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ANNALES DE L’INSTITUT FOURIER
S. FRANKEL
MICHAEL KATZ
The Morse landscape of a riemannian disk
Annales de l’institut Fourier, tome 43, no 2 (1993), p. 503-507
<http://www.numdam.org/item?id=AIF_1993__43_2_503_0>
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Ann. Inst. Fourier, Grenoble
43, 2 (1993), 503-507
THE MORSE LANDSCAPE OF A RIEMANNIAN DISK
by S. FRANKEL & M. KATZ
Consider Riemannian metrics g on the two-dimensional disk D such
that the length of the boundary S1 is less than 1, and the Riemannian
distance from every point of D to the boundary is less than 1.
QUESTION 1. — Does there exist a constant C > 0 such that for
every metric g as above, the boundary circle can be homotoped to a point
through a family of curves each of which has length less than C ?
The answer is negative.
THEOREM 2. — One can construct a sequence of metrics Qn on D
and a function f{n) tending to infinity with n, such that every homotopy
of S1 to a point in (D,^n) contains an intermediate curve of length bigger
than f{n).
In terms of the length functional on the loop space of D, one can say
that the boundary loop is separated from the constant loop by a mountain
range of height at least f(n). Hence the term "Morse landscape" used in
[1].
The above Question 1 was asked by M. Gromov in his talk at Luminy,
July 1992. It is related to questions in geometric group theory (see [1]).
A different counterexample was constructed independently by Jean-Marc
Schlenker.
Key words : Diameter - Area - Closed Curves - Hyperbolic metrics - Trees.
A.M.S. Classification : 53C23 - 53C20 - 05C05.
504 S. FRANKEL & M. KATZ
Note that a counterexample cannot be obtained by attaching spikes
to the standard disk, because the curve can pass over them one at a time. In
our example the basic construction is not a spike but a "wall" (see below).
The proof uses a topological property of trees in the disk. If the tree
is sufficiently complicated, then any homotopy of the boundary to a point
will contain an intermediate curve that crosses several edges. If we now
build a wall over the tree, the curve will have to climb over it several
times and become very long. To prevent the curves from traveling along an
entire edge, we make the edges of the tree long. This can be done either
by redefining the Riemannian metric in coordinates adapted to the tree by
making it large in the direction of the edges, or simply imbedding the edges
in the standard disk by means of a sufficiently wiggly zig-zag contained in
a thin neighborhood of the edge. It is conceptually simpler to reformulate
this construction in terms of hyperbolic geometry as follows.
PROPOSITION 2^. — Let T C D be a tree. There exists a metric
h on the disk with the following properties :
(i) h is hyperbolic on the complement ofT.
(ii) the circumference equals 1 and the Riemannian distance from every
point of D to the boundary is < 1.
(iii) the distance between any two non-adjacent edges ofT is at least 1.
Proof. — Let U = Uc be the unit ball in the hyperbolic plane of
curvature —c2. Let q > 0 and consider a regular 2g-gon with geodesic sides
inscribed in the unit ball. As c —^ oo, the length of the side tends to 2.
Choose c suffficiently large so the sides have length 1.
Every tree T with q edges can be thought of as a circle subdivided
into 2q arcs with a suitable identification among pairs of arcs (compare
with the projection of the boundary of an e-neighborhood of T C R2 to
T). Here two neighboring arcs are identified if and only if their common
point projects to a terminal vertex of T.
We partition the circle 9U by the vertices of the regular 2^-gon.
Let / : 9U —> T be the (locally isometric) projection. Then T Uf U is
homeomorphic to the sphere S2. Let e > 0 be chosen so that the concentric
ball Be C U is of circumference 1. We now set (D, h) = (TUy U) \B^. Then
the metric is hyperbolic on D \ T, and has infinite positive curvature along
the edges of T. Property (iii) is satisfied by construction.
MORSE LANDSCAPE OF A RIEMANNIAN DISK 505
Proof of Theorem 2. — We perturb the homotopy so as to satisfy the
hypothesis of Proposition 5 below. The perturbation does not substantially
increase the length of the curves, and does not increase the number of the
(closed!) edges crossed by each curve. Now Proposition 5 provides a curve
7 meeting a large number of edges. Since the valence at each vertex is
bounded (by 3), we can choose a long sequence of non-neighboring edges
crossed consecutively by 7. By Proposition 2^, the length of the sequence
is a lower bound for the length of 7, proving the theorem.
We give an inductive definition of a binary tree.
DEFINITION 3. — The binary tree T^ of height 1 is a point, which
is its own root. The binary tree Tn+i of height n + 1 is constructed by
choosing its root and joining it by two edges to the roots of two binary
trees of height n.
LEMMA 4. — The tree Tn has 271 — 1 vertices. Removing an edge
separates Tn into two connected components, one of which contains the
root, and the other is another binary tree Tn' of height n1 < n.
