## Is a function which is finitely multiple-valued in each variable separately, also finitely multiple-valued in all its variables jointly?

It is well known that under suitable conditions, a function which is:

1. a polynomial in each variable separately is a polynomial in all its variables jointly.

2. a rational function in each variable separately is a rational function.

3. a holomorphic function in each variable separately is holomorphic in all its variables.

A complete analytic function can be single-valued or multiple-valued according as it does not have, or does have, branch points. The algebraic functions are examples of the latter.

Here is my question: is a complete analytic function, which is finitely multiple-valued in each variable separately, also finitely multiple-valued jointly?

-
I don't understand: if for each $x$, the function $f(x,-)$ is finitely-valued, then equivalently for each $(x,y)$, $f(x,y)$ takes only finitely many values, but this is the statement you want, right? – Theo Johnson-Freyd Feb 11 2010 at 17:08
Just to check my understanding, by finitely multivalued, you mean that the function takes only a finite number of values at each point? Then in one sense the answer is trivially “yes”. In another sense, a “no” answer would mean that, even though $(f(x_0,y_0))$ could take an infinite number of values, if you were to change only one variable in a continuous manner (as in moving around a branch point) and returning to $(x_0,y_0)$, you can reach only a finite number of the values of $(f(x_0,y_0))$, whereas you can reach an infinite number by moving both coordinates together? – Harald Hanche-Olsen Feb 11 2010 at 17:11
no....it is possible that for each different y the number of values of f changes...the question asks if there is a uniform upper bound for the number of values of f regardless of the y...and vice versa – Mark B Villarino Feb 11 2010 at 17:13
I do not see how you can reach an infinite number of values for $f(x_0,y_0)$ by moving both coordinates together, if you ALWAYS obtain only a finite number of values by varying only one of them – Mark B Villarino Feb 11 2010 at 17:52
@Mark: I am nut sure I can see it either, or I would have provided an Answer to the question! But equally, I can't see how to rule it out. To state it more clearly (not really knowing how analytic continuation plays out in several variables, hoping it is like a single variable): There would be a connected surface $z=f(x,y)$, and a path in the $(x,y)$-plane lifts to this surface. Make a closed path, and it may lift to a non-closed path. Restricting your paths to keep $x$ or $y$ to be fixed may limit the end point possibilities from infinite to finite. Or why not? – Harald Hanche-Olsen Feb 11 2010 at 20:23

## 2 Answers

Engineer a function that has n>=2 values exactly in the set $S_n \backslash S_{n+1}$ where $S_n$={(x,y) | x>n, y>n, (x-n)*(y-n)>1}, and 1 value elsewehere. That should work as a counterexample, as far as $C^{\infty}$ functions go - analytic, not sure

-

Isn't the function x+y single-valued in each variable separately, but takes any value infinitely often? Or am I misunderstanding your definition?

-
A multi-valued analytic function is a function like the logarithm which can take multiple values at a particular point depending on branching. x+y takes a single value at any particular point. – Qiaochu Yuan Feb 11 2010 at 16:56
x+y is single-valued as a function of (x,y)...my question asks if for each $(x,y_0)$, $f(x,y_0)\equiv g(x)$ is finitely multiple-valued as a function of $x$, and $f(x_0,y)\equiv h(y)$ is a finitely multiple-valued function of $y$, then is $f(x,y)$ a finitely multiple-valued function of the pair $(x,y)$? If, for example, one writes "algebraic" in place of "finitely multiple-valued", then under certain conditions the joint function is algebraic. – Mark B Villarino Feb 11 2010 at 17:01