# Powell moves and the Goeritz group

Michael Freedman  Microsoft Station Q
University of California
Santa Barbara, CA 93106-6105
and
Mathematics Department
University of California
Santa Barbara, CA 93106-3080 USA<EMAIL_ADDRESS>and  Martin
Scharlemann  Martin Scharlemann
Mathematics Department
University of California
Santa Barbara, CA 93106-3080 USA<EMAIL_ADDRESS>
###### Abstract.

In 1980 J. Powell [Po] proposed that five specific elements sufficed to
generate the Goeritz group of any Heegaard splitting of $S^{3}$, extending
work of Goeritz [Go] on genus $2$ splittings. Here we prove that Powell’s
conjecture was correct for splittings of genus $3$ as well, and discuss a
framework for deciding the truth of the conjecture for higher genus
splittings.

Following early work of Goeritz [Go], the genus $g$ Goeritz group of the
$3$-sphere can be described as the isotopy classes of orientation-preserving
homeomorphisms of the $3$-sphere that leave the standard genus $g$ Heegaard
surface $T_{g}$ invariant. Goeritz identified a finite set of generators for
the genus $2$ Goeritz group; that work has been recently updated, extended and
completed, to give a full picture of the group (see [Sc], [Ak], [Cho]). In
1980 J. Powell [Po] extended Goeritz’ set of generators to a set of five
elements that he believed would generate the Goeritz group for any fixed
higher-genus splitting. Unfortunately, his proof that these suffice contained
a serious gap; here we prove that they do suffice for genus $3$ splittings,
using largely techniques that were unknown in 1980, and introduce methods that
could be helpful in deciding the conjecture for higher genus splittings.

Powell’s actual viewpoint on the Goeritz group, which we will adopt, is framed
somewhat differently. Following Johnson-McCullough [JM] (who extend the notion
to arbitrary compact orientable manifolds) consider the space of left cosets
$\mathop{\rm Diff}\nolimits(S^{3})/\mathop{\rm Diff}\nolimits(S^{3},T_{g})$,
where $\mathop{\rm Diff}\nolimits(S^{3},T_{g})$ consists of those orientation-
preserving diffeomorphisms of $S^{3}$ that carry $T_{g}$ to itself. The
fundamental group $P_{g}$ of this space projects to the genus $g$ Goeritz
group (with kernel $\pi_{1}(SO(3))=\mathbb{Z}_{2}$ [Po, p.197]) as follows: A
non-trivial element is represented by an isotopy of $T_{g}$ in $S^{3}$ that
begins with the identity and ends with a diffeomorphism of $S^{3}$ that takes
$T_{g}$ to itself; this diffeomorphism of the pair $(S^{3},T_{g})$ represents
an element of the Goeritz group as defined earlier. The advantage of this
point of view111 The $Z_{2}$ distinction (unimportant for our purposes) is
best illustrated by Powell’s generator $D_{\eta}$ in Figure 1. It has order
$g$ when viewed as a diffeomorphism $(S^{3},T_{g})\to(S^{3},T_{g})$ but is of
order $2g$ when viewed as an isotopy of $S^{3}$ to itself. is that an element
of the Goeritz group can be viewed quite vividly: it is a sort of excursion of
$T_{g}$ in $S^{3}$ that begins and ends with the standard picture of
$T_{g}\subset S^{3}$. These excursions are what are pictured by Powell in
Figure 1.

Figure 1. Powell’s proposed generators (from [Po])

The study of motion groups of three dimensional objects has a deep connection
to the emerging study of $3+1-D$ topological quantum field theory (TQFT) and
also to four dimensional questions such as the smooth Schoenflies problem. One
may think of $P_{g}$ as a higher dimensional analog of the braid group
$B_{n}$, the motion group of $n$-points in the plane. In the last 50 years
$B_{n}$ has been closely studied with its unitary ($2+1-D$ TQFT)
representations becoming the main focus in the last 30 years. With increasing
interest in $3+1-D$ TQFTs, $P_{g}$ and its unitary representations are a
natural target for condensed matter theorists.

We would like to thank Alex Zupan for helpful conversations; his upcoming [Zu]
interleaves the Powell Conjecture with the question of the connectivity of the
complexes of reducing spheres that arise from Heegaard splittings of $S^{3}$.

##  1\. Composing Powell generators

Let $T_{g}\subset S^{3}$ be the standard genus $g$ Heegaard surface in
$S^{3}$, dividing $S^{3}$ into the genus $g$ handlebodies $A_{g}$ and $B_{g}$.

######  Definition 1.1.

Any finite composition of Powell generators, illustrated in Figure 1, will be
called a Powell move.

Suppose $T\subset S^{3}$ is a genus $g$ Heegaard surface, and
$h,h^{\prime}:(S^{3},T)\to(S^{3},T_{g})$ are two orientation-preserving
homeomorphisms.

######  Definition 1.2.

$h,h^{\prime}$ are Powell equivalent if $h^{\prime}h^{-1}:T_{g}\to T_{g}$ is
isotopic in $T_{g}$ to a Powell move.

######  Conjecture 1.3 (Powell).

Any $h,h^{\prime}$ are Powell equivalent.

The genus two case $g=2$ is Goeritz’s Theorem.

Powell’s description of $T_{g}\subset S^{3}$ begins with a round $2$-sphere in
$S^{3}$, to which is then connect-summed at each of $g$ points a standard
unknotted torus. These summands will be called the standard genus $1$
summands; their complement is a $g$-punctured sphere. The two generators
$D_{\eta}$ and $D_{\eta_{12}}$ show that any braid move of the standard genus
$1$ summands around the complementary $g$-punctured sphere is a Powell move.
It follows that there will be no need to keep track of the labels of the
standard summands. Furthermore, $D_{\omega}$ allows us not to worry about the
orientation of the longitude of the $1$-handle within each summand.

Example: Suppose $S^{3}=A\cup_{T}\cup B$ is a genus $g$ Heegaard in $S^{3}$
and $\\{a_{1},...,a_{g}\\}$ and $\\{b_{1},...,b_{g}\\}$ are orthogonal sets of
disks in $A$ and $B$ respectively (orthogonal means $|a_{i}\cap
b_{j}|=\delta_{i,j})$. There is an obvious orientation preserving
homeomorphism of $S^{3}$ that carries $T$ to $T_{g}$ and each pair
$\mbox{$\partial$}a_{i},\mbox{$\partial$}b_{i}$ to the meridian and longitude
respectively of one of the standard summands of $T_{g}$. It follows from the
comments above that any two such homeomorphisms are Powell equivalent.

More broadly, but in a similar spirit we have:

######  Lemma 1.4.

Any braid move of a collection of standard genus $1$ summands over their
complementary surface is a Powell move.

Figure 1 shows an example in the case of two summands.

###### Proof.

Let $\mathcal{S}$ be the collection of standard genus $1$ summands which we
are isotoping. From the brief discussion above it suffices to assume that the
first Powell standard solid handle $H_{1}$ is in $\mathcal{S}$ and $H_{2}$ is
not, and then to show that any braid move of $H_{1}$ around the punctured
torus $T\cap H_{2}$ is a Powell move. The composition
$D_{\theta}D_{\omega}D_{\theta}D_{\omega}$ isotopes $H_{1}$ around the
longitude of $H_{2}$. The generator $D_{\nu}$ isotopes $H_{1}$ around the
meridian of $H_{2}$. But any braid move of a point in a punctured torus is a
composition of such circuits around the meridian and around the longitude. ∎

Let $\\{c_{1},...,c_{g-1}\\}$ be the disjoint separating circles on $T_{g}$
shown in Figure 2, with each $c_{i}$ separating the first $i$ standard
summands from the last $g-i$ standard summands. Note that each $c_{i}$ bounds
a disk in both $A$ and $B$ and so defines a reducing sphere $S_{i}$ for
$T_{g}$.

2pt $c_{1}$ at 110 140 $c_{2}$ at 167 133 $c_{g-1}$ at 240 135

Figure 2.

Extension 1: Suppose $S^{3}=A\cup_{T}\cup B$ is a genus $g$ Heegaard surface
in $S^{3}$ as above. Let $g_{1}+g_{2}=g$. Suppose $\\{a_{1},...,a_{g_{1}}\\}$
and $\\{b_{1},...,b_{g_{1}}\\}$ are orthogonal sets of disks in $A$ and $B$
respectively. There are (many) orientation preserving homeomormorphisms of
$S^{3}$ that take $T$ to $T_{g}$, and each
$\mbox{$\partial$}a_{i},\mbox{$\partial$}b_{i}$ to the meridian and longitude
respectively of one of the first $g_{1}$ standard summands of $T_{g}$.

######  Lemma 1.5.

If the Powell Conjecture is true for genus $g_{2}$ splittings of $S^{3}$ then
all such homeomorphisms $T\to T_{g}$ are Powell equivalent.

###### Proof.

Let $h,h^{\prime}:(S^{3},T)\to(S^{3},T_{g})$ be two such homeomorphisms. Let
$c=h^{\prime}h^{-1}(c_{g_{1}})\subset T_{g}$. Then $c$, like $c_{g_{1}}$,
separates $T_{g}$ into a genus $g_{1}$ surface containing the first $g_{1}$
handles and a genus $g_{2}$ surface disjoint from them. Also $c$, like
$c_{g_{1}}$, bounds disks in both $A$ and $B$.

If we cut off the first $g_{1}$ handles from $T_{g}$ then $c$ and $c_{g}$ both
bound disks, and so can be isotoped to coincide, in the resulting genus
$g_{2}$ surface. It follows then from Lemma 1.4 that a Powell move will carry
the first $g_{1}$ $1$-handles to themselves and take $c$ to $c_{g_{1}}$. It
follows that we may assume that $h^{\prime}h^{-1}(c_{g_{1}})$ takes
$c_{g_{1}}$ (and so the reducing sphere in which it lies) to itself. The lemma
then follows by operating separately on the genus $g_{1}$ and $g_{2}$ Heegaard
splittings determined by $S_{g_{1}}$. ∎

######  Definition 1.6.

Let $M=A\cup B$ be a Heegaard splitting of a compact manifold $M$. A
collection $\\{a_{1},...,a_{m}\\}$ of disjoint disks in $A$ is primitive if
there is a collection of disjoint disks $\\{b_{1},...,b_{m}\\}$ in $B$ so that
for every $i,j$, $|a_{i}\cap b_{j}|=\delta_{ij}$.

Extension 2: For $S^{3}=A\cup_{T}\cup B$ as above, suppose
$\\{a_{1},...,a_{g}\\}$ is a collection of $g$ primitive disks in $A$. One can
define a homeomorphism from $T$ to $T_{g}$ that takes each
$\mbox{$\partial$}a_{i}$ to the meridian of one of the standard summands of
$T_{g}$.

######  Lemma 1.7.

All such homeomorphisms are Powell equivalent.

###### Proof.

Let $\\{b_{1},...,b_{g}\\}$ and $\\{b^{\prime}_{1},...,b^{\prime}_{g}\\}$ be
different sets of orthogonal disks in $B$. For each $1\leq i\leq g$, let
$\mbox{$\beta$}_{i}$ (resp $\mbox{$\beta$}^{\prime}_{i})$ denote the arc
$\mbox{$\partial$}b_{i}-\mbox{$\partial$}a_{i}$ (resp
$\mbox{$\partial$}b^{\prime}_{i}-\mbox{$\partial$}a_{i}$).

Special Case: $\\{b_{1},...,b_{g}\\}$ and
$\\{b^{\prime}_{1},...,b^{\prime}_{g}\\}$ can be isotoped (leaving the $a_{i}$
invariant) so that the collection of curves
$\\{\mbox{$\beta$}_{1},...,\mbox{$\beta$}_{g}\\}$ and
$\\{\mbox{$\beta$}^{\prime}_{1},...,\mbox{$\beta$}^{\prime}_{g}\\}$ are
disjoint.

Let $P$ be the planar surface obtained by compressing $T$ along
$\\{a_{1},...,a_{g}\\}$. For each $1\leq i\leq g$
$\mbox{$\beta$}_{i}\cup\mbox{$\beta$}^{\prime}_{i}$ forms a circle $c_{i}$ in
$P$; if any $c_{i}$ bounds a disk in $P$ then the corresponding disks $b_{i}$
and $b^{\prime}_{i}$ can be isotoped to coincide; the proof now proceeds by
induction on the size of the set $\mathcal{S}$ of indices for which this is
true; we may as well take $\mathcal{S}=\\{1,...,k\\}$; when $k=g$ we are done.
Temporarily ignore the boundary components of $P$ that correspond to all
$a_{i},i\leq k$ and choose a circle $c_{\ell},\ell>k$ that is then innermost.
Then between $\mbox{$\partial$}b_{\ell}$ and
$\mbox{$\partial$}b^{\prime}_{\ell}$ there are only $1$-handle summands whose
indices lie in $\mathcal{S}$. Isotope these once around the curve $a_{\ell}$
(a sequence of moves each corresponding to Powell generator $D_{\nu}$). This
isotopy (Powell equivalent to the identity) moves $b_{\ell}$ to
$b^{\prime}_{\ell}$, adding another index to $\mathcal{S}$ and completing the
proof in this case.

