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\title{Comparing Heegaard and JSJ structures of orientable
		3-manifolds}  
\author{Martin Scharlemann}
\address{\hskip-\parindent
        Martin Scharlemann\\
        Mathematics Department\\
        University of California\\
        Santa Barbara, CA USA}
\email{mgscharl@math.ucsb.edu}

\author{Jennifer Schultens}
\address{\hskip-\parindent
        Jennifer Schultens\\
        Mathematics Department\\
        Emory University \\
        Atlanta, GA  30322 USA
        }
\email{jcs@mathcs.emory.edu}

\thanks{Research supported in part by NSF grants and MSRI}

\begin{abstract} 

The Heegaard genus $g$ of an irreducible closed orientable $3$-manifold
puts a limit on the number and complexity of the pieces that arise in
the Jaco-Shalen-Johannson decomposition of the manifold by its
canonical tori. For example, if $p$ of the complementary components
are not Seifert fibered, then $p \leq g-1$.  This generalizes work of
Kobayashi \cite{Ko}.  The Heegaard genus $g$ also puts explicit bounds
on the complexity of the Seifert pieces.  For example, if the union of
the Seifert pieces has base space $P$ and $f$ exceptional fibers, then
$$f - \chi(P) \leq 3g - 3 - p.$$

\end{abstract}
\maketitle

\section{Introduction}

Nearly a century ago, Heegaard noticed that quite a few
$3$-manifolds could be written as the union of two handlebodies 
(\cite{He}, see also \cite{Prz} for a translation of the relevant
parts) .  Later it was discovered that this first global structure
theorem applied in fact to all
$3$-manifolds.  Now called a {\em Heegaard splitting} of the
$3$-manifold, this structure has proven to be a deceptively simple
picture because, although the existence of the structure is easy to
prove, it is not unique.  A single manifold may have several Heegaard
splittings and the relationship between the various splittings has
been difficult to understand.

A modern and more useful structure theorem for $3$-manifolds, due to
Jaco-Shalen and Johannson, does not have the problem of
non-uniqueness.  In its simplest form the theorem states that, for any
irreducible orientable closed $3$-manifold $M$ there is a collection
$\Theta$ of incompressible tori (called the {\em canonical tori} of
$M$) so that each complementary component of
$M - \Theta$ is either a Seifert manifold (possibly without exceptional
fibers) or is both acylindrical (any properly imbedded incompressible
annulus is boundary parallel) and atoroidal (any properly imbedded
torus is boundary parallel).  Moreover, $\Theta$ is unique up to
ambient isotopy in $M$.

The connection between Heegaard structure and this ``JSJ structure''
has been poorly understood.  The only significant information comes
from two theorems of Kobayashi \cite[Theorems 1 and 2]{Ko}.  The first
states that if a closed orientable $3$-manifold has a
genus $g$ Heegaard splitting then $\Theta$ has at most $3g - 3$
complementary components.  Moreover, if it has exactly $3g - 3$, then
every complementary component is atoroidal (though not necessarily
acylindrical); the second theorem gives more detail about the
structure of these $3g - 3$ components, particularly the non-Seifert
pieces.

In part, we here expand on Kobayashi's theme.  For example, we show
that if $M$ has a genus $g$ Heegaard splitting, then at most $g-1$
complementary components are not Seifert manifolds.  Moreover, among
the Seifert pieces, if $n'$ fiber over the twice punctured projective
plane or, with one exceptional fiber, over the once punctured
projective plane, and $n$ other components are also toroidal, then the
number of complementary components is no more than
$3g - 3 - n - n'/2$. (See Corollary \ref{maincor}.) We know no
counterexamples to the stronger statement that there are no more than
$3g - 3 - n - n'$ toroidal complementary components.

In addition, we find limits on the complexity of the Seifert
pieces. In order of increasing generality, these limits occur as
Theorems
\ref{main1}, \ref{main2}, and \ref{main3}.  Mostly to obtain the
limits on the number of toroidal components mentioned above, the
results are a bit stronger than the following more easily stated
corollary: For $p$ the number of non-Seifert
components, $P$ the base surface of the Seifert parts, and $f$ the
number of exceptional fibers, we have 
$f - \chi(P) \leq 3g - 3 - p$. (See Corollary \ref{final}.)

Finally, Kobayashi's structure theorem (\cite[Theorem
2]{Ko}) for ``full" Haken manifolds is shown to have the following
beautiful explanation (see Corollary \ref{main1+}):  When an
irreducible Heegaard splitting of a full Haken
$3$-manifold is put in thin position (see \cite{ST}), then it is
strung out like an array of jewelry:  The setting consists of Seifert
pieces connected together by amalgamating tori. Embedded in this
setting are $g-1$ ``jewels", each homeomorphic to the complement of a
$2$-bridge link in $S^3$.

Here is an outline:  In Section 2 we briefly recount the theory of
generalized Heegaard splittings and untelescopings, mostly from
\cite{ST}.  The first of two core sections is Section 3, wherein we
explain the delicate process of positioning the canonical tori
optimally with respect to the surfaces that arise from a strongly
irreducible generalized Heegaard splitting.  In the end, the
Seifert pieces can have one of two positions, aligned or
non-aligned.  Section 4 explains the connection between the complexity
of Seifert pieces and how they intersect the compression-bodies of the
generalized splitting.  The results here apply only to the aligned
pieces, but they are the critical ones.  The remainder of the paper is
designed to incorporate the non-aligned pieces so that ultimately we
are able to make statements that do not require any knowledge about
which Seifert pieces are aligned and which are not.  So, as the
argument progresses, the statements get a bit weaker but the generality
with which they can be applied improves.  By the end of Section 6 the
distinction between the aligned and non-aligned positions of the
Seifert pieces no longer needs to be considered;
an inequality shows (roughly) that the vertical index, with which
the complexity of the aligned pieces is measured, bounds the horizontal
index, with which the complexity of the non-aligned pieces is measured.

\section{Heegaard splittings and their untelescopings}

\begin{defin}

A {\em compression body} $H$ is a connected 3-manifold obtained from a
closed surface $\bdd_- H$ by attaching 1-handles to $\bdd_- H \times \{
1 \} \subset \bdd_- H \times I$.  Dually, a compression body is
obtained from a connected surface $\bdd_+ H$ by attaching 2-handles to
$\bdd_+ H \times \{ 1 \} \subset \bdd_+ H \times I$ and 3-handles to any
2-spheres thereby created. The cores of the $2$-handles are called {\em
meridian disks}

For $H$ a compression body, define the index $J(H) = \chi(\bdd_- H)
- \chi(\bdd_+H)$.

\end{defin}

A {\em Heegaard splitting} $M = \aub$ of a compact orientable 3-manifold
consists of an orientable surface $S$ in $M$, together with two
compression bodies $A$ and $B$ so that $S = \bdd_+ A = \bdd_+B$ and 
$M = \aub$.  $S$ itself is called the splitting surface.  The genus of
the splitting is defined to be the genus of $S$.

A {\em stabilization} of \aub\ is
the Heegaard  splitting obtained by adding to $A$ a regular
neighborhood of a proper arc in $B$ which is parallel in $B$ to an arc
in $S$.  A stabilization has genus one larger and, up to isotopy, is
independent of the choice of arc in $B$.If  the
construction is done symmetrically to an arc in $A$ instead, the two
splittings are isotopic.  

Recall the following (see e. g. \cite{Sc}):  If there are meridian disks
$D_{A}$ and $D_{B}$ in $A$ and $B$ respectively so that $\bdd D_{A}$ and
$\bdd D_{B}$ intersect in a single point in $S$, then \aub\ can be
obtained by stabilizing a lower genus Heegaard splitting.  We then say
that \aub\ is {\em stabilized}. If there are meridian disks $D_{A}$ and
$D_{B}$ in $A$ and $B$ respectively so that $\bdd D_{A}$ and $\bdd
D_{B}$ are disjoint in $S$, then \aub\ is {\em weakly reducible}.  If
there are meridian disks so that $\bdd D_{A} = \bdd D_{B}$, then \aub\
is {\em reducible}.  It is easy to see that reducible splittings are
weakly reducible and that (except for the genus one splitting of $S^3$)
any stabilized splitting is reducible. 
It is a theorem of Haken \cite{Ha} that any Heegaard splitting of a
reducible 3-manifold is reducible and it follows from a theorem of
Waldhausen \cite{W} that a reducible splitting of an irreducible
manifold is stabilized.

\begin{defin}

Suppose $M$ is an irreducible closed orientable $3$-manifold. 
A {\em generalized} Heegaard splitting of $M$ is a structure

  $$M = (A_1 \cup_{S_1} B_1) \cup_{F_1} (A_2 \cup_{S_2} B_2) \cup_{F_2}
... 
\cup_{F_{m-1}} (A_m \cup_{S_m} B_m).$$ 
Here each $A_i$ and $B_i$ is a compression body, $\bdd_+ A_i = S_i =
\bdd_+ B_i$ (so $A_i \cup_{S_i} B_i$ is a Heegaard splitting), 
$\bdd_- B_i = F_i = \bdd_- A_{i+1}$, and each $F_i$ is incompresible
in $M$.  We say that a generalized Heegaard splitting is {\em strongly
irreducible} if each Heegaard splitting $A_i \cup_{S_i} B_i$ is strongly
irreducible. For a generalized splitting we will often denote
 $S = \cup_{i = 1}^{m} S_i$ and $F = \cup_{i = 1}^{m-1} F_i$. (See
Figure 1.)

\end{defin}

\begin{figure}
\centering
\includegraphics[width=.6\textwidth]{thin.eps}
\caption{}
\end{figure} 

The central theorem of \cite{ST}, via the calculation
of \cite[Lemma 2]{Sc3},  directly implies this:

\begin{thm} 

Suppose $M$ is an irreducible closed orientable $3$-manifold and $M$ has
a genus $g$ Heegaard splitting. Then $M$ has a strongly
irreducible generalized Heegaard splitting 
$$(A_1 \cup_{S_1} B_1) \cup_{F_1} (A_2 \cup_{S_2} B_2) \cup_{F_2}
... 
\cup_{F_{m-1}} (A_m \cup_{S_m} B_m)$$
so that $$\sum_{i = 1}^{m} J(A_i) = \sum_{i = 1}^{m} J(B_i) 
= \chi(F) - \chi(S) \leq 2g - 2.$$

\end{thm}

This structure is created by {\em untelescoping} a minimal genus
Heegaard splitting of $M$. Notice that the theorem is a tautology if $M$
has a genus $\leq g$ Heegaard splitting that is strongly irreducible.

\section{Heegaard splitting surfaces vs. canonical tori}

We would like to understand how the surfaces $S$ and $F$ of a
generalized strongly irreducible Heegaard splitting of $M$ intersect the
canonical tori $\Theta$ of $M$.  More generally, we would like to
simplify as much as possible the intersections of $S$ and $F$ with
those complementary components of $\Theta$ in $M$ that are Seifert.

We begin with a fairly easy argument:

\begin{defin}

A properly imbedded collection of annuli in a 3-manifold $M$ is {\em
essential} if it is incompressible and no component is $\bdd$-parallel.

\end{defin}

\begin{lemma}
\label{placement1}

Suppose $M$ is an irreducible closed orientable $3$-manifold with a
strongly irreducible generalized Heegaard splitting 
$$(A_1 \cup_{S_1} B_1) \cup_{F_1} (A_2 \cup_{S_2} B_2) \cup_{F_2}
... 
\cup_{F_{m-1}} (A_m \cup_{S_m} B_m)$$ and suppose $\Theta$ is the
set of canonical tori of $M$.  Then $\Theta$ may be isotoped so
that $F \cap \Theta$ consists
of curves essential in both $F$ and $\Theta$ and so that $\Theta$
intersects each compression body
$A_i$ and $B_i$ only in essential annuli and incompressible tori.

