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Unformatted text preview: Combinatorial aspects of the theory of qseries
P. R. Hammond
Submittedfor the degreeof D. Phil.
University of Sussex
June, 2006 C2oos] Declaration
I herebydeclarethat this thesishas not beensubmitted,either in the sameor different
form, to this or any other university for a degree. i_/ ý'y 1 Acknowledgements
Firstly I would like to express my deep gratitude to my supervisor, Dr. Richard Lewis.
Without his support, help and patience it could not have been written. I would also like
to thank the all other mathematicans who have helped me along the way, in particular
Roger Fenn, James Hirschfeld and Gavin Wraith, all of whom were given the task of
reading my yearly reports, came to my seminars on Partitions and were extremely
supportive all along the way. As well as these Sussex mathematicans, I would also like
to thank George Andrews for his encouragement with many aspects of my work. My
thanks also goes to Tom Armour, Sue Bullock, Richard Chambers and Christine
Glasson, all of whom were a great help. I would like to thank James, Jon, Linus,
Richard, Stevie, Toby and all the many others who helped ensure I had a memorable
time as a student at Sussex. My final word of thanks is to my parents, Keith and Helen
Hammond, for their constant support, understanding and love. This thesisis dedicatedto the memory of Colin and Diana Carothersand Leslie and
Muriel Hammond,my grandparents. Abstract
This thesis is concerned mainly with the interplay between identities involving power series (which are called qseries) and combinatorics, in particular the theory of partitions.
The thesis includes new proofs of some qseries identities and some ideas about the generating functions for the rank and crank, a new proof of the triple product identity and a
combinatorial proof of a qelliptic identity. Contents
1 Introduction 3 1.1 Partitions
1.1.1 1.2 3 .................................. Partitions and sets of partitions
....
...
1.1.2 The graph of a partition
.......................
Power Series
...
. .........
............ ... 3 ......... 4
.. ..... 1.2.1 Generatingfunctions and notation
.................
1.2.2 The qbinomial coefficients
.....................
1.2.3 Partitions and qseries
.......................
1.3 A bit of history
...............................
1.3.1 ElementaryIdentities
........................
1.3.2 Congruences the Euler product
in
.................
2 5 5
7
8
10
10
12 Franklin 14 2.1 14 and Sylvester
The Franklin bijection
2.1.1 2.2 3A .....
. ........
The generating function for p(' Sylvester's bijection .....
n)  p(D,, n)
, .. .......
.... ... .. 19 ............................. combinatorial approach to some partition identities
3.1
3.2
3.3
3.4 18 The three identities
..........
.......
A combinatorial proof of a result of Fine
......
A different approach
. .....
..........
The second and third identities
........... 3.4.1 22
...
...
... ....
.......
....... .......... Two qbinomial identities
.....................
3.4.2 Nestedpartitions
..........................
3.5 Two identities, one old and onenew
....................
3.5.1 An identity of Subbaroand Vidyasagar
..............
3.5.2 A new identity
........................... ... ..
..
..
.. 22
23
25
27 27
28
32
32
35 4 An involutive proof of the triple product identity 36 5 Ranks and biranks
5.1 A generalisationof an identity of Fine 46 I ................... 46 5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6A 5.1.1 The aboveidentity and partitions ..................
The rank generatingfunction ........................
The generatingfunction ...........................
Ranks
....................................
The Birank
.................................
The behaviourof the birank
.........................
RelatedIdentities
..............................
Other Ranks
.................................
Summary
.................................. 47
47
49
52
53
55
65
67
68
69 qelliptic identity
6.1 The identity. ................................
6.2 The caseN=3...............................
6.3 The caseN=4............................... 69
70
74
77 Bibliography 2 Chapter 1 Introduction 1.1 Partitions
1.1.1 Partitions and sets of partitions
A partition is a sequence of positive integers, for example each of the four sequences
(4,4,1), (8,5,3,2), (6,6,6) and (1) is a partition. Repetition, as in the first and third
of these four sequences, is allowed. What is not permitted is for the sequence to be
increasing, for instance (1,4,4) is not a partition, nor is (4,1,4). The Greek letter A is
A2 =4 and A3 =1.
used to denote a partition, if A= (4,4,1) then A=4, Thus a partition A is defined as being a finite, nonincreasingsequenceof positive
integers: A= (A1,A2, Ak), k is therefore(the letter usedto denote)the numberof parts,
..,
which can also be written as k= #(A), some authors use BAIfor the number of parts.
The partition A is said to be composedof, or consist of, the parts, or entries, Ai (where
1<i<
k). The weight of a partition, wt(A), is defined to be the sum of the parts of A,
someauthorsuse IIAII for the weight, or we may say simply "a partition of n" insteadof
"a partition of weight n" (this may also be written as A I n). For exampleA= (4,4,1)
is a partition of 9 having three parts. The empty partition, 0=(. ), has weight zero and is
composedof zero parts. 3 The set of all partitions, P, occurs frequently in the thesis and so do certain subsets of
it. Of particular interest are V. De, D,, and 0, where
{A EP: Ai > Ai+lthe
D
set of partitions into distinct parts,
{A ED De #(A) 0 mod 2}, the set of partitions into an even number of distinct parts, #(A)
Do := {A EV:
1 mod 2}, the set of partitions into an odd number of
distinct parts,
d) := {a EP: Ai 1 mod 2}, the set of partitions into odd parts.
The empty partition is an element of each of the sets P, D, De and 0 but is not in Do.
If H and H' are two sets of partitions and 0: H
H' then the map 0 is said to
+
be weight preserving if wt(q5(A)) = wt(A), for all AEH.
In this thesis any bijection
between two sets of partitions is assumed to be weight preserving, unless it is stated
otherwise.
Whenever H, say, is a set of partitions and there are m sets such that H is the disjoint
H1 U H2 U
U H, will be called a decomposition of H
union of the Hi then H=
...
(and where the union is not disjoint the word decomposition will not be used). Thus, for
example, the set of partitions into distinct parts can be decomposed into D= De U Do.
For any HCP,
the number of partitions in H of weight n will be written as p(H, n).
For n¢N,
the convention is p(H, n) :=0 (the set N is understood to include 0, also
used is the notation N* :=N\ {0}). The total number of partitions of weight n is written
simply as p(n), or p1 (n), as is explained in section 1.2.3. The function p(n) is called the
partition function.
sign, :, is used to indicate that an entry has been omitted.
Thus (5,5,3,3,2) = (5,5,3,2), (7) = 0, (Al) A2, ýj,
A) is A with ., omitted and
...,
...,
(A1, A2) £i, :., ýj,
A, ) is A with parts Ai to Aj omitted (which could, of course, be
...,
...,
written as (al, A2, """I Ai1) A +1, """, )k), but the hat is used to emphasise the omission).
The hat 1.1.2 The graph of a partition
Any nonempty partition can be visualised as the set of rows of coordinates in the bottom right quadrant of the plane where the ith row contains ) entries. This is called the
graphical representation or the Ferrers graph (or simply the graph), Ga, of the partition.
Before describing this explicitly, an example will help to illustrate the idea: If A= (4,4,1)
then 9a is
0o00
0e00
0 (in order to be completely unambiguousit would be necessary distinguish betweenthe
to
{(1, 1), (2) 1), (3, 1), (4, 1), (1, 2), (2, 2), (3, 2), (4,
(1, 3)} and the
set
2),
diagram above,but it is clear what is happening).
4 partition A= (Al, A2, ..., Ak), for each i in the range
given a nonempty
The formal definition is that
0<i <_Al let SA(i) be the maximum 1 for which a, i>0.
is the set of dots with integer coordinates (i, j) in the plane such that (i, j) E ýa
the graph
Now, Al and c5a(i) <j<0.
if and only if 0<i<
The dual, or conjugate, of a partition is the partition whose graph is the one
by rotating the graph of the original partition around the falling diagonal (the
obtained
(2,
}). Visually, the dual is obtained by interchanging each row with
{(1)
set
2), ...
1),
its corresponding column in 9a. In terms of the set, ga" :_{ (i, j) : (j, i) E 9a}. The
dual of A will be denoted by A'. In the case A= (4,4,1) the dual is A' = (3,2,2,2). It
follows immediately that the dual induces a weight preserving involution on P,
A" =A for any AEP.
If A is such that A' =A then A is said to
wt(A') = wt(A) and
be a selfdual, or selfconjugate, partition. Thus, for example, (4,4,1) is not selfdual but
(4,1,1,1)is. 1.2 Power Series
1.2.1 Generating functions and notation
Given a sequence {a}, it is often desirable to find an expression, F(q), whose coefficients
i. e. F(q) = a°+a, q+... +a; q' +.... When this happens
are the elements of the sequence,
F(q) is said to be the generating function for the sequence {a, '}. For example,
the qseries
100 and a,, =0 for all other n. The generating for
0<n<
suppose a =1 whenever
this sequence is (1  q10°)/(1 q). A power series in the parameter q is called a qseries.
The following standard notation will be used frequently throughout:
For z and qEC, define
n (z; q)n := f[(1 zq1), nE Iii"
 and 00
(z; q)... :=11(1  zý1)
i=l
(the condition lql <1 is assumedin a nonterminatingpower series,this ensuresconverIn this thesis, the subscriptn is assumed be a nonnegativeinteger,which need
to
gence).
actually be the case.
not 5 The identity
(z; q)n =
((q"q4). (1.1) holds for any nE N*, and leads by extensionto a definition of (z; q),, for any
clearly
In particular, (z; q)o = 1. Form E Z, the sequence (n) is definedby
p,.,,
real n.
, (q; q)«.
pm(n)g' := (1.2) nEZ It follows that, for any m, pm(n) =0 whenevern¢N
(in fact the subscript m need
not evenbe an integer, but in practice it always is). The product 1/(q; q)a, is sometimes
written asP(q) or simply as P (which should not be confusedwith the set P).
The squarebracket notation is defined,when z 0, as
[z; q] := (z; q)ao(z 14; 4)00. It is elementary,
provided that z#0, that
1[z; q]
z (1.3) and that
[zq; 4] =
z1[z; Q]. (1.4) Now, an expressionin z and q (such as (z; q),,, for somegiven n) can be seenas a power
in
seriesin q, whosecoefficientsare expressions z, for example,
(z; q)3 = (1 z) + (z + z2)q + (z
(z2 z3)g3
} z2)g2 I in
and it can also be seenas a power seriesin z whosecoefficients are expressions q,
(z; (1)3 =1 (1+q+g2)z+(q+q2+q3)z2g3z3. This suggestsit might prove helpful to find a general expressionfor (z; q)n as a power
seriesin z, so the questionis, what is the coefficient of z=in (z; q)n? With this in mind it
is now helpful to introduce the qbinomial coefficients. 6 1.2.2 The qbinomial coefficients
The qbinomial coefficients are defined as
[i (q; q)n q (1.5) = (q;
4)i(q; 4)ni The subscriptq can be droppedsafely from the expressionon the left, exceptin chapter3
wherethe subscriptis q2. The qbinomial coefficientsare introducedat this stagebecause
they appearin the expressionfor (z; q),,, but before addressingthat a few properties of
the qbinomial coefficients are worth mentioning: Clearly,
[n] IL (q; q)n
_ (q; q)i (q; q)ni Jq i+l; 9')i (q; 9')ni(qn (q; q),,
q)=(4; i [n] LJ _ (4=+1; 4)=i (4;
q)1 Now, it can easily be shownthat
[n
i] [n (1.6) 1]+qn_=[2Z 1]. (1.7) This is (3.3.3) in [2], and it follows from this identity that, when i and n are both nonnega[_] is in fact a polynomial, of degree
P. The qbinomial
integers and 0 <_i<n,
tive
ni [n"
Gaussian polynomials. Note that the identity [_
coefficients are also called
i]
follows from the definition (1.5) of the qbinomial coefficients.
Closely related to the role the qbinomial coefficients play in the expansions for (z; q)
n
and (z; q); 1, that is in identities (1.9) and (1.10) below, is the fact that they are the generating functions for partitions where both the largest part and the number of parts are
forbidden to exceed a given pair of nonnegative integer values, a and b, say. This is dealt with in the following
Lemma:
Let U(a, b) := {A : al < a, #(A) < b}, thus 1. b) is the set of all partitions whose
ß(a,
graphlies inside a rectangleof length a and height b. Then
Ia + bl
gwt(A) _
L
AEU(a,
b) aJ . (1.8) Briefly, the proof involves the inductive step that firstly [°+ä1] should be the generating
function for partitions having less than b parts, none of which exceed a (i. e. partitions in
U(a, b 1)), and secondly qb[ °4 6ý1] should be the generating function for partitions into
b parts, no part exceeding a (clearly (1.8) holds whenever either a or b is 0).
precisely
This is explained in (3.4.1. ) in [2] and also in [16]. 7 It follows from (1.8) that the qbinomial coefficients are symmetric, in the sense that
if [_]= a(O) + a(1)q +
+ a(ni  i2)q"h12 then a(k) = a(ni  i2  k). It has also
...
[16], that the they are unimodal, which is to say that if k< !L' ' then
been shown, in
a(k  1) < a(k).
Furthermore,
Qm[Z] ) _\ ji the familiar binomial coefficient, as maybe seenby applying l'Hopital's rule to (1.6).
The following identity, due to Euler, statesthat the coefficient of zi in (z; q)n is
(1)z [_] q(=a=)/2,
n
[n]
E(1)'z=
(z; q)n = i q(i2_i)/2 (1.9) =_o and the coefficient of z=in the reciprocal is given by a result of Rothe,
10 n+i1 (z; 4)n (1.10) Li. o For a proof of the above two identities, see Theorem 3.3 in [2], or see [3] where they are
derived from the qbinomial theorem.
Now, since (q; q), can be defined for any real n, it follows that [_] can be defined
a
for i and n any real numbers. In fact the sums on the right in (1.9) and (1.10) can be
replaced by sums with i ranging over all integers, for if n is a nonnegative integer and
{0,1, then a limiting argument applied to (1.5) and (1.1) gives zl=0
n}
...,
(which is to be expected, as (i) =0 for all such i).
The limiting case as q 4 1 in identity (1.9) is the familiar expansion for (1  z)" and
likewise, q 4 1 in identity (1.10) gives the expansion for (1  z)n. This illustrates a iEZ\ general principle, the limit as q  1 in any identity involving qbinomial coefficients is
an identity involving ordinary binomial coefficients. 1.2.3 Partitions and qseries
It is now possibleto give an outline of the connectionbetweenqseriesand partitions. For
instance,if H is the set {0, (1), (1,1), } CP of partitions with eachpart equal to 1 then
...
1=1+ q =1+q+g2+q3+...
ql + q1+1 + g1+1F1 + 1: ... q40'), AEH What this meansis that (1  q)1 is the generatingfunction for p(H, n),
E p(H, n)gn n>O S More generally, if H were to be the set of all partitions composedof elementsfrom a
{al, a2
all C N*, then it would follow that
given subset
i ...,
ý1
qýKa) 1
i=1 AEH which is to say that
ý T((1
n)qn = li=1 p(H, q°ý,)1
 n>O This includes the case H is the whole of P (as can be seenby taking d= oo and, for
example,ai = i) so
00
1E
gwt(A)
ý1
)
Q:
 i_1 AEP which is to say that
Ep(n)q' (4; 9)'
=
. (1.11) n>O Putting m= in (1.2) givesp(n) = p_1(n). In fact, pm(n) is the number of ordered
1
mtuples of partitions,
(4', 4)m,, AE
In particular, if m=
then p_2(n) is the number of ordered pairs of partitions. The
2
behaviour of p_2 (n), in particular some of its congruences (and an explanation for these
congruences, namely the birank) is the subject of the second half of chapter 5.
Some subsets of P have already been defined. Concerning these, it is not hard to show
that 00 II (1
i=1 which is to say that 1E =q
g211) wt(A) AEO (q; q2)P(O, n)4" =
00 (1.12) n>O and 00 T(ý1 +
qi) _
i1=11 gwt(a)
AED is to say that
which
Ep(D, (4; 4')00
n)4" = (1.13) n>O
finally, pl (n) = p(D., n) p(D0, n), or equivalently,
and
 (p(D,, n) P(Do, n))q"
(q; q)
(1.14)
=
.
n>O
So (q; q), maybe called, with a slight abuseof terminology, the generatingfunction for
partitions into distinct parts where thosehaving an odd number of parts are countednegatively. 9 1.3 A bit of history
1.3.1 Elementary Identities
In the mid eighteenthcenury Euler observedthat
00 (1qz)=1qg2+q5+q7q12q15+.... The product fl j° is sometimescalled the Euler product (the notation (q; q),, is a
1(1 q=)
twentieth century development).
He went on to demonstrate
that
00 (1.15)
i=1 MEZ This is called the pentagonal number theorem, because a number n is said to be pentagonal
if and only if n=m 32+1 for some integer m. Euler was probably aware that (q; q)oo is
the generating function for p(De, n)  p(D0, n) but there is no evidence to suggest that he
considered looking for a combinatorial proof that
(1)m'
p(D., n)  p(D0, n) =2 3m}1 if n=m (1.16) if n is not pentagonal 0 Franklin was the first to find one (Legendre apparently suggested that there might be a
and
combinatorial proof, though he did not give a proof himself). Franklin's proof is described
in section 2.1. Briefly though, Franklin's proof involves decomposing V into the union
of three sets. Between the first two sets there exists a bijection (or at least a bijection can
32+1 for
be defined) and the third set contains no partitions of weight n unless n=m
some integer m, whence it contains exactly one partition of weight n (which has an even
number of parts precisely when m is even, as will be explained).
Franklin's proof, as outlined above, involves pairing off elements of a given set and
counting those that remain is an instance of a proof by involution. In chapter 4a new proof
of the Jacobi triple product identity is presented. This proof also involves an involution.
The triple product identity states
[z;
f2 57(l)neq
q](q; q),,,, 2n = (1.17) nEZ This is not to say that involutions are the only approach to qseries identities. For
example given certain identities it is possible to find a bijection from the set whose generating function is the expression on the left to the set with generating function the set on
the right, as opposed to an involution which `stays on one side of the equation'. 10 As an illustration of this, consider
00 (q; q)00 = ý(1
ti + q`) 00(1_gai) l_ll(1q=)
°° (1 q2,
)
(1  qaii)(1  q21) 00 1 ýý1_g2i1)
so (4; Q)001
(4'3
4'2)00 (1.18) Thus by (1.13) and (1.12), p(D, n) = p(O, n). This suggests that there might a weight
preserving bijection between the set of partitions into distinct parts and the set of partitions
into odd parts. In fact, there is more than one. One of these, the bijection that was found
by Sylvester, is described in section 2.2 and is related to the bijection in section 3.1.
Now, having established that p(D, n) = p(O, n), it is clear that p(O, n) 0 mod 2
if and only if n is not a pentagonal number (by virtue of (1.16), and the obvious fact that
2). Hence it seems reasonable to ask if the set 0 can
p(D«, n)  p(D0, n)  p(D, n) mod
be decomposed into two sets in such a way as to ensure that the number of elements of
O (i. e. partitions, into odd parts) of weight n in one of the sets is the same as the number
in the other set, unless n is a pentagonal number (just as V can
of elements of weight n
be decomposed into two such sets, De and Do, which is where the identity (1.16) comes
from).
The question is then, what property of partitions in 0 is it that `almost half' of
them have that could give rise to an involution on 0? The Franklin involution reverses
mod 2 the parity of the number of parts of a partition in D. As, clearly, the first part of
any nontrivial partition in 0 has first part congruent to 1 mod 2, the proposed involution
on 0 clearly will not have this effect. The Franklin involution on V also has the effect
of reversing the parity mod 2 of the number of parts of any partition in D on which it
acts. This cannot be the case for the involution on 0, as for any partition aE0,
the number of parts in the partition has the same parity as the weight of the partition:
#(A) = wt(A) mod 2.
