If $y$ is a power of $x$ then you are done by your observation. If not then you get the case $n=1$ of the above question.

• I think you have made some mistake. I've posted a coubter example. Please have a look. – Anubhav Mukherjee Oct 25 '17 at 20:23 This is a counter exampleb of my question. (proposed by Dan Margalit)

• What is the curve $z$ in your pic? – Cusp Oct 27 '17 at 8:12
• Take a curve in the 3 rd hole away from one curve – Anubhav Mukherjee Oct 27 '17 at 12:54
• Sorry I am a bit confused. If you take the third hole (from left) and consider any curve around it as $z$ then $i(x,z)=0$. – Cusp Oct 27 '17 at 13:53
• @cusp take something transversal to the loop. And that own intersect $y$ . – Anubhav Mukherjee Oct 27 '17 at 15:25
• Any loop in and around the third hole has geometric intersection number 0 with $x$ and hence with $y$. It would be helpful if you update the pic with more details. – Cusp Oct 28 '17 at 5:04