# Elliptic functions, periodic lattice, equivalence classes

1. Mar 7, 2017

### binbagsss

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?

- 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. Mar 8, 2017

### pasmith

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 $[\cdot]$ as a function from $\mathbb{C}$ 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 $w \in [z]$.

For example:

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

3. Mar 8, 2017

### pasmith

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 $[\cdot]$ as a function from $\mathbb{C}$ 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 $w \in [z]$.

For example:

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

4. Mar 21, 2017

### binbagsss

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