Instead it transpires that the place to look is the residue of the first part, of a partition
in 0, mod 4. This to say that an involution on 0 is presented that has the effect of taking
partitions with Al 1
in section 3.1. mod 4 to those with al 3 11 mod 4, and vice versa. This is done 1.3.2 Congruences in the Euler product
In 1918 the Indian mathematican Ramanujan wrote a letter to Hardy. The letter included,
amongst other things, the following congruences in the partition function which he (Ramanujan) had conjectured, p(5n + 4) =0 mod 5, (1.19) p(7n + 5) =0 mod 7, (1.20) p(lln + 6) 0 mod 11, (1.21) of which he later gaveproofs, which can be found in [29].
The first two of theseare entirely elementary,the following is a sketchof a proof of
(1.19): Dividing both sidesof the triple product identity (1.17) by 1z gives
[Z; 4] E(1)nzn (4ýq')00=11z nEZ
1z 1z (z 1z2) 1z q+3 2z3ý lz 1z 3z4) (z g 6+ 1z which is equal to
1(z+1+z1)q+(z2+z{ 3)q6+... i+Zl+z2)q3(z3+Z2+Z+1+Zl+z2+z and it follows that
(z4; q)oo(z 1q; 4)ß(q; q)oo = E(1)"(z n+ zn1 +
+ zn)q
... n
ný2 (1.22) n>0 and then setting z=1 in (1.22) gives,
(q; q) 0_ ý(1)(2n 1)qn a
+ (1.23) n>o
Two power series, ao + alq + a2q2 +... and b0+ blq + b2q2+..., are said to be congruent
modulo r if and only if a= = b=mod r for all i. Now for any prime r, (1 q)'' =1r by the binomial theorem, so for
qr mod
(1  qz)r _ (1 q=r)mod r. Taking products over all i in the positive
i=1,2,...
'
integersgives
r is prime = (q; q) ,= 12 (q''; q")... mod r. (1.24) The generating function for p(n),
>p(r )qn =1 n>O (q;q),, togetherwith (1.24) gives,
Ep(n)gn _1 (qi n>0 (4; 4)00
(45' X5)00
Q)00 mod 5 so p(n)q" _
n>0 (4'; 4)ö0(4; q)ý.
(q5; q5)00 mod 5 and it follows, by (1.23) and (1.15), that
p(n)qn = (ao(g5) + gai(q5)) (/3o(95) + q, (q5) + g2#2(45))
81
(Q5i q5) n>0 mod 5 /.
where ao, al, 13o,31and Q2are five particular power series (which can of course be determined explicitly, but it's not necessary to actually do this). The point is that if t(n), say,
is the coefficient in the above expression then clearly t(5n + 4) 0 mod 5. Not only did Ramanujanconjecturethe congruences
(1.19), (1.20) and (1.21), he also
madeconjecturesfor higher powers of 5,7 and 11. He conjecturedthat
x=5or 11= p(n)=0mod x', if 24n=1mod (1.25) x'. Ramanujan was correct to conjecture (1.25). In fact he conjectured that (1.25) was true
for x=5,7
and 11, which is not actually the case. Chowla noticed that whilst p(243) is
divisible by 49, it is not divisible by 343 (so (1.25) fails for m=3,
The correct statement for powers of 7 is if
p(n) =0 mod 7Lm/21+1, 24n 1 if x= mod 7"`. 7). (1.26) The bracketshere denotethe floor function (the greatestinteger no more than m/2), i. e.
Lm/2i is m/2 if m is evenor (m 1)/2 if m is odd.
The corrected version of (1.25) was proved by Atkin, [4]. 13 Chapter 2
Franklin and Sylvester 2.1 The Franklin bijection This sectionoutlines Franklin's combinatorial proof of (1.16)
(1)m p De, n)  p(Do, n) =0 if n=m 3m+1
2 if n is not pentagonal. Franklin's approachwas to constructa function F definedon most partitions into distinct
parts that satisfiesthe following conditions
F is weight preserving:
F reversesthe parity of the numberof parts:
F is an involution: wt(F(A)) = wt(A),
#(F(A)) #(A) mod 2, F(F(A)) = A. The reasonfor the word most should becomeclear shortly.
It is necessary to introduce some more (standard) notation before outlining his idea,
which appears in 14.5 in [1], and as theorem 1.6 in [2], as theorem 19.5 in [6], in [7],
in chapter III of [8], as 19.10 in [20], as theorem 15.5 in [25], as 256 in [26], as § 100 in
[27], in chapter 5 in [31] and in [12]. For a partition into distinct parts, the slope, o (A) or
Also, for
simply a, of the partition is the maximum value of i for which Al  Aj =i1.
a partition into distinct parts, s(A), or simply s, is the smallest part of the partition. The
o, and s here defined (on partitions into distinct parts) are analogous to (but not the same
as) the E and S defined (on partitions into odd parts) in section 3.2 and to the u and s
defined (on partitions into nested parts) in subsection 3.4.2. The sets A and B are defined,
provisionally at this stage, as A= {X : Q(A) <3 (A)}
14 and B={A: a(A)>s(a}}.
It is now possible to tentatively define F: D 4 V as the map that removes the slope and
it as the smallest part if AEA, or if AEB then F does the opposite, removes the
places
smallest part and adds 1 to the appropriate number of entries at the start, ý, (ý) ._ (al  1, aQ1, aQ+l, ýºkiýýk, d) if u(A) < s(A)
...ý
..., (A, 1,..., + 1, ...Ak1)
ak)
As iºß+li'
+ if UM> s(A) where the hat, :, indicates omission.
Clearly F is weight preserving, and F(A) has either one part more than A or one part
less. Furthermore F reverses the inequality o,(A) < s(A) (meaning a(A) < s(A) if and
> s(F(A))) and from this it follows that F(F(A)) = A. Hence F satisfies
the above conditions. But there are some `exceptional' partitions on which it is impossible
to define F, it is necessary to establish precisely which partitions these are. only if a(F(A)) For any nonempty partition in D, the inequality < #(A)
a(A) will hold, since the slope of such a partition cannot exceed the number of parts in the
partition. If A= (Al) A2, ..., Ak) is such that a(A) < #(A), i. e. there exists some i such
A1_1 1, then there is no
for which there is a strict inequality A<
that 1<i<k
in defining F(A) (although F(A) itself might be a partition having slope equal to
problem
Thus it is only necessary to consider A= (Ar, A2, Ak) EV for which
number of parts).
...,
1). Any
#(A), i. e. partitions of the form A= (A1, At  1, Al  2,..., Al k+
o(A) =
k< al  k, k= Al  k, k= Al k+1
or k> Al k+1.
such partition satisfies either
These four cases are now considered:
1), and
al k+
Case 1: a(A) = #(A), k< Al  k: In this case A= (Al, Al  1,
...,
A1 k+1 and a(A) = k. Thus, since, \1 k>k, s(A) >o (A) and so AEA.
so s(A) =
Al  k, k) is well defined, since al k>k
Hence F(A) = (A1  1, al  2,
ensures
...,
that the penultimate entry exceeds the last entry (and k, being the number of parts, is a
positive integer implies that the last entry in F(A) is positive).
Al k+
1),
Case 2: o,(A) = #(A), k= Al  k: As in case 1, A= (A,, Al  1,
...,
Al k+1
and s(A) > Q(A),
and o,(A) = k. Thus, since Al k=k
and so s(A) =
if F(A) were indeed defined, then A would be an element of A. This would give
F(A) is not permitted
Al  k, k) but, since Al k=k,
F(A) = (A1  1, \1  2,
...,
on account of the last two entries being equal. Hence, in this case, F(A) is not defined.
Since these partitions are those having length k and \1 = 2k, they are of the form
(2k, 2k
k+ 1). Hence, if X is the set of partitions in V for which F is not
1,
...,
defined then 0, (2), (4,3), (6,5,4),
E X. Note that this illustrates why the set A was
...
initally defined only provisionally, there are some elements AEV for which a(A) < s(A)
but for which the map A p F(A) can not be defined. Case 3 now demonstrates that the
A= same is true for some of the partitions that should, it seems at first, belong in B. 15 Again A= (al, Al 1,
A1 k+ 1), and so
Case 3: a (A) _ #(A), k= al k+1:
...,
k. Since A, k+1=k,
it follows that s(A) =a (A) and
Al k+1
and a(A) =
s(A) =
so A should be an element of B, provided that F(A) is indeed defined. This would imply
that A has length k and so F(A) should have length k1, i. e. F(A) = (A1 + 1, al,, \1 kA1 (2) and similarly
Alk+2).
Now wt(A) = Al+(A11)+...
I,
+(Al k+l)
=
,
(k 1)A1 + 1 (k22). But this gives wt(A)  wt(F(A)) = Al  2k +2=1
wt(F(A)) =
Hence for these such partitions, F
2k  1) and so wt(F(A)) ; wt(\).
(since A=
is not defined. These partitions have length k and Al = 2k  1, so are of the form
)A= (2k 1,2k k). Hence (1), (3,2), (5,4,3),
2,
C X.
...,
... Case4: a(A) = #(A), k> A1 k+1: Again A= (A,, Al 1,
Al k+ 1), and so
...,
< v(A), henceACB.
s(A) = Ai k+1 and v(A) = k. Now A1 k+1<k
and so s(A)
alk+2).
There is no problem here,asF(A) = (A1+1, al, al1,
ais+2,. 11s,
...,
...,
Thus no problems arise from cases1 and 4, from cases2 and 3 it is seenthat the set
F is not defined is X= {O, (1), (2), (3,2), (4,3),... 1.
of exceptionalpartitions on which
Clearly V is the disjoint union of the set of nonexceptionalpartitions and X, the set of
exceptionalpartitions, and x={(2k, 2k1,...,k+l): k>o}u{(2k+1,2k1,..., k+l) : k>o}.
The set X can be split into Xe and Xo where Xe := De nX and Xo := Do fl X. Note
that Xe = Xe U Xe and Xe = Xö U Xö where
Xe = {(4k, 4k1,..., k>
: 2k+1) Xe = {(4k+3,4k+2,..., 2k+2) :k> 0}, 0}, Xi = {(4k + 2,4k + 1, 2k + 2) :k> 0},
...,
X2 = {(4k + 1,4k,..., 2k + 1) :k> 0}.
It follows that
E gwt(A)
=>q + 2k +1= E 6k2 + k) and gwt(a) g6k2+11k+51 =E gwt(A) 7k}2 (2.3) g6ka} =Z 7 k>O XEXö
gwt(J1) _
AEXX (2.2) k>O AEXý2
`'`
L (2.1) k>O AEXj (because 4k + 4k 1+... 6k2+k `ý`
kk>O 16 g6k215k+ 1 (2.4) The sets A and B having been provisionally defined above can now be properly defined, A: = JA : Q(a) < S(A): Ae X}
and B := JA: Q(A) > s(A) :A¢ X}.
Having thus definedA and B, it is correct to say that F: A 4 B is a bijection.
It might seem unfortunate that establishing which set, A or B, a particular nonexceptional partition is an element of is not the same as establishing whether the number
of parts in that partition is even or odd. Given that the motivation for the abovewas to
(1.16), and yet what has actually been accomplishedis the construction of a map
prove
that alters the relationship betweenthe slope and the smallestpart of the partition, what
was the point? Looking at the map A + F(.\) again, however,two observationscan be
made: Firstly it is clear that F reverses parity of the first part of the partition, because
the
the first part is either increasedor decreased 1, and secondly F reversesthe paritiy of
by
the number of parts, for again this number is either one more or one less in F(A) than
in the original A. The first of theseobservations,it will be shown leadsto identity (2.5)
below,but before that the train of thought in the secondobservationneedsto be followed
through.
The set A can be written as the disjoint union of AO and Al where A;: =JA EA: #(A)=imod 2} for B. Just as F: A + B is a bijection, so too is F: (AoUBo) + (AZUBI).
and similarly
(1)#F(a) and since \E Ao U Bo
As has been noted, (1)#(al
= +1
\ E Al U Bl = (i)#ýal = 1, it follows that to determine p(D,, n)  p(D0, n) for
and,
it suffices to consider only those partitions in V that are also in X (because
some given n
any nonexceptional partition, A, can be paired off with F(A) which has the same weight
and opposite sign). Thus
ý(1)#(a)gwt(a)
AED (_l)#(A)gwt(a)+ _>
AEAoUBo (_1ý#(a)gwt(a)+ý(_1ý#(a)gwt(a)
XEX )EAiUBi so ý(1)#(a)gwt(ý) gWt(a) _ AED 'ý AEAOUBo E'(1)#(A)qwt(A)
gwt(a)+
AEX aEAIUBI and the expressionreducesto a sum over X. Now the sum over X is equal to the (sum of
the) sums over the four subsets,
>(1)#(a)gwt(')
AEX et(a)  gwt(A)+E =j
AEXX XEX. 2 gwt(A)E
AEXp from (2.1), (2.2), (2.3) and (2.4) is equal to 1qwhich,
(1.16).
combinatorial proof of
17 gwt(A)
AEXö q2 + q5 + .... This is the Now, the first observation about F was that it reverses the parity of the first part. This
means that, as well as thinking of F as a map from the set of nonexceptional partitions (in
V) having an even number of parts to those having an odd number of parts, it is equally
valid to consider it as a map from the set of nonexceptional partitions having first part
even to those having first part odd. So if D. denotes the set of partitions in D having first
part even and D. the set of such partitions with first part odd, then
ýý1}aigwt(a) _ AED i1)Jlgwt()) + ý(1))lgwt(A) gwt(a)+ 1)
AEDý\X \EX AEDQ\X so ý(1)algwt(a) gwt(a)  JýED 1: (_l)a1Q+vt(A)
+
gwt(A)
AEX \EV. \X ýEDe\X and, similarly to the abovecase,this becomes
q"'t(, ') E
AEXl gwt(A)+
AEX. 2 q+Vt(a)
>
AEXö gwa(a)=1q+
AEXö q2  q5 + ... hence 1 +1
p (D' , n)  p(D', n) _ 2.1.1 1
0 ifn=3m+1
andm>0
if n= "' 32+1 and m<0
if n is not pentagonal (2.5) The generating function for p(D., n)
 p(D., n) Let D(k) be the set of partitions into distinct parts with first part k. The generating
function for p(D(k), n) is
E p(D(k), n) qn = qk(1 + qk1)
n>O
= qk(q; + qk2)... (1 + q) q)kI and summing over kE N'` gives
>p(D, n)4n =1+E 4k(q; Q')k1
k>O n>O (so the sum on the left is equal to (q; q),,., the generatingfunction for p(D, n), see(1.13)
above).Similarly,
E(p(V, E(1)kgk(q;
n)  p(V, n))qn =1 +
k>O n>O which, by (2.5) is equal to 1q+ q2  q5 + .... 18 4)ki (2.6) 2.2 Sylvester's bijection The equality between the generating functions for the sequences p(D, n) and p(O, n)
is identity (1.18). The formula for the difference between the number of partitions in
0 with first part congruent to 1 mod 4 and those with first part congruent to 3 mod 4,
p(Ol, n)  p(O3i n), will be given later (3.8). But first a combinatorial proof that
p(D, n) = p(O, n) is presented.
Sylvester constructed a weight preserving bijection from the set of partitions into odd
parts to the set of partitions into distinct parts. The bijection, T: D + 0, described here
is the inverse of Sylvester's.
A good way to illustrate the nature of the map T is through an example of how it acts
on the graph of a given partition. Let A= (19,18,17,16,15,12,9,5,2). Below is the graph of A, or at least what would be the graph if the letters, which are used here because
they make the description of T slightly clearer, were replaced by dots.
a a w w w b b b b x x x c c c y d z a a w w w b b b b x x x c c c y d z a a w w w b b b b x x x c c c y d a a w w w b b b b x x x c c c y a a w w w b b 6 6 x x x c c c a a w w w b b b b x x x a a w w w b b b b a a w w w a a. e So instead of dots the columns of odd height are composed of letters from the start of
the alphabet (each dot in the first such column is replaced by an a, each dot in the second
such column is replaced by a b, and so on). The columns of even height consist of letters
from the end of the alphabet, z for dots in the columns (or column in this case) of height
one, y for each dot in the columns of height three and so on.
Now, the `double block' of 9 as is transposed to give two columns of as, both of length 9. This can be written as (2,91a) 3 (9,2f a). Likewise (4,71b)  (7,41b),
(5,31c), (1,31d)
(3,51c)
(3,11d) and (1, lie)
(1, Ile). The block of 24 ws is
+
}
*
halved in height and doubled in length, (3,81w)
(6,41w), similarly (3,61x)
(6,31x),
3
+
(1,41y)
(2,21y) and (1,21z)
(2,11z). The resulting blocks are stacked in such a
+
+
way that the left most column has the e at the top, to its right are two columns, both with
ds at the top and so on to the last pair of columns with as at the top (see below). Also the 19 ws, xs, ys and zs are aligned so to produce one row starting with the letter z at its left, the
next row starts with y, the next row after this starts with x and the last one starts with w,
as shown below,
ezzyyxxxxxxwwwwww
dddyyxxxxxxwwwwww
cccccxxxxxxwwwwww
ccccc wwwwww ccccc bbbbbbb
bbbbbbb
bbbbbbb
bbbbbab
aaaaaaaaa
aaaaaaaaa and
Now, raise the column with e at the top to the point (1,
slide the z column along,
1)
it starts at (1, 2). Then raise the next two columns so the two top ds occupy the points
so
(3,
(2,
and
and slide the next row in so the first y is at (4, 2). Then raise the
2)
2)
fourth and fifth columns so the two top cs in these colums are at (4,
(5,
3) and
3) and
slide the row starting with x along so that the first x is at (6, 3). Then raise the next
two columns so that the two top bs occupy (6, 4) and (7, 4) and slide the row of six
along so the leftmost of these is at (8, 4) and finally insert the remaining four as as
ws
shown,
e z z y y x x x x x x w w w w w w d d d y y x x x x x x w w w w w w c c c c c x x x x x x w w w w w w c c c c c b b w w w w w w c
b c
b c
b c
b c
b b b a a b b a a b b b b b b b b b b b b a b b b b b a a
a a a a a a a a a a a This diagram (viewed as a collection of dots, not letters) is the graph of T(A),
so T(A) = (17,17,17,13,9,9,7,7,7,5,5). 20 The map T has some interesting properties. In the above example, the first part of
T (A) is 17 and T (A) has 11 parts. Now (17  1)/2 + 11 = 19, which is A1, the first
part of the original partition. This is true in general. It is also true that the number of
sequences of consecutive integers in A is the number of distinct odd numbers in T(A).
This is an exercise in chapter 7 of [17], which also defines the bijection of Sylvester. An
immediate consequence is that the number of partitions of n into odd parts (repetition
allowed), where there are k distinct parts is the same as the number of partitions of n
into distinct parts where the are k sequences of consecutive numbers, a proof based on
generating functions is given in [2]. As an aside, it is worth noting that the partitions for
which T (A) =A are those of the form (2n  1,2n  3,..., 3,1).
