# On stable commutator length of non-filling curves in surfaces

Max Forester  Mathematics Department
University of Oklahoma
Norman, OK 73019
USA<EMAIL_ADDRESS>and  Justin Malestein  Mathematics Department
University of Oklahoma
Norman, OK 73019
USA<EMAIL_ADDRESS>
###### Abstract.

We give a new proof of rationality of stable commutator length (scl) of
certain elements in surface groups: those represented by curves that do not
fill the surface. Such elements always admit extremal surfaces for scl. These
results also hold more generally for non-filling $1$–chains.

##  1\. Introduction

This paper concerns the computation of stable commutator length (scl) in
fundamental groups of closed orientable surfaces. It is of considerable
interest to know whether scl is always rational in these groups, or more
generally, whether the scl norm is piecewise rational linear (in the sense of
[Cal09c]).

In more concrete terms, let $\Sigma$ be a closed orientable surface and
$\gamma\colon\thinspace S^{1}\to\Sigma$ a null-homologous loop. We are
interested in identifying efficient maps of surfaces into $\Sigma$, with
boundary mapping to a positive power of $\gamma$. “Efficient” means that the
ratio of the topological complexity of the surface to the power of $\gamma$ on
the boundary is as small as possible. The infimal value of this ratio is the
stable commutator length of $\gamma$. In some cases, there may be a surface
realizing this infimal value; these are called _extremal surfaces_ for scl,
and when they exist scl is rational.

Calegari [Cal09a] has shown that one can always find an _immersed_ surface
with boundary a power of $\gamma$, and such surfaces are the most efficient
among those in their relative homology classes. However there are infinitely
many relative classes, and one cannot draw any immediate conclusions about the
existence of an extremal surface for $\gamma$.

In the present paper we give a new proof the following theorem of Calegari,
which establishes rationality in the special case of curves $\gamma$ that do
not fill $\Sigma$. It appears as Example 4.51 in [Cal09b], as an application
of Theorem 4.47. We state the result here for integral $1$–chains, i.e.
immersed multicurves.

######  Theorem 1.1 (Calegari).

Let $\Sigma$ be a closed orientable surface of genus $>1$ and
$\gamma\colon\thinspace\coprod S^{1}\to\Sigma$ a null-homologous integral
$1$-chain which does not fill $\Sigma$. Then the stable commutator length of
$\gamma$ is rational and computable. Moreover there exists an extremal surface
for $\gamma$.

It follows immediately that _rational_ $1$–chains supported on non-filling
multicurves also have rational stable commutator length. Indeed, we find that
scl is piecewise rational linear on “non-filling” rational subspaces of the
space of $1$–chains; that is, subspaces spanned by rational $1$–chains
$c_{1},\dotsc,c_{k}$ such that the union of their supports is a non-filling
multicurve.

### Other rationality results

The list of groups known to have rational scl, excluding those with
$\operatorname{scl}\equiv 0$, is fairly short. Such groups include:

  * •

free groups [Cal09c]

  * •

fundamental groups of non-compact Seifert fibered $3$–manifolds [Cal09c]

  * •

free products of free abelian groups [Cal11]

  * •

free products of cyclic groups [Wal13]

  * •

amalgamated products of free abelian groups [Sus15]

  * •

Baumslag–Solitar groups, alternating elements only [CFL16]

  * •

free products of groups in which $\operatorname{scl}\equiv 0$ [Che18]

  * •

generalized Baumslag–Solitar groups [Che20]

Surface groups are conspicuously absent from this list, despite being among
the simplest and most well-understood one-relator groups.

It is worth noting that finitely presented groups do not always have rational
scl; a counterexample was given by Zhuang in [Zhu08]. If one allows non-
finitely presented groups, then in fact every non-negative real number occurs
as scl of some element of a small cancellation group [Heu19].

### Methods

The general outline of the argument is similar to the case of free groups as
presented in [BCF11]. The stable commutator length of $\gamma$ is the infimum
of $\frac{-\chi(S)}{2n(S)}$ over all admissible surfaces $S$ mapping to
$\Sigma$. First we show that the infimum can be taken over the much smaller
subset ${\mathscr{T}}(\gamma)$ of “taut” admissible surfaces—these will be
described below and defined precisely in Section 3.

Taut surfaces can be encoded as integer vectors via a map
$v\colon\thinspace{\mathscr{T}}(\gamma)\to{\mathbb{R}}^{k}$, and the function
$\frac{-\chi(S)}{2n(S)}$ factors through this map as
${\mathscr{T}}(\gamma)\overset{v}{\to}{\mathbb{R}}^{k}\to{\mathbb{R}}$ where
the second map is a ratio $A(v)/B(v)$ of linear functions. The computation of
$\operatorname{scl}(\gamma)$ is then reduced to a linear programming problem
after showing that the image of $v$ in ${\mathbb{R}}^{k}$ fills (in a suitable
sense) the integer points of a finite-sided rational polyhedron in
${\mathbb{R}}^{k}$.

The two main steps are _encoding_ , which entails defining
${\mathscr{T}}(\gamma)$ and the function $v$; and _reassembly_ , in which
specific vectors are shown to be in the image of $v$. Our main assumption
throughout the paper is that $\Sigma$ contains a subsurface $\Sigma_{1}$ with
non-trivial topology that is disjoint from the image of $\gamma$. This
property is needed for both steps.

To make a surface $S$ taut, one needs to put it into a normal form that can be
encoded as a finite integer vector. Following Gabai [Gab85], we arrange that
away from a region containing the double points of $\gamma$, the preimage in
$S$ maps by a branched immersion. We arrange further that the remaining parts
of $S$ fall into finitely many configurations, and all branch points lie above
$\Sigma_{1}$. The branch points are now the main obstactle to having a finite
encoding, but these can be removed by passing to finite-sheeted covers, using
the non-trivial topology of $\Sigma_{1}$.

For reassembly we show that every integer vector in the rational polyhedron
has a multiple that encodes a taut surface. When building this surface out of
pieces, a phenomenon involving branch points arises. A taut surface cannot be
built directly, but one can build one with branch points. The presence of
branch points means that the linear functional on ${\mathbb{R}}^{k}$ reporting
$\chi(S)$ does not report this number correctly. Once again, using finite
coverings of $\Sigma_{1}$, we are able to eliminate these branch points and
construct a taut surface with the desired encoding.

In contrast, Calegari’s approach is quite different. He proves Theorem 1.1 by
viewing surface groups as fundamental groups of graphs of groups in which
vertex groups are free and edge groups are cyclic. If $\gamma$ is non-filling
then it avoids an essential annulus, and one may encode $\Sigma$ as a graph of
groups with one edge. The proof then proceeds by reducing the computation of
$\operatorname{scl}$ to computing $\operatorname{scl}$ in each vertex group,
using the assumption that $\gamma$ is supported in the vertex groups.

######  Acknowledgments.

The authors were partially supported by Simons Foundation awards #638026 (MF)
and #713006 (JM). They thank the referee for making helpful suggestions.

##  2\. Preliminaries

### Stable commutator length

We start with basic working definitions for stable commutator length of group
elements and of $1$–chains. See [Cal09b, Chapter 2] for more information.

######  Definition 2.1 (scl of integral $1$–chains).

Let $X$ be a space with fundamental group $G$. An _integral $1$–chain over
$G$_ is a finite formal sum $c=\sum_{i}g_{i}$ with $g_{i}\in G$. Let
$\gamma_{i}\colon\thinspace S^{1}\to X$ be a loop representing $g_{i}$ for
each $i$ and let $\gamma\colon\thinspace\coprod_{i}S^{1}\to X$ be the map
given by $\coprod_{i}\gamma_{i}$. An _admissible surface for $\gamma$_ (or
_for $c$_) is a compact oriented surface $S$ together with a map
$f\colon\thinspace S\to X$ such that

  * •

$\partial S\not=\emptyset$ and $S$ has no sphere or disk components

  * •

the restriction $f|_{\partial S}$ factors through $\gamma$; that is, there is
a commutative diagram

${\partial
S}$${S}$${\coprod_{i}S^{1}}$${X}$$\scriptstyle{f}$$\scriptstyle{\coprod_{i}\gamma_{i}\
}$

  * •

the restriction of the map $\partial S\to\coprod_{i}S^{1}$ to each connected
component of $\partial S$ is a map of _positive_ degree

  * •

there is a positive integer $n(S)$ such that for each component of
$\coprod_{i}S^{1}$, the preimage in $\partial S$ maps to it by total degree
$n(S)$.

