Existence and Uniqueness of Rectilinear Slit Maps
Author(s): Carl H. Fitzgerald and Frederick Weening
Source: Transactions of the American Mathematical Society, Vol. 352, No. 2, (Feb., 2000), pp.
485-513
Published by: American Mathematical Society
Stable URL: http://www.jstor.org/stable/118050
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TRANSACTIONS
OF THE
AMERICAN
MATHEMArICAL
SOCIETY
Volume 352, Nuimber 2, Pages 485-513
S 0002-9947(99)02538-6
Article electroniically published oni October
EXISTENCE
r,
1999
AND UNIQU1ENESS
OF RECTILINEAR
SLIT MAPS
CARL H. FITZGERALD AND FREDERICK WEENING
ABSTRACT.We consider a generalization of the parallel slit uniformization in
which the angle of inclination of each image slit is assigned independently.
Koebe proved that for domains of finite connectivity there is, up to a normalization, a unique rectilinear slit map achieving any given angle assignmnent.
Koebe's theorem is partially extended to domains of infinite connectivity. A
uniqueness result is shown for domains of countable connectivity and arbitrary
angle assignments, and ani existence result is proved for arbitrary domains under the assumption that the angle assigrnment is continuous and has finite
range. In order to prove the existence result a new extremal length tool, called
the crossing-module, is introduced. The crossing-module allows greater freedom in the family of admissible arcs than the classical module. Several results
known for the module are extended to the crossing-module. A generalization
of Jenkins' 0 module condition for the parallel slit problem is given for the
rectilinear slit problem in terms of the crossing-module and it is shown that
rectilinear slit maps satisfying this crossing-module condition exist.
1. INTRODUCTION
The theory of conformal uniformization attempts to generalize the Riemann
MappirngTheorem to multiply connected domains. The goal is to provide theorems
which allow the one-to-one, conformal mapping of a given domain onto a domain
with prescribed geometrical properties, and to determine conditions under which
such mappings are uniquely determined. A classical example is the parallel slit
uniformization. This uniformization is based on a theorem due to Gr6tzsch [Grl]
and de Possel [dP] which shows that an arbitrary domain can be mapped conformally onto a domain all of whlose complementary components are parallel slits
(i.e., straight line segments inclined at a fixed angle or points). Grbtzsch [Gr2]
showed that requiring a certain extremal length condition on the image produces a
unique normnalizedmnap.Jenkins [J] later reformulated this condition to obtain the
following theorem.
Theorem 1.1 (Grbtzsch and Jenkins). If Q is an arbitrary domain containing oo,
then for each angle 0 EC[0, -F) there exists a unique parallel slit rnap go(z) with
expansion
al
near cX
go (z)zz+-?r
z
Received by the editors July 3, 1995 and, in revised form, October 24, 1996 and February 19,
1999.
1991 Mathematics Subject Classification. Primary 30C35; Secondary 30C20, 31A15.
Key words and phrases. Conformal uniformization, slit maps, extremal length.
?
485
1999 American
Mathematical
Society
486
CARL H. FITZGERALD
AND FREDERICK
WEENING
that satisfies the following condition.
p)
(Po)
(For any sufficiently large square domain So, oriented symmetrically
and 0 + 7r/2, the module
) of the family of arcs in So n go(Q) which Join the sides of So at
inclination 0 + 7r/2 is equal to one.
J about the origin with sides at inclination 0
In particular this theorem shows that normalized parallel slit maps are uniquely
determined on domains of countable connectivity, but not in general on infinitely
connected domains without requiring a condition such as (Po) (For instance, if E
is a Cantor set of positive measure on [0,1] and Q is the domain whose complement
consists of the horizontal slits {x + iy: x C [0,1], y C E}, then both the identity
and go(z) are horizontal slit maps on Q.)
The rectilinear slit uniformization generalizes the parallel slit uniformization by
allowing the angle of inclination of each image slit to be assigned independently.
The main results of this paper can be summarized as follows.
* Normalized rectilinear slit maps are shown to be uniquely determined on
domains of countable connectivity (see Theorem 2.3).
* An example of a domain of countable connectivity and a (discontinuous) angle assignment for which no normalized rectilinear slit map achieving the
assignment exists is demonstrated (see the end of section 2).
* A new extremal length tool called the crossing-module is developed. The
crossing-module extends the definition of the classical module to allow arcs
to cross over specified boundary components of the domain. Useful results
regarding the cross-module of a domain whose outer boundary component is
a rectangle are obtained (see Theorems 3.3 and 3.6).
* The crossing-module is used to state an extremal length condition, called
(NP0-), which generalizes Jenkins' (Po) condition in the case of an angle
assignment which is continuous and has finite range (see Theorem 4.4).
* For an arbitrary domain it is shown that there exists a rectilinear slit map
achieving a given continuous angle assignment of finite range which satisfies
the (NP,0) condition (see Theorem 4.4).
In [W] it was conjectured that condition (NP0-) determined a unique rectilinear
slit map. This conjecture has recently been shown to be true by Maitani and
Minda [MM].
In the remainder of this section we give some precise definitions and provide some
of the history of the rectilinear slit uniformization and related uniformizations.
Definitions.
Let Q be a domain in the extended plane C.
(i) The space of complementary components of Q will be denoted by C(Q).
(ii) A mapping e: C(Q) -*> R is called an angle assignment on C(Q). Inasmuch
as Q and C(Q) determine each other, the phrase 0 is an angle assignment on
a domain Q will sometimes be used.
(iii) If f: Q -* Q' is a homeomorphism, then we shall use the notation fC to
denote the correspondence from C(Q) to C(Q') induced by f (z).
(iv) If () is an angle assignment on Q and if f: Q -* Q' is a conformal homeomorphism such that for each C e C(Q), fC(C) is a slit which lies on a line
of inclination ((C) to the positive real axis, then f (z) is said to achieve the
angle assignment 0 (see Figure 1).
EXISTENCE
AND UNIQUENESS
OF RECTILINEAR
SLIT MAPS
487
71~~~~~~~~~~~~~~~~~~~~-/
-n/6
f(z)
7r/2
/3/
3/
FIGURE
1. A rectilinear slit map achieving an angle assignmient.
(v) A function which achieves some angle assignment is called a rectilinear slit
map. The function is said to be normalized if cx belongs to -the domain and
the function has the expansion z + alz-1 + .. near oo.
For the case of domains of finite connectivity the rectilinear slit mapping problem
was completely solved in 1918 by Koebe [K]. He used the continuity method to
prove the following theoremn.
Theorem 1.2 (Koebe). If Q is a finitely connected domain containing ox and () is
any angle assignment on C(Q), then there exists a unniquerectilinear slit map f(z)
achieving E3which satisfies the normalization
near oo.
+
f (z)-+_
z
Various other uniformizations were found for domains of finite connectivity (see
[Go]), and ultimately very general theorems were proven. In particular, Harrington [Ha] showed that for any assignment of shapes to the boundary component of a
given domain there exists a conformal map on the domain such that the boundary
components in correspondence have the assigned shapes (allowing scaling but not
rotation). A new proof of this theorem and certain generalizations"have recently
been given by Schramm in [Scl].
Obtaining uniformtizationresults on domains of infinite connectivity has proven
to be much more difficult; for example, Koebe's circle uniformization problem while
recently solved for the case of countably connected domains (see [HS]) is still open
in general. Much attention has been given to slit mnappings,parallel and otherwise.
The original work in this area is due to Grotzsch circa 1930. In a series of articles
(see [J] for references) he proved various existence and uniqueness results using
extremal problem arguments in combination with his "method of strips" a precursor
to the modern notion of extremal length.
A number of Gr6tzsch's theorems and proofs have been refined over the years.
