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I'm looking for some easy compact, oriented 3-manifolds without boundary that have a nonabelian fundamental group.

It needn't be perfect. "Easy" means that it has an easy Heegard diagram, say, one that could be memorised and drawn by heart. The Poincaré Homology Sphere does not have an easy Heegard diagram, as far as I know, see http://www.its.caltech.edu/~wjiajun/compprog/hfhat/sigma235.html.

All the examples of 3-manifolds I know and I can find are either not closed ($\mathbb{R}^3$ with two $D^2 \times \mathbb{R}$ taken out, $D^2$ being a 2-disk) or have an abelian fundamental group ($S^3$, $\mathbb{RP}^3$ etc.) or are horribly complicated (Poincaré Homology Sphere).

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  • $\begingroup$ Probably I'm missing something, but won't $\mathbb{R}^3$ with two $B^2 \times \mathbb{R}$ taken out, where $B^2$ denotes an open two-dimensional disk, suit you? $\endgroup$
    – lisyarus
    Feb 10, 2015 at 17:16
  • $\begingroup$ As I've written (I'm using the convention $D^2$ for the disk), I need a closed manifold. $\endgroup$
    – Turion
    Feb 10, 2015 at 17:17
  • $\begingroup$ The space I'm talking about is closed - it is $\mathbb{R}^3$ minus two open subspaces. $\endgroup$
    – lisyarus
    Feb 10, 2015 at 17:19
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    $\begingroup$ Sorry, I'm using closed as "compact and without boundary". Of course, every topological space is closed as a subset of itself. But thanks, I've clarified the question. $\endgroup$
    – Turion
    Feb 10, 2015 at 17:21
  • $\begingroup$ Glue two genus-2 handlebodies using the "identity" map of their boundaries. That does work, no? $\endgroup$ Feb 10, 2015 at 18:01

2 Answers 2

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The product of a circle with a closed genus 2 surface $X$ has fundamental group $\mathbb{Z}\times\pi_1(X)$, where $\pi_1(X)$ is nonabelian with presentation $$\langle a,b,c,d | aba^{-1}b^{-1},cdc^{-1}d^{-1}\rangle$$

Edit: Or just take a genus 2 handlebody. This has fundamental group isomorphic to the free group on 2 generators. If I understand Heegaard diagrams correctly, its heegaard diagram should be trivial...

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  • $\begingroup$ What would be a Heegard diagram of $X$? $\endgroup$
    – Turion
    Feb 10, 2015 at 17:32
  • $\begingroup$ I don't know, but it can't be that complicated. You could also just take any genus $> 1$ handlebody. For example a genus 2 handlebody is just the thickening of a bouquet of 2 circles, which has fundamental group the free group on two generators (extremely nonabelian!) $\endgroup$
    – oxeimon
    Feb 10, 2015 at 17:45
  • $\begingroup$ Right, so for example $S^1 \times S^2 \# S^1 \times S^2$? $\endgroup$
    – Turion
    Feb 11, 2015 at 10:35
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Look at pushouts. Any null homologous essential surface in a 3-manifold gives you a decomposition as an amalgamated product $\pi_1(M)= \pi_1(M_1) *_{\pi_1(S)}\pi_1(M_2)$.

In particular you get connected sums $M_1\# M_2$ gives you free products on fundamental group since $S^2$ is simply connected: $\pi_1(M_1\# M_2) = \pi_1(M_1)*\pi_1(M_2)$.

So if you ask for the simplest example you might take your favorite simple non-simply connected closed 3-manifold and take the connected sum with itself.

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    $\begingroup$ That's particularly practical since the connected sum amounts to just taking the disjoint union of the Heegard diagrams. $\endgroup$
    – Turion
    Feb 12, 2015 at 9:17

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