\end{lemma}

Note that an incompressible torus in a compression body $H$ must be
parallel to a torus component of $\bdd_- H$.

\pf  Here is a sketch.  More detail can be found in e. g. \cite{Sc4}.
First note that, since both $F$ and $\Theta$ are incompressible, a
simple innermost disk argument can be used to remove all components of
$F
\cap \Theta$ that are inessential in either surface (hence both
surfaces).  So we can assume that all components of $F \cap \Theta$ are
essential in both $F$ and $\Theta$.  Now the surface $S_i$ can be used
to ``sweep out'' the region between $F_{i-1}$ and $F_i$, once certain 
$1$-complexes incident to $F_{i-1}$ and $F_i$ are removed (the spines
of $A_i$ and $B_i$ respectively).  At the beginning of the sweep-out,
each component of $S_i \cap \Theta$ is either essential or bounds a
tiny disk in $A_i$, each disk corresponding to an intersection point
of the spine of $A_i$ with $\Theta$.  Similarly, at the end  of the 
sweep-out, each component of $S_i \cap \Theta$ is either essential or
bounds a tiny disk in $B_i$, each disk corresponding to an intersection
point of the spine of $B_i$ with $\Theta$.  There cannot
simultaneously be disk components of intersection of $\Theta$ with
$A_i$ and with $B_i$, since $A_i \cup_{S_i} B_i$ is strongly
irreducible.  So in some positioning, all components of intersection
are essential.  

Repeat this argument for each $S_i, 1 \leq i \leq m$. 
Then each component of $(F \cup S) \cap \Theta$ is essential in
$\Theta$ and hence in $F \cup S$.  At this point we know that $\Theta$
intersects each compression body $A_i$ and $B_i$ only in incompressible
annuli and tori.  Now remove any boundary parallel annuli by isotopies.
\qed

\begin{defin}

Suppose $V$ is a Seifert manifold with base space $P$.  A surface $T
\subset V$ is {\em vertical} if it is a union of generic fibers and is
{\em horizontal} if it is transverse to each fiber.

\end{defin}

Note that a vertical surface then must have Euler characteristic
zero, and so is a union of annuli and tori.

\begin{defin}  Suppose $E$ is a possibly non-orientable surface and
$\xi$ is an $I$-bundle over $E$ whose total space is orientable. 
Then let $\bdd \xi$ denote the restriction of $\xi$ to $\bdd E$ and
let $\dot{\xi}$ denote the associated $\bdd I$-bundle of $E$.  The
boundary of the total space of $\xi$ is the union of the two.

\end{defin}

\begin{thm}
\label{vert/horiz}

{\rm \cite[Theorem VI.34]{Ja}} Any properly imbedded incompressible and
$\bdd$-incompressible $2$-sided surface in an orientable Seifert
manifold $V$ can be properly isotoped so that either it is vertical or
it is horizontal.  If it is horizontal, then $V$ is the union, along
$\dxi$, of two copies of an $I$-bundle $\xi$ over a surface $E$, and
the incompressible surface consists of parallel copies of $\dxi$.

\end{thm}

Of course, if $\xi$ is an orientable $I$-bundle (so $E$ is
orientable), then $V$ fibers over the circle with fiber $E$.

\begin{defin}

When a Seifert manifold $V$ is expressed as the union, along $\dxi$, of
two copies of an $I$-bundle $\xi$ over a surface $E$, we call $E$ the
associated $I$-base of this construction, and say that $V$ is an
$I$-bundle construct over $E$.  $E$ is a branched cover of
the Seifert base $P$ of $V$.

\end{defin}

For the purposes of this paper, we will always be able to assume that
$\chi(E) \leq -1$, since if $E$ is the M{\"o}bius band or the
annulus we could fiber $V$ differently so that $\dxi$ is vertical in
$V$. Also, for expository purposes, little would be lost by always
taking $\xi$ to be a product bundle, so that $(\xi; \dxi, \bxi) = (E
\times I; E \times \bdd I, \bdd E \times I)$.  

\begin{cor}
\label{placement2}

Under the hypotheses of Lemma \ref{placement1}, $F$ may be further
isotoped so that it intersects each Seifert component of $M - \Theta$
either in a horizontal surface or in vertical essential
annuli and incompressible tori.

\end{cor}

\pf  Let $V$ be a Seifert component of $M - \Theta$ and isotope
$\Theta$ as provided in Lemma \ref{placement1}.  Then each component
of $F \cap V$ is incompressible in $V$, since $F$ is incompressible in
$M$ and all curves of $F \cap \Theta$ are essential in $F$.  If any is
$\bdd$-compressible then, dually, some annulus $A$ in $\Theta - F$ $
\bdd$-compresses to $F$. The $\bdd$-compression cannot turn $A$ into
an essential disk, since $F$ is incompressible, so $A$ must be
$\bdd$-parallel in the submanifold $A_i \cup_{S_i} B_i$ in which it
lies.  It's easy to see that then either $A$ intersects either $A_i$
or $B_i$ in a $\bdd$-parallel annulus or $A_i \cup_{S_i} B_i$ would be
weakly reducible.  Either case contradicts our hypotheses, so $A$ is
also $\bdd$-incompressible.  The result then follows from Theorem
\ref{vert/horiz}.  \qed

We would like now similarly to simplify the positioning of $S$ in each
Seifert component of $M - \Theta$. We can come surprisingly
close.

\begin{thm}
\label{placement3}

Assume the hypotheses of Lemma \ref{placement2}, and let $V$ be the
union of Seifert components of $M - \Theta$.  Then
$S$ may be further isotoped so that for each $S_i$, every component
but at most one of $S_i \cap V$ is either horizontal or vertical. 
The one exception is obtained either from one or two horizontal
components by attaching a vertical tube between them, or from one or
two vertical components by attaching a single horizontal tube, i. e. a
tube whose core projects homeomorphically to an embedded arc in the
base space
$P$.

\end{thm}

This positioning of $\Theta$ with respect to the generalized Heegaard
splitting is called a {\em preferred positioning}.

\pf  If $S_i \cap V$ is incompressible in $V$, the proof mimics that of
Lemma \ref{placement2}.  So suppose it is compressible and, with no loss
of generality, suppose it compresses into $A_i \cap V$.  After
compressing maximally into $A_i \cap V$ it is standard to see (via
strong irreducibility) that the resulting surface $\hat{S_i}$ is
incompressible in
$V$, so we can reconstruct $S_i$ by starting with an incompressible
surface in $V$ and attaching tubes on one side (dual to the compressing
disks we've just used.) Now consider $B_i$. By strong irreducibility,
any meridian disk $D$ in $B_i$ must have a boundary that runs along each
tube (i. e. intersects each compressing disk for $A_i$ in $V$).  Either
$D$ lies in $V$ or an outermost arc cuts off a $\bdd$-compressing disk
$D' \subset B_i$ which $\bdd$-compresses an annulus of $\Theta - S_i$ to
$S_i$ through $B_i$.  Since the annulus cannot be $\bdd$-parallel, the
result is a compressing disk in $B_i$.  If the disk were outside $V$,
this would contradict strong irreducibility, so we deduce that it lies
in
$V$.  The upshot is that we may assume that $S_i \cap V$ compresses in
both $A_i \cap V$ and $B_i \cap V$ and, indeed, in the same component of
$V$.  So henceforth we may assume $V$ is connected.

Select a family of essential $2$-sided arcs in the base
space $P$ of $V$, chosen so that their complement consists entirely of
disks, each containing at most one exceptional point and sufficiently
plentiful that no disk lies on both sides of the same arc.  Let
$\Aaa \subset V$ denote the family of vertical annuli that covers these
arcs.  The complementary components of $\Aaa$ in $V$ are solid
tori $T_1,...,T_t$.  Let $T = \cup_{j = 1}^t T_j$. 

\begin{figure}
\centering
\includegraphics[width=.6\textwidth]{position.eps}
\caption{}
\end{figure}
 
{\bf Claim:}  $S_i$ can be isotoped so that it intersects each
$\bdd T_j$ only in essential curves.  (See Figure 2.)

{\it Proof of claim:}  This is an adaptation of the argument of Lemma
\ref{placement1}.  The details of the proof are a little more complex
than the somewhat condensed version about to be given here. A more
complicated argument in the same spirit, with full details, occurs in
\cite[Section 2.2]{RS}.

{\bf Step 1:} Parameterize the sweep-out of Lemma \ref{placement1} by
the interval.  Within it is a subinterval in which every curve in $S_i
\cap \bdd V$ is essential in both $S_i$ and $V$, but just before the 
subinterval a curve in $S_i \cap \bdd V$ cuts off a
meridian disk of $A_i$, say, lying in $\bdd V$ and just after the
subinterval a curve in $S_i \cap \bdd V$ cuts off a meridian disk of
$B_i$ lying in $\bdd V$.  Now consider how
$S_i$ intersects $\Aaa$ and $\cup_{j = 1}^t \bdd T_j$ during the
sweep-out (now meaning sweep-out through this subinterval).  There will
be no critical points of intersection in $\bdd V$ (since every
component of $S_i \cap V$ remains essential in $\bdd V$ throughout the
sweep-out), so in fact we may as well take the curves
$S_i \cap \bdd V$ to be fixed throughout the sweep-out, and minimally
intersecting the set of curves $\bdd \Aaa$.  We can also remove by an
isotopy all $\bdd$-parallel annuli in $S_i \cap V$. 

{\bf Step 2:} At the beginning of the sweep-out, no curve of
intersection of $S_i \cap \bdd T$ can be inessential in $\bdd T$ and
cut off a meridian of $B_i$ and at the end, no such curve can cut off
a meridian of $A_i$.  So again there is a subinterval where no
inessential curve of $S_i \cap \bdd T$ in $T$ can cut off a meridian
in either $A_i$ or $B_i$. Restrict to such a subinterval,
chosen to have the property that just before the beginning a meridian of
$A_i$ is cut off and just after the end, a meridian of $B_i$ is cut
off. 

\begin{figure}
\centering
\includegraphics[width=.7\textwidth]{slop.eps}
\caption{}
\end{figure}  

{\bf Step 3:} It is still possible that during the sweep-out there will
be curves in $S_i \cap \bdd T$ that are inessential in both surfaces;
those that lie entirely inside of $\Aaa$ are easily removed, so we
focus on those that slop across the annuli $\bdd V \cap \bdd T$.  (See
Figure 3.) Anytime during the sweep-out that there is such a curve in
$S_i \cap \bdd T$ (inessential in $T$ and $S_i$, and not lying
entirely in $\Aaa$) there will be an outermost inessential arc of $S_i
\cap \Aaa$, cutting off a disk that lies in $A_i$ or in $B_i$.  This
allows a $\bdd$-compression of $S_i$ into $\bdd V$ and thereby reveals
a meridian disk in $A_i$ or $B_i$ lying in $V$.  

