"Y-system and functional dilogarithm identities"의 두 판 사이의 차이

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imported>Pythagoras0
20번째 줄: 20번째 줄:
 
* Let $\epsilon : \mathbf{I}\to \{1,-1\}$ be the function defined by $\epsilon(\mathbf{i})=\pm 1$ for $\mathbf{i}\in \mathbf{I}_{\pm}$ and $P_{\pm} =\{(\mathbf{i},u)\in \mathbf{I}\times\mathbb{Z}| \epsilon(\mathbf{i})(-1)^u=\pm 1\}$.  
 
* Let $\epsilon : \mathbf{I}\to \{1,-1\}$ be the function defined by $\epsilon(\mathbf{i})=\pm 1$ for $\mathbf{i}\in \mathbf{I}_{\pm}$ and $P_{\pm} =\{(\mathbf{i},u)\in \mathbf{I}\times\mathbb{Z}| \epsilon(\mathbf{i})(-1)^u=\pm 1\}$.  
 
* Roughly speaking, we want our alternate bicoloring interchanges their colors as $u\in \mathbb{Z}$ changes by 1.  
 
* Roughly speaking, we want our alternate bicoloring interchanges their colors as $u\in \mathbb{Z}$ changes by 1.  
 +
 +
 +
==five-term relation==
 +
* $\mathbb{Y}(A_2,A_1)$ was explicitly worked out
 +
* we saw that $$S=\left\{x,y,\frac{y+1}{x},\frac{x+y+1}{x y},\frac{x+1}{y}\right\}$$ forms a half-period of $\mathbb{Y}(A_2,A_1)$.
 +
* So we have $r=2,h=3$ and $r'=1,h'=2$.
 +
* They are all Laurent polynomials in $x$ and $y$.
 +
* From this, one can get functional dilogarithm identities
 +
\begin{align}
 +
&\sum_{a\in S}L\left(\frac{a}{1+a}\right) \notag \\
 +
=&
 +
L\left(\frac{x}{1+x}\right)+L\left(\frac{y}{1+y}\right)+L\left(\frac{1+y}{x (1+\frac{1+y}{x})}\right)+L\left(\frac{1+x+y}{x y (1+\frac{1+x+y}{x y})}\right)+L\left(\frac{1+x}{(1+\frac{1+x}{y}) y}\right) \notag
 +
\\
 +
=& L\left(\frac{x}{x+1}\right)+L\left(\frac{y}{y+1}\right)+L\left(\frac{y+1}{x+y+1}\right)+L\left(\frac{x+y+1}{x y+x+y+1}\right)+L\left(\frac{x+1}{x+y+1}\right) \notag \\
 +
=&3L(1)=\frac{\pi^2}{2} \notag
 +
\end{align}
 +
and
 +
\begin{align}
 +
&\sum_{a\in S}L\left(\frac{1}{1+a}\right) \notag \\
 +
=&
 +
L\left(\frac{1}{x+1}\right)+L\left(\frac{1}{y+1}\right)+L\left(\frac{1}{\frac{y+1}{x}+1}\right)+L\left(\frac{1}{\frac{x+y+1}{x y}+1}\right)+L\left(\frac{1}{\frac{x+1}{y}+1}\right) \notag \\
 +
=& L\left(\frac{1}{x+1}\right)+L\left(\frac{1}{y+1}\right)+L\left(\frac{x}{x+y+1}\right)+L\left(\frac{x y}{x y+x+y+1}\right)+L\left(\frac{y}{x+y+1}\right) \notag \\
 +
=&2L(1)=\frac{\pi^2}{3} \notag.
 +
\end{align}
  
  

2013년 7월 15일 (월) 02:33 판

introduction


main results

  • Bloch group element

$$ \sum_{(\mathbf{i},u)\in S_{+}} Y_{\mathbf{i}}(u)\wedge (1+Y_{\mathbf{i}}(u))=0\in \Lambda^2 \mathbb{Q}(y)^{\times} $$ where $S_{+}=\{(\mathbf{i},u) |0\leq u \leq 2(h+h')-1,(\mathbf{i},u)\in P_{+}\}$.

