Journal of Algebraic Combinatorics 10 (1999), 173–188
c 1999 Kluwer Academic Publishers. Manufactured in The Netherlands.
°
Combinatorics of Necklaces
and “Hermite Reciprocity”
A. ELASHVILI
alela@rmi.acnet.ge
M. JIBLADZE
jib@rmi.acnet.ge
D. PATARAIA
dito@rmi.acnet.ge
A. Razmadze Mathematical Institute of the Georgian Academy of Sciences, M. Alexidze St. 1, Tbilisi 380093,
Republic of Georgia
Received December 5, 1996; Revised March 26, 1998
Abstract. Combinatorial proof of an explicit formula for dimensions of spaces of semi-invariants of regular
representations of finite cyclic groups is obtained. Using bicolored necklaces, a certain reciprocity law following
from this formula is also derived combinatorially.
Keywords: partition, necklace, semi-invariant, reciprocity
Introduction
The classical Hermite Reciprocity Law asserts the isomorphism
S m S n (k 2 ) ∼
= S n S m (k 2 )
of symmetric powers of representations of the Lie group SL2 (k) acting standardly on k 2 ,
for a characteristic zero field k (see [6], Remark 12 by Popov in Appendix 3 of the Russian
translation). In particular, the space of degree m polynomial invariants of the irreducible
(n + 1)-dimensional representation is equidimensional with the space of degree n invariants
of the irreducible (m + 1)-dimensional representation.
Recently in [3] there was obtained an explicit formula for the dimension a0 (n, m) of
the space of degree m homogeneous polynomial invariants of the regular representation
of the nth order cyclic group. This formula implies that a0 (n, m) = a0 (m, n). In the
present paper, we give a combinatorial explanation of a certain generalization of this fact
(see below), which we also call Hermite reciprocity.
Relationship with combinatorics stems from the observation that, as shown in [3], the
number a0 (n, m) coincides with the number of solutions of the system
n−1
X
j=0
jλ j ≡ 0
(mod n);
n−1
X
λi = m.
i=0
The first author is partially supported by the INTAS Grant #93-0893.
The second and third authors are supported by the ISF Grant MXH200 and by the INTAS Grant #93-2618.
(1)
174
ELASHVILI, JIBLADZE AND PATARAIA
Clearly this is the same as the total number of partitions of multiples of n into no more than
m parts not exceeding n − 1. Applying combinatorial arguments one can obtain a formula
for the number ak (n, m) of solutions of an even more general system
n−1
X
jλ j ≡ k
(mod n);
j=0
n−1
X
λi = m
(2)
i=0
where k is any nonnegative integer.
As a kind of illustration let us reproduce the first few values of ak (n, m) (computed using
the MAPLE package):
(1) a0 (n, m), for 1 ≤ n, m ≤ 10:
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
4
3
5
3
7
4
10
4
12
5
15
5
19
6
22
1
1
1
1
1
1
1
3
3
4
4
5
5
6
5 10
7 14
10 22
12 30
15 43
19 55
22 73
14
26
42
66
99
143
201
22
42
80
132
217
335
504
30
43
55
66
99
143
132
217
335
246
429
715
429
810 1430
715 1430 2704
1144 2438 4862
73
201
504
1144
2438
4862
9252
(2) a1 (n, m), for 1 ≤ n, m ≤ 10:
1
1
1
1
1
1
2
2
1
2
3
5
1
2
5
8
1
3
7
14
1
3
9
20
1
4
12
30
1
4
15
40
1
5
18
55
1
5
22
70
1
1
1
3
3
4
7
9
12
14
20
30
25
42
66
42
75
132
66
132
245
99
212
429
143
333
715
200
497
1144
1
4
15
40
99
212
429
800
1430
2424
1
1
5
5
18
22
55
70
143
200
333
497
715
1144
1430
2424
2700
4862
4862
9225
Derivation of ak (n, m), given below, is analogous to a proof for the a0 (n, m) communicated to us by G. Andrews. The expression obtained has all the advantages of an explicit
formula, in particular it immediately implies the equality ak (n, m) = ak (m, n), which too
COMBINATORICS OF NECKLACES AND “HERMITE RECIPROCITY”
175
may be called “Hermite reciprocity”. But the proof does not explain in any way the reason
of this reciprocity. In the second part of the paper we give one of the possible explanations
for the equality. Namely: to any solution of (1) we assign a necklace, i.e., a circular arrangement, consisting of n + m beads, n of them black and m white, together with a chosen
orientation and a basepoint somewhere between two adjacent beads. Thereafter, the equality ak (n, m) = ak (m, n) turns out to follow from the existence of an involution on the set
of such necklaces, acting by choosing opposite orientation and swapping black and white.