There is another weight preserving bijection from V to 0. It is defined by breaking
even part, n= alb of A, into 2b copies of a (where a is the largest odd divisor of
each
n). Thus (19,18,17,16,15,12,9,5,2) + (19,92 17,116,15,34,9,5,12)= (19,17,15,93,5,34,118),
where the powers denote multiplicities. 21 Chapter 3
A combinatorial approach to some
identities
partition 3.1 The three identities The following identities are to be proved in this chapter. The identities are not new, but
the proofs given here are.
n J: (1)n =1q2 2. q2)
n
q2 n>O qn
(qi m 9m1 )n
q2 (1  qm), (3.1) + g4m+2), (3.2) m>0
(1
(1)mg2m(3m+2) =1+q
m>0 n>0 E )n =E
qn(q; q2 n>0 (1)mgm(3m+2) (1 + q2m+l). (3.3) m>0 here involve consideringpartitions into odd parts andpartitions into
The proofs presented
distinct parts. Identity (3.1) is equivalentto identity (23.2) in [11], due to the fact that
In1
n>0 1ý2nL
\q; q2)n qm
m,>0 `q2+ q2)m . This is so because each side of the above equation is the generating function for p(Oi, n)p(O3i n) (see (3.8) below). Identity (3.2) is equivalent to (26.96) and (27.97) in [11].
The approach taken here is different from that of Fine. Whereas he uses the Hyperto prove (3.1) and (3.2) (and lots more besides), here all that is used are
geometric series
combinatorial arguments (adapting the Franklin bijection) and qbinomial coefficients. 22 3.2 A combinatorial proof of a result of Fine There is, as indicated earlier, an expression similar to (1.16) but involving partitions into
odd parts. It was found by Fine using hypergeometric series, see 23.7 in [11]. In this
a weight preserving involution, G, is constructed that acts on most of the partitions
section
into odd parts. It also has the property that (on all the partitions on which it is defined)
it reverses the residue mod 4 of the first part of the partition (this is the raison d'etre
of the map G). Just as the Franklin bijection depends on whether the partition satisfies
< s(A) or not, it is necessary to know if a certain inequality holds (for a given
u(A)
partition) in order to determine the action of G on that partition. In order to discuss the
terms in this inequality it is necessary to introduce some new notation.
For Aa given partition (into odd parts), let t be such that at > 2t 1 but At+l < 2t 1,
At  2t +1 and define d(A) as the maximum l such that at+i = 2t  1. Also for
e(A) :=
i from 1 to t let xa(2i  1) := #{l
r(A), E(A) and S(A) by 1i = 2i  1} and xa(2i) :=.,
:. 2t1
r(ý) ife(A)=0, 2t Ai+l. Now, define
 if e(A) > 0, E(A): =min(j:
d(a) +1
e(A)/2 xa(j)>0),
if e(A) = 0,
if e(X) > 0. In order to define G on a given partition AE0 it is necessary know if E(A) < S(A)
to
E(A) > S(. ). But, whereasin the case of the Franklin map there were only two
or
possibilities,for G there are eight possibilities. Any given partition, AE0, is in precisely
one of the following sets, Al := JA: E(A) < S(A),
A2 := {A : E(A) < S(A), E(A) = 2Q,
E(A) = 2a + 1, r(A)
r(A) 0 mod 21,
0 mod 21, E(A) = 2v,
A3 := {A : E(A) < S(A),
A4 :_ {A : E(A) < S(A), E(A) = 2Q + 1, r(A) 1 mod 2}, r(A) 1 mod 2}, Bl := {a : E(A) ? S(A), S(A) = 2s, r(A) S(\) = 2s + 1,
S(A) = 2s,
S(A) = 2s + 1, r(A) 0 mod 2},
0 mod 21, r(A) 1 mod 21, r(A) 1 mod 2}. B2:= {A: E(i1) S(A),
>_
B3 := (A: E(X) > S(A),
B4 := {A : E(A) >_S(A), 23 Depending on whether AE Al, A2, A3, A4, B1, B2, B3 or B4, G(A) is defined (for
most such partitions) as
Ak),
At  4a, 2t + 1,
2t + 1, At+1,
(A1 4o 2,
A,  4a  2, A, +l  4Q,
...,
...,
...,
...,
20 Ak1),
A
(al
At  4a  2,2t + 1,
2t + 1ý +1, 7
4a  2,
...
...,
...,
241 ak),
't+2a+i,
(Ai + 4Q  2, aQ+ 4o, 2, aQ+i + 4Q, At + 4a,
...,
......,
...,
...,
(Al (ill4s+2,... ) +4Q+ 4s+2, At+27+2,..., At + 4a+2, 2,..., Ak1), ...... 7 As+14s,..., At4s, 2t1,..., 2t1 At+,,..., Ak), tis
a (Al  4s  2, At  4s  2,2t 1,
)ki, Ak, 2s + 1),
2t  1ýAt+i,
...,
...,
...,
2sß1 (A, +4s+2,..., as+4s+2, As+i+4s,..., Ati+4s,...... IAt+2s,... ) Ak), (A, + 4s + 2,..., Ac, + 4s +2,......, At+2s+1) AkIý Ak, 2s + 1).
...,
The six dots
are used to emphasise the omission of some entries, which can be seen
,......,
for example,
by looking at the subscript. Writing A= (17,17,17,13,9,9,7,7,7,5,5),
A5 = 9, so t=5.
Now, e(A) =0 and d(A) = 1. The sequence x.\(j) starts
gives
(0,0,0,0,2,... ) and so has first nonzero entry at j=5,
giving E(A) = 5. Since e(A) = 0,
(and so s= 1). Hence E(A) > S(A), r(A) =9 is
it follows that S(A) = d(A) +1=2
(17 +
but S(A) is even. This is case B2, and G((17,17,17,13,9,9,7,7,7,5,5))
odd
=
6,17+4,17+4,13+4,9,9,7,7,7,5,5)= (23,21,21,17,7,7,7,5,5). Clearly G reverses al mod 4. It is also a weight preserving involution. Thus, to find
for p(01, n)  p(03, n) it is only necessary to consider the partitions for
an expression
G is not defined, these fall into four families, 0, (7), (132), (193),..., ((6n + 1)fz)
which
(1), (34), (57),..., ((2n+1)3n+l) and (12), (35), (58),..., ((2n+1)3n+a) and (5), (112),
and
(173),..., ((6n + 5)nß1). 24 Thus, if Ol :={. E0: Al 1 mod 4} and 03 := {A E 0: al =3 mod 4}, then
it follows that p(Oi, n)  p(03, n) =0 unlessn= m(3m + 1)/2 whence,
p(Oj, m(3m + 1)/2)  p(03, m(3m + 1)/2) = f1 1 if m=0,4,8,
or f1, f5, f9, ..., or 2, 6, 10, ...
...,
ifm=4,
orf3, f7, ±11,..., or2,6,10,...
8, 12,..., (3.4) 1
Whereasthe Franklin map is a fairly intuitive construction,ascan be seenby considering
the effect it has on the graph of the partition (this is done in all the referencesmentioned
in section 2.1), the map G seemssomewhatcontrived. The idea behind G is simply that
G (A) = TFT1 (A) (except for the exceptional partitions, for which the
for AE0,
Franklin bijection is not defined). In fact, for AED, #A = r(T(A)), a(A) = E(T(A))
and s(A) = S(T(A)).
Finally note that (3.4) implies
1: (P(Oi,
n)  P(0)3, n))qT =1+q+
n>o q2 + q5  q7 _ q12_ q'5  q22+ q26+... (3.5) 3.3 A different approach
In this section a, rather longwinded, proof of (3.1) is given. The proof involves the
conceptof the hook, definedbelow, of a partition into odd parts.
The following identity will be needed,it follows from (21.21) in [11],
qý
i=0 n
Z (qi _ Q)n. (3.6) 42 Now, recall that U(a, b) is the set of all partitions having no more than b parts, the
largest of which is no greater than a. Similarly, let U' = Li (2a, b) be the set of partitions
into not more than b even parts, the largest of which is no greater than 2a. It follows from (1.8) that
E [a + b] gwt(a)_ (3.7) La 4 AEU' The next stepis to obtain the generatingfunction for p(Oi, n)  p(03, n).
For AE0, let ea := (1)()11)/2. By considering the graph of all such A it can be
seen,where O(k) denotesthe number of partitions into precisely k odd parts, that
E
AEO(k) \gwt(A) = qk(1, q2k + q4k I ... )(1 q2(k1) + q4(k1) + ... )... (1  q2 + q4 + ... ) Qk
(42; 25 42)k hence(taking co = 1),
Qk Eagwt(A)
= ýZýk lq2;
k>O XEO(k) and so, qm j>(01, n)  p(ds, n))qn = (q2; i ý1
2)m. m>o n>o (3.8) As was seen in section 2.1.1., similar arguments give (2.6) which, to recap, states,
lp(De, n)  p(Do, n))qn ýl1)mgm(qi = q)m1. (3.9) m>0 n>O 2
Now the sum on the left of (3.8) is the sum over partitions in o of (1)
The number of parts of such a partition is congruent mod 2 to its weight. So q
identity (3.8) gives
E qm ý"#(a)gt0t(a)_ J)Alý AEO 1)m (q2; qwt())
in
q (3.10) q2)m. m>0 Now, define h(A) :_ (al  1)/2 + #(A). This is a sort of modified hook (the hook is
Al 1+
#(A), but here h(A) is only defined for partitions into odd parts). A
usually
for
A, has h(A) =m if and only if both (A) Al = 2i +1 and (B) #(A) =mi,
partition,
For a given i, summing over all partitions (into odd parts)
some i for which m>i>0.
which satisfy both (A) and (B) gives (using (3.7)),
[m
gwt(a) = qm+i summing over i gives from 0 to m1
and q2 gives
[m
qm+i _E h(A)=m Z M1
gwt(A) 11  11
J i=0 42 and by (3.6) this becomes
E gwt(A) = qm(q; 4')m1 h(a)=m the partitions with h(A) = 1,2, ... gives the following identity
considering
E zh(a)gwt(a) AEO* where 0* :=0\0. zmgm(_q; => 4)m1 m>0 Setting z= 1 and taking h(O) =1 gives
>(1)'
AEO (')qwt(") E(1)"`4'(4;
= 1I tl)m,. (3.11) m>O The sum on the right of (3.11) is that in (3.9). As was noted at the end of chapter 2, the
the left of (2.6) (and so (3.9)) is 1q+
sum on
q2  q5 + .... So, given (3.10), (3.1) is
proved. Note that, having done all this, (3.1) follows directly from q + q in (3.5) and
(3.8) (so actually there is no need to invoke (3.6) or the hook of a partition). 26 3.4 The second and third identities
In this section identities (3.2) and (3.3) are proved. The method of proof requires introducing the (new) concept of nested partitions. Then it is shown that the generating
function for nested partitions with those into an odd number of parts counted negatively
is the sum on the left hand side of (3.2). Next, using two elementary identities ((3.12) and
(3.16) below), it is shown that this simplifies to F, q2m+l (q2; q4)j which is essentially the
the left in (3.3). This being so means that if (3.3) were to be proved, then so
sum on
(3.2). The proof of (3.3) relies on the fact that the sum on the left of this identity
would
is also related to nested partitions: it is the generating function for nested partitions into
distinct parts, again those into an odd number of parts being counted negatively. Finally
an involution is defined for most such nested partitions, proving (3.3) (and so also proving
(3.2)). Having outlined the method of this section it is now necessaryto introduce the two
identities required here, this is done in the next subsection.Note
qbinomial elementary
that the two identities form part of theorem3.4 in [2] where they are proved, and attributed
Gauss(the proof given here of (3.12) is different to that in [2]).
to
3.4.1 Two qbinomial identities The first identity is m n+i ii=O
gLi]=[n+m+ll.
m (3.12) This is a consequence of the lemma in section 1.2.2. It follows from that lemma (identity
(1.8)) that the generating function for partitions with no more than n+1 parts and first
1], which is to say that
is [t
part no greater than in
E 1 n+m+ wt(A)
q=mý. (3.13) \EU(m, n+1)
. Now, if I4(m, n+ 1) is defined to be the set of such partitions having first part al =i
then it is clear that if (m, n+ 1) is the disjoint union of the sets U3(m, n+ 1) (as i ranges from 0 to m). So m gwt(a) =E>
\
.EU(m, n+1) i=0 gwt(a) (3.14) )ýEU; (m, n+1) a partition in U (m, n+ 1) has (by definition) a first part of size i and no more than n
and
parts. Hence
additional
gwt(A)= q'
AEU; (m, n { 1) 4"'t(a)
AEU(i, n) 27 (3.15) in the aboveequationis clearly equalto ["s'
the sum on the right
(3.15) becomes
[n + i]
gWt(a) qi
= (by (1.8), again). Thus AEU (m, n+1) togetherwith (3.14), implies that
which,
EE [iI gwt(A)_m qi n+i i=0 \EU(m, n+l)
, Finally, the fact that the sum on the right in the aboveequation is (by (1.8), once again)
["+m+l] implies the veracity of identity (3.12), as desired.
equal to
The secondidentity is
E E( z
_1 [n] :o (q; g2)ti if n= 2m, 0 if n is odd. (3.16) This, like the previous identity, is proved in [2] where it is part of theorem3.4.
3.4.2 Nested partitions
[Ah]) where (for i such
A nested partition is here defined as a sequence, A= Ni [A1],
...,
k) Ai > Ai11 0. For i>0,
the entry [Ai] signifies the pair (Ai, A,  1),
>
that 0<i<
be called a nested entry (so the only nonnested entry is A0). The weight of A is
and will
E= Ai where k, clearly, is the number of (nested) parts.
defined as wt(A) = Ao k+2
1
A typical nested partition is A= (9, [9], [9], [6], [4], [4], [1], [1], [1]). Now k=8 but there
14 rows in the graph, each nested entry greater than 1 is represented by two rows, the
are
in the second row has an empty dot in its place, thus the graph for A defined
second entry
above is
"""""""""
"""""00 "" "0""""""
"""""""""
"00""000"
""""""
"0""""
"""" "0""
"""" "0"0
"
"
" 28 71 (the number of black dots). Let A denotethe set of all (nonempty) nested
so wt(A) =
For AEA, define h(A) := A0+k (this is the hook for nestedpartitions). Now
partitions.
h(A) =a+b+c
where a= Ao, b= #{i >0: A> 1} and c= #{j >0: AA= 1}.
Sucha partition is said to havetype (a, b, c), for instance,in the aboveexampleA hastype
(9,5,3). Consideringall nestedpartitions of type (9,5,3)
121
qwt()) = q27 qs [1]),
(9, [2], [2], [2], [2], [2], [1], [1], To see this, consider the graph for
which has weight 27.
There is a7 by 5 rectangle that can be filled (by pairs of dots) in various ways. Every
type (9,5,3) arises this way. This is basically the same idea as that in
nested partition of
3.3 where it was observed that for a partition in 0 to have a given value for its
section hook, it must satisfy two conditions. Now
qn
n (4. q2)n  2n1 (1 q 4n2 }(1 I q+... implies
which 2n3 q " 4n6 } q 2 }... (1
} ...
q {q+... } ý } (q; 42)n a So the approach is: for a given 1, investigate all partitions with h(A) =l and sum over 1.
Then there are a dots in the top row (of the graph), 3b (nonempty)
Suppose 1=a+b+c.
dots for the b nested parts that are >1 (because there are three dots for each such part)
and c dots for the c parts that are equal to 1. In the above example a+ 3b +c= 27. Also
is, in general, a (a  2) by b rectangle that may be filled with pairs of dots. So,
there
those partitions where k is odd negatively gives
counting
(_1)kgwt(A) = ra +b2j (_1)b+cga+36+c \jbJ qa for a fixed, summingover b gives
ia
(1)kgwt(a)` /_1)1aql
_ Ia
> [a+b2]
q26 bE((E By (3.12) this can be written as
1a (1)kgwt(a))
b0 \ (_1)ýaq! IlJL4= Aa(a, b,c) a] 2 and summing over a gives
ta rI EE
a=1 b=0 (l)kgwt(a) (1)1gl
= Aº(a, b,c) (I)a
a=11 29 l 1
l a 4ý Now, by (3.16) the expressionon the right is q2m+l(q2;q4)y when l= 2m+1 and 0 when
1is even. Thus, for 1= 2m +1
(1)kgwt(A) _ q2m+l (q2; q4)m over l gives
so summing
1: (_1)kgwt(a) g2m+l(42; =E AEA q4)m. m>0 Thus identity (3.2) has been shown to follow from identity (3.3) (by q 4 q2 and then
multiplication by q). In order to prove (3.3), it is necessary consider nestedpartitions into distinct parts.
to
Ai > at+l} U {0}. Note
Let B be the set of thesepartitions, that is B= {a E Abi >0=
that ao = al is allowed. Now,
q2)n = qn(1  q(n)+(n1))(1qn(q; q(n1)+(n2))... (1  q(1)+(0)) (_1)kgwt(A) _>
AEB(n) B(n) := {a EB:
where Ao = n} (with B(O) = {0}). Summing over n gives
E qn(q; q2)n = E(1)kgwt(a). (3.17) AEB n>O This means that proving
>(_1)kgwt(A) = E(_1)mgm(3m+2)(1 + g2m11) (3.18) AEB is equivalent to proving (3.3) (and so (3.2)). In order to prove (3.18) it is necessary
A + A'. This map plays the same role as A
F(A) (Franklins'
to construct a map
*
bijection) defined on normal (i. e. not nested) partitions into distinct parts in chapter 2
A + G(A) defined on (normal) partitions into odd parts earlier in this chapter. In
and
particular, wt(A') = wt(A), u(A) < s(A) 4* a(\') > s(A) (a and s for nested partions
defined below), A 4 A' } A" = A, and A' is defined for most such partitons.
are The proof is similar to Franklin's. Define
2Ak1 ifk>0 00 ifk=O if i is defined to be 0 if A0 0 Ai and as the maximum j such that A0  A3 =j1
and,
A0 = Al then
Q=v(A): if =2i+1. Now, define A' by (Xo [Al1), ...,[Ai1], [A%+1], [i + 11) if a<s,
[Akl,
1,
...,
(Ao+ 1a 1a ++ 7 [Ak11) 4) if 1> s. Note that if A= (n) (which happensif k= 0) and n>1 then A' = (n, [1]). The only
k<1 and A' is undefinedis the empty partition.
other partition where
30 All that needs to be done is identify the exceptional partitions (for which k> 0). Now
2Ak 1 so i+1<
Ak. The only possible problem (in the
if o,(A) < s(A) then 2i +1<
Ak  1. For the penultimate entry in A' is either [Ah] or [Ak 1]
< s) is at i+1=
case o,
Ak. However if i+2=
Ak,
the last is [i + 1], which is not a problem if i+2<
and
the last entries in A' could be [Ak  1] followed by [i + 1]. This would occur if and
then
k1
which is equivalent to saying that )tp = Ap+1 +1 for all p=1,2,
only if i=k,
...,
(and that A, = A1). This reduces the set of possible exceptional (for the case Cr < s)
to those of the form (n, [n], [n  1], ..., [m]). In short, what is required is to
partitions
Ah  1. But (for such
find all such partitons satisfying both (a) Q<s and (b) i+1=
Ak =m so s= 2m
and a= 2n  2m + 3. Hence (a) gives
partitions), i=nm+1,
1
2m 2 and, more importantly, (b) becomes n= 2m  3. Clearly, for a<s,
the
n<
(3, [3]), (5, [5], [4]),..., (2m
[m]). These
3, [2m  3],
only problematic partitions are
...,
(m
1)(3m  5) (form > 2).
have k=m2
and wt(A) =
partitions
Now, if o,(A) > s(A) then 2.ßk1 < 2i +1 soak 1 < i. Clearly i<k with equality
at (and only at) those partitions mentioned above that have successive entries differing by
and (from the definition of A'), problems
precisely 1 (except Ao = A1). So Ah 1<k
when equality holds. This happens when m1=nm+1
occur precisely
and so
(for s< a) the exceptional partitions are those of the form (2, [2]), (4, [4], [3]),..., (2m 2, [2m  2],... [m]). These partitions have k=m1
and wt(A) = (m  1)(3m  1) (for
m> 1). Let B' be the set of non exceptional partitions. Now, A 4 A' is indeed a weight
involution on B' with a(A) < s(A) e* s(A) < a(A) and so
preserving
E(_t)kgwt(a)
XEB
. _E =E (_l)kgwt(a) +Z
l AEB\B' (_1)kgwt(A) AEB' (_1)kgwt(a) =1+q+1: (1)m2q(m1)(3m5)
m>2 AEB' m1Q(m1)(3m1).
m>1 This proves (3.3), and so (3.2) is proved too. 31 3.5 Two identities, one old and one new 3.5.1 An identity of Subbaro and Vidyasagar
In [30] Subbaroand Vidyasagarprove the following identity:
2n n2 fn
qq E(1)" z q,ý 2) n+1
` n>0 E(_1)mz3mg3m2+2m(1 = + zg2m+1) (3.19) m>0 (it is (1.5) in their paper, they use a and x instead of z and q). Their proof involves algebraic manipulation of power series such as these and relies on the triple product identity,
and (a version of) what is known as the quintuple product identity. What follows here is a
direct, combinatorial, proof of the above identity.
