# Geometric intersection with incompressible surfaces

Let $M$ be a oriented compact $3$-manifold, closed or with boundary. For any incompressible surface $F$, define a function $i_F$ on the set of homotopy classes of closed curves in $M$ by $$i_F (\alpha) = \alpha \cap F$$ the geometric intersection number of $\alpha$ with $F$.

Is it true that two incompressible, $\partial$-incompressible surfaces $F$ and $F'$ are isotopic if $i_F =i_{F'}$?

• This is true in the case of closed surfaces, for the ones with boundary you probably need to refine the intersection number definition. Mar 18 '15 at 1:25

Yes, this is true (with an appropriate definition of $i_F(\alpha)$).
An incompressible surface gives rise to an action of $\pi_1(M)$ on a tree (see for example Chapter 1 of Shalen's notes). For a homotopy class of loops $\alpha$, define $i_F(\alpha)=\|\alpha\|$ to be the translation length of $\alpha$ acting on this tree (well-defined up to conjugacy). In fact, $\alpha$ can be homotoped so that the geometric intersection number is $\|\alpha\|$. Then this defines a length function on conjugacy classes in $\pi_1(M)$, satisfying the Culler-Morgan axioms (1.11): Culler and Morgan show that their axioms uniquely characterize a minimal action of a group on a tree, if the action is semi-simple. It's not hard to see that if the surface is not a fiber or semi-fiber, that the action is semi-simple, essentially from the cocompactness of the action. In the fiber or semi-fiber case, the intersection number determines a homomorphism to $\mathbb{Z}$ or $D_\infty$, whose kernel is finitely generated and determines the surface group. In the semifibered case, I don't think $I_F(\alpha)$ distinguishes the two non-orientable surfaces. However, your assumption of incompressible surface usually means the surface is orientable, and thus this case won't appear.