A Note on the Orderability of Dehn Fillings of the Manifold $v2503$

TL;DR

This paper demonstrates that all Dehn fillings of the manifold v2503 with slopes in (-∞, -1) produce non-orderable spaces, supporting the L-space conjecture.

Contribution

It provides the first comprehensive analysis of the orderability of Dehn fillings for manifold v2503 across a range of slopes.

Findings

Dehn fillings with slopes in (-∞, -1) are non-orderable.

Supports the L-space conjecture for manifold v2503.

All relevant fillings in the specified interval are non-orderable.

Abstract

We show that Dehn filling on the manifold $v2503$ results in a non-orderable space for all rational slopes in the interval $(-\infty , -1)$. This is consistent with the L-space conjecture, which predicts that all fillings will result in a non-orderable space for this manifold.

Equations

$${\pi }_1(M)=⟨a,b|a^2{b}^{-2}a{b}^{-2}{a}^{2}b{a}^{2}bab{a}^{2}b=1⟩$$

$$m$$

$$l$$

$$\mu$$

$$\lambda$$

$$x$$

$$y$$

$$a^2{x}^{2}ab{a}^{2}bab{a}^{2}b$$

$$\mu$$

$$\lambda$$

$${\mu }^{-1}\lambda =\lambda {\mu }^{-1}$$

$$=y{a}^{-1}{y}^2{x}^{-1}$$

$$1$$

$$=a^2{b}^{-2}a{b}^{-1}{\left(a^{-2}{b}^{-1}{a}^{-2}b\right)}^{-1}bab{a}^{2}b$$

$$=a^2{b}^{-2}a{b}^{-1}{\lambda }^{-1}bab{a}^{2}b$$

$$b$$

$$b^2$$

$$x$$

$$x^{2}a{y}^2{a}^{-1}ba{y}^2=1$$

$$x{a}^{-1}{y}^2=1$$

$$x{a}^{-1}{x}^4=1$$

$$a=x^5$$

$$G=⟨a,b,x,y|a=x^5,b=x^2,x=b^{-2}a,y=b{a}^2,b^2=y^2,x=ba{y}^2⟩$$

$$G$$

$$=⟨x|1=x^{20},1=x^{30}⟩$$

$$=⟨x|1=x^{10}⟩$$

$$\cong \mathbf{Z}/10\mathbf{Z}$$

Full text

1 Introduction

This paper studies the orderability of a certain 3-manifold in view of an outstanding conjectured relationship between orderability and L-spaces.

A left-ordering on a group $G$ is a total ordering $\prec$ on the elements of $G$ that is invariant under left-multiplication; that is, $g\prec h$ implies $fg\prec fh$ for all $f,g,h\in G$. A group is said to be left-orderable if it is nontrivial and admits a left ordering. A 3-manifold $M$ is called orderable if $\pi_{1}(M)$ is left-orderable.

If $M$ is a rational homology 3-sphere, then the rank of its Heegaard Floer homology is greater than or equal to the order of its first (integral) homology group. $M$ is called an L-space if equality holds; that is, if \mathrm{rk}\big{(}\widehat{HF}(M)\big{)}=\left|H_{1}(M;\mathbf{Z})\right|.

This work is motivated by the following proposed connection between L-spaces and orderability, first conjectured by Boyer, Gordon, and Watson.

Conjecture 1** ([BGW13]).**

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

In [BGW13], this equivalence was shown to hold for all closed, connected, orientable, geometric three-manifolds that are non-hyperbolic.

If $M$ is a rational homology solid torus, then a framing of the boundary $(\mu,\lambda)$ is called a homological framing for $\partial M$ if $\lambda$ is (rationally) nullhomologous. Given a framing on $\partial M$ and a reduced fraction $\frac{p}{q}\in\mathbf{Q}\cup\{\infty\}$, we denote the $\frac{p}{q}$ Dehn filling by M\big{(}\frac{p}{q}\big{)}.

Culler and Dunfield [CD16] have remarked that the cusped hyperbolic manifold $v2503$ has the property that every non-longitudinal Dehn filling is an L-space (the longitudinal filling is $S^{1}\times S^{2}\#\mathbf{RP}^{3}$). Thus, if Conjecture 1 holds, one would expect none of the Dehn fillings of $v2503$ to be orderable (the longitudinal filling is non-orderable as its fundamental group has torsion). To that end, we prove the following partial result:

Theorem 1**.**

Let $M=v2503$. Then for a certain homological framing, $M(r)$ is not orderable for any rational slope $r\in(-\infty,-1)$.

