Left-orderability for surgeries on twisted torus knots (Q2418812)

The L-space conjecture claims that an irreducible rational homology $3$-sphere is an L-space if and only if its fundamental group is not left-orderable. A knot is called an L-space knot if it admits a positive Dehn surgery yielding an L-space. It is well known that $p/q$-surgery on an L-space knot $K$ yields an L-space if and only if $p/q\ge 2g(K)-1$, where $g(K)$ is the genus of $K$. \par For example, \textit{K. L. Baker} and \textit{A. H. Moore} [J. Math. Soc. Japan 70, No. 1, 95--110 (2018; Zbl 1390.57002)] showed that the $(-2,3,2n+1)$-pretzel knots with $n\ge 3$ and their mirror images are the only hyperbolic L-space knots among Montesinos knots. Recently, \textit{Z. Nie} [Topology Appl. 261, 1--6 (2019; Zbl 1421.57011)] showed that $p/q$-surgery on the $(-2,3,2n+1)$-pretzel knot with $n\ge 3$ yields a $3$-manifold whose fundamental group is not left-orderable if $p/q\ge 2n+3$, and left-orderable if $p/q$ is sufficiently close to $0$. We remark that its genus is $n+2$. \par The purpose of the paper under review is to extend the above result by Nie to a larger family of L-space knots. Let $K$ be the $(n-2)$-twisted $(3,3m+2)$-torus knot, where $m,n\ge 1$. More precisely, $K$ is obtained from the $(3,3m+2)$-torus knot by adding $(n-2)$-full twists on an adjacent pair of strings. Some authors use the notation $K(3,3m+2;2,n-2)$. We remark that $g(K)=n+3m-1$ and that if $m=1$ then $K$ is the $(-2,3,2n+1)$-pretzel knot. \par The main theorem states that $p/q$-surgery on $K$ yields a $3$-manifold whose fundamental group is not left-orderable if $p/q\ge 2g(K)-1$, and left-orderable if $p/q$ is sufficiently close to $0$. For the first part, the argument is the same as the one in [loc. cit.]. Using the presentation of the fundamental group $G$ of the resulting manifold by Dehn surgery, the assumption that $G$ is left-orderable implies the existence of a monomorphism from $G$ to $\mathrm{Homeo}^+(\mathbb{R})$ without global fixed point. This leads to a contradiction. For the second part, the author shows that the Alexander polynomial of $K$ has a simple root on the unit circle by a calculation. Then the conclusion follows from the result by \textit{M. Culler} and \textit{N. M. Dunfield} [Geom. Topol. 22, No. 3, 1405--1457 (2018; Zbl 1392.57012)] and \textit{C. Herald} and \textit{X. Zhang} [Proc. Am. Math. Soc. 147, No. 7, 2815--2819 (2019; Zbl 07073437)].