We must note that in a private conversation with the first author, N. Alon communicated
a proof involving necklaces, of a0 (n, m) = a0 (m, n), when n and m are coprime. In the
present paper this idea has been extended to the more general setting. The authors would
like to express their gratitude to Alon, Andrews and Stanley for valuable information and
interest to the paper. They are grateful to the referee for careful reading of the paper and
finding of several misprints in important formulae.
Everywhere in the sequel, for any integers n, m, k, . . . their greatest common divisor will
be denoted by (n, m, k, . . .); for n > 0, we denote by (k)n the residue of k modulo n, i.e.,
the number determined by 0 ≤ (k)n < n, (k)n ≡ k(mod n).
1.
Explicit formula
Let us start with a purely formal expression for ak (n, m). Therefore recall that p(N , M, s),
for any integers N , M, s, denotes the number of partitions of s into no more than M parts,
each not exceeding N . The generating function for these numbers,
G(N , M; t) =
X
p(N , M, s)t s
s
is the Gauss polynomial (see e.g., [1], 3.2).
Then, one obviously has
ak (n, m) =
X
p(n − 1, m, jn + k).
(3)
j
We shall also need the definition of Ramanujan sums (see e.g., [4], 17.6; for applications
in number theory see [5]). For any n and k, the Ramanujan sum cn (k) is the sum of kth
powers of all primitive nth roots of 1. In particular, cn (0) = ϕ(n) (the Euler function),
cn (1) = µ(n) (the Möbius function). It is known (and easily seen using Möbius inversion)
that
X µn ¶
d.
µ
cn (k) =
d
d|(n,k)
Also note that this last equality obviously implies cn (k) = cn ((n, k)), in particular, cn (−k) =
cn (k).
We then have the following:
176
ELASHVILI, JIBLADZE AND PATARAIA
Theorem 1 For any integers k, n, m,
Ã
!
X
n/d + m/d
1
cd (k)
;
ak (n, m) =
n + m d|(n,m)
n/d
(4)
in particular
ak (n, m) = ak (m, n).
Proof: By (3), ak (n, m) equals the sum of coefficients of G(n −1, m; t) at those powers
P of
t which are congruent to k modulo n. Now in general, given any polynomial f (t) =
fν t ν ,
one has
X
1 X −k
fν =
ζ f (ζ ),
n ζ n =1
ν≡k (mod n)
the sum on the right running over all nth roots of 1. This fact easily follows from the
equality
(
X
n if n | ν
ν
ζ =
0 otherwise.
ζ n =1
So in our case
ak (n, m) =
1 X −k
ζ G(n − 1, m; ζ ).
n ζ n =1
Values of Gauss polynomials at roots of 1 are known; see e.g., [7], Chapter 3, Exercise 45(b).
In particular, for any ζ n = 1 which is a primitive dth root of 1, for some d | n, one has
Ã
!

 m/d + n/d − 1
if d | m
G(n − 1, m; ζ ) =
m/d


0
otherwise.
Hence
1X X
ζ −k
ak (n, m) =
n d|n ord(ζ )=d|m
Ã
m/d + n/d − 1
m/d
!
,
where ord(ζ ) means order of the element ζ in the group of roots of 1. Now ζ has order d
iff ζ is a primitive root of order d. Hence
Ã
!
m/d + n/d − 1
1 X
ak (n, m) =
cd (−k)
n d|(n,m)
m/d
¡ m+n
¢
−1 !
1 X
d
¢
=
cd (k) ¡ n
n d|(n,m)
− 1 ! md !
d
COMBINATORICS OF NECKLACES AND “HERMITE RECIPROCITY”
=
177
d m+n
!
1 X
d
cd (k) m+n
d
m
n
n d|(n,m)
! !
n d d
1 X
n
=
cd (k)
n d|(n,m)
m+n
X
1
cd (k)
=
m + n d|(n,m)
Ã
Ã
m/d + n/d
!
n/d
m/d + n/d
n/d
!
