LXXXI.3 (1997)
Root systems and the Erd˝ os–Szekeres Problem
by
Roy Maltby (Burnaby, B.C.)
1. Introduction. An n-factor pure product is a polynomial which can be expressed in the form Q n
i=1 (1 − x α
i) for some natural numbers α 1 , . . . , α n . We will use two different norms, both defined in terms of the expansion of a pure product. Given a pure product Q n
i=1 (1 − x α
i) whose expansion as a polynomial is P ∞
r=0 a r x r , the 1-norm of Q n
i=1 (1 − x α
i) is denoted by k Q n
i=1 (1 − x α
i)k 1 and is defined to be the sum of the absolute values of the coefficients in the expansion, i.e.
X ∞ r=0
a r x r 1 :=
X ∞ r=0
|a r |.
The 2-norm is defined by
X ∞ r=0
a r x r 2 :=
X ∞
r=0
a 2 r
1/2 .
Two old problems to which norms of pure products are relevant are the Prouhet–Tarry–Escott Problem and the Erd˝os–Szekeres Problem. We use square brackets to delimit a list, and we call two lists (or, more precisely, k-lists) [a 1 , . . . , a k ] and [b 1 , . . . , b k ] equal if (a 1 , . . . , a k ) is a permutation of (b 1 , . . . , b k ). That is, a list is like a set except that repeated elements are allowed, and a list is like a tuple except that the order of entries does not matter. (Some authors call a list a multiset.) Suppose we have two unequal lists of integers [a 1 , . . . , a k ] and [b 1 , . . . , b k ] such that
X k i=1
a r i = X k
i=1
b r i
for r = 1, . . . , d. Then we say that [a 1 , . . . , a k ] and [b 1 , . . . , b k ] form a multi- grade of size k and degree d. The Prouhet–Tarry–Escott Problem is to find
1991 Mathematics Subject Classification: 11C08, 11F22.
Key words and phrases: root system, Erd˝os–Szekeres Problem.
[229]
multigrades of the smallest possible size for each degree. It is known that for any degree up to d = 9, there is a multigrade of size d + 1, and it is conjectured (cf. [BI94, p. 10]) that this holds for all degrees. The connection to pure products is that any n-factor pure product of 1-norm 2k can be used to construct a multigrade of degree n − 1 and size k.
Erd˝os and Szekeres [ES58] asked how small the ∞-norm of an n-factor pure product can be, where the ∞-norm of a polynomial is defined by
X d i=0
a i x i
∞ := sup
{z∈C:|z|=1}
X d i=0
a i z i ,
where C is the set of complex numbers and |z| denotes the modulus (i.e. absolute value) of the complex number z. At the start of Section 3, we give some old inequalities relating the three different norms we have mentioned.
Throughout this paper, N denotes the set of all natural numbers, and n denotes the set {1, . . . , n}. If P ∞
r=0 a r x r = Q n
i=1 (1 − x α
i), then for each r, a r =
n
I ⊆ n : X
i∈I
α i = r, |I| even o
− n
I ⊆ n : X
i∈I
α i = r, |I| odd o
. So to make the 1-norm of Q n
i=1 (1 − x α
i) as small as possible, we need to pick the exponents α 1 , . . . , α n so as to maximise the number of disjoint pairs {I, J} (that is, no two different pairs have the same I or the same J) such that P
i∈I α i = P
j∈J α j where I, J ⊆ n, |I| even, and |J| odd. The same criterion helps to make the 2-norm as small as possible, but one also has to consider the magnitude of individual coefficients. Among pure products having a given 1-norm, the ones with the smallest 2-norm will be those (if there are any) whose expansions have no coefficients other than 0, 1, and −1.
That is, we want to choose α 1 , . . . , α n so that for every r ≥ 0, the number of ways to express r as a sum of an even number of α 1 , . . . , α n and the number of ways to express r as a sum of an odd number of α 1 , . . . , α n differ by at most one.
For a definition of root systems, we refer the reader to Carter [C72]. For
the present, we will just describe the properties of root systems which are
useful in the context of constructing pure products of small norm. Positive
root systems are sets of vectors which have the property that a small num-
ber of vectors can be represented as a sum of positive roots in exactly one
way, and all other vectors can be represented as a sum of an even num-
ber of positive roots in exactly as many ways as they can be represented
as a sum of an odd number of positive roots. We construct pure products
of small norm essentially by projecting root systems down to one dimen-
sion and using the projections of the positive roots as exponents in pure
products.