If, at any stage
during the sweep-out, there are simultaneously such
$\bdd$-compressions via $A_i$ and via $B_i$ then we could complete the
proof of Theorem \ref{placement3} immediately:  If the
$\bdd$-compressions were into disjoint annuli in $\bdd V - S_i$ it
would contradict strong irreducibility, so in fact the annuli into
which they $\bdd$-compress must be adjacent in $\bdd V$.  More
generally, even if the annuli are adjacent, the $\bdd$-compressing
disks themselves may be taken to be disjoint. So the boundary
compressions can be done simultaneously, and the resulting surface
still intersects $\bdd V$ in essential curves. We can imagine $S_i$
isotoped so that both boundary compressing disks lie in a small collar
of a component $\bdd_0 V$ of $\bdd V$.   Put another way, we can
isotope $S_i$ so that it intersects a collar of $\bdd_0 V$ in an
easily described way:  There is a single horizontal or vertical tube
attached to a collection of annuli which are either spanning annuli or
$\bdd$-parallel in the collar $\bdd_0 V \times I$.  In particular, this
collar contains meridian disks of both $A_i$ and $B_i$. What remains of
$S_i$ in $V$ when this collar is removed must then be
incompressible, hence
$\bdd$-incompressible, by strong irreducibility.  Hence the surface
$S'$ obtained by compressing the single tube identified in the
collar $\bdd_0 \times I$ is either vertical or horizontal, and $S$ is
obtained from $S'$ by attaching either (respectively) a horizontal or a
vertical tube, as claimed by Theorem \ref{placement3}.

{\bf Step 4:}  Following the previous step, we may as well assume there
are not simultaneously disjoint $\bdd$-compressions into $A_i$ and
$B_i$, so there is a subinterval in which there are no $\bdd$
compressions at all, and this implies that in this subinterval, every
curve of intersection in $S_i \cap \bdd T$ is essential in $\bdd T$,
as was our claim.  

\bigskip

We now want to understand how $S_i$ intersects
each of the tori in $T$.  This question is well-understood (see
\cite{Sc2}, \cite{MR}).  There are four possibilities for each solid
torus $T_j \in T$:  

\begin{itemize}

\item $S_i$ intersects $T_j$ in meridian disks

\item $S_i$ intersects $T_j$ in meridian disks and exactly one
$\bdd$-parallel annulus, parallel to a meridinal annulus in $\bdd T_j$.

\item $S_i$ intersects $T_j$ in a family of incompressible annuli in
$T_j$

\item $S_i$ intersects $T_j$ in a family of incompressible annuli, plus
one other component obtained by tubing two incompressible annuli
together or one annulus to itself, via a $\bdd$-parallel tube.

\end{itemize}

If any $T_j$ contains a component of the second type, we
can, by the previous argument, push a tube to the outside of $V$ and
what remains inside will be incompressible, as required.  If any
$T_j$ contains a component of the  fourth type, we are done by
the same argument, unless the incompressible annuli are vertical,
since if the annuli are vertical we do not know that they will
$\bdd$-compress into $\bdd V$.  Just as we eliminated inessential
curves of intersection with $\bdd T$ earlier, a further
subinterval of the sweep-out can be found in which the third type does
not arise, except perhaps when the annuli are vertical and so do not
necessarily
$\bdd$-compress to $\bdd V$. So the only remaining possibilities are
the first, and also  the third and fourth when the annuli are
vertical.  Clearly meridian disks and vertical annuli cannot occur in
neighboring solid tori, since their boundaries would intersect in some
annulus of $\Aaa$.  It follows that either $S_i$ intersects each $T_j$
in meridian disks (i. e. $S_i$ intersects $V$ in a horizontal
incompressible surface) or
$S_i$ intersects each $T_j$ in vertical annuli, plus possibly somewhere
a single horizontal tube. (More than one would contradict strong
irreducibility.)  This last would mean that
$S_i$ intersects all of $V$ in vertical annuli and tori, with possibly
one horizontal tube attached. \qed

\begin{defin}

Suppose $V$ is a Seifert manifold.  Then any manifold obtained by
replacing a tubular neighborhood of a regular fiber by an
irreducible, $\bdd$-incompressible, acylyndrical, atoroidal manifold
(with torus boundary) is called a {\em scrambling} of $V$.

\end{defin}

\begin{cor}
\label{placement4}

Suppose, in the conclusion of Theorem \ref{placement3}, there is a
component of $S_i
\cap V$ that is obtained from vertical surfaces by
attaching a single horizontal tube.  Then we can scramble $V \subset M$
so that the resulting manifold $M'$ has a Heegaard splitting of
the same genus as $M$.  

\end{cor}

\pf  The core of the horizontal tube projects to an arc in $P$
and that arc has preimage a vertical annulus $\aaa \subset V$. 
Suppose the tube compresses via a disk $D_A$ in $A_i$, say.  Then
the complement of the tube in $\aaa$ is a compressing disk $D_B$ for
$B_i$ and $\bdd D_B$ intersects $\bdd D_A$ in two points.  (See Figure
4.)  A neighborhood of $\aaa$ in $V$ is a solid torus $T$ whose
boundary intersects $S_i$ in four vertical curves and both $A_i \cap
T$ and $B_i \cap T$ are genus two handlebodies.  Viewing these genus
two handlebodies as balls with two unknotted arcs removed, another
view of $(A_i \cup_{S_i} B_i) \cap T$ is that $T$ is naturally
homeomorphic to the complement of the unknot $u \subset S^3$, and, if
we put the unknot in $2$-bridge position, then $S_i$ is a $4$-punctured
equatorial sphere dividing $S^3 - u$ into two balls with two unknotted
arcs in each.  Replace $T$ with some
atoroidal acylyndrical $2$-bridge knot complement.  This scrambles $V$
but merely changes the attaching map of $\bdd_+ A_i$ and $\bdd_+ B_i$
along $S_i \cap T$.  \qed

\begin{figure}
\centering
\includegraphics[width=.45\textwidth]{prescram.eps}
\caption{}
\end{figure}


\section{First results}

We will be presenting three extended arguments, each one of greater
complexity than the preceding, but yielding more refined results.  In
this section we give the easiest of the three, whose conclusion
implies immediately (and says much more than) that $M - \Theta$ has at
most $3g - 3$ components, at most $g-1$ of which are not Seifert.

\begin{defin}
Suppose $\Aaa \subset H$ is a properly imbedded essential
collection of annuli in a compression body $H$.  A complementary
component $Y$ is {\em toral} if it is either a solid torus or is
homeomorphic to $torus \times I$.

If the toral component $Y$ is a solid torus, define the {\em complexity}
of $Y$ to be $c(Y) = |\bdd_+ H \cap Y| - \eee$, where $\eee = 1$ if the
annuli $\Aaa \cap \bdd Y$ are longitudes and $\eee = 0$ if they are
not.  In the former case, say that $Y$ intersects $\Aaa$ {\em
longitudinally} or $Y$ is longitudinal.  Otherwise say $\Aaa$
is  {\em twisted} in $Y$, or $Y$ is twisted. (See Figure 5.)

If $Y \neq H$ is homeomorphic to $torus \times I$ then define
$c(Y) = |\bdd_+ H \cap Y|$. 

In the degenerate case in which $Y = H$ is itself just
$torus \times I$, and $\Aaa = \emptyset$, say $c(Y) = 0$.

The complexity of the union of toral components is the sum of the
complexities of each individual component.
\end{defin}

\begin{figure}
\centering
\includegraphics[width=.7\textwidth]{complexity.eps}
\caption{}
\end{figure}

Note that no essential annulus can have both boundary components in
$\bdd_- H$, so it follows that any solid torus component
$Y \subset H$ intersects $\bdd_- H$ in at most one component,
an annulus.  Such an intersection occurs if and only if exactly two
components of  $\Aaa \cap \bdd Y$ are spanning annuli.  We then know
more:

\begin{lemma}

If a solid torus complementary component $Y$ of $\Aaa \subset
H$ intersects
$\bdd_-$ H in a single annulus, then $Y$
intersects $\Aaa$ longitudinally.  

\end{lemma}

\pf: Let $\aaa \subset \bdd_- H$ denote the core of the annulus of
intersection.  If $\aaa$ is non-separating, consider a spanning
annulus whose end in $\bdd_- H$ is a simple closed curve
which intersects $\aaa$ in a single point.  By standard innermost disk,
outermost arc arguments, we can choose the annulus so that it
intersects $Y$ in a single disk
$D$ and
$\bdd D$ crosses $\aaa$ exactly once.  This guarantees that $D$ is a
meridian and so $\aaa$ is a longitude of $Y$.

If $\aaa$ is separating, the argument is only slightly more complex: 
The same sort of innermost disk, outermost arc argument shows that one
can find spanning annuli $A_1$ and $A_2$ so that each curve $\aaa_i
= A_i \cap \bdd_- H$ is parallel to $\aaa$ but the two curves $\aaa_1,
\aaa_2$ lie on opposite sides of $\aaa$ in $\bdd_- H$, and so that
$A_i \cap Y = \emptyset$.  Then choose $\ggg$ to be a spanning arc for
the annulus between the $\aaa_i$ in $\bdd_- H$ and construct a
``spanning square'' $\Sss \subset H$ so that $\Sss \cap \bdd_- H =
\ggg$, $\Sss \cap A_i$ is a spanning arc of $A_i$ and the rest of
$\bdd \Sss$ lies in
$\bdd_+ H$.  Then an innermost disk, outermost arc argument on $\Sss
\cap \Aaa$ shows that $\Sss$ can be chosen so that $\Sss \cap Y$ is a
meridian disk $D$ as above. \qed


\begin{defin}

Suppose $\Aaa \subset H$ is a properly imbedded essential
collection of annuli in a compression body $H$.  A complementary
component $L$ is a {\em basic block} if it is a genus two handlebody
in which at least two components of $\Aaa \cap L$ are longitudes,
separated by a meridian disk of $L$.

A complementary component is a {\em spanning product} if, for some
surface $E$ it is of the form $$E \times (I; 0, 1) \subset (H; \bdd_-
H, \bdd_+ H).$$

\end{defin}

\begin{figure}
\centering
\includegraphics[width=.6\textwidth]{basicblock.eps}
\caption{}
\end{figure}

\begin{lemma}
\label{primitive}

Suppose $H$ is a compression body, and $\Aaa$ is a properly imbedded
essential collection of annuli in $H$.  Suppose no two adjacent
components of $H - \Aaa$ are toral, and there are $n$ non-toral 
components of $H - \Aaa$ that are disjoint from
$\bdd_- H$.  Then $$n \leq J(H)/2.$$  Moreover, if $Y$ is the union of
toral components of $H - \Aaa$ and $\aaa$ is the number of non-spanning
annuli which are not adjacent to toral components (on either side),
then
$$c(Y) + \aaa \leq J(H).$$

Finally, suppose in fact $c(Y) + \aaa = J(H)$.  Then 

\begin{itemize}

\item n = J(H)/2.

\item Each non-toral component is either a basic block or a spanning
product.  

\item If more than one annulus of $\Aaa$ lies on the same
side of a separating meridian disk $D$ for a basic block
$L$ then one of those annuli $A^0$ has either one non-toral
component of $H - \Aaa$ on each of its sides in $H$ or one annulus
component of $\bdd_+ H - \Aaa$ on each of its sides. (See Figure 7.)

\end{itemize}

\end{lemma} 

\begin{figure}
\centering
\includegraphics[width=.35\textwidth]{blockschem.eps}
\caption{}
\end{figure}
 
\pf  We will proceed by induction on the pair $(J(H), c(Y))$.

When $J(H) = 0$ then $H$ is a product.  In a product,
the only essential annuli are spanning annuli.  Then $J(H) = c(Y) =
\aaa = 0$, every complementary component is a spanning product, and
the lemma is true in this case.

Note that any toral component $Y^0$ of $H - \Aaa$ that has
$c(Y^0)$ = 0 is just a product neighborhood of a spanning annulus.  The
two annuli in $\Aaa \cap Y^0$ could be removed with no effect on the
argument.  So we may as well assume that any toral component has
positive complexity.

