베일리 격자(Bailey lattice)

이 항목의 수학노트 원문주소

베일리 격자(Bailey lattice)

개요

$\{\alpha_r\}, \{\beta_r\}$ 를 a에 대한 베일리 쌍이라 하고, 다음을 정의하자.

$\alpha_0'=\alpha_0$ , $\alpha_n'=(1-a)a^nq^{n^2-n}(\frac{\alpha_n}{1-aq^{2n}}-\frac{aq^{2n-2}\alpha_{n-1}}{1-aq^{2n-2}})$ $\beta_n'=\sum_{r=0}^{n}\frac{a^rq^{r^2-r}}{(q)_{n-r}}\beta_{r}$
$\{\alpha_r'\}, \{\beta_r'\}$ 는 $aq^{-1}$ 에 대한 베일리 쌍이 된다

comparison with Bailey chain

베일리 사슬(Bailey chain)
$\alpha^\prime_n= a^nq^{n^2}\alpha_n$

$\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r$
This does not change the parameter a of the Bailey pair.
lattice construction changes this

corollary

Let $\{\alpha_r\}, \{\beta_r\}$ be the initial Bailey pair relative to a. Then the following is true :

$\sum_{n_1\geq\cdots\geq n_{k}\geq0}\frac{a^{n_1+\cdots+n_{k}}q^{n_1^2+\cdots+n_{k}^2-n_1-n_2-\cdots-n_i}\beta_{n_{k}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}-n_{k}}}=\frac{1}{(a)_{\infty}}\left{[}\alpha_0+(1-a)\sum_{n=1}^{\infty}(\frac{a^{kn}q^{kn^2-in}\alpha_n}{1-aq^{2n}}-\frac{a^{k(n-1)+i+1}q^{k(n-1)^2+(i+2)(n-1)}\alpha_{n-1}}{1-aq^{2n-2}})\right{]}$

(proof)

apply Bailey chain construction k-i times 베일리 사슬(Bailey chain)

At the (k-i)th step apply Bailey lattice

apply Bailey chain construction i-1 times again.

Then we get a Bailey pair

$\{\alpha_r'\}, \{\beta_r'\}$ is a Bailey pair relative to $aq^{-1}$ .

If we use the defining relation of Bailey pair to $\{\alpha_r'\}, \{\beta_r'\}$ ,

$\beta_L'=\sum_{r=0}^{L}\frac{\alpha_r'}{(q)_{L-r}(q)_{L+r}}$

and take the limit L\to\infty ■

Example. Do this for k=5 and i=2

응용

앤드류스-고든 항등식(Andrews-Gordon identity) 의 증명
initial Bailey pair

$\alpha_{L}=(-1)^{L}q^{\binom{L}{2}}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}=(-1)^{L}q^{L(L-1)/2}\frac{(1-aq^{2L})(a)_{L}}{(1-a)(q)_{L}}$

$\beta_{L}=\delta_{L,0}$
In the corollay above, set a=q and replace i by i-1

$\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1+\cdots+n_{k-1}}q^{n_1^2+\cdots+n_{k-1}^2-n_1-n_2-\cdots-n_{i-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}$
On LHS, we get

$L=\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}$
On RHS, we get

$R=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-qq^{2n}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-qq^{2n-2}})\right{]}$

$=\frac{1}{(q)_{\infty}}\left{[}1+(1-q)\sum_{n=1}^{\infty}(\frac{q^{kn}q^{kn^2-(i-1)n}\alpha_n}{1-q^{2n+1}}-\frac{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}\alpha_{n-1}}{1-q^{2n-1}})\right{]}$

Now use the original Bailey pair,

$\alpha_{n}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})(q)_{n}}{(1-q)(q)_{n}}=(-1)^{n}q^{n(n-1)/2}\frac{(1-q^{2n+1})}{(1-q)}$

$\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}$

$R=\frac{1}{(q)_{\infty}}\left{[}1+\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}(-1)^{n}q^{n(n-1)/2}}-{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}(-1)^{n-1}q^{(n-1)(n-2)/2}}\right{]}$

$=\frac{1}{(q)_{\infty}}\left{[}1+(-1)^{n}\sum_{n=1}^{\infty}({q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}+{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}\right{]}$
first part in the summation is

$(-1)^{n}\sum_{n=1}^{\infty}{q^{kn}q^{kn^2-(i-1)n}q^{n(n-1)/2}}=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n}q^{n(n-1)/2}}$

$=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2+(k-i+1)n+n(n-1)/2}}=(-1)^{n}\sum_{n=1}^{\infty}q^{n(2kn+2(k-i+1)+(n-1))/2$

$=(-1)^{n}\sum_{n=1}^{\infty}q^{n((2k+1)n+2k-2i)+1)/2$
secont part in the summation is

$(-1)^{n}\sum_{n=1}^{\infty}{q^{k(n-1)+i}q^{k(n-1)^2+(i+1)(n-1)}q^{(n-1)(n-2)/2}}$

$=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+i+1)(n-1)+i}q^{(n-1)(n-2)/2}$
$=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in-i+i}q^{(n-1)(n-2)/2}$

$=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in}q^{(n^2-3n+2)/2}$

$=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-2nk+k+kn-k+n-1+in+\frac{n^2}{2}-\frac{3n}{2}+1)}$

$=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-nk+in+\frac{n^2}{2}-\frac{n}{2})}$

$=(-1)^{n}\sum_{n=-1}^{-\infty}q^{kn^2+nk-in+\frac{n^2}{2}+\frac{n}{2})}$

$=(-1)^{n}\sum_{n=-1}^{-\infty}q^{n((2k+1)n+2k-2i+1)/2}$
by summing two parts, we get

$R=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}$
Therefore we have proved the following are equal

$\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}$
You can use Jacobi triple product identity to get

$\sum_{n_1\geq\cdots\geq n_{k-1}\geq0}\frac{q^{n_1^2+\cdots+n_{k-1}^2+n_i+\cdots+n_{k-1}}}{(q)_{n_{1}-n_{2}}\cdots (q)_{n_{k-2}-n_{k-1}}(q)_{n_{k-1}}}=\prod_{n\neq 0,\pm i\pmod {2k+1}}(1-q^n)^{-1}$