2. A construction from root systems. Let Π be a set of fundamental roots of a root system Φ. Let Φ + be the corresponding system of positive roots. For r ∈ Φ, let H r denote the hyperplane through the origin perpen- dicular to r, and let w r denote the self-inverse bijection on points of the space consisting of reflection in the hyperplane H r . The Weyl group W as- sociated with Φ is the group of bijections generated by {w π : π ∈ Π} under composition of functions. A chamber is a connected subset of V \ ( S
r∈Φ H r ) where V is the whole space. The intersection of the boundary of a chamber with one of the reflecting hyperplanes is called a wall of the chamber. Define
S := 1 2
X
r∈Φ
+r.
The result that makes our construction of pure products useful is The- orem 1. Theorem 1 is essentially a well-known theorem of Weyl [C72, The- orem 10.1.8; M72, equation 0.1], which is usually expressed in the form of the equation
e −S Y
r∈Φ
+(1 − e r ) = X
w∈W
(−1) l(w) e w(S) .
In [M96], we state and prove the interesting part of this theorem in (we believe) a clearer presentation than those in the literature. In particular, we do not mention any ring, module, Lie algebra, or [C72] “alternating element of a rational group algebra”. We need one more definition before repeating our statement from [M96].
Call a chamber C odd if it satisfies the following two conditions: there is exactly one Ω ⊆ Φ + such that S − P
Ω ∈ C, and for every point p in a wall of C, exactly half the subsets Ω ⊆ Φ + such that S − P
Ω = p have even cardinality. Theorem 1 is the interesting part of Weyl’s theorem.
Theorem 1. Every chamber is odd.
For every v ∈ V, define a(v) :=
n
Ω ⊆ Φ + : X
Ω = v, |Ω| even o
− n
Ω ⊆ Φ + : X
Ω = v, |Ω| odd o
.
Corollary 2. There are exactly |W | points v ∈ V for which |a(v)| = 1, and a(v) = 0 for all other v ∈ V.
P r o o f. This is clear from Theorem 1 since there are exactly |W | cham- bers [C72, Corollary 2.3.3].
It is easy to see that for any natural numbers α 1 , . . . , α n and a, k Q n
i=1 (1 − x α
i)k = k Q n
i=1 (1 − x aα
i)k for the 1-norm, 2-norm, and ∞-norm.
Every n-factor pure product has 1-norm at least 2n. In [M94], S. Maltby an- swered the question: For which natural numbers β 1 , . . . , β n , γ 1 , . . . , γ n does k Q n
i=1 (1 − x bβ
i+cγ
i)k 1 = 2n for all b, c ∈ N? This question is equivalent to asking: In 2-dimensional space V with basis vectors π 1 , π 2 , for what sets of n vectors {p 1,1 π 1 + p 1,2 π 2 , . . . , p n,1 π 1 + p n,2 π 2 } does the construction in Lemma 3 produce a pure product of 1-norm 2n for all d 1 , d 2 ∈ N? S. Maltby found that, in fact, the construction produces such pure products if and only if the p i,j ’s come from a 2-dimensional root system.
Let l be the rank of Φ and label the fundamental roots π 1 , . . . , π l . Let n = |Φ + | and label the positive roots r 1 , . . . , r n . For i = 1, . . . , n and j = 1, . . . , l, let p i,j be such that P l
j=1 p i,j π j = r i . Since Π is a basis of V, the p i,j ’s are uniquely determined. And since Φ + is a positive root system, the p i,j ’s are all non-negative integers [C72, Proposition 2.1.6]. For any given α 1 , . . . , α n ∈ N, define for every k ∈ N ∪ {0},
b(k) :=
n
I ⊆ n : X
i∈I
α i = k, |I| even o
− n
I ⊆ n : X
i∈I
α i = k, |I| odd o
. It is easy to see that Q n
i=1 (1 − x α
i) = P ∞
k=0 b(k)x k .
Lemma 3 is intuitively clear if one thinks of the function p as a projection from the space V to 1-dimensional space. Remember that Π is a basis of V, so p in the lemma is well-defined.