For instance Jenkins uses the modern formulation of extremal length to state the
condition (Po) of Theorem 1.1 which is not trivially equivalent to t;he original
condition given by Gr6tzsch. In the case of circular and radial slit uniformization, a simple proof of existence and uniqueness was given in 1960 by Reich and
Warschawski [RW]. It is worth noting that the slit map which they found as the
488
CARL H. FITZGERALD
AND FREDERICK
WEENING
unique solution to anl extremal problem also has the property that its image is a
minimal slit domain in a sense analogous to that of condition (Po) Later Marden
and Rodin [MR] obtained an extremxallength result on a general Riemnannsurface
which in particular implied the resuilts of Reich and Warschawski.and provided a
new characterization of uniqueness. Also notable in these uniformization problems
as the sois the approach of using principal functions to obtaini various slit mn-aps
lutions to certain extremal problems within the framework of Riemann surfaces
(see [AS], [NS], or [RS])
Attention to the rectilinear slit mapping problem returned in the mid 1970's wher
Rodin began its study on domains of infinite connectivity in the hopes of obtaining
insight into Koebe's circle uniformization problem. In 1977 Rodin [ABB] suggested
that a continuity assumption on the angle assignment might be sufficienltto imply
the existence of a rectilinear slit map achie-vingan otherwise arbitrary assignment
on any domain. Roughly speaking, an angle assignment 0 is continuous if for every
sequence {C,} of complementary components which converges to a complementary
component C, the sequence {e0(C,) } converges to 0 (C) modulo -F. This continuity
condition can be described more easily in terms of the natural topology on C(Q).
Here a subset 0 c C(Q) is open if and only if there exists an open set U in C such
that
Q - tC C C(Q): C
c
U}.
Definition. For any domain Q in C, an angle assignment e) on C(Q) is said to be
is continuous.
continuous provided that the map ei2 H: C(Q) -> fz: Iz1}
This work shows that Rodin's continuity assumption is sufficient to imply the
existence of a rectilinear slit mnapachieving any angle assignment with finite range.
The question of existence under the continuity assumption for the case of angle
assignments with infinite range is still open.
2.
UNIQUENESS OF RECTILINEAR SLIT MAPS
The proof of the uniqueness theorem is by the argument principle. The main
difficulty is in choosing an appropriate finite collection of curves homologous to
zero and a function on which to apply the argument principle. The method is
motivated by He and Schramm's proof of the uniqueness of the circle normalization
for domains of countable connectivity [HS]. In their proof, ho-wever,they extended
maps continuously to a larger domain and applied a generalized argument principle;
we need only restrict the maps to a suitable finitely connected subdomain and apply
the standard argument principle.
In the course of the proof we shall use the following approximation lemmas. A
proof of the first may be found in [W], while the second follows immediately from
the first. Here, and elsewhere, we use the notation In [-y]to denote the bounded
domain determined by a Jordan curve a' lying in the finite complex plarne;similarly
Out [ry]will denote the unbounded domain determined by .
Lemma 2.1e If Q is a domain in C containing oc and C C C(Q), then for any
c > 0 there exists a Jordan curve -y in Q such that C C In [<y]and
z c Q n TI [
#]
=>
dist(z, C) < c.
EXISTENCE
AND UNIQUENESS
OF RECTILINEAR
SLIT MAPS,
489
Lemma 2.2. If Q is a domain containing oc and C c C(Q), then for any open set
U in C containing C there exists a Jordan curve ry in Q such that
C C In [-y]C U
Theorem 2.3. If Q is a countably connected domain which contains oc and 0 is
any angle assignment on C(Q), then there exists at most one rectilinear slit mapping
f(z) achieving 0 which satisfies the normalization
f(z) = z --
near oc.
+..
z
Proof. Suppose to the contrary that there were two distinct normalized rectilinear
slit maps ~o(z) and g/(z) achieving 0. By considering p o b- L(z) we rmayreduce to
the case where Q itself is a domain with each component C C C(Q) lying on a line
of inclination #(C), and where there exists a normalized map f(z) # z on Q that
achieves 0. We shall show that this situation is impossible.
The difference f(z) - z is analytic in Q and has the convergent expansion
aj
+
near oc
for some j > 1 and aj 74 0. Thus the image of a large circle under f(z) - z has
winding number -j about the origin. This follows by Rouche's Theorem, since we
may choose R sufficiently large such that {z: lzl> R} C Q, and
(1)
Jf(z)-Z-
?
Z3
<
22 z-
on
jzl
- R.
In fact inequality (1) shows that as z describes the circle CR ={Z
lz= R} once
in the positive direction, arg[f(z) -+ a - z] changes by -2-rj for any- aI < aa- R-.
Let fa(z) = f(z) + a for any a c C. Since fa(z) is simply a translation of f(z),
each of the maps fa (z) achieve the angle assignment 0. For a fixed a we imagine
that the image of fa(z) is in the same plane as the domain Q. For any C C C(Q),
C and faC(C) are parallel slits. Thus by choosing a appropriately we can achieve
(2)
CnfaC(C)o
forallCCC(Q)
=
and lal < 1aj R-3. Indeed, since C(Q) is countable there exists 00 C [0, ir) not in
the range of 0 (mod ir). If a is taken to be of the form a = rei0", then for each
C e C(Q) there is (because Oo74 0(C) (mod ir)) at most one value of r, say rc,
such that C n faC(C) 74 0. Hence we may take a = roeioo where ro < aj1 Rand ro does not equal any, of the countably many, rc.
Having fixed a, we now claim that there exists a finite collection {'Yk}k-i of
Jordan curves in QR= Q n {Z: Iz < R} satisfying
II
(3)
(4)
In[-yk n In [-y,] -- 0
In [yk]n In [faQ(Yk)]
(5)
In
C C U[LYk]
foirk ?41,
0
for each k, and
for each C C (Q).
To see this observe first that since Q" is compact, if there exists a sequence {'Yk} I
of Jordan curves in QR satisfying properties (3), (4), and (5), then the sequence
may be replaced by a finite collection as required. Thus, since C(Q) is countable,
it suffices to show that if {-yk}tL1 possess properties (3) and (4), and C C C(Q) is
such that C t U' 1 In [K]k then there exists a Jordancurve -ym+1 in QR enclosing
C and such that {Yk}ki+l possess properties (3) and (4).
490
CARL H. FITZGERALD
AND FREDERICK
WEENING
Suppose that {-yk}IL' and C are as just described. By Lemma 2.1 we may choose
a Jordan curve C in Q such that C c In [or]and
1
z C Qn In [or] X
dist(z, C) < - dist(C,fa(C)).
2
Likewise there exists a Jordan curve ca'in fa (S) such that fC (C) c In [c'] and
w E fa(Q) n In [a']
dist(w, faC(C)) < 2 dist(C,
faC(C))
The curves axand ', thought of as lying in the same plane, satisfy In [c]rInl [u']
0;
while In [ax]n In [fa-j (')] 5] 0 as C belongs to the intersection (this follows since
fa(0o) = 00). Thus, the set
-
m
U = In [a] n In [f'
(Cu')]C {z: zl <R} C
Out [-yk]
k=1
is open and contains C. By Lemma 2.2 we may take
Q such that
Cc
In[yrn?l]
'Ym+l
to be a Jordan curve in
CU.
The choice of axand or', the fact that fa(??) = oo, and the definition of U imply
that {fy4}km1 satisfies (3) and (4). This shows our claim.
We are now ready to apply the argument principle to the cycle
n
F = CR ->Tk
(all curves positively oriented)
k=1
and the function fa(z)z. Properties (3) and (5) of Q-Yk}kli show that 7 is
homologous to zero. We have already seen that arg[fa(z) - z] changes by -27rj as
z describes CR. Consider now the change of argument of fa (z) - z as z describes a
curve 'Yk for some fixed k. By composing with a homotopy we may, by property (4) 7
deform the image of -Yk under fa(z) to be a point in Out ['Yk] without altering the
change in argument. It is therefore clear that the change in arg[fa (z) - z] is zero as
z describes -yk. Our conclusion is that the change of argument of fa (z) - z on the
cycle F is -21Tj < 0. This implies that fa(z) - z has a singularity in QR. However
we know that fa(z) -- z is analytic in all of Q. This contradiction completes the
proof.
C7
Remarks. 1. The assumption that the domain Q is of countable connectivity cannot be entirely removed from the theorem. Indeed it is known that even for the
parallel slit uniformization there is a lack of uniqueness in the case of uncountable
connectivity [J, p. 84].
