Math 702 Enumerative Combinatorics Project: Introduction to a combinatorial proof of the Rogers-Ramanujan and Schur identities and an application of Rogers-Ramanujan identity Guo, He November 2, 205 Abstract In Cilanne Boulet and Igor Pak s article, they give a combinatorial proof of the first Rogers-Ramanujan identity by using two symmertries. These symmetries are established by direct bijections. Then I introduce an application of Rogers-Ramanujan identity in analysis, which I found in George E. Andrews book.
2 Algebraic Part The first Rogers-Ramanujan identity: + k= t k2 ( t)( t 2 ) ( t k ) = ( t 5i+ )( t 5i+4 ) Recall the Jacobi triple product identity: k= z k q k(k+)/2 = ( + zq i ) ( + z q j ) ( q i ) i= j=0 i= Set q t 5, z ( t 2 ) and divide both sides by we have ( t 5i+ ) ( t 5i+4 ), ( t 5i+ ) ( t 5i+4 ) = m= ( ) m t m(5m )/2 i= ( t i ) It remains to prove Schur s identity, which is equivalent to Rogers-Ramanujan identity: ( + k= t k2 ( t)( t 2 ) ( t k ) ) = ( t i ) i= m= ( ) m t m(5m )/2 Some combinatorial definitions: denote by P n the set of all partitions λ of n, and let P = n P n, p(n) = P n. Let l(λ), e(λ) be the number of parts and smallest part of the partition, respectively. Say λ is a Rogers-Ramanujan partition if e(λ) l(λ). Denote by R n the set of Rogers-Ramanujan partitions, and let R = n R n, q(n) = R n. We have P (t) := + n= p(n)tn = i=, ( t i ) Q(t) := + n= q(n)tn = + k= t k2 ( t)( t 2 ) ( t k ). 2
Let λ be conjugate of λ. For m 0, define an m-rectangle to be a rectangle whose height minus its width is m. Define the first m-durfee rectangle to be the largest m rectangle which fits in diagram [λ]. Denote by s m (λ) the height of the first m Durfee rectangle. Define the second m-durfee rectangle to be the largest m rectangle which fits in diagram [λ] below the first m Durfee rectangle, and let t m (λ) to be its height. We allow an m Durfee rectangle to have width 0 but never height 0. Finally, denote by α, β, γ the three partitions to the right of, in the middle of, and below the m Durfee rectangles. Define (2, m) rank,r 2,m (λ) of a partition λ by the formula: r 2,m (λ) := β + α sm(λ) t m(λ) β + γ Let H n,m,r be the set of partitions of n with (2, m) rank r. h(n, m, r) = H n,m,r. Note that (2,0)-rank is only defined for non-rogers-ramanujan partitions because otherwise β does not exist. Denote by h(n, m, r) the number of partitions λ of n with r 2,m (λ) = r. Similarly, let h(n, m, r) and h(n, m, r) be the number of partitions with the (2,m)-rank r and r, respectively. The following is from the definition: for all r Z and n. h(n, m, r) + h(n, m, r + ) = p(n), m > 0, h(n, 0, r) + h(n, 0, r + ) = p(n) q(n) The main part of the proof is: (first symmetry) h(n, 0, r) = h(n, 0, r) and (second symmetry) h(n, m, r) = h(n r 2m 2, m + 2, r) The first symmetry holds for any r and the second symmetry holds for m, r > 0 and for m = 0 and r 0. For j 0 let a j = h(n jr 2jm j(5j )/2, m + 2j, r j) and b j = h(n jr 2jm j(5j )/2, m + 2j, r j + ). We have a j + b j = p(n jr 2jm j(5j )/2), for all r, j > 0. 3