Suppose that some component $Y^0$ of $H - \Aaa$ is a solid
torus component of $H - \Aaa$ (with positive complexity) that intersects
$\Aaa$ longitudinally.  If there is a spanning annulus in  $\Aaa
\cap \bdd Y^0$ delete it from $\Aaa$; otherwise delete a non-spanning
annulus in $\Aaa \cap \bdd Y^0$ from $\Aaa$.  This attaches $Y^0$ to
a non-torus component of $H - \Aaa$, thereby lowering
$c(Y)$ by $c(Y^0)$.  On the other hand, every non-spanning component of
$\Aaa \cap \bdd Y^0$ (except, perhaps, the one removed) now contributes
to $\aaa$.  So $c(Y)$ is lowered, but $\aaa + c(Y)$ remains the same.
(See Figure 8.) The operation leaves both $J(H)$ and the number of
non-toral components not intersecting $\bdd_- H$ the same.  It cannot
create a new basic block or spanning product.  On any basic block the
operation may increase but cannot decrease the number of annuli
intersecting the basic block.  The operation can decrease but cannot
increase the number of annuli in $\bdd H - \Aaa$.  So the result
follows by induction.

\begin{figure}
\centering
\includegraphics[width=.8\textwidth]{trade.eps}
\caption{}
\end{figure}

There remains only the case in which
$J(H) > 0$ and no solid torus component of $H - \Aaa$ intersects $\Aaa$
longitudinally. In particular, no spanning annulus is adjacent to
any torus component.  Unless $\Aaa$ is 
$\bdd$-compressible, $\Aaa$ would consist entirely of vertical spanning
annuli, so $\aaa = c(Y) = 0$, and the result
follows immediately.  So we will suppose that $\Aaa$
is $\bdd$-compressible.  Let $A \in \Aaa$ be a component of $\Aaa$ on
which such a $\bdd$-compression can be done, and let $D$ be the disk
that results from $A$ by the $\bdd$-compression.  

First suppose there is such a choice of $\bdd$-compression so that $D$
is separating in H.  Cut $H$ open along $D$ to get two components:
$H_1$ which comes from the side of $A$ that contains $D$, and $H_2$
which comes from the opposite side.  Let $\Aaa_1 = \Aaa \cap H_1$ and
$\Aaa_2 = (\Aaa - A) \cap H_2$.  Note that no component of $\Aaa_2$ can
be an inessential annulus, for if any were, it would have been
adjacent to a solid torus component of $H - \Aaa$ in which $A$ is
longitudinal.  So, by induction, the lemma is true for
$\Aaa_2 \subset H_2$ (or, possibly, $H_2$ is simply a solid torus on
which $A$ is not longitudinal).  On the other hand, we cannot
immediately assume that the inductive hypothesis applies to $H_1$, for
two things could go wrong.  First, it's quite possible that exactly
one component $A_1$ of $\Aaa_1$ is inessential in $H_1$, exactly when
$A \cup A_1$ together cut off a genus two solid handlebody component
of $H - \Aaa$ (containing $D$) in which both annuli $A$ and $A_1$ are
longitudinal, i. e. a basic block $L$ with $|L \cap \Aaa | = 2$.
Another problem could be that the component $Y_0$ adjacent to $D$ in
$H_1$ is a solid torus, and also $Y_0$ is adjacent to other toroidal
components, in violation of the inductive hypothesis.  Note that
in this case $Y_0$ must be longitudinal and adjacent via only one
annulus (which we again call
$A_1$) to a toroidal component, since any annulus in $H$ is
$\bdd$-compressible and all solid torus components of $H - \Aaa$ are
twisted.  The component of $H - \Aaa$ containing
$D$ is a basic block $L$, but this time $|L \cap \Aaa | \geq 3$, and
furthermore $A_1$ intersects an annulus component of $\bdd_+ H - \Aaa$
on both sides.  In either case, removing $A_1$ from $\Aaa_1$ (call the
result $\Aaa_1'$) restores the inductive hypothesis to $\Aaa_1'
\subset H_1$.

\begin{figure}
\centering
\includegraphics[width=.5\textwidth]{H1H2.eps}
\caption{}
\end{figure}

If either failure of the inductive hypothesis arises (so that the
component containing $D$ is a basic block $L$, and an incident annulus
$A_1$ is either inessential in $H_1$ or adjacent to a toroidal
component $Y_1$ of $H -
\Aaa$) then, back in $H$, removing $A_1$ from $\Aaa$ would either
decrease $\aaa$ by one, or decrease $c(Y)$ by $c(Y_1)$ and increase
$\aaa$ by
$c(Y_1) - 1$.  Similarly, if the annulus $A$ is adjacent only to
non-toral components, then removing it would decrease 
$\aaa$ by one; if it is adjacent to a solid torus $Y_2$, removing $A$
would lower $c(Y)$ by $c(Y_2)$ and raise $\aaa$ by $c(Y_2) -
1$.  In either case, removing the annulus would decrease
$c(Y) + \aaa$ for $\Aaa \subset H$ by exactly one.  So  if and only if
$A_1$ exists (i. e. the inductive hypothesis fails) we get that $c(Y) +
\aaa$ drops by two in going from $\Aaa \subset H$ to
$\Aaa_1' \cup \Aaa_2$ in $H_1 \cup H_2$ (which is necessary if the
second inequality is an equality).   So if the second inequality is an
equality, $A_1$ exists.  

Note that $J(H) = J(H_1) + J(H_2) + 2$ and $H - \Aaa$ has at most
one more non-toral component not intersecting $\bdd_-H$ than the sum
of the number of such components in $H_1$ and $H_2$. It has exactly
one more unless the component adjacent to $D$ in $H_1$ is itself not
toral, with the consequence that $A_1$ is not defined.  So the proof
now follows by induction:  If the second inequality is an equality,
then $A_1$ exists, the component $L$ containing $D$ is a basic
block satisfying the required conditions (as we have seen), and also
the second inequality must remain an equality for
$\Aaa_1' \cup \Aaa_2 \subset H_1 \cup H_2$.  There we apply the
inductive hypothesis to show that all non-toral components other than
the one containing $D$ are spanning products or basic blocks
satisfying the required conditions, and the first inequality for
$\Aaa_1' \cup \Aaa_2 \subset H_1 \cup H_2$ is an equality.  We have
just shown that if $A_1$ exists, then this equality implies the
first equality for $\Aaa \subset H$.

Finally, suppose that such a $\bdd$-compression results in a 
{\em non-separating} disk $D$, so $D$ is incident to a non-separating
annulus
$A \subset \Aaa$.  Cut $H$ open along $A$ and, along the copy of $A$
that is still incident to $D$, attach a twisted solid torus to get a
new handlebody $H'$ (since the attaching annulus is longitudinal after
the cut) and a family of annuli $\Aaa' \subset H'$ with the same
number of elements as $\Aaa$. Indeed we can think of just replacing
$A$ with a copy of $A$.  It's easy to see that neither operation
affects $J(H)$ or the number or type of non-toral components.  The cut
along $A$ reduces one of $c(Y)$ or $\aaa$ by exactly one, depending on
whether or not on the other side of $A$ from $D$ is a (twisted) toral
component or a non-toral component.  On the other hand, attaching the
twisted torus raises $c(Y)$ by exactly one.  So the result for $\Aaa'
\subset H'$ would imply the result for $\Aaa \subset H$.  But a proof
for $\Aaa'
\subset H'$ is provided by the previous case, since $A$ is
separating in $H'$.  \qed

\begin{defin}

Let $V$ be a connected orientable Seifert manifold with non-empty
boundary, $f$ singular fibers, and orbit space the surface $P$.  Define
the {\em vertical index} of $V$ to be $$I_v (V) = f - \chi (P).$$ 
Define the augmented vertical index as

\[ I_{v}^+ (V) = \left\{ \begin{array}{ll}
				        1 + I_v (V) = 1 - \chi (P) & \mbox{if $V$ has no
exceptional fibers.}  \\
            1/2 + I_v (V) = 3/2 - \chi (P) & \mbox{if $V$ has
one exceptional fiber}  \\
            I_v (V) = f - \chi (P) & \mbox{if $V$ has
$f \geq 2$ exceptional fibers}
                       \end{array}
            \right. \]


Define the epsilon vertical index as

\[ I_{v}^{\eee} (V) = \left\{ \begin{array}{ll}
				        I_{v}^+ (V) & \mbox{when $|\bdd V| = 1$} \\
            I_{v}^+ (V) - 1/2 = I_{v} (V)+ 1/2  = 1/2 - \chi (P) &
\mbox{if $|\bdd V| = 2$ and $f = 0$}  \\
            I_v (V) = f - \chi (P)  & \mbox{otherwise}  
                       \end{array}
            \right. \]

If $V$ is not connected, the (resp. augmented, epsilon) vertical index
of $V$ is defined to be the sum of the (resp. augmented, epsilon)
vertical indices of its components.
\end{defin}

Note:

\begin{itemize}

\item If $I_{v}^+ (V) \leq 1$ or $I_{v}^{\eee} (V) \leq 1$ then $V$ is
atoroidal.  Indeed, the only toroidal manifolds for which either
index is $\leq 2$ are those for which $f = 1$ and $P$ is the M{\"o}bius
band, or $f = 0$ and $P$ is the once-punctured M{\"o}bius band.  

\item A scrambling of $V$ effectively removes from the Seifert piece
a vertical solid torus (and creates a non-Seifert piece).  So
scrambling deletes a disk from $P$ and so the process will increase
both
$I_v (V)$ and $I_{v}^+ (V)$ by exactly one.

\end{itemize}

The theorems below will put bounds on the various complexities,
based on the genus of the Heegaard splitting.  It would make the
definitions considerably easier, the aesthetics better, and the theorems
stronger, if the fractions in the definitions of augmented and
epsilon index above could be raised to the nearest integer.  We know of
no counterexamples to this hope.  The difference between the types of
complexity above are small, and would not be worth making,
except that, following Kobayashi's agenda, we would like to be able to
get a bound on the number of toroidal components and a few toroidal
manifolds have unaugmented vertical index $1$: a circle bundle over the
once punctured torus or Klein bottle, or over the punctured M{\"o}bius
band, or a Seifert bundle over the M{\"o}bius band with one
exceptional fiber.  For this reason we need to consider augmentation.

\bigskip
 
\begin{defin}

Suppose $M$ is a closed irreducible $3$-manifold, $V$ is a Seifert
component from the torus decomposition of  $M$, and $M$ has the
(generalized) Heegaard splitting 
  $$M = (A_1 \cup_{S_1} B_1) \cup_{F_1} (A_2 \cup_{S_2} B_2) \cup_{F_2}
... 
\cup_{F_{m-1}} (A_m \cup_{S_m} B_m).$$  Let $S = \cup S_i$ and $F =
\cup F_i$.  Then $V$ is {\em aligned} with respect to the Heegaard
splitting if $S \cup F$ intersects $\bdd V$ only in vertical fibers.

