5
$\begingroup$

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?

$\endgroup$

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
0

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.