C OMPOSITIO M ATHEMATICA
A RTHUR B ARAGAR
Asymptotic growth of Markoff-Hurwitz numbers
Compositio Mathematica, tome 94, n
o1 (1994), p. 1-18
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Asymptotic growth of Markoff-Hurwitz numbers
ARTHUR BARAGAR
Department of Mathematics, University of Texas, Austin, TX 78712
Received 12 November 1991; accepted in final form 8 April 1993
Introduction
In
1879,
Markoff[Ma]
made theequation
famous when he noted the connection between its
integral solutions,
classes ofquadratic forms,
anddiophantine approximation. Using
a descent argument, he showed all theintegral
solutions(except (0, 0,0))
can begenerated by
thefundamental solution
(1, 1, 1)
and a group ofautomorphisms W
of the surface(0.1).
The set ofpositive
orderedintegral
solutions to(0.1)
has a natural treestructure
(Fig. 1,
leftside).
Hurwitz
[Hu]
noted that Markoff’s descenttechnique
can beapplied
to theequation
Like the
Markoff equation,
the set ofpositive
orderedintegral
solutions to(0.2)
can be
separated
into a finite number of trees[Hu].
Zagier [Z]
did some beautiful work when he counted the Markoff numbers - thosepositive
numbers that appear inintegral
solutions to the Markoffequation.
He showed that the number of Markoff numbers less than H growsFig. 1.
asymptotically
likec(log H)2 (modulo
theunicity conjecture),
and gave arapidly converging
series for c z 0.1807. Histechnique
involves an idea(due
to Cohn
[Co])
ofcomparing
the Markoff tree with the Euclid tree(Fig. 1).
Suppose (p,
q,r)
with0 p q r is a
solution of the Markoffequation.
Then
and if p, q and r are
large,
thenHence the
branching operations
of the Markoff treecan be
compared
with thebranching operations
or more
simply
The tree thus
generated
is called a Euclid tree sincegoing
left down the tree isthe Euclidean
algorithm.
In Part 1 we make a similar’logarithmic’ comparison
between Hurwitz trees and an
(n - 1)-branch generalization
of the Euclid tree.Part 2 is devoted to
counting ya (x) - the
number of nodes withheight
lessthan x in a k-branch Euclid tree rooted at
a E (IR +)k.
The classical 2-branch Euclid tree isrelatively
easy to count, andZagier
showsThe k-branch Euclid tree rooted as 1
= (1,1, ...,1)
sitsnaturally
inC1L+t.
Hence
and it is
tempting
toconjecture ya(x)
is of order Xk. We show that for any G >0,
for all
sufficiently large
x, andfor an unbounded set of x’s. We also
give
bounds ona(k)
and show thesurprising
result that forlarge k, a (k)
grows likelog
k. For some small valuesof
k,
From this we conclude results about the
growth
of Hurwitz numbers.This paper is in part derived from results in the author’s thesis written under the
supervision
ofJoseph
H. Silverman at BrownUniversity [Bl].
1. A
comparison
of treesDefine a
height
It is clear that the number of
integral
solutions ofMa,n
with boundedheight
His finite. Hence it makes sense to define
Zagier actually
dealt with thequantity 93,3(H):
THEOREM 1.1
(Zagier.)
Withoutassuming
theunicity conjecture,
where c z 4.337.
Modulo the
unicity conjecture
and a factor of24, 93,3(H)
counts the numberof Markoff numbers less than H.
The aim of this paper is to
approximate ga,n(H)
for the Hurwitzequations Ma,n
which have non-trivialintegral
solutions.Deciding
whichpairs (a, n) yield
Hurwitz
equations Ma,n
that have non-trivialintegral
solutions is aninteresting problem partially
addressedby
Hurwitz[Hu]
andagain by Herzberg [He].
The Hurwitz
equations Ma,n
admitlarge
groups ofautomorphisms cga,n generated by
the deck transformationthe
sign change
and the
group Sn
ofpermutations
on the variables(xi, ... , xn}’
Note that theset of
integral
solutionsM(Z)
ofMa,n
is closed under the action ofW,,,.
Hence if we have a non-trivial
integral
solution s ofMa,",
then we can obtainmany more
integral
solutionsby considering
thecg a,n-orbit
of s:r;g a,n {s}.
In eachorbit
cga,n{s}
there exists apositive
ordered element r with minimalheight.
Wesay r is the
fundamental
solution or root ofcga,n{s}.
Hurwitz showed that r isunique
for each orbitr;g a,n {s},
and the set of rootsffa,n
inMa,n
is finite.Investigating
thepossible
size ofJ%,n
is anotherinteresting question
studied in
[B2].
