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Riemann surface equation RB ( f(z) ) = f( p(z) )
#1
Let f(z) be an analytic function with a riemann surface that has branches.

Let RB be shorthand for going from one specific Riemann Branch to another specific one.

Then the equation that holds locally or globally :

RB ( f(z) ) = f( p(z) )

with p(z) a degree 1 or 2 polynomial.

fascinates me.

For instance f(2z+1) is another branch of f(z).
That is fascinating.

But how to solve such a thing ?

in particular

RB ( f(z) ) = f(z + C)

This relates to tetration and dynamics.

Not sure if its in the books.

Btw do not confuse with a simple invariant :

p (f(z)) is another branch of f(z).

which is different !!


regards

tommy1729
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#2
It is tempting to do the following

by definition :
solve f(z) = g(z)
solve p(z) = q(z)

solve RB ( f(z) ) = solve f( p(z) )

=> RB solve f(z) = solve f( p(z) )

=> RB g(z) = q( g(z) )

and then work with that q invariant.

Not sure if that is both formal and usefull ...

Also there are branch issues perhaps ...

g(z) = q ( g(z) )

we might continue like ...
so g(z) is the superfunction of T(z) where q(T(z)) is another branch of T(z). In other words T(p(z)) = T(z).

And then we get the weird fact that we have a similar equation.

SO if Q(z) is a solution then it appears superf(Q^[-1](z)) is also a solution !
When done twice : Q(z) -> superf(Abel(Q^[-1](z))) or when done C times : superf(Abel^[C](Q^[-1](z))).

This © reminds me of the " half superfunction " We have talked about before !!

Hmm.

regards

tommy1729
Reply


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