4. Integer partitions and q -binomial coefficients
4.3. Jacobi’s triple product identity
here, we have replaced mby n−wk
in the sum, since the conditionm+wk =n forcesmto be n−wk
=
∑
k∈Z
p(n−wk)·(−1)k
here, we have extended the range of summation; this does not change the sum, since p(n−wk) =0 whenever n−wk <0
=
∑
k∈Z
(−1)kp(n−wk)
= (−1)0
| {z }
=1
p
n− w0
|{z}
=0
+
∑
k∈Z;
k6=0
(−1)k p(n−wk)
= p(n) +
∑
k∈Z;
k6=0
(−1)kp(n−wk).
But the xn-coefficient on the right hand side of (4.2.3) is 0 (since n is positive).
Hence, comparing the coefficients yields p(n) +
∑
k∈Z;
k6=0
(−1)k p(n−wk) = 0.
Solving this for p(n), we find p(n) = −
∑
k∈Z;
k6=0
(−1)kp(n−wk) =
∑
k∈Z;
k6=0
(−1)k−1p(n−wk).
Corollary 4.2.3 is thus proved.
4.3. Jacobi’s triple product identity
Instead of proving Theorem 4.2.2 directly, we shall prove a stronger result:
Jacobi’s triple product identity. This identity can be stated as follows:
n
∏
>0
1+q2n−1z 1+q2n−1z−1 1−q2n
=
∑
`∈Z
q`2z`. (136)
What are q and z here? It appears that (136) should be an identity between multivariate Laurent series (in the indeterminates q and z), but I have never defined such a thing48. However, it is not hard to see that any negative power of z in (136) is “compensated” by a positive power of q. For example, the q2n−1z−1 term on the left hand side will always have total degree ≥ 0, and so will theq`2z`term on the right hand side. So this is not as bad as Laurent series in general.
One natural ring in which the identity (136) can be placed is (Z[z±]) [[q]]
(that is, the ring of FPSs in the indeterminate q whose coefficients are Lau-rent polynomials over Z in the indeterminate z). In other words, we state the following:
Theorem 4.3.1 (Jacobi’s triple product identity, take 1). In the ring (Z[z±]) [[q]], we have
n
∏
>0
1+q2n−1z 1+q2n−1z−1 1−q2n
=
∑
`∈Z
q`2z`.
However, we aren’t just planning to view this identity as a formal iden-tity between power series; instead, we will later evaluate both sides at certain powers of another indeterminate x (i.e., we will set q = xa and z = xb for some positive integers a and b). Alas, this is not an operation defined on the whole ring (Z[z±]) [[q]]. For example, setting q = x and z = x in the sum
`∈∑Z
q`z−` ∈ (Z[z±]) [[q]]yields the nonsensical sum ∑
`∈Z
x`x−` = ∑
`∈Z1.
Thus, Theorem 4.3.1 is not a version of Jacobi’s triple product identity that we can use for our purposes. Instead, let us interpret the identity (136) in a different way: Instead of treating q and z as indeterminates, I will set them to be powers of a single indeterminate x(more precisely, scalar multiples of such powers). This leads us to the following version of the identity:
Theorem 4.3.2 (Jacobi’s triple product identity, take 2). Let a and b be two integers such that a > 0 and a ≥ |b|. Let u,v ∈ Qbe rational numbers with v 6=0. In the ringQ((x)), setq =uxa and z=vxb. Then,
n
∏
>0
1+q2n−1z 1+q2n−1z−1 1−q2n
=
∑
`∈Z
q`2z`.
Before we start proving this theorem, let us check that the infinite product on its left hand side and the infinite sum on its right are well-defined:
48Multivariate Laurent series can indeed be defined, but this is not as easy as the univariate case and involves some choices. See [ApaKau13] for details.
• The infinite product is this product is multipliable, because
– (2n−1)a+bgrows linearly when n →∞(since a>0);
– (2n−1)a−bgrows linearly when n →∞(since a>0);
– 2nagrows linearly when n →∞(since a>0).
Thus, the infinite product is well-defined.
