Puiseux series expansion for space curves? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:56:38Z http://mathoverflow.net/feeds/question/88867 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88867/puiseux-series-expansion-for-space-curves Puiseux series expansion for space curves? ssquidd 2012-02-18T23:42:37Z 2012-03-16T23:06:00Z <p>This result is apparently well known and used by many people. I am, however, quite frustrated that I cannot seem to find a proof that I can understand. For plane algebraic curves, this is not too hard. For an irreducible polynomial $F(x,y) \in \mathbb{C}[x,y]$ nonconstant in both $x$ and $y$ with $F(0,0) = 0$, one can formally expand $y$ as a Puiseux series in $x$ $$y(x) = \sum_{i=0}^\infty y_i x^{\alpha_i}$$ The Puiseux theorem actually states that such a series converges (in some sense) near $x=0$. Alternatively, we can construct Riemann surface over the point $(0,0)\in \mathbb{C}^2$ of the algebraic function $y(x)$ which has the normal representation as holomorphic element $$\begin{eqnarray} x &amp;=&amp; t^m \cr y &amp;=&amp; \sum_{i=1}^\infty y_i t^i \end{eqnarray}$$ With some abuse of notation, I think we can even say the two are the same. I can find many sources, among which I like Walker's representation the best.</p> <p>Now how about space algebraic curves (of one complex dimension) in $\mathbb{C}^n$? For a polynomial system $F(x_0,\ldots,x_n) = (f_1(x_0,\ldots,x_n),\ldots,f_n(x_0,\ldots,x_n))$ where each $f_i \in \mathbb{C}[x_0,\ldots,x_n]$ with $F(0,\ldots,0) = (0,\ldots,0)$, and lies only on a one (complex) dimensional irreducible component of $V(F)$, if we fix a place of the this algebraic curve centred at $(0,\ldots,0)$, we should be able to find a parametrization $$\begin{eqnarray} x_0 &amp;=&amp; t^m \cr x_k &amp;=&amp; \sum_{i=1}^\infty c_{k,i} t^i \end{eqnarray}$$ with convergent power series. Or equivalently, we could express $x_1,\ldots,x_n$ as Puiseux series in $x_0$ that converge in certain sense.</p> <p>I could only find "proofs" that reference Hironaka's resolution of singularity, which I don't think I can understand any time soon. I'm hoping to find a proof using only complex geometry or basic complex algebraic geometry. In particular, I was thinking maybe I can repeatedly apply Weierstrass preparation theorem together with Puiseux theorem, however, I'm not quite sure how to continue after the first step.</p> http://mathoverflow.net/questions/88867/puiseux-series-expansion-for-space-curves/91430#91430 Answer by Samuele for Puiseux series expansion for space curves? Samuele 2012-03-16T23:06:00Z 2012-03-16T23:06:00Z <p>If $(C,0)$ is a germ of complex curve in $\mathbb{C}^n$, then you can find coordinates $(z_1, z')$ and a polydisc $V=V_1\times V'$ centered in $0$ such that the canonical projection $V\ni (z_1,z')\mapsto \pi(z_1,z')=z_1\in V_1$ is a ramified covering when restricted to $C$ with $p$ sheets. Let $S$ be the ramification locus, then $S$ is an analytic set in $V_1$, that is, a discrete set of points. Therefore, upon taking a smaller $V_1$, we can think that $S={0}$ or it is empty. Let us suppose $S$ is not empty, otherwise the projection is a local biholomorphism with the unit disc and this gives the thesis with $z_1=t$, $z_k=f_k(t)$, $f_k\in\mathcal{O}(D )$. Now, $C^*=C\cap \pi^{-1}(V_1\setminus S)$ is a $p-$sheeted covering of $V_1\setminus S=D\setminus{0}$, therefore it is isomorphic to the standard $p-$sheeted covering: $$D\setminus{0}\ni t\mapsto t^p\in D\setminus{0}.$$ This means that there exists a map $g:D\setminus{0}\to C^*$ such that $\pi\circ g (t)=t^p$; this map extends clearly to a holomorphic bijection by setting $g(0)=0$ (it is continuous in $0$ and holomorphic outside). Therefore, we have $$z_1=t^p$$ $$z_k=g_k(t)=\sum_{j=0}^\infty a_{kj}t^j.$$</p> <p>The existence of such a system of coordinates is a standard result in local complex geometry, sometimes called local parametrization theorem.</p>