,
2
and the theorem follows.
Remark Note that the formula obtained implies, due to the mentioned properties of
Ramanujan sums, that there are many equalities between ak (n, m) for fixed n, m and
various k. Namely, one has
ak (n, m) = a(k,n,m) (n, m).
For the sequel, let us fix two positive integers n, m and denote (n, m) by d.
2.
Combinatorial proof of reciprocity
In this section, we are going to give another, purely combinatorial proof of the equality
ak (n, m) = ak (m, n).
Consider the set
(
)
n−1
X
n
3n,m = (λ0 , . . . , λn−1 ) ∈ N | ∀i λi ≥ 0 and
λi = m .
i=0
We define an action of the cyclic group Cn = Z/nZ of order n on 3n,m by setting, for
r ∈ Z/nZ and λ = (λ0 , . . . , λn−1 ) ∈ 3n,m ,
r λ = (λn−r , λn−r +1 , . . . , λn−1 , λ0 , . . . , λn−r −1 ).
For any λ ∈ 3n,m , denote by t (λ) the minimal positive integer with (t (λ)1)λ = λ,
i.e., the number of elements in the orbit of λ under the action of Cn (1 is the element of
Z/nZ = {0, 1, . . . , n − 1}). Denote n/t (λ), i.e., order of the stabilizer of λ under this
action, by s(λ).
Let 3̃n,m be the quotient 3n,m /Cn and denote by πn,m the quotient map πn,m : 3n,m →
3̃n,m .
−1
−1
(α). Note that #(πn,m
(α)) = t (λ). Since s(λ) does not
Take any α ∈ 3̃n,m and λ ∈ πn,m
depend on the choice of the inverse image of α, we may denote by s(α) the number s(λ)
−1
(α).
for any λ ∈ πn,m
178
ELASHVILI, JIBLADZE AND PATARAIA
Let K n,m be the mapping from 3n,m to Cn determined by
K n,m (λ0 , . . . , λn−1 ) = (0 · λ0 + 1 · λ1 + · · · + (n − 1) · λn−1 )n .
−1
Then it is clear that ak (n, m) equals #(K n,m
(k)), for any k ∈ Cn .
Proposition 1 For any r ∈ Cn , λ = (λ0 , . . . , λn−1 ) ∈ 3n,m one has K n,m (r λ) =
K n,m (λ) + r · (m)n .
Proof: The case r = 0 is obvious.
For r = 1, one has
K n,m (1 · (λ0 , . . . , λn−1 )) = K n,m (λn−1 , λ0 , . . . , λn−2 )
= (0 · λn−1 + 1 · λ0 + · · · + (n − 1)λn−2 )n
= (0 · λ0 + 1 · λ1 + · · · + (n − 2)λn−2 + (n − 1)λn−1
+ (λ0 + · · · + λn−2 − (n − 1)λn−1 ))n
= K n,m (λ) + (λ0 + · · · + λn−2 + λn−1 − nλn−1 )n
= K n,m (λ) + 1 · (m)n .
For 1 < r < n, one has
K n,m (r λ) = K n,m (1 · ...(1· λ)...)) = K n,m (λ) + (m)n + · · · + (m)n
|
{z
}
| {z }
r times
= K n,m (λ) + r · (m)n
r times
2
Denote by pdn the natural projection from Cn to Cd ; for r ∈ Z/nZ = {0, . . . , n − 1},
= (r )d . Since d divides n, this is clearly a group homomorphism.
Define now the mapping
pdn (r )
K̃ n,m : 3̃n,m → Cd
by assigning to α ∈ 3̃n,m the element pdn (K n,m (λ)) ∈ Cd , where λ is any element of
−1
(α). Since d divides m, by Proposition 1 the element pdn (K n,m (λ)) does not depend on
πn,m
the choice of λ.
So one obtains a commutative diagram
K n,m
3n,m −→
πn,m ↓
Cn
pdn ↓
K̃ n,m
3̃n,m −→ Cd .
(5)
COMBINATORICS OF NECKLACES AND “HERMITE RECIPROCITY”
179
Proposition 2 Let α ∈ 3̃n,m and K̃ n,m (α) = r ∈ Cd . Then for any r 0 ∈ ( pdn )−1 (r ) there
−1
(α) ⊂ 3n,m which are mapped to r 0 by K n,m .
are exactly d/s(α) elements in πn,m
−1
(α). By commutativity of (5), clearly pdn K n,m (λ) = r .