The _stable commutator length (scl) of $c$_ is defined by

$\operatorname{scl}(c)\ =\ \inf_{S}\frac{-\chi(S)}{2n(S)}$

where the infimum is taken over all admissible surfaces for $\gamma$. If no
admissible surface exists, we define $\operatorname{scl}(c)=\infty$.

Note: the third bullet is not included in Calegari’s definition of admissible
surface in [Cal09b]. However, Proposition 2.13 of [Cal09b] shows that
including it does not change the meaning of scl.

Using homogeneity properties of $\operatorname{scl}$, the definition extends
to rational chains. If $c=\sum_{i}c_{i}g_{i}$ with $c_{i}\in{\mathbb{Q}}$ then
$mc$ is integral for some $m$, and we may define
$\operatorname{scl}(c)=\frac{1}{m}\operatorname{scl}(mc)$. Extending by
continuity, it is also defined for real chains. See [Cal09b, Section 2.6].

######  Definition 2.2 (scl of group elements).

For any $g\in G$ we define $\operatorname{scl}(g)$ by considering $g$ as an
integral $1$–chain consisting of one element. The above definition, in this
case, specializes to the standard definition of $\operatorname{scl}(g)$. Note
that admissible surfaces for $g$ exist if and only if $g^{k}\in[G,G]$ for some
$k\not=0$, and so $\operatorname{scl}(g)$ is finite exactly when the latter
occurs.

######  Definition 2.3.

Let $c$ be an integral $1$–chain over $G$. An _extremal surface for $c$_ is an
admissible surface $S$ for $c$ that realizes the infimum in the definition of
$\operatorname{scl}(c)$. If $c$ is a rational $1$–chain, an extremal surface
for $c$ is just an extremal surface for any $mc$ that is integral. Extremal
surfaces need not exist, but when they do, $\operatorname{scl}(c)$ must of
course be rational.

Let $C_{1}(G)$ be the space of $1$–chains, which is a vector space over
${\mathbb{R}}$ with basis $G$. Let $B_{1}(G)\subset C_{1}(G)$ be the subspace
of boundaries of $2$–chains, i.e. the kernel of the quotient map $C_{1}(G)\to
H_{1}(G;{\mathbb{R}})$. Let $B_{1}^{H}(G)$ be the quotient of $B_{1}(G)$ by
the subspace $H$ spanned by elements of the form $g^{n}-ng$ and $g-hgh^{-1}$
for all $g,h\in G$, $n\in{\mathbb{Z}}$. Then the function $\operatorname{scl}$
is a pseudo-norm on $B_{1}^{H}(G)$. Moreover, whenever $G$ is hyperbolic,
$\operatorname{scl}$ is a geniune norm on $B_{1}^{H}(G)$ [CF10].

### Surfaces

Let $\Sigma$ be a closed surface. We say that a multicurve
$\gamma\colon\thinspace\coprod S^{1}\to\Sigma$ is _non-filling_ if it is
homotopic to an immersion in general position such that some component of the
complement of the image is not a disk (or equivalently, $\Sigma$ contains an
essential simple closed curve disjoint from the image of $\gamma$).
Correspondingly, an integral $1$–chain or group element in $G=\pi_{1}(\Sigma)$
is _non-filling_ if it is represented by a non-filling multicurve.

######  Definition 2.4.

A map $f\colon\thinspace S\to\Sigma$ of orientable surfaces is _compressible_
if there is an essential simple closed curve $C\subset S$ such that $f(C)$ is
nullhomotopic in $\Sigma$. If $f$ is not compressible it is called
_incompressible_.

When $f$ is compressible, one can replace an annular neighborhood of $C$ by
two disks and extend $f$ over these disks, thus obtaining
$f^{\prime}\colon\thinspace S^{\prime}\to\Sigma$ of smaller complexity.

### Coverings

Throughout this paper curves and surfaces typically have orientations, and by
the _degree_ of a map we always mean the homological degree.

Suppose $p\colon\thinspace X\to Y$ is a finite-sheeted covering of oriented
manifolds. Partition the connected components of $X$ as $X^{+}\sqcup X^{-}$
such that $p$ is orientation-preserving on $X^{+}$ and orientation-reversing
on $X^{-}$. We define the _positive degree_ of $p$ to be the number of sheets
of $X^{+}$ and the _negative degree_ of $p$ to be the number of sheets of
$X^{-}$. Thus the sum of these numbers is the total number of sheets of $p$
and their difference is the homological degree.

Next we record some basic covering lemmas for surfaces.

######  Lemma 2.5.

Let $S$ be a compact connected oriented surface having $p\geqslant 1$ boundary
components, with $S\not\cong D^{2}$.

  1. (1)

For any $q<p$ boundary components $C_{1},\dotsc,C_{q}$ of $S$ and integers
$n_{1},\dotsc,n_{q}\geqslant 1$, there is a connected regular finite-sheeted
cover $S^{\prime}\to S$ such that every component of $\partial S^{\prime}$
mapping to $C_{i}$ covers with degree $n_{i}$.

  2. (2)

For any integer $n\geqslant 1$ there is a connected finite-sheeted cover
$S^{\prime}\to S$ such that every component of $\partial S^{\prime}$ covers
its image with degree $n$.

###### Proof.

For (1), let $x_{i}\in H_{1}(S)$ be the element represented by $C_{i}$, for
$1\leqslant i\leqslant q$. This set extends to a basis
$\\{x_{1},\dotsc,x_{n}\\}$ of $H_{1}(S)$, because $q<p$. Let $N$ be the least
common multiple of the numbers $n_{i}$. Let $\varphi\colon\thinspace
H_{1}(S)\to{\mathbb{Z}}/N{\mathbb{Z}}$ send $x_{i}$ to $N/n_{i}$ for
$1\leqslant i\leqslant q$ (the values on the other basis elements are
irrelevant). Now $S^{\prime}\to S$ is the cover corresponding to the kernel of
the composition $\pi_{1}(S)\to H_{1}(S)\to{\mathbb{Z}}/N{\mathbb{Z}}$. Since
$\varphi(x_{i})$ has order $n_{i}$ in ${\mathbb{Z}}/N{\mathbb{Z}}$ and the
cover is regular, the boundary $\partial S^{\prime}$ behaves as desired.

For (2), suppose first that $p>1$. Choose boundary components
$C_{1},\dotsc,C_{p-1}$ of $\partial S$ and a basis $\\{x_{i}\\}$ of $H_{1}(S)$
as before. Let $\varphi\colon\thinspace
H_{1}(S)\to({\mathbb{Z}}/n{\mathbb{Z}})^{p-1}$ send $x_{i}$ to the $i$th
generator $(0,\dotsc,1,\dotsc,0)$ for $1\leqslant i\leqslant p-1$. Then
$\varphi$ sends the class represented by the last boundary component of $S$ to
$(-1,\dotsc,-1)$. Since every boundary component maps to an element of order
$n$, the cover $S^{\prime}\to S$ corresponding to the kernel of $\pi_{1}(S)\to
H_{1}(S)\to({\mathbb{Z}}/n{\mathbb{Z}})^{p-1}$ has the desired property.

If $\partial S$ has one component, first let $S^{\prime\prime}\to S$ be any
connected $2$–sheeted cover (which exists since $S\not\cong D^{2}$). It will
have $2$ boundary components, each mapping by degree $1$, because
$\chi(S^{\prime\prime})$ is even. Apply the preceding case to get
$S^{\prime}\to S^{\prime\prime}$, and take the composition $S^{\prime}\to
S^{\prime\prime}\to S$. ∎

It will be convenient to have a version of this lemma for disconnected
surfaces. By a finite-sheeted cover of a disconnected surface, we mean a cover
for which the number of sheets is the same over every component.

######  Lemma 2.6.

Let $S$ be a compact oriented surface where each component has at least $1$
boundary component and no component is homeomorphic to $D^{2}$. For any set of
boundary components $C_{1},\dotsc,C_{q}$ of $S$ which does not contain all the
boundary components of any component of $S$ and integers
$n_{1},\dotsc,n_{q}\geqslant 1$, there is a finite-sheeted cover
$S^{\prime}\to S$ such that every component of $\partial S^{\prime}$ mapping
to $C_{i}$ covers with degree $n_{i}$.