The general case is proven via induction on the number of components of
$b_{i}\cap b^{\prime}_{j},i,j\in\\{1,...,g\\}$. We can assume that all
components of intersection are arcs and say that such an arc is outermost in
$b^{\prime}_{i}$, say, if it cuts off a subdisk of $b^{\prime}_{i}$ that
contains neither another arc of intersection nor the point $b^{\prime}_{i}\cap
a_{i}$. Choose an arc of intersection $\gamma$ that is outermost in, say,
$b^{\prime}_{i}$ and let $b_{j}$ be the other disk containing $\gamma$.
Replace the subdisk of $b_{j}$ that is cut off by $\gamma$ (on the side
disjoint from the point $b_{j}\cap a_{j}$) by the outermost subdisk of
$b^{\prime}_{i}$ cut off by $\gamma$. This converts $\\{b_{1},...,b_{g}\\}$ to
a disjoint family (so an equivalent family, per the Special Case) that
intersects $\\{b^{\prime}_{1},...,b^{\prime}_{g}\\}$ in fewer arcs, as
required. ∎

It will later be shown (see Corollary 3.6) that the choice of just a single
primitive disk determines the Powell equivalence type.

Cautionary Note: Suppose $\\{a_{1},...,a_{g_{1}}\\}$ is a non-separating
collection of disks in $A$ and $c$ is a simple closed curve $c$ in $T$ that
separates $T$ into one component containing the curves
$\\{\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{g_{1}}\\}$ and the other a
genus $g_{2}$ surface. Then $c$ automatically bounds a disk in $A$; if $c$
also bounds a disk in $B$ then there is an orientation preserving
homeomorphism $h:(S^{3},T,c)\to(S^{3},T_{g},c_{g_{1}})$,. The disks bounded by
$c$ in $A$ and $B$ divide each into handlebodies, so the two pieces into which
$T$ is divided are each Heegaard splittings of the $3$-balls in which they
lie. Moreover, if the Powell Conjecture is true in genus $g_{1}$ and genus
$g-g_{1}$, then any two such homeomorphisms are Powell equivalent. But there
is no obvious reason why different choices of $c$ will give Powell equivalent
homeomorphisms, since not all braid moves on the $a_{i}$ (even if the $a_{i}$
are a primitive set) are Powell moves. For example, if we rotate $a_{1}$
around a $\partial$ reducing disk for $B$ whose boundary lies on the genus
$g_{2}$ side of $c$, $c$ will be taken to a curve $c^{\prime}$ that satisfies
the same conditions. Yet it is not immediately apparent that such an isotopy
of $T$ to itself is a Powell move.

##  2\. Powell-like moves in a more general setting

Following Lemma 1.4 and the remarks that precede it we can focus on four types
of Goeritz elements, each of which is a Powell move, but less stringently
defined than Powell’s originals:

  1. (1)

Let $S\subset S^{3}$ be a reducing sphere for $T_{g}$ bounding a ball $B$
containing $g_{1}$ of the standard genus $1$ summands. A standard bubble move
is an isotopy of $B$ through some path in $T_{g}-B$ that returns $B,B\cap T$
to itself, see Figure 3.

Figure 3. Standard bubble move

  2. (2)

Let $B$ be a bubble containing just a single standard genus $1$ summand. A
standard flip is the homeomorphism $(B,B\cap T)\to(B,B\cap T)$ shown in Figure
1 as Powell’s move $D_{\omega}$.

Figure 4. Standard flip

  3. (3)

Let $B_{1},B_{2}$ be disjoint bubbles, each containing just a standard genus
$1$ summand, and let $v\subset T_{g}-(B_{1}\cup B_{2})$ be an arc connecting
them. Let $B$ be the reducing ball obtained by attaching the $1$-handle
regular neighborhood of $v$ to $B_{1},B_{2}$. A standard switch is the
homeomorphism $(B,B\cap T_{g})\to(B,B\cap T_{g})$ shown in Figure 1 as
Powell’s move $D_{\eta_{12}}$.

Figure 5. Standard switch

  4. (4)

Again let $B_{1},B_{2}$ be disjoint bubbles, each containing just a standard
genus $1$ summand. For $i=1,2$ let $\mu_{i}\subset B$ be a meridian disk for
$T_{g}\cap B_{i}$ and $\ell_{i}\subset A$ be a longitudinal disk for
$T_{g}\cap B_{i}$. Let $v\subset T_{g}$ be an embedded arc connecting
$\mbox{$\partial$}\mu_{1}$ to $\mbox{$\partial$}\ell_{2}$ with the interior of
$v$ disjoint from $\mu_{1},\mu_{2},\ell_{1},\ell_{2}$. Then
$\mu_{1}\cup\ell_{2}\cup v$ is called a standard eyeglass in $T_{g}$ see
Figure 6. The disks $\mu$ and $\ell$ are called the lenses of the eyeglass,
$v$ is the bridge. The eyeglass defines is a natural automorphism
$(S^{3},T_{g})\to(S^{3},T_{g})$ called an eyeglass twist which is supported on
a $3$-ball regular neighborhood of the eyeglass, see Figure 7. Powell’s
generator $D_{\theta}$ is a standard eyeglass twist, in fact the model for
this discussion.

Figure 6. Standard eyeglass Figure 7. Eyeglass twist

Each of these moves can be put in a more general setting. Suppose $M$ is any
compact oriented $3$-manifold and $T$ is a stabilized Heegaard splitting, with
some genus $1$ summands of $T$ designated as standard. Then all the
definitions above make sense. More generally, if we drop the word ’standard’
all the concepts make sense even when specific genus $1$ summands have not
been designated as standard. In fact, the notion of eyeglass, and so eyeglass
twist, can often apply even when the splitting $M=A\cup_{T}B$ is not
stabilized.

######  Definition 2.1.

An eyeglass is the union of two disks, $\ell_{a},\ell_{b}$ ( the lenses ) with
an arc $v$ (the bridge) connecting their boundaries. Suppose an eyeglass
$\eta$ is embedded in $M$ so that the $1$-skeleton of $\eta$ (called the
frame) lies in $T$, one lens is properly embedded in $A$, and the other lens
is properly embedded in $B$. The embedded $\eta$ defines a natural
automorphism $(M,T)\to(M,T)$, as illustrated in Figure 7, called an eyeglass
twist.

Remark: It will be useful later to note that a $\partial$-compression of one
of the lenses of an eyeglass $\eta$ into $T$ breaks the eyeglass twist around
$\eta$ into a composition of two eyeglass twists $\eta_{1}\eta_{2}$, where in
each $\eta_{i}$ the $\partial$-compressed disk is replaced by one of the disks
that is the result of the $\partial$-compression. See Figure 8.

Figure 8.

##  3\. Weak reduction

Recall ([CG]) that a Heegaard splitting $M=A\cup_{T}B$ is weakly reducible if
there is an essential disk in $A$ and one in $B$ so that their boundaries are
disjoint in $T$. Since the two lenses in an eyeglass are disjoint, it follows
that only weakly reducible splittings can support non-trivial eyeglass twists.
In this section we consider families of weakly reducing disks in Heegaard
splittings of $S^{3}$ and describe conditions which guarantee that an eyeglass
twist is a Powell move. Of course the Powell Conjecture would assert that any
eyeglass twist is a Powell move.

######  Definition 3.1.

Suppose a non-empty collection $\\{a_{1},...,a_{m}\\}$ of disks embedded in
$A$ is disjoint from a non-empty collection $\\{b_{1},...,b_{n}\\}$ of disks
in $B$. Then the pair of collections is called weakly reducing. If the
complement in $T$ of their boundaries
$\\{\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{m}\\}\cup\\{\mbox{$\partial$}b_{1},...,\mbox{$\partial$}b_{n}\\}$
consists of planar surfaces in $T$ then the pair of disk collections is
complete. If the complement is a single surface then they are called non-
separating.

Suppose $\\{a_{1},...,a_{g_{1}}\\}$, $\\{b_{g_{1}+1},...,b_{g}\\}$ is a
complete non-separating weakly-reducing pair of disk collections for a genus
$g$ splitting $S^{3}=A\cup_{T}B$. Let $c\subset T$ be a simple closed curve
that is disjoint from the boundaries of the two sets and separates one set
from the other. Then $c$ bounds a disk $a_{c}$ in $A$ and a disk $b_{c}$ in
$B$. The union $a_{c}\cup b_{c}$ is a sphere, dividing $S^{3}$ into two balls,
and $T$ intersects each ball in a Heegaard surface. Waldhausen’s theorem
applied to each then shows that there is an orientation preserving
homeomorphism $(S^{3},T,c)\to(S^{3},T_{g},c_{g_{1}})$. Moreover, if the Powell
Conjecture is true for genus $g_{1}$ and for genus $g-g_{1}$ splittings of
$S^{3}$ then all such homeomorphisms are Powell equivalent.

Consider the standard genus $g$ Heegaard splitting $S^{3}=A\cup_{T_{g}}B$, in
which we denote the meridians of the standard summands by
$\\{a_{1},...,a_{g}\\}$ and the longitudes by $\\{b_{1},...,b_{g}\\}$. In
particular $|a_{i}\cap
b_{j}|=|\mbox{$\partial$}a_{i}\cap\mbox{$\partial$}b_{j}|=1,1\leq i,j\leq g$.
The curve $c_{g_{1}}\subset T_{g}$ separates $T_{g}$ into two components,
$T_{A}$ containing
$\\{\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{g_{1}}\\}$ and $T_{B}$
containing $\\{\mbox{$\partial$}b_{g_{1}+1},...,\mbox{$\partial$}b_{g}\\}$.
Let $P_{A}$ (resp $P_{B}$) be the planar surface
$T_{A}-\\{a_{1},...,a_{g_{1}}\\}$ (resp $T_{B}-\\{b_{g_{1}+1},...,b_{g}\\}$)
and let $P$ be the combined planar surface $P_{A}\cup_{c}P_{b}$.

######  Lemma 3.2.

Suppose $\eta$ is an eyeglass in $T_{g}$ whose lenses consists of some
$a_{i},i\leq g_{1}$ and some $b_{j},j\geq g_{1}+1$ and whose bridge $v$
intersects $c_{g_{1}}$ exactly once. Then an eyeglass twist along $\eta$ is a
Powell move.

###### Proof.

Observe that, after perhaps some conjugating Powell moves as described at the
start of Section 1, Powell’s generator $D_{\theta}$ can be viewed as an
eyeglass twist around the same lenses and a bridge $v^{\prime}\subset P$ which
also intersects $c_{g_{1}}$ once. Now use Lemma 1.4 on the $1$ handles in
$T_{B}$ to slide them around $T_{A}$ so that afterwards $v\cap T_{A}$
coincides with $v^{\prime}\cap T_{A}\subset P_{A}$. Then symmetrically slide
the $1$-handles of $T_{A}$ around $T_{B}$ until the entire
$v=v^{\prime}\subset P$. ∎

######  Lemma 3.3.

Suppose $\eta$ is an eyeglass in $T_{g}$ whose lenses consist of a disk
$a\subset A$ with $\mbox{$\partial$}a\subset P_{A}$ and a disk $b\subset B$
with $\mbox{$\partial$}b\subset P_{B}$. Suppose further that the bridge $v$
intersects $c_{g_{1}}$ exactly once. Then an eyeglass twist along $\eta$ is a
Powell move.

###### Proof.

As in the proof of Lemma 3.2 we can assume that the bridge $v$ lies in
$P=P_{A}\cup P_{B}$. Then $\mbox{$\partial$}a$ is coplanar with a number of
boundary components of $P_{A}$ (informally, $\mbox{$\partial$}a$ bounds a disk
containing copies of some of the $a_{i}$). Noting the remark following
Definition 2.1, we can view the twist around $\eta$ as a composition of twists
around each of these components of $P_{A}$. Do the same for
$\mbox{$\partial$}b$. The result follows now from Lemma 3.2. ∎

######  Lemma 3.4.

Suppose $\eta$ is an eyeglass in $T_{g}$ whose lenses consist of a disk
$a\subset A$ with $\mbox{$\partial$}a\subset T_{A}$ and a disk $b\subset B$
with $\mbox{$\partial$}b\subset T_{B}$. Suppose further that the bridge $v$
intersects $c_{g_{1}}$ exactly once. Then an eyeglass twist along $\eta$ is a
Powell move.

###### Proof.

As before, we may assume that $v$ lies in $P$. The proof is by induction on
the number of arc components of $a\cap(a_{1}\cup...\cup a_{g_{1}})$ and
$b\cap(b_{g_{1}+1}\cup...\cup b_{g})$ . If there are none, the result follows
from Lemma 3.3. Otherwise choose an arc of intersection that is outermost in,
say, $a_{i}$. Use the arc to $\partial$-compress $a$, breaking up the twist
around $\eta$ into the twist around two eyeglass curves as illustrated in
Figure 8. (If one of the new $A$ lenses is inessential then $v$ could have
been extended past the $\partial$-compressing disk, replacing the arc of
intersection by a new point of intersection of $a_{i}$ with $v$. This can be
removed as usual.) The result now follows by induction. (But, cautionary note,
one of the new eyeglasses has a bridge that intersects $a_{i}$. Fortunately,
having $v$ disjoint from $\\{a_{1},...,a_{g_{1}}\\}$ is not part of our
hypothesis but was achieved by argument. This condition is needed, else the
boundary compression used in the argumentwould intersect $v$.) ∎

Here is an application:

######  Proposition 3.5.

Suppose $b\subset B$ is a disk orthogonal to $a_{1}$. Then there is a Powell
move that leaves $a_{1}$ unchanged and carries $b_{1}$ to $b$.