Proof: Consider any λ ∈ πn,m
By Proposition 1, one has K n,m ((l · 1)λ) = K n,m (λ) + l · (m)n for any l.
Those l for which K n,m ((l · 1)λ) = K n,m (λ), are precisely those for which l · (m)n = 0 in
Z/nZ. Since (n, m) = d, those l must be divisible by n/d. So the minimal nonzero l with
K n,m ((l · 1)λ) = K n,m (λ) is n/d. Hence for any distinct l1 , l2 from the interval [0; n/d[
one has K n,m ((l1 · 1)λ) 6= K n,m ((l2 · 1)λ).
This implies, firstly, that t (λ) is divisible by n/d, i.e., t (λ)d/n = d/s(λ) is integer, and
secondly, since the inverse image of r under pdn has n/d elements, that when l runs over
the interval [0; n/d], then K n,m ((l · 1)λ) will become each element of the inverse image of
r under pdn exactly once. This means that, for each r 0 ∈ ( pdn )−1 (r ), the number of those
0 ≤ l < n with K n,m ((l · 1)λ) = r 0 equals d.
−1
(α) with K n,m (λ) = r 0 equals d/s(α).
2
Hence the number of those λ from πn,m
n
, whose elements are circular arrangements (“necklaces”) of
Consider now the set Wn+m
n + m beads, n of them black and m white. There has to be fixed orientation of the circle,
as well as a “basepoint” located somewhere between two adjacent beads.
n
n
be the subset of Wn+m
consisting of those arrangements for which the first bead
Let ϒn+m
along the orientation after the basepoint is black.
n
n
as follows: for β ∈ ϒn+m
and r ∈ Cn , rβ will be
Let us define an action of Cn on ϒn+m
the same arrangement as β but with the basepoint shifted counterorientationwise exactly
by the amount needed for the number of passed black beads to become r .
n
n
n
/Cn by ϒ̃n+m
; so ϒ̃n+m
is the set of arrangements as above, without any
Denote ϒn+m
basepoint, and considered up to rotation.
n
n
n
to ϒ̃n+m
will be denoted by πn+m
.
The natural projection from ϒn+m
n
For β ∈ ϒn+m , we denote by t (β) the minimal positive integer with (t (β) · 1)β = β.
n
and r ∈ Cn . The number n/t (β) will be denoted
Clearly t (β) = t (rβ) for any β ∈ ϒn+m
by s(β).
n
n
, the number s(β), for β ∈ (πn+m
)−1 (γ ) does not depend on the choice
For any γ ∈ ϒ̃n+m
of β; we will denote this number by s(γ ).
n
n
n
: ϒn+m
→ Cn . Take β ∈ ϒn+m
and suppose that numbers of
Let us construct a map gn+m
black beads in β, counted orientationwise from the basepoint, are 1, r2 , . . . , rn (by definition
n
(β) to be the element
the first bead is black). Then one defines gn+m
(1 + r2 + · · · + rn )n − (1 + 2 + · · · + n)n
of Cn .
n
n
n
Now construct the map g̃n+m
: ϒ̃n+m
→ Cd . Given any γ ∈ ϒ̃n+m
, choose a basepoint on
it somewhere between two adjacent beads. Suppose the numbers of the black beads counted
orientationwise w. r. t. this basepoint are r1 , . . . , rn . Consider the number r1 + · · · + rn .
If one would choose a different basepoint, each of the ri would change to ri0 in such a way
180
ELASHVILI, JIBLADZE AND PATARAIA
that (ri − ri0 )n+m would be the same for all i; denote this residue by 1. Then
(r1 + · · · + rn )n+m = (r10 + · · · + rn0 )n+m + n1
and pdn+m ((r1 + · · · + rn )n+m ) = pdn+m ((r10 + · · · + rn0 )n+m ) + pdn+m (n1) = pdn+m ((r10 +
· · · + rn0 )n+m ).
So we may define
n
= (r1 + · · · + rn )d − (1 + · · · + n)d .
g̃n+m
It is clear that the diagram
n
gn+m
n
−→ Cn
ϒn+m
n
πn+m
↓
pdn ↓
n
g̃n+m
n
−→ Cd
ϒ̃n+m
commutes.
