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1

$begingroup$

Suppose the matrices $alpha, beta$ with the given commutating relations:

$$
alpha^2=a I\
beta^2=b I\
alphabeta=c I\
betaalpha=d I
$$

where $a,b,c,d $ are elements of $mathbbR$ and where $I$ is the identity matrix.

How can I find specific examples of $alpha, beta$, or even solve for the general case?

matrices lie-groups

share|cite|improve this question

asked Mar 17 at 23:08

Alexandre H. TremblayAlexandre H. Tremblay

364213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $frac1calphacdotbeta=IRightarrow frac1calpha=beta^-1Rightarrow betacdotfrac1calpha=IRightarrow c=d.$
  $endgroup$
  – Minz
  Mar 18 at 1:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @minz I am actually fine with $c=d$, but I cannot afford $a=b$. Can we say that $a$ is not necessarily equal to $b$?
  $endgroup$
  – Alexandre H. Tremblay
  Mar 18 at 1:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  surely. Consider $alpha =I, beta=2I$ then $alpha^2 =I, beta^2=4I, alpha beta= betaalpha =2I $
  $endgroup$
  – Minz
  Mar 18 at 1:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @minz, my apologies, what I am looking for is actually a matrice where a, b and c are independent of each other. c and d can be equal this is fine.
  $endgroup$
  – Alexandre H. Tremblay
  Mar 18 at 2:07
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose the matrices $alpha, beta$ with the given commutating relations:

$$
alpha^2=a I\
beta^2=b I\
alphabeta=c I\
betaalpha=d I
$$

where $a,b,c,d $ are elements of $mathbbR$ and where $I$ is the identity matrix.

How can I find specific examples of $alpha, beta$, or even solve for the general case?

matrices lie-groups

share|cite|improve this question

asked Mar 17 at 23:08

Alexandre H. TremblayAlexandre H. Tremblay

364213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $frac1calphacdotbeta=IRightarrow frac1calpha=beta^-1Rightarrow betacdotfrac1calpha=IRightarrow c=d.$
  $endgroup$
  – Minz
  Mar 18 at 1:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @minz I am actually fine with $c=d$, but I cannot afford $a=b$. Can we say that $a$ is not necessarily equal to $b$?
  $endgroup$
  – Alexandre H. Tremblay
  Mar 18 at 1:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  surely. Consider $alpha =I, beta=2I$ then $alpha^2 =I, beta^2=4I, alpha beta= betaalpha =2I $
  $endgroup$
  – Minz
  Mar 18 at 1:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @minz, my apologies, what I am looking for is actually a matrice where a, b and c are independent of each other. c and d can be equal this is fine.
  $endgroup$
  – Alexandre H. Tremblay
  Mar 18 at 2:07
  

  
  
  
  

add a comment | 

$begingroup$

Suppose the matrices $alpha, beta$ with the given commutating relations:

$$
alpha^2=a I\
beta^2=b I\
alphabeta=c I\
betaalpha=d I
$$

where $a,b,c,d $ are elements of $mathbbR$ and where $I$ is the identity matrix.

How can I find specific examples of $alpha, beta$, or even solve for the general case?

matrices lie-groups

share|cite|improve this question

asked Mar 17 at 23:08

Alexandre H. TremblayAlexandre H. Tremblay

364213

$endgroup$

share|cite|improve this question

asked Mar 17 at 23:08

Alexandre H. TremblayAlexandre H. Tremblay

364213

share|cite|improve this question

asked Mar 17 at 23:08

Alexandre H. TremblayAlexandre H. Tremblay

364213

asked Mar 17 at 23:08

Alexandre H. TremblayAlexandre H. Tremblay

364213

asked Mar 17 at 23:08

Alexandre H. TremblayAlexandre H. Tremblay

364213