\end{defin}

Note that if the canonical tori $\Theta$ have been put  in
preferred position with respect to the splitting, then a Seifert
component $V$ is aligned if {\em any} boundary component of $V$
intersects any surface $F_i, S_i$ in a vertical fiber.  If $F \cup S$
is disjoint from a boundary component $T$ of $V$ then, since $\Theta$
is incompressible, it must be that $T$ is parallel to a component of
$F$. If this happens for any component of $\bdd V$ then this,  too,
implies that $V$ is aligned and we say that
that component of $\bdd V$ is {\em strongly
aligned}.  

\begin{thm}
\label{main1}

Suppose $M$ is a closed orientable $3$-manifold that has a genus $g$
irreducible Heegaard splitting.   Let $\Theta$ be the collection of
canonical tori for $M$, put in preferred position with respect to
a strongly irreducible generalized Heegaard splitting for $M$, and
suppose no component of $\Theta$ is strongly aligned.  Denote by $V$ the
union of aligned Seifert components and let
$p$ be the number of components of $M - \Theta$ that are not aligned
Seifert components.  Then 

\begin{itemize}

\item  $p \leq g - 1$ and 

\item $I_{v}^+ (V) \leq 2g - 2.$

\end{itemize}

\end{thm}

\pf Suppose that $$M = (A_1 \cup_{S_1} B_1) \cup_{F_1} (A_2
\cup_{S_2} B_2) \cup_{F_2} ...  \cup_{F_{m-1}} (A_m \cup_{S_m} B_m)$$ 
is a strongly irreducible splitting, of genus $\leq g$, with respect to
which the canonical tori $\Theta$ can be put in preferred
position with no components of $\Theta$ strongly aligned.  There may be some of the $S_i$
which intersect the aligned Seifert manifolds in vertical surfaces to
which a horizontal tube has been attached, as described in
\ref{placement3}.  The proof will be by induction on the number of such
$S_i$.  The inductive step is easy, so we present it before examining
the case when there are no such horizontal tubes.

If there is a component of $S_i \cap V$ which contains a horizontal
tube, apply Corollary \ref{placement4} to scramble $V$, replacing $M$
with a manifold $M'$ of the same genus but one for which the number of
horizontal tubes in aligned Seifert pieces has been reduced by one, the
number of components which are not aligned Seifert pieces is increased
by one, as is $I_{v}^+ (V)$.  The theorem for $M'$ then implies the
theorem for $M$.  

It remains to prove the theorem under the assumption that there are no
horizontal tubes of any $S_i$ in $V$, so all components of $(F \cup
S) \cap V$ are vertical surfaces in $V$, either incompressible tori or
essential annuli.  Hence each component of each $A_i \cap V$ and $B_i
\cap V$ is toral.

We first count the number of components of $M - \Theta$ that cannot be viewed as
aligned Seifert manifolds.  Suppose $N$ is such a component.   Then no
component of intersection of $N$ with any $A_i$ or $B_i$ is toral, for
if one were then $N$ would contain an essential annulus and it could
not be vertical; this would imply that $N$ is a Seifert piece (since
it contains an essential annulus) and the annulus would be horizontal 
(by Theorem \ref{vert/horiz}).  Hence $N$ is an $I$-bundle construct
with $I$-base the M{\"o}bius band or the annulus.  In either case,
it could be fibered differently and so be viewed as aligned. 

The upshot is that, to get a bound on the number of components that
are not aligned Seifert components, we need to count exactly those
that do not intersect any compression body in any toral
component. Moreover no two adjacent components of $A_i - \Theta$ (or
$B_i - \Theta$) can be toral, since if they were, then the component of
$\Theta$ between them would be incident to aligned Seifert pieces on
both sides, and so should have been eliminated from $\Theta$.  It
follows then from 
\ref{primitive} that the number of components of $A_i -
\Theta$ that are not part of an aligned Seifert piece and which don't
intersect $\bdd_- A_i$ is at most $J(A_i)/2$.  So this is an upper
bound to the number of components of $M - \Theta$ that simultaneously
are not aligned Seifert pieces, intersect $A_i$, and
 are disjoint from all
$A_j, B_j, j \leq i$.  On the other hand, any component of $M -
\Theta$ which is not an aligned Seifert component must intersect some
$A_i$ and some $B_i$ and do so in non-toral pieces.  So starting at
$A_1$ and summing across each part of the generalized Heegaard
splitting, we conclude that there are at most 
$\sum_{i=1}^{m} J(A_i)/2 = g-1$  components which are not aligned
Seifert components.  This verifies the first claim.

Now we count the index of the aligned Seifert piece $V$.
Let $P$ be the base surface.  We construct a graph $\Ggg$ which is a
deformation retract for $P$, noting exceptional fibers as we go (See
Figure 10):  Include a vertex for every solid torus component of $A_i
\cap V$ and
$B_i \cap V$.  (Call the vertex {\em longitudinal} if the annuli in
which the torus intersects $S_i$ are longitudinal.)  Include a loop
for every $torus \times I$ component of the intersection.  Include an
edge in $\Ggg$ for every annulus in $F_i
\cap V$ and $S_i \cap V$, with ends of the edge on the vertices
corresponding to the solid torus (or $torus \times I$) component on
either side of the annulus.  The exceptional fibers correspond
precisely to the $f$ torus components whose intersection with $S_i$ is
not longitudinal, i. e. the $f$ non-longitudinal tori.  If $v$ is the
number of vertices in $\Ggg$ and $e$ is the number of edges (ignoring
loops in both counts), then  $I_{v} (V) = e - (v - f)$. 

\begin{figure}
\centering
\includegraphics[width=.6\textwidth]{Gamma.eps}
\caption{}
\end{figure}

Now $e$ is half the sum of the
valences of the vertices, so a local formula for $I_{v} (V)$ can be
gotten as follows:  For each vertex $u$ in $\Ggg$ not the base of a
loop, note that $c(\ttt) =  valence(u)$ if the corresponding solid
torus $\ttt$ (in some $A_i \cap V$ or $B_i \cap V$) is not longitudinal;
$c(\ttt) =  valence(u) - 1$ when $\ttt$ is longitudinal and disjoint
from the $F_i$; and $c(\ttt) =  valence(u) - 2$ when $\ttt$ intersects
the $F_i$.  For each loop representing a component $\aaa \cong torus
\times I$, $c(\aaa)$ denotes the number of ends at the base of the loop
not belonging to the loop itself. Abusing notation a bit, for $u$ a
vertex (not the base of a loop) in $\Ggg$ representing $\ttt$, let
$c(u) = c(\ttt)$ and for $\aaa \subset \Ggg$ a loop, let $c(\aaa)$
denote the complexity of the corresponding component $\cong (torus
\times I)$.  Then $$2I_{v} (V) =  \sum_{u \in \Ggg} valence(u) +
\sum_{\aaa
\subset \Ggg} c(\aaa) - 2 (v - f)) = \sum_{u \in \Ggg} c(u) +
\sum_{\aaa \subset \Ggg} c(\aaa) - \Lll,$$ where $\Lll$ is
the  total number of longitudinal components occuring in all $A_i \cap
V$ or $B_i \cap V$ that are disjoint from $F$.  

To summarize, if we let $a_i$ (resp. $b_i$) represent the sum of the
complexities of all the toral components of $A_i \cap V$ (resp. $B_i
\cap V$) then 
\begin{equation}
\label{defect}
2I_{v} (V) = \sum_{i=1}^{m} (a_i + b_i) - \Lll.
\end{equation}  

No boundary component of $V$ is totally aligned, by
assumption.  This means that for any component $V^0$ of $V$ and the 
lowest $i$ for which $A_i \cap V^0$ is non empty, in fact $A_i \cap
V^0$ must be a solid torus disjoint from $\bdd_- A_i$.  Similarly
for the largest $i$ for which $B_i \cap V^0$ is non empty.  So the
component $V^0$ contributes at least $2$ to $\Lll$ if it has no
exceptional fibers and contributes at least $1$ to $\Lll$ if it has
only one exceptional fiber.  The upshot is that we have the
inequality:

$$2I_{v}^+ (V) \leq  \sum_{i=1}^{m} (a_i + b_i) = \sum_{i=1}^{m} a_i +
\sum_{i=1}^{m} b_i$$

On the other hand, Lemma \ref{primitive} above shows that
$$\sum_{i=1}^{m} a_i + \sum_{i=1}^{m} b_i \leq \sum_{i=1}^{m} J(A_i) +
\sum_{i=1}^{m} J(B_i) = 4g - 4.$$  Hence we get the second
inequality $$I_{v}^+ (V)  \leq 2g - 2.$$ \qed

\begin{cor}
\label{main1+}

If, in the conclusion of Theorem \ref{main1}, we have $I_{v}^+ (V) = 2g
- 2$, then each of $M - \Theta$ is either an aligned Seifert manifold
or a $2$-bridge link exterior.

\end{cor}

\pf In order to achieve equality, all inequalities in the proof must
be equalities.  This implies that when Lemma \ref{primitive} is
applied to each compression body $A_i$ (or $B_i$) we have $\aaa =
\emptyset$, and each non-toral component is a basic block or a spanning
product. Since $\aaa = 0$ it follows immediately from the proof of
Lemma \ref{primitive} (in which longitudinal solid tori are traded in
immediately for annuli in $\aaa$) that the only longitudinal solid
tori are spanning products. In particular, any annulus of $A^0 \in \Aaa
= \Theta \cap A_i$ that is incident to a basic block is incident on the
other side to a twisted solid torus.  In particular, since $A^0$ must
$\bdd$-compress, it will $\bdd$-compress through the basic block.  This
boundary compressing disk $D$ can be chosen to avoid all other
components of $A_i - \Aaa$, since of those that it might intersect,
that one adjacent to $L$ cannot be non-toral (since $\aaa = 0$), nor a
longitudinal torus (we have eliminated that possibility too), or a
twisted torus (since $D$ intersects one of its boundary annuli in
exactly one arc).  Hence each annulus in each basic block of $A_i$
$\bdd$-compress within the basic block to $\bdd_+ A_i$.  

In particular we conclude, (or could also from the last
claim of \ref{primitive}) that each basic block $L$ in $A_i$ has
exactly one component of $\Aaa$ on each side of a separating meridian
disk.  Since $L$ was chosen so that one of the annuli
$\bdd$-compresses through $L$, $\bdd L \cap \bdd H$ is a $4$-punctured
sphere.

Any $\bdd$-compression in a basic block in $A_i$ must be into the same
component of $\bdd_+ A_i - \Theta$ as a $\bdd$-compression into
$\bdd_+ B_i - \Theta$, since $A_i \cup_{S_i} B_i$ is strongly
irreducible.  This means that there can be at most one basic block in
$A_i$ and one in $B_i$ and they are glued together along the
$4$-punctured sphere on which they coincide along $S_i$.  All other
components must be toral.  \qed

\begin{cor} 
\label{main1++}

Suppose $M$ is a closed orientable $3$-manifold that has a genus $g$
irreducible Heegaard splitting.   Let $\Theta$ and $V$ be as in
Theorem \ref{main1} but allow the possibility that some
components of $\bdd V$ are strongly aligned.  Then 
$I_{v}^{\epsilon} (V) \leq 2g - 2.$

\end{cor} 

\pf  Suppose a component $T$ of $\bdd V$ is parallel to a component of
$F$.  The proof is by induction on the number $t$ of strongly aligned
components of $\bdd V$.  Since $I_V (V) \leq I_{v}^{\epsilon} (V) \leq
I_{v}^+ (V)$, the case $t = 0$ is proven in Theorem
\ref{main1}.  So suppose some component $T \in \Theta$ of $\bdd V$ is
parallel to a component of $F$.  Let $V^0$ denote (one of the)
Seifert components on which $T$ lies.

Construct a new manifold $M'$ from $M$ by cutting $M$ open along $T$
and gluing on twisted solid tori to each copy of $T$.  These attaching
maps can be chosen so that if, on the other side of $T$ there is an
acylindrical atoroidal piece of $M -
\Theta$, it will remain so in $M' - \Theta'$.  The effect on $V$ is to
replace one or two boundary components by essential fibers.  This has
no effect on $I_v{V}$ but may cause a loss of augmentation in $V^0$. 
Hence we have immediately that $I_{v} (V) \leq 2g - 2.$
What we need to show is that this loss can be limited if $|\bdd V^0|
\leq 2$.  