Let us now construct a map
n
.
w : 3n,m → ϒn+m
For λ = (λ0 , . . . , λn−1 ) ∈ 3n,m choose an orientation and a basepoint on a circle; start
moving from the basepoint orientationwise and put the first black bead. Then put next λn−1
white beads and the next black one; again put λn−2 white beads and the next black one and
so on. On the (n − 1)-th step, when we will put λ0 white beads there will be n + m beads
arranged—the next one will be the black bead we started with. So one obtains an element
n
which we define to be w(λ0 , . . . , λn−1 ).
of ϒn+m
It is easy to see that w is a bijection compatible with the action of Cn . Moreover one has
the following proposition.
Proposition 3
The diagram
w
n
3n,m −→ ϒn+m
n
K n,m &
.gn+m
Z/nZ
commutes.
Proof: Take (λ0 , . . . , λn−1 ) ∈ 3n,m . Then in w(λ0 , . . . , λn−1 ), numbers of black beads
counted from the basepoint orientationwise will be
1, λn−1 + 2, λn−2 + λn−1 + 3, . . . , λ1 + · · · + λn−1 + n.
181
COMBINATORICS OF NECKLACES AND “HERMITE RECIPROCITY”
n
(w(λ0 , . . . , λn−1 )) =
Hence gn+m
= (1 + (λn−1 + 2) + (λn−2 + λn−1 + 3) + · · · + (λ1 + · · · + λn−1 + n))n
− (1 + · · · + n)n
= ((n − 1)λn−1 + (n − 2)λn−2 + · · · + λ1 )n
= (0 · λ0 + · · · + (n − 1)λn−1 )n
= K n,m (λ0 , . . . , λn−1 ).
2
Since the bijection w commutes with the action of Cn , it induces a bijection w̃ : 3̃n,m →
n
ϒ̃n+m
, and moreover by Proposition 3 there is a commutative diagram
πn,m
3n,m
A w
≈ AA
U
n
ϒn+m
K n,m
n
πn+m
Proposition 4
w̃ ≈
- ϒ̃ n
n+m
K̃ n,m
(*)
C
n
C
g̃n+m
pdn
Cn
3̃n,m
C
n
gn+m
? -
C
CW
-
?
Cd
For any r ∈ Cn ,
X
¡ −1 ¢
(r ) =
# K n,m
n
γ ∈(g̃n+m
d
.
s(γ )
)−1 ( p n (r ))
d
Proof: This is obvious from Proposition 2 and commutativity of (*).
2
Let us now construct the map
n
m
x : ϒ̃n+m
→ ϒ̃n+m
n
m
as follows: for γ ∈ ϒ̃n+m
, define the element x(γ ) ∈ ϒ̃n+m
by reversing the orientation
and changing black beads by white ones and vice versa.
182
ELASHVILI, JIBLADZE AND PATARAIA
Proposition 5
The diagram
x
n
m
ϒ̃n+m
→ ϒ̃n+m
n
m
&
.
g̃n+m
g̃n+m
Z/dZ
commutes.
n
Proof: Take any γ ∈ ϒ̃n+m
and choose in it a basepoint between some adjacent beads.
Choose the same basepoint in x(γ ). Suppose that in γ the numbers of black beads are
b1 , . . . , bn and those of white ones are c1 , . . . , cm . Since each bead is either white or black,
b1 + · · · + bn + c1 + · · · + cm = 1 + · · · + (n + m),
i.e.,
c1 + · · · + cm = 1 + · · · + (n + m) − (b1 + · · · + bn ).
(6)
Numbers of black beads in x(γ ) w.r.t. the new orientation will be n + m + 1 − c1 , . . . , n +
m + 1 − cm . Hence
m
g̃n+m
(x(γ )) = (n + m + 1 − c1 + · · · + n + m + 1 − cm )d − (1 + · · · + m)d
= (m(n + m + 1) − c1 − · · · − cm )d − (1 + · · · + m)d
= −(c1 + · · · + cm )d − (1 + · · · + m)d
(d | m)
= (b1 + · · · + bn )d − (1 + · · · + (n + m))d − (1 + · · · + m)d (by (6))
¶
µ
(n + m)(n + m + 1) m(m + 1)
+
= (b1 + · · · + bn )d −
2
2
d
µ
¶
n(n + 1)
= (b1 + · · · + bn )d − m(n + m + 1) +
2
d
¶
µ
n(n + 1)
= (b1 + · · · + bn )d −
(d | m)
2
d
= (b1 + · · · + bn )d − (1 + · · · + n)d
2
n
= g̃n+m
(γ ).
