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I'm trying to see that a compact 3-manifold $M\subset\mathbb{R}^3$ with $H_1(M)=0$ has $\pi_1(M)=0$.

So far: since $M$ is compact, let $\mathcal{T}$ be a smooth triangulation of $M$. I want to be able to say that we can push any loop $\gamma\subset M$ into the 1-skeleton of this triangulation, I'm pretty sure this is possible. From there, what does $H_1(M)=0$ do for us? Does it tell us that $\gamma$ bounds some collection of $2-$simplices?

Thanks

J. Moeller
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    The $H_1=0$ condition tells you that $\pi_1(M)$ is generated by commutators. In other words it is a perfect group. – Arnaud Mortier Jan 31 '18 at 15:21
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    To address your comments, you can indeed push any loop into the $1$-skeleton. This is a special case of the Simplicial Approximation Theorem. And yes you are right, the loop will bound a collection of simplices. – Cheerful Parsnip Jan 31 '18 at 20:16

2 Answers2

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First observe that $M$ cannot be a closed manifold[Alexander Duality is an obstruction for that]. Also observe that $M$ is orientable.[the same duality works again] Now claim is that all it's boundary components are sphere. Observe that $rk(H_1(M,\mathbb Z))\geq rk(H_1(\partial M)/2$ [ basically half of the boundary homology dies]. So $H_1=0$ forces that $H_1(\partial M)=0$. By classification of surfaces, it can only be $S^2$ (since it has a induced orientation from $M$]. Now use higher dimension Jordan separation theorem. Since $M$ is in $\mathbb R^3$, all the boundary sphere should lie inside one of the componenet, called it as outer most component. Since as we observe that $M$ is orientable and compact, so either $M$ is $D^3$ or $D^3-$balls removed. In both the cases $\pi_1=0$. Hence complete the proof.

Observe that if $M$ is non-compact, then it is not true. In Hatcher you can see that unbounded open region of the complement of Alaxander horned sphere is a counter example.

Anubhav Mukherjee
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  • How can one use Alexander Duality to see that M cant be closed and has to be orientable? – Womm Jul 15 '24 at 12:12
  • @Womm Honestly, I do not remember what exactly I had in my mind when I wrote this. But there are many ways one can prove this claim (maybe think on your own before seeing it here https://math.stackexchange.com/questions/210408/embedding-compact-boundaryless-n-manifolds-in-n-dimensional-real-space). And codim-0 submanifold of an orientable manifold is orientable (as you consider induced smooth structure). Sorry for me being dumb but sometimes I am very surprised to see my old answers which I probably cannot prove if I am asked today hahahaaha – Anubhav Mukherjee Jul 16 '24 at 17:33
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If you don't ask for $M$ to be embedded in $\mathbb{R}^3$ the statement is not true. The Poincare homology $3$-sphere is a counterexample: its homology is by definition that of a $3$-sphere (therefore its $H_1$ trivial), but its $\pi_1$ is non-trivial.

Arnaud Mortier
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