Firstly, for nEN consider the set S(n) defined by
S(n) := Ili = (7r0,7r1,7r1,7r2,7f2i irn; 7rn) : 1Cp i ..., 7f1 i ... ý! ?fn i0} (the reason why an arbitrary element of this set is denoted by µ, whilst its parts are labelled
become clear shortly). The elements of S(n) can be viewed as partitions having
ir= will
the property the second part is equal to the third, the fourth part equal to the fifth, and so
Viewed in this light it is seen that the dual of an element of S(n) is a partition into
on.
the biggest of which does not exceed 2n +1 (the largest part of the dual of an
odd parts,
0). It follows that
S(n) is 2n +1 if, and only if, irn
element pE
1= gWOO) (qi q2)n
Fl 1ES(n) (where the weight is the sum of the parts of µ, wt(p) = wo + 21r1+ 2ir2 +
+ 2irn).
...
In fact, since the first part (i. e. 7ro) of an element of S(n) is the number of parts of the
element's dual,
1
(zq; z7pgwt(R)
q2)n
DES(n) hence
zY
(zq; zarp{2nQwt(µ){n2}n
Q2)n+
pES(n) which implies that
ý(1)n
n>0 2n n2}ýn
zq_ (zq; q2)n+l ý(l)n E n>O z7ro+2ngwt(ic)+n2+n 7rES(n) 3 20
. Now if T (n) is defined to be the set of all ordered pairs with first part an element of
5(n) and secondpart the sequence(2n, 2n  2,..., 4,2),
T(n) := {ir = (/c, (2n, 2n  2,..., 4,2)) :µE S(n)}
32 (so T(O) consists of all pairs ((iro), 0), T(1) consists of all pairs of the form ((iro, irl, 7rl), (2))
for 7r = (Oro, 7rl, 7r1, it irn), (2n, 2n  2,... 4,2)) E T(n), the statistics k (ir)
etc) and,
..., a,
'
and 1(7r) are defined as l(ir) := 7ro+ 2n k(ir) :=n,
the Weight, Wt(7r) as
and Wt(7r) := wt(7r) + n2 + n.
then it follows that
E z7'o+2ngwt(ir)+n2+n 111n =E IrES(n) 't(7r). (1)k(ir)zl(7r)q (3.21 aET(n) Hence,defining T to be the union of all T (n),
(2n, 2n
T: = {7r = (? 1r1,7r1,
2,..., 4,2):
r.,
7'n, 7n),
..., n> O} it follows that, after summing over n in (3.21), (3.20) becomes
E(i)k(7r)zi(7r)gwt(7r) _'ý`(_11n ET
?r nn>>OO z2ngn2+n
1 (zq;
q2)n+1 So, in order to prove (3.19), it suffices to show that
E(1)k(ir)zt(ir)gWt(lr) = E(_1)mz3mg3m2+2m(1 +zg2m+1) (3.22) m>0 7rET Proof of (3.22): An involution r + ir' is to be defined on (most of) T. The involution will
be Weight preserving, i. e. Wt(ir) = Wt(ir). The involution will also have the property
that l(ir) = l(ir), but k(ir') 0 k(7r) mod 2 (ensuring that (1)k(7r) A (1)k( r)). As
there will be a small subset XCT
containing the exceptional elements on which it
usual
is impossible to define the involution, and the sum on the left of (3.22) reduces to a sum
over X (on account of most elements 7r being paired off with a corresponding 7r').
As usual the involution depends on whether or not an inequality of the
form a(7r) < s(ir) is satisfied. In this case the two statistics are given, for
(2n, 2n
((70,71, irl,
2,..., 2)), by
7ri, 7rn),
7r _
...,
Q(7r) := max(r : ir,. = gyro) and s(ir) := 7r,
ß define the sets A an B as
and
A: ={irET: Q(ir) <s(ir)} and B: ={irET : or(ir) > s(7r)}. A bijection betweenthesetwo sets,or at least bijection that works for all elementsexcept
thosethat give rise to the expressionon the right side of (3.22)) is required. This would
be an involution on T. The proposedinvolution, A 4 Aý,is defined as
33 ((iro 2,7r1  2,7r1  2,
it
...,
(2n+2,2n,...,
2)) 2,7r,  2,7rQ+l  1,7Ta+l  1,
ir,  1, firn  1, a, a),
...,
ifitEA ((iro + 2, irl + 2, ir1 + 2,
7r9+ 2, ir8 + 2, ic, +l + 1, i'8+1 + 1, ..., in1 + 1, irn1 + 1),
...,
(2n
if 7r EB
2,2n  4,..., 2))
It is now necessary to establish for which 7r ET the map it 4 7r is not defined on:
Firstly, for ir EB it is clear that if o,(7r) <n then the map ir + 7r is defined
(by definition the slope a(ir) cannot be greater than the number of parts, n). If ir is
(and a (ir) = n) then 7r =
that a (ir) =n then irtz < n, for if in =m>n
such
((m, m) m, m, m), (2n, 2n  2,... 4,2)) and then a(ir) =n<m=
s(7r) and so ir E
...
A. For it such that o,(7r) = n, it + 7r' is defined if and only only if 7r,, < n. Thus
((0), 0), ((1,1,1), (2)), ((2,2,2,2,2), (4,2)),
the elements
are those in B for which the
...
is undefined. These form X0 where
map
Xo = {((n, n, n,..., n, n), (2n, 2n  2,
4,2))}.
...,
Secondly if 7r EA then the definition of o, implies that 7rß > 7r,+1 and so 7rQ2>
1, which is required (at least if Q(7r) < n). Furthermore, for it E A, s(7r) > a(7r)
7rQ+1
>_ a, so there is no problem unless Q(7r) = n. If ir EA is such that
and so irn 1
then the first half of the sequence ir' ends (..., in  2, a, a) = (..., ir z2, n, n).
a(lr) =n
So suppose that 7r is
Thus, for ir E A, there is a problem if a(ir) =n and 7rn <n+2.
in this case Q(7r) =n>
such that a(ir) =n and 7rn <n+1,
7rn = s(7r) so a(ir) > s(ir)
so s(7r) =n+1
and a(ir) =n then
and is E B. If, on the other hand, 7r =n+1,
> a(7r) and 7r E A, and it is impossible to define the map it 4 7r'. These are the
s(ir)
((1), 0), ((2,2,2), (2)), ((3,3,3,3,3), (4,2)),.... Defining
elements
Xl :={((n+1, n+1, n+1,..., n+1, n+1), (2n, 2n2,..., 4,2))}, it is clear these are the elements of A for which there is a problem. If X is the set of
for which the map is undefined then X= X0 U X1.
elements it
Now
E(1)k(1r)zr(')gwt(1r) =E irET (1)k(ir)z' @)gwc(n) E(1)k(7r)zl(1)gWt(1r)
+ 7rE7AX the sum over T\X
and irEX is 0, so D1) k(it)zt(hr)gwt(ir) E(_1)k(ir)zt(it)gwt(it)
= irET irEX which is equal to
E
irEXO (i)k(")zt(")gN't(") +E (1)k(w)zt(ir)gwt(1r) rEXI +
the first sumis equal to 1 z3q5+ zsg16 ... and the secondto zq  z4q8+z7 q21+
and
....
This proves (3.19).
34 3.5.2 A new identity The following identity is, I believe, new:
D1)ri
n>o znqn2
=1
(zq; 42)n+i (3.23) It is provedin a similar way to (3.19) above:
Let it = (iro)7r1,
7r,,,
7rn)as above.The set U is now defined to be
irl, ...,
U={ir=(p, (2n1,2n3,..., 3,1)): µES(n), nEN}. For example,((0), 0), ((4), 0) and ((5,5,5,3,3), (3,1)) are all elementsof U.
The statisticsK(ir) and L(7r) are defined,for it E U, as
K(7r) = n, L(ir) = iro + n. The weight, W (7r)is definedto be
W (ir) = iro + 21r1 2ir2 +
+
+ 27rfz n2.
+
...
Thus (3.23) is equivalentto
E(1)x(")zL(")qw(") 1.
= (3.24) 7rEU What is required is a weight preserving involution on U, it+ir', for which K(am) =K(ir)
L(7r) # L(7r) mod 2. The only element of U on which the map is undefined should
and
be it = ((0), 0). With this in mind the setsA and B are defined by
A: = {i EU: 7rn= 0} and B := fir U: irn 7t O}. Once this has been done, the map it 3 it can be defined (the trick here is to notice that
the first entry of the secondpart of any elementit is 2n 1 and the length of the first part
is 2n + 1).
(2n
So, for 7r = ((iro, 7r1,irl,
7rn,7rn),
1,2n
...,
ir is given by
as 7r + it where 3,..., 3,1)) E U, the map is defined
 ((7ro+ 1,7ri + 1,1r, ++1,7rn_1
3,1)) if 1r E A,
+ 1), (2n 3,2n  5,
...,
(2n + 1,2n
((Iro  1,7r1  1,7ri  1,
1,7rn 1,0,0),
3,1)) if it E B.
ern 1, ...,
...,
(5,3,1)) then irEA and 7r'= ((8,7,7,7,7), (3,1)).
For instance, if it= ((7,6,6,6,6,0,0),
It is clear that the only element for which the map is undefined is ,7r = ((0), 0). It is also
that the map has all the desired properties and so (3.24) is proved, and so therefore
clear is (3.23).
35 Chapter 4
An involutive proof of the triple product
identity The triple product identity statesthat
[z; q] (q; q)... = Ez" z_
qn 2n (4.1) nEZ (this is (1.17), but with z replaced by z).
Now, the set D has already been defined as the set of strictly decreasing sequences
of positive integers, i. e. D is the set of partitions into distinct parts. The set Do is now
defined to be the set of decreasing sequencesof nonnegative integers (the set Do is not to
be confused with Do, the set of partitions into an odd number of distinct parts, which does
in this chapter). Clearly DE Do. The weight of a sequence in Do is defined
not appear
be the sum of the parts of the sequence and for a= (al, a2)
to
ak) E Do, #(a)
...,
the number of parts (these defintions are the same as for sequencesin D).
It follows that
E
z*(a)gt(a) = [z; q]
,
aED0 and that E [zlq;
z#(6)gwt(15)= 6ED
, 36 q]. :=k, This
(a,, 0= a= a2, ... ak), and for it = (a, #, 'y) EJ J := Do xDxD, that if
means (ßi, #(ry) =m and d(ir)
c(7r) :=
(4.1), to be statedas $2, $j)
... ý #(a)  then and by (where
setting (ß) =k 1 allows the triple product identity >(1)c(lr)zd(lr)gwt(n) (4.2) znq(n2n)/a =E
nEZ irEJ the weight of it is defined as the sum of the weights of its
where
Wt(7r) := wt(a) + wt(ß) + wt(y). three parts: A crucial step in this proof of the triple product identity, as in the case of Franklin's
of the pentagonalnumber theorem, is finding (for any given 7r E J) two statistics
proof
(7r) and a map ir > it (which is a weight preserving involution) suchthat the
s(ir) and o,
inequality s(ir) >o (ir) is satisfiedprecisely when s(7r) < a(ir ).
It will be seen shortly that whether or not the inequality is satisfied is, in some cases,
(because this is s(ir) >o (7r)). It will
dependent on whether or not ßi > yj +1k
be seen that, in some cases, it will be necessary to know whether or not ak =0
also
#1 = 1. Because of this it is necessary to adopt the following two
and whether or not
Firstly, the first part of the empty sequence 0 is 0 (it is perhaps better to say
conventions:
the first part of 0 is 0). For example, if /j =0 (and ry is some other
that the weight of
is equivalent to k> 7yl, because =0=
#l =0 and
then 81 > ryl +1k
sequence)
ý8
1=0, /31being the first part of ß and l its length. Secondly, that the last part of an empty
is neither 0 or 1, i. e a=0=
ak i6 0 or 1.
sequence
Now, for 7r = ((ai, a2, ..., ak), (01,01), (71,72,
..., 7m)) E J, the two statistics are definedby s(7r) :_ Co if#, >'yl+lk,
'Ym if ßl > ryl +lk,
/31 else ak=0, ß=0, ak = 0, ß=0, 'y=o,
0,
'Y and I if Pi >71+lk, max(r: a, a,. =r1)
yl +lk ak=0, ß=0, else. Thus, if the set T is defined as T := {ir : 01 > 'Yl +lk,
aA;= 0, ß= 0} then it ET
implies that s (7r) is either oo or the last entry in y, dependingon whether or not 'y = 0,
(so a (7r) is the slope of the
that a(7r) is the maximum r such that ai  a,. =r1
and
first part of ir, when it E T). Conversely,it ýT=
s(ic) = ßi and or(7r) = y1+lk. 37 The action of the map zr + it is dependent on which subset of J it is that ir is a
the subsets being defined as
member of, Al :={ir: s(ir)>a(7r), ck9`O,Qj4 0}, A2:= fir : s(ir)> Q(lr), o,ß= 01,
cEk
A3: ={ir: s(lr)>o(7r), 01,
ß 34ak=0, A4 := fir : s(7r) > a(ir), ak = 0, ß= 01, 81 := {ir : s(ir) < a(ir), ßi =1},
B2: ={ir: s(7r)=Q(ir), ß=O, 7reT}, B3 := {ir : s(ir) < Q(7r), '3 = 0,7r e T} U {s(7r) < Q(7r), ßi 1ß ; 01, {ir : S(7r) <_Q(ir), ir E T}. B4 The set B3 is defined as the union of two sets,specifically B3 = B3 U B** where
B3 := {rr : s(ir) < a(ir), ß=0, it %T}
and
B3* :_ {7f : s(7r) < 6(7f), ßl 7' 1, h' 7' 0I Now, it is clear from the abovedefinitions that if
A := {ir EJ: s(7r) > u(7r)} J: and then J=AUB
and A= Al U A2 U A3 U A4. It needsto be shown that B can be likewise
decomposed.
Q} U jr EJ: s(ir) <o (ir), Q= O} and the
< or), f
s(ir)
first of these two sets is B1 U B3*. The second set is {7r EJ:
s(ir) < v(7r), Q= 0} =
ý T} U fir EJ: s(ir) < Q(ir), Q=0, it E T} and the
{7r EJ: s(7r) < a(ir), ß=0,7r
first of these two sets is B2 U B. The second set is B4, because it ET
so the
,B=0
{ir EJ: Clearly B= part of any element in B4 is empty. Hence B=
second B1 U B2 U B3 U B4. It is now possibleto define,provisionally at least, an involution 7r 4 7ri on J. Depend(ß1,
81)) ('yi,
(a,
((al)
J it is that it =
7m))
ing on which subsetof
ak),
... ),
... '
...,
is defined as: is a member of, ((al  1,
ak  1)eß2 + 1, ..., ß1+ 1,1), ßßl 1 + ký'Yl, ... ý'Ym)) ir E Al
...,
((al
1), 0, (k, 71, 'Ym))
1, ..., ak ir E A2
...,
«al
Qt + l), (Ql
1, eak1  1), (ß2 + lý
k7'1,
7'm)) 7i c A3
l+
...
"ý
..., I
((al 7i =  11 ... au  1, au+1) 7t' E A4 ((al + 1,
+ 1), (71 + 1k,
ak
...,
«al + 1, ((a1 + 1,
«al + 1, ..., ak + 1) 07 (72) 2 ... 2 ß1  1, 7m)) ... ) ßt1 1), (72) 7m))
..., it E B1
E B2
it ý'Y2, ý'm)) ir E B3
+Zk, Äi 1,
+ 1, ý), ý'Y1
ak
1)i
..., 1ßt
...,
"..,
O, (111,
it E B4
Ym1))
+ 1, as+1,..., ak) >
... 3
..., as
38 The element 7r may be written For instance if it is such that as 7f' = (6, p, T) _ ((S11
... 5K), 0T then s(Tr') =p (/1, AL);
... ) (Ti, rM))
..., " and a (mm) rl +LK.
= Now, by checking each of the eight casesit is easy to see that wheneverit is defined
the map it ý lr is weight preserving, wt(7r) = wt(ir). It is also easy to see that
#(r) = #(, y) ±1 so c(7r) = c(7r) f1 which implies that it 4 it reversesthe sign,
(1)c(ß) ; (1)*). It is also easy to seethat the difference in length between the first
is unalteredin each of the eight cases,i. e. that KL=kI
so
and secondsequence
d(7r) = d(7r).
'_ ir, that it E A;
What remains to be checked is that this is actually an involution 7r
implies that 7r' E Bz, and to find the exceptional elements of J on which the map is not
defined. This can be done by looking at each of the eight cases one at a time, before doing
this it is shown that T= A4 U B4.
0), Tn Al = 0. Similarly an element
(and B
Note that, since ir ET=
ak ,00
0. Now suppose
A2 has ak #0 so Tn A2 = 0, and Tn A3 =0 because
of
,B 5E
Then
that, for some ir E A4, the element it were to be such that ,ßi < ryl +lk.
0 T, so (again, since ßl < 'yl +Ik) s(ir) < a(7r), which can't
ßl < yj +lk=
7r
happen for rE A4. So if it E A4 then 01 > yl +Ik,
which together with ak =0 and
A4 = A4. Similarly if it ET then
implies that it E T. Hence TnA=Tn
,B=0
,B=0,
B4 = B4
is not in Bl or B**, and neither is it in B2 or B. Hence TnB=Tn
so it
(because irEB4=>. 7rET=ß=0),
soT=A4UB4. Now let S be a subsetof J having the property that if zr ES then it > is is defined
is defined on all elements of S). Let 9 be the set defined by
(i.e. the map
Now for the eight
S' := {7r' : ir E S}, i. e. S' is the image of S under the map it *
.
cases:
Case 1,7r E Al: From the definition of Al, any element it in this set is such that the
last entry of its first part is nonzero, i. e. ak ; 0. Hence for 7r E Al either a=0, in which
the first part 6 of 7r' is also empty, or ak 1>0
and 6E Do (because aEV
and
case
in S is 1 less than the corresponding entry in a).
each entry
0, and if ß is a singleton ß= (ßl) then µ= (1) E D.
For any 7r E Al ß
ßi) where
If, on the other hand, ß has more than one element then ß=
ßl > ß2 >
> ßi so ß2+1 > ß3+1 > ... > ßi+1
>1 and it follows that
...
(ß2 +1, ßs+1,..., ßt+1,1) ED.
