PARTITIONS, YOUNG DIAGRAMS AND BEYOND

Lecture 6 (03/26/2016).

Third proof of the Jacobi triple product formula. Set u = −q/z and v = −qz to rewrite the Ja-cobi’s identity as

∞∏k=1

(1− ukvk)(1− ukvk−1)(1− vkuk−1) =

= 1 +∞∑m=1

(−1)m(um(m+1)/2vm(m−1)/2 + vm(m+1)/2um(m−1)/2

).

For a partition λ = (λ1, . . . , λ`) denote by λ− the partition (λ1 − 1, λ2 − 1, . . . , λ` − 1). If λ isempty, set λ− to be the empty partition as well. The coefficient of unvs in the left hand side is∑

µ,λ,ν∈D: |µ|+|λ|+|ν|−`(ν)=n, |µ|+|λ|+|ν|−`(λ)=s

(−1)`(µ)+`(λ)+`(ν)

We need to show that it is (−1)m, if n = m(m+ 1)/2, s = m(m− 1)/2, or n = m(m− 1)/2,s =m(m + 1)/2 for m = 0, 1, 2, . . ., and 0 otherwise. The proof uses an idea similar to that of theFranklin’s involution. Since the sum is symmetric, it suffices to consider only the case n ≥ s.Define involution on the set of such triples of partitions in the following way (use the conventions(∅) =∞):

(1) If b(λ) ≥ s(µ), remove the bottom row of µ and add a right diagonal of length s(µ) to λ.(2) If b(λ) < s(µ) and λ is not triangular (i.e. of the form (m,m − 1, . . . , 2, 1), this also

includes the case of the empty partition for m = 0), remove the right diagonal of λ andadd a bottom row of length b(λ) to µ.

µ λ ν µ λ ν

Figure 1. Jacobi’s involution. Cases (1)-(2).

(3) If λ is triangular of the form (m,m− 1, . . . , 2, 1), and m < s(µ) < m+ s(ν), remove thebottom row of µ, and add a right diagonal of length m + 1 to λ and a bottom row oflength s(µ)−m to ν.

(4) If λ is triangular of the form (m,m− 1, . . . , 2, 1), m+ s(ν) ≤ s(µ), and ν 6= ∅, remove theright diagonal of λ and the bottom row of ν, and add a bottom row of length m+s(ν)−1to µ.

(5) If µ = ν = ∅, and λ is triangular, do nothing.

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1

PARTITIONS, YOUNG DIAGRAMS AND BEYOND 2

µ

µ

λ ν µ λ ν

Figure 2. Jacobi’s involution. Cases (3)-(4).

Theorem 0.1 (Rogers-Ramanujan identities).

1 +

∞∑k=1

tk2

(1− t)(1− t2) · · · (1− tk)=

∞∏i=0

1

(1− t5i+1)(1− t5i+4), (0.1)

1 +

∞∑k=1

tk(k+1)

(1− t)(1− t2) · · · (1− tk)=

∞∏i=0

1

(1− t5i+2)(1− t5i+3). (0.2)

Combinatorial Interpretations (why?):

(1) The partitions of n into summands that differ from each other by at least 2 are equinu-merous with the partitions into parts of the forms 5i+ 1 and 5i+ 4.

(2) The partitions of n into summands each larger than 1 which differ from each other by atleast 2 are equinumerous with the partitions into parts of the forms 5i+ 2 and 5i+ 3.

Indeed, left hand sides of the Rogers-Ramanujan identities can be rewritten as respectively∑λ

t|λ|+λ1(λ1−1) =∑λ

t|λ|+`(λ)(`(λ)−1) and∑λ

t|λ|+λ21 =

∑λ

t|λ|+`(λ)2,

and to a partition λ = (λ1, . . . , λ`) correspond the partition (λ1+2`−2, λ2+2`−4, . . . , λ`−1+2, λ`)into summands that differ from each other by at least 2 and the partition (λ1 + 2`− 1, λ2 + 2`−3, . . . , λ`−1 + 3, λ` + 1) with the same property and also all summands larger than 1.

Exercise 0.2 (One problem per week). Show that partitions of n into summands that differ fromeach other by at least 2 are equinumerous with the partitions λ ` n with s(λ) ≥ `(λ).

These latter two identities appeared in 1894 in a paper by L. J. Rogers that appeared inthe Proceedings of the London Mathematical Society, hardly an obscure journal. The paperwas entitled, Second Memoir on the Expansion of Certain Infinite Products. Apparently themathematical public had lost interest somewhere in the first memoir on the Expansion of SomeInfinite Products. In any event, the paper was quite forgotten when nineteen years later anunknown Indian clerk, S. Ramanujan, sent these identities to G. H. Hardy. At first glance,these formulas look very much like (??) ones that Euler had found. However, Hardy found thathe was completely unable to prove them. Let us now turn to Hardy’s account of the momentof illumination: ”The mystery was solved, trebly, in 1917. In that year Ramanujan, lookingthrough old volumes of the Proceedings of the London Mathematical Society, came accidentallyacross Rogers’s paper. I can remember very well his surprise, and the admiration which heexpressed for Rogers’s work. A correspondence followed in the course of which Rogers was ledto a considerable simplication of his original proof. About the same time I. Schur, who was thencut off from England by the war, rediscovered the identities again. Schur published two proofs,one of which is “combinatorial” and quite unlike any other proof known.” We will present herethis combinatorial proof due to Schur.