2. The connectivity assumption was used in two different ways in the proof
of Theorem 2.3. First it was used in order to obtain a sufficiently small complex
number a such that (2) held. Then it was used to show the existence of a finite
collection {-yk}i=1 of Jordan curves satisfying properties (3), (4), and (5). This
latter argument can be replaced with an argument relying only on the compactness
of QC
3. In the case of a parallel assignment Jenkins' weaker condition (Po) concerning
the extremal length of the image, characterizes a unique parallel slit map (see
Theorem 1.1). We shall formulate a generalization of this condition called (NPo-)
(see Theorem 4.4) for continuous angle assignments with finite range. It has recently
EXISTENCE
AND UNIQUENESS
OF RECTILINEAR
SLIT MAPS
491
been shown by Maitani and Minda [MM]that (NP0) implies uniqueness in the case
of such a rectilinear assignment. Their proof uses the condition (NPeq) to show
that the rectilinear slit map has a certain boundary behavior and then they apply
a uniqueness result of Shiba [Sh]
4. Our proof using the argument principle could be extended to the case where
the connectivity assumption is replaced with the assumption of condition (NPe) if
one could show that (NPe) is sufficient to imply the existence of a sufficiently small
complex constant a such that (2) holds for a domain of arbitrary connectivity.
Theorem 2.3 can be used to demonstrate an example which shows that not all
angle assignments can be achieved by normalized maps. The example we provide
is of a discontinuous angle assignment on a domain of countable connectivity. This
shows that the existence portion of Koebe's theorem is sharp, and also motivates
using Rodin's continuity assumption for proving existence.
Example. For each nonzero integer n, let hn be the horizontal segment w:hose
imaginary part is 1/n and whose real part ranges between -1 and 1; and let ho
be the segment [-1,1] on the real axis. Let Q be the domain whose complement
consists of the line segments hn, and consider the angle assignment 0 on C(Q)
defined by
h
{T/2,
n
0.
Claim. No normalized mapping can achieve the assignment 0 on C(Q).
The proof of this claim uses the uniqueness theorem in two different ways: first to
conclude that the image of a supposed map achieving the assignment has a certain
symmetry, and second to rule out one possible form of the image. With regard to
the latter use, we remind the reader that a point component is a slit whose angle
of inclination may be taken to be any desired real number.
Proof of Claim. Suppose to the contrary that f(z) achieves 0 and is normalized
with expansion z+alz-1 +. - near o. The uniqueness theorem and the symmetry
of Q imply that f(z) - -f(-z).
It follows that for each n :4 0, fC(hn) is a
horizontal line segment syrnmetric to the imaginary axis. Further fC (ho) is either
(i) the point {0}, or (ii) a vertical line segment {w: Rew = 0, 1ImwI < p} for some
p > 0. The first case is impossible since then both the maps f(z) and the identity
would achieve the assignment 6 -0, contradicting Theorem 2.3. In the second
case, by considering boundary correspondence, we deduce that if w ---e ( E fC (ho)
through {w : Rew < 0} then f-1(w) -> -1. By the reflection principle this implies
that f1'(w) is a constant, also impossible. Since both cases lead to contradictions,
LZ
we conclude that no such f(z) exists.
3. A
GENERALIZED
MODULE
TECHNIQUE
In this section we describe a variation of the classical module which generalizes the standard notion to allow greater freedom in the family of arcs. A new
module-like quantity, called the crossing-module, will be defined in Section 3.1.
The crossing-module shall not be a conformal invariant; however useful estimates
for the variation of quantities related to the crossing-module under a conformal
map will be obtained in Section 3.2. Explicit calculation of the crossing-module for
492
CARL H. FITZGERALD
_
AND FREDERICK
el
71 >,
FIGURE
WEENING
2. A chain of arcs in a domain.
a certain type of arc family will be given in Section 3.3. The crossing-module is the
key tool used in Section 4 to prove the existence of rectilinear slit maps.
Recently Schramm [Sc2] has introduced a different (but similar) generalization
which he calls the transboundary extremal length. His notion is a conformal invariant, and he has developed a general theory in which the transboundary extremal
length is used to prove many types of uniformizations. His theory is suited toward
proving the existence of uniformizations in which the complementary components
of the image are all "fat" sets. It does not apply to our mapping problem because
a slit is the antithesis of a fat set.
3.1. The crossing-module.
The crossing-module is motivated as an attempt to
develop a tool to show that some of the complementary components of a domain
are slits lying at the same inclination, while other components are known not to
be slits at this inclination. Essentially one would like to ignore the fact that these
other components are in the complement and allow arcs to pass through them. The
behavior of such arcs in the interior of the complement is of no interest, it is only
important to keep track of where the boundary is crossed.
We shall define the crossing-module for domains in the extended complex plane.
A module-like quantity will be defined by modifying the standard notions of plength and p-area. The modification of the length will be to add a term representing
the Euclidean distance "jumped" for each boundary crossing. While the area will
be modified by adding the Lebesgue area of all the complementary regions which
may be crossed.
Definitions. Let Q be a domain in C.
(i) A rectifiable arc 'y: [0,1] -+ C is said to join boundary components of Q if the
interior of -y is contained in Q and the endpoints of -y belong to 9Q.
(ii) A chain in Q is a finite sum of =
arcs ?j each joining boundary
-iof
components of Q, and such that the terminal point of yj and the initial point
of ?j+1 belong to the same boundary component of Q for j = 1, 2,.. .n - 1
(see Figure 2).
(iii) The crossing-number of a chain o = En 1>j is defined to be the number n
of arcs comprising a and will be denoted by c#(a).
(iv) The initial and terminalpoints of a = En11'y3are definedrespectivelyas the
initial and terminal points of yi and Ayn.
_
EXISTENCE
AND UJNIQUENESS OF RECTILINEAR
SLIT MAPS
493
(v) The space of boundary components of Q will be denoted by B3(Q).
(vi) A fixed subset B of 13(Q) is called a set of crossing-boundaries provided that
the Lebesgue area of the complementary components of Q corresponding to
the elements of B is finite. This area will be denoted by Ab(B).
If
1Bis a set of crossing-boundaries, then the set of all chains in Q which, aside
(vii)
for initial and terminal points, are contained in Q U (U B) will be denoted by
CQ(B3).
(viii) A non-negative func-tionp(z) defined on Q is called a metric on Q provided
that p(z) is lower semi-continuous.
(ix) If p(z) is a metric on Q, and F is a family of chains contained in CQ(B), then
the p-length of a chain a = i
yj E F is defined by
n-1
'rb
Ip(c)
=
J
p(z)ldzl +
E
|I
(O)-'Yj (l)
(x) The p-lengthof a family F is definedby
Lp(F) =inf lp(a)
(xi) The p-area of Q with respectto B is definedby
Ap(Q, B)
p(x,y)2 dx dy +
J
Ab(B)
Q
B.
for any set of crossing-boundaries
of Q, then the crossing-module
(xii) If B is a set of crossing-boundaries
of a family
of chains F in C? (B) is defined by the expression
,u~~(F)
(17) =inpf
[42,
if)
AP (Q)B
where the infimum is taken over all metrics p(z) on Q, and indeterminate
ratios are defined to be equal to oo. When clear from context; we shall use
the notation ,u(F).
(xiii) Sets Q, B, and F for which a crossing-module is defined shall be referred to
collectively as a crossing-module problem; specifically as the crossing-module
problem (Q, 1, F).
Remarks. 1. If 1B 0, then the definition of the crossing-module coincides with
that of the module.
2. The crossing-module will not be a conformal invariant; we will, however,
obtain a useful estimate (see Theorem 3.1) for the change in the area to lengthsquared ratio for a fixed metric under the action of a conformal map.
3. It will also be the case that the area to length-squared ratio will not be
homogeneous in the metrics. This is not viewed as a deficiency, but rather as a
normalization of the problem (see Theorem 3.3).