We have reached the goal of this section.
Proof of ak (n, m) = ak (m, n):
By Proposition 4,
¡ −1 0 ¢
(n ) =
# K n,m
X
n
γ ∈ ϒ̃n+m
n
g̃n+m
(γ )=d 0
Take any n 0 ∈ Cn and m 0 ∈ Cm with (n 0 )d = (m 0 )d = k.
d
.
s(γ )
COMBINATORICS OF NECKLACES AND “HERMITE RECIPROCITY”
183
n
m
→ ϒ̃n+m
one obtains
Hence using the isomorphism x : ϒ̃n+m
X
¡ −1 0 ¢
(n ) =
# K n,m
n
γ ∈ ϒ̃n+m
n
g̃n+m
(γ )=d 0
d
=
s(γ )
X
=
n
γ ∈ ϒ̃n+m
m (x(γ ))=d 0
g̃n+m
X
n
γ ∈ ϒ̃n+m
n
g̃n+m
(γ )=d 0
d
=
s(x(γ ))
d
s(x(γ ))
X
m
δ ∈ ϒ̃n+m
m (δ)=d 0
g̃n+m
¡ −1 0 ¢
d
= # K m,n
(m ) .
s(δ)
It follows that
¡ −1 0 ¢
¡ −1 0 ¢
ak (n, m) = # K n,m
(n ) = # K m,n
(m ) = ak (m, n).
3.
2
Another proof of the formula
Our next task will be to obtain another derivation of the formula 4,
ak (n, m) =
n+m
1 X
0 !
cd 0 (k) n d m ,
n + m d 0 |d
! !
d0 d0
of a more combinatorial nature.
n
, under which r ∈ Cn+m acts by shifting the
Consider the action of Cn+m on the set Wn+m
n
(k), for k ∈ Cd , be the subset of
basepoint by r beads counterorientationwise. Let Wn+m
n
Wn+m consisting of those elements with
Ã
n
X
ri −
i=1
n
X
i=1
!
= k,
i
d
where r1 , . . . , rn are numbers of black beads counted orientationwise from the basepoint.
Proposition 6
n
.
Wn+m
n
n
The subset Wn+m
(k) ⊂ Wn+m
is invariant under the action of Cn+m on
n
Proof: Take r ∈ Cn+m , and let γ be an element of Wn+m
(k), numbers of black beads of γ
n
will be r −ε1 (n +m)+r1 ,
being r1 , . . . , rn . Then the numbers of black beads in r γ ∈ Wn+m
r − ε2 (n + m) + r2 , . . . , r − εn (n + m) + rn , where each εi is 0 or 1 depending on whether
ri + r < n + m or not (in other words, numbers of those beads not passed by the basepoint
will grow by r , while of those passed—by r − (n + m)). Hence the sum of numbers of
black beads will become
n
X
i=1
r − εi (n + m) + ri = nr − (n + m)
n
X
i=1
εi +
n
X
i=1
ri .
184
ELASHVILI, JIBLADZE AND PATARAIA
So since both n and n + m are divisible by d, the sum of numbers of black beads will remain
the same modulo d.
2
n
n
n
n
n
(k)/Cn+m by W̃n+m
(k). Clearly W̃n+m
(k) = {γ ∈ ϒ̃n+m
| g̃n+m
(γ ) = k}.
Denote Wn+m
n+m
n
n
The inverse image in Wn+m (k) of each γ ∈ W̃n+m (k) has s(γ ) elements, since the pattern
and there are n+m
possibilities to choose
of the arrangement γ is periodic with period n+m
s(γ )
s(γ )
the basepoint. Hence
¢
¡ n
(k) =
# W̃n+m
X
n
γ ∈ ϒ̃n+m
m (γ )=k
g̃n+m
n+m
.
s(γ )
Comparing this with the formula from Proposition 4,
¡ −1 0 ¢
# K n,m
(n ) =
X
n
γ ∈ ϒ̃n+m
n
g̃n+m
(γ )=k
d
,
s(γ )
one concludes
¡ −1 0 ¢
(n ) =
# K n,m
¢
¡ n
d
(k) .