###### Proof.

We can assume $\mbox{$\partial$}b$ intersects a bicollar neighborhood
$Y\subset T_{g}$ of $\mbox{$\partial$}a_{1}$ in an arc parallel to but
disjoint from $\mbox{$\partial$}b_{1}\cap Y$.

Special Case: $b$ and $b_{1}$ are disjoint.

Band $\mbox{$\partial$}b$ to $\mbox{$\partial$}b_{1}$ together using one of
the two bands they cut off from the bicollar $Y$, creating a new disk $b_{+}$.
Push the interior of this band into $B$ to properly embed $b_{+}$ in $B$.
Consider $a_{1}$ and $b_{+}$ as lenses of an eyeglass $\eta$ whose bridge $v$
is one of the small arcs $(\mbox{$\partial$}b\cap Y)-\mbox{$\partial$}a_{1}$.
It is easy to see (Figure 9) that an appropriate eyeglass twist of $\eta$ will
move $b_{1}$ to $b$. The eyeglass visibly satisfies the criterion required by
Lemma 3.4 : Let $c$ be the boundary of a regular neighborhood of
$\mbox{$\partial$}a_{1}\cup b_{1}$. Then $c$ bounds a punctured torus in
$T_{g}$ (in fact the first standard summand) that contains $a_{1}$, is
disjoint from $b_{+}$, and intersects $v$ in a single point. See Figure 10.

2pt $\mbox{$\partial$}b_{1}$ at 235 70 $\mbox{$\partial$}b$ at 300 30
$\mbox{$\partial$}b_{+}$ at 250 10 $\mbox{$\partial$}a_{1}$ at 110 65 $v$ at
138 25

Figure 9.

2pt $c$ at 155 125 $\mbox{$\partial$}b_{+}$ at 250 10 $\mbox{$\partial$}a_{1}$
at 110 65 $v$ at 135 25

Figure 10.

The general case now follows much as in the proof of Lemma 1.7, via induction
on the number of components of $b\cap b_{1}$: Suppose $b$ is a disk
intersecting $b_{1}$ in $n$ arcs and the proposition is true for disks that
intersect $b_{1}$ in fewer than $n$ arcs. Let $\gamma$ be an arc of $b\cap
b_{1}$ that is outermost in $b$ (with reference to the point $b\cap a_{1}$)
and let $b_{o}\subset b$ be the disk $\gamma$ cuts off from $b$. Replace the
disk in $b_{1}$ cut off by $\gamma$ with $b_{o}$ to get a disk $b^{\prime}$
orthogonal to $a_{1}$ that is disjoint from $b_{1}$ and intersects $b$ in
fewer than $n$ arcs. By the special case just proven there is a Powell move
$\rho$ that carries $b^{\prime}$ to $b_{1}$ and so carries $b$ to a disk
$b^{\prime\prime}$ intersecting $\rho(b^{\prime})=b_{1}$ in fewer than $n$
arcs. By inductive assumption, there is a Powell move $\rho^{\prime}$ carrying
$b"$ to $b_{1}$. Then
$\rho^{\prime}\rho(b)=\rho^{\prime}(b^{\prime\prime})=b_{1}$ as required. ∎

######  Corollary 3.6.

Suppose the Powell conjecture is true for genus $g-1$ splittings of $S^{3}$
and $T\subset S^{3}$ is a genus $g$ splitting. Then the choice of a single
primitive disk $a\subset A$ (or primitive $b\subset B$) defines a Powell
equivalence class of homeomorphism $(S^{3},T)\to(S^{3},T_{g})$.

###### Proof.

Since $a$ is primitive, there is a disk $b\subset B$ whose boundary intersects
the boundary of $a$ in a single point. Choose a homeomorphism $h:T\to T_{g}$
which carries the pair $(a,b)$ to the pair $(a_{1},b_{1})$. By Proposition 3.5
the Powell equivalence class of $h$ does not depend on $b$. The proof then
follows by applying the inductive assumption to the genus $g-1$ side of the
reducing sphere $S_{1}$ for $T_{g}$. ∎

Returning now to the general discussion, we drop the assumption that the
$a_{i}$ and $b_{j}$ are primitive, but otherwise maintain the notation above.
Suppose $\\{a_{1},...,a_{g_{1}}\\}$ and $\\{b_{g_{1}+1},...,b_{g}\\}$ are a
(not necessarily primitive) pair of non-separating weakly-reducing disk
collections for $T\subset S^{3}$. Let $c\subset T$ be a simple closed curve
that separates the two sets and $\eta$ is an eyeglass in $T$ whose lenses
consists of a disk $a\subset A$ with $\mbox{$\partial$}a\subset T_{A}$ and a
disk $b\subset B$ with $\mbox{$\partial$}b\subset T_{B}$.

######  Lemma 3.7.

Suppose that the bridge $v$ for $\eta$ intersects $c$ exactly once. Then an
eyeglass twist along $\eta$ does not change the Powell equivalence class of
any homeomorphism $h:(S^{3},T,c)\to(S^{3},T_{g},c_{g_{1}})$.

###### Proof.

First note that there is such a homeomorphism: the hypothesis guarantees that
$c$ bounds a disk in both $A$ and $B$ and so is part of a reducing sphere for
$T$. Then Waldhausen’s theorem applied to both $T_{A}$ and $T_{B}$ provides a
homeomorphism $h$. (This homeomorphism is unique up to Powell equivalence if
the Powell Conjecture is true for lower genus splittings. We do not need that
inductive assumption here.) Now apply Lemma 3.4. ∎

######  Theorem 3.8.

Suppose $\\{a_{1},...,a_{g_{1}}\\}$ and $\\{b_{g_{1}+1},...,b_{g}\\}$ are a
(not necessarily primitive) pair of non-separating weakly-reducing disk
collections for $T\subset S^{3}$. Then

  * •

there is a homeomorphism $h:T\to T_{g}$ so that
$\\{h(a_{1}),...,h(a_{g_{1}})\\}$ lie on the side of the separating sphere
$S_{g_{1}}$ that contains $c_{1}$ and $\\{h(b_{g_{1}+1}),...,h(b_{g})\\}$ lie
on the other side and

  * •

If the Powell Conjecture is true for genus $g_{1}$ and genus $g-g_{1}$
splittings, any two such homeomorphisms are Powell equivalent.

###### Proof.

Choose any simple closed curve $c\subset T$ that lies between the collections
$\\{\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{g_{1}}\\}$ and
$\\{\mbox{$\partial$}b_{g_{1}+1},...,\mbox{$\partial$}b_{g}\\}$. As noted in
the proof of Lemma 3.7 there is a homeomorphism
$h:(S^{3},T,c)\to(S^{3},T_{g},c_{g_{1}})$ as required, and its Powell
equivalence class depends only on the choice of $c$. The goal then is to show
that the Powell equivalence class does not even depend on $c$.

As before, let $P$ be the connected planar surface
$T-\\{a_{1},...,a_{g_{1}}\cup b_{g_{1}+1},...,b_{g}\\}$, so that $c\subset P$
separates the $2g_{1}$ components of $\mbox{$\partial$}P$ corresponding to
$\\{\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{g_{1}}\\}$ from the
$2(g-g_{1})$ components corresponding to
$\\{\mbox{$\partial$}b_{g_{1}+1},...,\mbox{$\partial$}b_{g}\\}$. Call the
former component $P_{A}$ and the latter $P_{B}$.

Suppose $c^{\prime}\subset P$ is another such simple closed curve. Picturing
$P$ as a $2g$ punctured sphere, a braid automorphism will move $c^{\prime}$ to
$c$. So we need only show that any braid automorphism $\rho$ that moves the
boundary components $\mbox{$\partial$}P\cap P_{A}$ back to themselves (and so
the boundary components of $\mbox{$\partial$}P\cap P_{B}$ back to themselves)
does not change the Powell equivalence class of $h$.

It is a classic result that the “mixed” braid group
$\mbox{$\mathcal{B}$}_{g_{1},g_{2}}$ on the $g$-punctured sphere $P_{A}\cup
P_{B}$ can be generated by a set of $g_{1}-1$ half-twists in $P_{A}$ (call
this subgroup $\mbox{$\mathcal{B}$}_{a}$), $g_{2}-1$, half-twists in $P_{B}$
(call this subgroup $\mbox{$\mathcal{B}$}_{b}$) and a single full twist
$\sigma$ along a chosen arc $\gamma$ connecting a specific component $a_{0}$
of $\mbox{$\partial$}P_{A}$ to a specific component $b_{0}$ of
$\mbox{$\partial$}P_{B}$. This is an eyeglass twist with lenses $a_{0}$ and
$b_{0}$ and bridge $\gamma$. Choose $\gamma$ to be an arc crossing $c$ once.

Clearly the subgroups $\mbox{$\mathcal{B}$}_{a}$ and
$\mbox{$\mathcal{B}$}_{b}$ commute. The proof that
$\rho\in\mbox{$\mathcal{B}$}_{g_{1},g_{2}}$ is Powell equivalent to one that
lies in $\mbox{$\mathcal{B}$}_{a}\times\mbox{$\mathcal{B}$}_{b}$ proceeds by
induction on the number $n_{\sigma}$, of occurrences of $\sigma$ in $\rho$
when expressed as a product of these generators. If $\sigma$ does not appear,
then $c$ is preserved by the braid. With no loss the initial segment of $\rho$
can be written $\alpha\beta\sigma$, where $\alpha\in\mbox{$\mathcal{B}$}_{a}$
and $\beta\in\mbox{$\mathcal{B}$}_{b}$. That is,
$\rho=\alpha\beta\sigma\omega$, where
$\omega\in\mbox{$\mathcal{B}$}_{g_{1},g_{2}}$ has one less occurence of
$\sigma$. Then
$\rho=(\alpha\beta\sigma\beta^{-1}\alpha^{-1})\alpha\beta\omega$. But it is
easy to see that $(\alpha\beta\sigma\beta^{-1}\alpha^{-1})$ is an eyeglass
twist along an eyeglass whose bridge is $\alpha\beta(\gamma)$, an arc that
still crosses $c$ once. The proof then follows from Lemma 3.7. ∎

To state the implication of Theorem 3.8 a little less formally:

######  Corollary 3.9.

If the Powell Conjecture is true for genus $g_{1}$ and genus $g-g_{1}$, then a
pair $\\{a_{1},...,a_{g_{1}}\\}$, $\\{b_{g_{1}+1},...,b_{g}\\}$ of non-
separating weakly-reducing disk collections for $T\subset S^{3}$ determines a
Powell equivalence class of homeomorphisms $h:T\to T_{g}$.

In fact, only one set of disks is needed, so long as we know it has at least
one complementary set:

######  Theorem 3.10.

Suppose the collection $\\{b_{g_{1}+1},...,b_{g}\\}$ of compressing disks in
$B$ can be extended to two possibly different non-separating complete
collections $\\{a_{1},...,a_{g_{1}}\\},\\{b_{g_{1}+1},...,b_{g}\\}$ and
$\\{a^{\prime}_{1},...,a^{\prime}_{g_{1}}\\},\\{b_{g_{1}+1},...,b_{g}\\}$ of
weakly reducing disks for $T$. If the Powell Conjecture is true for genus
$g_{1}$ and genus $g-g_{1}$, then the Powell equivalance classes determined by
each extension via Theorem 3.8 are the same.

###### Proof.

Let $H\subset S^{3}$ be the genus $g_{1}$ handlebody obtained from $A$ by
attaching $2$-handles along $\\{b_{g_{1}+1},...,b_{g}\\}$.

Special Case: $\\{a_{1},...,a_{g_{1}}\\}$ and
$\\{a^{\prime}_{1},...,a^{\prime}_{g_{1}}\\}$ are disjoint.

