Definition of cluster algebras.
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I've been doing some reading on cluster algebra from surfaces and in Schiffler I encountered this Definition:
Let $mathcal{X}$ be the set of all cluster variables obtained by mutation from $(textbf{x}, textbf{y}, Q)$.
The cluster algebra $mathcal{A} = mathcal{A}(textbf{x}, textbf{y}, Q)$ is the $mathbb{ZP}$-subalgebra of $mathcal{F}$ generated by
$mathcal{X}$.
By definition, the elements of $mathcal{A}$ are polynomials in $mathcal{X}$ with coefficients
in $mathbb{ZP}$, so $mathcal{A} ⊂ mathbb{ZP}[mathcal{X} ]$. On the other hand, $mathcal{A} subset mathcal{F}$, so the elements of $mathcal{A}$ are
also rational functions in $x_{1},dots , x_{n}$ with coefficients in $mathbb{QP}$.,
where $mathbb{P}$ is a tropical semifield, $mathbb{ZP}$ is its group ring and $mathcal{F}=mathbb{QP}(x_{1},dots , x_{n})$ is the field of rational functions in $n$ variables and coefficients in $mathbb{QP}$.
I'm not sure about the part when we say that the elements of $mathcal{A}$ are polynomials. For instance, looking at the elements of the cluster algebra from this example , how would one see them as polynomials? Is this just about introducing some new abbreviations for some of the cluster variables (if so, how do we pick the 'right' cluster variables?)? So that we can write every (other) cluster variable as a polynomial expression in those new abbreviations (can we always do that)? Any hints or explanations are much appreciated.
abstract-algebra
asked Dec 12 '18 at 16:48
amator2357amator2357
428
$endgroup$
add a comment |
$begingroup$
I've been doing some reading on cluster algebra from surfaces and in Schiffler I encountered this Definition:
Let $mathcal{X}$ be the set of all cluster variables obtained by mutation from $(textbf{x}, textbf{y}, Q)$.
The cluster algebra $mathcal{A} = mathcal{A}(textbf{x}, textbf{y}, Q)$ is the $mathbb{ZP}$-subalgebra of $mathcal{F}$ generated by
$mathcal{X}$.
By definition, the elements of $mathcal{A}$ are polynomials in $mathcal{X}$ with coefficients
in $mathbb{ZP}$, so $mathcal{A} ⊂ mathbb{ZP}[mathcal{X} ]$. On the other hand, $mathcal{A} subset mathcal{F}$, so the elements of $mathcal{A}$ are
also rational functions in $x_{1},dots , x_{n}$ with coefficients in $mathbb{QP}$.,
where $mathbb{P}$ is a tropical semifield, $mathbb{ZP}$ is its group ring and $mathcal{F}=mathbb{QP}(x_{1},dots , x_{n})$ is the field of rational functions in $n$ variables and coefficients in $mathbb{QP}$.
I'm not sure about the part when we say that the elements of $mathcal{A}$ are polynomials. For instance, looking at the elements of the cluster algebra from this example , how would one see them as polynomials? Is this just about introducing some new abbreviations for some of the cluster variables (if so, how do we pick the 'right' cluster variables?)? So that we can write every (other) cluster variable as a polynomial expression in those new abbreviations (can we always do that)? Any hints or explanations are much appreciated.
abstract-algebra
asked Dec 12 '18 at 16:48
amator2357amator2357
428
$endgroup$
1
$begingroup$
Any cluster variable is a polynomial in the set $mathcal{X}$ of cluster variables. Or perhaps I misunderstand your question?
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:09
$begingroup$
Is that somehow obvious, the fact that that any cluster variable is a polynomial in the set X of cluster variables?
$endgroup$
– amator2357
Dec 12 '18 at 17:31
$begingroup$
There may be a misunderstanding of terminology. In any ring $A$, any element $a$ is a polynomial in $a$: it is the polynomial $P(X)=X$ evaluated in $a$. More generally, if $mathcal{X}$ is any subset of $A$, then any element of $mathcal{X}$ is a polynomial in some elements of $mathcal{X}$.
