"Bailey lattice"의 두 판 사이의 차이

2012년 8월 26일 (일) 07:52 판

non-unitary c(2,2k+1) minimal models

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-<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">introduction</h5>
-Let <math>\{\alpha_r\}, \{\beta_r\}</math> be a Bailey pair relative to a and set
-<math>\alpha_0'=\alpha_0</math>, <math>\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}})</math><math>\beta_n'=\sum_{r=0}^{n}\frac{a^rq^{r^2-r}}{(q)_{n-r}}\beta_{r}</math>
-Then <math>\{\alpha_r'\}, \{\beta_r'\}</math>  is a Bailey pair relative to <math>aq^{-1}</math>
-<h5 style="line-height: 2em; margin: 0px;">comparison with Bailey chain</h5>
-* [http://pythagoras0.springnote.com/pages/9408994 베일리 사슬(Bailey chain)]
-* <math>\alpha^\prime_n= a^nq^{n^2}\alpha_n</math><br><math>\beta^\prime_L = \sum_{r=0}^{L}\frac{a^rq^{r^2}}{(q)_{L-r}}\beta_r</math><br>
-*  This does not change the parameter <em>a</em> of the Bailey pair.<br>
-*  lattice construction changes this<br>
-<h5 style="line-height: 2em; margin: 0px;">corollary</h5>
-Let <math>\{\alpha_r\}, \{\beta_r\}</math> be the initial Bailey pair relative to a. Then the following is true :
-<math>\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{]}</math>
-(proof)
-apply Bailey chain construction k-i times  [http://pythagoras0.springnote.com/pages/9408994 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
-<math>\{\alpha_r'\}, \{\beta_r'\}</math>  is a Bailey pair relative to <math>aq^{-1}</math>.
-If we use the defining relation of Bailey pair to <math>\{\alpha_r'\}, \{\beta_r'\}</math>,
-<math>\beta_L'=\sum_{r=0}^{L}\frac{\alpha_r'}{(q)_{L-r}(q)_{L+r}}</math>
-and take the limit L\to\infty ■
-Example. Do this for k=5 and i=2
-<h5 style="line-height: 2em; margin: 0px;">application</h5>
-*  the proof of [[Andrews-Gordon identity]]<br>
-*  initial Bailey pair<br><math>\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}}</math><br><math>\beta_{L}=\delta_{L,0}</math><br>
-*  In the corollay above, set a=q and replace i by i-1<br><math>\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{]}</math><br>
-*  On LHS, we get<br><math>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}}}</math><br>
-*  On RHS, we get<br><math>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{]}</math><br><math>=\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{]}</math><br> Now use the original Bailey pair,<br><math>\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)}</math><br><math>\alpha_{n-1}=(-1)^{n-1}q^{(n-1)(n-2)/2}\frac{(1-q^{2n-1})}{(1-q)}</math><br><math>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{]}</math><br><math>=\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{]}</math><br>
-*  first part in the summation is<br>  <math>(-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}}</math><br><math>=(-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</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{n((2k+1)n+2k-2i)+1)/2</math><br>
-*  secont part in the summation is<br><math>(-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}}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+i+1)(n-1)+i}q^{(n-1)(n-2)/2}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in-i+i}q^{(n-1)(n-2)/2}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{k(n-1)^2+(k+1)(n-1)+in}q^{(n^2-3n+2)/2}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-2nk+k+kn-k+n-1+in+\frac{n^2}{2}-\frac{3n}{2}+1)}</math><br><math>=(-1)^{n}\sum_{n=1}^{\infty}q^{kn^2-nk+in+\frac{n^2}{2}-\frac{n}{2})}</math><br><math>=(-1)^{n}\sum_{n=-1}^{-\infty}q^{kn^2+nk-in+\frac{n^2}{2}+\frac{n}{2})}</math><br><math>=(-1)^{n}\sum_{n=-1}^{-\infty}q^{n((2k+1)n+2k-2i+1)/2}</math><br>
-*  by summing two parts, we get<br><math>R=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}</math><br>
-*  Therefore we have proved the following are equal<br><math>\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}}</math><br>
-*  You can use Jacobi triple product identity to get<br><math>\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}</math><br>
-<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">history</h5>
-* http://www.google.com/search?hl=en&tbs=tl:1&q=
-<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">related items</h5>
-* [[Andrews-Gordon identity]]<br>
 * [[non-unitary c(2,2k+1) minimal models]]<br>
-<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles</h5>
-* [http://www.liafa.jussieu.fr/%7Elovejoy/lattice.pdf A Bailey Lattice]<br>
-** Jeremy Lovejoy, Proceedings of the American Mathematical Society, Vol. 132, No. 5 (May, 2004), pp. 1507-1516
-*  The Bailey lattice<br>
-**  David Bressoud, an introduction, pp. 57--67 in Ramanujan Revisited. G. E. Andrews et al. eds., Academic Press, 1988.<br>
-*  The Bailey Lattice<br>
-**  A. Agarwal, G.E. Andrews, and D. Bressoud,  J. Indian Math. Soc. 51 (1987), 57-73.<br>

"Bailey lattice"의 두 판 사이의 차이

2012년 8월 26일 (일) 07:52 판

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