Suppose $\bdd V^0$ has only one boundary component $\bdd V^0$ and it
is strongly aligned. The manifold $M'$ then consists of two components,
$V_+$ (of some genus $g_v$) which contains the Dehn filled $V^0$, and
$M_-$ of some genus $g_-$ with $(g - 1) = (g_- - 1) + (g_v -
1)$.  Since $V^0$ is entirely Seifert, some compressible component of
some $S_i$ lay entirely in $V^0$.  This means, via \ref{placement4},
that
$V^0$ could have been scrambled without affecting genus $M$ but raising
$I_{v}^+ (V_+)$ by $1$.This would counterbalance any loss of
augmentation; in fact we must have had
$I_{v}^{\epsilon} (V) = I_{v}^{\epsilon} (V - V^0) + I_{v}^{+} (V^0)
\leq I_{v}^{\epsilon} (V - V^0) + (I_{v} (V_+) + 1) \leq 2(g_- - 1) +
2(g_- - 1) = 2g - 2.$

If $\bdd V^0$ has two boundary components and they are both
totally aligned, the same induction argument shows that
$I_{v}^{\epsilon} (V) \leq 2g - 2,$ so we need only consider the case
in which exactly one boundary component is totally aligned.  If $V^0$
has exceptional fibers then, since $I_{v}^{\eee} (V^0) = I_{v} (V^0)$
so there is not augmentation and the proof follows immediately by
induction from Dehn filling at the strongly aligned component, as
above.  If $V^0$ has no exceptional fibers, then after the Dehn
filling, the augmentation drops from $1$ to $1/2$, but also
$I_{v}^{\eee} (V^0) = I_{v}^+ (V^0) - 1/2$, so again the argument
follows by induction. \qed

\section{Non-aligned Seifert pieces}

In this section we are able to refine further so that we also have
information about the Seifert components that are not aligned.  The
argument is, in outline, parallel to that of the previous section.

There is this technical refinement of \ref{primitive}.


\begin{lemma}
\label{numerics1}

Suppose $H$ is a compression body, $\Aaa$ is a properly imbedded
essential collection of annuli in $H$, and no two adjacent
components of $H - \Aaa$ are toral.  Suppose $(E, \bdd E) \subset
(H, \bdd_+ H)$ is a compact properly imbedded subsurface, with $s$
components and
$d$ boundary components, and, among the components of $H - \Aaa$, is a
collection homeomorphic to an $I$-bundle $\xi$ over $E$, with $\dxi
\subset \bdd_+ H$ and $\bxi \subset \Aaa$.   Then there are at most
$$J(H)/2 + \chi(E) + s$$ non-toral components of $H - \Aaa$ that don't
intersect $\bdd_- H$.  Moreover, if $Y$ is the union of the toral
components of $H - \Aaa$, then
$$c(Y) \leq J(H) + 2 \chi(E) + d.$$

\end{lemma} 

\pf   Let $H'$ be the compression body (or bodies) obtained by
deleting $\xi$ from $H$, and $\Aaa' = \Aaa - (\bxi)$.  Remove any
further components of $\Aaa$ that are
$\bdd$-parallel in $H'$.  This latter move has no effect on the
non-toral components of $H'$.  Then $J(H') = J(H) + \chi(\dxi) = J(H)
+ 2 \chi(E)$ and, since $\xi$ contains at most $s$ non-toral
components, the assertion that there are at most
$J(H)/2 +  \chi(E) + s$ non-toral components of $H -
\Aaa$ that don't intersect $\bdd_- H$ then follows from
Lemma \ref{primitive}.  

For the second result, first note that we may as well assume no
component of $E$ is an annulus or M{\"o}bius band.  For we could regard
such a component as toral and the theorem in that case would suffice,
for in no longer regarding the component as toral, $c(Y)$ goes down by
$1$, $s$ goes up by one, $J(H) + 2 \chi(E)$ remains the same, and $d$
goes up by two.

Achieving the inequality $c(Y) \leq J(H) + 2 \chi(E) +
d$ is relatively simple.  For this, let $H'$ be the compression body
obtained by removing $\xi$ plus all toral components whose
boundary intersects $\Aaa$ exactly in $\bxi$. Further remove from
$\Aaa'$ any annulus which is $\bdd$-parallel in $H'$; each 
corresponds to a longitudinal toral component of $H - \Aaa$ whose
boundary contains only a single annulus in $\Aaa'$.  The effect is to
reduce $c(Y)$ by at most $d$ and to set  $J(H') =
J(H) + 2 \chi(E)$. ($c(Y)$ is not necessarily reduced
exactly by $d$, because of longitudinal tori whose boundaries are
contained in $\bxi$ and components of $\bxi$ which
are not adjacent to toral components.)    Then Lemma \ref{primitive}
applied to  the compression body (or bodies) $H'$ completes the
proof.     \qed

\begin{defin}

Let $V$ be an orientable Seifert manifold with non-empty
boundary.  Suppose $(E, \bdd E) \subset (V, \bdd V)$ is an
incompressible possibly $1$-sided surface in $V$ such that 
$V$ is an $I$-bundle construct over 
$E$).  For $\rrr$ the set of slopes in $\bdd V$ determined by $\bdd E$,
call $I_{\rrr} (V) = |V| - 2 \chi (E) - |\bdd E|$ the {\em horizontal
index} of $V$ (with slope $\rrr$).  
\end{defin}

\begin{thm}
\label{main2}

Suppose $M$ is a closed orientable $3$-manifold that has a genus $g$
irreducible Heegaard splitting.  Let $\Theta$ be the
collection of canonical tori for $M$, put in preferred position
with respect to a strongly irreducible generalized Heegaard splitting
for $M$.  Suppose $M - \Theta$ has
$p$ non-Seifert components. Divide the Seifert components
into two classes: the non-aligned or {\em horizontal} components
$U_1,..., U_h$ whose union we denote $U$, and the aligned Seifert
components whose union we denote $V$. We know that $U$  can be written
as as an $I$-bundle construct over a surface $E$ (each component
of which has negative Euler characteristic). Denote the  slope of
$\bdd E$ in $\bdd U$ by $\rrr$. Then

\begin{itemize}

\item  $p \leq g-1 + \chi(E)$

\item  $I_{\rrr}(U)  + I_{v}^{\eee} (V) \leq 2g - 2 + h$.

\end{itemize}

\end{thm}

\pf  Suppose that $$M = (A_1 \cup_{S_1} B_1) \cup_{F_1} (A_2
\cup_{S_2} B_2) \cup_{F_2} ...  \cup_{F_{m-1}} (A_m \cup_{S_m} B_m)$$ 
is the strongly irreducible splitting with respect to which $\Theta$
has been put in preferred position.  By an inductive argument as in the
proof of Theorem \ref{main1}, we may as well assume that each
component of each $A_i \cap V$ and $B_i \cap V$ is toral.  Following
the argument of Corollary \ref{main1++}, we may as well aim to prove
the stronger inequality $I_{\rrr}(U)  + I_{v}^{+} (V) \leq 2g - 2 +
h$ but also assume that no component of $\bdd V$ is strongly aligned.

We first do a refined count of the number of components of $M - \Theta$
that are not aligned Seifert manifolds.  As in the proof of Theorem
\ref{main1}, we have to count exactly those components that do not
intersect any compression body torally.  

It follows from \ref{placement3}
that, with possibly the exception of one component, the horizontal
Seifert part intersects each $A_i$ in an $I$-bundle $\xi$.  The
possible exception, called a {\em tubed} component, starts with the
form
$\xi$, but then either a vertical tube along an $I$-fiber is deleted,
or a
$1-handle$ is attached to 
$\dxi$.  Ignore for the moment the possibility of
a tubed component, and note then that the components of intersection of
the horizontal Seifert pieces with each
$A_i$ can be divided into two types: those that are spanning (i. e. are
homeomorphic to $surface \times (I: 0, 1) \subset (A_i; \bdd_- A_i,
\bdd_+ A_i)$) and those that are non-spanning and so are homeomorphic
to $(\xi, \dxi) \subset (A_i, \bdd_+ A_i)$.  The union
of the latter can be written in the form $(\xi_i, \dxi)
\subset (A_i, \bdd_+ A_i)$, where $\xi_i$ is an $I$-bundle over a
surface $E_i$, just as in the hypothesis of Lemma \ref{numerics1}.  

For each component $U_j$ of $U$, choose an $i$ (there must be one, since
$M$ is closed) for which at least one component of $U_j \cap A_i$
is an $I$-bundle as in Lemma \ref{numerics1}.  Similarly
choose an $i'$  for which at least one component of $U_j \cap
B_{i'}$ is an $I$-bundle as in Lemma
\ref{numerics1}.  Then for each $1 \leq i \leq m$ let $h_i$ denote the
number of components of $U$ that are assigned to $i$, $E_i$ denote the
union of $I$-bases of those components of $U$ assigned to
$i$ and $d_i$ represent the number of boundary components of $E_i$.
 Similarly let $h_{i'}$ denote the number of components of $U$ that are
assigned to $i'$, $E_{i'}$ the union of the $I$-bases of those
components and $d_{i'}$ the number of boundary components of
$E_{i'}$.  We repeat, each component of $U$ has its $I$-base surface 
assigned to exactly one of the $E_i, 1 \leq i \leq m$ and one of the
$E_{i'}, 1 \leq i' \leq m$.  With this set-up we are in a position to
use the fact (from \ref{numerics1}) that there are at most $J(A_i)/2 +
\chi(E_i) + h_i$ non-toral components of $A_i -
\Theta$ that don't intersect $\bdd_- A_i$.  Note that this means
there are at most $J(A_i)/2 + \chi(E_i)$ components of $A_i - \Theta$
which are not parts of Seifert pieces, aligned or unaligned.  We apply
this fact, as we did the simpler inequality from Lemma \ref{primitive}
in the proof of \ref{main1} and conclude that there are at most 
$$\sum_{i=1}^m (J(A_i)/2 + \chi(E_i)) = g - 1 + \chi(E)$$
non-Seifert components, the first inequality.

If there is a tubed component of $U \cap A_i$, it only improves the
situation.  Suppose first that there is a component of intersection of
$A_i$ with a horizontal Seifert piece and it has the form $\xi
\cup (1-handle)$.  Then $\bdd$-reducing the $1$-handle returns
us to the situation analyzed above, and reduces $J(A_i)$ by two. Lemma
\ref{numerics1} after the $\bdd$-reduction implies the required
inequality before the $\bdd$-reduction.  Similarly, if a component is
of the form $\xi - (fiber)$ then removing the
component from $A_i$ reduces $J(A_i) + 2 \chi(E_i)$ by two and both
$h_i$ and the number of non-toral components by one.  So again Lemma
\ref{numerics1}, applied after the removal of the component, implies
the required inequality before the removal. 

The second inequality is obtained in exactly the same way, using the
inequality $c(Y) \leq J(H) + 2 \chi(E) + d$ from Lemma
\ref{numerics1}.  Again, we assume that all components of $U - (F \cup
S)$ are $I$-bundles over an $I$-base for one of the components of $U$,
since, just as we have seen, exceptional components only make the
inequalities stronger.  Arguing as we did in the calculation of
Theorem \ref{main1} we have

\begin{eqnarray*} 
2(I_{v}^+ (V)) \leq
\sum_{i=1}^{m} (J(A_i) + 2 \chi(E_i) + d_i) & + & \sum_{i'=1}^{m} 
(J(B_{i'})  + 2 \chi(E_{i'}) + b_{i'}) = \\  \sum_{i=1}^{m} (J(A_i)
+ (2 \chi(E_i) + d_i - h_i) + h_i) & + & \sum_{i'=1}^{m} (J(B_{i'}) +
(2
\chi(E_{i'}) + b_{i'} - h_{i'}) + h_{i'}) = \\  \sum_{i=1}^{m} J(A_i)
+
\sum_{i'=1}^{m} J(B_{i'}) - 2 I_{\rrr}(U) + 2 h  & = & 4g - 4 - 2
I_{\rrr}(U) + 2 h
\end{eqnarray*}

Divide by $2$ and add $I_{\rrr}(U)$ to both sides and the inequality
becomes
$$I_{\rrr}(U) + I_{v}^+ (V) \leq 2g - 2 + h$$ as required. \qed

\section{Advanced computation}

A further improvement in the last inequality is
possible.  It is based on a more sophisticated version of the argument
in Lemma \ref{numerics1}.  A few preliminary observations are needed.

\begin{lemma}
\label{outermost}

With the notation of Lemma \ref{numerics1}, some annulus $A_0$ in
$\bxi$ is either not adjacent to a toral component or, if it is, the
toral component $Y$ that it is adjacent to is longitudinal and $Y
\cap (\bxi) = A_0$.  