3.2. The conformal variance of the crossing-module.
Let (Q, B, F) be a crossing-module problem. A conformal homeomorphism f (z) on Q which extends continuously to the crossing-boundaries B E 1Binduces a natural crossing-module problem
CARL, H. FITZGERALD
494
AND FREDERICK
WEENING
(Qf Sf, rFf) where
Qf - f (Q),
3f = f{f(B): B c B3}, and
n
Tt
pf J{5f~=fQyj):
:(,yjE
},
L
ji='
j=l
The crossing-modules 1u(Ff) and t(F) are clearly equal when.f (z) is a translation
or rotation, bnt in general they need not be equal. The following theorem provides
an estimate for the difference
APf(Qf ,_)
L f(pf)2
AP(Q, B)
L
(r) 2
where p(z) is any metric on Q and pf(w) is the induced metric on f(Q) defined by
)
(6)
pf(w) - p(f -- (w) ) f' (f ' M(w)
Theorem 341I. Let (Q, B, F) be a crossing-moduleproblem, and f (z) be a conformal
homeomorphism on Q which extends continuously to UB. Assume that there exists
an integer N such that each a CiF has crossing-number at most N. If there exists
e > 0 such that for each B c3B
Z1,Z2 C B
(7)
If(Zl)
- f(z2)1
- IZ1-xZ211 < ?,
then for each metric p(z) on Q,
eN .
|Lpf(Ff) -- Lp(F) I
(8)
If additionally there exists e' > 0 such that
(9)
- Ab(B) I<<'
|Ab(Bf)
then for each metric p(z) on Q such that Lp(I) > eN,
(10)
_
Apf(Qf,B1)
Lpf(rf )2
AP(Q,B) <
'-Lp(r)2
1'-
e)
1
(Lp (F)
-N)2
+
_
Ap Q
+
2
LL
(-)
Proof. Assume that (7) holds for some e > 0 and each B c3B. Let p be an arbitrary
metric on Q. For a given u
I cjiC F, denote the initial and terminal points
of each -yj by aj and bj respectively. Since ff( ) pf(w) Idwl -j
p(z) ldzl for each
j, it follows from the definition of p-length that
n--i
(11)
19f(af)
-
l-p()
E
( If(aj+I) - f (b3) I -
laj+j
-
b3
)
jlI
Thus
n--
Lpf(f)
< lpfy(of) < Ip(a) +
E
( If(aj+i) -- f (bj)
j=l
and the assumption (7) yields that
Lp1(P)
< lp(oT) +- eN .
-
aj+1 - bj )
AND UNIQUENESS
EXISTENCE
OF RECTILINEARt
495
SLIT MAPS
Since this is true for any a E F, it follows that
Lpf(Ff)
-
Lp(F) <
(N
Similarly, starting from equation (11), one can show that
Lp(F)
-
Lpf(Ff) < cN
Hence (8) holds.
Now assume additionally that (9) holds for some 6' > 0. To simplify calculations
define
A
Jpf(u,
--
v)2
du dv
JJp(x, y)2 dx dy .
f (Q)
The definition of p-area and elementary algebra implies that
Apf(Qf, F'f)Ap
_
(Q, B3)
Lp(r)2
Lpf(rf )2
A + Ab(3f)
A + Ab(3)
Lpf(f f)2
Lp(-)2
L=f(Ff )2 [A + Ab(,f)
f
L_
(1 )_L
L,f(Ff)2
[Ab(1f)
-
-
(A + Ab(3)) +
1-
Lf
Ff
))(2 (A + Ab (3))]
r
Ab(1)
F
(Lp(F) - tLpf(Ff)) (Lp(r) + Lpf(Ff))
? Lp((r))2
1
AP(Q,23)]
Inequality (10) follows from this equation, under the assumption that Lp(F)-N
0, by taking absolute values and using inequalities (8) and (9).
>
LII
Remarks. 1. The important feature of estimate) (10) to note is that the right
hand side depends only on the original crossing-module problem and on e and C'.
Moreover it tends to zero as c and (E'tend to zero.
2. Theorem 3.1 can be used to obtain a lower bound for p(F) even when there
is no bound to the crossing-numbers of chains in F. One considers the subfamily
fN _{
E F C#(u) < N}; since Fr C F it follows that p(F) > II(pN).
The most natural setting in which to apply Theorem 3.1 is when the conformal
map f(z) is close to the identity on the crossing-boundaries. This indeed will be
the case in our application in Section 4. In this case, the choice of c in (7) is clear.
The choice of C'in (9) is less obvious. We now state a lemma which estimates how
small 6' may be taken in (9) under the assumption that there are only finite many
crossing-boundaries each of which is a rectifiable Jordan curve.
Lemma 3.2. Let Q be a domain and suppose that 23 C I3(Q) is a finite collection
{BI, B2, ... ,B13 of rectifiable Jordan curves. Suppose that f(z) is a conformal
homeomorphism on Q which extends continuously to U 3. If If(z) - zI < on U B
and 6 < Lk= arclength(Bk) for each k = 1, 2,... , n, then
(f
IAbS
where L
-
Z=1
Lk
)-
AbS(B)I < 81L
6
CARIL l1. FITZGER11ALD AND FREDERICK
496
WEENING
Proof. It suffices to assume that n = 1. Let in(t) parameterize B1 by arc length
on [0, L] and extend io(t) periodically. Consider, for each rn = 1, 2,. .., rL/l1 the
disk A,, of radius 26 and center to(rnm). Since any point within e of Bt is within
26 of one of these centers, f (Bt) is contained in the union of these disks. Hence it
follows that
Ab(lf
area(Ak)
Ab (B)l <?
<
(-+
of a punctured
3.3. The crossing-module
I)
< 8wrL 6 .
r(2c)2 = 4(L + >)wFE
rectangular
domain. Recall that
the module of a rectangle (i.e., of the arcs joining a pair of opposite sides of a rectangular domain) is equal to the ratio of the length of the sides not being joined to
the length of the joined sides. Further recall that this result still holds if the rectangular domain is replaced by a rectangular domain minus a collection of parallel
slits, whose projection onto a line parallel to the sides joined has linear measure
zero. These results are generalized by allowing the rectangular domain to have
complementary components over which arcs are allowed to cross. Theorem 3.3 will
show that the resulting crossing-module is equal to the appropriate ratio of the side
1.
lengths of the rectangle and that the only extremal metric is p(z)
in
which the decase
of
the
At least as important to our application is an analysis
and
a
is
not
crossing-boundary
scribed domain has some boundary component that
constructed
is
metric
An
explicit
which is also not a slit in the preferred direction.
to show that the crossing-module in this case is strictly less than the relevant ratio
of the side lengths of the rectangle, see Theorem 3.6.
We begin by making some definitions. Since the crossing-module is invariant
under translations and rotations, we may restrict our consideration to rectangular
domains whose boundary sides are parallel and symmetric to the axes.
Definitions.
Let
Ra,b =
{z:
IRe zl < a/2, IIm zl < b/2}
for a, b > O.
(i) A domain Q is said to be a punctured rectangular domain if there exists a
domain D such that
Q=
Ra,b
n D
and
Dc C Ra.b
for some a, b > 0; see Figure 3. The numbers a, b are called respectively the
length and height of Q.
(ii) For a fixed a and b, let Q be a punctured rectangular domain of length a
and height b and suppose that (Q, B, F) is a crossing-module problem. If the
initial and terminal points of every CT E F lie on opposite vertical sides of
&Ra,b) then F is said to join the vertical sides of Q.
(iii) In the same situation, the crossing-module problem is said to be horizontally
sectionable if for almost every y C (-b/2, b/2) there exists a chain (T, E F
joining the vertical sides of Q, whose image is Q n {z Imz = y}. Such a
chain at is called a horizontal chain.
Note that if Q is a punctured rectangular domain of length a and height b that
is horizontally sectionable, then the linear measure of the projection of &Q' (U B U
&Ja.b) onto the imaginary axis is zero.
EXISTENCE
AND UNIQUENESS
OF RECTILINEAR
SLIT MAPS
497
V
4L.
FIUR
3. A pucue
Theorem E 3. LeA
~ ~~ ~ ~ ~ ~ ~ ~ ~ ~ ~
retnua
doai
of legt
a,;
4
and heightb
bpunctured rectangular domain of length a and height b.
Suppose that (Q~,
B, F) is a crossing-module problem with F joining the vertical sides
of Q. If (Q, B, F) is horizontally sectionable, then
H(F) =b/a.
Fusrtherp* (z)-1
is the only extremal metric provided that B
#&0.
