# Obstructions to realizing a balanced presentation as a 3-manifold group

I am thinking of 3-manifolds as arising from Heegaard splittings which I am thinking about in terms of Heegaard diagrams. I know that 3-manifold groups are rather special in the class of all finitely presentable groups.

I know how to go from a Heegaard diagram ($g$ red curves and $g$ blue curves each forming a cut system on a genus $g$ closed orientable surface) to a presentation for $\pi_1$ - namely I have $g$-generators coming from the red handlebody $g$ relations coming from the disks attached along the blue curves. To find the relations I just orient all of the curves and the surface and for each blue curve I read off the word in the red curves (with the signs of the intersections determining the sign of the words).

I would like to know some explicit balanced group presentations that I can not get in this way - and I would like to see why I can't embed the curves if possible. Of course $< a ,b | [a,b] , [a,b]^2>$ for example fits the bill, since $\mathbb{Z}^2$ is not the fundamental group of a compact 3-manifold. But I would like a more "curves on surfaces" type explanation.

Ideally, I would like some explicit necessary and/or sufficient condition for a word or collection of words to be realizable by embedded curves as above.

Thanks!

• Have you looked at the software "Heegaard" ? It's quite effective at turning presentations of 3-manifold fundamental groups into actual 3-manifolds. I believe there is a rough description of what it does, by Googling-around and likely also in some Ian Agol posts, here on this forum. The strongest obstructions I suppose would be in terms of geometrization but you border on being kind of tautological if you go in that direction: free product decomp, amalgamated free product decomp, then there's a list of finite groups that arise, and a less compact list of infinite groups. . . – Ryan Budney Mar 8 '18 at 3:08

$\newcommand{\ZZ}{\mathbb{Z}}$For your first question: the Baumslag-Solitar group $BS = BS(2,3)$ is not residually finite, so cannot embed in a three-manifold group. Thus $BS \times \ZZ$ has a balanced presentation, but is not a three-manifold group. A nice two-generator example is $\ZZ/3\ZZ \times \ZZ$. (For both examples, my proofs rely on geometrization. But see HJRW's comments below.)