We first claim that, in this case, we can proceed from one family to the other
by a sequence of substitutions of a single disk at a time. This is obvious
(indeed there is nothing to prove) if each $a^{\prime}_{j}$ is parallel to one
of the $a_{i}$, so we induct on the number of $a^{\prime}_{j}$ that are not
parallel to any $a_{i}$. With no loss of generality, say $a^{\prime}_{1}$ is
not parallel to any $a_{i}$. Since the set $\\{a_{1},...,a_{g_{1}}\\}$ is non-
separating in $H$, $H-\\{a_{1},...,a_{g_{1}}\\}$ is a ball containing the
disjoint collection of properly embedded disks
$\\{a^{\prime}_{1},...,a^{\prime}_{g_{1}}\\}$. Each $a_{i}$ gives rise to a
pair of disks (called twins) on the boundary of the ball. Since
$H-\\{a^{\prime}_{1},...,a^{\prime}_{g_{1}}\\}$ is connected, there is at
least one $a_{i}$, say $a_{1}$, which is parallel to no $a^{\prime}_{j}$ in
$H$ and whose twins in the boundary of the ball lie on opposite sides of
$a^{\prime}_{1}$. Then replacing $a^{\prime}_{1}$ by $a_{1}$ changes only a
single disk in
$\\{a^{\prime}_{1},...,a^{\prime}_{g_{1}}\\},\\{b_{g_{1}+1},...,b_{g}\\}$ and
provides a non-separating complete collection
$\\{a_{1},a^{\prime}_{2},...,a^{\prime}_{g_{1}}\\},\\{b_{g_{1}+1},...,b_{g}\\}$
with more disks parallel in $A$ to disks in $\\{a_{1},...,a_{g_{1}}\\}$,
completing the inductive step. So, following the claim, we may as well assume
that $a_{i}=a^{\prime}_{i}$ for all $2\leq i\leq g_{1}$.

Let $W$ be the solid torus obtained from $H$ by deleting each
$a_{i}=a^{\prime}_{i},2\leq i\leq g_{1}$. The remaining disks
$a_{1},a^{\prime}_{1}$ are disjoint meridian disks of $W$, dividing it into
two cylindrical components $W_{\pm}$, each topologically a ball. Now consider
the $g-g_{1}$ properly embedded arcs $\\{\beta_{g_{1}},...,\beta_{g}\\}$ in
$W$, each arc $\beta_{j}$ dual to the disk $b_{j}\subset B$. If, say, $W_{-}$
contains none of the $\beta_{j}$, then all the $\beta_{j}$ lie in $W_{+}$.
Then there is a curve $c\subset\mbox{$\partial$}A\cap W_{+}$ that separates
$\\{\mbox{$\partial$}a^{\prime}_{1},\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{g_{1}}\\}$
from $\\{\mbox{$\partial$}b_{g_{1}+1},...,\mbox{$\partial$}b_{g}\\}$. Then in
$S^{3}$, $c$ bounds disks $a_{c}\subset A$ and $b_{c}\subset B$, and so
provides a reducing sphere $a_{c}\cup b_{c}$ for the splitting $T$ that
determines, as shown in Theorem 3.8, the same Powell equivalence class for the
pair of weakly compressing collections
$\\{a_{1},...,a_{g_{1}}\\},\\{b_{g_{1}+1},...,b_{g}\\}$ as it does for
$\\{a^{\prime}_{1},a_{2}...,a_{g_{1}}\\},\\{b_{g_{1}+1},...,b_{g}\\}$,
completing the argument in this case. (In effect, $\\{a_{1},...,a_{g_{1}}\\}$
and $\\{a^{\prime}_{1},...,a^{\prime}_{g_{1}}\\}$ are merely different choices
of complete collections of $\partial$-reducing disks for $A$ in the side of
the sphere $a_{c}\cup b_{c}$ on which they lie).

The remaining possibility in this special case is that some of the $\beta_{j}$
lie in each of $W_{\pm}$. There is a curve $c\subset T$ that separates
$\\{\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{g_{1}}\\}$ from
$\\{\mbox{$\partial$}b_{g_{1}+1},...,\mbox{$\partial$}b_{g}\\}$ and intersects
$\mbox{$\partial$}a^{\prime}_{1}$ in two points. (So $c$ bounds a disk
$a_{c}\subset A$ intersecting the disk $a^{\prime}_{1}$ in a single arc
$\gamma$.) Via Theorem 3.8 this determines a Powell equivalence class for the
pair of weakly compressing collections
$\\{a_{1},...,a_{g_{1}}\\},\\{b_{g_{1}+1},...,b_{g}\\}$. The union of a
component of $a_{c}-\gamma$ and a component of $a^{\prime}_{1}-\gamma$ cuts
off a bubble containing all the arcs $\beta_{j}$ lying in, say, $W_{-}$. A
bubble move around a longitude of $\mbox{$\partial$}W$ will push
$a^{\prime}_{1}$ into a new position so that all the $\beta_{j}$ lie in
$W_{+}$, where the previous argument applies. The bubble itself may not be
standard in the given Powell equivalence class but, invoking the inductive
assumption that the Powell Conjecture is true for genus $g-g_{1}$, the summand
contained in the bubble may be made standard without changing the Powell
equivalence class. Then the bubble move itself does not change the Powell
equivalence class, completing the proof in this special case.

General Case The general case now proceeds classically, by induction on
$|\\{a_{1},...,a_{g_{1}}\\}\cap\\{a^{\prime}_{1},...,a^{\prime}_{g_{1}}\\}|$,
the number of arcs in which the two systems of $A$ disks intersect. Consider
an outermost arc of intersection in $a^{\prime}_{1}$, say, cutting off an
outermost disk $D^{\prime}\subset a^{\prime}_{1}$. . With no loss assume the
arc also lies in $a_{1}$. The correct choice of subdisk $D\subset a_{1}$, when
attached to $D^{\prime}$ along the arc of intersection will give a disk
$a_{D}\subset A$ so that $\\{a_{D},a_{2},...,a_{g_{1}}\\}$ also satisfies the
hypotheses of the theorem. Since $a_{D}$ is disjoint from $a_{1}$ it follows
from the special case above that $\\{a_{D},a_{2},...,a_{g_{1}}\\}$ and
$\\{a_{1},a_{2},...,a_{g_{1}}\\}$ determine the same Powell equivalence class.
Since $|a_{D}\cap a^{\prime}_{1}|<|a_{1}\cap a^{\prime}_{1}|$ (and no other
pairs of disks have an increased number of intersection arcs) the inductive
hypothesis implies that $\\{a_{D},a_{2},...,a_{g_{1}}\\}$ and
$\\{a_{1},a_{2},...,a_{g_{1}}\\}$ determine the same Powell equivalence class.
∎

##  4\. Towards a proof of the Powell Conjecture

###  4.1. The philosophy

For the purposes of this section, let $T_{0}$ denote a copy of the standard
genus $g\geq 2$ Heegaard surface $T_{g}$ in $S^{3}$. Here is the philosophy
behind our (only partially successful) strategy to prove the Powell
Conjecture. Start with an element of the Goeritz group, represented by a path
$\tau_{\theta}:S^{3}\to S^{3},0\leq\theta\leq 2\pi$ in $\mathop{\rm
Diff}\nolimits(S^{3})$ that starts out as the identity and has
$\tau_{2\pi}(T_{0})=T_{0}$. For brevity denote each surface
$\tau_{\theta}(T_{0})\subset S^{3}$ by $T_{\theta}$.

We would like to find such a representative $\tau$ so that for some
$0=\theta_{0}<\theta_{1}<\theta_{2}<....<\theta_{n}<2\pi$

  1. (1)

for each $\theta\notin\\{\theta_{i},1\leq i\leq n\\}$ we can extract
information from the surface $T_{\theta}\subset S^{3}$ sufficient to determine
a Powell equivalence class of trivializations
$h_{\theta}:(S^{3},T_{\theta})\to(S^{3},T_{g})$.

  2. (2)

ensure that the information is unchanged throughout each interval in
$[0,2\pi]-\\{\theta_{i},1\leq i\leq n\\}$ so that
$h_{\theta}\tau_{\theta}:(S^{3},T_{0})\to(S^{3},T_{g})$ is a well-defined
Powell equivalence throughout each interval

  3. (3)

for each $1\leq i\leq n$ ensure that the information for $\theta_{i}-\epsilon$
gives the same Powell equivalence class as the information for
$\theta_{i}+\epsilon$ so that
$h_{\theta}\tau_{\theta}:(S^{3},T_{0})\to(S^{3},T_{g})$ is a well-defined
Powell equivalence class as we move from one interval to the next.

It would follow then that for all $0\leq\theta\leq 2\pi$,
$h_{\theta}\tau_{\theta}:(S^{3},T_{0})\to(S^{3},T_{g})$ are Powell equivalent.
In particular $h_{0}\tau_{0}$ is Powell equivalent to $h_{2\pi}\tau_{2\pi}$.
But since the Powell equivalence class of $h_{\theta}$ is determined entirely
by the surface $T_{\theta}$, and $T_{2\pi}=T_{0}$, it follows that we may take
$h_{2\pi}=h_{0}$ so that $\tau_{0}$ (the identity) is Powell equivalent to
$\tau_{2\pi}$, as required.

Although this is the philosophy, the outcome of our argument is not so neat.
Sadly, the information we will be able to extract does not rise to the level
(as exhibited, say, in Corollary 3.9 above) that is sufficient to determine
Powell equivalence class, at least as far as we have been able to determine.
(But it does suffice for the case of genus $3$, see Section 5 below.)

###  4.2. The complex $2C(T)$

First we describe the topological information that we will extract. Thinking
of $C(T)$ as the curve complex of $T=T_{0}$, define a 1-complex $2C(T)$ as
follows: the vertices are disjoint ordered pairs of simple closed curves in
$T$ (so each corresponds to an edge in $C(T)$, with an orientation). The
1-simplices in $2C(T)$ are of two types: pairs of pairs
$((a,b),(a,b^{\prime}))$ with the property that the curves
$\\{a,b,b^{\prime}\\}$ are pairwise disjoint and pairs of pairs
$((a,b),(a^{\prime},b))$ where the curves $(a,a^{\prime},b)$ are pairwise
disjoint. (So we can think of each edge as a 2-simplex in $C(T)$ in which two
of the edges have been oriented to share a head or a tail.) We speak of
“shuffling” between $(a,b)$ and $(a,b^{\prime})$ and between $(a,b)$ and
$(a^{\prime},b)$.

Now let $T_{\theta}=\tau_{\theta}(T_{0}),\theta\in S^{1}$ be a parameterized
Heegaard surface in $S^{3}$ representing an element of the Goeritz group, as
described above. By a circle $\gamma$ of weak reductions (cwr) representing
$T_{\theta}$ we mean an edge path $\gamma_{t}$, $0\leq t\leq 2\pi$ in $2C(T)$
so that

  * •

each vertex on $\gamma_{t}$ is a pair $(a,b)$ where $a$ compresses in $A$ and
$b$ compresses in $B$,

  * •

$\gamma_{2\pi}=\tau_{2\pi}(\gamma_{0})$, where $\tau_{2\pi}$ acts via the
natural action of $MCG(T)$ on $2C(T)$.

We will show that any such parameterization $\tau_{\theta}$ can be deformed
into one that somewhat naturally presents a cwr representing $\tau_{\theta}$.
Specifically, after the deformation, there will be successive values
$\theta_{1},\theta_{2},....,\theta_{n}\in(0,2\pi)$ so that for each
$\theta\notin\\{\theta_{i},1\leq i\leq n\\}$ there is a pair of weakly
reducing disks $(a_{\theta},b_{\theta})$ associated to the topological surface
$T_{\theta}\subset S^{3}$ so that

  * •

the isotopy class of the pair $(a_{\theta},b_{\theta})$ is unchanged
throughout each interval in $S^{1}-\\{\theta_{i},1\leq i\leq n\\}$,

  * •

for each $1\leq i\leq n$ the pair
$(a_{\theta_{i}+\epsilon},b_{\theta_{i}+\epsilon})$ differs from the pair
$(a_{\theta_{i}-\epsilon},b_{\theta_{i}-\epsilon})$ by a shuffle

Technically, the cwr is then the pull-back of this sequence of disk pairs
under the parameterization $(S^{3},T_{g})\to(S^{3},T_{\theta})$ that defines
$T_{\theta}$. Before we show how to construct the cwr, we describe how a proof
of the following conjecture would then lead to a proof of the Powell
Conjecture.

######  Conjecture 4.1.

There is a method of associating to any vertex $(a,b)$ in $2C(T)$ for which
$a$ bounds a disk in $A$ and $b$ bounds a disk in $B$, a Powell equivalence
class of homeomorphisms $h_{(a,b)}(S^{3},T)\to(S^{3},T_{g})$ with these
properties:

  * •

The method is topological. That is, for any homeomorphism
$\sigma:(S^{3},T)\to(S^{3},T)$ and vertex $(a,b)\in 2C(T)$,
$h_{(\sigma(a),\sigma(b))}\sigma=h_{(a,b)}$ and

  * •

the Powell equivalence class associated to the ends of any edge in $2C(T)$ are
the same.

Combining Conjecture 4.1 with the construction that precedes it, we get a
sequence of homeomorphisms
$h_{(a_{0},b_{0})},h_{(a_{1},b_{1})},...,h_{(a_{n},b_{n})}:T\to T_{g}$ which
all have the same Powell equivalence class and for which
$(a_{n},b_{n})=(\tau_{2\pi}(a_{0}),\tau_{2\pi}(b_{0}))$. This implies that
$h_{(a_{0},b_{0})}\tau_{2\pi}$ is Powell equivalent to
$h_{(\tau_{2\pi}(a_{0}),\tau_{2\pi}(b_{0}))}\tau_{2\pi}=h_{(a_{0},b_{0})}$ so
$\tau_{2\pi}$ is Powell equivalent to the identity, as required.

