A Proof of the Jacobi Triple Product Formula [expository notes]

3 Pages·2017·0.159 MB·English

by  Andrew O’Desky

A Proof of the Jacobi Triple Product Formula
Andrew O’Desky
March 23, 2017
Jacobi Triple Product Formula [1829]:
∞ ∞
(cid:89)(1−q2n)(1+q2n−1z)(1+q2n−1z−1)= (cid:88) qm2zm (1)
n=1 m=0
The proof given here of this remarkable identity used to be on Wikipedia
a few years ago. It uses physics to motivate the quantities introduced and
ultimately is a combinatorial proof which derives the formula as an equality
of generating functions. The main ideas were there attributed to Borcherds.
The proof on Wikipedia skipped over a portion of the argument, however, so I
augmented it and wrote it up. Since the proof was later taken off of the page I
decided to make this available since I think it’s neat and deserves to be known.
Consider the set of half-integers Z = 1/2+Z. Define the vacuum state v
to be the subset of negative half-integers: v = {−1/2,−3/2,−5/2,...} ⊂ Z.
Define a state to be a subset of Z whose symmetric difference with v is finite.
For such a state |S(cid:105), the elements of |S(cid:105) which are not in v are thought of as
particles; the elements of v which are not in |S(cid:105) are antiparticles.
Define the energy of |S(cid:105) by
(cid:88) (cid:88)
E = {x∈|S(cid:105):x>0}− {x∈/ |S(cid:105):x<0}
The energy of a state is always an integer or half-integer. We define also the
net number of particles of |S(cid:105) by N = n −n where n is the number of
+ − +
positive elements of |S(cid:105) while n is the number of negative elements. Note
−
that both E and N are well-defined for states and that the energy of a state is
non-negative with equality if and only if the state is the vacuum state. Because
of this property, the number of states with energy below some bound is always
finite.
Letg((cid:96),m)bethenumberofstateswithenergy(cid:96)andnetnumberofparticles
m. We have the following identity of generating functions:
antiparticlewithenergyn−1/2
∞ (cid:122) (cid:125)(cid:124) (cid:123) ∞ ∞
(cid:89) (1+qn−21z) (1+qn−12z−1) = (cid:88) (cid:88) g((cid:96),m)q(cid:96)zm
n=1 (cid:124) (cid:123)(cid:122) (cid:125) 2(cid:96)≥0m∈Z
particlewithenergyn−1/2
(2)
1
The exponent of the z variable indicates the net particle number and the expo-
nent of q indicates the energy. We now consider states with the distinction of
having minimal energy for a fixed net particle number. For any integer m∈Z
there is exactly one such state with minimal energy with net particle number
m:
|m(cid:105)(cid:44){ν ∈Z :ν <m}
Note that this definition works without modification also for negative m. Sup-
pose|m,(cid:96)(cid:105)isastatewithnetparticlenumbermandenergy(cid:96). Forconcreteness,
suppose m > 0. Let us label the (anti)particles with (negative) positive sub-
scripts according to their position in the state |m,(cid:96)(cid:105)⊂Z:
p <p <···<p <0<p <p <···<p <q <q <···<q
−k −k+1 −1 1 2 k 1 2 m
where k ≥0 accounts for particle / antiparticle pairs which may have cancelled
out. “Push” the m particles of |m(cid:105) up, respectively, to q ,...,q . If λ is the
1 m i
number of pushes necessary for the ith particle, then λ : 0 ≤ λ ≤ λ ≤ ··· ≤
1 2
λ is a partition depending only on the state |m,(cid:96)(cid:105). Let us denote the process
m
of pushing by P so that we have, symbolically,
λ
P |m(cid:105)=v∪{q ,...,q }
λ 1 m
The energy of |m(cid:105) is m2/2 and each push adds one to the energy so that the
resulting energy is m2/2+n, where n=|λ|=(cid:80)λ .
i
Define the shifting operator φ|S(cid:105)=|S(cid:105)+1={ν+1:ν ∈|S(cid:105)}. If E is the
energy of |S(cid:105) and N its net particle number, then the energy of φ|S(cid:105) is
(cid:40)
E+N + 1, −1 ∈|S(cid:105)
E(φ|S(cid:105))= 2 2
E+N − 1, −1 ∈/ |S(cid:105)
2 2
p−k − 21 is a negative integer and φ−p−k+12(|m,(cid:96)(cid:105)) is a state with no an-
tiparticles and −p + 1 +m particles. We now apply the pushing procedure
−k 2
described above to |−p−k+ 12 +m(cid:105) using the data of φ−p−k+12(|m,(cid:96)(cid:105)) to gen-
erate the partition λ : 0≤λ ≤···≤λ . We have
1 −p−k+21+m
φp−k−12Pλ|−p−k+1/2+m(cid:105)=|m,(cid:96)(cid:105)
Because 21 ∈/ φ−p−k+12 |m,(cid:96)(cid:105), it must be that λ1 > 0 so that λ has exactly
−p + 1 +m nonzero parts. Also note that because the energy of φ|S(cid:105) is
−k 2
linear in E(|S(cid:105)) it follows that
E(φp−k−12Pλ|−p−k+1/2+m(cid:105))=E(φp−k−21 |−p−k+1/2+m(cid:105))+|λ|
=E(|m(cid:105))+|λ|
m2
= +|λ|,
2
But this equals E(|m,(cid:96)(cid:105)) = (cid:96), so it follows that (cid:96)− m2 = |λ| and that λ has
2
exactly −p + 1 +m nonzero parts. We have now shown that for any fixed
−k 2
2
m > 0, n ≥ 0, there is a bijection between the partitions of n and those states
with energy (cid:96)= m2 +n and net particle number m. This implies an equality of
2
generating functions for arbitrary m>0,
∞ ∞
zmqm22 (cid:88)p(n)qn =zm (cid:88) g((cid:96),m)q(cid:96)
n=0 2(cid:96)=0
The same argument works for m < 0, and trivially for m = 0, so we may sum
the above over all m∈Z,
(cid:32) (cid:33)(cid:32) ∞ (cid:33) ∞ ∞
(cid:88) qm22zm (cid:88)p(n)qn = (cid:88) (cid:88) g((cid:96),m)q(cid:96)zm
m∈Z n=0 2(cid:96)≥0m∈Z
Putting this together with (2) we have that
∞ (cid:32) (cid:33)(cid:32) ∞ (cid:33)
(cid:89)(1+qn−21z)(1+qn−12z−1)= (cid:88) qm22zm (cid:88)p(n)qn
n=1 m∈Z n=0
Using the well-known identity (cid:80)∞ p(n)qn =(cid:81)∞ 1 ,
n=0 n=1 1−qn
∞
(cid:89)(1−qn)(1+qn−21z)(1+qn−21z−1)= (cid:88) qm22zm,
n=1 m∈Z
and replacing q (cid:55)→q2 yields equation (1).
3