Acknowledgements

The author would like to thank Professors Zoltán Szabó and Peter Ozsváth for suggesting this problem as well as for providing feedback on drafts of this paper.

2 Ordering

We note the following useful facts, which hold for any left-ordered group $(G,\prec)$:

• •

For each $g\in G$, $1\prec g\Leftrightarrow g^{-1}\prec 1$

• •

For all $a,b\in G$, $1\prec a,b\Rightarrow 1\prec ab$ and similarly $a,b\prec 1\Rightarrow ab\prec 1$.

We also call any element $g$ of $G$ positive whenever $1\prec g$, and similarly, $g$ is said to be negative if $g\prec 1$.

Let $M$ be a compact, connected, oriented irreducible 3-manifold with icompressible torus boundary, and let $(\mu,\lambda)$ be a framing for $\partial M$. In [CW10], Clay and Watson describe a criterion for obstructing left-orderability of Dehn fillings of $M$. One corollary of that criterion is:

Theorem 2** ([CW10]).**

Let $\frac{p}{q},\frac{p_{0}}{q_{0}},\frac{p_{1}}{q_{1}}$ be rational numbers satisfying \frac{p}{q}\in\big{(}\frac{p_{0}}{q_{0}},\frac{p_{1}}{q_{1}}\big{)} such that $q,q_{0},q_{1}>0$ and $p,p_{0},p_{1}<0$. Suppose that $\pi_{1}(\partial M)$ is not sent to 1 by the quotient map \pi_{1}(M)\rightarrow\pi_{1}\big{(}M\big{(}\frac{p}{q}\big{)}\big{)} and that for each left ordering $\prec$ of $\pi_{1}(M)$, $\mu^{p_{0}}\lambda^{q_{0}}\prec 1$ implies $\mu^{p_{1}}\lambda^{q_{1}}\prec 1$. Then \pi_{1}\big{(}M\big{(}\frac{p}{q}\big{)}\big{)} is not left-orderable.

Remark.

This is essentially Corollary 2.2 in [CW10] except in that paper, $p,p_{0},p_{1}$ are all required to be positive; however, their proof works just as well assuming they are all negative instead. Alternatively, one can simply replace $\mu$ with $\mu^{-1}$ and apply their theorem directly, noting that the only necessary property of $\mu$ and $\lambda$ is that they generate $\pi_{1}(\partial M)$. ∎

3 The Manifold $v2503$

Now let us turn our attention to the manifold named $v2503$ in the SnapPy census [CDGW], which we denote $M$ for the rest of this section. It is also known as $M7_{2459}$ in the nomenclature of [CHW99]. $M$ is a hyperbolic 3-manifold with one toroidal cusp, and $M$ is also a rational homology solid torus. Indeed, SnapPy gives that $H_{1}(M;\mathbf{Z})\cong\mathbf{Z}\oplus\mathbf{Z}/10\mathbf{Z}$.

Fundamental Group

According to SnapPy, the fundamental group of $M=v2503$ has the following presentation:

[TABLE]

In addition, SnapPy also gives that the “meridian” $m$ and “longitude” $l$ are:

[TABLE]

We follow the convention of Culler and Dunfield [CD16] for the homological framing. In particular, our homological meridan $\mu$ and homological hongitude $\lambda$ correspond to $(0,1)$ and $(-1,0)$ respectively in SnapPy’s framing . That is:

[TABLE]

Notice that, by considering the abelianisation of (1), the generator $a$ corresponds to a generator of the torsion subgroup of $H_{1}(M;\mathbf{Z})$, whereas $b$ is a free generator. Moreover, $[\mu]=[a]^{2}[b]^{-5}\in H_{1}(M;\mathbf{Z})$ and $[\lambda]=[a]^{-4}$, and so $\lambda$ is rationally nullhomologous, which is consistent with its being a homological longitude.

For convenience, let us put:

[TABLE]

We record for later the following:

[TABLE]

Apart from (9), these are straightforward consequences of (1)–(5). To see why (9) holds, observe that the group relation in (1) can be rewritten as:

[TABLE]

where (3) was used in the last step to substitute for $\lambda$. Now the desired expression follows by isolating $\lambda$ in the equation above.

