What is the relationship between holomorphic maps and elliptic functions?

[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!
Please Help

 

[matt grime]

what is wrong with the identity function?

 

[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!

 

[matt grime]

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

 

[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!

 

[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.

 

[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.

 

[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.