  • functional dilogarithm identity

$$ \sum_{(\mathbf{i},u)\in S_{+}}L\left(\frac{Y_\mathbf{i}(u)}{1+Y_\mathbf{i}(u)}\right)=h r r' L(1) $$


bicoloring

  • what's $P_{+}$?
  • We give an alternate bicoloring on the pair of Dynkin diagrams. Let us fix bipartite decompositions of $I$ and $I'$.
  • Let $ \mathbf{I}= I\times I'$ and $\mathbf{I}=\mathbf{I}_{+}\sqcup \mathbf{I}_{-}$ where $\mathbf{I}_{+}=(I_{+}\times I'_{+}) \sqcup (I_{-}\times I'_{-})$ and $\mathbf{I}_{-}=(I_{+}\times I'_{-}) \sqcup (I_{-}\times I'_{+})$.
  • Let $\epsilon : \mathbf{I}\to \{1,-1\}$ be the function defined by $\epsilon(\mathbf{i})=\pm 1$ for $\mathbf{i}\in \mathbf{I}_{\pm}$ and $P_{\pm} =\{(\mathbf{i},u)\in \mathbf{I}\times\mathbb{Z}| \epsilon(\mathbf{i})(-1)^u=\pm 1\}$.
  • Roughly speaking, we want our alternate bicoloring interchanges their colors as $u\in \mathbb{Z}$ changes by 1.


five-term relation

  • $\mathbb{Y}(A_2,A_1)$ was explicitly worked out
  • we saw that $$S=\left\{x,y,\frac{y+1}{x},\frac{x+y+1}{x y},\frac{x+1}{y}\right\}$$ forms a half-period of $\mathbb{Y}(A_2,A_1)$.
  • So we have $r=2,h=3$ and $r'=1,h'=2$.
  • They are all Laurent polynomials in $x$ and $y$.
  • From this, one can get functional dilogarithm identities

\begin{align} &\sum_{a\in S}L\left(\frac{a}{1+a}\right) \notag \\ =& L\left(\frac{x}{1+x}\right)+L\left(\frac{y}{1+y}\right)+L\left(\frac{1+y}{x (1+\frac{1+y}{x})}\right)+L\left(\frac{1+x+y}{x y (1+\frac{1+x+y}{x y})}\right)+L\left(\frac{1+x}{(1+\frac{1+x}{y}) y}\right) \notag \\ =& L\left(\frac{x}{x+1}\right)+L\left(\frac{y}{y+1}\right)+L\left(\frac{y+1}{x+y+1}\right)+L\left(\frac{x+y+1}{x y+x+y+1}\right)+L\left(\frac{x+1}{x+y+1}\right) \notag \\ =&3L(1)=\frac{\pi^2}{2} \notag \end{align} and \begin{align} &\sum_{a\in S}L\left(\frac{1}{1+a}\right) \notag \\ =& L\left(\frac{1}{x+1}\right)+L\left(\frac{1}{y+1}\right)+L\left(\frac{1}{\frac{y+1}{x}+1}\right)+L\left(\frac{1}{\frac{x+y+1}{x y}+1}\right)+L\left(\frac{1}{\frac{x+1}{y}+1}\right) \notag \\ =& L\left(\frac{1}{x+1}\right)+L\left(\frac{1}{y+1}\right)+L\left(\frac{x}{x+y+1}\right)+L\left(\frac{x y}{x y+x+y+1}\right)+L\left(\frac{y}{x+y+1}\right) \notag \\ =&2L(1)=\frac{\pi^2}{3} \notag. \end{align}


length of diagonals

 

  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]

 

 

relation to 5-term relation

 

 

five-term relation of dilogarithm

 

  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

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

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


 

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