# W̃n+m
n+m
(7)
n
(k)).
Let us calculate #(Wn+m
Proposition 7
¢ 1X
¡ n
(k) =
cd 0 (k)
# Wn+m
d d 0 |d
Proof:
Ã
n+m
d0
n
d0
!
.
Consider the polynomial
P(x, t) = x
n(n+1)
−k
2
(t + x)(t + x 2 ) · · · (t + x n+m ).
After expanding P(x, t) and collecting the terms, the coefficient at the monomial t m x l will
n(n+1)
be the number of representations of t m x l = x 2 −k t m x r1 x r2 · · · x rn , where 0 < r1 < · · · <
rn ≤ n + m with
X
n(n + 1)
r j = l.
−k+
2
j
Rewrite the last equality as
X
j
rj −
n(n + 1)
− k = l − n(n + 1);
2
COMBINATORICS OF NECKLACES AND “HERMITE RECIPROCITY”
185
since d divides n(n + 1), it is clear that the sum of these coefficients at t m x l over all l with
d | l will equal the number of those sequences 0 < r1 < · · · < rn ≤ n + m with
Ã
X
j
n(n + 1)
rj −
2
!
= k.
d
n
Assigning to such a sequence an arrangement from Wn+m
with r1 , . . . , rn as numbers of
n
(k)).
black beads, shows that the sum of these coefficients equals #(Wn+m
2πi
Let ζ = e d (or any other primitive dth root of 1). Since for any integer l one has
d
X
(
if d | l
otherwise,
d
0
ζ =
lj
j=1
n
(k)) equals the coefficient at t m of the polynomial
it follows that #(Wn+m
P(t) =
d
1X
P(ζ j , t).
d j=1
Consider now the polynomials P(ζ j , t) separately.
Since d | n, one has
(
ζ
i.e., ζ j
j
n(n+1)
2
n(n+1)
2
=
1
−1
= (−1)
if d is odd or both d and
if d is even and
(d−1)nj
d
P(ζ j , t) = (−1)
nj
d
nj
d
are even,
is odd,
. Hence
(d−1)n j
d
ζ − jk (t + ζ j )(t + ζ 2 j ) · · · (t + ζ (n+m) j ).
Denoting −t by s, one obtains
j)
¡
¡
¢¢ (n+m)(d,
d
d
ζ − jk (−1)n+m (s − ζ j ) · · · s − ζ (d, j) j
j)
¢ (n+m)(d,
¡ d
(d−1)n j
d
= (−1) d +n+m ζ − jk s (d, j) − 1
.
P(ζ j , t) = (−1)
(d−1)n j
d
Now collect together the terms of P(t) =
d. One obtains
P(t) =
1
d
P
P(ζ j , t) whose j’s have the same gcd with
X
¢ (n+m)d 0
(d−1)nj
(−1)n+m X ¡ d0
(−1) d ζ −jk .
sd −1 d ·
d
d 0 |d
( j,d)=d 0
186
ELASHVILI, JIBLADZE AND PATARAIA
The sign term in the last sum is
(
(−1)
(d−1)nj
d
−1
=
1
(
−1
=
1
= (−1)
if d is even and both
n
d
and j are odd,
n
d
and d 0 = ( j, d) are odd,
otherwise,
if d is even and both
otherwise,
n
(d−1) d/d
0
.
Hence
P(t) =
X
(d−1)n
(n+m)
(−1)n+m X
0
(−1) d 0 (s d − 1) d 0 ·
ζ − jk
d
d
d 0 |d
( j,d)= d 0
=
=
(−1)
d
n+m
X
(−1)
(d−1)n
d0
0
(s d − 1)
(n+m)
d0
cd 0 (−k)
d 0 |d
(d−1)n
(n+m)
(−1)n+m X
0
0
(−1) d 0 cd 0 (−k)((−1)d t d − 1) d 0 .
d
d 0 |d
So the coefficient at t m in P(t) will be
¡
#
Since
d
d0
n
(k)
Wn+m
¢
(d−1)n
(−1)n+m X
n
0 m
=
(−1) d 0 cd 0 (−k)(−1)d d 0 · (−1) d 0
d
d 0 |d
!