Lemma 3. Let d 1 , . . . , d l ∈ N. Define p : V → R by p
X l
i=1
c i π i
:=
X l i=1
c i d i . For i = 1, . . . , n, let α i = p(r i ) = P l
j=1 p i,j d j . Then for every k ∈ N ∪ {0},
b(k) = X
p(v)=k
a(v).
We are now able to prove our desired result.
Theorem 4. Let d 1 , . . . , d l ∈ N. For i = 1, . . . , n, let α i = P l
j=1 p i,j d j .
Then
Y n i=1
(1 − x αi)
1 ≤ |W | where W is the Weyl group of Φ.
P r o o f. We have
Y n i=1
(1 − x αi)
1 =
X ∞ k=0
b(k)x k 1 =
X ∞ k=0
|b(k)|.
Since a(v) = 0 whenever p(v) 6∈ N ∪ {0}, Lemma 3 tells us that
X ∞ k=0
|b(k)| = X ∞ k=0
X
p(v)=k
a(v) ≤
X ∞ k=0
X
p(v)=k
|a(v)| = X
v∈V
|a(v)|.
And Corollary 2 tells us that X
v∈V
|a(v)| = |W |.
There is a reason why one could allow the d j ’s in Theorem 4 to be any integers (not necessarily positive) which do not yield any α i = 0, but there is another reason why the resulting new permitted values of d j ’s would be redundant. Suppose one allows polynomials to include terms with negative exponents, and suppose also that one adjusts the definition of the 1-norm and 2-norm to include coefficients of negative exponents the same as coeffi- cients of positive ones. Then, for any natural numbers α 1 , . . . , α n , it is easy to see that the following equalities hold for the 1-norm, the 2-norm, and the
∞-norm:
(1 − x −α1) Y n i=2
(1 − x αi)
=
x −α1(x α1− 1) Y n i=2
− 1) Y n i=2
(1 − x αi)
=
− x −α1
Y n i=1
(1 − x αi)
=
Y n i=1
(1 − x αi)
. From this we see that if we allow negative exponents in the factors of a pure product, we can replace them with their absolute values and the expansion will have the same coefficients, but associated with different exponents, and perhaps with reversed signs, so that the norms are not affected. Because of this, it would make sense to allow the d j ’s to be any integers not yielding any α i = 0. The resulting pure product Q n
i=1 (1 − x α
i) would be equivalent, for our purposes, to the pure product Q n
i=1 (1−x |α
i| ). Notice that while replacing some α i ’s with the corresponding −α i ’s does not change the expansion in any important way, replacing some d j ’s with the corresponding −d j ’s is a more complicated matter and it may appear that allowing negative d j ’s would result in pure products not attainable with only positive d j ’s. We now explain why the new pure products that this would allow are not different in any important way from the ones attainable using only positive d j ’s.
Theorem 5. Suppose d 1 , . . . , d l ∈ Z are such that for i = 1, . . . , n, α i 6= 0 where α i = | P l
j=1 p i,j d j |. Then there exist d 0 1 , . . . , d 0 l ∈ N such that if α 0 i = P l
j=1 p i,j d 0 j for i = 1, . . . , n, then (α 0 1 , . . . , α 0 n ) is a permutation of (α 1 , . . . , α n ).
P r o o f. We define dot products by
(x 1 , . . . , x k ) · (y 1 , . . . , y k ) :=
X k i=1
x i y i .
Since some of the tuples in this proof represent vectors relative to the non- orthonormal basis Π, dot products involving these tuples do not satisfy the usual lengths-cosine relationship. So, for instance, saying that (x 1 , . . . , x k ) · (y 1 , . . . , y k ) = 0 is not equivalent to saying that x 1 π 1 + . . . + x k π k is perpen- dicular to y 1 π 1 + . . . + y k π k .
For i = 1, . . . , n, if (p i,1 , . . . , p i,l ) · (d 1 , . . . , d l ) > 0 then let ε i = 1; and if (p i,1 , . . . , p i,l ) · (d 1 , . . . , d l ) < 0 then let ε i = −1. So we have the equation
ε 1 p 1,1 . . . ε 1 p 1,l ε 2 p 2,1 . . . ε 2 p 2,l
.. . .. . ε n p n,1 . . . ε n p n,l
d 1
.. . d l
=
α 1 α 2 .. . α n
.
