8
$\begingroup$

Given an irreducible polynomial $P(X,Y) \in \mathbb{C}[X,Y]$, one obtains by analytic continuation a Riemann surface $M$ with a branched covering $f \colon M \to \mathbb{P}^1_{/\mathbb{C}}$, corresponding to $P(z,f) = 0$.

Of particular interest in this setup is the monodromy, i.e. the action of the fundamental group of $\mathbb{P}^1(\mathbb{C}) \setminus B$ (where $B$ is the set of branch points) on a fiber above some generic point $p$.

To compute the relevant local monodromies (around each individual branch point), one can compute Puiseux series around the branch point and read off the monodromy data.

How exactly does one compute, given only the polynomial $P(X,Y)$, what the different Puiseux series look like? Is there a more intuitive way of seeing what the cover looks like without delving into lengthy calculations with power series?

$\endgroup$
7
$\begingroup$

The potential ramification points are $X_0$ s.t. either $P$ and $\partial P/\partial Y$ have a common zero at some $(X_0,Y_0)$, or the degree of $P(X_0,Y)\in\mathbb{C}[Y]$ is less than $n$, or $X_0=\infty$. Let us consider the first case, i.e. suppose that $Y_0\in\mathbb{C}$ is a $k$-fold root of $P(X_0,Y)\in\mathbb{C}[Y]$ (the remaining cases are similar). For $X_1$ close to $X_0$ this multiple root will split into $k$ different roots. Generically we will get a $k$-cycle in the local monodromy at $X_0$ (the roots undergo a cyclic permutation when we go around $X_0$). Sometimes we may, however, get a different permutation of these $k$ roots. It can be determined using Newton polygon as follows.

We may suppose that $(X_0,Y_0)=(0,0)$ (by shifting the variables). For every monomial $Y^aX^b$ in $P$ we draw the point $(a,b)$ in the plane and then we take the convex hull of these points. We consider only the part of the boundary of the resulting polygon from which $(0,0)$ is visible. Let $s_1,\dots,s_q$ be these sides, and let $k_1,\dots,k_q$ be the lengths of the projections of $s_i$'s to the $x$-axis; we clearly have $k_1+\dots+k_q=k$. If none of the sides $s_i$'s contain any integer points inside then $(k_1,\dots,k_q)$ are the lengths of the cycles of the local monodromy.

For example, for the polynomial $X^3+ 5XY^3-8Y^7+iY^{10}+(2+i)X^4-X^2Y^4$ we get $-8Y^7+iY^{10}$ when we set $X=0$, i.e. $Y=0$ is a $7$-fold root, and the Newton polygon is

Newton polygon

so that we get a $3$-cycle and a $4$-cycle.

(If some of $s_i$ contain an integer point inside then things become more complicated: essentially we know that we have a Puiseux series solution $Y=cX^t+\dots$ of $P(X,Y)=0$, where $-t$ is the slope of $s_i$, and we need to determine what denominators appear in the exponents of this series. We basically pose $Z=Y-cX^t$, get an equation for $Z$, and use again Newton polygon to see what happens.)

Newton polygon is equally useful for determining ramifications for finite extensions of $p$-adic numbers (the method is the same).

$\endgroup$
  • 1
    $\begingroup$ Thanks for the added details. What do you need to do if you want to tell which are the exact permutations occurring, and not just the cycle type (needed to piece together the different local monodromy)? $\endgroup$ – Will May 13 '11 at 23:40
  • $\begingroup$ Is there any proof or reference of proofs for this method? $\endgroup$ – Yai0Phah Dec 28 '13 at 11:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.