###### Proof.

Let $S_{1},\dots,S_{m}$ be the components of $S$. We first apply Lemma 2.5 (1)
to each $S_{i}$ to obtain an $N_{i}$-fold cover $S_{i}^{\prime}\to S_{i}$ with
the correct degree on boundary components. Let
$N=\operatorname{lcm}(N_{1},\dots,N_{m})$. Then the cover of $S$ consisting of
$N/N_{i}$ copies of $S_{i}$ has the desired properties. ∎

We are also interested in the case where we allow branched covers with
prescribed degrees on the boundaries. The flexibility of degrees on boundaries
is much larger for branched covers.

######  Lemma 2.7.

Let $S$ be a compact connected oriented surface having $p\geqslant 1$ boundary
components. Let $C_{1},\dots,C_{p}$ be the boundary components of $S$, let $n$
be a natural number, and let $\lambda_{1},\dots,\lambda_{p}$ be partitions of
$n$. Then there is an $n$-fold branched cover $S^{\prime}\to S$ such that the
degrees of the components over $C_{i}$ are precisely the values in the
partition $\lambda_{i}$.

Note that $S^{\prime}$ is not necessarily connected.

###### Proof.

We start with the cover $S^{\prime\prime}\to S$ where $S^{\prime\prime}$ is
$n$ copies of $S$. Let $A,B$ be distinct copies of $C_{i}$ in
$S^{\prime\prime}$. Let $\alpha,\beta$ be copies of the same small arc from a
point on $C_{i}$ into the interior of $S$. Cut $S^{\prime\prime}$ along
$\alpha$ and $\beta$ and glue the right side of $\beta$ to the left side of
$\alpha$ and vice versa. This new surface, in a canonical way, admits a
branched cover to $S$ where the degrees over $C_{i}$ are $2,1,1,\dots,1$. By
using this kind of surgery to connect boundary components of
$S^{\prime\prime}$, it is clear that we can construct a branched cover
$S^{\prime}\to S$ where the degrees over $C_{i}$ are an arbitrary partition of
$n$. ∎

##  3\. Mapping surfaces to surfaces

### Setup

Let $\Sigma$ be a closed oriented surface of genus greater than one. Let
$\gamma\colon\thinspace\coprod S^{1}\to\Sigma$ be a null-homologous $1$–chain
that does not fill $\Sigma$. We can put $\gamma$ into general position
(smoothly immersed, transverse intersections, no triple points) such that at
least one complementary component is not a disk.

Let $\Sigma_{0}$ be one such complementary component. Let $B\subset\Sigma_{0}$
be an embedded closed disk and let $\lambda_{1},\dotsc,\lambda_{k}$ be
disjoint properly embedded arcs in $\Sigma_{0}-\operatorname{int}(B)$ with
endpoints on $\partial B$ such that the inclusion of
$B\cup\bigcup_{i}\lambda_{i}$ into $\Sigma_{0}$ is a homotopy equivalence. Let
$L\subset(\Sigma_{0}-\operatorname{int}(B))$ be the closure of a tubular
neighborhood of $\bigcup_{i}\lambda_{i}$, so that $\Sigma_{1}=B\cup L$ is
compact subsurface homotopy equivalent to $\Sigma_{0}$.

Next let $\mu_{1},\dotsc,\mu_{\ell}$ be disjoint properly embedded arcs in
$\Sigma-\operatorname{int}(\Sigma_{1})$ with endpoints on $\partial B-L$ such
that the $\mu_{i}$ cut each component of $\Sigma-\Sigma_{1}$ into a single
disk and the $\mu_{i}$ do not cross any double points of $\gamma$. We also
require that the arcs $\mu_{i}$ are transverse to the immersed submanifold
$\gamma(\coprod S^{1})$. Let $M\subset(\Sigma-\operatorname{int}(\Sigma_{1}))$
be the closure of a tubular neighborhood of $\bigcup_{i}\mu_{i}$, disjoint
from $L$, so that $\Sigma_{2}=\Sigma_{1}\cup M$ contains no double point of
$\gamma$ and $\gamma(\coprod S^{1})\cap M$ consists of fibers of $M$. Let $D$
be the closure of $\Sigma-\Sigma_{2}$; this is a disjoint union of embedded
closed disks. Finally, define
$\Sigma_{3}=\Sigma-\operatorname{int}(\Sigma_{1})=D\cup M$.

Each component of $\coprod S^{1}$ maps by $\gamma$ to a path which alternates
between embedded oriented arcs in $D$, called _turn arcs_ , and arcs crossing
$M$. Let $\tau_{1},\dotsc,\tau_{m}$ be the turn arcs. See Figure 1.

$B$$\alpha_{1}$ $\alpha_{2}$ $\alpha_{3}$ $\alpha_{4}$ $\alpha_{5}$
$\alpha_{6}$ $\alpha_{7}$ $\alpha_{8}$ $\alpha_{9}$ $\alpha_{10}$
$D_{1}$$D_{2}$$\alpha_{1}$$\alpha_{2}$$\alpha_{3}$$\alpha_{4}$$\alpha_{5}$$\alpha_{6}$$\alpha_{7}$$\alpha_{8}$$\alpha_{9}$$\alpha_{10}$
Figure 1. A surface $\Sigma$ and its handle decomposition $\Sigma=B\cup(L\cup
M)\cup D$. The shaded part is the subsurface $\Sigma_{1}=B\cup L$. The null-
homologous $1$–cycle $\gamma$ is shown in blue. All self-intersections of
$\gamma$ lie inside $D=D_{1}\cup D_{2}$. Outside of $D$, $\gamma$ meets only
the $1$–handles $M$ and crosses them transversely. The arcs of $\gamma$
crossing $D$ are called turn arcs. The arcs $\alpha_{i}$ are the maximal sub-
arcs of $\partial D$ that meet $\partial\Sigma_{1}$.

######  Definition 3.1 (Turn paths).

A _turn path_ is a loop $S^{1}\to D$ which is a concatenation of paths,
alternating between turn arcs and immersed paths in $\partial D$. Immersions
$S^{1}\to\partial D$ are included. We consider two turn paths to be the same
if they differ by an orientation-preserving reparametrization of $S^{1}$. See
Figure 2.

Each component of the oriented boundary of $D$ is a concatenation of arcs,
alternating between arcs in the boundary of $M$ and arcs in the boundary of
$\Sigma_{1}$. Let $\alpha_{1},\dotsc,\alpha_{n}$ be the arcs of $\partial D$
which lie in the boundary of $\Sigma_{1}$.

A turn path is _taut_ if it uses each turn arc at most once and each arc from
$\\{\alpha_{i}\\}\cup\\{\overline{\alpha}_{i}\\}$ at most once (where
$\overline{\alpha}_{i}$ is the reverse of $\alpha_{i}$). Note that there are
only finitely many taut turn paths.

$\alpha_{2}$$\alpha_{6}$$\tau_{1}$$\tau_{2}$ Figure 2. A taut turn path in
$D_{1}$ which runs over $\tau_{1}$, $\overline{\alpha}_{6}$, $\tau_{2}$, and
$\alpha_{2}$. Every turn path extends to a map of a disk; in this case the
disk is twisted.

######  Definition 3.2.

An admissible map $f\colon\thinspace S\to\Sigma$ is taut if

  1. (1)

$f\colon\thinspace f^{-1}(\Sigma_{2})\to\Sigma_{2}$ is an immersion

  2. (2)

$f^{-1}(D)$ is a disjoint union of disks,

  3. (3)

for each component $E$ of $f^{-1}(D)$, the boundary map $f|_{\partial E}$ is a
taut turn path.

The set of taut admissible maps for $\gamma$ will be denoted by
${\mathscr{T}}(\gamma)$. The disks in (2) and (3) will be called _turn disks_.

The main result of this section is the following.

######  Proposition 3.3.

Suppose $f\colon\thinspace S\to\Sigma$ is an admissible map. Then there exists
a taut admissible map $f^{\prime}\colon\thinspace S^{\prime}\to\Sigma$ such
that
$\frac{-\chi(S^{\prime})}{2n(S^{\prime})}\leqslant\frac{-\chi(S)}{2n(S)}$.

