"Rank 2 cluster algebra"의 두 판 사이의 차이
| 33번째 줄: | 33번째 줄: | ||
Put b=c=1 | Put b=c=1 | ||
| − | y_1,y_2 | + | y_1,y_2 |
| + | |||
| + | <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> | <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> | ||
| 49번째 줄: | 51번째 줄: | ||
Put b=1, c=3 | Put b=1, c=3 | ||
| − | y_1,y_2 | + | y_1,y_2 |
| + | |||
| + | <math>y_3y_1=y_2^3+1</math>. so <math>y_3=\frac{y_2^3+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_2^3+1}{y_1y_2}</math> | <math>y_2y_4=y_3+1 </math>implies <math>y_4=\frac{y_3+1}{y_2}=\frac{y_1+y_2^3+1}{y_1y_2}</math> | ||
| − | <math>y_3y_5=y_4^3+1</math> implies <math>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}</math> we are getting Laurent | + | <math>y_3y_5=y_4^3+1</math> implies <math>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}</math>[http://www.wolframalpha.com/input/?i=%28%28x%2By%5E3%2B1%29%5E3%2B%28xy%29%5E3%29/%28x%5E2y%5E3%28y%5E3%2B1%29%29 ][http://www.wolframalpha.com/input/?i=%28%28x%2By%5E3%2B1%29%5E3%2B%28xy%29%5E3%29/%28x%5E2y%5E3%28y%5E3%2B1%29%29 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} | + | <math>y_6=\frac{(y_1+1)^2+y_2^3}{y_1y_2^2}</math> |
| − | y_7=\frac{(y_1+1)^3+y_2^3}{y_1y_2^3} | + | <math>y_7=\frac{(y_1+1)^3+y_2^3}{y_1y_2^3}</math> |
| − | + | ||
y_9=y_1 | y_9=y_1 | ||
2011년 1월 19일 (수) 12:14 판
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_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 \(y_1=1\)
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_9=y_1
y_{10}=y_2
matrix formulation
\(B=\begin{bmatrix} 0 & -b\\ c &\,0 \end{bmatrix}\)
\(\mu_{1}(B)=\begin{bmatrix} 0 & b\\ -c &\,0 \end{bmatrix}\)
\(\mu_{2}(B)=\begin{bmatrix} 0 & b\\ -c &\,0 \end{bmatrix}\)
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
history
encyclopedia
- http://en.wikipedia.org/wiki/
- http://www.scholarpedia.org/
- http://www.proofwiki.org/wiki/
- Princeton companion to mathematics(Companion_to_Mathematics.pdf)
books
expositions
articles
- http://www.ams.org/mathscinet
- http://www.zentralblatt-math.org/zmath/en/
- http://arxiv.org/
- http://www.pdf-search.org/
- http://pythagoras0.springnote.com/
- http://math.berkeley.edu/~reb/papers/index.html
- http://dx.doi.org/
question and answers(Math Overflow)
blogs
- 구글 블로그 검색
- http://ncatlab.org/nlab/show/HomePage
experts on the field