# Incompressible surface

In mathematics, an incompressible surface, in intuitive terms, is a surface, embedded in a 3-manifold, which has been simplified as much as possible while remaining "nontrivial" inside the 3-manifold.

For a precise definition, suppose that S is a compact surface properly embedded in a 3-manifold M. Suppose that D is a disk, also embedded in M, with

$D \cap S = \partial\!D.$

Suppose finally that the curve $\partial\!D$ in S does not bound a disk inside of S. Then D is called a compressing disk for S and we also call S a compressible surface in M. If no such disk exists and S is not the 2-sphere, then we call S incompressible (or geometrically incompressible).

Note that we must exclude the 2-sphere to get any interesting consequences for the 3-manifold. Every 3-manifold has many embedded 2-spheres, and a 2-sphere embedded in a 3-manifold never has a compressing disc.

Sometimes one defines an incompressible sphere to be a 2-sphere in a 3-manifold that does not bound a 3-ball. Thus, such a sphere either does not separate the 3-manifold or gives a nontrivial connected sum decomposition. Since this notion of incompressibility for a sphere is quite different from the above definition for surfaces, often an incompressible sphere is instead referred to as an essential sphere or reducing sphere.

There is also an algebraic version of incompressibility: Suppose $\iota: S \rightarrow M$ is a proper embedding of a compact surface. Then S is $\pi_1$-injective (or algebraically incompressible) if the induced map on fundamental groups

$\iota_\star: \pi_1(S) \rightarrow \pi_1(M)$

is injective.

In general, every $\pi_1$-injective surface is incompressible, but the reverse implication is not always true. For instance, the Lens space $L(4,1)$ contains an incompressible Klein bottle that is not $\pi_1$-injective. However, if $S$ is a two-sided properly embedded, compact surface (not a 2-sphere), the loop theorem implies $S$ is incompressible if and only if it is $\pi_1$-injective.