\end{lemma}

\pf At least one component, $A_0 \subset \bxi$, has the
property that it $\bdd$-compresses in the complement of $\Aaa$. A
$\bdd$-compressing disk intersects $A_0$ in a single spanning arc, so
if $A_0$ is adjacent to a toral component, that component is
longitudinal.  \qed

\begin{lemma}
\label{graph}

Suppose $\Ggg$ is a bipartite graph with red and blue
vertices.  Then there is a subgraph $\Ggg_- \subset \Ggg$ so that  

\begin{enumerate}

\item $\Ggg_-$ contains all red vertices;

\item  each component of $\Ggg_-$ contains at most one more red vertex
than blue vertices;

\item  each blue vertex not in $\Ggg_-$ is incident to at least three
components of $\Ggg_-$.

\end{enumerate}

\end{lemma}

\pf  The subset consisting of all red vertices satisfies the first two
conditions.  Let $\Ggg_-$ be a maximal subgraph satisfying these
conditions.  If any blue vertex is incident to one or two components
of $\Ggg_-$, it could have been added without violating either
condition.  The result follows. \qed

\begin{lemma}
\label{numerics2}

Under the hypotheses of Lemma
\ref{numerics1},   the last inequality can be improved
to $c(Y) + s
\leq J(H) + 2
\chi(E) + d$.

\end{lemma} 

\pf:  We would like to apply
the argument of Lemma \ref{numerics1} one component of $E$ at a time. 
A difficulty is that removing a single component $\xi^0$ may
result in an annulus which is boundary parallel in the new compression
body and also lies in $\bxi - \bxi^0$.  Deleting the annulus
would destroy the inductive assumption that $(\chi - \chi^0)|\bdd
\subset \Aaa$, but leaving it in violates the hypothesis that no
annulus is boundary parallel. In either case, the inductive hypothesis
isn't attained. 

To circumvent this difficulty, construct an abstract bipartite graph
$\Ggg$ as follows:  Choose a red vertex for each component of $\xi$ and
a blue vertex for every longitudinal toral component which intersects
$\Aaa$ only in annuli that lie in $\bxi$.  Add an edge between a red
and blue vertex for each annulus through which the corresponding
components of $H - \Aaa$ are adjacent. Let $\Ggg_-$ be the subgraph
identified by Lemma \ref{graph}, and let
$\Ggg_0$ be the component of $\Ggg_-$ which contains the red vertex
corresponding to the component of $\xi$ that contains the
annulus $A_0$ from Lemma
\ref{outermost}. So, on the other side of $A_0$ is either a non-toral
component or a longitudinal toral component $Y$ adjacent to no other
component of $E$.  

Let $l$ be
the number of longitudinal toral components whose boundaries intersect
$\Aaa$ entirely in $\bxi$ (i. e. the number of blue
vertices in $\Ggg$).  Remove from $H$ the submanifold
$W$ consisting of all components represented by vertices in $\Ggg_0$.
Call the new handlebody $H'$, and examine how the numbers have
changed.   The change lowers $s$ by at most one more than it lowers
$l$.  On the other hand, it lowers $c(Y)$ no more than
it lowers $d - l$.  $J(H) + 2 \chi(E)$ is unchanged.  So, in either
case, $c(Y) + s$ is lowered at most one more than $$J(H)
+ 2 \chi(E) + (d - l) + l = J(H) + 2 \chi(E) + d.$$ In other words,
(using primes to denote the relevant numbers in $H'$) we can proceed
from the inductive assumption that in $H'$
$$  c(Y') + s' \leq J(H') + 2 \chi(E') + d'$$ 
only to the conclusion that in $H$
$$ c(Y) + s \leq (J(H) + 2 \chi(E) + d) + 1$$
and this is not quite sufficient for the inductive step.

But the following simple alteration of $H'$ can raise $c(Y')$ by one and
have no effect on any of the other numbers or on the fact that $H'$ is a
handlebody.  Note that after removing $W$ to obtain $H'$, the annulus
$A_0$ is parallel to a longitude of $H'$.  This means that if a
twisted solid torus is attached to $H'$ along $A_0$ the result is
still a handlebody.  But the operation raises $c(Y')$ by one, either
because the new torus becomes a new component of $Y'$ of complexity
one, or, if the component of $H' - \Aaa$ to which the torus is
attached is itself toroidal (hence longitudinal), it becomes twisted
and so its complexity goes up by one.  After this maneuver we are able
to inductively conclude from $$c(Y') + s' \leq J(H') + 2 \chi(E') +
d'$$ that in $H$
$$ c(Y) + s \leq J(H) + 2 \chi(E) + d$$ as required.

To ensure that the inductive hypothesis is intact, first disregard any
(necessarily twisted) toral component which had been adjacent only to
components of $W$.  Any toral component that was adjacent to $W$ but
still remains either is twisted, or is adjacent to a component not in
$\xi$ or is adjacent to at least two other components of $\xi$ that
remain. (The last corresponds to a blue vertex not in
$\Ggg_-$ above.)  In any case, no $\bdd$-parallel annulus is adjacent
to a remaining component of $\xi$, so each can be removed
without changing the inductive hypothesis.  \qed

\begin{thm}
\label{main3}

The last inequality in \ref{main2} can be improved to $I_{\rrr}(U) +
I_{v}^{\epsilon} (V) \leq 2g - 2$.

\end{thm}


\pf We argue in a manner similar to that used in \ref{main2}, using
the refinement of \ref{numerics2}: $\sum_{k=1}^{t} c(Y_k) + s
\leq J(H) + 2 \chi(E) + d$.  Again, we will assume that no component
of $\bdd V$ is strongly aligned and aim to show that $I_{\rrr}(U) +
I_{v}^{+} (V) \leq 2g - 2$.

For each $i, 1 \leq i \leq m$, consider $U \cap A_i$. The components
can be divided into two sorts: those that are spanning
(i. e. are homeomorphic to $surface \times (I: 0, 1) \subset (A_i;
\bdd_- A_i,
\bdd_+ A_i)$) and those that are non-spanning and so are homeomorphic
to $(\xi, \dxi) \subset (A_i, \bdd_+ A_i)$.  Suppose here that
$\xi$ lies in a non-aligned Seifert piece $U^0$ with $I$-base
$E^0$.  Then the  $I$-base for
$\xi$ is either $E^0$ or, if $\xi$ is a collar of
the boundary of the $I$-bundle over $E^0$ (which is connected if
$E^0$ is non-orientable), then the $I$-base for $\xi$ is the
orientable double cover of $E^0$.  In any case, denote by $\xi_i$
the union of all components of $U \cap A_i$ that appear in the form
$(\xi, \dxi)
\subset (A_i, \bdd_+ A_i)$.  Denote the $I$-base of $\xi_i$ by $E_i$. 
It is a union of
(possibly multiple copies) of (possibly orientable double
covers of) components of $E$, the $I$-base of
$U$.  Let $s_i = |E_i|$ and $d_i = |\bdd E_i|$. For each $B_i, 1 \leq i
\leq m$, similarly define $E'_i, s'_i$ and $d'_i$.  

For each $i, 1 \leq i \leq m$ let $a_i$ denote the complexity of the
union of all toral components of intersection of Seifert pieces with
$A_i$ and similarly define $b_i$ using $B_{i}$.  Then Lemma
\ref{numerics2} gives 
\begin{eqnarray}
\label{inequality}
a_i + s_i & \leq & J(A_i) +  2\chi(E_i) + d_i \\
b_i + s'_i & \leq & J(B_{i}) + 2\chi(E'_i) + d'_i.
\end{eqnarray} 

Notice that if we decide to ignore one component $(\xi^0, \dot{\xi^0})
\subset (\xi_i, \dxi_i) \subset (A_i, \bdd_+ A_i)$ the effect is to
reduce
$s_i$ by one and to lower $2 \chi(E^0) + |\bdd E^0|$ by at most one
(since $2 \chi(E^0) + |\bdd E^0| \leq 1$, with equality only when $E^0$
is a pair of pants).  So the two inequalities remain true even if we
ignore components of $\xi_i$.  Now since
$M$ is closed, each component of $U$ intersects some
$A_i$ in the form $(\xi^0, \dot{\xi^0})
\subset (A_i, \bdd_+ A_i)$, where $\xi^0$ has the same $I$-base as
$U$. Each component of $U$ intersects some $B_i$ similarly.  Ignore all
other components when defining $E_i$.  With this alteration, we have
seen that the inequality remains true, and now there is exactly one
$I$-base of each component of $U$ in $\cup_{i=1}^m E_i$. Hence 
$$\sum_{i= 1}^m (2\chi(E_i) + 2\chi(E'_i) + d_i +  d'_i -  s_i -  s'_i)
= -2 I_{\rrr}(U).$$

Now sum both inequalities of (\ref{inequality}) over $1 \leq i \leq
m$.  We have calculated in the proof of \ref{main1} (see equation
\ref{defect}) that
$$\sum_{i= 1}^m c_i +
\sum_{i = 1}^m c'_i = 2 I_v (V) + \Lll,$$ where $V$ is the
union of aligned Seifert pieces and $\Lll$ is the  total number of
longitudinal torus components occuring in all
$A_i \cap (V)$ or $B_i \cap (V)$ that are disjoint from $F$. 
We also know that $$\sum_{i= 1}^m J(A_i) +  \sum_{i = 1}^m J(B_i) = 4g
- 4.$$ So we have
$$2 I_v (V) + \Lll + \sum_{i= 1}^m (s_i + s'_i) \leq 4g -
4 + \sum_{i= 1}^m (2\chi(E_i)+ 2\chi(E'_i) +d_i + d'_i).$$

Now subtract from both sides $\sum_{i= 1}^m (s_i +
s'_i)$.  Rearranging a bit we have

$$2 I_v (V) + \Lll \leq 4g - 4 + \sum_{i= 1}^m (2\chi(E_i) +
2\chi(E'_i) + d_i +  d'_i -  s_i -  s'_i) = 4g - 4 - 2I_{\rrr}(U).$$

So we have $$2 (I_{\rrr}(U) + I_v (V)) +
\Lll \leq 4g - 4.$$  Each component of $V$ with no
exceptional fibers contributes at least two to $\Lll$, and those with
one exceptional fiber contribute at least one each to $\Lll$. So for
each component $V^0$ in $V$ that has no exceptional fibers, we can
shift two from $\Lll$ into $I_v (V^0)$ to get
$I_{v}^+ (V^0)$, and for each component with one exceptional fiber,
we can shift one.  Drop all the rest of
$\Lll$ from the inequality to get 
$$2 (I_{\rrr}(U) + I_{v}^+ (V)) \leq 4g - 4$$
as required.  \qed

\begin{defin}

A toroidal Seifert piece is called {\em weakly toroidal} if each
incompressible torus is either boundary parallel or bounds an
$I$-bundle over the Klein bottle.  

