# One-sided incompressible surface in 3-manifolds

Let $$M^3$$ be a closed orientable $$3$$-manifold. If $$H_2(M,\mathbb Z)=0$$ and $$H_2(M, \mathbb Z_2)\ne 0$$, can we show that $$M$$ contains a 1-sided incompressible surface?

• What is the definition of 1-sided incompressible surface? Is $P^2\times \{x\}$ incompressible inside $P^2 \times S^1$? Aug 24 at 9:19
• @BrunoMartelli A surface is 1-sided if its normal bundle in the ambient 3-manifold is not trivial. $P^2 \times {x}$ is incompressible in $P^2 \times S^1$. Aug 24 at 10:14

This follows from two exercises.

Exercise 1. Every class in $$H_2(M, \mathbb{Z}_2)$$ is represented by a smooth embedded surface. (See Exercise 4.5.12(b) in "4-Manifolds and Kirby Calculus" by Gompf and Stipsicz.)

Suppose that the surface is $$S \subset M$$. If $$S$$ is orientable, then it bounds modulo $$\mathbb{Z}$$ and thus modulo $$\mathbb{Z}_2$$, a contradiction. Thus $$S$$ is not orientable. Since $$M$$ is orientable, any such surface is one-sided. We now "compress".

Exercise 2. Compressing a surface does not change the homology class represented. If the compression separates, then at least one component will be non-trivial in $$H_2(M, \mathbb{Z}_2)$$ and so will again be one-sided. (See the proof of Lemma 6.3 of "3-manifolds" by Hempel for a very similar technique.)

Thus we eventually arrive at the desired one-sided incompressible surface.

• Non-orientable does not imply odd Euler characteristic, but you don't need that to deduce that it is 1-sided. On the other hand, I am a bit confused on the definition of incompressible surface in the 1-sided setting. Aug 24 at 9:17
• I see. The surface $S$ obtained in Ex.1 is nonorientable and 1-sided since $M$ is orientable. If $S=P^2$, then we are done by Ex.2. If $S$ is compressible, one can do surgery along the compressing disk to obtain a new surface $S'$ with $\chi(S') \ge \chi(S)+1$. Eventually one obtains an incompressible surface. Aug 24 at 10:23
• A compression can turn a 1-sided surface into a 2-sided surface, eg. think about the Klein bottle in $S^2\times S^1$. Aug 24 at 16:57
• Here $S$ is non-separating. After compressing as much as possible, some component will be incompressible and non-separating. That component cannot be 2-sided since $H_2(M;\mathbb{Z})=0$. Aug 24 at 16:59
• @JoshHowie - Ok, this time I was not wrong, exactly... I have clarified. Aug 24 at 17:29