###  4.3. The Rieck background and its refinement

Recall the classical sweep-out technology applicable to any Heegaard splitting
of a closed $3$-manifold $M=A\cup_{T}B$ (see [RS]): Pick a spine in each
handlebody $A$, $B$, that is a $1$-complex $X$ in $A$ (say) whose complement
is homeomorphic to $T\times(0,1)$, with $T\times\\{1\\}$ corresponding to
$\mbox{$\partial$}A$. This gives rise to a mapping cylinder structure on $A$,
$A\cong\partial A\times[0,1]/$$(a\times 0)\equiv f(a)$, some $f:\partial
A\rightarrow X$. In its classical application, the mapping cylinder structures
on $A$ and $B$ can be combined to parameterize the entire complement of the
spines in $M$ as $T\times(-1,1)$ describing how most of $M$ is swept-out by
copies of $T$.

In the case that $M=S^{3}$ there is another natural sweep-out (actually the
genus $0$ version of the sweep-out just described). Viewing $S^{3}$ as the
standard $3$-sphere in $4$-space, a height function in $R^{4}$ describes a
sweep-out of $S^{3}$ from south-pole to north-pole. That is, we can view
$S^{3}$ also as $S^{2}\times[-1,1]$ with $S^{2}\times\\{0\\}$ crushed to the
south pole and $S^{2}\times\\{1\\}$ crushed to the north pole.

In [R] (based on arguments in [RS]) Rieck proves Waldhausen’s theorem by
comparing these two “sweep-outs” of $S^{3}$ by surfaces, one parameter
$s\in[-1,1]$ for the sweep out by spheres $S$ and one parameter $t\in[-1,1]$
for the sweep-out by a genus $g\geq 2$ Heegaard surface $T$. A “graphic”
$\Gamma$ of this $(t,s)$-square is analyzed to find a weak reduction of $T$,
that is a pair of compressing disks, one in $A$ and the other in $B$, whose
boundaries are disjoint in $T$. Then [CG] implies that the Heegaard splitting
is reducible, and that finishes the proof by induction.

The graphic consists of open regions $R_{i}$ where $S_{s}$ and $T_{t}$
intersect transversely, edges or “walls” where the two have a tangency, and
cusp points where two types of tangencies cancel. As argued in [RS] only
domain walls corresponding to saddle tangencies need to be tracked. Cusps and
tangencies of index 2 or 0 can be erased as they amount only to births/deaths
of inessential simple closed curves of intersection in $S_{s}\cap T_{t}$. The
most interesting event which occurs are transverse crossings of saddle walls;
at this point two independent saddle tangencies occur. It will be useful to
very briefly review Rieck’s analysis of this graphic:

Each region $R_{i}\subset I\times I\backslash\mbox{$\Gamma$}$ is labeled by
$I$ or $E$: $I$ if every component of $S_{s}\cap T_{t}$, $(t,s)\in R_{i}$ is
inessential in $T$, $E$ otherwise. Call an $I$ region green, labeled $I_{g}$,
if $T_{t}\cap(\bigcup_{s^{\prime}\geq s}S_{s^{\prime}})\subset
T_{t}\backslash$disks, that is most of $T_{t}$ lies north of $S_{s}$.
Similarly we call an $I$ region yellow, labeled $I_{y}$, if
$T_{t}\cap\bigcup_{s^{\prime}<s}S_{s^{\prime}})\subset T_{t}\backslash$ disks,
that is most of $T_{t}$ lies south of $S_{s}$. [R, Lemma/Definition 3.4]
asserts that every $I$-region is labeled exactly once as $I_{g}$ or $I_{y}$.
Furthermore, [R, Proposition 3.3] states that no two $I$ regions bearing
different color labels can touch, even at a corner. (This is where requiring
genus $g\geq 2$ comes in; see for example Lemma A.4 in the Appendix.)

It is clear that for $s$ sufficiently near -1, $(t,s)$ must lie in an $I_{g}$
region (since most of $T_{t}$ lies north of $S_{s}$) and that for $s$
sufficiently near +1, $(t,s)$ must lie in an $I_{y}$ region. It follows that
between and $I_{g}$ and $I_{y}$ region there is an entire strip of $I\times
I$, running from the side $t=-1$ to the side $t=1$, in which each region is
labelled $E$. Moreover, when $t$ is near $-1$, $T_{t}$ is close to the spine
of $A$, so the essential curves bound disks in $A$ and when $t$ is near $+1$,
$T_{t}$ is close to the spine of $B$, so the essential curves bound disks in
$B$. It follows that there must be some region (or two adjacent regions) in
such a strip in which both sorts of disks occur, and this provides a weak
reduction.

Rieck’s result can be refined. In the Appendix A we show that the lowest of
the strips that appear in Rieck’s argument is actually monotonic in $t$. By
this we mean that there is a function $r:[-1,1]\to[-1,1]$ whose graph lies in
the $1$-skeleton of the reduced graphic and, for small $\epsilon$,
$\\{(t,r(t)+\mbox{$\epsilon$})\\}|t\in[-1,1]\\}$ lies entirely inside the
lowest strip. To put it another way, the graph of $r+\mbox{$\epsilon$}$ in
$I\times I$ is transverse to the reduced graphic and, for any $(t,s)$ lying in
a region that the graph intersects, among the circles in $S_{s}\cap T_{t}$,
there is at least one circle that is essential in $T_{t}$. This fact has no
real importance in Rieck’s argument, but its analogue in our context will be
quite useful (though not mathematically essential).

###  4.4. Adding the parameter $\theta$

The fundamental idea now will be to add a third parameter to Rieck’s proof,
namely the parameter $\theta$. We first need to show that the choice of spines
in $A$ and $B$ above is unimportant, even in the general context of a closed
manifold $M=A\cup B$. One can think of a spine for $A$ (or $B$) as a 1-complex
$X\subset\text{int }A$ together with a dual cell structure on $A$: a proper
disk associated with (and normal to) each 1-simplex of $X$ and a 3-ball
containing each vertex of $X$. But such a presentation of the spine involves
no choice in the sense that its space of parameters is contractible. The
argument for this has two parts: It follows from [Mc] that the choice of disks
defining the spine is unimportant since the disk space of $A$ is contractible.
(Here the disk space is the simplicial complex whose $n$-cells are $n+1$
pairwise disjoint properly embedded disks, which together divide $A$ into
balls.) Once these disks are chosen, Hatcher has shown that the exact
placement of the disks in $A$, and then the ensuing parameterization of the
complementary $3$-balls, also involves no choice. Putting together these
results it follows that once a Heegaard surface $T\subset M$ is determined,
$M$ is foliated by a family of surfaces $M=\bigcup_{t\in[-1,1]}T_{t}$ which
degenerate at $t\in\\{-1,+1\\}$. Here $T_{0}$ is of course the original
Heegaard surface $T=\partial A=\partial B$. Moreover, this foliation is
canonical, up to choices coming from a weakly contractible parameter space.
The disk complex is contractible and the corresponding function spaces whose
simplicial realization is the disk complex is therefore weakly contractible.

Now add the circular parameter $\theta\in[0,2\pi]/$$0\equiv 2\pi$ and consider
the family of Heegaard surfaces $T_{\theta}\subset S^{3}$. Choose spines for
$A_{0}$ and $B_{0}$, propagate them along $\theta$ by isotopy until at
$2\pi-\epsilon$ we see that the initial and final spines are not matching up.
However using just the $\pi_{1}$-aspect of ”canonical” above we may isotope
the spines to match at $2\pi=0$. This yields a $\theta$-parameter family of
singular foliations $T_{t,\theta}$,
$(t,\theta)\in[-1,1]\times[0,2\pi]/$$0\equiv 2\pi\coloneqq I\times S^{1}$.

######  Theorem 4.2.

Let $T_{\theta}$ be a loop of genus $g$ Heegaard surfaces, $g\geq 2$ as
described above. There is a circle $\gamma$ of weak reductions (cwr)
representing $T_{\theta}$.

###### Proof.

We study the 3-parameter family $T_{t,\theta}\cap S_{s}$,
$(t,s,\theta)\in[-1,1]\times[-1,1]\times[0,2\pi]/$$0\equiv
2\pi\coloneqq[-1,1]\times[-1,1]\times S^{1}$. These intersections may be
regarded as the level set at height = $s$ of a 2-parameter family of functions
on $T$. All told this means we are looking for a 3-parameter family of germs
of smooth functions $(R^{2},0)\rightarrow(R,0)$. According to Thom’s Jet-
transversality Theorem any such family can be perturbed a generic one with
local simulation types as listed below. In these dimensions the generic local
germs 222All other germ types such as hyperbolic umbilic,
$f(x,y)=x^{3}+y^{3}$, have codimensions $\geq 4$ and need not be considered.
[HW] are represented by:

1. | $f(x,y)=x$ |  | codim = 0
---|---|---|---
2a. | $f(x,y)=x^{2}+y^{2}$ | source / sink | codim = 1
b. | $f(x,y)=x^{2}-y^{2}$ | saddle | codim = 1
3a. | $f(x,y)=x^{3}+y^{2}$ | birth/death of $(1,2)$-handle pair | codim = 2
b. | $f(x,y)=x^{3}-y^{2}$ | birth/death of $(1,2)$-handle pair | codim = 2
4a. | $f(x,y)=x^{4}+y^{2}$ | dovetail singularity | codim = 3
b. | $f(x,y)=x^{4}-y^{2}$ | dovetail singularity | codim = 3
|  | (  ) |

These are the germs. We also may invoke genericity to ensure that where a
single function on $T$ restricts to singular germs on several different points
of $T$, the local unfoldings are transverse. These is no difficulty making
this precise as the stratification of the unfoldings obey the two Whitney
conditions [W] are are in fact piecewise smooth submanifolds of the parameter
space.

So what does the singular set of a graphic in $[-1,1]\times[-1,1]\times S^{1}$
look like? It consists of 2D-manifold sheets of two types - saddle and sink -
meeting along birth/death 1D strata, which are allowed to have dovetail
singularities at isolated points. The 1D and 2D strata may pass transversely
through each other at points and the 2D strata may have transverse 1-manifolds
of intersection and isolated standard triple points.

The picture we must study is much simpler. Just as in [RS], the 2D sheets
labeled by source/sink tangencies may be discarded forming the reduced
graphic, as crossing these walls only changes the intersection $T\cap S$ by
inessential simple closed curves. The remaining saddle-labeled walls (it does
not matter for connectivities, but for convenience take their closures) divide
$[-1,1]\times[-1,1]\times S^{1}$ into complementary open regions
$\\{R_{i}\\}$, each of which has a constant topological pattern $T\cap S$, up
to birth and death of inessential sccs. Thus the 3D reduced graphic $G_{3}$ is
a straightforward generalization of the 2D one: 2-manifold sheets (perhaps
with borders) crossing in double curves and triple points.

###  4.5. The graphic on an annular surface

After having defined the 3D reduced graphic, we now replace it with a 2D
graphic by restricting it to the sub-surface
$\Sigma\subset[-1,1]\times[-1,1]\times S^{1}$, where $\Sigma$ is the graph of
the function described in Proposition A.5. Its structure as a graph provides
$\Sigma$ with a natural diffeomorphism to an annulus $[-1,1]\times S^{1}$,
parameterized by $(t,\theta)$, and that is how we will view it. This structure
as a graph allows us, when thinking of the intersection $T_{t,\theta}\cap
S_{s}$ that corresponds to a point in $\Sigma$, to take the sphere as fixed,
so henceforth we drop the subscript $s$. (When there is little risk of
confusion, we will also drop the subscript on $T$.) Each region of the graphic
in $\Sigma$ corresponds to a transverse intersection of $T_{t,\theta}$ with
$S$ in which some of the curves of intersection are essential in
$T_{t,\theta}$ and bound disks in $A_{t,\theta}$ or $B_{t,\theta}$, perhaps
both. For regions near $t=-1$, all such disks lie in $A_{t,\theta}$ (since $T$
is near the spine of $A_{t,\theta}$, which we may take transverse to the
height function $s$) and we label these regions $A$. Similarly, near $t=+1$,
all such disks lie in $B_{t,\theta}$ and we label such regions $B$.

Let us establish the following labeling and “artistic” convention for the
general region $R$ of our annular graphic. If $T_{t,\theta}\cap S$ contains a
scc $a$ essential in $T_{t,\theta}$ and compressing in the handlebody
$A_{t,\theta}$ label the region containing $(t,\theta)$ by $A$. (Do this
regardless of whether $T_{t,\theta}\cap S$ contains an essential scc
compressing in $B_{t,\theta}$). If $T_{t,\theta}\cap S$ does not contain any
essential scc compressing in $A_{t,\theta}$, but does contain a scc
compressing in $B$, label that Region $B$.

Artistic convention: If labels $A$ and $B$ alternate around a saddle wall
double point render the boundary between $A$-regions and $B$-regions to be a
1-manifold favoring $A$. We label this 1-manifold $M$, it separates $A$
regions from $B$ regions. We recall both this labeling and rendering
convention with the phrase: “favor $A$.”