\end{defin}

A weakly toroidal piece either 

\begin{itemize}

\item has base the once punctured M{\"o}bius band and has no
exceptional fibers or 

\item has base the M{\"o}bius band and has one exceptional fiber.

\end{itemize}

In particular, a Seifert piece $V^0$ is weakly toroidal if and only
if $I_{v}^{\eee} (V^0) = 3/2$

\begin{cor}
\label{maincor}

Suppose $M$ is a closed orientable irreducible $3$-manifold that has a
genus $g$ Heegaard splitting.  Let $\Theta$ be the
decomposing collection of tori for $M$.  Suppose there are $n'$
weakly toroidal Seifert pieces of $M - \Theta$ and $n$ other toroidal
pieces.

Then the number of non-Seifert components of $M - \Theta$ is at most
$g - 1$ and the total number of components is at most $3g - 3 - n -
n'/2$.

\end{cor}

\pf:  The first inequality of \ref{main2} implies that $p + h \leq
g - 1$.  Now examine the inequality of \ref{main3}.  Note first that if
$V^0$ is a Seifert component with $I_{v}^{\eee} (V^0) = 1$, then $V^0$
is atoroidal. So if
$v$ of the $r$ components of $V$ are toroidal but not weakly toroidal
and $n'$ are weakly toroidal, then the inequality gives 
$$I_{\rrr}(U) + r \leq 2g - 2 - v - n'/2.$$

Now for each component $U_0$ of $U$, $I_{\rrr}(U_0) \geq 1$ unless
$U_0$ has $I$-base a pair of pants (and so is atoroidal).  So if $h'$
of the components of $U$ are toroidal we have
$$r \leq 2g - 2 - v - n'/2 - h'.$$

Combining, we get $$p + h + r \leq 3g - 3 - v - n'/2 - h' = 3g - 3 -
n - n'/2,$$ as required.
\qed

A defect of Theorems \ref{main2} and \ref{main3} is that
one doesn't know {\it ab initio} whether a Seifert piece
of $M - \Theta$ will be aligned or not.  So it would be
useful to have as broad a statement as possible which does not require
knowing which pieces are aligned.  Corollary  \ref{maincor} is an
example, but it can be generalized.  

\begin{lemma}
\label{numerics4} 

Suppose $W$ is a connected orientable Seifert manifold with non-empty
boundary and suppose $W$ is an $I$-bundle construct over $E$, with
$\chi(E) \leq -1$. Let
$\rrr$ denote the slope of $\bdd E$ in $\bdd W$. Then either
$W$ is a product bundle and $E$ is a pair of pants or 
$$I_{v}^+ (W) \leq \Irr (W) - \chi(E).$$

\end{lemma}

\pf  Recall the definitions:  $\Irr (W) = 1 - 2\chi(E) - d$ where $d$
is the number of components in $\bdd E$. If the base space of $W$ is
$P$ and there are $f$ exceptional fibers, then $I_{v}^+ (W)$ is
either $f - \chi(P)$ if there is more than one exceptional fibers,
$1/2 - \chi(P)$ if there is exactly one and
$1 - \chi(P)$ if there are no exceptional fibers.  So, as an
alternative formulation of the inequality, it suffices to show that 
$$ 1 - d - f + \chi(P) - 3\chi(E) \geq 0$$ if $f \geq 1$ and 
$$ -d + \chi(P) - 3\chi(E) \geq 0$$ when $f = 0$.  

First suppose there are no exceptional fibers, so $E$ is a cover of
$P$ of degree $\mu$, say.  Furthermore $\chi(E) \leq -1$, so $\chi(P)
\leq -1$, and $-d - \chi(E) = -\chi(\hat{E})$, where $\hat{E}$
is the closed surface obtained by capping off all boundary components
of $E$.  So in order to show that $$ -d + \chi(P) - 3\chi(E) \geq 0,$$
it suffices to show that 
\begin{equation}
\label{inequality2}
\chi(P) - 2 \mu \chi(P) = (1 - 2 \mu) \chi(P)
\geq \chi(\hat{E}).
\end{equation}
 Noting that  $\chi(\hat{E}) \leq 2$, this
inequality follows immediately unless $\mu = 1$ (so $P = E$), and even
then it is immediate unless $\chi(P) = -1$. When $\mu = 1$ and $\chi(P)
= -1$ there are three possibilities:  If $P$ is a once-punctured torus
or Klein bottle, then $\chi(\hat{E}) = 0$ and the inequality
(\ref{inequality2}) still follows. Similarly, if $P$ is a
twice-punctured projective plane, then $\chi(\hat{E}) = 1$ and the
inequality (\ref{inequality2}) again follows. This leaves only the
final case, when $P = E$ is a pair of pants and $W$ is a product
bundle, which is a case allowed by Lemma \ref{numerics4}. 

Now consider the case in which there are exceptional fibers.  $E$ is a
manifold cover of an orbifold whose underlying surface is the base
$P$.  The Riemann-Hurwitz formula describes the relation between
the Euler characteristics (\cite[p. 427]{Sco}) of $P$ and $E$.  Suppose
that the covering map $E \rightarrow P$ is of degree $\mu$, and that
the exceptional fibers are of type $p_i/q_i, 1 \leq i \leq f$. Then 

$$\chi(E) = \mu (\chi(P) - \sum_{i = 1}^f (1 - 1/q_i)).$$  So we are
asked to verify that 

$$1 - d - f + \chi(P) - 3 \mu (\chi(P) - \sum_{i = 1}^f (1 - 1/q_i))
\geq 0.$$  

First notice that this expression is minimized when each $q_i$ is as
small as possible so in order to prove the inequality it suffices to
set $q_i = 2$, i. e. $\sum_{i = 1}^f (1 - 1/q_i) = f/2$.  So it
suffices to prove

$$1 - d - f + \chi(P) - 3 \mu (\chi(P) - f/2) \geq 0.$$

We distinguish three cases:

{\bf Case 1: $\chi(P) = 0$.}  Then $P$ is an annulus or a M{\"o}bius
band.  If $f = 1$ then we can calculate directly:  $$\Irr (W) - \chi
(E) \geq -
\chi (E)
\geq 1 = I_{v}^+ (W)$$ as required.  So we may as well assume that $f
\geq 2$. 

$P$ has one or two boundary components.  This
implies that $d \leq \mu |\bdd P| \leq 2 \mu$, so, using also $\mu \geq
2$ we have
\begin{eqnarray*}
1 - d - f + \chi(P) - 3 \mu (\chi(P) - f/2) & \geq &
1 - 2 \mu - f + 3 \mu f/2 = \\ 1 - 2 \mu + (3\mu /2  - 1)f
& \geq & 1 - 2 \mu + (3\mu /2 - \mu/2)f = \\ 1 - 2 \mu + \mu f 
& = & 1 + (f - 2) \mu \geq 0
\end{eqnarray*}
as required.

{\bf Case 2: $\chi(P) \leq -1$.}  We don't know how many
components $\bdd P$ has, but we do know that $d \leq \mu |\bdd P|$. 
Then it suffices to prove that 
$$1 - \mu |\bdd P| - f + \chi(P) - 3 \mu \chi(P) + 3\mu f /2 \geq 0.$$
The left expression can be rearranged to become
\begin{eqnarray*}
1 - \mu (|\bdd P| + \chi(P)) + \chi(P) - 2 \mu
\chi(P) + (3\mu /2 - 1) f & \geq & \\ 1 - \mu \chi(\hat{P}) + (1 - 2
\mu)
\chi(P) + (3\mu /2 -
\mu/2)f & \geq & \\ (1 - 2 \mu)(1 + \chi(P)) + \mu f & \geq & 0
\end{eqnarray*}
as required.

{\bf Case 3: $\chi(P) = 1$.}  Then $P$ is a disk, so $f \geq 2$.  If $f
= 2$ then we can calculate directly:  $$\Irr (W) - \chi (E) \geq - \chi
(E) \geq 1 = I_{v}^+ (W)$$ as required.  So we may as well assume that
$f \geq 3$. 

Since $P$ is a disk, it has one boundary component.  This
implies that $d \leq  \mu$, so
$$1 - d - f + \chi(P) - 3 \mu (\chi(P) - f/2) \geq
2 - \mu - f - 3 \mu (1 - f/2) =
2 - f - \mu (4 - 3f/2).$$  Since $f \geq 3, 4 - 3f/2 \leq 0$ and the
expression $2 - f - \mu (4 - 3f/2)$ is minimized when $\mu = 2$.  In
that case we have $$2 - f - \mu (4 - 3f/2) \geq -6 + 2f \geq 0,$$ 
as required. \qed

\begin{cor}
\label{cor2}
Suppose $M$ is a closed orientable irreducible $3$-manifold that has a
genus $g$ Heegaard splitting.  Let $\Theta$ be the
decomposing collection of tori for $M$.  Suppose $p$ of the components
of $M - \Theta$ are non-Seifert, and $s$ are of the form $S_3 \times
S^1$, where $S_3$ is the three-punctured sphere.  Let $W$ denote the
union of the remaining Seifert pieces.  Then $$p + s +
I_{v}^{\eee}(W) \leq 3g - 3.$$
\end{cor}

\pf As in the proofs of Theorems \ref{main1}, \ref{main2},
and \ref{main3}, untelescope a minimal genus Heegaard splitting for
$M$.  Let
$V$ be the union of aligned Seifert pieces of $M - \Theta$.  Let $s'$
be the number of non-aligned components that are of the form $S_3
\times S^1$,  let $U$ denote the union of the remaining
non-aligned Seifert components, and let $E$ be the $I$-base of $U$.
(Recall from the note before \ref{placement2} that we can assume each
component of $E$ has negative Euler characteristic.)  Then from Theorem
\ref{main2} we have $$p + s' - \chi(E) \leq g - 1.$$

From Theorem \ref{main3} we have 
$$\Irr(U) + I_{v}^{\eee}(V) \leq 2g-2.$$

Apply Lemma \ref{numerics4} and add the inequalities to get
$$p + s' + I_{v}^+(U) + I_{v}^{\eee}(V) \leq p + s' - \chi(E) + \Irr(U) +
I_{v}^{\eee}(V) \leq 3g-3.$$

Now let $W$ be the union of $U$ and all components of $V$ which are
not of the form $S_3 \times S^1$, and suppose $s''$ of the components
of $V$ are of the form $S_3 \times S^1$.  Then, since
$I_{v}^{\eee}(U) \leq I_{v}^{+}(V)$ the previous inequality can be
written

$$p + s' + 2s'' + I_{v}^{\eee}(W) \leq 3g-3,$$
 
and this certainly implies that 

$$p + s + I_{v}^{\eee}(W) = p + s' + s'' + I_{v}^{\eee}(W) \leq 3g-3$$
as required.
\qed

Note that this generalizes Corollary \ref{maincor}, since any toroidal
Seifert piece has epsilon vertical index $\geq 2$, except the weakly
toroidal pieces with epsilon vertical index $= 3/2$.

\begin{cor}
\label{final}

Suppose $M$ is a closed orientable irreducible $3$-manifold that has a
genus $g$ Heegaard splitting.  Let $\Theta$ be the
decomposing collection of tori for $M$.  Suppose $p$ of the
components of $M - \Theta$ are not Seifert fibered spaces, and the
union of Seifert components of $M - \Theta$ has base space $P$ and $f$
exceptional fibers.  Then
$$f - \chi(P) \leq 3g - 3 - p.$$

\end{cor}

\pf  Apply Corollary \ref{cor2}:  $-\chi (S_3) = 1$, so $$f - \chi(P)
= s + I_{v} (W) \leq s + I_{v}^{\eee}(W) \leq 3g - 3 - p.$$
\qed



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