######  Corollary 3.4.

There is an equality

$\operatorname{scl}(\gamma)\ =\
\inf_{S\in{\mathscr{T}}(\gamma)}\frac{-\chi(S)}{2n(S)}.$

The next lemma provides the first step toward Proposition 3.3. It achieves all
the desired properties except for tautness of the turn paths in item (3).

######  Lemma 3.5.

Suppose $f\colon\thinspace S\to\Sigma$ is an admissible map which is
incompressible. Then $f$ is homotopic to an admissible map $g\colon\thinspace
S\to\Sigma$ such that

  1. (1)

$g\colon\thinspace g^{-1}(\Sigma_{2})\to\Sigma_{2}$ is an immersion

  2. (2)

$g^{-1}(D)$ is a disjoint union of disks,

  3. (3)

for each component $E$ of $g^{-1}(D)$, the boundary map $g|_{\partial E}$ is a
turn path.

The boundary maps $f|_{\partial S}$ and $g|_{\partial S}$ differ only by
reparametrization.

###### Proof.

We proceed as in Step 1 of [Gab85, Theorem 2.1]. Homotope $f$ rel boundary to
$f_{1}$ which is transverse to the disk $B$. We then have that each component
of $f_{1}^{-1}(B)$ is a disk mapping homeomorphically onto $B$. Next, let
$T=S-\operatorname{int}(f_{1}^{-1}(B))$. Since $f_{1}|_{\partial T}$ is
transverse to the arcs $\lambda_{i}$ and $\mu_{i}$, we may homotope $f_{1}$
rel $\partial S\cup f_{1}^{-1}(B)$ to be transverse to these arcs. Among all
maps $f_{2}$ homotopic to $f$ rel $\partial S$ that are transverse to $B$ and
transverse to the arcs $\lambda_{i}$ and $\mu_{i}$ on
$S-\operatorname{int}(f_{2}^{-1}(B))$, let $f_{2}$ minimize the _complexity_ ,
which is the pair
$\left(\left\lvert{\pi_{0}\bigl{(}f_{2}^{-1}\bigl{(}B\bigr{)}\bigr{)}}\right\rvert,\left\lvert{\pi_{0}\bigl{(}f_{2}^{-1}\bigl{(}\bigcup_{i}\lambda_{i}\cup\bigcup_{i}\mu_{i}\bigr{)}\bigr{)}}\right\rvert\right)$
ordered lexicographically.

Since $f_{2}$ is incompressible, no component of $f_{2}^{-1}(\lambda_{i})$ or
$f_{2}^{-1}(\mu_{i})$ is an essential closed curve. No component is an
inessential closed curve either. Suppose $C$ is such a curve mapping to
$\lambda_{i}$, say. For a sufficiently close arc $\lambda_{i}^{\prime}$ that
is parallel to $\lambda_{i}$, there is a curve $C^{\prime}$ parallel to $C$
mapping to $\lambda_{i}^{\prime}$ that bounds a disk $E\subset S$, with $C$ in
the interior of $E$. By re-defining the map on $\operatorname{int}(E)$ to map
into $\lambda_{i}^{\prime}$ one obtains a map of smaller complexity; the new
map is homotopic to $f_{2}$ rel $\partial S$ because $\pi_{2}(\Sigma)=0$.

Thus each component of $f_{2}^{-1}(\lambda_{i})$ is an arc with endpoints
mapping to $\partial\lambda_{i}$, and each component of $f_{2}^{-1}(\mu_{i})$
is an arc with endpoints mapping to
$\partial\mu_{i}\cup\\{\mu_{i}\cap\gamma(\coprod S^{1})\\}$. Note that in the
second case, both endpoints cannot map to the same point
$z\in\\{\mu_{i}\cap\gamma(\coprod S^{1})\\}$, by orientation considerations.
There is a consistent transverse orientation on $f_{2}^{-1}(\mu_{i})$, but
$f_{2}|_{\partial S}$ always passes through $z$ in the same direction, being a
positive power of $\gamma$.

Since complexity is minimized, both endpoints cannot map to the same point of
$\partial\mu_{i}$; otherwise,
$\left\lvert{\pi_{0}\bigl{(}f_{2}^{-1}\bigl{(}\bigcup_{i}\lambda_{i}\cup\bigcup_{i}\mu_{i}\bigr{)}\bigr{)}}\right\rvert$
could be reduced, and then
$\left\lvert{\pi_{0}\bigl{(}f_{2}^{-1}\bigl{(}B\bigr{)}\bigr{)}}\right\rvert$
as well by homotopy rel $\partial S$ to a new map which is still transverse to
$B\cup\\{\mu_{i}\\}\cup\\{\lambda_{i}\\}$. Hence the endpoints map to distinct
points. The same reasoning applies to components of $f_{2}^{-1}(\lambda_{i})$.
Hence the endpoints of every arc in $f_{2}^{-1}(\lambda_{i})$ or
$f_{2}^{-1}(\mu_{i})$ map to distinct points.

Now, by a further homotopy rel $\partial S\cup f_{2}^{-1}(B)$, the components
of $f_{2}^{-1}(\lambda_{i})$ and $f_{2}^{-1}(\mu_{i})$ can be tightened so
that each maps by an immersion; let $f_{3}$ be the resulting map. There are
compatible tubular neighborhoods on which $f_{3}$ is a bundle map, and by
another homotopy to $f_{4}$ (which expands along fibers of the tubular
neighborhoods $L$ and $M$) we can ensure that each component of
$f_{4}^{-1}(L)$ and $f_{4}^{-1}(M)$ maps by an immersion. Now
$f_{4}\colon\thinspace f_{4}^{-1}(\Sigma_{2})\to\Sigma_{2}$ is an immersion,
and $f_{4}$ satisfies (1).

Next consider a component $E$ of $f_{4}^{-1}(D)$. Each boundary component of
$\partial E$ maps by a concatenation of paths that alternate between turn arcs
and paths in $\partial D$; moreover these latter paths are part of the
boundary map of $f_{4}\colon\thinspace f_{4}^{-1}(\Sigma_{2})\to\Sigma_{2}$
and hence are immersions. Thus $\partial E$ maps to $D$ by turn paths and
$f_{4}$ satisfies (3).

Next, suppose some component $E$ of $f_{4}^{-1}(D)$ is not a disk. Then there
is a simple closed curve $C\subset\operatorname{int}(E)$ which is essential in
$E$. Since $f_{4}(C)$ is nullhomotopic in $\Sigma$, we have that $C$ bounds a
disk $E^{\prime}\subset S$. Since $\pi_{2}(\Sigma)=0$ there is a homotopy rel
$S-\operatorname{int}(E^{\prime})$ from $f_{4}$ to $f_{5}$ such that
$f_{5}(E^{\prime})\subset D$. By performing finitely many homotopies of this
type, we arrive at $g$ such that each component of $g^{-1}(D)$ is a disk and
$g$ agrees with $f_{4}$ on
$g^{-1}(\Sigma-\operatorname{int}(D))=g^{-1}(\Sigma_{2})$. In particular, $g$
satisfies (1) and (2).

Lastly, $f_{4}^{-1}(\partial D)\supseteq g^{-1}(\partial D)$ and $f_{4}$ and
$g$ agree on $g^{-1}(\partial D)$ (and on $\partial S$), so the boundary paths
of $g^{-1}(D)$ are unchanged and $g$ satisfies (3). For the last statement,
note that $f|_{\partial S}=f_{3}|_{\partial S}$ and $f_{4}|_{\partial
S}=g|_{\partial S}$, and the change from $f_{3}|_{\partial S}$ to
$f_{4}|_{\partial S}$ simply reparametrizes along $\partial S$. ∎

The map $g$ given by the previous lemma may be quite badly behaved on
$g^{-1}(D)$, with branch points, folding, twisting, etc. This will mostly not
be a concern for us; what matters most is the boundary maps on $g^{-1}(D)$.
The next lemma achieves tautness of these boundary maps at the expense of
creating branch points over $\Sigma_{1}$.