For i = 1, . . . , n, let r i 0 = ε i r i . So for i = 1, . . . , n, r i 0 = P l
j=1 ε i p i,j π j and r i 0 is a root in the root system Φ; r 0 i is either r i or −r i —whichever one has positive dot product with (d 1 , . . . , d l ). That is, all the r i 0 ’s lie on the same side of the hyperplane of vectors having zero dot product with (d 1 , . . . , d l ) (which are not necessarily perpendicular to (d 1 , . . . , d l ) since the coordinates (p i,1 , . . . , p i,l ) of each r i are with respect to the basis Π which is not orthonormal). This means that {r i 0 : i ∈ I} is a positive root system of the same type as Φ + —call this positive root system Φ 0+ . Denote the fundamental roots of Φ 0+ by π 1 0 , . . . , π l 0 and put Π 0 = {π 1 0 , . . . , π 0 l }. For i = 1, . . . , n, let (p 0 i,1 , . . . , p 0 i,l ) be the coordinates of r i 0 relative to Π 0 . That is, each r i 0 = P l
j=1 p 0 i,j π 0 j . Since Φ + and Φ 0+ are positive root systems of the same type, we know that the matrices
p 1,1 . . . p 1,l p 2,1 . . . p 2,l
.. . .. . p n,1 . . . p n,l
and
p 0 1,1 . . . p 0 1,l p 0 2,1 . . . p 0 2,l .. . .. . p 0 n,1 . . . p 0 n,l
have the same rows, but not necessarily in the same order. So we prove the theorem if we show that there exist d 0 1 , . . . , d 0 l ∈ N satisfying
p 0 1,1 . . . p 0 1,l p 0 2,1 . . . p 0 2,l .. . .. . p 0 n,1 . . . p 0 n,l
d 0 1
.. . d 0 l
=
α 1 α 2
.. . α n
.
For j = 1, . . . , l, let i 0 j be so that r 0 i0
j
= π 0 j . So for j = 1, . . . , l, p 0 i0
j
,1 , . . . , p 0 i
0 j,l
are all 0 except for p 0 i0
j
,j = 1. Hence, the only assignment of d 0 1 , . . . , d 0 l that can satisfy the matrix multiplication above is to let each d 0 j = α i
0j
since we require α i0
j
= (p 0 i0
j
,1 , . . . , p 0 i
0j
,l ) · (d 0 1 , . . . , d 0 l )
= (0, . . . , 0, 1, 0, . . . , 0) · (d 0 1 , . . . , d 0 j−1 , d 0 j , d 0 j+1 , . . . , d 0 l ) = d 0 j . Notice that this means we are choosing d 0 1 , . . . , d 0 l to be natural numbers.
Now we just need to verify that for i = 1, . . . , n, α i = P l
j=1 p 0 i,j d 0 j . This is actually fairly easy to see from equalities we have already established. The details are as follows. For i = 1, . . . , n, r 0 i = P l
j=1 p 0 i,j π j 0 , so (p 0 i,1 , . . . , p 0 i,l )
= P l
j=1 p 0 i,j (p 0 i
0j
,1 , . . . , p 0 i
0j
,l ) and (ε i p i,1 , . . . , ε i p i,l ) = P l
j=1 p 0 i,j (ε i
0j
p i0
j
,1 , . . . . . . , ε i
0j
p i0
j
,l ). Now X l
j=1
p 0 i,j d 0 j = (p 0 i,1 , . . . , p 0 i,l ) · (d 0 1 , . . . , d 0 l )
=
X l
j=1
p 0 i,j (p 0 i0
j
,1 , . . . , p 0 i
0 j,l )
· (d 0 1 , . . . , d 0 l )
= X l j=1
p 0 i,j ((p 0 i0
j
,1 , . . . , p 0 i
0j
,l ) · (d 0 1 , . . . , d 0 l ))
= X l j=1
p 0 i,j α i0
j
=
X l j=1
p 0 i,j ((ε i0
j
p i0
j
,1 , . . . , ε i
0j
p i0
j
,l ) · (d 1 , . . . , d l ))
=
X l
j=1
p 0 i,j (ε i0
j
p i0
j
,1 , . . . , ε i
0j
p i0
j