Equivalence relation on C

  • Thread starter Zorba
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  • #1
Zorba
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I was checking that the following is an equivalence relation on [tex]\mathbb{C}[/tex]

[tex]xRy[/tex] iff [tex]x\bar{y}=\bar{x}y[/tex]

It is an equivalence relation and so by letting x=a+bi and y=c+di, then it is equivalent to a/b=c/d so I was viewing it as partitioning points in [tex]\mathbb{C}[/tex] by drawing lines through the origin and equivalent points lie on the line, but rearranging a/b=c/d gives the 2x2 determinant formula (zero case) so I was wondering whether I'm missing something here, is there some other way to think about this, some other possible insight? It seems rather curious that it comes out like the determinant...
 

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  • #2
Tinyboss
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Is this meant to be on the nonzero complex numbers? Else 0Rz for every complex z, and transitivity fails. Assuming that, then you can view the quotient as RP^2, if your equivalence is correct (I didn't verify).
 
  • #3
Zorba
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Yea, you're right I noticed that myself later on. I'm not familiar with "RP2" but I think that I see the (what seems obvious now) connection, if you just take two lines then if both these lines are the same (like two points satisfying the relation) then the system is over determined and the matrix of the lines isn't invertible etc. hence determinant is zero.
 

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