######  Lemma 3.6.

Suppose $g\colon\thinspace S\to\Sigma$ is an admissible map which satisfies
the conclusions of Lemma 3.5. Then there is an admissible map
$g_{1}\colon\thinspace S_{1}\to\Sigma$ with $n(S_{1})=n(S)$ and
$\chi(S)\leqslant\chi(S_{1})$ such that

  1. (1)

$g_{1}\colon\thinspace g_{1}^{-1}(\Sigma_{2})\to\Sigma_{2}$ is a branched
immersion with all branch points in the interior of $g_{1}^{-1}(\Sigma_{1})$,

  2. (2)

$g_{1}^{-1}(D)$ is a disjoint union of disks,

  3. (3)

for each component $E$ of $g_{1}^{-1}(D)$, the boundary map $g_{1}|_{\partial
E}$ is a taut turn path,

  4. (4)

$g_{1}$ is an immersion on every component of $g_{1}^{-1}(\Sigma_{3})$ that
does not meet $\partial S_{1}$.

###### Proof.

Let $E$ be a component of $g^{-1}(D)$. There are two types of moves we will
use to achieve tautness in item (3). Suppose first that $\partial E$ contains
two or more instances of the turn arc $\tau_{i}$. Delete the interior of $E$,
cut the midpoints of the two instances of $\tau_{i}$ and glue the first half
of one to the second half of the other and vice versa. This splits $\partial
E$ into two components, which can be filled by two disks mapping to $D$,
increasing $\chi(S)$. After finitely many moves of this type, there are no
repeated instances of turn arcs in components of $\partial g^{-1}(D)$. The new
boundary of $S$ still maps by positive powers of $\gamma$ with degree $n(S)$.

Now suppose that the arc
$\alpha\in\\{\alpha_{i}\\}\cup\\{\overline{\alpha}_{i}\\}$ appears two or more
times in $\partial E$. These two occurrences are also part of $\partial
g^{-1}(\Sigma_{1})$; denote them by $\alpha^{\prime}$ and
$\alpha^{\prime\prime}$. Let $U$ be an open neighborhood of $\alpha$ in
$\Sigma_{1}$ that is evenly covered by $g$, and let $U^{\prime}$,
$U^{\prime\prime}$ be the corresponding neighborhoods of $\alpha^{\prime}$ and
$\alpha^{\prime\prime}$ in $g^{-1}(\Sigma_{1})$. Let $\beta$ be a small arc in
$U$ from the midpoint of $\alpha$ to an interior point of $U$. Let
$\beta^{\prime}$, $\beta^{\prime\prime}$ be the lifts of $\beta$ to
$U^{\prime}$ and $U^{\prime\prime}$ respectively. Now delete the interior of
$E$ and cut along $\beta^{\prime}$ and $\beta^{\prime\prime}$. Rejoin the left
side of $\beta^{\prime}$ to the right side of $\beta^{\prime\prime}$ and vice
versa. This creates an index two branch point in $g^{-1}(\Sigma_{1})$ and
splits $\partial E$ into two components, which can be filled with two disks
mapping to $D$. The Euler characteristic $\chi(S)$ does not change, and
neither does $n(S)$, because this move takes place in the interior of $S$.
After finitely many moves of this second type, we arrive at
$g_{0}\colon\thinspace S_{1}\to\Sigma$ satisfying (1)–(3).

Let $E$ be a component of $g_{0}^{-1}(D)$ which does not meet $\partial S$.
Then the map $g_{0}|_{\partial E}$ is a homeomorphism to $\partial D^{\prime}$
for some component $D^{\prime}$ of $D$ by (3). For each such $E$, we replace
$g_{0}|_{E}$ by a homeomorphism $E\to D^{\prime}$ extending $g_{0}|_{\partial
E}$ to get a new map $g_{1}\colon\thinspace S_{1}\to\Sigma$. Note that $g_{1}$
still satisfies (1)–(3). By (1) and the above, $g_{1}$ also satisfies (4). ∎

For the last step of the argument, we use finite covers to eliminate the
branch points over $\Sigma_{1}$. This step depends on $\Sigma_{1}$ having non-
trivial topology. Note that conclusions (1) and (2) of this lemma are not
required for Proposition 3.3 but will be useful in a later section.

######  Lemma 3.7.

Suppose $g_{1}\colon\thinspace S_{1}\to\Sigma$ is an admissible map which
satisfies the conclusions of Lemma 3.6. Let $r$ be the number of sheets of the
branched cover $g_{1}\colon\thinspace g_{1}^{-1}(\Sigma_{1})\to\Sigma_{1}$.
Then there is a taut admissible map $g_{2}\colon\thinspace S_{2}\to\Sigma$ and
a positive integer $N$ such that

  1. (1)

$n(S_{2})=Nn(S_{1})$,

  2. (2)

$\chi(g_{2}^{-1}(\Sigma_{3}))=N\chi(g_{1}^{-1}(\Sigma_{3}))$ and
$\chi(g_{2}^{-1}(\Sigma_{1}))=Nr\chi(\Sigma_{1})$,

  3. (3)

$\frac{-\chi(S_{2})}{2n(S_{2})}\leqslant\frac{-\chi(S_{1})}{2n(S_{1})}$.

###### Proof.

Let $T_{3}^{\prime\prime}$ be the union of all components of
$g_{1}^{-1}(\Sigma_{3})$ which meet $\partial S_{1}$. Let

$d=\operatorname{lcm}\\{\text{degree of }C^{\prime}\overset{g_{1}}{\to}C\mid
C^{\prime},C\text{ are components of }\partial
T_{3}^{\prime\prime},\partial\Sigma_{3}\text{ resp.}\\}$

By Lemma 2.6, there is a finite-sheeted cover $T_{3}^{\prime}\to
T_{3}^{\prime\prime}$ such that every component of $\partial T_{3}^{\prime}$
covers a component of $\partial\Sigma_{3}$ with degree $\pm d$ under the
composition $T_{3}^{\prime}\to
T_{3}^{\prime\prime}\overset{g_{1}}{\to}\Sigma_{3}$. (The orientation of a
covering surface will be induced by the base unless indicated otherwise.) Let
$\Sigma_{3,1},\dots,\Sigma_{3,\ell}$ be the components of $\Sigma_{3}$. By
Lemma 2.5(2), there are connected finite-sheeted covers
$T_{3,i}^{\prime}\to\Sigma_{3,i}$ and $T_{1}^{\prime}\to\Sigma_{1}$ such that
every component of $\partial T_{3,i}^{\prime}$ and $\partial T_{1}^{\prime}$
covers a component of $\partial\Sigma_{3}$ with degree $\pm d$.

Let $s_{i}^{+},s_{i}^{-}$ be the positive and negative degrees by which
$g_{1}^{-1}(\Sigma_{3,i})-T_{3}^{\prime\prime}$ covers $\Sigma_{3,i}$. Let
$r^{+}$ and $r^{-}$ be the positive and negative degrees respectively of the
branched cover $g_{1}\colon\thinspace g_{1}^{-1}(\Sigma_{1})\to\Sigma_{1}$. By
taking multiple copies of $T_{3,i}^{\prime}$ for all $i$ and multiple copies
of $T_{3}^{\prime}$ and $T_{1}^{\prime}$, and by orienting the copies of
$T_{3,i}^{\prime}$ and $T_{1}^{\prime}$ appropriately, we can find covers
$h_{3}\colon\thinspace T_{3}\to T_{3}^{\prime\prime}$,
$h_{3,i}\colon\thinspace T_{3,i}\to\Sigma_{3,i},$ and $h_{1}\colon\thinspace
T_{1}\to\Sigma_{1}$ such that, for some positive integer $N$,

  * •

the positive (resp. negative) degree of $h_{3,i}$ is $Ns_{i}^{+}$ (resp.
$Ns_{i}^{-}$),

  * •

$h_{3}$ has degree $N$,

  * •

$h_{1}$ has positive degree $Nr^{+}$ and negative degree $Nr^{-}$.

