# Reduction of self-intersections without reducing the geometric intersection

Let $F$ be a hyperbolic surface. Given a closed curve $a$, let $\bar{a}$ denotes the free homotopy class of $a$. Let $i(\bar{a},\bar{b})$ denotes the geometric intersection number and $i(\bar{a})$ denotes the self intersection number.

Suppose $a$ and $b$ are two given closed curves in $F$ with the following properties:

1) $a$ is a simple closed curve.

2) $i(\bar{b})=k\neq 0$.

3) $i(\bar{a},\bar{b})=m\neq 0$.

Q) Does there exists a cover $\widetilde{F}$ of $F$ such that:

1) $a$ and $b$ lifts to closed curves $\alpha$ and $\beta$ respectively in $\widetilde{F}$.

2) $i(\bar{\beta})<k$.

3) $i(\bar{\alpha},\bar{\beta})=m$.

In other words, can we lift a pair of curves (one of which is simple) to a cover without reducing the geometric intersection number but reducing the self intersection number of one of them.

PS 1: I have tried LERF property and double coset separabilty of surface groups, but was unable to prove it.

PS 1: Any reference of any kind will be helpful. Thanks in advance.

I don't think this is possible in general. The point is that if $i(\alpha, \beta)= m$, then all the arcs of $b-a$ lift uniquely to $\tilde{F}$ to arcs of $\beta-\alpha$. In fact, the graph consisting of the union of $a$ and $b$ along their points of intersection lifts from $F$ to $\tilde{F}$. One may then close these arcs to closed (basepointed) loops which also must lift. But choosing $b$ complicated enough, one may ensure that these loops generate the fundamental group, so no cover is possible. I give an example in the figure, where the closed-off arcs of $b$ give standard generators for the fundamental group of the surface.
Maybe it's easier to understand the case that $b$ is embedded. Consider the graph $a\cup b$ in the following diagram:
If there is a cover $\tilde{F}\to F$ that $a$ and $b$ lift to so that the intersection number is the same, then the entire graph $a\cup b$ lifts. But $\pi_1(a\cup b)\twoheadrightarrow \pi_1(F)$ surjects, so $\pi_1(\tilde{F})\twoheadrightarrow \pi_1(F)$ surjects, i.e. is a trivial (connected) cover.
• @Cusp : I think it's a bit easier to think about the graph when $b$ is embedded. I've added an explanation in this case to clarify, and one may forget about lifting closed arcs unless it helps. – Ian Agol Apr 24 '15 at 4:32
• @agol sorry but I am confused now. My question is valid only when $b$ is not embedded. But do you mean for the same reason as embedded the lift is possible? – Cusp Apr 24 '15 at 6:09