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I have been reading about Puiseux series in the context of the Newton–Puiseux algorithm for resolution of singularities of algebraic curves in $\mathbb{C}^2$. Given a curve $f(x,y)=0$ with $f$ a convergent power series and such that the curve has a singularity at the origin, the algorithm produces convergent Puiseux series $\varphi_1, \dotsc, \varphi_m$ verifying $f(x,y) = (y-\varphi_1(x))\dotsm(y-\varphi_m(x))$.

All of the books I have consulted start by constructing the Puiseux series as a formal power series and then invoking the Implicit Function Theorem to show that there exist analytic solutions to the problem, and therefore the constructed series must be convergent (in a small enough neighbourhood of the origin). Then they continue by studying the properties of the singularity using $\varphi_1, \dotsc, \varphi_m$ as parametrizations of the curve.

Here is where I have a problem. Say for instance that one of the resulting Puiseux series is $\varphi_1(x) = x^{3/2} + x^{7/4}$. This is not a well-defined complex function, is it? For $x=1$ we could have $x^{3/2}$ equal to $1$ or $-1$, and $x^{7/4}$ equal to $\pm1$ or $\pm i$, so there are eight different determinations of this particular Puiseux series. In general, if the Puiseux series has infinitely many terms, there would be infinitely many determinations.

As far as I understand, only one of these determinations of $\varphi_1$ is the actual function you want to consider, which is the function that the Implicit Function Theorem gives you — but because of the nature of the proof, you can't really know which one it is! It is not clear to me how do these books account for this problem, but their treatment of the series is as if they were well-defined functions (see for instance Brieskorn–Knörrer).

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One standard way to bring actual functions in the picture is the following formulation of the existence + convergence results of Puiseux roots: "Given an irreducible power series $f \in \mathbb{C}[[x, y]]$ which is a Weirstrass polynomial in $y$ of degree $d$ (i.e. $f = y^d + \sum_{i=1}^d f_i(x)y^{d-i}$ with $f_i(x) \in x\mathbb{C}[[x]]$), there is $\phi(t) \in \mathbb{C}[[t]]$ such that $$f(t^d, y) = \prod_{i = 1}^d (y - \phi(\zeta^it))$$ where $\zeta$ is a $d$-th primitive root of $1$. In addition, and here is where you have actual functions, if $f$ is convergent, then so is $\phi$."

Added later (prompted by OP's comment about choices for $\phi$): in this formulation the uniqueness of Puiseux roots is the statement that if $\psi(t)$ is any other power series such that $y - \psi(t)$ divides $f(t^d, y)$, then $\psi(t) = \phi(\zeta^i t)$ for some $i$. In other words, there are precisely $d$, not infinitely many, Puiseux roots of $f$, namely $\phi(\zeta^it)$, $i = 1, \ldots, d$.

This is covered, e.g. in chapter one of Singularities of Plane Curves by Casas-Alvero.

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  • $\begingroup$ I don't see how this solves the problem. The proof in that book follows the same outline as I described in my question. And thus, even if you know that one of the determinations of $\phi$ is the actual function you are looking for, there is no way of knowing which determination it is. $\endgroup$
    – Gutiérrez
    Aug 16 at 10:36
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    $\begingroup$ I did not explicitly write it, but the uniqueness of Puiseux roots is the statement that once you have any solution $\phi_0(t)$ for the series $\phi(t)$, all other possible choices of $\phi(t)$ are of the form $\phi_0(\zeta^it)$. So there are precisely $d$ choices for $\phi$, not infinitely many. $\endgroup$
    – pinaki
    Aug 16 at 11:20
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    $\begingroup$ Aaah! This makes sense! Once you choose the determination of $\varphi_1(x)$, the rest of the series $\varphi_2(x), ..., \varphi_d(x)$ are already determined, and they are precisely the other possible determinations of $\varphi_1(x)$ (so the choice does not matter). Thanks! $\endgroup$
    – Gutiérrez
    Aug 16 at 17:23
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Responding to the question somewhat tangentially: there is no obligation to think of a Puiseux series as any sort of "global" pointwise-valued function. Luckily so, because, as you observe, it's impossible. :)

But/and the degree of impossibility is probably not as severe as I suspect you are worried about, namely, a power series in $x^{1/N}$ does not have ambiguities of which $N$th root in every term, but only one choice. So, for example, $x^{1/2}+x^{3/2}=x^{1/2}+(x^{1/2})^3$, in its role as Puiseux series, has just two possible values, depending on the choice of sign for $x^{1/2}$. Not four choices.

Of course, we could declare that we interpret such a series so that it has all possible ambiguities in terms independently of each other.

But the way that Puiseux series arise, as elements of finite algebraic extensions of convergent (or formal) power series, dictates that we only need a single choice. This is a consequence of a theorem, that needs some proof. Namely, ... maybe up to changes of coordinates..., all the algebraic extensions of $\mathbb C[[z]]$ are of the form $\mathbb C[[z^{1/N}]]$... with a single "function" $z^{1/N}$, ... not different ones in different terms.

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  • $\begingroup$ I've come across the same type of problem in the question of fractionally iterates of the exponential (aka "tetration") with complex bases. My interpretation of the derivation of a "principal" solution has always included a similar approach for the resolve of the occuring powerseries like your example $x^{1/2}+x^{3/2}=x^{1/2}+(x^{1/2})^3$ (for instance when implementing the "Schroeder's method"). To add references into my drafts/essays - could you provide some examples in literature where this has been discussed? $\endgroup$ Aug 16 at 10:31
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    $\begingroup$ Thank you, that is indeed reasuring! Going from infinitely many determinations to a finite amount is nice. If my question is not fully answered soon, I will mark this as the answer. $\endgroup$
    – Gutiérrez
    Aug 16 at 10:32
  • $\begingroup$ @GottfriedHelms, I don't know a reference off-hand, unfortunately. $\endgroup$ Aug 16 at 16:57

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