Let
9Jla,n{ r}
be the Hurwitz tree for anequation Ma,n
rooted at r =(r 1’.." rn)
with 0
r 1 - - -
rn, andgenerated by
thebranching operations
for j
=1, 2,..., n -
1(for
apositive
ordered solution s,4Jo
is thedescending
branch when s is not a
root.)
The hat " indicates that that element is omitted.This definition does not demand that r be a root, or even that
rEMa,nCZ).
However,
if r is a root, then9Jla,n {r}
u9Jla,n {cpo(r)}
is the subset ofpositive
ordered
integral
solutions inrga,n {r}.
We compare the Hurwitz tree with
0152n-1{b},
the Euclid-like tree rooted atb c- (R ’)" - ’
and definedby
thebranching operations
for j
=1,..., n -
1. We use thesubscript n -
1 toemphasize
the number of branches. Both9Jla,n { r}
andOE,, - 1 (b)
are(n - 1)-branch
trees.Conceptually,
these trees are easy tounderstand,
but the notation we needto describe them is cumbersome. It would be convenient if we could describe each node
by
the vector at thatplace,
but since we have allowed thepossibility
of
duplication,
we cannot.For a set S of k
letters,
let us denoteby (S)
the free monoid on the k letters in S. SoS>
is the set of formal words written with letters in S and can bethought
of as a formal k-branch tree, rooted at the empty word(the
word withno
letters.)
For the Hurwitz trees and Euclid-like trees, the elements{cp1’"’’ CPn-1}
and{E 1’"’’ En-1}
are free as elements of a monoid(but
not aselements of a
group),
so the trees can berepresented by
andE J, * - - , En-1). However,
since we areprimarily
interested in the action of the elements in each monoid on the root solutions r andb,
let us denote these treesby
We then have the obvious map between these trees
that
pairs matching
nodes in each tree. Sometimes we will abuse ournotation,
and write
0(q)
= c where q = wr and0(w, wr) = (W, c).
LEMMA 1.2.
Suppose (w, q)e MR (r), 0(w, q) = (W, c) c- OE,, - 1 b,
andProof. (By
induction on m, the number of letters inw.)
We need
only
check that ifLEMMA 1.3.
Suppose
The
proof
is identical to theproof
of Lemma1.2,
except we useFor a Hurwitz tree
Wla,n{r},
defineg,(H)
is notquite
the same asg,,,@,,(H),
but it is close. We have omitted the solutions ofMa,n
obtained from qby sign changes
andpermutations
of thevariables. This accounts for a factor of
roughly n !2n -1.
We have also allowed thepossibility
that(w, p) =1= (cv’, q)
yet p = q, but this isonly possible
if r hascomponents ri and ri
suchthat ri
= ri. The number ofduplications
that canoccur is easy to
determine,
and is bounded for every q. Another différence isthat we have not chosen r to be a fundamental solution of
Ma,nClL).
A furthercomplication
is thatMa,nClL)
may have more than one fundamentalsolution;
however,
it isimportant
that we knowMa,n(Z)
has a finite number of roots.For a Euclid tree
OEN -
1{b},
define theheight
and the
counting
functionTHEOREM 1.4.
Suppose
Then
The
proof is straightforward
from Lemma 1.2 and Lemma 1.3.We still need a few more results before Theorem 1.4 can be useful. For
instance,
we need to know such a k exists. If we chooseq E Wla,n {r}
and q :0 r,then we can descend from q. Hence
and k =
al2
works. We can do better:LEMMA 1.5.
Suppose
The claim is clear for q = r, so assume
q =
r. Then we can descend from q, andq[ # qn-1 qn where qn
=a(Q1...qn-1) - qn (see
Cassels[Ca],
p.27,
for aclever
proof
ofthis,
or[B2]).
HenceWe also need to know there exist vectors b and c that
satisfy (ii)
and(iii)
and have
positive
components. If b or c has anon-positive
component, theny(x)
may not be finite. We can guarantee b and c havepositive
components if a >, 3. A more delicate argument is needed for a = 1 and2,
but we leave it tothe reader.
Lastly,
we need to know whatYb (X)
looks like. If we knowthis,
then we may be able to get agood approximation
forgr(H).
By
Lemma1.5, k
is near a if r 1 ... r n - 2 islarge. ln that
case, choosehi
=log(a1/(n-2)rJ
and c = eb so that c satisfies(iii).
Since k is near a, we canchoose e near 1. And since
Theorem 1.4 must
give
agood approximation
togr(H).