Proof. First note that A_*(Q, B) < ab and Lp (F) > a. Thus kt(F) < b/a, and p*(z)
will be shown to be extremal if the reverse inequality is shown. Let CB denote the
complementary component corresponding to a boulndarycomponent B E B(2). Let
p(z) be any metric on Q~,and define p(z) on Q= U (U{CB B E B}) by
{1Xi
Z E (U{CB: B e
B}).
Consider the family r consisting of all arcs oyin Q joining the vertical sides of Q
such that zyfl Q is the imageof some chaina E F. In other words,each chainaf E F
is extended in all possible ways into the complementary components corresponding
to the boundaries it crosses.
The definition of the metric p(z), and of area and length in a crossing-module
problemimply that Ap(Z, B) = Ap() and that Lp(F) < Lp(F) Hence
Since (Q, B3,F) is horizontally sectionable, the horizontal segment from -a/2 ? iy
to a/2 ? iy belongs to r for almost every y E (-b/2, b/2). Thus it follows from the
classical theorem on the module that
(13)
AP( Z) > b/a.
CARL 1i. FITZGERALD
498
AND FREDERICK
WEENIN-G
By combining (12) and (13), and taking the infirnumover all metrics p(z) C P one
obtains that b(F) > b/a as desired.
Since it is known that the only extremal metrics in the module problem are
the constants, it follows from the above extension procedure that no metric except
lI
p*(z) can be extremal.
If, with the notation of Theorem 3.3, some B* C 3(Q) x (3 U {&Ra,b}) is not a
horizontal slit, then one might expect that the crossing-module would be less than
b/a. The intuitive reason for this is that the boundary component B* would serve
as a barrier, impeding chains from joining the vertical sides of Q. We shall show
in Theorem 3.6 that this intuition is correct, under the additional assumption that
the distance from B* to B is strictly positive. This theorem will then provide a
test to see if a particular group of boundary components in an arbitrary domain
consists of parallel slits.
A boundary component B* as just described can be characterized as a connected
set which contains two points whose imaginary parts are different. The following
lemma allows this global characterization to be transformed into a local property.
Roughly speaking, the local property obtained is that there exists a square region S
with sides parallel to the axes and of arbitrary small length, such that within S the
boundary component B* separates a point on the left side of S from all points on
the right side of S. Additionally, this geometric property is then used to produce
an inequality involving a specific metric suitable for a crossinlg-moduleproblem.
By way of notation, if S is the square region
S
b<Imz<b+s},
{z:a<Rez<a+-s,
then the left and right sides of S are defined by
(&S)i
{z: Rez =
a} nOS
and
(OS)r
-{z:
Rez-a
+ s} n OS.
Lemma 3.4. Let D :&-C be a simply connected domain containing oo such that
AD is not a horizontal slit. Then for any so > 0,
(14)
(there exists a closed square region S with sides parallel to the axes and
length less than so, such that there exists a point z1 C (OS), n 1,
and such that any rectifiable arc from z1 to any point Z2 E (&S)r n D
which is contained in S necessarily intersects OD.
J of
Moreover, for any fixed so > 0 and square region S as described in (14), there exists
a disk A* C S such that, if W1, W2 C OS and ay is a rectifiable arc in D n S joining
w1 to W2 then
(15)
Jp*(z)
Idz > I Rew, -- Rew21
where
(16)
P* (z)
{ob
z
E
-D
n,, A*.
So as not to interrupt the proof of Lemma 3.4 we first state the following simple
result, whose proof is left to the reader, regarding the replacement of a rectifiable
curve with a Jordan curve.
AND UNIQUENESS
EXISTENCE
OF RECTILINEAR
SLIT MAP'S
499
Lemma 3.5. Suppose that D is a domain, K is a convex region, and y is a rectifiable arc in D 0 K joining points z1 to Z2. Then there exists a Jordan arc in D n K
jointng z1 to Z2-
Proof of Lemma 3.4. The first assertion will be proven by contradiction. Suppose
that there exists so > 0 such that
for every closed square region S with sides parallel to the axes and
fof length less than so and every z1 C (aS), n D, there exists a point
{Z2 C (OS), n D and a rectifiable arc -y from z1 to Z2 with -y C S
and -yrnhD 0.
(14')
Choose points (1, (2 E OD with Im((4 - (2) > 0. We construcl, an infinite row
of square regions of side length s min{so,Im((4 - ?2)}/4 which will separate (I
from ?2 Define b = Im((1 + (2)/2 -s/2, and let
Sk
{z: ks <Rez
< (k+
)s, b < Imz < b+s}
for all integers k.
Since D contains oo, there exists an integer jo such that OD C {z: lzl < jo}.
Hence we may choose a point w_jo C (0s-jo)l n D. By repeatedly applying the
supposition (14') for k =-jo, -jo + 1, .. . , jo - 1, we find points Wk e (OSk) n D(9Sk+I)l n D and rectifiable arcs -k in Sk joining Wk to Wk+l. It follows that
-k iS
Ek=-io
a rectifiablearcin D
-jo Sk)
=(U
By Lemma3.5
joiningw_ jotowj
we may replace this arc with a Jordan arc F, having the same properties. The points
w-jo andwjo can clearly be connected by a Jordan arc F2 in D - {z: IzI < jo}. But
then F
=
Fl
U F2 is a Jordan curve separating the points (I and
(2;
this contradicts
the fact that OD is connected. Therefore the first assertion of the lemma is shown.
To show the second claim, let so be given and assume that S = {z: a < Re z <
a +s, b < Im z < b+ s} is as described in (14). Choose a point z1 E (OS)1n D such
that for any point Z2 C (aS) n D, there does not exist a rectifiable arc in D n s
joining z1 to Z2.
We now construct the disk A*. Choose r > 0 such that
A1 = {z:
z-
zil < r} c D,
and choose a point z* C A1 n S. Choose r* > 0 so small that
*
2r* <
= {z :z
min {[(x
xE[a,a?s]
-- zfl < r*} C A1 nS
- Rez*)2
+ dist(z*,S)2]L/2
and
_
x-Rez*}
J
Define p*(z) as in (16).
Let w1, w2 C OS and suppose that -y is a rectifiable arc in D n S joining w1 to
w2. It may be assumed that -yn A* 7? 0. By construction, any point in A* can
be joined to z, by a rectifiable arc in D n s, thus it follows that neither w1 nor w2
can belong to (&S)r. Inequality (15) is trivial if both w1 and w2 belong to (OS)1,
CAIRL H. FITZGERALD
500
AND FREDERICK
WEENING
hence assume that w2 X (&S)1. In this case we have
Jp*(z)
>
ldzl ?w1-z
--
*-r
+ Jw2 z*I -- r*
I
Re Z*)2 + (Im w2
+ [(Re w2
- Z*|
>
Re wI-Re
>
Rewi-Rez*
>
Rewi-Rew21
z* + [(Re W2-Re
+
Rew2
-
Imz*)21l/2
Z*)2 + dist(z*,&S)2]1/2
-
2*
- 2r *
Rez*
which shows that (15) holds and finiishes the proof of the lemma.
D:
Theorem 3.6. Let Q be a punctured rectangular domain of length a and height b,
and suppose that (Q, B, r) is a crossing-module problem with F joining the vertical
stides of Q. If there exists B* CE 3(Q) K (3 U {&Rab}) that is not a horizontal slit
(* C B*, ( C (U B)} > 0, then
and such that dist(B*,13) = inf{ (*p(F)
Proof. Suppose that B* C B(Q)
x
< b/a.
(X3U
{0Ra,b}) is not a horizontal slit and that so
1 dist(B,
13) > 0. Apply Lemma 3.4 with the domain D equal to the unbounded
component of B*C, to obtain a square region S satisfying (14) and a disk A* c S
such that (15) holds where p*(z) is defined by (16). Since Q c D, p*(z) restricts
to a metric on Q.
In order for (14) to hold it must be the case that B* n S 7 , thus
(~C S
??
dist((,,B*)
<
2so < dist(B
,B)
This shows that S does not contain any points on any crossing-boundaries,
thus that
and
Ap*(Q, 13)< ab - area(A*)
We shall use inequality (15) of Lemma 3.4 to obtain a lower bound for LP*(F).