Since near one boundary component of the annulus $\Sigma$ the regions are all
labelled $A$ and at the other they are all labelled $B$, there is at least one
component of $M$ that is essential in $\Sigma$. Such a component will be
homotopic in $\Sigma$, fixing a point, to a core curve $\\{pt\\}\times
S^{1}\subset[-1,1]\times S^{1}$. Here then is the plan for the proof of
Theorem 4.2. We will show

  * •

Each component of $M$ in $\Sigma$ naturally describes an edge-path in $2C(T)$
through vertices $(a,b)$ in which the first curve bounds an essential disk in
$A$ and the second an essential disk in $B$

  * •

For any component of $M$ that is essential in $\Sigma$ the associated edge-
path is a cwr representing $\tau$.

ABBA$\longrightarrow$renderABBA Figure 11.

The crux is to understand the intersection locus where two saddle walls cross,
what the four possible resolutions of the two saddles look like, and which
curves among these resolution could be labeled $a$ or $b$. (Again a curve is
labeled $a$ (resp $b$) if it is essential in $T$ and compresses in $A$ (resp
$B$).)

The singular pattern $\xi$ associated with a saddle wall crossing is present
simultaneously in the sphere $S$ and the surface $T$. Thus the pattern $\xi$
as a 4-valent graph is the same in $S$ and $T$. Furthermore the tangential
information is also preserved so at a saddle tangency we know which legs are
opposite (this is not quite the information of a cyclic ordering; it might be
called a ”dihedral ordering” as the information is a coset of $S(4)/D(4)$).
But interestingly, it is not true that the pairs
$(\mathcal{N}_{S}(\xi),\xi)\subset S$ and $(\mathcal{N}_{T}(\xi),\xi)\subset
T$ are necessarily homeomorphic. (Here $\mathcal{N}_{S}$ and $\mathcal{N}_{T}$
denote neighborhood in $S$ and $T$ respectively.) The important example is
Figure 12; it is a torus of revolution intersected with a sloping plane to
produce two circles of intersection. In the plane they have one positive and
one negative intersection, whereas in the torus they have two positive
intersections.

$\xi\subset$ Torus(plane not rendered) Figure 12.

For our analysis we need to list all possible pairs $(\mathcal{N}(\xi),\xi)$
up to homeomorphism. Let us start by enumerating the possibilities on $S$.
There are 4:

1234 Figure 13.

These four possible $\xi$s can yield neighborhood pairs on $T$ as shown in
Figure 14. (The small circles in the figures represent boundary components of
the neighborhood, typically essential circles in $T$.)

1234ororor(unique)Case 5 Figure 14.

So, overall, on $T$ we have 5 cases to consider: 1, 2, 3, 4, and 5 above.

Figure 15 displays the four local resolutions within $\mathcal{N}(\xi)$ of
$\xi$ in each of the 5 cases.

Arrows indicate sccs that must intersect Figure 15.

Most of $M\subset\Sigma$ runs along arcs in the graphic, representing saddle
walls for which one side is labelled $A$ and the other is labelled $B$. A
neighborhood of the corresponding saddle is a $3$-punctured sphere and
crossing the saddle wall splits one curve into two: $c\rightarrow
c^{\prime}\mathbin{\rotatebox[origin={c}]{180.0}{$\Pi$}}c^{\prime\prime}$.

We start with a simple lemma.

######  Lemma 4.3 (two of three).

If crossing a saddle wall transforms $c\rightarrow
c^{\prime}\mathbin{\rotatebox[origin={c}]{180.0}{$\Pi$}}c^{\prime\prime}$ and
if two of the three curves compress in $A$ (resp $B$) then so does the third.

###### Proof.

Consider capping off two boundaries of the 3-punctured sphere. ∎

Of course the third curve may be inessential in $T$, so crossing from a region
labelled $A$ to one labelled $B$ may represent two curves bounding essential
disks in $A$ fusing at the saddle into a curve inessential in $T$, leaving
other unrelated curves bounding disks in $B$. (In this case the two fused
curves at $A$ are isotopic in $T$.) Or passing through could represent a curve
that bounds a disk in $A$ fusing with a curve that is essential in both $A$
and $B$ to create a curve bounding a disk in $B$, or a similar fission. In
each case there is a unique essential intersection curve $a$ bounding a disk
in $A$ associated with the saddle wall and it is disjoint from every essential
intersection curve bounding a disk in $B$.

To expedite the discussion we introduce:

######  Definition 4.4.

  * Some conventions:

  * •

We call a scc $a$ (resp $b$) if it is essential in $T$ and compresses in $A$
(resp $B$).

  * •

An edge-path in $2C(T)$ in which each vertex is of type $(a,b)$ (as just
defined) is an admissible path.

  * •

Near a saddle wall crossing (swc) point $p$ we call a scc of type $a$ or $b$
local if it is a scc of one of four resolutions of $\xi_{P}$. Otherwise we
call it far.

With these conventions, we have just observed that any saddle wall that is
part of $M$ is associated to a ‘cloud’ of vertices in $2C(T)$ of the form
$(a,b)$, with $a$ uniquely defined, and $b$ any curve of that type coming from
the adjacent graphic region labelled $B$. Any two vertices in such a cloud are
connected via the relation

$(a,b)\rightarrow(a,b^{\prime})\>\textrm{since $a$, $b$, $b^{\prime}$ are all
pairwise disjoint.}$

Recalling our artistic rendering convention for defining the 1-manifold $M$,
we now consider how these clouds of vertices in $2C(T)$ are related as we pass
from one saddle wall to another, either through or around a corner of a vertex
in the graphic representing a saddle wall crossing.

######  Lemma 4.5.

For all cases (i) $\begin{array}[]{c|c}A&A\\\ \hline\cr B&B\end{array}$ , (ii)
$\begin{array}[]{c|c}A&B\\\ \hline\cr B&A\end{array}$ , (iii)
$\begin{array}[]{c|c}A&A\\\ \hline\cr A&B\end{array}$ , and (iv)
$\begin{array}[]{c|c}A&B\\\ \hline\cr B&B\end{array}$ there is an admissible
path in $2C(T)$ from some vertex in one cloud to some vertex in the other.

###### Proof.

Since each pair of vertices in a cloud are connected by an edge in $2C(T)$,
and such an edge is obviously an admissible path, it suffices to replace both
occurences of ‘some’ with ‘every’ in its two occurences in the lemma.

Case (i) $\begin{array}[]{c|c}A&A^{\prime}\\\ \hline\cr
B&B^{\prime}\end{array}$ : If any $b$ (or any $b^{\prime}$) is far it will be
unaffected by passing through any saddle wall, so pick $b=b^{\prime}=$ far.
Moreover, passing through the saddle wall between the regions $A$ and
$A^{\prime}$ will change the $a$ curve to a disjoint $a^{\prime}$. Then
$(a,b)\rightarrow(a^{\prime},b)=(a^{\prime},b^{\prime})$ is an admissible path
from one cloud to the other.

If $b$ and $b^{\prime}$ are both local there is a subtlety. In all cases at
most one diagonal pair of quadrants contain intersecting curves (see Figure
15); call this a dangerous diagonal. But the appropriate choice of a two-step
process avoids the dangerous diagonal, so all curves are disjoint and the path
is admissible:
$(a,b)$$(a,b^{\prime})$$(a^{\prime},b)$or$(a^{\prime},b^{\prime})$

Case (ii) BABA : It suffices to consider one arc of $M$. Regardless of whether
$b$ is local or far leave it fixed. The only problematic possibility for a
transition $(a,b)\rightarrow(a^{\prime},b)$ is if $a$ and $a^{\prime}$ are
intersecting curves from a dangerous diagonal. In this context case (5) cannot
occur since there $a$ and $a^{\prime}$ would have intersection number = 1. But
where it is possible, cases (2), (3), and (4), the topology is the same: $a$
and $a^{\prime}$ are crossing non-boundary-parallel sccs on a 4-punctured
sphere:

Figure 16.

But an innermost circle argument reduces the geometry to precisely the case
pictured in Figure 16, where we see that all four boundary curves also
compress in $A$. By Lemma 4.3 at most two of these boundary curves are
inessential in $T$; let $\partial$ be a boundary curve that is essential in
$T$. Then $(a,b)\rightarrow(\partial,b)\rightarrow(a^{\prime},b)$ is an
admissible path from one cloud to the other. (Note that the intermediate
vertex $(\partial,b)$ is not in either cloud.) Call this the lantern trick.

Case (iii) AA$\prime$A$\prime\prime$B : This is similar to case (ii).
Regardless of whether $b$ is local or far, leave it fixed. The only
problematic case in passing from $(a,b)$ to $(a^{\prime\prime},b)$ is if $a$
and $a^{\prime\prime}$ are intersecting sccs from a dangerous diagonal. As in
case (ii), case (5) cannot occur. Cases (2), (3), and (4) are also handled as
in case (iii) by the lantern trick.

Case (iv) BAB$\prime\prime$B$\prime$ : If any $b$, $b^{\prime}$,
$b^{\prime\prime}$ is far, select it and keep it constant, and keep $a$
constant as well. Then the corresponding vertex in $2C(T)$ lies in both clouds

Now assume all $b$, $b^{\prime}$, $b^{\prime\prime}$ are local.

Case (1): If $a,b,b^{\prime},b^{\prime\prime}$ are all disjoint, then
$(a,b)\rightarrow(a,b^{\prime\prime})$ is an admissible path.

Case (2), (3), (4): The only non-disjoint situation is when $b$ and
$b^{\prime\prime}$ are intersecting curves from a dangerous diagonal pair.
Again use the lantern trick.

Case (5): This case cannot occur: By intersection number, the three $B$
regions cannot contain a dangerous diagonal pair, so $A$ must be one of the
dangerous diagonal pair of quadrants. Then the other quadrant in that pair
must be $B^{\prime}$. The other two quadrants then are represented by the
boundary curves in the twice-punctured torus shown in Figure 15, one
compressible in $A$ and the other in $B$. In order to avoid a label $A$ in one
of these quadrants, the former curve must be inessential in $T$ so the twice-
punctured torus is actually only once-punctured. But in that case, the label
$B$ (or $B^{\prime\prime}$) for this quadrant implies that there is a (far)
$b$ curve, contradicting assumption. ∎

We now complete the proof of Theorem 4.2 as described in the plan above. Pick
an essential component of the $1$-manifold $M$ in $\Sigma$ and a saddle wall
edge in it. Beginning at that edge and moving around the component of $M$,
construct an admissible path through $2C(T)$ as described above, until one
returns to the original edge. Since the component is homotopic in $I\times
I\times S^{1}$, rel its beginning and ending point, to a simple loop
$\\{pt\\}\times S^{1}$ which represents $\tau$, the cwr given by the
admissible path also represents $\tau$. ∎

##  5\. The genus $3$ case

We are now in a position to prove the Powell Conjecture for genus $3$
splittings of $S^{3}$, by verifying Conjecture 4.1 in this case.

We need a bit of terminology. Suppose $H$ is a genus $3$ handlebody and
$D\subset H$ is a separating disk. Then $D$ divides $H$ into a solid torus
$H_{D}$ and a genus $2$ handlebody; the meridian disk $D^{\prime}$ of $H_{D}$
is well-defined up to proper isotopy in $H$ and is non-separating in $H$; call
$D^{\prime}$ the surrogate of $D$ in $H$.

######  Lemma 5.1.

Suppose $S^{3}=A\cup_{T}B$ is a genus $3$ splitting, and disks $a\subset
A,b\subset B$ are a weakly reducing pair. Then at least one of the following
$4$ disks is primitive:

  * •

$a$, if $a$ is non-separating in $A$,

  * •

the surrogate of $a$ if $a$ is separating in $A$

  * •

$b$, if $b$ is non-separating in $B$

  * •

the surrogate of $b$ if $b$ is separating in $B$

###### Proof.

Case 1: $a$ is separating. One possibility in this case is that
$\mbox{$\partial$}b$ lies on the boundary of the solid torus $A_{a}\subset A$.
In that case, if $b$ is non-separating, then $\mbox{$\partial$}b$ must be a
longitude of $A_{a}$, verifying that the surrogate of $a$ (and incidentally
also $b$) is primitive. If $b$ is separating, then $\mbox{$\partial$}b$ must
be parallel to $\mbox{$\partial$}a$ in $T$, so together $a$ and $b$ constitute
a reducing sphere for $T$, cutting off a genus $1$ Heegaard splitting whose
splitting surface is $\mbox{$\partial$}A_{a}$. It follows again that the
surrogate of $a$ (i. e. the meridian of $A_{a}$) is primitive (and
incidentally that the surrogate of $b$ is also primitive).

The other possibility is that $\mbox{$\partial$}b$ is essential in the
punctured genus $2$ component of $T-\mbox{$\partial$}a$. This implies that the
surrogate of $b$ (if $b$ is separating) is disjoint from the surrogate of $b$,
so we may as well jump immediately to Case 2.