We now show that we can glue the above pieces together to get a taut
admissible surface $S_{2}$. Note that every component of
$\partial\Sigma_{3,i}$ is covered by $\partial T_{3}^{\prime\prime}$ with
positive and negative degrees $r^{+}-s_{i}^{+}$ and $r^{-}-s_{i}^{-}$
respectively. Thus, under $g_{1}\circ h_{3}$, every component of
$\partial\Sigma_{3,i}$ is covered by $\frac{N(r^{+}-s_{i}^{+})}{d}$ curves by
degree $d$ and $\frac{N(r^{-}-s_{i}^{-})}{d}$ curves by degree $-d$. Each
component of $\partial\Sigma_{3,i}$, under $h_{3,i}$, is covered by
$\frac{Ns_{i}^{+}}{d}$ curves by degree $d$ and $\frac{Ns_{i}^{-}}{d}$ curves
by degree $-d$. Each component of $\partial\Sigma_{3,i}$, under $h_{1}$, is
covered by $\frac{Nr^{+}}{d}$ curves by degree $-d$ and $\frac{Nr^{-}}{d}$
curves by degree $d$. Thus, by gluing $T_{1}$ to $T_{3}\cup\bigcup_{i}T_{3,i}$
along the preimage of $\partial\Sigma_{3}$ appropriately respecting
orientations, we obtain an oriented compact surface $S_{2}$ and a map
$g_{2}\colon\thinspace S_{2}\to\Sigma$, which, by construction, is a taut
admissible map.

It is clear that $n(S_{2})=Nn(S_{1})$, so it remains to compute the Euler
characteristic of various surfaces.

$\begin{array}[]{rcl}\chi(g_{2}^{-1}(\Sigma_{3}))&=&\chi(T_{3})+\sum_{i}\chi(T_{3,i})\\\
&=&N\chi(T_{3}^{\prime\prime})+\sum_{i}N(s_{i}^{+}+s_{i}^{-})\chi(\Sigma_{3,i})\\\
&=&N\chi(T_{3}^{\prime\prime})+\sum_{i}N\chi(g_{1}^{-1}(\Sigma_{3,i})-T_{3}^{\prime\prime})\\\
&=&N\chi(g_{1}^{-1}(\Sigma_{3}))\end{array}$
$\chi(g_{2}^{-1}(\Sigma_{1}))=N(r^{+}+r^{-})\chi(\Sigma_{1})=Nr\chi(\Sigma_{1})$

Finally, note that $\chi(g_{1}^{-1}(\Sigma_{1}))\leqslant r\chi(\Sigma_{1})$
since $g_{1}\colon\thinspace g_{1}^{-1}(\Sigma_{1})\to\Sigma_{1}$ is a
branched cover, and so

$\chi(S_{2})=\chi(g_{2}^{-1}(\Sigma_{3}))+\chi(g_{2}^{-1}(\Sigma_{1}))=N\chi(g_{1}^{-1}(\Sigma_{3}))+Nr\chi(\Sigma_{1})\geqslant
N\chi(S_{1}).$

∎

###### Proof of Proposition 3.3.

Start with an admissible surface $f\colon\thinspace S\to\Sigma$. We may assume
that $f$ is incompressible, since performing a compression reduces $-\chi(S)$
without changing $n(S)$ or the property of being admissible. Now apply Lemmas
3.5, 3.6, and 3.7. ∎

##  4\. Encoding surfaces

Let $\operatorname{TP}$ denote the set of all taut turn paths. For each taut
turn path $p\in\operatorname{TP}$, we define the sides of $p$. Recall that
$p\colon\thinspace S^{1}\to D$ is an immersed path which alternates between
turn arcs and immersed paths in $\partial D$. Since the intersection of
$\gamma$ with $\Sigma_{2}$ lies in $M$, the turn path $p$ alternates between
arcs from $\\{\tau_{i}\\}\cup\\{\alpha_{i}\\}\cup\\{\overline{\alpha}_{i}\\}$
and arcs in $\partial M-\Sigma_{1}$. A side of $p$ is an ordered pair
$s\in(\\{\tau_{i}\\}\cup\\{\alpha_{i}\\{\cup\\{\overline{\alpha}_{i}\\})^{2}$
such that $p$ traverses the first arc in the ordered pair, then a part of
$\partial M-\Sigma_{1}$, and then the second arc in the ordered pair. For any
side $s$, we let $m_{s}\subseteq\partial M$ denote the middle arc. See Figure
3. Note that since $p$ is taut, it traverses any side at most once.

Let $\mathcal{S}$ denote the set of all ordered pairs of arcs from
$\\{\tau_{i}\\}\cup\\{\alpha_{i}\\}\cup\\{\overline{\alpha}_{i}\\}$ which are
a side for some taut turn path $p$. Note that $\mathcal{S}$ does not contain
all ordered pairs. E.g., $\mathcal{S}$ cannot contain any pairs of the form
$(\alpha_{i},\overline{\alpha}_{j})$ since a taut turn path maps as an
immersion to $\partial D$. E.g., any pair $(\tau_{i},\tau_{j})\in\mathcal{S}$
must have one arc oriented into $M$ and one oriented out from $M$ on the same
side of $M$.

For any side $s\in\mathcal{S}$, there is a unique side $\hat{s}\in\mathcal{S}$
such that $m_{s},m_{\hat{s}}$ and two unique arcs from
$(\gamma\cup\partial\Sigma_{1})\cap M$ bound a subrectangle of $M$. Call
$\hat{s}$ the dual side of $s$, and let $M_{s}\subseteq M$ be the bounded
subrectangle. Note that, here, we are using the fact that $M$ contains no
self-intersections of $\gamma$. Also, note that $M_{s}=M_{\hat{s}}$. See
Figure 3 again.

Figure 3. Two examples of dual pairs of sides $(s,\hat{s})$. Geometrically,
each side encodes a portion of the boundary of a possible turn disk (shaded
gray in the figure) mapping to $D$.

### The encoding

Let $V={\mathbb{R}}^{\operatorname{TP}}\oplus{\mathbb{R}}^{2}$ with
coordinates $\\{x_{p}\\},\\{r^{+},r^{-}\\}$. We define a function

$v\colon\thinspace{\mathscr{T}}(\gamma)\to V$

as follows. Let $f\colon\thinspace S\to\Sigma$ be a taut admissible map. For
each $p\in\operatorname{TP}$, define $x_{p}(S)$ to be the number of components
$E$ of $f^{-1}(D)$ such that $f|_{\partial E}=p$. Let
$S_{1}=f^{-1}(\Sigma_{1})$. As mentioned above, since $f(\partial
S)\cap\Sigma_{1}=\emptyset$, the map $f|_{S_{1}}\colon\thinspace
S_{1}\to\Sigma_{1}$ is a covering map. Define $r^{+}(S)$ to be the positive
degree of $f|_{S_{1}}$ and $r^{-}(S)$ the negative degree. Finally let
$v(S)=((x_{p}(S)),r^{+}(S),r^{-}(S))$.

### Matching equations

The vector $v(S)$ will satisfy two kinds of equations. Broadly speaking, for
each turn path $p$ with a side $s$, there must be a corresponding turn path
$p^{\prime}$ (possibly equal to $p$) with a matching dual side $\hat{s}$.
Moreover, $r^{+}$ and $r^{-}$ should match with the number of $\alpha_{i}$’s
and $\overline{\alpha}_{i}$’s appearing in turn paths of $S$. We define some
linear maps. For each side $s\in\mathcal{S}$, we define $d_{s}\colon\thinspace
V\to{\mathbb{R}}$ as

$d_{s}((x_{p}),r^{+},r^{-})\ =\sum_{p\text{ has }s}x_{p}.$

For each $\alpha\in\\{\alpha_{i}\\}\cup\\{\overline{\alpha}_{i}\\}$, define
$d_{\alpha}\colon\thinspace V\to{\mathbb{R}}$ as

$d_{\alpha}((x_{p}),r^{+},r^{-})\ =\sum_{\begin{subarray}{c}\alpha\text{ is
a}\\\ \text{subarc of }p\end{subarray}}x_{p}.$

For each $\tau_{i}$, define $d_{\tau_{i}}\colon\thinspace V\to{\mathbb{R}}$ by

$d_{\tau_{i}}((x_{p}),r^{+},r^{-})\ =\sum_{\begin{subarray}{c}\tau_{i}\text{
is a}\\\ \text{subarc of }p\end{subarray}}x_{p}.$

######  Lemma 4.1.

Let $(f,S)$ be a taut admissible surface. Then, $((x_{p}),r^{+},r^{-})=v(S)$
satisfies the following matching equations:

  1. (1)

$d_{s}((x_{p}),r^{+},r^{-})=d_{\hat{s}}((x_{p}),r^{+},r^{-})$ for all
$s\in\mathcal{S}$.

  2. (2)

$d_{\alpha_{i}}((x_{p}),r^{+},r^{-})=r^{+}$ for all $i$.

