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Sam Nead
Sam Nead
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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 (nor. If the number of sides)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.

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 (nor the number of sides).  (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.

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.

fixed bug pointed out by Bruno.
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Sam Nead
Sam Nead
  • 28.2k
  • 5
  • 72
  • 131

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. Thus $S$ has odd Euler characteristic. Since $M$ is orientable, any such surface (with odd Euler characteristic) is one-sided. We now "compress".

Exercise 2. If $M$ containsCompressing a surface does not change the homology class represented (nor the number of odd Euler characteristic, then any such with maximal Euler characteristic is incompressiblesides).   (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.

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. Thus $S$ has odd Euler characteristic. Since $M$ is orientable, any such surface (with odd Euler characteristic) is one-sided.

Exercise 2. If $M$ contains a surface of odd Euler characteristic, then any such with maximal Euler characteristic is incompressible. (See Lemma 6.3 of "3-manifolds" by Hempel.)

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 (nor the number of sides).   (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.

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Sam Nead
Sam Nead
  • 28.2k
  • 5
  • 72
  • 131

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. Thus $S$ has odd Euler characteristic. Since $M$ is orientable, any such surface (with odd Euler characteristic) is one-sided.

Exercise 2. If $M$ contains a surface of odd Euler characteristic, then any such with maximal Euler characteristic is incompressible. (See Lemma 6.3 of "3-manifolds" by Hempel.)