µ=
For any 7r E Al, i0T
so s(7r) =, 61+1k and Q(ir) = yl. The fact that s(ic) > Q(7r),
71 (which ensures that ßl l+k>0,
together with it 0 T, implies that ßl l+k>
even
if ry = 0) and so r= (ßl l+k,
yl, .... 'ym) E D. Thus 6E Do, µED and r ED and
(6, p, r) E J, i. e. for each it E Al, it is defined. What remains to be established
so ?r=
is in which subset of J it is that 7r' lies. Clearly pr, = 1, so 7r' 0 T, and ßl > 92 so 02 +1< ßi and so (since, as always is the
k1 =K L) #2+1 < (ßi l+ k) +LK
which is to say that µ1 < Ti +LK.
case,
It follows that, in this case,s(7r') < Q(mm) ir' E B1, i. e. Al C B1, and so that 7r' = ir.
so
39 Case 2, it E A2: Firstly note that, for any it E A2, the first part is nonempty. For
Thus s(7r) > or(7r) implies that
A2 = 7r ý T, so s(ir) = ßl and v(or) = ryl +lk.
ir E
ßl > ryl +l  k, which is k> ryi (since ß=0, for ir E A2). If there were to be an element
k=0,
A2 with a=0,
then k> ryl would imply 0> ryl. This can't happen as
so
ir E
>0 (with equality if and only if y=
ryl 0), so any element of A2 has nonempty first part. (al,
(but not Do, sinceak A 0) and so S= (a, 1,
Now, a=
ED
ak 1) E DO.
ak)
...,
...,
Clearly µ=0 and since k> yi, T= (k, ryl, ..., y, ) EV (as in case 1, if ry =0 then
(k) E D). Hence,for any it E A2,7r' is defined.
k> ryl =k>0
and so T=
(the length of the first sequence is unaltered). So, since
Clearly it =0 and K=k
(because µl =0 and L= 0). The fact that µl = Ti +LK
k, µl = ri +LK
Ti =
(W), and it E B2. It is clear that
that 7r ý T, so µl = ri +LK=
s(ir) =o
ensures
A2 C B2, and also that . _ 7r.
7r" is
It
Case3i, it E A3 and ß= (ßl):
that
necessary to
check
0, (ßl l+k,
((a, 1,
ryl, ..., 7m)) is actually defined (i. e. it is in J)
ak_11),
7r =
...,
for all such elementsir:
0, so
(note that ak =0 ensuresthat a
Now 7r E A3 = ak =0=
ak_1 1>0
there is no problem in removing the last entry from a).
Hence
Since it E A3, it follows that it 0 T. Thus s(ir) = ßl and a(ir) = 7i +lk.
= ßl l +k > yl (as in case 1, this ensures that if y=0 then ßl l +k >0
s(ir) > Q(7r)
(ßl l+
k) E D). When y 5L 0 this ensures that r= (ßl l+k,
7i, ..., 'ym) ED.
and so Thus for all ir E A3 with ßa singleton, the map 7r } it is defined. Now it is shown
that for suchir, 7r E B3:
k) +LK
0< (ßl l+
so
), hence s(ir)
+LK=
o(ir
s(7r) = µ1 =0<r,
B;. It is clear that W1= ir.
so it E
Clearly ßl >0 T, means that
which, since I
< a(ir ). Clearly µ=0 (and i0 T) Note that in the abovecaseß has length 1 so I=1.
Case3ii, Ir E A3 and ß; (ßl): As in case3i, 5= (at  1, ..., ak_1 1) E Do and b
0). Also as in case3i, r= (ßi
is always defined (because,
again, a0
1+k, 71, ..., 7m)
is a partition in to distinct parts, evenif y=0.
ßi) and t>2,
it must be the case that
Since, in this case, ß= (ßi,
...,
0T
ßz + 1) ED and that µ0
(ßa + I,
so 1r' 0 T. Now µL = ßi +1>1
and 7r
µ=
,
ßl 1 +k+LK=
ßl.
implies that s (7r) = µi =, 82+1 and that Q(ir) = r1+LK=
ßl and so s(ir) < o(ir ). The last inequality,
Since ßl > ß2, it follows that ß2 +1<
E B**.
together with µL > 1, implies that Hence, any it E A3 with two or more entries in the second part is such that I is
definedand an elementof B3*. It is clear that ir' = 7r.
Hencefor any it E A3, it is the casethat iT' E B3 so A3 C B3, and that ir" = ir. 40 Case 4i, ir E A4 and y=0: An element ir in the set A4 for which y=0
must be of
0), 0,0). Conversely, any element which is of this form has
form ir = ((al, a2,
the
... )
> a(ir) and is thus in A4.
s(7r) = oo
Now any such element it of A4 has u(7r) < k, because the slope cannot
it is the case that
the number of parts. For it such that o=o, (ir) <k
exceed
0, (a)). Note that 0> a(ir)
(ir) +lk
((al
1, aQ+l,
ak),
7r =
k=o,
1, ..., aQ ... )
Thus pi > Tl +LK,
together
and so, since pi =0 and r1 = a(ir), pi > Tl +LK.
SK = ak =0 and p=0, gives it E T.
with
Now s(7r) = TM =U=
U(7r) which is the slope of a. The slope of a is less than or
to the slope of 6 which is u(i'). Thus s(7r) <o (7r ), which together with ir ET
equal
implies that 7r E B4. For such it it is clear that ir' = Tr.
If, on the other hand such an element it of A4 is such that the slope of the first part is
1,0), 0,0) for some k>0.
Clearly the map
its length, then ir = ((k, k1,
equal to
...,
1,0) has
is not defined on these elements (note that the sequence (k, k1,
ir'
it +
...,
length k+ 1).
Case 4ii, 7rE A4 and y 0: For such 7; it is the case that s (7r) = ym and that a= o(7r)
is the slope of a. Since s(7r) > a(ir), it follows that T=
o) E V. Note that
there are at least two elements in r, as opposed to case 4i above where T was a singleton. It appearsthat there is a problem, if u(7r) =ki. e. if the slope of a were to equal its
length then, for sinceak = 0, it would follow that 8K = 1. It is necessary investigate
to
there are any elementsfor which this happens:
whether
Suppose it is such that, for some k>0,7r = ((k  1, k2,
1,0), P), E A4 (and
y)
...,
0). Since 7r e A4 = ir E T, it must be the case that ß1 > yl +lk
('y,..., y
y=
,)#
is to say that k> yl. Clearly yl > ym (with equality if and only if
which, as ,ß=0,
1). But s(ir) > a(7r) = y,,, > k. So this would give k> yl > y,,, > k, which
m=
Thus no such memebers of A4 exist, i. e. if it E A4 is
implies the contradiction k>k.
then there must be some entry in a which exceeds by more than 1 the
that y00
such
following entry (the slope is is less than the length).
The fact, in this case, or(7r) <k (as has just been shown) ensures that 6K = ak = 0.
As was shown above, k> yl. Thus 0> yi +0k
Clearly p=0.
which is to say that
> rl +LK
and it follows that ir' E T. As in case 4i, the slope of 7r' is at least
p,
that of ir and so it E B4, and it is easy to see that ir' = 7r. As stated earlier, an element
A4 having y=0 maps to an element 1E B4 for which r= (Ti) is a singleton, and
7r E
E A4 having 70
maps to an element 7r' E B4 for which there are at least
an element it
two entries in T. 41 Case 5i, iv E Bl and y=0: The action of the map it + 7' involves, for, 7r E B1, the
first entry in y. Thus the map map is not defined for iv E Bl when y=0.
removal of the
Thus, it is necessary to estabish which elements of Bl have third part empty: =
Bi has y=0 then s(ir) < Q(mm) ßi <lk,
as yl = 0. Now since Q is
If it E
into distinct parts, it follows that its first part is at least the number of parts:
a partition
#l > 1.Thus problematicelementsof Bl satisfy both #l <1k and j1 > 1,i. e. are such
k=0 so a=0. This implies that /31= 1,which happensif and only if the number of
that
in ß is the number of parts, ß= (l, l1, ..., 2,1) so it = (0, (l, l 1, ..., 2,1), 0)
entries
for 1a positive integer. It is clear that the map it + 7r' is not defined on such elements.
0: Firstly (al,
Case 5ii, iv E Bl and y
0 and S= (al
S=0, or a= (al,
ak)
...,
last part of J is nonzero).
Since it 0 T, it follows that s(ir) = #l
ßl 1 and µED.
implies yl +1k> in which case
ak) E Do so either a=0,
...,
+ 1,
+ 1). Either way SE Do (and the
ak
..., Now s(ir) < a(7r)
and a(7r) = yl +lk.
Note that yj +lk>ß1
ensures that
if the second part of 7r is a singleton, ß= (1) then µ= (yl +l  k).
and that yl +l k>0
Clearly p is nonempty. Now r= (y2, ym) EV (so r is empty if and only if y is a singleton).
..., 0, ir' exists and has 5K
0 and p 54 0 (either
Bl which has y
Thus for each iv E
imply it 0 T). Now, for such ir, yl > y2 (if y is a singleton then y2 = 0) so
of which
+L K which, since it 0 T, implies s(ue) > a(7r').
it, > rl
y2 +L K=
yl +l k>
Thus i' E Al, and it is clear that 7r" = ir.
Thus
Case 6, it E B2: Since 7r ý T, it follows that s(7r) = ßl and a(ir) = yl +Ik.
0 then S= (a, + 1,
0). Now if y= (yi,
k= yl (since
'y)
a, + 1) G.Da
...,
...,
,ß=
last entry, SK = ak +1>0,
clearly u=0 and 7 = (y2i ..., ym} E D. Hence the
and the
is defined and iv' E A2. It is clear that iv" = ir.
map iv + it Suppose, the other hand, that an element7r E B2 is such that the third part y=0.
on
Then, as above,k= yl so k=0, i. e. a=0. Since the other other two parts are empty
follows that the only such element of B2 is it = (0,0,0) and that this is the only
too, it
B2 for which the map ar4 it is undefined.
elementof
Case 7i, 7rE B3*:Firstly it 0 T, so s(7r) =01 and Q(7r) = yl +lk.
(so ßl =0 and l=
it follows that ßl < 71 +Ik,
and since 1=0 Since s(ue)< 0(7r),
0) this implies that yik>0.
Now it is impossible for an element it E B3 to have y ='O. For, as shown in the
Case 4i, any element it EJ of the form it = ((a,,..., 0), 0,0) has
remarks at the start of
thus s(ir) > o(7r) (in fact it was shown that such elements are necessarily
s(ir) = oo and
0. Thus there is no problem in removing the first entry in
in A4). Hence it E B3 =y;
for any 7r E B.
the third part, 42 any such 7r the map is defined, specifically it + it where,
((al + 1)
+ 1,0), (71  k), (72i ..., Y,.))  Clearly 5K = 0. It also follows that
ak
it =
... )
Thus for 0 T. Thus s(7r) = ryl k= ryl +1 k and u(7r) = rye
+LK
it
(as in case5ii, if y= (ryl) then rye= 0). Since yj > rye,it follows that it E A3 and it is
clear that iv" = iv.
CaseIii, it E B**: For any it E B3, the map is + 7r involves removing the first entry
in
from y,ashappened case5i (andcase6). This is a problem if 7=0. It is thus necessary
there are any elementsit of B** which have third part empty:
to establishwhether
(I, Suppose iv E B** is such that ry = 0. Then
so s(ir) = ßl and
it ET
Thus the fact that s(7r) < a(ir) ensurses that ßl <1k.
+lk=lk.
a(7r) = yj
,
As noted in case 5i, a partition into distinct parts has first part no less than the number
ßj > 1, it follows that in this case ßl > 1. Thus
Bl >1 but since 7r e B3*
of parts, ,
,
must satisfy both ßl <lk
and ßl > 1, which is
of B3* having 7=0
an element
impossible. Thus it E B3* . ry ,E0. It remainsto investigatewhat happenswhen it E B3* and ry 560:
+ 1) 0) E Do, and so SK = 0. Since it 0 T, it follows that
ak
...,
Ql  1. From the
Thus s(ir) <o (ir)
ßl and a(7r) = yj +lk.
+lk>
'yl
s(ir) =
0,
ßt
01  1,
definition of B3*, #1 >1 and so I.c = (yl +lk,
E D. Clearly µ
..., 1)
0 T. Note that third part of it is empty if and only if ry = (7yl).
so it
Clearly a= (a, + 1, V T, it follows that
Thus 7r is defined for all it E B3*. For these such it, since
(7r) = ri = 'Y2+LK
(as usual y=0
+lk
and u
rye= 0).
p, = yl
s(ir) =
Clearly yj > ßy2 so s(ir) >o (7r). Hencefor thesesuch it, ir' E A3 and it" = iv.
and
is defined
Looking at case 7i and case Iii together, it is seen that for any it E B3,
is an element of A3. Hence BB C A3, and it is clear that 7r" = i.
and
For if there were such an
Case 8, iv E B4: Firstly, no element it E B4 has y=0.
then the fact that it ET would imply that s(ir) = oo > a(ic) (this
element with y=0
in the remarks in case 4i, and also in case 7i).
was explained
0, it follows that s= s(7r)
Since any it E B4 is an element of T and has y;
= ym
is the slope of a. Since it E T, it must be the case that flu > 7i +lk.
Hence
or(7r)
and
(with equality if and only if m= 1). Thus
(since /3 = 0) k> yl, and clearly yl
k> So it = ((a, + 1, ...,
defined for all it E B4. ay + 1, a3+1, ..., ak), 0) (717
... ' 7`m1)), i. e. the map 7i + it is Clearly the length of the first part of iv is left unchanged, i. e. 6 has the same number of
does. Hence k=K
and this, together with 71= yl (unless y=(, yl)) implies
as a
parts
K> 71.If, on the other hand, y=(, yl) then r1 = 0, so clearly K> r1. Either way
that
+L K (because µ= 0). Now k> ym means that
K> tj =0>
+0 K so µl > 71
rl
last element of a is left unchanged, so 5K = ak = 0. Thus it E T. All that remains
the
do is show that s(7r) > u(ir ):
to 43 Now if ir' has last part empty, i. e. T=0,
then (since 7r' E T)s(l)
oo > Q(ir'). If,
=
then s(ir) = TM = y,
the other hand, T0
and Q(ir') is the slope of 5, which is
on
1
S='irºi. Thusrym1>'Ym=s(7r')>U(7r'). Thus, for all 7r E B4, it is defined and since s(ir) > o(ir ), SK =0 and p=0
follows that 7r' E A4. HenceB4 C A4 and it is clear that ir" _ jr. it Looking at the above eight cases it can be seen that whenever it EJ is such that 7r' is
defined then so is 7r' and in fact ir" _ ir. From case3 it is seenthat A' C B3 and from case7 that B3 C A3. This, together
implies that A3 = B3 and B' = A3. This can't be done for the other three
with ir" = ir,
becauseBi, B' and A'4are not defined (since they contain elementson for
pairs of sets,
the map is not defined). To remedy this, the following setsare now defined:
which
X+ := {7r EJ: 7r= ((k, k 1,
1,0), 0,0), k? 0},
..., xo := {(r, 0,0)},
X: = {ir EJ: it = (0, (1,1 1,
Now, given that 7r" = it (wheneverir` is defined), it follows from cases1 and 5 that
Ai = Bl \ X_ and that (B1 \ X_)ý = Ai.
r"
Similarly, again given that
_ ir (whenever it is defined), it follows from cases 2
6 that A'3 = B3 \ Xo and that (B3 \ X0)ß = A3.
and Finally, given that i=
(A4\X4'=B4 and that ir (whenever7r is defined), it follows from cases4 and 8 that
B4=A4\X+. 44 What all this means is that if iýX
:= X+ U Xo U X_ then 7r and it can be paired off
Recall that the map it } it is weight preserving, i. e. wt(7r) = wt(ir),
with each other.
d(am) = d(ir) but (1)°('r') # (Thus
the map preserves everything apart
and also
from the sign, so the contribution made by a given it in the sum
E(1)c(")zd(")gwt(")
7rEJ is cancelled out by the contribution made by it and so the above sum reduces to a sum
,
X. To be precise,
over
E(1)c(a)zd(a)gwt(lr) (l)c(a)zd(1r)gwt(lr) + E(l)c(7r)Zd(?
l =E aEX irEJ\X irEJ zd(7r)gwt(rr) E
rEJ\X
c(, )odd 7rEX c(7r) is evenif and only c(ir') is odd, the first two sumscancel since, for 7r EJ\X,
and
out and so (_ 1)c(a) zd(a) gwt(lr) zd(ir) gwt(Tr) +E wEEJJ\\X
c(a)even r)gwt(lr) E( 1)c(r)zd(ir)gwt(w) E(_i)c(1r)zd(7r)gwt(7r)
l _ 7rEX irEJ 1)c(7r)zd(7r)gwt(Tr) _j
IrEXouX+UX_ _ nEX+ 7rEXo (1)c(a)zd(7r)Qwt(a) (_1)c(s)zd(r)gwt(7r) +E (_1)c(?r)zd(?
r)gwt(r) +> irEX_ for all 7r E X, becomes
=0
which, since c(am)
E zd(a)gwt(lr) +E +E zd(a)gwt(lr)
lrcx+ 7rEXo zk+1 qk+k1+... =1+E zd(7r)gwt(7r)
7rEX_ +l+o lql+t1+... +z
lj>>11 k>O
2 1 a +Ez =1 +Ezk+lq
k>O q2 l>1 (k+1)2(k+,
)
+z
zk+1q2q2 =1+ 1 k>O lq
1>1 1+
.ý nq RZA 2q2 n>O f32n
+ Zn n<0 and so
_1)c(7r)zd(a)gwt(ac)
n>O irEJ This proves (4.2), and so (4.1). 45 znnq"22 ". +2+1 Chapter 5
Ranks and biranks 5.1 A generalisation of an identity of Fine
The following new identity is a generalisationof an identity due to Fine:
ý2 ý (1)m(Lmtmxm2q... (tlxngi q)ntn
(q; On _. (q; q)m(txm; m>O n>o (5.1) q). is
Convergence ensuredby stipulating that jai < 1, Iil < 1, Ist <1 (and, as usual for
infinite series,tqI < 1).
This identity is now proved by showing that the coefficient of t' on the right is the
as that on the left:
same
Firstly the coefficient of tk on the right is the coefficient of tk in the finite sum
kE (1)'namtmx'nýq
(txm; (q; q) zm
2
q),,,, m=0 Now, putting n txtmin the identity of Rothe, (1.10), gives
oo and z=
tlxml 1
(txm; (q; q)
1
1>0 q) tk on the right of (5.1) is the coefficient of tk in
and so the coefficient of
(t (1)mamT xml .k
(q;
m)L_/0 q)m (q;
1>0 q)t k n2gZa'
(q; mL_`0 q),. M tlxmiim)
(q;
! >m the coefficient of tk in this is
and
k m\ý_o` !
_
l1 ilmamxm'q'"22
(qi q)m m xm(km)
(qi q)km 46 k Pf_1`mamxkmqm!
\
(q; q)m(q; Q)km
m\ý_`0
, q), m in terms of qbinomial coefficients, is
which,
k Z(1)mamx
([k]4)k
Finally, by putting z=
the left of (5.1) is km ^'2 m (5.2) q2m m=0 in identity (1.9), it can be seen that the coefficient of tk on
axkq ýaý k4i q}k
_
(4'i 4)k 1kz IC ýý_1)mamxkmgrn2 (Qi q)k (5.3)
m ,
n10 Identity (5.3) shows that the coefficient on the right of (5.1) equals the coefficient of tk
left of (5.1), proving (5.1) as required.
on the 5.1.1 The above identity and partitions
When x=q is put into in (5.1) the resulting special caseis
(aqn+l; (_1)'amtmg3m: q)ntn (q; q) m (tqm; q)
00 is due to Fine, it is identity 25.94 in [11]. Fine uses this identity to find the generwhich
function for the rank. This is done here too, but in more detail than in Fine's book.
ating
Before doing this it is of course necessary to define the rank of a partition, this is done
in the next section, but first it should be noted that putting t= q2 into the above identity
gives lagn+li
4)ng2n
l
(qi ýi)n
n>0 1 EC1)mamn8m2+5m
(qi q),,,,
Yl
m>0 rl  q+') for ha nonnegativeinteger putting a= qh and rearranging gives
and
E
n>O 5.2 (2n± h]q2n
L E(1)mggm
(4'' 4')°° z , am+hm(1
qm+1). (5.4) m>o The rank generating function The partition function, p(n), is known to satisfy certain congruences: including (1.19),
(1.20) and (1.21) above which, to recap, state p(5m + 4) =0 mod 5,
p(7m + 5) 0 mod 7, p(11m + 6) =0 mod 11.