The proof is obvious.
PROPOSITION 5. — Imbed the binary tree Tn in the disk D.
Consider a homotopy of the boundary S1 to a point such that each
intermediate curve passes through at most one vertex of Tn. Then some
( n \intermediate curve meets at least 0[ -— \ edges ofTn.\lnn/
Proof. — Given a curve 7 C D not passing through any of the
vertices of T^, we may consider the number of vertices of Tn with respect to
which 7 has even winding number. Such vertices will be called even (with
respect to 7), and the others odd.
Every integer between 1 and 2n — 1 must occur as the number of
even vertices of some intermediate curve. Indeed, at the beginning of the
homotopy this number is 0, at the end it is 272 — 1, and the vertices are
passed one at a time, hence change their parity one at a time.
The idea of the proof is to find a number "poorly representable" as a
sum (with ± signs) of numbers of the form 271 — 1. It is technically easier
to show that such a number exists than to exhibit the number.
LEMMA 6. — Let Tn C D be a binary tree and let 7 C D be a
506 S. FRANKEL & M.KATZ
curve not passing through any vertex ofTn. Suppose 7 meets Tn in k edges
e\,..., Cfc. Let rii be the height of the binary tree Tm separated by removing
Ci (see Lemma 4), and di the winding number of 7 with respect to the root
of Tm. Then the number of even vertices of Tn can be written in the form
po+^-l)0^-!).
ies
Here po is 0 if the root of Tn is odd, and 271 — 1 if it is even, while
S C { 1 , . . . , k} is a subset.
Proof. — The proof is in the spirit of de Morgan's law in set theory.
The binary trees Tm are partially ordered by inclusion. The unique maximal
element is Tn itself. Trees nearest for the partial order will be called
neighbors. Clearly each T^ has a unique bigger neighbor.
First we suppose that the curve meets an edge of Tn no more than
once, and that the intersection is transverse. Then the roots of a pair of
neighbors always have opposite parity, and 6 '={1, . . . ,A;} .
Assume that the root of Tn is odd. Then the set of even vertices is
the disjoint union of the smaller neighbors of Tyi, minus the disjoint union
of the smaller neighbors of these neighbors, plus the third neighbors, etc.
This proves the formula in the case of odd root. Otherwise we have to start
with Tn itself and so get the extra summand po = 271 — 1.
Without the assumption of unique transversal intersection, the parity
of the roots of two neighbors may be the same. We delete from the partial
order those trees whose bigger neighbor has a root of the same parity. The
remaining trees define the set S C {! , . . . , k} of the lemma.
Returning to the proof of Proposition 5, we note that each cut ei gives
rise to 2n possibilities (the height of the separated tree T^ and the parity
of the root of T^). Thus there are at most (2r^)A;+l possible numbers of
even vertices that can be represented by k cuts. Since we need to represent
271 — 1 of them, we must have the inequality (2n)A;+l > 2n — 1, which implies
that of Proposition 5.
With an additional assumption on the "monotonicity" of the homo-
topy, we can provide a linear lower bound for the number of crossings. If the
curve sweeps out the disk without coming back upon itself, we can think
of the curves as level sets of a real valued function.
PROPOSTION 7. — Every continuous real-valued function on a
binary tree of height n has a level set with at least n/2 points.
MORSE LANDSCAPE OF A RIEMANNIAN DISK 507
Proof. — We argue by induction on the height n of the tree. For
n = 1 and 2 the statement is immediate from the fact that the complement
of a point has nontrivial fundamental group.
Define numbers a,b e R by a = m!f(Tn), b = supf{Tn), where
/ : Tn -» R is the function. Choose preimages a,b G Tn. Consider the
imbedded path a C Tn joining a with b. The complement Tn \ a contains
a binary tree B of height at least n - 2. Indeed, if a does not pass through
the root of Tn then a stays away from a branch of height n — 1. If a passes
through the root, then at the next vertex it has to go either right or left,
and the remaining edge leads to a tree of height n - 2. By the inductive
hypothesis applied to B, there is a point x C R with at least n—— == n -1
preimages in B. Since a <: x <: &, the point x has an additional preimage
on a, proving the proposition.
QUESTION 8. — Consider a binary tree Tn of height n imbedded
in the 2-disk D, and a homotopy shrinking the boundary to a point. Does
there exist an intermediate curve meeting Tn in at least n — 1 points ?
QUESTION 9. — What is the answer to the original question of
Gromov with the additional bound on the area of the Riemannian metric
on the disk ?
BIBLIOGRAPHY
[1] M. GROMOV, Asymptotic invariants of infinite groups, IHES preprint, 1992.
S. FRANKEL,
Columbia University and Universite de Nantes
&
M. KATZ,
URA 750 (CNRS)
Departement de Mathematiques
Universite de Nancy I
B. P. 239
54506 Vandoeuvre (Prance).


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