Let $M$ be a closed non-Haken hyperbolic $3$-manifold. Are there two, nonhomotopic, $\pi_1$-injective closed surfaces in $M$, and which do not intersect?
1 Answer
No, this is impossible. Assume that the surfaces $\Sigma_1, \Sigma_2$ are immersed and disjoint realized by immersions $f_i:\Sigma_i\to M$. Take a region $N$ in the complement of these surfaces, and whose boundary intersects both surfaces (such a region must exist if the manifold is connected, which is implicit in your question). Then this submanifold has non-trivial $H_2$, and hence contains an embedded $\pi_1$-injective orientable surface which is homologically non-trivial in $N$ and separating the two immersed surfaces. Any sphere component must bound a ball since hyperbolic manifolds are irreducible, and this ball cannot contain either immersed $\pi_1$-injective surface, hence it will bound a ball in $N$. Hence there must be a component $\Sigma$ of genus at least one which is incompressible in $N$ and still separates the two surfaces. Assume that $\Sigma$ has minimal genus with respect to these properties. Let $D$ be a compressing disk for $\Sigma$, say on the side containing $\Sigma_1$. Then we may homotope $\Sigma_1$ off of $D$. There is an immersion $f_1:\Sigma_1\to M$ such that $f_1$ is transverse to $D$. Then the preimage $f_1^{-1}(D)$ will be curves in $\Sigma_1$ which are homtopically trivial in $D$. Since $f_1$ is $\pi_1$-injective in $M$, these also bound embedded disks in $\Sigma_1$. Take an innermost disk and surger $\Sigma$ along this disk to remove the component of $f_1^{-1}(D)$, which may be accomplished by a homotopy by irreducibility. One may repeat this until $f_1^{-1}(D)=\emptyset$. Then we may compress $\Sigma$ along $D$ to obtain a surface $\Sigma'$ which still separates $f_1(\Sigma_1)$ from $f_2(\Sigma_2)$, a contradiction to the minimality assumption of $\Sigma$. Thus, we may find an incompressible surface in $M$, contradicting the assumption that it was non-Haken. These arguments are fairly standard in classical 3-manifold topology going back at least to Waldhausen, and probably earlier.