Thesecan be proved by considering the generatingfunction, (q; q); of p(n), how to do
for (1.19) was outlined in section 1.3.2. Now proving that p(5m + 4) =0 mod 5
this
47 by invoking the generating function is an instance of a noncombinatorial proof: It has
been established that 5 divides p(4), p(9), p(14), ... but the proof does not describe how to
the set of partitions of weight 4,9 or 14 etc as the disjoint union of five equinumerous
write subsets.
Suppose that there were to be a map g: P 4 Z5 with the property that for
5, it were the casethat
n4 mod
Z5 =>: #{, \ EP:
xe wt()) and g(\) = x} = 5p(n) =n then there would indeedbe five equinumeroussets.
The first person to actually invent a statistic with the properties described above was
Dyson. He defined, in [9], the rank of a partition as the first part minus the number of
parts: rk(A) := A1 #(A) (5.5) Al k where A= (A1,A2P )ºk). The rank of the empty
or, equivalently, rk(A) :=
...,
is taken to be zero, rk(o) = 0. The rank is therefore a map from P to Z, not Z5,
partition
but reduction mod 5 remediesthis.
To do this it helps to introduce the following standardnotation:
N(m, n) := #{A EP: wt(A) =n and rk(A) = m} (5.6) and
N(r, m, n) := #{a EP: wt(A) =n and rk(A) r mod m}. (5.7) Thus N(m, n) is, by definition, the number of partitions of weight n having rank m
N(r, m, n) is defined to be the the number of partitions of weight n whose rank is
and
congruent to r mod m.
If r' =r mod m then N(r', m, n) = N(r, m, n), so it suffices to find N(r, m, n) for
By considering the dual, Aý of a given partition A, it is easy to see that
0 <_ r<m.
Hence N(m, n) = N(m, n) and so N(m  r, m, n) = N(r, m, n).
rk(A') = rk(A).
Lm/2j.
Hence it transpires that it suffices to determine N(r, m, n) in the range 0<r<
It is clear that
N(r, m, n) N(t, n)
t_r mod m and that
N(m, n) p(n) =E
mEZ
n1 E N(m, n) m=n+l and also that m1 N(r, m, n) p(n)
ro 48 As an example, consider the partitions of weight 9: There are six such partitions
having rank congruent to 0 mod 5;
(3,3,3), (2,2,1,1,1,1,1), (4,3,1,1),
(7,2). (5,1,1,1,1), (4,2,2,1), 6. Likewise N(1,5,9)
=
9 having rank congruent to 1 mod 5; = 6, because there are six partitions of weight Hence N(0,5,9) (4,3,2), (2,2,2,1,1,1), (3,1,1,1,1,1,1), (5,2,1,1), (4,4,1), (8,1). it also follows that N(4,5,9) = 6, as the dual of any of the above partitions has
and so
to 4 mod 5. Finally N(2,5,9) =6 (and so, by looking at the duals,
rank congruent
N(3,5,9) = 6) because the following six partitions;
of
(3,2,1,1,1) 1), (1,1,1,1,1,1,1,1,1), (5,3,1), (5,2,2), (2,2,2,2,1), (6,1,1,1). So
N(0,5,9) = N(1,5,9) = N(2,5,9).
Dyson, noticing equalities such as the one above, conjectured that
N(0,5,5m + 4) = N(1,5,5m + 4) = N(2,5,5m + 4) (5.8) is both a proof, and a combinatorial interpretation of the statement p(5m + 4)
which
0 mod 5. Dyson also conjectured that N(0,7,7m+5) N(1,7,7m+5)
= N(2,7,7m+5)
= N(3,7,7m+5)
= (5.9) is likewise related to the congruence p(7m + 5) 0 mod 7. Dyson stated his
which
in [9]. Later they were proved by Atkin and SwinnertonDyer, [5].
conjectures 5.3 The generating function Let P denote the set of all partitions. The generating function for the Dyson rank alluded
to earlier is Ez
AEP a
rk Oq wt a (1 _  Z) (qi q),,,, E(1)
nEZ n n(3n+1)/2 1zn q (5.10) function for the rank was first presentedin [5], in a slightly different form
The generating
to (5.10) above.
49 Identity (5.10) is now proved by writing the expression on the right as a power series
in z, where the coefficients are qseries, and then using (5.4) to show that this is the same
the left of (5.10).
as the expression on Now, the expressionon the right of (5.10) is
(1 1 z)  n(3n+I)/2 (1)n (q, q)oo 1z+ 1 zqn nEZ\{o} the sum in the aboveexpressionis
and
(1)nqn(sri+i)/z
1 zqn  nEZ\10) ncý
(n#o) gn(311+1)/2 Egn(3n+1)/2 n>o
n even gn(sn+_
1 zqn + 1 zqn n<O gn(3n+1)/2
 1 zqn n>o 1 zqn  zqn n<0 q(3fl2+n)/2 +1  zqn 1 zqn n odd q(3n2n)/2 q(3fl2+n)/2 n>0 1 gn(3fl+l)/2 n odd n even E gn(3nt_
1 zqn
n odd q(3n2n)/2 +1 1 zqn zqn n odd n even is equal to
which
q(3n2+n)/2 q(3n2+n)/2 E1  zqn z1 1 q(3n2+n)/2 zign  g6ma+m h1 1 zg2m z1 g6m2+m g6m25m}1 zlg2m j_. /t h>O m>0 +Z' zhg6m25m+1+h(2m1)»
m>0
m> O) h>0 j_. zhg6mz5m+I+h(2m1) `(Z
m>0 h>0
(E h>O h>0 z1g2m1 ) E ))
j> g6m2+(2h5)mh+1 g6m2+(2h1)m h>O m>O (> ý Z h g6m2+(2h+1)m) zh g6m25m}1 h>0 (EZh(E E zign zhg6m2+m+2hm m>0 (Z 1 1 (E [ý
z1 z1 1 zg2m1 z1 m>O (zhg6m2+m+2hml
ý'` ý"` 1 zqn =2m1
m> O m> O q(3n2+n)/2 m>O
(Z + zh
h>0 m>0 50 g6m2+(2h7)mh+2
m>0 Multiplying through by 1z gives (1 gn(3n+I)/2 (1)n z) 1 zqn nEZ\{0} E Zh h>0 ( h>0 g6m2+(2h+1)m) L
+^ m>O
[ý ( _Ez (1  z) 1 (E
g6m2+(2h7)mh+2) 1\ h>0 and so g6m2+(2h1)ml
mf>>O h+l /% m>0 zh11([ß
t\ hh>>00 g6m2+(2h5)mh+11 j_, h>0 /J J m>0 ý` + g6m2+(2h+1)ml J g6m2+(2h7)mh+21 jý h>0 hh>OO zh+1 zh fý ( +E m>0 1 (`ý"` m>0 _E {\ h>0 /l g6m2+(2h5)mh+11 zh+l 1\ 6m2+(2h1)m
Q, Z h m>0 (JE zh E Z m>0 n(3n+1)/2 E(1)ng 1 zqn nEZ
(Eq (
g6m2+(2h+1)ml zh =1+E
h>0 6m2+(2hI)ml _zh J
h>0 m>0 m>0 (ý (Eq6m2+(2h5)mh+l`
zh _E
h>00 1 m>0 _ zh
h>0 l g6m2+(2h7)mh+21 J m>0 ( (f
zh `m>0 +E + g6m2+(2h1)m) zh g6m2+(2h+1)m1 J h>0 h>0 m>0
(E +E zh
h>0 E g6m2+(2h7)mh+21 h _Ez
h>0 m>0 g6M2+(2h5)mh+ll . (5.11) m>0 the whole
The above expression is clearly invariant under z 3 z1. When z=1
is equal to 1 and so, dividing by (q; q),. implies that, the left of (5.10) is invariant
thing
(which is to be expected as the rank of each partition is minus the value
under z + z1
its dual) and equal to P(q) _ (q; q) when z=1.
of the rank of
Now, for h>0, the coefficient of zh in the expression on the right of (5.11) is
E g6m2+(2h+1)m g6m2+(2h5)mh+1 _> _
m>0 m>0 rn>>O g6m2+(2h1)m +E g6m2+(2h7)mhß2
m>0 if the h in the aboveexpressionwere replacedby IhI, then it would also hold
(note that
for negativeh). This is equal to
C'` g6m2+(2h7)mh+2(1 l  q2m1  gsm+h2 mj>>O 51 + q8m+h2) (1
`J q g6m2+(2h+5)m+h =qE 2m+l  q6m+h+4 + 48mth6 m>>0 _ qh+l E (1
g2mh. q6m2+5m `  g2m+1J 1E m>>0 ýh q(2m+1)h. (1
`J  g6m2+llm+4 2m+2\
ß, m>0 Fl 58 (1 qhg+3e22 qg+l)  qhg s=2m
m>0 { 3x22 ba (1  qs+1) s=2m+1
m>O _ qh+l ä as(1 y(_l)sghsýs  q'+1) s>ol the coefficient of z° on the right of (5.11) is 1 2q2+ 2q7+ + (1)82g3eý '+
and
...
It hasbeenshownthat the coefficient of zh in
(1 n(3n+1)I2 z (qi q)) (1
nEý
nEz  zqn) is (1 qn+l E1)Sgrs+ý(4; q),,,,
8>0
1 (1+2(_l)8q) (q, q)oo 5.4 if h>0, q'+1) if h=0. (5.12) (5.13) s>o Ranks If a partition A has rk(A) = h, where h>0 then it must have first part Al=n+h+1
parts some n>0. Conversely whenever it is the case
number of parts #(A) =n+1
and
that, for some fixed h, there exists some n such that Al =n+h+1
and #(A) =n+1
then it must be the case that rk (A) = h. Thus the graph of a partition having rank h has
n+h+1 dots in the top row and h+1
" " dots in the left column, as shown
" " " " " " 0 this leaves and n+h by n rectangle `inside' that is to be filled with some partition. Thus,
by lemma (1.8) summing over all such partitions gives
[2n + hi
ý`
2n+h+l
f' gwt(a) = q,
n and summing over all n gives
E gwt(A) = qh+1 E
n>o rk(A)_h 52 r2n +h
q2nLnJ (5.14) For partitions having rank 0 there is a problem as the empty partition has not been
This is easily remedied; for h=0 identity (5.14) becomes
counted.
I"2n l 1, + q 1:
gwt(A) =
rk(.1)=0 (5.15) q2n n>O Now the expression on the right of (5.14) is qh+l multiplied by the expression on the left of (5.4), so for h>0
22 Sm+ýttFl(1 qh+l
gwt(A)  qm+l). (rk(A)=h
qi q). (5.16) m(>_"i0 by (5.12), it is seen that for h>0, it is indeed the case that summing gTDt(A)
over all A
and
having rank h is the coefficient of zh in the expression on the right of (5.10). This being
it must also hold for h<0, for it has been shown that this expression is invariant
the case,
under z pz1.
It remains to check the case h=0:
qýoo E(1)mqSm 1 +(
`ýý Because 2ism (1 qm+1) _q1,.
`ý m>0 (1+
+2 (lýsqý2 ) s>0 in (5.4) and (5.13) imply that qu(a) summed
it follows that (5.15) together with h=0
having rank zero is indeed the coefficient of z° in (5.10). This proves
over all partitions
(5.10). 5.5 The Birank
The rank was invented to explain congruences in the sequence p_1(n), that it does so is
from (5.8) and (5.9). There are similiar congruences in sequences having the form
seen
(n) where k; 1. In particular, for x=2,3
or 4,
pk p_2(5n + x) 0 mod 5 (5.17) that p_2(n) is the coefficient of qn in (q; q) , can be proved by an argument
which, given
to that used to prove (1.19).
similar
There is, so far at least, a relative shortage of Dysontype ranks related to such sebut [14] looks at the sequencep_24(n). What follows here is about the sequence
quences,
a birank is introduced related to the congruences in (5.17). The behaviour of
p_2(n):
birank modulo 2,3,4,6 and 8 is investigated, as well as modulo 5 which is perhaps
the
the most interesting case as this gives rise to actual congruences in the sequence p_2(n).
A paper on the mod 5 case, which is dealt with in theorem 4 below has recently been
[19].
published 53 The birank of an ordered pair of partitions, ir = (A, iz) is here defined to be the number
of parts of the first partition minus the number of parts of the second partition: b(7r) := #(ir)  #(u) (5.18) A= (A1, A27 Ah) and it = (µl, µ2)
b(7r) :=k1
where
µt)"
or, equivalently,
...,
...,
The following notation is analogous to that for the rank ((5.6) and (5.7)), R(m, n) := #{7r E P2 : wt(7r) =n and b(7r) = m}, (5.19) the weight of the ordered pair it = (A, µ), is defined to be wt(ir) := wt(A) +wt(µ),
where
and R(r, m, n) := #{ir E P2 : wt(7r) =n and b(ir) r mod m}. (5.20) Clearly b(A,p)=b(µ, A) and so R(m, n)=R(m, n) and R(m  r, m, n)=R(r, m, n).
In this respectthe birank introducedhere is similar to the rank of Dyson.
As with the rank, it is necessaryto look at the generating function for the birank.
Fortunately this is easierthan for the rank. It is clear that
E zb(") qwt(") 1
_ (zq; (z1q;
q),,,, q),,. zrEP2 (5.21) which is to say that
E R(m, n)zmq' =/1
lzgi mEZ n>0 q)oo(z1Qi q)oo (5.22) In the next section, the behaviour of the birank is investigated by substituting suitable
for z into (5.22). The approach is (for a specified value of k) to put z=w, where
values
2xi
is some root of unity, usually w=ek
in which case it follows that
w R(m, n)wmq" = R[0, k] + wR[1, k] +
+ wk1R[k  1, k]
...
mEZ n>O R[r, k] is defined as
where
R[r, k] :=E
R(r, k, n)q"
n>O
R(s, n)ßn =>>
s_kr n>O b(ar)=kr (here k means congruent mod k). 54 (5.23) 5.6 The behaviour of the birank The following theorems hold for n any natural number, bar the exceptions listed.
Theorem 1: R(0,2,2n) > R(1,2,2n), R(0,2,2n + 1) < R(1,2,2n + 1). (5.24) Theorem 2: R(0,3,3n) > R(1,3,3n),
R(0,3,3n+1) <R(1,3,3n+1), The only exception is R(0,3,5)
Theorem 3:
R(0,4,4n) <R(1,3,3n+2). (5.25) = R(1,3,5). > R(2,4,4n), R(0,4,2n) R(0,3,3n+2) R(2,4,4n > R(1,4,2n), + 2) > R(0,4,4n R(2,4,2n) + 2), > R(1,4,2n), R(0,4,2n + 1) = R(2) 4,2n + 1) < R(1,4,2n + 1). (5.26) With two exceptions,R(0,4,2) = R(1,4,2) and R(0,4,4) = R(2,4,4).
Theorem 4:
R(O, 5,5n) > R(1,5,5n) = R(2,5,5n),
R(1,5,5n + 1) > R(0,5,5n + 1) = R(2,5,5n + 1),
the following equalities (which imply (5.17)),
and R(0,5,5n + 2) = R(1,5,5n + 2) = R(2,5,5n + 2),
R(0,5,5n + 3) = R(1,5,5n + 3) = R(2,5,5n + 3),
R(0,5,5n+4)
The only exception is R(0,5,6)
Theorem 5: R(1,5,5n+4)
= R(2,5,5n+4).
= (5.27) R(1,5,6).
= R(0,6, n) > R(2,6, n),
R(0,6,2n) > R(3,6,2n),
With exceptions:n=1,2,4= R(0,6,2n + 1) < R(3,6,2n + 1). R(0,6, n) = R(2,6, n) and n=1,3,5,7=>. (5.28)
R(0,6, n) R(3,6, n).
Theorem 6: R(0,8, n) > R(4,8, n),
The first inequality failing only at n=1, R(l, 8, n) > R(3,8, n). the second only at n=0. 55 (5.29) In the following proofs, repeated use will be made of the triple product identity,
[z; q](q; q). = n2n E(1)nq (5.30) 2 nEZ This is identity (4.2), proved in chapter4.
The triple product is useful here becauseit be used to obtain dissectionsof qseries.
A dissectionof a series,F(q), is an expressionof the form
+ ... {qni Fni (qn)" F(q)n)
For example, 1_1q ++ 1q3 1q q2
1_q3 q3 is the 3dissection of 1/(1  q).
When looking at dissections of certain qseries, the following notation will prove useful: [a, b,..., x; q] := [a; q][b; q]...[x; q]"
Theorem 1. It is necessary consideronly R(0,2, n)  R(1,2, n). Putting z= 1 in (5.22) gives
to
ZER(m, n) (1)mq' =/1 mEZ n>O and so
R(m, n)(1)mgn (q; q2)0
= mEZ n>O is to say that
which
R(m,
m evenn>O by (5.23) (with k=2,
so, n)qn R(m,  n)qn (q; g2)o
= m odd n>0 sincew= 1),
R[O, 2]  R[1,2] = (q; q2)'.. (5.31) Thus it is necessary to show that the coefficient of qn in (q; q2)ß is positive if n is
if n is odd. Now
even and negative
(q; g2)ö0 = [Q; q2]
so [q; q2] (q2; g2)oo
(q; g2)öo =
(q2; q2)"" by the triple product identity (q + q2in (5.30) and then set z= q),
and so,
E(_1)nnz
(qi g2)oo
=1 (ql;
)00
q2
ncz 56 (qn2 1
ff
q' _qn2' 2)
X1' oo n odd Haven (qn2 1E
(q2; g n=2m
mEZ n=2m]
mEZ (2m1)2 q(2m)2 _1E
`q2; _E g2)OO
MEZ
1EQ, `q2; Qns _> 2 )00 mEZ
4m2 q2)O° g4m24m _qE mEZ mEZ This is the 2dissectionof (q; q2),,. This, together with (5.31), gives
E[ R (O 2 ), ,n R(1 2 n)] q n=
,, n>0 E( (92, Q2)c q 4m2 q ,>  MEZ g 4m24m mEZ is equivalentto
which
E[R(0,2,2n) =1 R(1,2,2n)]g2'z n>U >q4"',
4Z' qa)°°
. mEz and
E[R(0,2,2n + 1)  R(1,2,2n + 1)]q n>O 1=q` g4m24m (q2; q2)°° mjEEZ it is clear that the above two equations may be written as
and
E[R(O, 2,2n)  R(1,2,2n)]qn 1
(4i 4)00 n>0 E qMs MEZ and
Y[R(1,2,2n + 1)  R(0,2,2n + 1)]q" _ (q;
)0
q1 n>0 2M2 _21n q.
mEZ All the coefficients of both these qseries are positive, which proves theorem 1.