So to count a Hurwitztree
Wla,n {r},
first take a subtree whose endpoints
are a number of solutions{rk}k=l
1 with the property thatrk,l ... r k,n - 2 is large
for each k. Thenapproxi-
mate the
counting
of each tree rooted at rkusing
Theorem1.4,
and sum up.Zagier
works out the details forWl3,3{(1, 1, 1)}
in[Z].
The first detail is toapproximate Yb(X),
whichZagier
does rathersuccessfully
with Theorem 1.1.Approximating Yb(X)
for n >, 4 is thesubject
of thefollowing
section.2. The k-branch Euclid tree In this
section,
we show:THEOREM 2.1. For a k-branch Euclid-like tree rooted at b =
(b J,, * , bk) C (R +)k
with k a3,
thequantity
is
finite. Furthermore,
a =a(k) depends only
onk,
andsatisfies
For small
k,
In other
words,
for any s >0,
where
a (k)
grows likelog k,
and the’big omega’
notation meansyb (x)
> x’(k) - E for all x in some unbounded subset of thepositive
reals.Zagier
was able todirectly
compute anasymptotic
formula forY(a,blx).
Forthe k-branch Euclid trees with k >
2,
ourapproach
is much more convoluted.We
begin by defining
another Euclid-like tree0152k’{r}
rooted at r andgenerated by
thebranching operations
0152{l}
and0152k{l}
are relatedby
the linear mapand in
general, @{r}
ismapped by
L to@{L(r)}.
Note thatwhere A = La. Hence
counting
the elements inOE[ (r)
withheight
less than x isthe same as
counting
the elements in@{L(r)}
withheight
less than x. The treeG{r}
is easier to deal with because it has the classical two branch Euclid treenaturally
imbedded in it.We now define a different sort of
counting
function:Our
goal
is to show thatfr(t)
converges for all t > a anddiverges
for allt a. The
following
result willhelp
us determine whenf
converges:LEMMA 2.2.
If fr (t)
converges at t,then fcr(t)
converges at t andfor
all iand
The
proofs just require
nodeby
nodecomparisons
of the trees inquestion.
COROLLARY 2.3. For every
rc-(R’)’, 1»(t)
converges at tif
andonly if fi(t)
converges at t.
Proof.
Without loss ofgenerality,
we may set 0r 1 ....
rk . Thenfrom which the result follows.
Consequently,
we mayregard
a =a(k)
asdepending only
on the dimensionof r. We may also fix r, and choose r =
(1, 1,...,1) =
1. As isimplied by
ournotation, a(k)
is in fact thelimsup
defined in Theorem 2.1. This result will be shown in Theorems 2.8 and 2.9. Let us first find bounds fora(k):
THEOREM 2.4
fl(t)
convergesfor
allt > a(k)
anddiverges for
allta(k)
where
a(k) satisfies
and 0
is the Eulerphi function.
As mentioned
before,
the property that makes0152k{l}
nice to work with is that it has the 2-branch Euclid tree(which,
with a little abuse ofnotation,
we also call@) naturally
imbedded in it:and we have a very
good description
of the elementsW(l, 2,..., 2)
EOE2.
Sincewe start at
(1, 2,..., 2), OE2
has the property that ifW(l, 2,...,2) = W’
(1, 2,...,2),
then W = W’. So each node(W, W(l, 2,..., 2»
iscompletely
characterized
by W( 1, 2, ... , 2). Also,
we know:LEMMA 2.5. a =
W ( 1, 2,..., 2) for
someW E Tl, T2> if and only if a 1 and a2
are
relatively prime,
a1 1 a2, andfor all j>3.
Proof. Only
the last needs somejustification.
We can show ai = a + a2 - 1using induction,
orby noting
that a3 = ... = ak andwhere W’ E
(LTIL -1, LT2L -1) = E1, E2>.
ButE1 and E2
leave the first k - 2 components of c fixed for every c. Hencefor j >
3.Hence we can write
Also imbedded in
01522
is the trivial one branch treeNote also that if a
so
Thus,
So,
if t >a(k),
then the series converges and we must haveand with a little
help
from a computer, thisgives
the lower bounds ona(k).
