Lebesgue identity

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imported>Pythagoras0님의 2013년 3월 13일 (수) 13:50 판 (→‎search for other b's)
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introduction

\[f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\]

 

fermionic form expression

\[f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+2j^2-i}{2}}}{(q)_{i}(q)_{j}}\label{faz}\]

(proof)

We use the q-binomial identity (see useful techniques in q-series) \[(-z;q)_{n}= \sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r(r-1)/2}z^r\] and  \[(-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r.\]

Then the LHS of \ref{faz} can be written as \[ \begin{aligned} f(a,z)& =\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}\\ {}&=\sum_{k\geq 0}\frac{a^kq^{k(k-1)/2}}{(q)_{k}}\sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r \end{aligned} \] By putting \(j=r\) and \(k=i+j\), \[ \begin{aligned} {}&=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{(i+j)(i+j-1)/2+j(j+1)/2}}{(q)_{i}(q)_{j}}\\ {}&=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+2j^2-i}{2}}}{(q)_{i}(q)_{j}} \end{aligned} \]  ■

\[ \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\]

 

 

Lebesgue's identity

  • Put a=q, c=z. we get Lebesgue's identity.

\[f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq)_{k}}{(q)_{k}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}=\prod_{m=1}^{\infty} (1+zq^{2m})(1+q^{m})\]

  • special case : we get a rank 2 form of Lebesgue's identity

\[f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{z^{j}q^{\frac{i^2+2ij+j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}\]

 

 

specializations

  • we expect to find five vectors for linear terms

$$\vec{b}=(1/2,-1),(0,0),(1/2,0),(1/2,1),(1,1)$$

Theorem

For $\vec{b}=(1/2,0)$, \[f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2},\] For $\vec{b}=(1/2,1)$, \[f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\]

proof

Let us use the following identities from useful techniques in q-series \[(-q)_{n}=\frac{(q^2;q^2)_{n}}{(q;q)_{n}}\] \[(-q;q^{2})_{n}=\frac{(-q;q)_{n}}{(-q^{2};q^{2})_{n}}=\frac{(q^{2};q^{2})_{n}(q^{2};q^{2})_{n}}{(q^{4};q^{4})_{n}(q;q)_{n}}=\frac{(q^{2};q^{4})_{n}}{(q^{1};q^{4})_{n}(q^{3};q^{4})_{n}}\] . \[(-q^2;q^{2})_{n}=\frac{(q^4;q^4)_{n}}{(q^2;q^2)_{n}}=\frac{1}{(q^2;q^4)_{n}}\]

Therefore \[f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\] \[f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}.\]■

search for other b's

\[f_{A,B,0}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\]

\[f_{A,B,0}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\]

$$ \frac{\left(Q^3;Q^8\right){}_{\infty } \left(Q^5;Q^8\right){}_{\infty } \left(Q^8;Q^8\right){}_{\infty }}{\left(Q;Q^4\right){}_{\infty } \left(Q^3;Q^4\right){}_{\infty } \left(Q^4;Q^4\right){}_{\infty }} =\frac{1}{\left(Q;Q^8\right){}_{\infty } \left(Q^7;Q^8\right){}_{\infty } \left(Q^4;Q^8\right){}_{\infty }}=\frac{1}{\left(q^{1/2};q^4\right){}_{\infty } \left(q^{7/2};q^4\right){}_{\infty } \left(q^{2};q^4\right){}_{\infty }} $$ where $Q^2=q$

$$ f_{A,B,0}=\frac{2}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}} $$

$$ \frac{1}{\left(Q^3;Q^8\right){}_{\infty } \left(Q^4;Q^8\right){}_{\infty } \left(Q^5;Q^8\right){}_{\infty }}=\frac{1}{\left(q^{3/2};q^4\right){}_{\infty } \left(q^{2};q^4\right){}_{\infty } \left(q^{5/2};q^4\right){}_{\infty }} $$ where $Q^2=q$

  • $\frac{\eta(4\tau)}{\eta(\tau)}$

continued fraction expression

  • rank 2 continued fraction
  • [Alladi&Gordon1993] 277-278p
  • Let \(f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\) as above
  • consider the following continued fractions

\[F(a,c)=\frac{f(a,c)}{f(aq,c)}=1+a+\frac{acq}{1+aq} {\ \atop+} \frac{acq^2}{1+aq^2}{\ \atop+} \frac{acq^3}{1} {\ \atop+\dots}\] \[R(a,b)=\frac{f(a,a^{-1}b)}{f(aq,a^{-1}b)}-a=\frac{R^{N}(a,b)}{R^{D}(a,b)}=1+\frac{bq}{1+aq} {\ \atop+} \frac{bq^2}{1+aq^2}{\ \atop+} \frac{bq^3}{1} {\ \atop+\dots}\]
where \[R^{N}(a,b)=f(q,a^{-1}b)-af(aq,a^{-1}b)=f(aq,a^{-1}bq^{-1})=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}b)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}}}{(q)_{i}(q)_{j}}\] and \[R^{D}(a,b)=f(aq,a^{-1}b)=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}bq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}\]

  • applications

\[R^N(1,1)=f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q^1;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\] \[R^{D}(1,1)=f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\]

  • continued fraction

\[R(1,1)=\frac{R^{N}(1,1)}{R^{D}(1,1)}=1+{q \over 1+q + } {q^2 \over 1+q^2+} {q^3 \over 1+q^3} \cdots=\frac{(q^2;q^4)_{\infty}^2}{(q^1;q^4)_{\infty}(q^3;q^4)_{\infty}}\]

 

 

related dilogarithm identity

 

 

 

comparison with Rogers-Selberg identities

\[AG_{3,3}(q)=\sum_{n_1,n_{2}\geq0}\frac{q^{n_{1}^2+2n_1n_2+2n_{2}^{2}}}{(q)_{n_1}(q)_{n_{2}}}=\prod_{r\neq 0,\pm 3 \pmod {7}}\frac{1}{1-q^r}=\frac{(q^3;q^7)_\infty (q^4; q^7)_\infty(q^7;q^7)_\infty}{(q)_\infty}\] \[A(q)W(q)=AG_{3,3}(q)\]
where \[W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}\]

  • Lebesgue's identity

\[\frac{W(q)^2}{W(q^2)}=\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}\]

 

(proof)

Note that from useful techniques in q-series \[(-q;q^{2})_{\infty}=\frac{(-q;q)_{\infty}}{(-q^{2};q^{2})_{\infty}}=\frac{(q^{2};q^{2})_{\infty}(q^{2};q^{2})_{\infty}}{(q^{4};q^{4})_{\infty}(q;q)_{\infty}}=\frac{W(q)}{W(q^2)}\]

Therefore \[(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{W(q)^2}{W(q^2)}\]. ■

 

 

\[W(q)=\frac{\eta(2\tau)}{\eta(\tau)}\] \[W(q^2)=\frac{\eta(4\tau)}{\eta(2\tau)}\] \[\frac{W(q)^2}{W(q^2)}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{\eta(2\tau)^3}{\eta(\tau)^2\eta(4\tau)}=\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}\] \[W(q^2)W(q)=\frac{\eta(4\tau)}{\eta(\tau)}=q^{1/8}\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=\frac{q^{1/8}(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{q^{1/8}}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\]

 

 

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related items

 

 

encyclopedia

 

articles

  • New Proofs of Identities of Lebesgue and Göllnitz via Tilings
    • DP Little, 2007
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