"Rank 2 cluster algebra"의 두 판 사이의 차이

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<h5>introduction</h5>
  
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* cluster algebra defined by a 2x2 matrix
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<h5>cluster variables and exchange relations</h5>
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Fix two positive integers b and c.
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Let y_1 and y_2 be variable. Define a sequence {y_n}.
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<math>y_{m-1}y_{m+1}=y_m^b+1</math> if m odd
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<math>y_{m-1}y_{m+1}=y_m^c+1</math> if m even
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We call this ''''exchange relation''''
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y_m's are called ''''cluster variable''''
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<math>\{y_i,y_{i+1}\}</math> "cluster"
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Note : we can use the exchange relation any y_m in terms of arbitrary cluster {y_i,y_{i+1}} (rational expression)
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<h5>example 1</h5>
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Put b=c=1
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y_1,y_2,<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_1=1</math>
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<h5>example 2</h5>
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Put b=1, c=3
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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>
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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_2^3+1}{y_1y_2}</math>
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<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 polynomials 
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y_6=\frac{(y_1+1)^2+y_2^3}{y_1y_2^2}
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y_7=\frac{(y_1+1)^3+y_2^3}{y_1y_2^3}
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y_8=\frac{y_1+1}{y_2}
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y_9=y_1
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y_{10}=y_2
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<h5>matrix formulation</h5>
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<math>B=\begin{bmatrix} 0 & -b\\ c  &\,0 \end{bmatrix}</math>
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<math>\mu_{1}(B)=\begin{bmatrix} 0 & b\\ -c  &\,0 \end{bmatrix}</math>
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<math>\mu_{2}(B)=\begin{bmatrix} 0 & b\\ -c  &\,0 \end{bmatrix}</math>
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<h5>observations</h5>
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(FZ) For any b,c, y_m is a Laurent polynomial.
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Positivity conjecture: coefficients of these Laurent polynomials are positive (numerator and denomonator always have )
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In this example, 
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<math>bc\leq 3</math> iff the recurrence is periodic
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<h5>history</h5>
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
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<h5>related items</h5>
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">encyclopedia</h5>
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* http://en.wikipedia.org/wiki/
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* http://www.scholarpedia.org/
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* http://www.proofwiki.org/wiki/
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* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])
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<h5>books</h5>
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* [[2011년 books and articles]]
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* http://library.nu/search?q=
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* http://library.nu/search?q=
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<h5>expositions</h5>
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles</h5>
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* http://www.ams.org/mathscinet
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* http://www.zentralblatt-math.org/zmath/en/
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* http://arxiv.org/
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* http://www.pdf-search.org/
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* http://pythagoras0.springnote.com/
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* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html]
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* http://dx.doi.org/
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<h5>question and answers(Math Overflow)</h5>
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* http://mathoverflow.net/search?q=
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* http://mathoverflow.net/search?q=
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<h5>blogs</h5>
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*  구글 블로그 검색<br>
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**  http://blogsearch.google.com/blogsearch?q=<br>
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** http://blogsearch.google.com/blogsearch?q=
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* http://ncatlab.org/nlab/show/HomePage
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<h5>experts on the field</h5>
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* http://arxiv.org/
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<h5>links</h5>
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* [http://detexify.kirelabs.org/classify.html Detexify2 - LaTeX symbol classifier]
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* [http://pythagoras0.springnote.com/pages/1947378 수식표현 안내]
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
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* http://functions.wolfram.com/

2011년 1월 19일 (수) 11:57 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 \(y_5=\frac{y_4^3+1}{y_3}= \frac{(y_1+1)^3+y_2^3(y_2^3+3y_1+2)}{y_1^2y_2^3}\) 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

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

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

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

 

 

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

 

 

related items

 

 

encyclopedia

 

 

books

 

 

 

expositions

 

 

 

articles

 

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links
원본 주소 "https://wiki.mathnt.net/index.php?title=Rank_2_cluster_algebra&oldid=36832"