  3. (3)

$d_{\overline{\alpha}_{i}}((x_{p}),r^{+},r^{-})=r^{-}$ for all $i$.

  4. (4)

$d_{\tau_{i}}((x_{p}),r^{+},r^{-})=d_{\tau_{j}}((x_{p}),r^{+},r^{-})$ for all
$i,j$.

###### Proof.

Let $M^{\prime}$ be a component of $M$ and let $N^{\prime}$ be a component of
$N=f^{-1}(M)$ mapping to $M^{\prime}$. By tautness, the map $f\colon\thinspace
N^{\prime}\to M^{\prime}$ is an immersion. Since $M^{\prime}$ is simply
connected, $N^{\prime}$ is compact, and $\partial S$ maps exclusively to
$\gamma$, the surface $N^{\prime}$ is a topological rectangle with two edges
mapping to $\partial D$ and two edges mapping to
$(\gamma\cup\partial\Sigma_{1})\cap M$. The map $f|_{N^{\prime}}$ may be
orientation reversing or preserving.

Let $E^{\prime},E^{\prime\prime}$ be the two components of $f^{-1}(D)$
bordering $N^{\prime}$. Then $N^{\prime}$ borders
$E^{\prime},E^{\prime\prime}$ of $E$ in the middle of sides in $f|_{\partial
E^{\prime}},f|_{\partial E^{\prime\prime}}$. Since $f$ is an immersion from
$f^{-1}(\Sigma_{2})$ to $\Sigma_{2}$ and from $\partial S$ to $\Sigma$ (as a
$1$-manifold), the map $f|_{\partial E^{\prime}}$ (resp. $f|_{\partial
E^{\prime\prime}}$) is determined near the intersection of $\partial
N^{\prime}$ and $\partial E^{\prime}$ (resp. $\partial N^{\prime}$ and
$\partial E^{\prime\prime}$). Thus, $f|_{N^{\prime}}$ determines precisely one
side each of the turn paths $f|_{\partial E^{\prime}},f|_{\partial
E^{\prime\prime}}$ and those sides must be dual. Conversely, the side of
$f|_{\partial E^{\prime}}$ at $\partial E^{\prime}\cap\partial N^{\prime}$
determines the image of $f(N^{\prime})$ and the side of $f|_{\partial
E^{\prime\prime}}$ at $\partial E^{\prime\prime}\cap\partial N^{\prime}$ which
must be dual. Since $f|_{\partial E}$ is taut for any component $E$ of
$f^{-1}(D)$, each arc $\tau_{i},\alpha_{j},\overline{\alpha}_{j}$ is traversed
at most once and thus each side appears in $f|_{\partial E}$ at most once.
Thus, $d_{s}(v(S))$ is the total number of times $s$ appears as a side in $f$,
and the above argument shows this is equal to the total number of times
$\hat{s}$ appears as a side in $f$ which is $d_{\hat{s}}(v(S))$.

For the second equality, first note that $r^{+}(S)$ equals the positive degree
of the map $f\colon\thinspace f^{-1}(\alpha_{i})\to\alpha_{i}$ where we view
$f^{-1}(\alpha_{i})$ as a subarc of the boundary of $f^{-1}(\Sigma_{3})$. This
is equal to the number of times that the boundary of $f^{-1}(D)$ traverses
$\alpha_{i}$, and since $f$ is taut, this is $d_{\alpha_{i}}(v(S))$. The
argument for the third equation is similar. Again by tautness,
$d_{\tau_{i}}(v(S))=n(S)$ for all $i$, and the last equality holds. ∎

### Degree and Euler characteristic

Recall from the preceding proof that $d_{\tau_{i}}(v(S))\ =\ n(S)$ for each
turn arc $\tau_{i}$ and all taut admissible surfaces $S$. Thus we have the
factorization

${\mathscr{T}}(\gamma)\overset{v}{\to}V\overset{d_{\tau_{i}}}{\to}{\mathbb{R}}$

of the degree function $n(S)$ through $v$ via any of the linear maps
$d_{\tau_{i}}$.

The Euler characteristic $\chi(S)$ also factors through $v$ in a similar
fashion. First we need a formula for $\chi(S)$ based on the combinatorics of
turn disks.

######  Definition 4.2.

For $p\in\operatorname{TP}$ let

$\kappa(p)\ =\ 1-\frac{1}{2}(\\#\text{ sides of }p).$

Let
$F_{\chi}\colon\thinspace{\mathbb{R}}^{\operatorname{TP}}\oplus{\mathbb{R}}^{2}\to{\mathbb{R}}$
be the linear map

$F_{\chi}((x_{p}),r^{+},r^{-})\ =\
\Bigl{(}\sum_{p\in\operatorname{TP}}\kappa(p)x_{p}\Bigr{)}\ +\
\chi(\Sigma_{1})(r^{+}+r^{-}).$

######  Lemma 4.3.

If $S$ is a taut admissible surface, then $\chi(S)=F_{\chi}(v(S))$.

###### Proof.

Given a taut admissible surface $f\colon\thinspace S\to\Sigma$, the subsurface
$S_{3}=f^{-1}(\Sigma_{2}-\operatorname{int}(\Sigma_{1}))$ consists of turn
disks mapping to $D$ and $1$-handles mapping by immersions to $M$ where each
$1$-handle attaches to the middle arcs of a dual pair of sides. It is clear
then that

$\chi(S)\ =\ \Bigl{(}\sum_{p\in\operatorname{TP}}\kappa(p)x_{p}(S)\Bigr{)}.$

Let $S_{1}=f^{-1}(\Sigma_{1})$. By definition of taut surface, $S_{1}$ maps to
$\Sigma_{1}$ by an immersion and since $S_{1}\cap\partial S=\emptyset$, the
map is a covering map. Therefore
$\chi(S_{1})=\chi(\Sigma_{1})(r^{+}(S)+r^{-}(S))$. Since $S=S_{1}\cup S_{3}$
and $S_{1},S_{3}$ meet along full boundary components of $S_{1}$ and $S_{3}$,
we have $\chi(S)=\chi(S_{1})+\chi(S_{3})$ and the lemma follows. ∎

######  Remark 4.4.

Having explained our encoding of surfaces, we can relate this encoding to that
used to determine stable commutator length in a free group. Our $1$–chain
$\gamma$ lies in the union of the $D_{i}$ and bands $M$. If we collapse each
$D_{i}$ to a point and the bands in $M$ to edges, we obtain a $1$–chain in a
graph. However, only the turn paths which traverse no $\alpha_{i}$ correspond
to turn circuits as defined in [BCF11], and of course, there is nothing
analogous to the role of $\Sigma_{1}$ and its matching equations with turn
paths.

##  5\. Reassembly

Given a taut admissible surface, we associate the vector $v(S)$ which encodes
it and satisfies matching equations. Now, we would like to show that we can
construct a taut admissible surface with the right Euler characteristic and
degree ($n(S)$) from a vector satisfying the matching equations. In general,
it’s not clear that this is possible for arbitrary positive integral vectors
satisfying the matching equations. However, it is possible after multiplying
the vector by a sufficiently large integer. In fact, we make the statement for
positive rational vectors.

######  Lemma 5.1.

Suppose a nonzero, nonnegative vector
$((x_{p}),r^{+},r^{-})\in{\mathbb{Q}}^{\operatorname{TP}}\oplus{\mathbb{Q}}^{2}$
satisfies the matching equations (1)-(4) in Lemma 4.1. Then, there is a taut
admissible surface $S$ such that

$\frac{\chi(S)}{n(S)}\ =\
\frac{F_{\chi}((x_{p}),r^{+},r^{-})}{d_{\tau_{1}}((x_{p}),r^{+},r^{-})}.$

###### Proof.

We can, without loss of generality, assume that the vector is integral. The
basic idea is as follows. We take $x_{p}$ turn disks for each $p$ whose
boundary will map to the turn path $p$. We then attach $1$-handles to dual
sides. This surface, call it say $S_{3}$, then has boundary components some of
which map to powers of components of $\gamma$ and some to components of
$\partial\Sigma_{3}$ with varying degrees. We then want to glue in a (probably
disconnected) surface to latter boundary components which will map to
$\Sigma_{1}$ by a covering map. The difficulty is that the part of $\partial
S_{3}$ mapping to $\partial\Sigma_{3}=\partial\Sigma_{1}$ may not extend to a
covering map of $\Sigma_{1}$ depending on the topology of $\Sigma_{1}$ and the
degrees of the maps from components of $\partial S_{3}$ to components of
$\partial\Sigma_{1}$. Lemmas 2.7 and 3.7 will fix this problem at the cost of
scaling $((x_{p}),r^{+},r^{-})$ by some positive integer.

