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Elliptic functions, periodic lattice, equivalence classes

  1. Mar 7, 2017 #1
    1. The problem statement, all variables and given/known data

    ##\Omega = {nw_1+mw_2| m,n \in Z} ##

    ##z_1 ~ z_2 ## is defined by if ##z_1-z_2 \in \Omega ##

    My notes say ##z + \Omega## are the cosets/ equivalence classes , denoted by ##[z] = {z+mw_1+nw_2} ##

    2. Relevant equations

    above

    3. The attempt at a solution

    So equivalance classe form a partition, i.e. two elements that are equivalent are within the same equivalence class,

    But if I consider ##\Omega ## with basis ##w_1 = i ## ##w_2 =1 ##

    ##z_1 = 1/2 + i/2 ##, ##z_2 = 1/2-i/2##
    Then ##z_1 - z_2 = i \in \Omega ## , ##(m=1, n=0)##

    And so ##z_1 ~ z_2 ## so these two are in the same equivalent class right?

    However my notes say that ##[z_1]## and ##[z_2]## are each equivalence classes, or is this definition not ##\all z \in C##, what is an efficient way to look, from the basis, which ##z \in C## are equivalent so how many equivalence classes there will be?

    thanks in advance

    - also a notation question:
    Say ##~## is defined by the difference between ##x \in Z## being ##2##, then equivalence classes are odd and even numbers, and we use the notation ##[1],[7],[3]..## represent the same element for ##Z/\~##

    Can you equally use the notation ##[1]=[7]## (mod 2) ?

    So above I can either say:
    - ##[z_1]=[z_2] ## (mod ##\Omega##)
    OR
    - ##[z_1],[z_2] ## represent the same element for ##C/\Omega##

    and these mean the same thing?
     
  2. jcsd
  3. Mar 8, 2017 #2

    pasmith

    User Avatar
    Homework Helper


    I think you are looking for "two equivalence classes are either equal or disjoint".

    The latex for ~ is \sim.

    [z] is the equivalence class of z. [w] is the equivalence class of w. If z ~ w then [z] and [w] are different names for the same set. You should think of [itex][\cdot][/itex] as a function from [itex]\mathbb{C}[/itex] to the set of equivalence classes.

    Frequently we want to define a function from the set of equivalence classes to some other set where the image of [z] is given by a formula involving z; here we must check that the formula gives the same result for every [itex]w \in [z][/itex].

    For example:

    Since [itex]\{w_1,w_2\}[/itex] is a basis for [itex]\mathbb{C}[/itex], for each [itex]z \in \mathbb{C}[/itex] there exists a unique [itex](n,m) \in \mathbb{Z}^2[/itex] and a unique [itex](u,v) \in [0,1)^2[/itex] such that [itex]z = (n + u)w_1 + (m + v)w_2[/itex]. Then [itex]z_1 \sim z_2[/itex] if and only if [itex]u(z_1) = u(z_2)[/itex] and [itex]v(z_1) = v(z_2)[/itex]. Hence we may define a bijection [itex]\phi[/itex] from the set [itex]E[/itex] of equivalence classes to [itex][0,1)^2[/itex] by [itex]\phi([z]) = (u(z),v(z))[/itex].
     
  4. Mar 8, 2017 #2

    pasmith

    User Avatar
    Homework Helper


    I think you are looking for "two equivalence classes are either equal or disjoint".

    The latex for ~ is \sim.

    [z] is the equivalence class of z. [w] is the equivalence class of w. If z ~ w then [z] and [w] are different names for the same set. You should think of [itex][\cdot][/itex] as a function from [itex]\mathbb{C}[/itex] to the set of equivalence classes.

    Frequently we want to define a function from the set of equivalence classes to some other set where the image of [z] is given by a formula involving z; here we must check that the formula gives the same result for every [itex]w \in [z][/itex].

    For example:

    Since [itex]\{w_1,w_2\}[/itex] is a basis for [itex]\mathbb{C}[/itex], for each [itex]z \in \mathbb{C}[/itex] there exists a unique [itex](n,m) \in \mathbb{Z}^2[/itex] and a unique [itex](u,v) \in [0,1)^2[/itex] such that [itex]z = (n + u)w_1 + (m + v)w_2[/itex]. Then [itex]z_1 \sim z_2[/itex] if and only if [itex]u(z_1) = u(z_2)[/itex] and [itex]v(z_1) = v(z_2)[/itex]. Hence we may define a bijection [itex]\phi[/itex] from the set [itex]E[/itex] of equivalence classes to [itex][0,1)^2[/itex] by [itex]\phi([z]) = (u(z),v(z))[/itex].
     
  5. Mar 21, 2017 #3
    Okay so equivalence classes may be either equal or disjoint, fine.
    Now I think my confusion is coming from this... cosets form a partition of the set right?
    To be a partition they must be disjoint,##z_1## and ##z_2## are not, they are equal? But my notes say ##[z] ## are cosets?
    Or cosets do not form a partition?
    Or when my notes say ##[z] ## are cosets it is reffering to non-equivalent ##z_i ## s ?

    thanks
     
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