Note that this also shows
«(k)
> 2 for all k > 3 a result we will need later.We find the upper bounds in a similar
fashion,
butunfortunately
mustcontend with an ’error’ term. To deal with this term, we want
fl (t)
to belarge,
so weexploit
thedivergence
forf, (t)
with ta(k),
and instead consider thepartial
sumsThese
partial
sums haveproperties
similar to those outlined in Lemma 2.2:LEMMA 2.6. For c >
0,
We can
decompose
thepartial
sumf, (t, y)
the same way wedecomposed
the convergent infinite sum
f (t):
This time we note
so
The second term on the
right-hand
side is what we have referred to as the’error’ term. We note that
where (t)
is the Reimann Zeta function.Thus,
this term converges for allt > 2. Since
a(k)
> 2 for all k >,3,
we may choose t so that 2 ta(k).
Then,
afterdividing through by f, (t, y)
andletting
y go toinfinity,
we getwhich
gives
us our upper bounds fora(k).
We now show
a(k)
grows likelog
k:THEOREM 2.7.
Hence
and
and
using
theinequality
establishedabove,
We are now
ready
to relate these results to the functionYr(x):
THEOREM 2.8.
Suppose rE(IR+)k. Then for
every B >0,
there exists aC(e)
suchthat
for
every x >C(e).
Proof. Suppose
not. Then there exists a sequence{xn}n=
1 that goes toinfinity,
and has the property thatfor every n. But then
for every n.
So,
choosea(k)
ta (k)
+ a. ThenfL-1(r) (t)
converges, but theright-hand
sidediverges
as Xnapproaches infinity - a
contradiction.Thus,
if xis
sufficiently large,
thenTHEOREM 2.9.
Suppose re(R+)k.
Thenfor
every e >0,
there exists an unbounded set X such thatfor all x in X.
Proof. Suppose
not. Then there exists a sequence{Xn}n=
1 and a constantc > 0 such that 0
(xn+
1 -xn)
c; Xnapproaches infinity
as n goes toinfinity;
and
for every n. So set
a(k) -
e ta(k)
and considerfL -l(r) (t),
which we knowdiverges
at t. Butwhere xo =
h(r).
We canjustify
thereordering
in the infinite sum, since thedifference
of the nthpartial
sums iswhich goes to zero as Xn goes to
infinity
for t >a(k) -
s.Continuing
with(2.1):
where
g(t)
is the first term, and hence converges for all t. Since X2 > c, we know(1
-c/xn)
is bounded away from zero, and since ta(k),
we canreplace
it witha constant, so there exists a constant C such that
which converges if t >
a(k) - E,
a contradiction.Hence,
we could not have had these ’fenceposts’
ofinequalities,
and there exists an unbounded set X such thatfor all x in X.
COROLLARY 2.10
(of
Theorems 2.8 and2.9).
Gathering
Theorems2.4,
2.7 andCorollary
2.10gives
us Theorem 2.1.Using
Theorem
1.4,
the remarksfollowing
Theorems 1.3 and 1.5 and Theorem2.1,
our main result follows:
THEOREM 2.11.
If Ma,n
has a non-trivialintegral solution, then for
every e >0,
where the exponent
a(n - 1) depends only
on n, andsatisfies
and for
small n.The constants and subset
implied by
the’big
oh’ and’big omega’ depend
on e, aand n.
References
[B1] A. Baragar: The Markoff Equation and Equations of Hurwitz, Ph.D. Thesis, Brown University, 1991.
[B2] A. Baragar: Integral solutions of Markoff-Hurwitz equations, J. Number Theory (to appear).
[Ca] J. W. S. Cassels: An Introduction to Diophantine Approximation, Chap. II, Cambridge
Univ. Press, Cambridge, 1957.
[Co] H. Cohn: Growth types of Fibonacci and Markoff, Fibonacci Quart. 17(2) (1979) 178-183.
[He] N. P. Herzberg: On a problem of Hurwitz, Pac. J. Math. 50 (1974) 485-493.
[Hu] A. Hurwitz: Über eine aufgabe der unmbestimmten analysis, Archiv. Math. Phys. 3 (1907) 185-196; Mathematisch Werke, Vol. 2, Chap. LXX (1933 and 1962) 410-421.
[Ma] A. A. Markoff: Sur les formes binaires indéfinies, Math. Ann. 17 (1880) 379-399.
[Mo1] L. J. Mordell: On the integer solutions of the equation
x2 + y2 + z2
+ 2xyz = n, J. LondonMath. Soc. 28 (1953) 500-510.
[Mo2] L. J. Mordell: Diophantine Equations, Chap. 13.4. Academic Press, New York, London, 1969, pp. 106-110.
[S] J. H. Silverman: The Markoff equation
x2 + y2 + z2 = axyz
over quadratic imaginaryfields, J. Number Theory 35(1) (1990) 72-104.
[Z] D. Zagier: On the number of Markoff numbers below a given bound, Math. Comp. 39 (1982) 709-723.