Case 2: Both $a$ and $b$ are non-separating. Let $T^{\prime\prime}\subset
S^{3}$ be the torus obtained by simultaneously compressing $T$ along $a$ and
$b$. $T^{\prime\prime}$ bounds a solid torus $W$ on one side or the other, say
$W$ lies on the side in which $b$ lies. The surface $T^{\prime}$ obtained from
$T$ by compressing only along $a$, when pushed into $W$, describes a genus $2$
Heegaard splitting $W=A^{\prime}\cup_{T^{\prime}}B^{\prime}$ in which
$\mbox{$\partial$}W=\mbox{$\partial$}_{-}B^{\prime}$ and $b$ is the unique
boundary-reducing disk for the compression body $B^{\prime}$. It follows that
$b$ is primitive in $T^{\prime}$, hence in $T$. ∎

Following Lemma 5.1 and Corollary 3.6 there is an obvious candidate for
assigning a Powell equivalence class of homeomorphisms to the pair $(a,b)$: if
$a$ (or its surrogate) is primitive, assign the class of homeomorphisms that
carries that disk to the disk $a_{1}$ in the standard splitting
$A\cup_{T_{3}}B$. If $b$ (or its surrogate) is primitive, assign the class of
homeomorphisms that carries that disk to $b_{3}$ in the standard splitting. It
is only necessary to check that if two of the disks listed in Lemma 5.1 are
primitive, the assignments we have made coincide. So suppose both $a$ (or its
surrogate) and $b$ (or its surrogate) are primitive.

If both $a$ and $b$ are non-separating (so surrogates are not involved) and
therefore primitive, a straightforward outermost arc argument will find a disk
$b_{\perp}\subset B$ that is orthogonal to $a$ and disjoint from $b$.
Similarly, one can then find a disk $a_{\perp}\subset A$ that is dual to $b$
and is disjoint from both $a$ and $b_{\perp}$. There is then an obvious
homeomorphism $(S^{3},T)\to(S^{3},T_{3})$ (well-defined up to the $2$-strand
braid group of the torus) that takes the four disks $a,b_{\perp},b,a_{\perp}$
to, respectively $a_{1},b_{1},b_{3},a_{3}$. The Powell class of this
homeomorphism then coincides both with that determined by $a$ and that
determined by $b$.

There remains the possibility that one or both primitive disks are surrogates
of $a$ or $b$ and intersect, in which case they are orthogonal. There is then
a homeomorphism $(S^{3},T)\to(S^{3},T_{3})$ that carries one disk to $a_{1}$
and the other one to $b_{1}$. But a further Powell move (an exchange) carries
the pair to $a_{3}$ and $b_{3}$. The first homeomorphism is in the Powell
equivalence class determined by $a$ or its conjugate and the second by $b$ or
its conjugate. Since there is a Powell move taking one to the other, the
equivalence classes coincide.

The second and final step is to show that any two vertices in $2C(T)$ that are
connected by an edge determine the same Powell equivalence class. Given the
simple rule above for assigning a Powell equivalence class, this will follow
immediately from the following lemma.

######  Lemma 5.2.

For the genus $3$ Heegaard splitting $S^{3}=A\cup_{T}B$, suppose $a\subset A$,
$b_{1},b_{2}\subset B$ are three pairwise disjoint essential disks . Then
either

  * •

$a$ is primitive

  * •

$a$ is separating and its surrogate is primitive

  * •

$b_{1}$ and $b_{2}$ are parallel in $B$ or

  * •

$b_{1}$ is the surrogate of $b_{2}$ or vice versa.

The symmetric statement is true for three pairwise disjoint essential disks
$(a_{1},a_{2},b)$

###### Proof.

Case 1: $a$ is separating. If the boundary of either $b_{i}$ lies on the same
side of $a$ as $a^{\prime}$ then the disk is either orthogonal to $a^{\prime}$
or the disk together with $a$ is a reducing sphere cutting off a genus $1$
summand with $a^{\prime}$ as meridian. In any case, $a^{\prime}$ is primitive
and we are done. So we henceforth assume that the boundary of both $b_{i}$ lie
on the genus $2$ side of $T-\mbox{$\partial$}a$, and neither is parallel to
$\mbox{$\partial$}a$.

If, together, $b_{1}$ and $b_{2}$ separate $T$ then the component containing
$\mbox{$\partial$}a$ is either a torus in which $a^{\prime}$ is a meridian,
and so is primitive, or a genus $2$ surface. In the latter case, the other
component must be attached either by a single disk that cuts off a torus with
the other disk as meridian (so one of $b_{1},b_{2}$ is surrogate to the other)
or attached by two disks, one copy of each $b_{i}$, that cut off a sphere, in
which case $b_{1}$ and $b_{2}$ are parallel in $B$.

Case 2: $a$ is non-separating. One of the components of $B-(b_{1}\cup b_{2})$
contains $\mbox{$\partial$}a$ and has connected boundary; attaching a
$2$-handle to that component along $a$ creates a $3$-manifold $N$ that will
still be connected.

Also $\mbox{$\partial$}N$ is still connected, for if it were disconnected,
with boundary components $\mbox{$\partial$}_{1}N,\mbox{$\partial$}_{2}N$, each
containing a copy of $a$, then there would be a path through that component of
$B-(b_{1}\cup b_{2})$ from $\mbox{$\partial$}_{1}N$ to
$\mbox{$\partial$}_{2}N$. But there is already a path from
$\mbox{$\partial$}_{1}N$ to $\mbox{$\partial$}_{2}N$ in $T-a$, since $a$ is
non-separating. Together the paths would give a circle that intersects the
surface $\mbox{$\partial$}_{1}N$ in a single point, which is impossible in
$S^{3}$.

So let $\gamma\subset\mbox{$\partial$}N$ be a path from one copy of $a$ in
$\mbox{$\partial$}N$ to the other. Tubing the copies of $a$ together along the
boundary of a neighborhood of $\gamma$ gives a separating disk $a^{+}$ in $A$,
disjoint from the $b_{i}$, and $a$ is the surrogate for $a^{+}$. Now apply
case 1 to the three disks $a^{+},b_{1},b_{2}$. ∎

######  Corollary 5.3.

The Powell Conjecture is true for genus $3$ splittings.

###### Proof.

Suppose two vertices in $2C(T)$ are connected by an edge. With no loss of
generality, the two vertices are $(a,b_{1})$ and $(a,b_{2})$ with the three
disks $a,b_{1},b_{2}$ pairwise disjoint. In each case given by Lemma 5.2 the
procedure described for assigning a Powell equivalence class assigns the same
equivalence class to both cases. ∎

##  6\. Concluding remarks

The methods here seem to hold great promise for the general case, genus $g\geq
4$. But there are many difficulties: Presented with a pair of weakly reducing
disks $a\subset A,b\subset B$, the methods of Casson-Gordon ([CG]) describe
how to proceed to exhibit a complete non-separating weakly reducing collection
of disks $a_{1},...,a_{g_{1}},b_{g_{1}+1},...,b_{g}$ for $T$. But many choices
are involved in creating such a collection, and it is not clear that different
choices will lead, via Theorem 3.8, to the same Powell equivalence class of
homeomorophisms to $T_{g}$, even under strong inductive assumptions.

To begin with, the first step in the Casson-Gordon recipe, following the
choice of $a,b$, is to use the pair to maximally weakly reduce the splitting.
But this involves choices: for example, one could choose first to expand $a$
to a maximal collection of compressing disks in $A$ that are disjoint from
$b$. (Or one could do the reverse!) The resulting compression body lying in
$A$ may itself be well-defined, via the arguments of [Bo], but there are many
possible families of compressing disks for the compression body, and, to apply
Theorem 3.8, a specific collection of compressing disks in $A$ is required.

The next step in [CG] is to maximally compress the surface $T_{A_{b}}$ just
defined towards the side that contains $B$. The same difficulty arises: the
actual compression disks into the $B$ side that one uses involves a choice. On
the positive side, the resulting surface $T_{BA_{b}}$, which is well-defined,
can be shown to bound a handlebody $M_{BA_{b}}$ in which $T_{A_{b}}$ is a
Heegaard surface, providing perhaps an inductive foothold. Furthermore,
Powell-like moves (bubble-moves, eyeglass twists, etc) in the Heegaard surface
of $T_{A_{b}}\subset M_{BA_{b}}$ can be ’shadowed’ by similar moves in the
original splitting surface $T$. But again choice is involved in how to shadow
the moves. Also, even once a primitive disk is chosen for the splitting
$T_{A_{b}}\subset M_{BA_{b}}$ (inductively we might assume that this choice is
unique up to Powell-like moves), such a disk does not pick out a unique
primitive disk in our original surface $T$.

All along one could hope to show that different choices made lead to Powell
equivalent trivializations, but we have failed to find such arguments.

Perhaps a good intermediate question is this:

######  Conjecture 6.1.

Any eyeglass twist on $T_{g}$ is a Powell move.

Of course any eyeglass twist is a Goeritz element, so this conjecture is
weaker than the Powell Conjecture. An affirmative solution would bode well for
an affirmative solution of the Powell Conjecture as well. A weaker but perhaps
still challenging conjecture is this:

######  Conjecture 6.2.

Any eyeglass twist on $T_{g}$ is Goeritz conjugate to a Powell move.

That is, given an eyeglass twist $\rho:(S^{3},T_{g})\to(S^{3},T_{g})$ there is
a homeomorphism $h:(S^{3},T_{g})\to(S^{3},T_{g})$ so that $h^{-1}\rho
h:(S^{3},T_{g})\to(S^{3},T_{g})$ is a Powell move.

Observe the following cautionary example about eyeglass twists: In the
argument of Section 4.5 a crucial weakly reducing pair that may well appear
can be viewed as follows: One of the pair, $b\subset B$ say, is an innermost
disk of $B\cap S$ for some level sphere $S$; $b$ lies inside another disk $D$
that is also bounded by a curve of $S\cap T$, a curve to which $b$ is tangent.
Finally, the complement of $b$ in $D$ lies entirely in $A$ and becomes the
disk $a$ when the point of tangency is resolved. See Figure 17.

2pt $b$ at 100 75 $a$ at 135 140 $\mbox{$\partial$}D$ at 45 110

Figure 17.

Now imagine rolling the disk $b$ around the inside of $D$ so that the point of
tangency circumnavigates $\mbox{$\partial$}D$. It is easy to see that this is
a Goeritz move and only a bit harder to see that it is an eyeglass twist of
$b$ around $a$. Yet, the methods of Section 4.5 would provide the same pair of
disks $a,b$ throughout the Goeritz move. This suggests that, as it stands,
Section 4.5 is not in itself powerful enough to resolve even the weaker
Conjecture 6.1. Other tools may be needed to solve the problem for at least
some types of eyeglass twists.

In that spirit, we conclude with one rather limited class of eyeglass twists
which we are able to show are Powell moves. These are described using the
notation that precedes Lemma 3.7.

######  Definition 6.3.

An eyeglass $\eta\subset T$ satisfying the following conditions will be called
a short eyeglass.

  * •

There is a complete non-separating pair of weakly reducing disk collections
$\\{a_{1},...,a_{g_{1}}\\}$ and $\\{b_{g_{1}+1},...,b_{g}\\}$ for $T\subset
S^{3}$.

  * •

There is a simple closed curve $c\subset T$ that separates the two sets
$\\{\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{g_{1}}\\}$ and
$\\{\mbox{$\partial$}b_{g_{1}+1},...,\mbox{$\partial$}b_{g}\\}$ in $T$.

  * •

The lens $a\subset A$ of $\eta$ has $\mbox{$\partial$}a\subset T_{A}$.

  * •

The lens $b\subset B$ of $\eta$ has $\mbox{$\partial$}b\subset T_{B}$.

  * •

The interior of the bridge $v\subset\eta$ intersects at least one of $T_{A}$
or $T_{B}$ entirely in $P_{A}$ or $P_{B}$.

######  Proposition 6.4.

Suppose $\eta\subset T$ is a short eyeglass as above, with $c\subset T$ the
separating curve. Then an eyeglass twist along $\eta$ does not change the
Powell equivalence class of any homeomorphism
$h:(S^{3},T,c)\to(S^{3},T_{g},c_{g_{1}})$.

The central difference between this lemma and Lemma 3.7 is that we no longer
require the bridge $v$ to intersect $c$ in exactly one point; on the other
hand, we do require that the bridge of the eyeglass be disjoint either from
$\\{\mbox{$\partial$}a_{1},...,\mbox{$\partial$}a_{g_{1}}\\}$ or from
$\\{\mbox{$\partial$}b_{g_{1}+1},...,\mbox{$\partial$}b_{g}\\}$

###### Proof.

The proof is by induction on $|v\cap c|$; the case $|v\cap c|=1$ is Lemma 3.7.

