Steve Boyer, Cameron Gordon, and Liam Watson have an interesting new preprint out today on the arXiv. In it, they posit:

**Conjecture.** *An irreducible rational homology 3-sphere is an L-space if and only if its fundamental group is not left-orderable.*

The motivation here is as follows: An L-space is something whose Heegaard Floer homology is as simple as possible; such 3-manifolds have no taut foliations. A nice type of taut foliation are those that are **R**-covered, and in this case, the fundamental group of the 3-manifold inherits a left-order from the action of the leaf space. (I’m always assuming here that foliations are co-orientable.)

Of course, it’s not known whether every non-L-space has a taut foliation, and there are certainly non-**R**-covered foliations, so a reasonable initial reaction is that this conjecture isn’t very plausible. However, their paper outlines a surprising amount of evidence for it, and in this post I’ll give some more data in that direction.

For instance, Danny and I gave a list of 44 small hyperbolic 3-manifolds with non-left-orderable fundamental groups; it turns out all of these are indeed L-spaces.

More systematically, some years ago I ran some experiments on the 11,031 small hyperbolic 3-manifolds in the Hodgson-Weeks census and found:

- There are at least 205 that have left-orderable fundamental groups. These were certified by reps to PSL(2,
**R**) whose euler class vanishes. Hence the rep lifts to an action on the line.
- At least 3038 (27%) of the census manifolds are L-spaces. These were certified using the exact triangle, starting with lens spaces and other elliptic manifolds as known examples of L-spaces.

Surprisingly, consistent with the conjecture, these two sets were disjoint, and the naive odds of this happening is 10^{-29}!

Now another sort of order associated to every taut foliation is a circular order coming from its universal circle. Thus one could also ask:

**Conjecture.** *An irreducible rational homology 3-sphere is an L-space if and only if its fundamental group is not circularly-orderable.*

However, this conjecture is false; I found some 256 L-spaces in the above sample that had irreducible reps to PSL(2, **R**), and hence circular orders.

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Can you say a little more on what probability assumption you used to get 10^(-29)?

Comment by Mayer A. Landau — July 26, 2011 @ 10:04 pm |

Take a set of 11031 elements, and pick two subsets at random, one of size 205 and the other of size 3038. The odds that they are disjoint is 10^(-29). For instance, view the larger subset as fixed so that the probability that a randomly chosen element is not in that set is 0.73. The probability all elements in the smaller set are not in the larger one is thus about 0.73^205 = 10^(-28).

Comment by Nathan Dunfield — July 26, 2011 @ 10:54 pm |

If M1 and M2 are manifolds with left orderable fundamental group, is the fundamental group of the connected sum left orderable? How about for two manifolds glued along boundary tori?

Comment by Mayer A. Landau — July 27, 2011 @ 8:25 pm |

Yes and no respectively. The first corresponds to the fact that a free product of left-orderable groups is again left-orderable. For the second, the group Z is LO but Z/nZ is not, so think of a lens space as the result of gluing together two solid tori.

Comment by Nathan Dunfield — July 27, 2011 @ 8:44 pm |

What about gluing along irreducible boundary tori?

Comment by Henry Wilton — August 24, 2011 @ 5:59 am

Henry,

At least some of the time, you can order the group of such an amalgam, but I’m not sure if this is true in general. See this paper for more.

Comment by Nathan Dunfield — August 24, 2011 @ 11:34 am

This is cool! Do you still have the data – e.g. can you say what some of the simplest of the hyperbolic L-spaces were, and perhaps some of the simplest of the L-spaces that also had a circular ordering? How close is the relationship between Rep(pi_1, PSL(2,R)) and orderability – do you know some examples which are left-orderable, but have no PSL(2,R) reps with non-vanishing euler class?

Comment by lewallen — July 31, 2011 @ 9:35 am |

Sure, I still have the data — I never throw anything electronic away. Here’s the first few L-spaces:

m003(-3,1) # The Weeks manifold

m007(3,1)

m003(-4,3)

m003(-4,1)

m003(-5,3)

m007(1,2)

m007(4,1)

m007(3,2)

Here’s the first few L-spaces with a rep to PSL(2, R):

m006(4,1)

m016(3,2)

m017(1,3)

m011(-2,3)

m011(4,3)

m017(-5,1)

m016(-5,1)

m016(-1,4)

Here, m016 is the exterior of the (-2, 3, 7) pretzel knot with some non-standard framing on boundary torus.

——-

I’m in the process of redoing all these calculations. I did them a while ago, and hopefully my shiny 12-core Mac Pro will be able to find some additional reps.

Comment by Nathan Dunfield — August 1, 2011 @ 7:06 am |