Let o = E= IYj C F be given. Since each -yj is rectifiable we may break -yj into
finitely many subarcs which alternately avoid A* and enter A*. For each j, choose
m3 ? 0 and collections {j,k}jIo
and {/j,k}ijlm
of arcs such that
mi
-Yj =aj()
+
(/3j,k + aj,k)
k=1
where for each k:
?2>
(i) aj,k is an arc in Q
(ii) /3j,k contains a point of A*,
(iii) the endpoints of /j,k belong to AS, and
(iv) the interior of /3j,k is contained in S.
EXISTENCE
AND UNIQUENESS
OF RECTILINEAR
SLIT MAPS
501
Let aj,k and bj,k denote respectively the initial and terminal points of each cj,k,
and note that bjk-l and ajk are the initial and terminal points of ,j,k. By inequality (15) and definitions, we have that
n
n-i
lp* (or) =E
dzl + E
Jp*(Z)
j
aj+i,o
,I,Yj
n
bj,mj
min-1,m
! >z[J
+
Idz+
Idzl + (p*(z)
k--l Ij
k,
O,
j=l
O
n
> E
-
jlI
laj+i,o--bj,TbI
ldz)+ZE
j=l
k
rnf
[
Rebj,o
-
Reaj,ol +
j--l
>3(I Reaj,k
- Reebj,k
I + IRe bj,k - Rea,kl)]
kl1
n-I
+3I
Reaj+i,o-Rebj,mj2.
j=1
Hence, by repeated use of the triangle inequality, we find that ip*(a) > a. Since
a C F was arbitrary we have that Lp<(F) > a. Therefore,
AP*(Q, B)
ab - area(AZ*)
Lp*(]F)2
a2
_
b
area(A*)
b
a
a2
a
which shows u(F) < b/a and completes the proof.
1
4. EXISTENCE OF RECTILINEAR SLIT MAPS
In this section the existence of rectilinear slit maps achieving continuous angle
assignments with finite ranges will be shown for arbitrary domains. Additionally it
will be shown that the image of the rectilinear slit maps satisfy a certain crossingmodule condition called (NP9) (see Theorem 4.4) which generalizes Jenkins' (Po)
condition. The combination of Theorem 4.4 with the uniqueness result of Maitani
and Minda [MM]provides a complete generalization of Grotzsch and Jenkins' Theorem (Theorem 1.1) for the case of rectilinear slit maps achieving continuous angle
assignments with finite ranges.
In order to facilitate the statement and proof of Theorem 4.4 we introduce some
notation and give some approximation lemmas.
Definition. If Q is a domain containing oo and -yis a Jordan curve in Q then the
open subset {C C C(Q): C C In [y]} of C(Q) shall be denoted by K(Qy).
Proofs of the following lemmas may be found in [W].
Lemma 4.1. Let Q be a domain containing oo. If {Ck}Vn_,is a collection of disjoint
closed subsets of C(Q) which partition C(Q), then there exists disjoint Jordan curves
1, 7Y2,* -Yn inQ such that Ck = JJ(k) for k-1, 2,..,n,
Lemma 4.2. Let Q be an n-connected domain containing oo with boundary comnponents BI B2, ... , Bn. If e > O, then
(i) There exists disjoint piecewise linear Jordan curves Ji, J2,... , Jn in Q such
that Bfor 1, 2i ...c n
a
n,
Bk C In [Jl]
if and onlxy if
k = l, and
CARL ff. FITZGERALD
502
ANI) FREDERICK
WEENING
(ii) Moreover, if B1, B2,,
B,, are known to be rectifiable Jordan curves, then
J1, J2, *. , Jn can be chosen to additionally satisfy
n
area(In [Jk] 'N, In [Bkl) < '
for any e' > 0.
Lemma 4.3. Suppose that Q is a bounded domain such that the unbounded complementary component, CO, is an isolated component. Then there exists a piecewise
linear Jordan curve J in Q which encloses every complementary component of Q
except for C0.
The condition (NP(0) involves a collection of crossing-module
problems. To de-
scribe such a collectioni, we first show how a conitinuous angle assignment with finite
range can induce a set of crossing-boundaries on a domain.
Consider a domain D containing HO, and suppose that 1 is a continuous angle
1, 2,...
n,
,... 7, ,n}. For each j
assignment on C(D) which has range {$i,
(17)
is a closed su'bset of C(D), and the collection {Cj}>8=1 forms a partition of C(D).
Hence Lemmas 4.1 and 4.2 imply the existence of disjoinit, rectifiable Jordan curves
B1, B2,... , Brt in D satisfying
C
(18)
forj
A((B1)
1, 2,...,
n.
If D is a domain containing oc and '1 is a continuous angle assignment
. ,Bn I} of
{B1,B2,.
{qS1l,q2....
IqnSI} then, any collection B3
disjoint rectifiable Jordan curves in D satisfying (18) where Cj is defined by (17) is
said to be a '1-induced crossing-boundary collection in D.
Definition.
on D with range
Theorem 4.4. Suppose that Q is a domain containing oo, and that 3 is a conThen there exists a map
tinuous angle assignment of finite range {fO0,0
I, O,}.
I2,...
f (z) achieving 0 with the expansion
nearoo
f (z) = z + a--t+
z
that satisfies the following condition which will be referred to as (NP).
crossing-boundary collection
Bn} is any 00 (f-)C-induced
If 3
{Bl,7 B2,...
in f (Q), and s is sufficiently large so that
(f (Q)cU (J
(19)
r))C Ss
for
b
all q$E
{O1, 02, .
n}
where S,,4
{ei'z:z
Rezl < s, Imzl < s}, then for each j
crossing-Tnodule problemn (Dj, l3;, Fj) defined by
(f (Q) n Ssj)j
rD}
(20)
1 2.
In, the
Uk#a In [Bkl,
L
B
j
{Bj}, and
LFj = {o- C CDj (L3j) uf joins the sides of SS,o at inclination Oj + 7r/2},
has crossing-module
-- 1 .
A((Se)
(See Figure
4
for an iilus'tration of SUch a crossing-module problem.)
EXISTENCE
AND UNIQUENESS
OF RECTILINEAR
SLIT MAPS
503
/
B3
/-AX
FIGURE
4. Illustration of a crossing-module problem (D2,
12,
172).
Proof. It suffices to assume that n > 2, for in the case of t- 1 (i.e., a parallel
assignment) the crossing-module may be replaced with the module and then the
statement of the theorem coincides with part of an established theorem (Theorem 1.1) proven in [J, pp. 81-84] (see specifically Lemma 5.8 on p. 82).
We first construct a candidate map f (z). Since 0 is a continuous angle assignment with finite range, Lemmas 4.1 and 4.2 imply the existence of disjoint Jordan
curves JI,J2,...
Since
,J, in Q such that &'1(0k) = JV(Jk)for k - 11,2,...,n.
C:\ Ut1 In [Jk] is a compact subset of Q, we may choose an exhaustion {Qm} such
that
n
C
(21)
U In [Jk]C Qrm
k=1
for all m.
Now fix m and consider a component C of Q' . The inclusion (21) shows that
C is enclosed by precisely one of the curves J, .J2...., Jn. Thus we may define an
angle assignment Om on C(Qm) by requiring that for each C E C(Qrn),
O3m(C)
=64k
where k is the unique integer such that C C In [Jk]. Since Qm is finitely connected,
Koebe's Theorem (Theorem 1.2) gives the existence of a map fm(z) achieving the
angle assignment OJmand having expansion
X
(z) _
+ al,m +
z
near c.
504
CARL. H. FITZGERALD
AND FREDERICK
WEENING
B2
_ _
FIGURE
5. A crossing-module problem (D. L3n, PrtF)
It is easy to show (see [W, p. 671) that {frn(z)} forms a normal family and has
a subsequential limit f(z) that is univalent on Q with expansion
f(z)z=z+
- +
z
near o.
By passing to the subsequence, it may be assumed that the entire sequence {ffn(z)}
converges to f(z).