By second symmetry, a j = h(n jr 2jm j(5j )/2, m + 2j, r j) = h(n jr 2jm j(5j )/2 r j, 2m 4j 2, r j) = b j+ We have h(n, m, r) = a 0 = b = b + (a b 2 ) (a 2 b 3 ) + (a 3 b 4 ) = (b + a ) (b 2 + a 2 ) + (b 3 + a 3 ) (b 4 + a 4 ) + = p(n r 2m 2) p(n 2r 4m 9) + p(n 3r 6m 2) = j= ( )j p(n jr 2jm j(5j )/2). H m, r (t) := n= h(n, m, r)tn = n= j= ( )j p(n jr 2jm j(5j )/2)t n jr 2jm j(5j )/2 t jr+2jm+j(5j )/2 = n= ( t n ) j= ( )j t jr+2jm+j(5j )/2. In particular, H 0, 0 (t) = n= H 0, (t) = n= ( t n ) j= ( )j t 5j /2, ( t n ) j= ( )j t 5j+ /2. By first symmetry, H 0, 0 (t) + H 0, (t) = H 0, 0 (t) + H 0, (t) = P (t) Q(t). Therefore ( t n ) ( ( ) j t 5j /2 + n= = n= j= which proves the Schur indentity. ( ) j t 5j /2) j= ( t n ) ( + t k2 ( t)( t 2 ) ( t k ) ), k= 4
3 Proof of First and Second Symmetry 3. Proof of the first symmetry Define an involution φ : H n,0,r H n,0, r., which preserves the size of partitions as well as their Durfee squares. Let λ be a partition with two Durfee square and partitions α, β, γ. Denote s = s 0 (λ), t = t 0 (λ). φ : λ λ by first mapping (α, β, γ) to (µ, ν, π, ρ, σ), then to (ˆα, ˆβ, ˆγ). () Firstly, let µ = β. Secondly, remove the following parts from α : α s t βj +j for j t. Let ν be the partition comprising of parts removed from α and π be the partitions comprising of the parts which were not removed. Thirdly, for j t, let k j = max{k s t γ j k π s t k+ }. Let ρ be the partition with parts ρ j = k j and σ be the partition with parts σ j = γ j k j. (2) Let ˆγ = ν + µ be the sum of partitions (σ j = γ j k j ). Secondly, ˆα = σ π be the union of partitions. Thirdly, ˆβ = ρ. Note that k is the unique integer k which satisfies π s t k+ γ j k π s t k. To show φ is an involution, check:. ρ is a partition. 2. σ is a partition. 3. ˆλ = φ(λ) is a partition. 4. φ 2 is the identity. 5. r 2,0 (ˆλ) = r 2,0 (λ). Proof.. If k j k j+, then π s t kj + + k j π s t kj+ + + k j+ γ j+ γ j π s t kj + k j. Thus π s t kj + γ j+ k j π s t kj. The uniqueness implies that k j = k j+. Therefore k j k j+, then ρ is a partition. 5
2. If k j > k j+, then we have s t k j + s t k j+ and therefore π s t kj+ π s t kj +. We conclude that γ k k j γ j+ k j+. If k j = k j+, since γ is a partition, γ j k j γ j+ k j+. Therefore σ is a partition. 3. By definition ˆα, ˆβ, ˆγ are also partitions and fit into s, t Durfee rectangles. 4. Let ˆλ = φ(λ). Similarly, let ˆµ, ˆν, ˆπ, ˆρ, ˆσ be the partitions occuring in the intermediate stage of the second application of φ and let α, β, γ be the partitions to the right of, in the middle of and below the Durfee squares of φ 2 (λ) = φ(ˆλ). We have ˆµ = ˆβ = ρ. It can be shown that α = µ π = α, β = µ = β, and (γ ) = ρ + σ = γ 5. We have r 2,0 (λ) = β + α s t β + γ = µ + ν ρ σ. r 2,0 (ˆλ) = ˆβ + ˆα s t ˆβ + ˆγ = ρ + σ µ ν. Therefore r 2,0 (ˆλ) = r 2,0 (λ). 