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:58
$begingroup$
Of course, I see now. Thank you for taking your time to comment, I appreciate it.
$endgroup$
– amator2357
Dec 12 '18 at 18:51
add a comment |
$begingroup$
I've been doing some reading on cluster algebra from surfaces and in Schiffler I encountered this Definition:
Let $mathcal{X}$ be the set of all cluster variables obtained by mutation from $(textbf{x}, textbf{y}, Q)$.
The cluster algebra $mathcal{A} = mathcal{A}(textbf{x}, textbf{y}, Q)$ is the $mathbb{ZP}$-subalgebra of $mathcal{F}$ generated by
$mathcal{X}$.
By definition, the elements of $mathcal{A}$ are polynomials in $mathcal{X}$ with coefficients
in $mathbb{ZP}$, so $mathcal{A} ⊂ mathbb{ZP}[mathcal{X} ]$. On the other hand, $mathcal{A} subset mathcal{F}$, so the elements of $mathcal{A}$ are
also rational functions in $x_{1},dots , x_{n}$ with coefficients in $mathbb{QP}$.,
where $mathbb{P}$ is a tropical semifield, $mathbb{ZP}$ is its group ring and $mathcal{F}=mathbb{QP}(x_{1},dots , x_{n})$ is the field of rational functions in $n$ variables and coefficients in $mathbb{QP}$.
I'm not sure about the part when we say that the elements of $mathcal{A}$ are polynomials. For instance, looking at the elements of the cluster algebra from this example , how would one see them as polynomials? Is this just about introducing some new abbreviations for some of the cluster variables (if so, how do we pick the 'right' cluster variables?)? So that we can write every (other) cluster variable as a polynomial expression in those new abbreviations (can we always do that)? Any hints or explanations are much appreciated.
abstract-algebra
asked Dec 12 '18 at 16:48
amator2357amator2357
428
$endgroup$
I've been doing some reading on cluster algebra from surfaces and in Schiffler I encountered this Definition:
Let $mathcal{X}$ be the set of all cluster variables obtained by mutation from $(textbf{x}, textbf{y}, Q)$.
The cluster algebra $mathcal{A} = mathcal{A}(textbf{x}, textbf{y}, Q)$ is the $mathbb{ZP}$-subalgebra of $mathcal{F}$ generated by
$mathcal{X}$.
By definition, the elements of $mathcal{A}$ are polynomials in $mathcal{X}$ with coefficients
in $mathbb{ZP}$, so $mathcal{A} ⊂ mathbb{ZP}[mathcal{X} ]$. On the other hand, $mathcal{A} subset mathcal{F}$, so the elements of $mathcal{A}$ are
also rational functions in $x_{1},dots , x_{n}$ with coefficients in $mathbb{QP}$.,
where $mathbb{P}$ is a tropical semifield, $mathbb{ZP}$ is its group ring and $mathcal{F}=mathbb{QP}(x_{1},dots , x_{n})$ is the field of rational functions in $n$ variables and coefficients in $mathbb{QP}$.
I'm not sure about the part when we say that the elements of $mathcal{A}$ are polynomials. For instance, looking at the elements of the cluster algebra from this example , how would one see them as polynomials? Is this just about introducing some new abbreviations for some of the cluster variables (if so, how do we pick the 'right' cluster variables?)? So that we can write every (other) cluster variable as a polynomial expression in those new abbreviations (can we always do that)? Any hints or explanations are much appreciated.
abstract-algebra
abstract-algebra
asked Dec 12 '18 at 16:48
amator2357amator2357
428
asked Dec 12 '18 at 16:48
amator2357amator2357
428
asked Dec 12 '18 at 16:48
amator2357amator2357
428
asked Dec 12 '18 at 16:48
amator2357amator2357
428
asked Dec 12 '18 at 16:48
amator2357amator2357
428
428
1
$begingroup$
Any cluster variable is a polynomial in the set $mathcal{X}$ of cluster variables. Or perhaps I misunderstand your question?
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:09
$begingroup$
Is that somehow obvious, the fact that that any cluster variable is a polynomial in the set X of cluster variables?
