One Powell generator is redundant

TL;DR

This paper demonstrates that one of Powell's proposed generators for the Goeritz group of certain 3-manifold splittings is redundant, simplifying the understanding of the group's generating set.

Contribution

It provides a short proof that one of Powell's five generators is unnecessary, reducing the generating set for the Goeritz group.

Findings

One generator is redundant in Powell's generating set.

The redundancy is a consequence of three other generators.

Simplifies the structure of the Goeritz group for genus g ≥ 4.

Abstract

In 1980 J. Powell proposed that five specific elements sufficed to generate the Goeritz group of any Heegaard splitting of $S^3$. This conjecture remains unresolved for genus $g \geq 4$. Here a short argument shows that one of his proposed generators is redundant, in fact a consequence of three of the other four.

Full text

I am grateful to Michael Freedman for many inspiring and helpful conversations on various aspects of the Powell Conjecture.

Powell proposed generators for the genus $g$ Goeritz group, $g\geq 3$ as shown in Figure 1.

It was noted in [FS, Lemma 1.4] that the first standard genus $1$ summand of the standardly embedded genus $g$ Heegaard surface of $S^{3}$ can be passed around a longitude of the second standard summand by the use of Powell’s generators. Since Powell’s element $D_{\nu}$ similarly passes the first summand around the meridian of the second generator, it followed almost immediately that an isotopy of any standard summand around its complementary surface, and back to its initial position, is a product of Powell’s generators. What was not noticed there, and is shown here, is that a version of the same argument, appropriately dualised, shows that Powell’s $D_{\nu}$ is itself also a consequence of the other generators, so that Powell’s actual conjecture would imply that only four generators suffice.

The proof is mostly a sequence of pictures, modeled on those underlying the proof of [FS, Lemma 1.4]. It will be convenient to display how to create not $D_{\nu}$ itself, but rather the conjugate of $D_{\nu}$ by generator $D_{\eta_{12}}$, which we will denote $D^{\prime}_{\nu}$. This is the move which isotopes standard summand $H_{2}$ around a meridian of $H_{1}$, instead of vice versa and is illustrated in Figure 2.

The first image shows the meridians and longitudes of the two standard summands. The second image shows in blue the trajectory through which the top standard summand is moved with respect to the right summand: it is passed behind and then around the meridian of the right summand. The effect on the meridians and the longitudes of this isotopy is shown in the third image.

Note in Figure 1 that the Powell generator $D_{\omega}$ can be described as a half-twist of the right summand shown in the images in Figure 2. The conjugate $D_{\eta_{{}_{12}}}D_{\omega}D^{-1}_{\eta_{{}_{12}}}$ is then the half-twist of the top summand. Denote this move by $D^{\prime}_{\omega}$. Recall also from Figure 1 that $D_{\theta}$ denotes the move that slides the right handle over the top handle.

The top row shows in blue, in the first image, the trajectory of $D^{-1}_{\theta}$ and then in the next image the result of this action on the meridians and longitudes. The further effect of $D^{\prime}_{\omega}$ is shown in the last image in the row. The second row repeats the process for $D^{-1}_{\theta}$ again, and then $D^{\prime-1}_{\omega}$. ∎