• The infinite sum is
`∈
∑
ZWe will give a proof of Jacobi’s triple product identity that works equally for both versions of it (Theorem 4.3.1 and Theorem 4.3.2). But first, let us see how it yields Euler’s pentagonal number theorem as a particular case.
Proof of Theorem 4.2.2 using Theorem 4.3.2. Set q = x3 and z = −x in Theorem
The left hand side of this equality simplifies as follows:
since each positive integerkcan be uniquely represented as 3n−1 or 3n−2 or 3nfor some positive integer n.
Comparing this with (137), we obtain
∏
k>0 (becausew`is definedas(3`−1)`/2))
Now, let us “substitute x for x2” in this equality (see below for how this works). As a result, we obtain
k
∏
>0This is Euler’s pentagonal number theorem (Theorem 4.2.2).
What did I mean by “substituting x for x2”? I meant using the following simple fact:
Proof. This is easy: Write f and g as f = ∑
Lemma 4.3.3 justifies our “substituting x for x2” in the above proof; indeed, we can apply Lemma 4.3.3 toK=Qand f = ∏
k>0
1−xk
andg= ∑
k∈Z(−1)kxwk (because (138) says that these two FPSs f andgsatisfy f
x2 using Theorem 4.3.2. It therefore remains to prove the latter.
The following proof is due to Borcherds, and I have taken it from [Camero16,
§8.3] (note that [Loehr11, §11.2] gives essentially the same proof, albeit in a different language).
Proof of Theorem 4.3.1 and Theorem 4.3.2. The following argument applies equally to Theorem 4.3.1 and to Theorem 4.3.2. (The meanings ofqandzdiffer between these two theorems, but all the infinite sums and products considered below are well-defined in either case.)
We will use a somewhat physics-inspired language:
• Alevelwill mean a number of the form p+1
2 with p∈ Z. (Thus, there is exactly one level midway between any two consecutive integers.)
• Astatewill mean a set of levels that contains – all but finitely many negative levels, and – only finitely many positive levels.
Here is an example of a state:
−5 Visually, it can be represented as follows:
· · · ·negative levels· · · · only s left of here
0
· · · ·positive levels· · · ·
only s right of here
where
• A white (=hollow) circle means a level that is contained in the state (you can think of it as an “electron”).
• A black (=filled) circle means a level that is not contained in the state (think of it as a “hole”).
For any stateS,
• we define theenergyofS to be energyS :=
∑
p>0;
p∈S
2p
|{z}>0
−
∑
p<0;
p/∈S
2p
|{z}<0
∈N
(where the summation index pin the first sum runs over the finitely many positive levels contained inS, while the summation index pin the second sum runs over the finitely many negative levels not contained inS).
• we define theparticle number ofSto be
parnumS := (# of levels p >0 such that p∈ S)
−(# of levels p <0 such that p ∈/S)∈ Z.
For instance, in the above example, we have
energyS =1+3+7+13−(−3)−(−7) = 34 and
parnumS =4−2=2.
We want to prove the identity
n
∏
>0
1+q2n−1z 1+q2n−1z−1 1−q2n
=
∑
`∈Z
q`2z`.
We will first transform this identity into an equivalent one: Namely, we move the 1−q2n
factors from the left hand side to the right hand side by multiply-ing both sides with ∏
n>0
1−q2n−1
. Thus, we can rewrite our identity as
n
∏
>0
1+q2n−1z 1+q2n−1z−1
=
∑
`∈Z
q`2z`
!
n
∏
>0
1−q2n−1
. We will prove this new identity by showing that both of its sides are
Sis a state
∑
qenergySzparnumS.
Left hand side: We have
pis a positive level
here, we have substituted p+1
2 forn in the first product, and have substituted −p+1
2 fornin the second product
(here, we have expanded both products using (121))
=
∑
(by the definition of energyS)
z(number of positive levels inS)−(number of negative levels not inS)
| {z }
=zparnumS
(by the definition of parnumS)
here, we have combined Pand N into a single stateS:= P∪N,
Right hand side: Recall that
n