3.2 Proof of the second symmetry We present a bijection ψ m,r : H n,m, r H n r 2m 2,m+2, r for m, r > 0 and for m = 0, r 0. Describe the action of ψ := ψ m,r by giving the sizes of the Durfee rectangles of ˆλ := ψ m,r (λ) = ψ(λ) and ˆα, ˆβ, ˆγ. () If λ has two m Durfee rectangles of height s := s m (λ), t := t m (λ). Then ˆλ has two (m + 2)-Durfee rectangles of height s := s m+2 (ˆλ) = s + and t := t m+2 (ˆλ) = t +. (2) Let k = max{k s t γ r k α s t k+ }. Obtain ˆα from α by adding a new part of size γ r k, ˆβ from β by adding a new paet of size k, and ˆγ from γ by removing its first column. Since k j is the unique integer k which satisfies π s t k+ γ k k π s t k. By considering k = β, we see k is defined and indeed have k β. And k is the unique k such that α s t k+ γ r k α s t k. Check the map ψ = ψ m,r defined above is a bijection by 4 parts. 6
. we prove that ˆλ = ψ(λ) is a partition. Note that λ has m Durfee rectangles of nonzero width, ˆλ may have (m+2) Durfee rectangles of width s and t. Also, the partitions ˆα and ˆβ have at most s + and t + parts. While the partitions ˆβ and ˆγ have parts of size at most s t and t. This means they can sit to the right of, in the middle of, and below the two m + 2 Durfee rectangles of ˆλ. 2. The size of ˆλ is n r 2m 2. Note the sum of the sizes of the rows added to α and β is r less than te size of the column removed from γ, and that both the first and second (m + 2) Durfee rectangles of ˆλ have size m + less than the size of the correpnding m Durfee rectangle of λ. 3. r 2,m+2 (ˆλ) r The part we insert to β will be the largest part of the resulting partition, i.e. ˆβ = k. We have α s t k + γ r k α s t k. Therefore, we have ˆα s t ˆβ + = ˆα s t k + = γ r k. We have chosen k in the unique way so that the rows we insert into α and β are ˆα s t ˆβ + and ˆβ, respectively. Therefore r 2,m+2 (ˆλ) = ˆα s t ˆβ + + ˆβ ˆγ = γ r k +k ˆγ r. 4. Present the inverse map ψ. The above characterization of k also shows us that to recover α, β, γ from ˆα, ˆβ, ˆγ, we remove ˆα s t ˆβ + from ˆα, remove ˆβ from ˆβ, and add a column of height ˆα s t ˆβ + + ˆβ + r to ˆγ. Since we can also easily recover the sizes of the previous m Durfee rectangles, we conclude that ψ is a bijection between the desired sets. 7
4 An Application in Analysis We wish to consider in as simple a manner as possible a two-variable generalization f(q, t) that has the following properties: Let. f(q, t) = n=0 D n(q)t n, where D n (q) are polynomials. 2. lim n D n (q) = n=0 /( q5n+ )( q 5n+4 ) 3. f(q, t) satisfies a first-order nonhomogeneneous q-difference equation. f(q, t) = n=0 t 2n q n2 ( t)( tq) ( tq n ) Firstly, ( t)f(q, t) = + t 2 qf(q, tq), thus 3 is satisfied. Next we note f(q, t) = n=0 m Thus D n (q) = ( N n ) 0 2n N qn2 n m=0 t2n q n2 t m( n+m lim n D n (q) = lim t ( t)f(q, t) = n=0 We have 2. q )q = ( N n ) N=0 tn 0 2n N qn2 n, we have. q. q n2 = ( q)( q 2 ) ( q n ) n=0. ( q 5n+ )( q 5n+4 ) 8
5 Reference () C. Boulet, I. Pak, A combinatorial proof of the Rogers-Ramanujan and Schur identities, Journal of Combinatorial Theory, Series A 3 (2006) 09-030 (2) G.E. Andrews, q-series: Their development and application in analysis, number theory, combinatorics, physics, and computer algebra, in: CBMS Reg. Conf. Ser. Math., vol. 66, Amer. Math. Soc., Providence, RI, 986 (3) The theory of partitions, Encyclopedia of Mathematics and Its Applications (Rota, Editor), Vol. 2, G.-C, Addison-Wesley, Reading, 976 9