Ã
n+m
(−1)n X
dn
d0
0
d
=
.
(−1) cd 0 (−k)
n
d d 0 |d
d0
n+m
d0
n
d0
!
dn
| n, the sign term in that last sum is (−1) d 0 = (−1)n . Hence
1X
n
# Wn+m
(k) =
cd 0 (k)
d d 0 |d
¡
Ã
¢
Ã
n+m
d0
n
d0
!
(we have used the evident equality cd 0 (−k) = cd 0 (k)).
2
Remark A similar argument for deriving a0 (n, m) has been communicated to us by Alon
and Stanley: for the role played by our P(x, t), they use the function (1 − t x)−1 (1 −
t x 2 )−1 · · · (1 − t x n−1 )−1 .
187
COMBINATORICS OF NECKLACES AND “HERMITE RECIPROCITY”
−1
(k)) one
We have reached the desired formula: indeed, from (7) and ak (n, m) = #(K n,m
immediately obtains
1 X
cd 0 (k)
ak (n, m) =
n + m d 0 |d
4.
Ã
n+m
d0
n
d0
!
.
Some Final Remarks
Generating functions for a0 (n, m) mentioned in [3] can be easily generalized: for any
integer k one has
∞
X
ak (n, m)x n y m = 1 −
n,m=0
∞
X
cd (k)
d=1
d
log(1 − x d − y d )
and
ζk (s + t)
∞
X
ak (n, m)n −s m −t = ζ (s + t + 1)
n,m=1
∞
X
(n + m − 1)! −s −t
n m ,
n! m!
n,m=1
P
where ζk (x) = d|k d −x (in particular ζ0 (x) = ζ (x) is the Riemann zeta function).
Let us note also a relationship of ak (n, m) to free Lie algebras. Consider a free Lie
algebra Lie(x, y) on two generators x, y over a characteristic zero field. Then Th. 2(b) in
[2], Ch. II, Section 3, no. 3 immediately implies that for any n, m,
a1 (n, m) = dim(Lie(x, y)n,m ),
where ( )n,m denotes the homogeneous component of bidegree n in x and m in y. We could
not find an explicit correspondence between generators of the above Lie algebra and our
“necklace” interpretation of a1 (n, m). This connection looks even more interesting in view
of the fact that in a sense, all the ak may be reduced to a1 .
Proposition 8
For any n, m, and k,
X
a1 (n/d 0 , m/d 0 ).
ak (n, m) =
d 0 |(n,m,k)
Proof:
One calculates directly:
X
X
1
a1 (n/d , m/d ) =
0 + m/d 0
n/d
d 0 |(n,m,k)
d 0 |(n,m,k)
0
0
00
µ(d )
d 00 |(n/d 0 ,m/d 0 )
X
1
=
d 0 µ(d/d 0 )
n
+
m
0
d|(n,m)
d |(d,k)
X
Ã
X
Ã
n+m
d
n
d
!
n+m
d 0 d 00
n
d 0 d 00
!
188
ELASHVILI, JIBLADZE AND PATARAIA
X
1
=
cd (k)
n + m d|(n,m)
Ã
n+m
d
n
d
!
= ak (n, m).
2
We do not know a combinatorial explanation of this fact, either.
References
1. G.E. Andrews, “The theory of partitions,” Encyclopedia of Mathematics and Its Applications, G.-C. Rota,
(Ed.), Addison-Wesley, Reading, MA, 1976, Vol. 2.
2. N. Bourbaki, Groupes et algébres de Lie, Hermann, Paris, 1971–1972.
3. A. Elashvili and M. Jibladze, “Hermite Reciprocity for regular representations of cyclic groups,” Indag Mathem.,
N.S., 9(2), 1998, 233–238.
4. Higher Transcendental Functions, Bateman Manuscript Project, A. Erdélyi (dir.), McGraw-Hill, New York,
1955.
5. S. Ramanujan, “On certain trigonometrical sums and their applications in the theory of numbers,” Trans. Camb.
Phil. Soc., XXII(13) (1918), 259–276
6. T. A. Springer, Invariant theory, Lecture Notes in Math., Vol. 585, Springer-Verlag, Berlin and New York, 1977.
7. R. Stanley, Enumerative Combinatorics, Vol. I, Wadsworth & Brooks/Cole, Monterey, CA, 1986.