Now we verify that (NP0) is satisfied for this f(z). Let B ={Bt1, B2, -, B.} be
any Oo(fl-')C-induced crossing-boundary collection in f(Q), and let s be sufficiently
large so that (19) holds. Consider the crossing-module problem (Dj, 3j, Fj) defined
by (20) for some fixed j E {1, 2, ... , n}. For simplicity of notation it may be assumed
that j
n; further, since the crossing-module is invariant under rotation we may
assume that On- 0 (see the illustration in Figure 5). We shall also use the notation
S= S8,o, and denote the left and right sides of the square OS, by (OS,)> and (3S,),
respectively.
For the metric p*(z) - 1 on D. it is clear that A* (Dn, 23n)< 82 and that
Lp*(Fn) > s, hence the inequality
F(rn) < 1
is immediate. Showing the reverse inequality shall require much more work. A succession of five crossing-module problems will be constructed. The value of crossingmodule of the final problem will be found by using the theorem giving the crossingmodule of a horizontally sectionable, punctured rectangular domain (Theorem 3.3).
While the necessary chain of inequalities relating the crossing-module problems will
follow from various estimates and the theorem on the conformal variance of the
crossing-module (Theorem 3.1).
AND UNIQUENESS
EXISTENCE
OF RECTILINEAR
SLIT MAPS
505
In order to apply Theorem 3.1 the crossing-number of all chains in the crossingmodule problem under consideration must be uniformly bounded. For each positive
integer N, let
FN
-{ t
E
F>: c#(u) < N}.
The inclusion1N C Fn impliesthat LP(FN) > Lp(F72)for any metric p(z) on Dn,
and hence that
M (Fn)
-
H(n)-
Therefore, in order to prove that b(Fn) > 1, it suffices to show that for each 3 > 0
there exists N such that
I_(FN) > 1-
Let a > 0 be given. The choice of N will be determined by approximating the
curves Bk from the inside and outside by piecewise linear curves. The inner curves
shall serve as the crossing-boundaries in a yet to be specified crossing-module problem. By choosing N sufficiently large this problem will be assured to be horizontally
sectionable. For each k -1, 2, ... , n, apply Lemma 4.3 to the domain In [Bk] n f (Q)
to obtain a piecewise linear Jordan curve Ik in f(Q) that satisfies
C In [Bk]
Ik
Af(Ik) =
and
(Bk)-
Next apply Lemma 4.2 to the domain f (Q) \. Un IIn [Bk] and
dist(U 13,Ss),
to obtain a collection {OkI}n., of disjoint piecewise linear Jordan curves in f(Q)
Uk=1In [Bk] each contained in S8 that satisfies
Bk c In [0j]
for k -1,
if and only if j-k,
2, ... , n, and also the condition
n
(22)
2,
area(In [0k] -. In [Bkl) < -Ab(Sn)
k~1
This construction is illustrated in Figure 6. Define N to be the total number of
segments comprising all of the curves Ok for k -1,
2, .. ., n.
We shall show the crossing-module problem (Dn,Bn, 1N) has crossing-module
greater than or equal to 1 - 3. Since n > 2 the area bounded by the crossingboundaries Ab(Sn) is strictly positive. This observation allows one to limit the
class of metrics for which analysis must be performed; in particular we note that
for any metric p(z) on Dr,
(23)
Lp(TFN)
n < - Ab(l3n) l/2
Ap(Dn, l3n) > Ab(Bn)
LP(FN)2~~~
Ab(Bn)
Thus it suffices to consider a metric p(z) on Dn such that LP(PnN) > Ab(Bn)1/2.
Let p(z) be such a metric and let (TN be a chain in ijN such that I((TN) < 2LO(1jN).
Since {Qm} exhausts Q and f(z) is a homeomorphism onto f(Q), the sequence of
domains {f(Qm) } exhausts f(Q). Hence a compactness argument shows that there
In [Ik] are contained in f(Q,r) for
exists M such that aN and the domain C . UkV1
each m > M.
Consider, for each mrr 1, 2 ... ., the map g9 f(Qm) -f> fm(Qm) defined by
gm(z)
= fm o f
l (z)
CARL H. FITZGERALD
506
AND FREDERICK
WEENING
12
/
//02
13
0 ?13
1
w_
FIGURE 6. The curves Ok and Ik
Since {frn (z) } converges normnallyto f (z), the sequence {grn(z) } conivergesnormrally
to g(z) -z on f(Q). Let c > 0 be less than min{1 60 C):2,
where
3,64},
6062
min dist (Ik, Bk)
and the minimums
61 = mil arclength(Bk)
A6([3,)1/2/(4N)
C3 --
,
dlst (Bk7 Ok) 6,
xthin
are taken over k
1 2,
< C
on
o.
T
Choose
4 -_
min dist (0
k,
9q)
rno > iM stuch that
rn
grno(Z)-Z
C
LJIn [1kj
k=1
By choice of M, each of the crossing-boundaries in Bn is a boundary component
of Dn, n f(Qmo) Thus -we may consider the crossing-module problem (Dr,B3F3),
illustrated in Figuire 7, where the domain f (Q) is replaced by f (Qrn);that is where
Do? = Dnrl fl Pin?o ) I
o
-- B,
and
r? --- ta E 1N
o07
CDO)(BO)}
Let po be the metric on Do defined by restricting p(z),
P
PO(z) -/(Z)
)D
Since Do c Dn and Bo-t Bn it is clear that AP (])0 B0) < A4(Dr,,Br,), while the
inclusion F0 C F,N implies that LPO(Fr)> L6,(F') Hence
(24)
Ap (.Dn Bn) > APO(D, BT
3)
L p (rnN )2
Lp
(]
)
2
Also, since mo > M, the chain aN belongs to FPnwhich imnpliesthat
? 2Lp(1TN)
(25)
L6P0(Fo<)
EXISTENCE
AND UNIQUENESS
OF RECTILINEAR
SLIT MAPS
507
B2
B
FIGURE 7. A crossing-module
i3 5F).
problem (Do}, B,
Consider next the crossing-mod-uleproblem (DX, B, Frl) induced by-the mapping
Ymo (z) on the crossing-module problem (DO, B?, P)F, that is where
D
= grno
(Do))
.,gm0 (Bn- i)}, and
13= {gmo (Bi),gSmo (B2),.
1n =
:o
{grn0(()
n}.
This problem is illustrated. in F'igtire 8. Sinrceg,no (z) is a homneomnorphism
of Do}
onto DX, the set Fl consists of all chains a E CDI(Al) joining grmo((&Ss)i) to
gYio((aS,),) whose crossing-numbers are bounded by N.
Let p1(z) = pogno(z) be the induced metric on DX defined as in equation (6).
We shall use Theorem 3.1 to obtain a lower bound for Ap (D2 130)/L (0 )2 in
terms of A,1 (DX,BX)/Lp1(P1)2 The choice of mo implies that
(26)
gm((ZI)
for z1, Z2 e Bk, and k = 1, 2,...
(27)
Since c <
indeed,
(28)
-gmo
,
Ab(Qn)1/2/(4N)
-
>
Lp(rN), this inequality is:lontrivial;
2cN > 1Ab(b )12
Also observe that c < cl and Lemmaa3.2 imply that
(29)
2c
> Lt0(I7)-2cN
and Lpo(Fr)
LPO(F)
Z211 <
n. Hence inequality (8) implies that
LPI(F')
co -
Z -
(Z2)1
|Ab(B,) --- Ab(B3)1 < fi
CARL H. FITZGERALD
508
AND FREDERICK
WEENING
-
1
g~~~mO(B2)
gm B 1)
FIGURE 8.
A crossing-moduleproblem(D', 83X,
Fr).
is a constant independent of C. Now estiwhere KI = En= I arclength(Bk)/(8w7)
mate (10), the fact that c < 1, and inequalities (24) through (29) yield that
Ap0(D?,L3?)
L,(FO)2
>
-
A n1(Dn,L3S) _
L 1(Fj)2
4
rL+ 2N(4L(F) n
r
Ab(nLn)
N Ap(D BTn)1
Lp(N)2j6
Hence
(30)
A
(D(,L3)
> A2(DC,L3)
2)
where K2 > 0 is a constant independent of c.
We now examine the geometry of DX. The outer boundary component
is a Jordan curve which is close to the square OS, in the sense that
Ss-e
gi0
(aSs)
C In[gmo(0SS)] C S,+c.