Theorem 2. Again, there is only one thing to look at, namely R(0,3, n)  R(1,3, n). Putting
z=w, where w=e9 znc (5.22) gives
in L, Z R`m, n)wmgn 1
= mEZ n>0 (wq; q)oo(wlq; q)oo by (5.23) (here k= 3),
so,
(q; q)°°
R[0,3] + wR[1,3] + w2R[2,3] _
(q3;q3)00
57 (5.32) Now
1: (q; 4)00
= n (1)"g3n22 nEZ is identity (1.15) and is equivalentto (1.16), the latter having beenprovedby using
which
bijection of Franklin in section2.1. The aboveidentity may be written as
the
[q; g3](q3; q 3w2 3)00 n _ (5.33) nEZ is also an instance of the triple product identity, specifically q + q3 and then set
which
in (5.30). The sum on the right, over all integers n can be split into three seperate
z=q
i. e. (5.33) may be written as
sums,
[q; g3] (q3; g3)00 n Sn2 n+E( n 1)nq3nß + 27m2Sm `ý(1)"`q 47mß____ E(1)mq 2q 2 q2 q ý(1)mq 27mß s _m mEZ mEZ MEZ n n=3mß1 n=3m1 n=3m (1) n Sn2 in (5.30), followed by the appropriate value of z (z = q12, z=
and putting q 4 q27
for the first, second and third sums) gives
and z= q3 q6 (427' 427)0o
[q; q'3](g3; g3)oo= [q'2; g27](q27;
427)00 q[4'6;4'27](g27'
g27)0o 92[43;9'27]
and so
[q; 43] = 1_2
[93,99,412; [Qs, 4s, q9; q27] [q6,49,9'12; q27] q27] together with (5.32), implies that
which, R[0,3] + wR[l, 3] + w2R[2,3] =Iq,3 6) s 27 ;q]
,qq 2
3 9)
[q 276[q [q q12
1q ,q] q9, q271
q12;
, but w2 = 1 w and R[2,3] = R[1,3], so
E[R(0,3, R(1,3, n)]qn =1qn) [q3, qs, q9; 4'271 [q3,9'9, q12; g271 n>o q2
[q6,49, q12;927] implies the following three results, from which (5.25) follows,
which E[R(0,3,3n)
n>0 >[R(1,3,3n 1 R(1,3,3n)]gn =
[q, q2,43;q9]'
1 + 1)  R(0,3,3n n>o
j, "[R(1,3,3n+2)R(0,3,3n+2)]q'
n>o 58 + 1)]qn =
[q, q3, q4; q9] ,
=2 [4 31q4; s
9,4]
, Theorem 3. The quintuple product identity can be statedin the form,
[z; q] [qz 2; g21(q;q)00 =E z3ngn(3n1)/2ý1  zqn). (5.34) n A proof (of an equivalentversion of (5.34)) can be found in 3.2 in [18].
The quintuple product identity will be usedto find the 2dissectionof the Euler prodwhich is neededwhen looking at the behaviour of the birank mod 4.
uct,
Startingwith (1.15),
22n
3n (q; q),,,,
nEZ
n (1) = 3n2 (1) q a+ n 3n2 nn (1) q 2+ n=4m+1 n=4m
E g24m22m 321 E(g2)3m(q16)m
_ (1 q2 { +> mEZ (q 2) (q16)m) (1)"q 3nß n s n=4m}2 g24m2+10m+1
ýL 1 mEZ mEZ n n=4m1
g24m2+14m+2 E 3n2 g24m2+26m+7.
rnnERZ
3m1  gE(q6)3m(q16)m (1 (g6)(q16)m) mEZ mEZ and so, putting q 4 q16and z=
(q; q).. q2 and then z= q6 in (5.34), gives [q2; 4'16][9'12;
[14;
g16)00 q[q6; 4161 432](9'16;
g32}(q16;
gls)oo.
)= Now, putting z=i (5.35) in (5.22) gives
EE R(m, n)imgn =( 1
(iq; mEZ n>O q)00(i'q; q)00 (q2; q')...
(g4; 44)00 but, 1: 1: R(m,
n)imgn = R[0,4] + iR[l, 4]  R[2,4]  iR[3,4]
m n>O R[0,4]  R[2,4]
=
thus
E[R(O, 4, n)  R(2,4, n)]q' = (q2; 94),.. (5.36) n>O Now (q; q2)<, is the generating function for partitions into distinct odd parts, those partitions having an odd number of parts being counted negatively. Such a partition of an even
integer will have an even number of parts, and conversely a partition of an odd number
have an odd number of parts. Furthermore the only positive integer for which no
will
R(0,4,4n) > R(2,4,4n) and that
exists is 2. This proves that n1=
such partition
R(0,4,4n) < R(2,4,4n). It also proves that R(0,4,2n 59 + 1) = R(2,4,2n + 1). It has already been seen, by putting z= 1 into (5.22), that R[0,2]
(q; q2)ß.
R[1,2] =
Since R[0,2] = R[0,4] + R[2,4] and R[1,2] = R[1,4] + R[3,4] = 2R[1,4], it follows
that E[R(0,4, ,.
+ R(2,4, n)  2R(1,4, n)]qn = (q; (12)
n) (5.37) n>O Adding (5.36) to (5.37) gives
((q2; E[R(0,4, n)  R(l, 4, n)]qn 1
g4)00 + (q; 42)2 = 21 / n>O
(q; 92)00 ((_4;
` + (q; 42)0)
g2)00 2 (4; 4'2)00 [4;
2 R[1> 4] =1
 (q4; 44)00 (q4;
q4)00 g4](q4; 4'4)00 + [q; 4'4 (4'4; 4'4)00
(q4; q4)oo Putting q + q4, and then z= q and z=q
R[0> 4] f into (5.30), it is seenthat (q; g2)oo (q2n2ný(_1)nq2n2_n)
(q4; 44)00
nEZ
nEZ
nEZ and so R[0,4]  R[l, 4] = (q; g2)oo
(q4; E )O° (5.38) g8na2n nEZ which, together with q 4 q16 and z=
R[0,4]  in (5.30), shows that
qs R[1,4] and so
R[0 >>  R[1,41 4] (q; q)ý[q6;
g16](q16; q16).
(q2; 4)00(q4; q4)00 (q; q)o0[g12'g32](q16;q16)ß
(q2; g2)o, (q4;
g4)oo[q6; qls] (5.39) the expression on the right is (q; q),,. multiplied by a function of q2, so with the
now
2dissection of the Euler product (5.35), this becomes R[0,4]  R[1,4] =
[412;g32]2
[q2; 4'16]
(q1 4l6)ä0
[4'6;416][4'4;432][412;q32](ql6'
gl6)2
(q2; 92)0044; 44)00[46;q16] q
(q2; q2)00(ß4;44)co[g6;
ql6]
so R[0,4]  R[l, 4] =1
[q6; 4i6]2[44,48; 9sa aq
This proves identities (5.42) and (5.43) below. 60 (416' gls).
(q2; qa). (4si4a). (5.40) For the two other identities, the process is very similar. Subtracting (5.37) from (5.36),
instead of adding them together, gives
E[R(1,4, R(2,4, n))4'" = 2((42; 44).  (q; 42)2
n)  ) n>o
(q; g2)00((q;
2`
(q; g2)00 [q; ) (q4; 94)00
(q; q2),
42)00(q4; q4). g4](q4; g4)ao [q; q4] (q4;
g4)00  (q¢,q4)0 2
to (5.38), this gives
and so, similarly 2ý
R[1,4]  R[2,4] =q (q4; E g8n26n )o ,,o nEZ in (5.30), showsthat
which, togetherwith q + qls and z= q2
(q; q)00[g2; gls](q4 ; ql6)ý
R[l, 4]  R[2,4] =q
(4,4 )00(4
)co
;4 (5.41) be done,as above,is replace (q; q)... by the right hand side of (5.35)
Now all that needsto
This results in
and simplify.
(q16 16)0q
.
R[104]  R[2,4] =q
q21 [4a.
(42;g2)00(98;
,
q8)0
gl6]2[q8,q12;
q32]2
from which (5.44) and (5.45) below follow.
E[R(0,4,2n) R(1) 4,2n)]q'
 1
== [q3; n>0 L[R(l, 4,2n + 1)  R(0,4,2n n>0
E[R(2,4,2n) E[R(1,4,2n (qa' qa)O° + 1)]qn =
(qi q)oo(g4i g4)oo R(1,4,2n)]gn
 n>o
+ 1)  R(2,4,2n n>0 4s]z[g2,44; 416]7 1
= q([q; az4 [q q6; 4]
is
q
, + 1)]qn = (qa' q4)2
4.
)oo
lqý 4)oo(q iq (5.42) (5.43) (5.44) (5.45) This concludes the proof of theorem3 (5.26), and also shows directly that
R(0,4,2n + 1) = R(2) 4,2n + 1). 61 Theorem 4. in
Putting w= e27ri/5 (5.22) gives
R(m, 1
n)wmg __ mEZ n>0
(w2gi wq in the other version of the triple product (1 + (w ++ (q5; q5)ß q)00 (q5; g5)00 the term on the right, by q 4 q5 and z=
identity, namely (1.22), is equal to 1 q)oo(gi q)oo(w2qi _ 1 (wq; Q)oo(w1q; q)oo ((_1rq25n2+1on wi)qs )
 qio... (w + wi)q E(_1)ng25n2+5nl
+
l (q5; g5)00 nEZ nEZ by suitable values in (1.22) is equal to
which
25
(q25' . qq5 )co ([qlo; q251) q25]+(w+w_1)q[q5; = q5; i' 51 25 lW 'i' W1) 10_ 25 but,
ZE R(m, n)wmgn = R[0,5] + wR[1,5] + w2R[2,5] + w3R[3,5] + w4R[4,5]. rnEZ n>0 Thus (since w2 = 1 w and R[l, 5] = R[4,5] and R[2,5] = R[3,5]),
w3  w4, R[0,5]  R[2,5] + (w + w4) (R[l, 5]  R[2,5]) = r 1 + (w + w1)
[4104425]
ß425] This (and the irrationality of w+ w1) proves (5.46) and (5.47), and so proves (5.27),
1 n>o
E
n>o [R(0,5, n)
R(2,5, n)] qn =5
ýq 42s , (5.46) [R(1,5, n)
R(2,5, n)] q" =
[qlo; q25] (5.47) Theorem 5.
The proof uses the well known expansion for (q; q),,,,, i. e. Euler's Pentagonal Number
Theorem. Also needed is that q 4 q3 and then z= q in (5.30) gives
(g2i 42)oo(Q3i g3)oo (q; q)00(g6i q6) 00 ºý` gn(3n1)/2 =' (5.48) nEZ then z= q in (5.30) gives
and that q 4 q2 and
2 loo = E(_1)nq'ý2
(q2;42)00
nEZ
62 (5.49) finally q * q6 and then z=
and
(g2i g2)oo(q3i
(q; 4)0o(g6i in (5.34) gives
q
g3)oo E(_1)ng9na(1 _ 6nf 1).
l + 46)00 (5.50) nEZ Now, putting w= e2'r='6in (5.22) gives
(g2; 1
(wq; q)ao(W1q; q2)oo(Q3iq3)00 (q; q),,. (q6; g6)oo q),,. mEZ n>O but
E R(m, n)wmgn = R[O,6]+wR[l, 6]+w2R[2,6]+w3R[3,6]+w4R[4,6]+w5R[5,6] tnEZ n>O R[0,6] + R[1,6]  R[2,6]  R[3,6]
=
(q2' q2) (q3;q3)(5.51)
""
R[l, 6]  R[2,6]  R[3,61 _
R[0,6] + (q;
.
q)ý(q6;
q6). Putting w= e2"'/3 in (5.22) gives
E mqn =1=
n)W
(wq; Q),,. (wlq; R(m, (4'; 4).
(q3; g3)00 q),,. mEz n>0 but
EE R(m, n)wmgn = R[0,6] +wR[1,6]+w2R[2,6]+w3R[3,6]+w4R[4,6]+w5R[5,6] mEZ n>O R[0,6]  R[1,6]  R[2,61 + R[3,61
=
(q; q). R[0,6]  R[1,6]  R[2,61 + R[3,61
Also
R(m, (q; q3). "
(qýq) 1
n)(1)mgn (q; (5.52) (q2; 42)00 q)00 rnEZ n>0 (q; q) 00
R[0,6]  2R[1,6] + 2R[2,61  R[3,61 =
(q q2)2 (5.53) The identity
E[R(0,6, ýnEi
n)  R(2,6, n)]qn = n>o g6n2}n (q3i 93)00 follows from multiplying both equations (5.51) and (5.52) by (q3;q3),, adding the sum,
(5.48) (and the pentagonal number theorem), and dividing by 2(q3;q3).. This
using
first part of (5.28), the rest comesfrom
provesthe
1: [R(0,6, (1)ng9n2
n)  R(3,6, n)]qn n>0 63 = .
(q2; floc follows in a similar way from (5.51) and (5.53) and then (5.49) and (5.50).
which
Theorem 6. Putting w= e27i/8in (5.22) gives mqn
n)W R(m' (44;
(wq; 8q))0 ((w3q;
q)oo(W3qi ýi)oo(Wlei q)oo q)oo(gi 4)00) mEZ n>_0 by (1.22) is equal to
which,
(R'4;
48)"ý (q;
q)0 l 1+(ý1)q(ýI)q3q6q10{... (q4; q8)ß (>
a(n)gn(n+ß)l2 +E
_ (q;
`L
q)ý
n>_o ß(n)gm(n+1)/2
n>o mod 8 else = 1, and ß(n) =1 if n=1,6
a(n) =1 if n0,2,5,7
where
mod 8,
if n2,5 mod 8, else= 0. But, by q + q2 and z=q in the triple product identity
= 1
this becomes
(q4; gl)oo ([q3;[q; 48] [46; 416](48; q
co + gV2[q2; 4'161(416;
4'16)00l
(q; q
q8l(q4; q8).,, (ql6; ql6)co
00 But EE R(m, n)wgn mEZ n>O R[0,8] + wR[1,8] + iR[2,8] + w3R[3,8]  R[4,8] + w5R[5,8]  iR[6,8] +W7R[7,8]
=
R[0,8]  R[4,8] + v(R[1,8]  R[3,8]).
=
This implies
Y[R(0,8,
n>o (q4; 4'3)
82
n)  R(4,8, n)]4n = [q2;
q ý[4 4 and
E[R(1,8, n)  R(3,8, n>o togetherimply (5.29).
which q n)lq'= (q; (g8;
q2), gls)ý[q6;
q16] 64 (5.54) (5.55) Related Identities 5.7 There is another approach to the birank, based on the following observation: E z#(a)gwa(a)
=E XE,
p
. (5.56) (zngn
q)n
`T n>0 This implies that, summing over all ordered pairs having birank congruent to r mod k, E h(s, k)h(t, k) qwt(")=E b(vr)kr (5.57) s+t=hr where qkn+9 h(s, k) (q; q)kn+s
n>o This suggests
that it may be worthwhile looking at the sum of zm/(q; q)m, summedover
the (positive) valuesof m that satisfy a certain congruence.This in turn leads to the well
known identity m>0 z"`
(Q; 4)m (z; 4)00 follows from n + oo in (1.10). It follows from this that for n any natural number,
which
and w := e2ait",
ri ( j wjmzm d)
(zd, qd)1L("I (q; q)m (5.58) din m>0 the product on the left is over all positive integerscoprime to, and not greaterthan,
where
am
a:
instance, if n=6
(and p is the mobius function). For
then w=e
i and (5.58)
n
becomes C 1+(4; Wz W2Z2 4)2+... 4)i+(q; Now, considerthe casen=3.
E
lrEPE C w5z (z2; g2)oo(z3; W10Z2 1+ (4;
+ (4;
+
4)1
4)2 (z; Q)oo It follows from (5.56) that zb(7r) gwt(7r) = zngn 1:
n>0 (q; q)n ) (,:
n>0 (5.59) (q; 4)n is equal to
which
(h(0,3) + zh(1,3) + z2h(2,3)) (h(0,3) + zlh(1,3) +
z2h(2,3)) 65 q6)00 (z6; znqn (this is true in general, but there's no point writing it like so unless n= q3)o, 3). Now, putting z=q and expandingout gives the expression
h(0,3)2 + h(1,3)2 + h(2,3)2  h(0,3)h(l, 3)  h(1,3)h(2,3)  h(2,3)h(0,3)
by (5.58) is equal to [q; q3]. This is equivalent to (5.32), it can be written as
which
2
q3n
n>0  }12
q3n + (qi q)
3n n>0 q3n 1:
0 (T q)3n+1 n>0 g3nt1 (q; q)3n q3n
 q3n (qi q)
3n+2  (qi q)3n g3fl+2 o
n>,) (q; q)3n+2 (q; q)
00
(q3; g3)00 n>0 into (5.58) gives Putting z=q Now for the case n=4. 1 n>o g3nt2
n>0 (Qi Q)3n
}2 (qi q)3n+1 n>o
 q3n {2 + (Qi q)3n
11 2 (h(0,4) + ih(1,4)
h(2,4)  ih(3,4)) (h(0,4) + ih(1,4)  h(2,4)  ih(3,4))
 [q2; q4]
= is (5.36). It may be written as
which
h(0,4)2 + h(1,4)2 + h(2,4)2 + h(3,4)2  2h(0,4)h(2,4)  2h(1,4)h(3,4)
(42i
 or as
2n
q
E((_1)fl(qq)2)
;n n>0 q 4)00, 22 (E(,
+ ). n>0 (42; 42)00 Q2n+1
 (q; Q)2n+1 (q4; g4)oo . So (5.58) can be used to help understandthe birank, the above two identities are maybe
because(5.58) is probably not very interesting (but, as
interestingin themselves
not very
just explained,can be a way of tackling the birank).
Now for the interesting part: It follows from (5.40), together with (5.57), that
h(0,4)2 + h(1,4)2 + h(2,4)2 + h(3,4)2
h(0,4)h(1,4) h(1,4)h(2,4)  h(2,4)h(3,4)  h(3,4)h(0,4)
(4; 42)00
[4s; 416}(416'gls)oo, _ (q4;
q4)00 from (5.41), togetherwith (5.57), that
and
h(0,4)h(1,4) 2h(0,4)h(2,4) + h(1,4)h(2,4) + h(2,4)h(3,4) 2h(l, 4)h(3,4) =q
 (q' q2 [q2; gls](qis' gis),, (ql,g4)oo two identities do not follow from (5.58).
and these
66 \ + h(3,4)h(0,4) 5.8 Other Ranks In this section, the 5 dissection of the Euler product is required. It is
25.25
_(q [q10.25
;4 \qi q)oo _ _q )O° [q5i q25] [q5;
(q25.25
q2,51 'q).
25)".