PARTITIONS, YOUNG DIAGRAMS AND BEYOND 3

Proof. It follows from the Jacobi’s identity with q = t5/2 and z = −t1/2 that∞∏i=0

1

(1− t5i+1)(1− t5i+4)

∞∏r=1

(1− tr) =∞∏i=1

(1− t5i)(1− t5i−2)(1− t5i−3) =∞∑

m=−∞(−1)mtm(5m−1)/2.

So to show the first Rogers-Ramanujan identity we need to prove(1 +

∞∑k=1

tk2

(1− t)(1− t2) · · · (1− tk)

) ∞∏r=1

(1− tr) =∞∑

m=−∞(−1)mtm(5m−1)/2.

Let D′ denote the set of partitions into summands that differ from each other by at least 2. Thenthis equality can be rewritten as∑

λ∈D

∑µ∈D′

(−1)`(λ)t|λ|+|µ| =

∞∑m=−∞

(−1)mtm(5m−1)/2.

Define the tangential line of µ ∈ D′ as the set of rightmost cells in the first u rows of µ, where uis the largest integer, such that µu = µ1 − 2u + 2 and u ≤ `(µ). We denote such u by u(µ) andcall it the length of the tangential line.

Figure 3. Tangential line.

To prove the last equality we define the Schur’s involution on D ×D′:(1) If λ1 ≥ µ1 + 2, remove the first row of λ and add a first row of length λ1 to µ.(2) If λ1 < µ1, remove the first row of µ and add a first row of length µ1 to λ.(3) If (λ, µ) is one of the pairs indicated on the Figure (4) or λ = µ = ∅, do nothing.

Figure 4. Fixed points of the Schur’s involution.

All other cases split into the following classes:

(1) λ1 = µ1 and s(λ) = min{s(λ), b(λ), u(µ)}. Apply the Franklin’s involution to λ and leaveµ as it is.

(2) λ1 = µ1 + 1 and b(λ) = min{s(λ), b(λ), u(µ)} < s(λ). Apply the Franklin’s involution toλ and leave µ as it is.

(3) λ1 = µ1 and u(µ) = min{s(λ), b(λ), u(µ)} < s(λ). Remove the tangential line of µ andadd a bottom row of length u(µ) to λ.

PARTITIONS, YOUNG DIAGRAMS AND BEYOND 4

Figure 5. Schur’s involution. Cases (1)-(2).

(4) λ1 = µ1 + 1 and s(λ) = min{s(λ), b(λ), u(µ)}. Remove the bottom row of λ and add atangential line of length s(λ) to µ.

Figure 6. Schur’s involution. Cases (3)-(4).

(5) λ1 = µ1 and b(λ) = min{s(λ), b(λ), u(µ)} < s(λ), u(λ). Remove the first row of µ and theright diagonal of λ, then add a first row of length µ1 to λ and a tangential line of lengthb(λ) to µ.

(6) λ1 = µ1 + 1 and u(µ) = min{s(λ), b(λ), u(µ)} < s(λ), b(λ). Remove the first row of λ andthe tangential line of µ, then add a first row of length λ1 to µ and a right diagonal oflength u(µ) to λ.

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PARTITIONS, YOUNG DIAGRAMS AND BEYOND 5

Figure 7. Schur’s involution. Cases (5)-(6).

Exercise 0.3. Prove the second Rogers-Ramanujan identity.

Theorem 0.4 (Ramanujan continued fraction).

1 +q

1 +q2

1 +q3

1 +q4

1 + · · ·

=

∞∏i=0

(1− q5i+2

) (1− q5i+3

)(1− q5i+1) (1− q5i+4)

.

Proof. Let

F (x, q) := 1 +∞∑k=1

xkqk2

(1− q)(1− q2) · · · (1− qk), c(x, q) :=

F (x, q)

F (xq, q).

Then

F (x, q)− F (xq, q) =∞∑k=1

xkqk2

(1− q)(1− q2) · · · (1− qk−1)=

= xq∞∑k=1

(xq2)k−1q(k−1)2

(1− q)(1− q2) · · · (1− qk−1)= xqF (xq2, q),

hence c(x, q) = 1 + xqc(xq,q) . It follows that

c(x, q) = 1 +xq

1 +xq2

1 +xq3

1 +xq4

1 + · · ·

.

Now just apply it for x = 1 together with the Rogers-Ramanujan identities. �