Further note that gmo (OS,) encloses all of the curves Ok, since 6 < 64. Also, since c
is less than 62 and 63, each of the crossing-boundaries gmo (Bk) E L3 is a perturbed
version of Bk which lies in the region between the curves 'k and Ok. In other words,
we have the inclusions
(31)
Ik
c In [gmo(Bk)]
c
In[Ok]
for k =
2,S,... , n- 1. Finally, it follows from the definition of fmo (z), that all
the remaining boundary components of Dn are horizontal slits enclosed by On and
that there are only finitely many of these slits.
Consider next the crossing-module problem (D2, L32,F2), illustrated in Figure 9,
obtained by replacing the outer boundary component gmo (OS,) by a (nearly square)
EXISTENCE
AND UNIQUENESS
OF RECTILINEAR
:~~~~~~
SLIT MAPS
509
,B
4B
i)
g~~~~~~~g
X
FIGURE 9. A crossing-module problem (D2,
L32, T2).
rectangle. Define
U SC-ev
s ) n R3
D2
3n =
7,
and
a joins the vertical sides of &R8su6},
whereRs+e,s- is the rectangulardomain
T2
{u E
CD2 (B2)
c#(u) < N,
Ilmzl < s-e}.
Rs+c,s-e --{z: IRezl < s+E,
Let P2(z) be the metric on D' defined by
z E
P2(Z)
JpI(z),
Di1
It is clear that Ap2(D2,132) < Ap1(Di, L3l). On the other hand, for each
thereexists a al E Fl, containedin
U2
n DX.Sincep1(z)=
of IPI(a) over a E FT is less than that of
1p2 (a)
P2 (z)
U2
C
p2
on a1, the infimum
over a C F2. Thus
LP2(F2) > Lp1(Fl)
(32)
and therefore
Api (D 1,,61 )
L
(Fl )2
Ap (D2,, 62l)
-
L
(]p2 ) 2
One final modificationof the crossing-moduleproblemmust be made beforethe
area to length-squaredratio can be estimated by Theorem3.3. The need for this
last modificationis due to the fact that the problem (D2, 32, F2) m-ightnot be
horizontallysectionablesince n2 only containschainswhose crossing-numbersare
less or equalto N whilethe crossing-boundaries
gno (Bk) may oscillateup and down
CARL H. FITZGERALD
510
AND F'REDERICK
k)
III
0
FIGURE
WEENING
10. Illustration of the possible oscillation of thle curves
gin0
(Bk)
frequently (see Figure 10).- To avoid this possibility we shall enlarge the domain
- 1). In the new
D 2 by expanding up to the boundary curves Ik (k _ 1, 2,. .
chains.
hor-izontal
the
it
be
sufficient
to
consider
shall
only
crossing-module problem
Let
(D L33
B~ F) betecosn-modu-le
problem defined by
D3 D2 U (In[gmo (Bk)V"-,,Inl{1)k
B3
{c3rto
D
and
__},
{Il,If2,...,I
3
a
(I3~)
is a horizontal chain joining the vertical sides of D 31.
Consider the metric p3 (z) defined oni D 3 by
z E Dn
P2(z)
{3()
max{tP2(Z),
1.7
~
I
z
0
- 9
Bg K-I,
where
72-1
0
U' I
0ko]1
Bg -
~~~~n-1
n-1
In[gmo(Bk)1 and
I2: UmInKk.k
It follows from the definition Of P3(z) and p-area that
A 3(D
,3"6)
<?A (D2,L32) + area(O_,B-9)
is small; indeed the inclusions (31), the condition (22), and
The term areaP(0".B)
the inequality (29), show that area(O- -, BY9) < ~-Ab(13n)+ tlc-. Therefore
4
EXISTl5ENCE AN'D UNIQUENESS
OF RECTILINEAR
SLIT MAPS
511
shall now estimate Lp2(Fr2) in terms of Lp3(Fn). It is useful to keep Figure 10
AWe
in mind during the construction involved in the estimate. Let the horizontal chain
a3 C F3 be given. Consider the horizontal chain U4 with the same imaginary
0. In other words, a4 is a subchain
part as a , but whose image lies in Dn
of ao3 which crosses the curves Ok. Since N is greater than the total number of
1,2,.. ,n - 1, the
segments comprising the piecewise linear curves Ok for k
crossing-nurnberof aJ4is less than N. Thus
for some I < N
a4 _ Zya
j-l
where each 'yj joins boundary components of D 3 0. Now, successively, extend
the ends of all segments ?yjhorizontally to the left and right until either they first
intersect a crossing-boundary in B2 or otherwise until they meet the next segment
rYj+1.Let a2 denote the resulting chain and note that a 2 C F2. It follows from the
definition of p3(Z) that IP2l(2) < IP3(a3). Since L 2(F2) < l12( 2) and o3 C F3
was arbitrary, we have that
L
(35)
j(F)
2
L 3(J)
We now consider the area to length-squared ratio. The combination of the
inequalities (34) and (35) yields that
(D 2,5
Ap
L
(F2
Ap (D3,i33)
>
)
L
-2
+ /<I6
-6Ab(13n)
)(LF(22)2
(3
2
Since inequalities (32), (27), and (28) imply L 2(Fn)
6 Ab(3n
(13n ) + KI-C
6aAb
LP (F2
)
2
> 2Ab(I3)n/2,
we have
I
) + /'s1
(Ab(13n ) I/2 /2) 2
3
where K3 > 0 is a constant independent of c. Therefore
(36)
~
A
L
~~~~(Fa2)2)>
3
( (83)2)d
We now apply Theorem 3.3 to the crossing-module problem (D3, 133,F3). We
need only remark that application of this theorem is permissible since there are
only finitely many horizontal slit components in OD3 and all other boundary components, except the outer coinponent, are crossing-boundaries which do not oscillate
arbitrarily. The theorem gives that
By the definition of crossing-module we thus have that
(37)
(37)
AP3 1nv
LL3(F3)2
>
-
s + C
Putting together the inequalities (24), (30), (33), (36), and (37), we obtain the
following estimate on the original crossing-module ratio:
Ap(Dn,Bn)
B?)>
L_FN+2
8 -S-
-
-
K3C -
d-26e
512
CARL H. FITZGERALD
AND FREDERICK
for any positive number e less than min{1,
that
WEENING
60, 61, 62, 63, 64}.
Lp(FnT7Un)
Therefore we have shown
>-d
holds for an arbitrary metric p(z) on Dn satisfying Lp(F4N) > Ab(3n)I/2.
The
combination of this inequality and implication (23) yields the desired result that
,(pN) > 1 - 6, and hence that
n(Un-1.
Finally, we show that f(z) achieves e. Suppose to the contrary that some
component B* C 13(f(Q)) were not a slit at the desired inclination 0J. Choose
o (f-1)C-induced crossing-boundary collection 13 and sufficiently large s.
any
Define (Dj,Bj,B, F) by (20), and consider the crossing-module problem (D', 139,
V)
e
induced by the map z " c-iSjz. The set B*' - e-'OjB* is a boundary component
of Y. which is not a horizontal slit, and satisfies dist(B*', 1) > 0. Thus, by
Theorem 3.6, p(F) < 1. On the other hand, since the crossing-module is invariant
1. This contradiction shows that f(z) must
under rotations, Pt(F) = p(F)
LZ
achieve 0.
ACKNOWLEDGMENTS
We are grateful for communications from Zheng Xu He, Fumio Maitani, David
Minda, Burt Rodin, and Oded Schramm.
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[W]
DEPARTMENT
IFORNIA
OF MATHEMATICS,
UNIVERSITY
OF CALIFORNIA
AT SAN DIEGO,
LA JOLLA,
CAL-
OF CALIFORNIA
AT SAN DIEGO,
LA JOLLA,
CAL-
92093
E-mail address: cfitzgerald@ucsd.edu
DEPARTMENT
OF MATHEMATICS,
UNIVERSITY
92093
Current address: Department of Mathematics and Computer Science, Doucette Hall, Edinboro
University of Pennsylvania, Edinboro, Pennsylvania 16444
E-mail address: fweening@edinboro .edu
IFORNIA