2
[gloi
_ q,
_ q(g25.q,
q25] (5.60) follows from lemma 6 in [5]. This chapter has dealt with the existence of a `2
which
dimensional rank', namely the birank. The aim here is to briefly describe a4 dimensional
i. e. one defined on members of P4. This is not an unnatural avenue to persue, given
rank,
the following easily proved congruence: (5n + 3)  p4(5n + 4) 0
p4 mod 5. (5.61) So what `superrank' can be used to explain this congruence? One answer is: For
(A(1), A(2), A(3), A(4)) any ordered 4tuple of partitions, define the superrank s(rc)
=
) = 2#A(1) + #A(2)  #A(3)  2#A(4). If w= e2"/5 then
as s(r. E
mEZ 1: R* (m' n)Wmqn =
(W24;q)(wq; 4) (1
w14; 4) (W24; q)
n>0 _ (g; )
g (45; q5) R* (m, n) denotesthe number of membersof P4 having weight n and whose superwhere
is m. So, by (5.60),
rank
E R* (m, n)wmgt
mEZ n>0 1
[q10; g25}(q25; q25)
1l
l [q5; q25](45; 4'5) q (q25; q25)
\
(q5; Q5) [q5; q25] (qý25.q25) _ q2 [q1O;
(q5; i5)
q25] But, for this superrank, R* [0,51  R* [2,5] + (w + w4)(R* [l, 5]  R* [2,5]) _
25.25 [g10,.
g25) (g , g)
g25](g25,.q (q5;
[q5; g25] (q5; ßi5)
45) (g5, .  (q25;
(j2L q25]
[410; . g25) 425](4'5;
45) Hence,for all n, R* (1,5, n) = R* (2,5, n) (which is not immediately obvious) and
E[R* (0,5,5n) R* (1,5,5n)] qn =
[q; 12 rio 1: [R*(1,5,5n + 1) R*(0,5,5n + 1)]q" 11
=
(1
n>O s16 1
q=) (where the product is over all i coprime to 5)
ý[R*(1,5,5n + 2)  R*(0,5,5n + 2)]qn =
[q2ýq5]2 n>o
implying that R* (0,5,5n) > R* (1,5,5n), R* (1,5,5n + 1) > R* (0,5,5n + 1) and
R* (1,5,5n + 2) > R* (0,5,5n + 2) except for R* (0,5,2) = R* (1,5,2). The equality
R*(0,5,5n + j) = R*(1,5,5n + j) for j=3
or 4 also follows. This equality, together
R* (1,5, n) = R* (2,5, n), implies (5.61).
with 67 5.9 Summary It is worth noting that the only tools used in this paper have been the Jacobi Triple Product
Identity, the Quintuple Product Identity and elementary results about algebraic independence (\ is irrational, for example). This appears to be perhaps somewhat ironic, given
the superranks are a generalisation of Dyson's rank, and there is no equally elementhat
his identities (see [5]). The appearance is deceptive. Whereas the Dyson
tary proof of
involves both the first part and the number of parts of a partition, the superrank of
rank
is defined only in terms of the number of parts (of each partition in the ordered
this paper
ktuple). An example of a proper generalisation would be: for an ordered pair of
pair or
define a rank as "Dyson rank of first partition plus twice Dyson rank of second
partitions,
Does this rank explain the congruence p_2(5n + 2) 0 mod 5? Just as there
partition".
similar to the rank, the crank for example, for ordinary partitions, there
are other statistics
be other super ranks for ordered pairs (or triples or whatever) of partitions.
may well
Finally, is there some combinatorial proof of these equalities and inequalities? Either
bijection that proves (say) R(1,5,5n) = R(2,5,5n) or indeed some argument showing
a
that R(0,5,5n) > R(1,5,5n). 68 Chapter 6
A qelliptic identity 6.1 The identity. is to present a new approach to the following identity, which is
The aim of this chapter
stated as a
Theorem Suppose that that al, ..., aN and bi, ..., bN are nonzero complex numbers
: for which blb2...bN
ala2...aN = (6.1) b% qequivalent, which is to say that the ratio of two distinct bis is not an
are
and no two
integer power of q, i. e. ijýb, 0 bbgt (6.2) integer t. When theseconditions are satisfiedthe following identity holds;
for any
N [ai' br; q] [a2 I br; q]... [a1 br; q]
=0
L=1"[bi ibri
[brbr li q]... [brb . ql
4'][b21briq]... (6.3) b,
indicates that the term [b,. .'; q] is omitted). The standard proof (which can be
..
(the
found in [18]) of this involves analytic arguments, but my aim here is to present an invoFor both these cases a
and N=4.
lutive approach to this identity for the cases N=3
identity that follows from (6.3) and the Jacobi triple product identity
bijective proof of an
is given. The triple product identity, which was proved in chapter 4, states
[z; 4](Q; q)0 ý(1)nz"gn s 2 (6.4) nE? Now, let it = (irl, 72i ..")irn) be an ordered ntuple of integers, so it can be viewed as an
V" or as a row matrix. The transposeof will be written in bold, ir. In this
element of
69 be defined as o,(ir) :_ 7rl + 7r2+
+ 7r, the sum of the parts and the
chapter a(7r) will
ti,
...
weight of it is defined as
wt(7C) 6.2 222
7ri  7rl := + 7ý2 2+... + 7rn  7rn 222 The case N=3. The identity obtained from (6.3) by choosing particular values for (al, a2, a3)
(b1, b2, b3) is the same as that which is obtained by chosing (Aal, Aa2, Aas) and
and
(tbl, Abe,ab3), where A is any nonzero scalar. In particular, we can divide through by bl,
Having done this (6.1) ensures that when four of the five other entries are set, the last
say.
is also determined. So if instead of (al, a2, a3) and (b1, b2, b3) in the above theorem
one
(al, a2, a3) and (1, bi, b2) then (6.1) ensures that a3 = blb2/(ala2). It follows
one picks
that there is the following equivalent version of the theorem (in the case N= 3): Wheneverno two of 1, bl and b2are qequivalent,the following identity holds;
[ai'; g][ai 1;q][ala2bi' b21;q] [ai' bi; q][a l bi; q][a1a2 '; 9'J
+
[bi; q][bib21;q]
[bi'; q][bz1;q]
l; 4']
[a,
'b2; q][a21b`i; g][a, a2bi
= 0.
[b2;q] [bi 1b2;q] (6.5) By using [z1; q] = z1 [z; q], i. e. identity (1.3), this can be written as
[a, '; 4][a2 bl 1; ib2 1; 9][ala2b1 b2 [dl q]
_ [bi; q][b2;q] lbi; bi [aa lbi;
4] [aiaab21;
q] q] [bi; q][bi' l; q] [a,
'b2; q] [az 1b2;q] [al a2b11; q]
+
=Q
[b2;q] [bi b2;9] (6.6) by [bi; q][b2;q][bl1b2;q] gives
and multiplication
bi ba[ai1; q][aa1;q][Qia2bilbi1; 4][b1b2;4)
bi
 [allbi; q][a 1b1,q][al a2bz'; q][b2;q] lb2; q][a21b2;
1; q][b1;q] 0.
+[ai
q][ala2bi
= (6.7) Multiplying through by (q; q), and using the Triple Product Identity (1.17) gives
(1)f bibs ßl f4 E fEZ i E(1)fai Ibfq
1 fEZ ý(1)hal E(1 ) a0. gbs q II2, f bfq ) hah hbh
ia2 2q "I hEZ 70 (1)k bi hbZq F(1)kb2 hEZ
E(i)haha OEZ E 22j= kEZ F(1 gEZ
'ý a2bl hbZ h, hEZ gEZ fEZ
+E(i)fal °q' (190.2 kEZ
b] hqhý I. E(1)kbi9ýkEZ = 0. (6.8) This in turn can be written as an expression involving three sums, each of which is over
V. To do this, let W1, W2 and W3 be sets of ordered 4tuples (so
all elements of
each W=
is the set Z4). Thus (6.8) becomes (with, for it E Wi, W2 or W3,7r1 = f, ire = g, 713 h,
=
7r4= k), 1
3 (r)b, 44(1r)gwt(R) E 2 7rEWl (ý')b, (ý')b2a(ý)qýc(,
ý) [ý (_l)'()ai'()a
 EW
IrEW2 (1}(ýr)ali(ý)a(ý)bls(ý)b2a(ý)gwt(ý) =0 ý (6.9) lrEW3 where + h, pa(ir) = 9 + h, P3(7r)= h k+1,114(7x)
/ti(7r) = f
(ir)=f+h,
vi
(lr)
T1 =f+ va(g)=g+P,
hý (7r)
T2 = v3(ir)=f+g1,
g + h, (7r)=
73 = h +k+1, v4(ir)=h+k+1,
h + k, 'r4 (7r) =f+ /g. What is required is a means by which an element of one of the three sets can be paired
a particular element of one of the other sets in a `nice' way. What this means
off with
will become clear later, but first it is helpful to decompose each of the three sets
precisely
by defining Wj' := {ir E W2 : a(ir) x mod 2} (x =0 or 1). It is now possible to define
six maps (three pairs of maps), x:
Xl W2 + W °,
Xä : W° + W2
,
: Wl + W3 X3 : W3 7 X2 : W2 4 W3 , * Wl X3 : W3 > W2 . (6.10) Thus, the notation x has been used to denote a map from a subset of Wi to a subset of Wj.
In fact, for x' (ir) to be defined it is necessary that it E W; and that either u(7r) =0 mod 2
(i, j) E {(1,2), (2,1), (3,1)} or a(7r) 1 mod 2 and (i, j) E {(1,3), (2,3), (3,2)}.
and
Also, X' (ir) may be written as it and similarly, the transpose of X?(1r) may be written in
bold, ir'. 71 The maps are defined by
7c := Aiir + c; (6.11) the Ai's are four by four matricesand the ct's are column matrices. They are
where
1
1 Al z1_ A2 1 1
1
2 1 1
1
1 1
2 1 1 2 c2
i1
ci =1 1
2 1 11 1
11' 1 1 1
1 1 1
1 1
1
1111 11 1 a_1
C1 1
21'
3 1 1
1 1 1
1 1 1
111
1 A3_A3= 1 1
1
1 1 1A3 1 2
1 Al s1 1 1
1 1 1 11
C3 =1' 1
0 1
11 1 1 1
1121 11
cZc31 Thus, for example, Xi ((f, g, h, k)) = (2 (f g  h+ k+ 1), 2 ( f +g  h+ k+ 1), 2 ( fg+h+k+1),
2(f ghk+3)).
As an example, take it := (3,7,10,2) E W°. Now, ir' = Xi(zr) = (10,0,3,
5).
What is nice is that wt(7r) =6+ 21 + 45 +1= 73 and wt(7r) = 55 +0+3+
15 = 73,
is weight preserving. Furthermore, µl (ir) = (3)
+ 10 = 13 and
the map
(7r) _ (10)
13. In fact vT(ir) = AT(r) for TE {1,2,3,4}.
The other nice
+3=
vl
is that Xä(Xi (ir)) = Xä(10) 0,3, 5)
7r, or 7r = 7r. What all this means is that
thing
(3,7,10,2)
E W° and (10,0,3, 5) E W2 can be paired off with each other. From
(6.9), (3,7,10,2)
contributes )d'1(ý)aP2(ý)bigtý) 4(ý)Qý,r) +Qi3a2bi i'b2rg73
t(, _
_1)a(
2
in the first sum, whilst (10,0,3,
ý_lýQ(ý )a1 )a2 5) gives
)qwt(ý)
b24(N =+ Q13Q3 "b2 7g73 in the secondsum (and so, since the secondsum in (6.9) is preceededby a minus sign,
canceleachother out).
the expressions 72 is nothing unusualin the choice of it in the above paragraph. It is always the
There
case that E W° = wt(Xi(ir)) = wt(ir).
it
This follows from the fact that ((f__h_k+1)2_ (f__h_k+1)) (_f+_h_k+1))
+1 +1 +1( C
C'(f
(_i
2(f +ghk)+1 a
2((_f_+h_k+1)) g+hk)+1 ((_f_9_h+k)+1))
1
2 gh+k)+1)2 f2 f
=2+2+2+2 g2g h2h k2k defined on 7r is weight preserving; wt(x? (ir)) = wt(7r). FurIndeed, for any ir, the map
thermore, since Aj (Ailr + c; ) + cj = ir
(d (ir)) = ir, or 7rß`_ ir so the maps are involutions.
it follows that that Xj'
4),
are (for 1<D<
straightforward
VD(X1(7r)) = D(7r)) /tD(X2(7r)) = VD(7r), TD(Xl(*7r)) = {LD(7r), ILD(X3(7r)) = TD(7r), TD(X2(7r)) = VD(E), VD(X3(7r)) = Equally TD(lr)" It remains to investigate the effect of the relevant map on the parity of the sum
4tuple. This has been dealt with implicity in (6.10). In
of the parts of a given
for example, that Xi : W° + W2 and not just X1 : W° _+ W2, it is implicit
stating,
2
(X2(ir)) = a(ir) mod 2. In fact, it is easy to verify that the maps Xi, X2, X3 and Xs
that o
reverse parity. Or more succintly,
preserve parity whereas Xi and X3 ij mod 2 4=* 7((1r))Q (7r) mod 2. (6.12) implies cancellation occurs, since the expression in (6.9) involves sums that are
This
by plus and minus signs.
alternately preceeded 73 6.3 The case N=4.
be used in the proof of the theorem when N=4. The following notation will (1r1,7213,74
,
7r= , 15,76 , 77 For )E Z7, define
81(7r) := 21r5 2ir6 + 21r7
1, S2 (7r) := 7r, 1,  7r2  ir3  ir4 + ir6 + 7i7 + lt3  74 + 7f5  lt7 I 2ý 7r3  7r4  7r6 + 63(ir) :_ Tfl  7r2  84(7r) := 1! '1  7r2  7r5  3" Sy(7r) =0 mod 3}. It can be shown that for any
#{y E {1,2,3,4}
:
:=
E Z7, either #(ir) =4 or #(7r) = 1. Without specifying at this stage what the subscript
is
i is, define U, to be the set {ir E Z7 : bl (7r) =0 mod 3}. For y=2,3
or 4, define UjY
öb(ir) =0 mod 3,5i(ir) #0 mod 3}, the last condition ensuring
be the set {7r EV:
to Let #(ir) that no it occurs in more than one of the four sets. Thus, for a given it E Z7, either
#(7r) =4 and so it E Ul and none of the other three sets, or there is precisely one y
(ir) 0 mod 3 (because ß(7r)
#(7r) = 1) and this is the y for which
4
that S,,
such
E UY. So any lr EZ is a member of precisely one UjY.
it
the theorem can be stated as: whenever no two of 1, bi, b2 and b3 are
the following identity holds;
qequivalent
When N=4, [a, '; q] [a2 1; Q] [d3 1; q] [ai a2a3b 1 b2 1 b3 [bi1;q][b21; [b31;
g]
R]
1bi; 4'][a21bl;q][a31bl;q][aia2a3b21bs
1; 4)
+[a,
[bi; 4}[b1b21; [bib3'; q]
4)
1b2;q][a21b2; q][a3lb2; q][aia2a3b11b31;
q]
+[al
[1k; q] [bi ib2; q] [b2b3'; q]
ib2';
1b3;
[a313; q][ala2a3bj
[a,
q][a2lb3; q]
+0.
[b3;q][bi 1b3; [b2ib3; 9)
q] 4] (6.13)  (1.3), can be rearranged to give
which, using
[a1; q][a2 l; b1b2b3 q][a31i q][a1a2a3b1 lb21b31; q] [bi; q][b2;q][b3;4'] b2b3[ai'bi; q] [a2'bi; q] [a3 bi; q}[a, a2a3b2ib '; q]
[bi; q] [bi 1b2;q] [bi l b3;q]
bi
lb2; q][a21b2; [a,
_b3 b2 LQl lb3; q][a2 +=0. q][a3 lb2i q][aia2a3b1'b [b2;4][bi 'b2; q][b21 4]
b3;
lb3;
lb21; q]
g1[a3 %;
g][aja2a3bl [b3; [bi'b3;q][b21b3;
q]
4]
74 '; q] (6.14) by [bi; q][b2; q][b3; q][bi lb2; q][bi 1b3;q][b21b3;q] gives
Multiplication
1; q][a21; q][as 1; q][ala2a3bl ib21b3'; 4][bi lb2; q][bi ib3; q][b 'b3; q]
bib2b3[ai bbb3
b31;
[ailbi; q][ai' bi; q][a "bi; q][a, a2a3b21 q][b2;Q][b3; [bzl b3;q]
q]
+
[a11b2;q][aa1b2;q][asl b2;q][al aza3bi l bs1; q][bi; q][b3;q][bi i b3;q]
L
[a 'b3; q][ala2a3bi lb21; q][bl; 9' [62;q][bi'b2; q] = 0.
+[al'b3; q][a2'b3; q] (6.15) Multiplying through by (q; q)7 and using the triple product identity (1.17) gives
r_1ýQ(ýýaiý(ý)a22(ý)a33(ý)4(ar)Lpb(ý)b3
7rEEUU1l
(ß)Q3 (7)bi4(ý)b25(7r)V36(7r)wt(7r) ()a3 (ý)b14(1r)b2 ai()Q2
IrEU2` al(r)a2 _ (7')b38(7r)ýwt(7r) irEUsl
(_1)a( +E )a'l'( )a2a( )Q3s()bia('r)bPa(")bgs(")Qwt(a) =0 IrEU4 Ul, U2, U3 and U4 are four sets of ordered 7tuples, it = (k, 1,m, n, p, r, s) E Z7,
where
and
jtl(ir) µa(lr) = t+n, = k+n, p5(ir) = n+ps+1, µ4(7r) =npr+1,
vl (7r) = k v4(7r)=k+l+m2, + n, (7r)
v2
= l + n, v5(7r)=n+ps+1, P 3(7r) = m+n,
p6(ir) =n+r+s+1,
v3 (7r) = m { n, v6(lr)=n+r+s+1, (7r)
(ir)
(7r) = k + n, 7.2 = 1 + n, 113 = m + n,
71
T4(lr)=n+ps, T5(ir)=k+l+m1, r6(ar)=n+r+s+1, (ir) = k + n, P2(7r)= l + n, Ps(7r) m + n,
Pi
=
p4(1r) _ n+ps, ps(ir) _ n+r+s, p6(lr) =k+l+m. Now, consider an element in any one of the four sets, it E U. By earlier remarks there is
that it E U= . For this y define Sy :=J . (7r)/3.
a unique y such 75 There are sixteen maps,
(i = 1,2,3 or 4), U1}U;
X=: 44 24 X2: U2 X,  ui 3U2"
x: U1 > U3) 423 X2 U2 u33 42 u1
U2 , .3
32 U4
4: U4
3 : U3 ý U4i X3 Ül . X4"U4 XýU1U4,
U2
2' Ü343 U333+ X3 424 3
X2 Ul U2
, U4 U3. X4: They are given by
(7r) _ (k, 1,m, n, pX! 51,
rI61, s81), (k + 84,1 + 54, m+ 54, n + 84, p + 64, r + ä4, s),
Xi(7r) = X2(7r) =
(r) = (k+S3, l+S3, m+63,
x n+63, s+63, r), p+d3+1, (k+S4, l+S4, m+S¢, n+S4, pS4+1,
X3(7') _
(k+S2, l+S2, m+S2, n+d2,
X14(7r)
=
(7r) = (k+S4, l+64,
X41
X2(7r) _ =X3(7r) r+S2+1, s, r+S4), m+S4, n+S4, s, pS4+1, (k+S3, Z+S3, m+S3, n+S3, p), s+S2+1,
r64+1), pS3, r, 8+S3), (7r) _ (k+S2, l+S2, m+S2, n+S2, rS2, p, s+S2+
X4
(k+S3, l+ 53,m+S3, n+63, r, p63, s63+
X4(7r)= 1),
1), (k+S2, l+S2, m+S2, n+S2, p, rS2, sS2). X3(7r)=X3(7r) It is a straightforward task to show that the maps are weight preserving involutions and
that
(7r),
AD AD(Xl(7r))  (X3 (10)
TD (70
ý TD (Xl (7r))
I'D
(x
TD (7r)) PD (Xl (1r)) TD (X2 (7r)) (X2(7r))
PD
(X3 (1r))
PD = AD (lr)  (7r),
VD = (7r)
TD VD(x2(7r))
(X44 (70)
PD
iID(X2(7T)) = VD(7r), = ILD (7r) pD(X3(1r)) = TD(7r), = (7r),
AD pD(X4(1r)) = PD(lr), (7r),
VD VD(X3(7r)) = TD(TC), (7r)
VD , W (7r))
VD (lr),
= TD , TD(X4(1r)) PD(7r),
= PD(7r). It remains only to check that cancellation occurs. This is so because for any
(i, j) E {1,2,3,4} the statement(6.12) is true.
76 Bibliography
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