# Holomorphic map

1. May 11, 2007

### Ant farm

grrr, so annoyed, can't see the wood from the trees on this problem!!
I'm trying to get a holomorphic map from C/(Z+iZ) -> C/(Z+iZ) where C=complex numbers and Z=integers.
Does this function have to be doubly periodic?
Are doubly periodic functions the same as elliptic functions?
Are all elliptic functions meromorpic but not holomorphic in which case I'm obviously not looking for an elliptic function!!

2. May 11, 2007

### matt grime

what is wrong with the identity function?

3. May 12, 2007

### Ant farm

It Can't be the Identity, must be something else with f(0)=0
I've been thinking of mayb a piecewise function involving the Weirestrass P function... or a rotation... I just can't see where I'm working in my head!

4. May 12, 2007

### matt grime

What's wrong with z-->2z? It just needs to be some function that maps Z+iZ into itself.

5. May 12, 2007

### Ant farm

Ah, ok, sorry, I was looking for something that was doubly periodic, but since the function is going form Z+iZ into itself, that condition will automatically be satisfied?!! And that is Holomorphic too.
It just seems too simple!

6. May 12, 2007

### matt grime

If you want something 'interesting' then you will need elliptic functions, or doubly periodic ones, for sure - but there are simple ones too.

7. May 12, 2007

### mathwonk

i think these are actually all group homomorphisms, so think in those terms, i.e. any group map C-->C that takes Z+iZ to itself.

i fact the map C-->C is complex linear I believe.

An elliptic function is a holomorphic map from C/Z+iZ-->P where P is the projective line C U {pt}.

they form a field. the holomorphic maps from the torus to itself form a ring, and of course the units in that ring, the automorphisms of the torus, form a group. this group may be always as simple as Z/2Z/, Z/4Z, or z/6Z.

Last edited: May 12, 2007
8. May 12, 2007

### mathwonk

a holomorphic map f:C/Z+iZ-->C/Z+iZ, defines a holomorphic map g(z) =
f(z)-f(z+i) with values in the discrete set Z+iZ, hence g is constant.

then g' is zero, so f is periodic with period i, and simiklarly 1.

then f' is bounded hence constant, so f is linear and since f(0) = 0, f(z) = ax for some a in C.