Suppose with no loss of generality that $v\cap T_{A}\subset P_{A}$ and that
$|v\cap c|>1$. Consider the arcs of $v\cap P_{A}$ that are not incident to
$\mbox{$\partial$}a$. Each such divides $P_{A}$ into two sub-planar surfaces,
one of which cuts off a subarc of $c$ intersecting $v$ in an even number of
points. By picking an outermost such arc we may find a properly embedded
segment $v_{A}\subset v$ that cuts off a planar surface whose interior is
disjoint from $v$. The union of $v_{A}$ and the subarc $c_{A}\subset c$ it
cuts off is a simple closed curve $a_{c}$ which bounds a disk in $A$. Create
an eyeglass $\eta^{\prime}$, one of whose lenses is $a_{c}$, the other is
$b\subset B$ and whose bridge intersects $c$ in at least two fewer points. By
induction, a twist around $\eta^{\prime}$ does not change the Powell
equivalence class of $h$, and it vacates entirely everything in $P_{A}$ that
lies between $v_{A}$ and $c_{A}$, putting it on the other side of $v_{A}$.
Then $v_{A}$ can be isotoped across $c$, reducing $|v\cap c|$. ∎

##  Appendix A Refining the Rieck argument

We begin with a well-known result. Suppose $X$ is a compact topological space
and $f:X\to\mbox{$\mathbb{R}$}$ is continuous. Then $f$ has a unique minimal
value $r\in\mbox{$\mathbb{R}$}$. Now let $Z$ be a topological space,
$\mbox{$\mathbb{R}$}^{X}$ denote the space of continuous functions from $X$ to
$\mathbb{R}$ with the compact-open topology and consider a function
$f:Z\to\mbox{$\mathbb{R}$}^{X}$. Denote the image of any $z\in Z$ by
$f_{z}:X\to\mbox{$\mathbb{R}$}$. Then $f$ gives rise to a function
$r:Z\to\mbox{$\mathbb{R}$}$ by defining $r(z)=min\\{f_{z}(x)|x\in X\\}$.

######  Proposition A.1.

If $f$ is continuous then so is $r$.

###### Proof.

The standard proof is this: Since $X$ is compact, the compact-open topology
coincides with the topology of uniform convergence, in which the conclusion is
obvious. There is also a proof directly from the definition of compact-open
topology that is analogous to (but much easier than) the proof of Proposition
A.2 below. ∎

Notice that the function $r$ could equivalently (but more obscurely) be
defined this way: $r(z)$ is the least $r\in\mbox{$\mathbb{R}$}$ so that the
image of $H_{0}(f_{z}^{-1}(-\infty,r])\to H_{0}(X)$ is non-trivial. Viewed in
this way, the proposition generalizes. In particular, we will be interested in
an analogous statement for $H_{1}$.

For this version, assume also that $H_{1}(X)$ is non-trivial. Then given a
continuous function $f:X\to\mbox{$\mathbb{R}$}$ and a value
$y\in\mbox{$\mathbb{R}$}$, let $G_{y}\subset H_{1}(X)$ be the image under the
inclusion-induced map of $H_{1}(f^{-1}(-\infty,y])\to H_{1}(X)$. There are
obvious properties:

  * •

for $y$ below the minimum of $f(X)$, $G_{y}$ is trivial,

  * •

$y\leq y^{\prime}\implies G_{y}\subset G_{y^{\prime}}$, and

  * •

for $y$ above the maximum of $f(X)$, $G_{y}=H_{1}(X)\neq 0$.

So it makes sense in this situation to define a value

$r=\inf\\{y\in\mbox{$\mathbb{R}$}|G_{y}\neq 0\\}.$

With this definition then, in analogy to the discussion above, a function
$f:Z\to\mbox{$\mathbb{R}$}^{X}$ gives rise to a function
$r:Z\to\mbox{$\mathbb{R}$}$ by setting $r(z)$ to be the value $r$ just defined
for the function $f_{z}$. That is,

$r(z)=\inf\\{y\in\mbox{$\mathbb{R}$}|image(H_{1}(f_{z}^{-1}(-\infty,y])\to
H_{1}(X))\neq 0\\}$

######  Proposition A.2.

Since $f$ is continuous, so is $r$.

###### Proof.

The proof (not the underlying mathematics) is made a bit more complicated
because we we do not know when $G_{r}$ is trivial.

Pick any $z_{0}\in Z$ and any $\mbox{$\epsilon$}>0$. Let $r=r(z_{0})$.

Claim 1: There is an open neighborhood $N$ of $z_{0}$ so that $z\in N\implies
r(z)<r+2\mbox{$\epsilon$}$.

Proof of Claim 1: First observe that
$f_{z_{0}}^{-1}(-\infty,r+\mbox{$\epsilon$}]$ is closed in $X$, hence compact,
so $\\{z\in
Z|f_{z}(f_{z_{0}}^{-1}(-\infty,r+\mbox{$\epsilon$}])\subset(-\infty,r+2\mbox{$\epsilon$})\\}$
is an open neighborhood $N$ of $z_{0}$. Put another way, for $z\in N$,

$f_{z_{0}}^{-1}(-\infty,r+\mbox{$\epsilon$}]\subset
f_{z}^{-1}(-\infty,r+2\mbox{$\epsilon$}).$

We know that the image of
$H_{1}(f_{z_{0}}^{-1}(-\infty,r+\mbox{$\epsilon$}])\to H_{1}(X)$ is non-
trivial, and this factors through
$H_{1}(f_{z}^{-1}(-\infty,r+2\mbox{$\epsilon$}))\to H_{1}(X)$, so the image of
the latter is non-trivial. Hence $r(z)<r+2\mbox{$\epsilon$}$ as required.

Claim 2: There is an open neighborhood $M$ of $z_{0}$ so that $z\in M\implies
r(z)>r-2\mbox{$\epsilon$}$.

Proof of Claim 2: Paralleling the argument in Claim 1, there is an open
neighborhood $M$ of $z_{0}$ so that for $z\in M$,

$X-f_{z_{0}}^{-1}(-\infty,r-\mbox{$\epsilon$})=f_{z_{0}}^{-1}[r-\mbox{$\epsilon$},\infty)\subset
f_{z}^{-1}(r-2\mbox{$\epsilon$},\infty)=X-f_{z}^{-1}(-\infty,r-2\mbox{$\epsilon$}].$

Hence

$f_{z}^{-1}(-\infty,r-2\mbox{$\epsilon$}]\subset
f_{z_{0}}^{-1}(-\infty,r-\mbox{$\epsilon$})$

By definition of $r$, $H_{1}(f_{z_{0}}^{-1}(-\infty,r-\mbox{$\epsilon$}))\to
H_{1}(X)$ has trivial image; it follows that
$H_{1}(f_{z}^{-1}(-\infty,r-2\mbox{$\epsilon$}])\to H_{1}(X)$ has trivial
image, so $r(z)>r-2\mbox{$\epsilon$}$ as required.

Hence for $z\in M\cap N$, $|r(z)-r(z_{0})|<2\mbox{$\epsilon$}$, showing that
$r$ is continuous at $z_{0}$. ∎

Now return to the setting of Section 4.3: $T$ is a genus $g\geq 2$ Heegaard
surface in $S^{3}$ and $T_{t},t\in[-1,1]$ describes a sweep-out of
$S^{3}=A\cup_{T}B$ from a spine of $A$ to a spine of $B$. Similarly
$S_{s},s\in[-1,1]$ describes the sweep-out of $S^{3}$ by $2$-spheres, from the
south pole no the north pole of $S^{3}$. The intersection patterns of the two
surfaces give rise to a reduced graphic $\Gamma\subset[-1,1]\times[-1,1]$ for
which a complementary region $R$ is labelled $E$ if and only if for each
$(t,s)\in R$, $T_{t}\cap S_{s}$ contains a circle that is essential in
$T_{t}$.

######  Proposition A.3.

There is a function $r:[-1,1]\to[-1,1]$ whose graph
$\\{(t,r(t))\\}\subset[-1,1]\times[-1,1]$ is transverse to $\Gamma$ and each
region that the graph intersects is labelled $E$.

###### Proof.

Apply Proposition A.2 to $X=T$ and $Z=[-1,1]$ by taking for $f_{t}$ the height
function $s$ restricted to $T_{t}\subset S^{3}$. Denote by $T_{t,(-\infty s]}$
the subsurface of $T_{t}$ that lies in $\cup_{s^{\prime}\leq
s}S_{s^{\prime}}.$, that is, in the part of $S^{3}$ that is south of the
sphere $S_{s}$. We then have a function $r:[-1,1]\to[-1,1]$ with the property
that for values of $s<r(t)$, none of the first homology of $T_{t,(-\infty,s]}$
persists into all of $T_{t}$, but for any values of $s>r(t)$ some does. It
follows that the topology of $T_{t,(-\infty s]}$ changes as $s$ rises through
$r(t)$, so the graph of $r(t)$ must lie in the $1$-skeleton of the reduced
graphic.

We can say more: For a generic value of $t$, the point $(t,r(t))$ will lie in
an edge of the reduced graphic, so at that point there is a saddle tangency of
$T_{t}$ with $S_{r(t)}$. That is, $T_{t,(-\infty,r(t)+\mbox{$\epsilon$}]}$ is
obtained from $T_{t,(-\infty,r(t)-\mbox{$\epsilon$}]}$ by attaching a
$1$-handle.

At a non-generic value of $t$, the point $(t,r(t))$ will be a vertex in the
graphic. In that case, one can get from the region just below $(t,r(t))$ to
the region just above by passing through two edges in the graphic, so
$T_{t,(-\infty,r(t)+\mbox{$\epsilon$}]}$ is obtained from
$T_{t,(-\infty,r(t)-\mbox{$\epsilon$}]}$ by attaching two $1$-handles. By
definition of $r(t)$, $T_{t,(-\infty,r(t)-\mbox{$\epsilon$}]}$ corresponds to
a subsurface of $T_{t}$ whose first homology is trivial in all of $T_{t}$. In
particular, $T_{t,(-\infty,r(t)+\mbox{$\epsilon$}]}$ contains no non-
separating simple closed curves i. e.,
$T_{t,(-\infty,r(t)+\mbox{$\epsilon$}]}$ is planar.

We now invoke an easy lemma:

######  Lemma A.4.

Suppose $P$ is a compact possibly disconnected planar surface and $P_{+}$ is
the surface obtained from $P$ by attaching no more than three $1$-handles.
Then either genus $P_{+}\leq 1$ or some boundary component of $P$ is non-
separating in $P_{+}$.

###### Proof.

If any of the $1$-handles has its ends attached to different components of
$P$, it is wasted: The resulting surface $P^{\prime}$ is still planar, and the
new boundary component created is non-separating in $P_{+}$ if and only if at
least one of the boundary components to which it is attached is non-separating
in $P_{+}$. So we could juat replace $P$ with $P^{\prime}$ in the argument,
but now with one fewer handle to attach.

So we may as well assume that all $1$-handles are attached to the same
component of $P$ and so we can take $P$ to be connected. In that case, if any
$1$-handle has exactly one end on any component of $\mbox{$\partial$}P$ then
that component becomes non-separating, finishing the argument. So we may as
well assume that no $1$-handle has a single end on an original boundary
component of $P$.

Now imagine attaching the $1$-handles sequentially, say $h_{1},h_{2},h_{3}$.
The first handle $h_{1}$ must have both ends on the same component of
$\mbox{$\partial$}P$ so the result is still a planar surface $P^{\prime}$,
with exactly two components of $\mbox{$\partial$}P^{\prime}$ not appearing as
components of $\mbox{$\partial$}P$. If, after attaching $h_{2}$, the surface
is still planar, attaching $h_{3}$ can raise the genus to at most $1$,
completing the proof. So we need to have the attachment of $h_{2}$ raise the
genus, so it must connect the two new boundary components of $P^{\prime}$. The
resulting surface $P^{\prime\prime}$ now has genus $1$ and only a single
boundary component not among the components of $\mbox{$\partial$}P$. It
follows that both ends of $h_{3}$ must lie on the same component of
$\mbox{$\partial$}P^{\prime\prime}$, either the new one or one of the original
components of $\mbox{$\partial$}P$. In any case, attaching $h_{3}$ does not
raise the genus any further, completing the proof. ∎

Following Lemma A.4, if every boundary component of
$T_{t,(-\infty,r(t)+\mbox{$\epsilon$}]}$ is inessential in $T_{t}$ then
$genus(T_{t})\leq 1.$ So under the hypothesis that $genus(T)\geq 2$ the
function $r(t)+\mbox{$\epsilon$}$ for small enough $\epsilon$ will be the
function we seek: its graph in $[-1,1]\times[-1,1]$ will lie entirely in the
regions labelled $E$. ∎

The argument is readily enlarged to include the third parameter $\theta$. That
is,

######  Proposition A.5.

In the parameter space $[-1,1]\times[-1,1]\times S^{1}$ as described in
Section 4.3, there is a function $r:I\times S^{1}\to I$ so that the graph
$(t,r(t,\theta)+\mbox{$\epsilon$},\theta)\subset[-1,1]\times[-1,1]\times
S^{1}$ of $r+\epsilon$ is transverse to the graphic and intersects only those
regions of the graphic in which $S_{s}\cap T_{t,\theta}$ contains circles that
are essential in $T_{t,\theta}$.

###### Proof.

The only additional case to consider is that in which $(t,r(t,\theta),\theta)$
is a (codimension $3$) vertex in the graphic. But in that case, the region
just below the vertex can be connected to the region just above the vertex by
passing through three saddle “walls”. Thus Lemma A.4 is still applicable, now
with $n=3$. ∎

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