Rank 2 cluster algebra

수학노트
http://bomber0.myid.net/ (토론)님의 2011년 1월 26일 (수) 06:30 판
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introduction
  • cluster algebra defined by a 2x2 matrix

 

 

 

cluster variables and exchange relations

Fix two positive integers b and c.

Let y_1 and y_2 be variable. Define a sequence {y_n}.

\(y_{m-1}y_{m+1}=y_m^b+1\) if m odd

\(y_{m-1}y_{m+1}=y_m^c+1\) if m even

We call this 'exchange relation'

y_m's are called 'cluster variable'

\(\{y_i,y_{i+1}\}\) "cluster"

Note : we can use the exchange relation any y_m in terms of arbitrary cluster {y_i,y_{i+1}} (rational expression)

 

 

example 1

Put b=c=1

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

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

 

 

example 2

Put b=1, c=3

y_1,y_2

\(y_3y_1=y_2^3+1\). so \(y_3=\frac{y_2^3+1}{y_1}\)

\(y_2y_4=y_3+1 \)implies \(y_4=\frac{y_3+1}{y_2}=\frac{y_1+y_2^3+1}{y_1y_2}\)

\(y_3y_5=y_4^3+1\) implies \(y_5=\frac{y_4^3+1}{y_3}= \frac{(y_1+1)^3+y_2^3(y_2^3+3y_1+2)}{y_1^2y_2^3}\)[1]http://www.wolframalpha.com/input/?i=((x%2By^3%2B1)^3%2B(xy)^3)/(x^2y^3(y^3%2B1))

Note that we are getting Laurent polynomials.

\(y_6=\frac{(y_1+1)^2+y_2^3}{y_1y_2^2}\)

\(y_7=\frac{(y_1+1)^3+y_2^3}{y_1y_2^3}\)

\(y_8=\frac{y_1+1}{y_2}\)

\(y_9=y_1\)

\(y_{10}=y_2\)

 

 

matrix formulation

\(B=\begin{bmatrix} 0 & -b\\ c &\,0 \end{bmatrix}\)

\(\mu_{1}(B)=\begin{bmatrix} 0 & b\\ -c &\,0 \end{bmatrix}\)

\(\mu_{2}(B)=\begin{bmatrix} 0 & b\\ -c &\,0 \end{bmatrix}\)

For \(k\in \{1,2,\cdots, n\}\),  \(x_kx_k' = \prod_{b_{ik}>0} x_i^{b_{ik}}+\prod_{b_{ik}<0} x_i^{|b_{ik}|}\)

x_1x_1'=x_2^c+1 call x_1'=x_3

x_2x_2'=x_1^b+1 call x_2'=x_4

 

\(\mu_k(B)\)

\(-b_{ij}\) if k=i or j

\(b_{ij}\) if \(b_{ik}b_{kj}\leq 0\)

 

\(b_{ij}+b_{ik}b_{kj}\) if \(b_{ik}, b_{kj}>0\)

\(b_{ij}-b_{ik}b_{kj}\) if \(b_{ik},{b_{kj}< 0\)

\(\mu_{1}(B)=\begin{bmatrix} 0 & b\\ -c &\,0 \end{bmatrix}\)

 

 

 

observations

(FZ) For any b,c, y_m is a Laurent polynomial.

Positivity conjecture: coefficients of these Laurent polynomials are positive (numerator and denomonator always have )

In this example, 

\(bc\leq 3\) iff the recurrence is periodic

 

 

 

1/25/2011

lecture following http://arxiv.org/abs/math/0307082v2

y_m cluster variables
\{y_m,y_{m+1}\} clusters
\{y_m^py_{m+1}^q\} cluster monomials (supported on a given cluster)
Goal : define and construct 'canonical basis' B in A(b,c) for bc\leq 4
By "Leurant phenomenen" each element in A(b,c) can be uniquely expressed as Laurent polynomial in y_m and y_{m+1} for
B.F.Zelevinsky 's result :
A(b,c) =\cup_{m\in\mathbb{Z}\mathbb{Z}[y_n^{\pm 1,\y_{m+1}^{\pm 1]=\cup_{m=0}^{\alpha}\mathbb{Z}[y_n^{\pm 1,\y_{m+1}^{\pm 1]
standard monomial basis : the set \{y_0^{a_0}y_1^{a_1}y_2^{a_2}y_3^{a_3} : a_{m}\in\mathbb{Z}_{\geq 0}, a_0a_2=a_1a_3=0\} is a \mathbb{Z}-basis of A(b,c).
Here support of any such monomial is \{y_0,y_1\},\{y_1,y_2\},\{y_2,y_3\},\{y_0,y_3\}.

A(b,c) is finitely generated,
A(b,c)=\mathbb{y_0,y_1,y_2,y_3]/<y_0y_2-y_1^b-1,y_1y_3-y_2^c-1>

Finite type classification :

A(b,c) related to root systems of Cartan matrix

rank 2 case

\( \begin{bmatrix} 2 & -b \\ -c & 2 \end{bmatrix}\)

Say A(b,c) is of finite/affine/indefinite type if bc\leq 3/bc=4/bc>4

when bc\leq 3

y_m=y_n if and only if m\equiv n mod (h+2) where h is coxeter number

bc=1,

 

 

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