# 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

## 1 Answer

$\newcommand{\ZZ}{\mathbb{Z}}$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.)

For your second question: As Ryan says, the program Heegaard is our friend here. There is extensive documentation explaining the algorithm. It operates on the Whitehead graph of the given presentation. The algorithm performs a sequence Whitehead moves until it either gives up or it makes the Whitehead graph planar. If there is now a good "pairing" of the vertices then the algorithm has produced a Heegaard splitting of the desired three-manifold.

Finally: see page 47, item C.4 of the book 3-manifold groups for a list of interesting references.

• Tiny nitpick: the fact that 3-manifold groups are residually finite also relies on geometrisation. But Jaco and Shalen were I think able to prove that Baumslag--Solitar groups don't embed into 3-manifold groups pre-geometrisation. In fact, the Kneser--Milnor and sphere theorems together show that torsion subroups of 3-manifold groups split off as free factors, so there's a geometrisation-free proof of your second example too. – HJRW Mar 9 '18 at 9:32
• Dear HJRW - Ah, the sentence "My proofs rely on geometrization..." was referring to both examples. Will fix. Nice to know that there are geometrization-free proofs. May I ask - what is the reference for the Jaco-Shalen result? They have a few papers together... I will have to think about your second remark on the way home. – Sam Nead Mar 9 '18 at 18:43
• I think the reference for Jaco--Shalen's proof that Baumslag--Solitar groups don't arise as subgroups of 3-manifold groups is their 1979 Memoir of the AMS "Seifert fibered spaces in 3-manifolds". But I don't have it in front of me, so I can't confirm. – HJRW Mar 11 '18 at 8:12