Y-system and functional dilogarithm identities

introduction

• 체비셰프 다항식
  \(U_n(x)^2=1+U_{n-1}(x)U_{n+1}(x)\)
  
• 정다각형의 대각선의 길이
  \(r_i^2=1+r_{i-1}r_{i+1}, 1\leq i \leq n-3\)
  
• Question : for what values of \(r_1=x\), is the recurrence \(r_i^2=1+r_{i-1}r_{i+1}\) periodic? (\(r_0=1\))

 

1. A := RecurrenceTable[{a[n] a[n - 2] + 1 == a[n - 1]^2, a[1] == x,
      a[2] == y}, a, {n, 10}]
   Simplify[A]

• Laurent phenomenon is true
• total positivity is broken
• 정오각형의 경우
• \(r_i^2=1+r_{i-1}r_{i+1}\), \(r_0=1,r_3=1\)
• 3가지 점화식의 해가 존재
• \(\{1,-1,0,1\}\), \(\{1,\frac{-\sqrt{5}+1}{2},\frac{-\sqrt{5}+1}{2},1 \}\) , \(\{1,\frac{\sqrt{5}+1}{2},\frac{\sqrt{5}+1}{2},1 \}\)

 

1. A := RecurrenceTable[{a[n] a[n - 2] + 1 == a[n - 1]^2, a[1] == 1,
      a[2] == 2 y}, a, {n, 10}]
   Simplify[A]
   NSolve[-4 y + 8 y^3 == 1, y]
   {1, 2 y, -1 + 4 y^2, -4 y + 8 y^3,
      1 - 12 y^2 +
       16 y^4} /. {{y -> -0.5`}, {y -> -0.30901699437494745`}, {y ->
        0.8090169943749475`}} // TableForm

 

 

 

total positivity

• \(r_{i-1}r_{i+1}=r_i^2+1\)

1. A := RecurrenceTable[{a[n] a[n - 2] - 1 == a[n - 1]^2, a[1] == x,
      a[2] == y}, a, {n, 10}]
   Simplify[A]

• rank 2 cluster algebra examples

 

 

relation to 5-term relation

• 5항 관계식 (5-term relation)
  \(1-x_{i}=x_{i-1}x_{i+1}\)
  

 

 

five-term relation of dilogarithm

• 5항 관계식 (5-term relation)
  
• 로저스 다이로그 함수 \(L(x)\)에 대하여 다음이 성립한다
  \(0\leq x,y\leq 1\) 일 때,
  \(L(x)+L(1-xy)+L(y)+L(\frac{1-y}{1-xy})+L\Left( \frac{1-x}{1-xy} )\right)=\frac{\pi^2}{2}\)
  
• \(1-x_{i}=x_{i-1}x_{i+1}\) 를 만족시키는 다섯개의 수
  
• is this also an example of a cluster variable?
• asymptotic analysis of basic hypergeometric series

 

1. f[{x_, y_, z_, w_}] := Simplify[(x - z)/(x - w)*(y - w)/(y - z)]
   A := Permutations[{0, 1, w, z}]
   Table[Limit[f[Ai], w -> \[Infinity]], {i, 24}]
   B := Subsets[{0, x*y, 1, y, z}, {4}]
   g[i_] := Table[
     Limit[f[n], z -> \[Infinity]], {n, Permutations[Bi]}]
   Table[f[Bi], {i, 1, 5}]
   Table[g[i], {i, 5}]

 

 

rank 2 example

• rank 2 cluster algebra

\(y_{m-1}y_{m+1}=y_m+1\)

Start with two variables \(y_1,y_2\).

\(y_3y_1=y_2+1\). so \(y_3=\frac{y_2+1}{y_1}\)

\(y_2y_4=y_3+1 \)implies \(y_4=\frac{y_3+1}{y_2}=\frac{y_1+y_2+1}{y_1y_2}\)

\(y_3y_5=y_4+1\) implies \(y_5=\frac{y_4+1}{y_3}= \frac{y_1+1}{y_2}\) we are getting Laurent polynomials

\(y_4y_6=y_5\) implies \(y_6=\frac{y_5+1}{y_4}= \frac{\frac{y_1+1}{y_2}+1}{\frac{y_1+y_2+1}{y_1y_2}}=\frac{y_1(y_1+1)+y_1y_2}{y_1+y_2+1}=y_1\)

 

 

history

• http://www.google.com/search?hl=en&tbs=tl:1&q=

 

 

related items

• cluster algebra
• Nahm's equationcluster algebra and dilogarithm identities

• 3 central charge of CFT, L-values, volume of threefolds and dilogarithm
• dilogarithm and dilogarithm identities
• Bloch group, K-theory and dilogarithm