For each turn path $p\in\operatorname{TP}$, we let $E_{p}$ be an oriented disk
and $f_{p}\colon\thinspace E_{p}\to D$ a map such that $f_{\partial E_{p}}=p$
(where $\partial E_{p}$ has its boundary orientation). If $p$ does not
traverse any turn arcs, then $p\colon\thinspace S^{1}\to\partial D$ is an
immersion; in this case, we insist $f_{p}$ is a homeomorphism. Note that if
$p$ traverses $\overline{\alpha}_{i}$’s, then $f_{p}$ is orientation-
reversing. For each side $s\in\mathcal{S}$, we let $N_{s}$ be an oriented disk
which is a copy of $M_{s}$ and $f_{s}\colon\thinspace N_{s}\to M_{s}$ the
canonical homeomorphism. We orient $N_{s}$ as follows. If $s$ has a turn
$\tau_{i}$, then we orient $N_{s}$ so that the boundary orientation is
consistent with the orientation of $\tau_{i}$. If $s$ has an $\alpha_{i}$,
then we orient $N_{s}$ such that $f_{s}$ is orientation-preserving and if $s$
has an $\overline{\alpha}_{i}$, then we orient $N_{s}$ such that $f_{s}$ is
orientation-reversing.

Now, we construct our initial surface. We take $x_{p}$ copies of $E_{p}$ and
$d_{s}((x_{p}),r^{+},r^{-})$ copies of $N_{t}$. (Recall that
$M_{s}=M_{\hat{s}}$ and thus $N_{s}=N_{\hat{s}}$. By the matching equations,
we are taking a well-defined number of copies of $N_{s}$.) We attach the
$E_{p}$ together using the $N_{s}$ in the obvious way such that $f_{p}$ and
$f_{s}$ extend to a map on the oriented surface constructed. Call this surface
$S_{3}$ and the map $f_{3}$. Some boundary components of $S_{3}$ map to
$\gamma$ and others to $\partial\Sigma_{3}=\partial\Sigma_{1}$. By the
matching equations, the positive degree of the map $\partial S_{3}$ to any
component of $\partial\Sigma_{3}$ is $r^{+}$ and the negative degree is
$r^{-}$. By applying Lemma 2.7 separately to the positive and negative
degrees, there is a branched cover $S_{1}\to\Sigma_{1}$ such that the degree
partition over any component of $\partial\Sigma_{3}$ is precisely the negative
of that of $S_{3}$. By gluing boundary components of $S_{3}$ to $S_{1}$ which
map to the same component of $\partial\Sigma_{3}$ with matching degrees (i.e.
equal but opposite), we obtain a surface $S^{\prime}$ and a map
$g^{\prime}\colon\thinspace S^{\prime}\to\Sigma$ satisfying the hypotheses of
Lemma 3.6.

By Lemma 3.7, there is a positive integer $N$ and a taut admissible surface
$g\colon\thinspace S\to\Sigma$ satisfying
$n(S)=Nn(S^{\prime})=Nd_{\tau_{1}}((x_{p}),r^{+},r^{-})$ and
$\chi(g^{-1}(\Sigma_{3}))=N\chi(S_{3})$ and
$\chi(g^{-1}(\Sigma_{1}))=N(r^{+}+r^{-})\chi(\Sigma_{1})$. We then have

$\chi(S)=\chi(g^{-1}(\Sigma_{3}))+\chi(g^{-1}(\Sigma_{1}))=N\chi(S_{3})+N(r^{+}+r^{-})\chi(\Sigma_{1})=NF_{\chi}((x_{p}),r^{+},r^{-}).$

The lemma follows. ∎

##  6\. Linear programming and scl

We are now ready to prove that the stable commutator length of the nonfilling
$1$-chain is the solution of a linear program. We define the region
$R\subseteq{\mathbb{R}}^{\operatorname{TP}}\oplus{\mathbb{R}}^{2}$ to be the
subset defined by the following linear equations and inequalities.

  * •

the matching equations from Lemma 4.1

  * •

$d_{\tau_{1}}((x_{p}),r^{+},r^{-})=1$

  * •

$x_{p}\geqslant 0$ for all $p\in\operatorname{TP}$ and $r^{+}\geqslant 0$ and
$r^{-}\geqslant 0$

######  Lemma 6.1.

The stable commutator length of $\gamma$ is equal to

$\inf\left\\{-\frac{1}{2}F_{\chi}((x_{p}),r^{+},r^{-})\mid((x_{p}),r^{+},r^{-})\in{\mathbb{Q}}^{\operatorname{TP}}\oplus{\mathbb{Q}}^{2}\cap
R\right\\}$ (6.2)

###### Proof.

Recall from Corollary 3.4 that

$\operatorname{scl}(\gamma)\ =\
\inf_{S\in{\mathscr{T}}(\gamma)}\frac{-\chi(S)}{2n(S)}.$

Let $S$ be a taut admissible surface. Then, by Lemma 4.1,
$v(S)\in{\mathbb{Z}}^{\operatorname{TP}}\oplus{\mathbb{Z}}^{2}$ satisfies the
matching equations. Moreover, $n(S)=d_{\tau_{1}}(v(S))$ and, by Lemma 4.3,
$\chi(S)=F_{\chi}(v(S))$. By linearity,
$d_{\tau_{1}}\left(\frac{v(S)}{n(S)}\right)=1$ and

$-\frac{\chi(S)}{2n(S)}=-\frac{1}{2}\frac{F_{\chi}(v(S))}{n(S)}=-\frac{1}{2}F_{\chi}\left(\frac{v(S)}{n(S)}\right).$

Thus,
$\operatorname{scl}(\gamma)\geqslant\textup{\maketag@@@{(\ignorespaces{\ref{linproginf}}\@@italiccorr)}}$.
On the other hand, by Lemma 5.1, for any vector $((x_{p}),r^{+},r^{-})$
satisfying the matching equations and $d_{\tau_{1}}((x_{p}),r^{+},r^{-})=1$,
there is a taut admissible surface $S$ such that

$-\frac{\chi(S)}{2n(S)}=-\frac{1}{2}\frac{F_{\chi}((x_{p}),r^{+},r^{-})}{d_{\tau_{1}}((x_{p}),r^{+},r^{-})}=-\frac{1}{2}F_{\chi}((x_{p}),r^{+},r^{-}).$

Thus
$\textup{\maketag@@@{(\ignorespaces{\ref{linproginf}}\@@italiccorr)}}\geqslant\operatorname{scl}(\gamma)$.
∎

We are now ready to prove the main theorem.

###### Proof of Theorem 1.1.

Since $v(S)\in R$ for any taut admissible surface $S$, the region $R$ is non-
empty. Since $R$ is defined by equations and inequalities with rational
coefficients, $R\cap{\mathbb{Q}}^{\operatorname{TP}}\oplus{\mathbb{Q}}^{2}$ is
dense in $R$. By Lemma 6.1 and the fact that stable commutator length is
nonnegative, $-\frac{1}{2}F_{\chi}$ is bounded below on $R$. Since $R$ is
finite-dimensional and defined by finitely many equations, minimizing the
linear function $-\frac{1}{2}F_{\chi}$ on $R$ is a linear program and
therefore has an optimal solution at a vertex of $R$. Since $R$ is defined by
equations with rational coefficients, all vertices are rational points, and so
by Lemma 6.1, $\operatorname{scl}(\gamma)$ is rational. ∎

######  Remark 6.3.

The proofs of Theorem 1.1 and Lemma 5.1 also establish that $\gamma$ has an
extremal surface.

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Claims 8 and 10 show that the instance $(G,k)$ of Multicolored Clique is
equivalent to the instance $(G^{\prime},k^{\prime})$ of Bounded-Width 1-Antler
Detection. This concludes the proof of Theorem B.1.

Observe that the proof of Theorem B.1 shows that the variant of Bounded-Width
1-Antler Detection where we ask for the existence of a 1-antler of width
_exactly_ $k$, is also W[1]-hard.