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

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imported>Pythagoras0
imported>Pythagoras0
23번째 줄: 23번째 줄:
  
  
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==five-term relation==
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==an example==
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* $\mathbb{Y}(A_2,A_1)$ was explicitly worked out
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* compute $\mathbb{Y}(A_2,A_1)$ explicitly
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* <math>y_{m-1}y_{m+1}=y_m+1</math>
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* Start with two variables <math>y_1,y_2</math>.
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* <math>y_3y_1=y_2+1</math>. so <math>y_3=\frac{y_2+1}{y_1}</math>
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* <math>y_2y_4=y_3+1 </math>implies <math>y_4=\frac{y_3+1}{y_2}=\frac{y_1+y_2+1}{y_1y_2}</math>
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* <math>y_3y_5=y_4+1</math> implies <math>y_5=\frac{y_4+1}{y_3}= \frac{y_1+1}{y_2}</math> we are getting Laurent polynomials
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* <math>y_4y_6=y_5</math> implies <math>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</math>
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* [[rank 2 cluster algebra]]
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===observations===
 
* 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 $\mathbb{Y}(A_2,A_1)$.  
 
* 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 $r'=1,h'=2$.  
 
* So we have $r=2,h=3$ and $r'=1,h'=2$.  
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* They are all Laurent polynomials in $x$ and $y$.  
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* They are all Laurent polynomials in $x$ and $y$.
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===dilogarithm identities===
 
* From this, one can get functional dilogarithm identities
 
* From this, one can get functional dilogarithm identities
 
\begin{align}
 
\begin{align}
45번째 줄: 58번째 줄:
 
=&2L(1)=\frac{\pi^2}{3} \notag.
 
=&2L(1)=\frac{\pi^2}{3} \notag.
 
\end{align}
 
\end{align}
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==length of diagonals==
 
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* [http://pythagoras0.springnote.com/pages/4682477 체비셰프 다항식]<br><math>U_n(x)^2=1+U_{n-1}(x)U_{n+1}(x)</math><br>
 
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* [http://pythagoras0.springnote.com/pages/6904713 정다각형의 대각선의 길이]<br><math>r_i^2=1+r_{i-1}r_{i+1}, 1\leq i \leq n-3</math><br>
 
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* Question : for what values of <math>r_1=x</math>, is the recurrence <math>r_i^2=1+r_{i-1}r_{i+1}</math> periodic? (<math>r_0=1</math>)
 
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# A := RecurrenceTable[{a[n] a[n - 2] + 1 == a[n - 1]^2, a[1] == x,<br>    a[2] == y}, a, {n, 10}]<br> Simplify[A]
 
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* Laurent phenomenon is true
 
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* total positivity is broken
 
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* 정오각형의 경우
 
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* <math>r_i^2=1+r_{i-1}r_{i+1}</math>, <math>r_0=1,r_3=1</math>
 
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* 3가지 점화식의 해가 존재
 
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* <math>\{1,-1,0,1\}</math>, <math>\{1,\frac{-\sqrt{5}+1}{2},\frac{-\sqrt{5}+1}{2},1 \}</math> , <math>\{1,\frac{\sqrt{5}+1}{2},\frac{\sqrt{5}+1}{2},1 \}</math>
 
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# A := RecurrenceTable[{a[n] a[n - 2] + 1 == a[n - 1]^2, a[1] == 1,<br>    a[2] == 2 y}, a, {n, 10}]<br> Simplify[A]<br> NSolve[-4 y + 8 y^3 == 1, y]<br> {1, 2 y, -1 + 4 y^2, -4 y + 8 y^3,<br>    1 - 12 y^2 +<br>     16 y^4} /. {{y -> -0.5`}, {y -> -0.30901699437494745`}, {y -><br>      0.8090169943749475`}} // TableForm
 
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==total positivity==
 
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* <math>r_{i-1}r_{i+1}=r_i^2+1</math>
 
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# A := RecurrenceTable[{a[n] a[n - 2] - 1 == a[n - 1]^2, a[1] == x,<br>    a[2] == y}, a, {n, 10}]<br> Simplify[A]
 
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* [[rank 2 cluster algebra examples]]
 
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==relation to 5-term relation==
 
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* [http://pythagoras0.springnote.com/pages/5956565 5항 관계식 (5-term relation)]<br><math>1-x_{i}=x_{i-1}x_{i+1}</math><br>
 
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==five-term relation of dilogarithm==
 
 
* {{수학노트|url=5항_관계식_(5-term_relation)}}
 
* {{수학노트|url=5항_관계식_(5-term_relation)}}
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* [http://pythagoras0.springnote.com/pages/5956565 5항 관계식 (5-term relation)]<br>
 
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*  로저스 다이로그 함수 <math>L(x)</math>에 대하여 다음이 성립한다<br><math>0\leq x,y\leq 1</math> 일 때,  :<math>L(x)+L(1-xy)+L(y)+L\left(\frac{1-y}{1-xy}\right)+L\left(\frac{1-x}{1-xy}\right)=\frac{\pi^2}{2}</math><br>
 
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* <math>1-x_{i}=x_{i-1}x_{i+1}</math> 를 만족시키는 다섯개의 수<br>
 
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* is this also an example of a cluster variable?
 
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* [[asymptotic analysis of basic hypergeometric series]]
 
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# f[{x_, y_, z_, w_}] := Simplify[(x - z)/(x - w)*(y - w)/(y - z)]<br> A := Permutations[{0, 1, w, z}]<br> Table[Limit[f[A[[i]]], w -> \[Infinity]], {i, 24}]<br> B := Subsets[{0, x*y, 1, y, z}, {4}]<br> g[i_] := Table[<br>   Limit[f[n], z -> \[Infinity]], {n, Permutations[B[[i]]]}]<br> Table[f[B[[i]]], {i, 1, 5}]<br> Table[g[i], {i, 5}]
 
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==rank 2 example==
 
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* [[rank 2 cluster algebra]]
 
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<math>y_{m-1}y_{m+1}=y_m+1</math>
 
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Start with two variables <math>y_1,y_2</math>.
 
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<math>y_3y_1=y_2+1</math>. so <math>y_3=\frac{y_2+1}{y_1}</math>
 
−
 
−
<math>y_2y_4=y_3+1 </math>implies <math>y_4=\frac{y_3+1}{y_2}=\frac{y_1+y_2+1}{y_1y_2}</math>
 
−
 
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<math>y_3y_5=y_4+1</math> implies <math>y_5=\frac{y_4+1}{y_3}= \frac{y_1+1}{y_2}</math> we are getting Laurent polynomials
 
−
 
−
<math>y_4y_6=y_5</math> implies <math>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</math>
 
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==history==
 
==history==
143번째 줄: 77번째 줄:
 
* [[Bloch group, K-theory and dilogarithm]]
 
* [[Bloch group, K-theory and dilogarithm]]
  
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==expositions==
 
==expositions==

2013년 8월 16일 (금) 15:41 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.


an example

  • compute $\mathbb{Y}(A_2,A_1)$ explicitly
  • \(y_{m-1}y_{m+1}=y_m+1\)
  • 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 \(y_6=\frac{y_5+1}{y_4}= \frac{\frac{y_1+1}{y_2}+1}{\frac{y_1+y_2+1}{y_1y_2}}=\frac{y_1(y_1+1)+y_1y_2}{y_1+y_2+1}=y_1\)
  • rank 2 cluster algebra


observations

  • 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 $y$.


dilogarithm identities

  • 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}


history

 

 

related items


expositions

 

 

articles

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