$endgroup$
– amator2357
Dec 12 '18 at 17:31
$begingroup$
There may be a misunderstanding of terminology. In any ring $A$, any element $a$ is a polynomial in $a$: it is the polynomial $P(X)=X$ evaluated in $a$. More generally, if $mathcal{X}$ is any subset of $A$, then any element of $mathcal{X}$ is a polynomial in some elements of $mathcal{X}$.
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:58
$begingroup$
Of course, I see now. Thank you for taking your time to comment, I appreciate it.
$endgroup$
– amator2357
Dec 12 '18 at 18:51
add a comment |
1
$begingroup$
Any cluster variable is a polynomial in the set $mathcal{X}$ of cluster variables. Or perhaps I misunderstand your question?
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:09
$begingroup$
Is that somehow obvious, the fact that that any cluster variable is a polynomial in the set X of cluster variables?
$endgroup$
– amator2357
Dec 12 '18 at 17:31
$begingroup$
There may be a misunderstanding of terminology. In any ring $A$, any element $a$ is a polynomial in $a$: it is the polynomial $P(X)=X$ evaluated in $a$. More generally, if $mathcal{X}$ is any subset of $A$, then any element of $mathcal{X}$ is a polynomial in some elements of $mathcal{X}$.
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:58
$begingroup$
Of course, I see now. Thank you for taking your time to comment, I appreciate it.
$endgroup$
– amator2357
Dec 12 '18 at 18:51
1
1
$begingroup$
Any cluster variable is a polynomial in the set $mathcal{X}$ of cluster variables. Or perhaps I misunderstand your question?
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:09
$begingroup$
Any cluster variable is a polynomial in the set $mathcal{X}$ of cluster variables. Or perhaps I misunderstand your question?
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:09
$begingroup$
Is that somehow obvious, the fact that that any cluster variable is a polynomial in the set X of cluster variables?
$endgroup$
– amator2357
Dec 12 '18 at 17:31
$begingroup$
Is that somehow obvious, the fact that that any cluster variable is a polynomial in the set X of cluster variables?
$endgroup$
– amator2357
Dec 12 '18 at 17:31
$begingroup$
There may be a misunderstanding of terminology. In any ring $A$, any element $a$ is a polynomial in $a$: it is the polynomial $P(X)=X$ evaluated in $a$. More generally, if $mathcal{X}$ is any subset of $A$, then any element of $mathcal{X}$ is a polynomial in some elements of $mathcal{X}$.
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:58
$begingroup$
There may be a misunderstanding of terminology. In any ring $A$, any element $a$ is a polynomial in $a$: it is the polynomial $P(X)=X$ evaluated in $a$. More generally, if $mathcal{X}$ is any subset of $A$, then any element of $mathcal{X}$ is a polynomial in some elements of $mathcal{X}$.
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:58
$begingroup$
Of course, I see now. Thank you for taking your time to comment, I appreciate it.
$endgroup$
– amator2357
Dec 12 '18 at 18:51
$begingroup$
Of course, I see now. Thank you for taking your time to comment, I appreciate it.
$endgroup$
– amator2357
Dec 12 '18 at 18:51
add a comment |
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1
$begingroup$
Any cluster variable is a polynomial in the set $mathcal{X}$ of cluster variables. Or perhaps I misunderstand your question?
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:09
$begingroup$
Is that somehow obvious, the fact that that any cluster variable is a polynomial in the set X of cluster variables?
$endgroup$
– amator2357
Dec 12 '18 at 17:31
$begingroup$
There may be a misunderstanding of terminology. In any ring $A$, any element $a$ is a polynomial in $a$: it is the polynomial $P(X)=X$ evaluated in $a$. More generally, if $mathcal{X}$ is any subset of $A$, then any element of $mathcal{X}$ is a polynomial in some elements of $mathcal{X}$.
$endgroup$
– Pierre-Guy Plamondon
Dec 12 '18 at 17:58
$begingroup$
Of course, I see now. Thank you for taking your time to comment, I appreciate it.
$endgroup$
– amator2357
Dec 12 '18 at 18:51