Orderability constraints for $v2503$

We now use the information about the fundamental group of $v2503$ to prove the following observations, which are the basic ingredients for the proof of the main theorem.

Lemma 1**.**

Let $\prec$ be a left ordering of $\pi_{1}(v2503)$. If $\mu^{-1}\lambda\prec 1$ then $\mu^{-n}\lambda\prec 1$ for all $n\geq 1$.

Proof.

Suppose that $\mu^{-1}\lambda\prec 1$. There are four cases, depending on the signs of the generators $a$ and $b$.

Case I: $b\prec 1\prec a$. In this case, $1\prec\mu$ since, by (2), $\mu$ can be expressed as the product of positive terms. Hence, $\mu^{-1}\prec 1$ and so for each $n\geq 1$, $\mu^{-n}\lambda\prec 1$ as it is the product of negative terms.

Case II: $a,b\prec 1$. Notice that, by (6), it must hold that $1\prec x$ for otherwise, 1 would be expressed as the product of negative terms. Now by (7), we see that $mu$ is the product of positive terms, and hence $1\prec\mu$. As in Case I, we once again have $\mu^{-n}\lambda\prec 1$ for all $n\geq 1$.

Case III: $1\prec a,b$. In this case, we see from (5) that $1\prec y$ as $y$ is the product of positive terms. On the other hand, we have that $x\prec 1$ for otherwise, 1 would be expressed as the product of positive terms in (6). But then, by (10), we see that $\mu^{-1}\lambda$ is expressed as a product of positive terms, contradicting the hypothesis that $\mu^{-1}\lambda\prec 1$. So this case cannot happen.

Case IV: $a\prec 1\prec b$. In (4), we see $x$ expressed as the product of negative terms, and so $x\prec 1$. Now, by ($10^{\prime}$), we conclude that $y\prec 1$ as otherwise, $\mu^{-1}\lambda$ would be the product of positive terms, contradicting the hypothesis that $\mu^{-1}\lambda\prec 1$. Now, by (8), $1\prec\lambda$ because $\lambda$ is expressed as the product of positive terms. The hypothesis that $\mu^{-1}\lambda\prec 1$ implies, by invariance under left-multiplication, that $\lambda\prec\mu$. Hence, $1\prec\mu$, but, from (2) we see $\mu$ as a product of positive elements, a contradiction. So this case, too, cannot happen. ∎

Lemma 2**.**

Let $r\in\mathbf{Q}$. If $\pi_{1}(\partial M)$ is sent to 1 by the quotient map $\pi_{1}(M)\rightarrow\pi_{1}(M(r))$, then $M(r)$ is not orderable.

Proof.

If the subgroup $\pi_{1}(\partial M)$ of $\pi_{1}(M)$ is sent to 1 by the quotient map, then that map factors as: $\pi_{1}(M)\rightarrow\left\langle\pi_{1}(M)|\mu=1,\lambda=1\right\rangle\rightarrow\pi_{1}(M(r))$. Let us examine the group $G=\left\langle\pi_{1}(M)|\mu=1,\lambda=1\right\rangle$. By (7), (8), and (10), we see that the following relations hold in $G$:

[TABLE]

Notice further that (6) can be re-written as

[TABLE]

This becomes, using (11) and (13):

[TABLE]

Using (11) and (12), this becomes:

[TABLE]

Hence, recalling (4) and (5), $G$ has the following presentation:

[TABLE]

This can be simplified to:

[TABLE]

Therefore, as $\pi_{1}(M(r))$ is the quotient of a finite group, it is finite as well, and hence not left-orderable (recall that, by convention, the trivial group is considered not left-orderable). ∎

We are now ready to prove our main result.

Proof of Theorem 1.

Let $r\in\mathbf{Q}\cap(-\infty,-1)$. By Lemma 2, we may assume that $\pi_{1}(\partial M)$ is not sent to 1 by the quotient map $\pi_{1}(M)\rightarrow\pi_{1}(M(r))$. Furthermore, as $M$ is hyperbolic, it is irreducible and has incompressible torus boundary. Then, since $r\in(-n,-1)$ for some integer $n\geq 1$, Lemma 1 together with Theorem 2 tells us that $M(r)$ is not orderable, as required. ∎