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This is a literature request for (hopefully) an English version to a rigorous proof that a complex algebraic curve cannot abruptly end.

That is, if the algebraic curve enters a closed region it must also leave it.

This has a historic significance because Gauss's proof in his Phd thesis assumed this property holds. From looking around it seems that A Ostrowski rigorously proved the result around the 1920's. Is this correct? I am unable to find the title of the paper.

Is there also a proof that a real algebraic curve does not end abruptly?

I don't regard this property as obvious, but it doesn't seem to be well commented in the literature. Maybe, I'm wrong.

Thanks in advance.

Abruptly end: Given an irreducible polynomial $p$, we define $V(p)$ to be the complex algebraic curve associated to $p$. I say that $V(p)$ does not abruptly end at $(x,y)\in\mathbb{C}^2$ with $p(x,y)=0$ if there is a disc small enough so that the boundary contains exactly two points in $V(p)$.

(This is a first attempt and maybe needs some corrections.)

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Can you define mathematically what you mean by "abruptly end" ? If a curve (complex or real) curve is assumed to be {\bf smooth}, then one knows what it looks like locally. –  Paul Broussous Nov 8 '11 at 9:33
    
I am not assuming the curves to be smooth. For instance, the algebraic curve associated to $y^2 = x^3$ has a singularity at $0$ but does not abruptly end. I try and come up with a mathematical description in the question. –  alext87 Nov 8 '11 at 9:42
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Under your definition $p=f(x,y)+xy(x-y)=0$ "abruptly ends" at the origin where $f$ is chosen for $p$ to be irreducible and have low order terms as displayed. You need look at branches (i.e. work with power series) and the result should follow from Puiseux's theorem. Have you looked at Walker's book? –  Felipe Voloch Nov 8 '11 at 11:54
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It is the simplest instance of D. Sullivan's theorem, saying that the Euler characteristic of the link of a point in a real algebraic set is always even. Here is a linkshort text on real algebraic geometry perso.univ-rennes1.fr/michel.coste/polyen/RASroot.pdf (strangely enough, Sullivan is absent from the references) –  BS. Nov 8 '11 at 12:05
    
... plus the fact (due to Milnor ?) that -- at least for isolated singularities -- this link is isomorphic to the intersection of the algebraic set with small spheres centered at the point. –  BS. Nov 8 '11 at 12:11
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1 Answer

up vote 6 down vote accepted

Following the idea of Felipe Voloch, I try to give a simple proof based on Puiseux series expansion. Let $C$ be a real algebraic curve at the origin. Look at the Puiseux series expansion (say in terms of $x$) of $C$ near $O$. By assumption one of the branches (over $\mathbb{C}$), call it $C_1$, has the form $$y = a_1x^{r_1} + a_2x^{r_2} + \cdots \quad\quad\quad (1)$$ for $a_i \in \mathbb{R}$. Let $q$ be the least common multiple of the denominators of $r_i$'s. If $q$ is odd, then the branch expands to both sides of the origin and therefore $C_1$ does not end abruptly. So assume $q$ is even. Let $\zeta := e^{2\pi i/q}$. For each $j$, $1 \leq j \leq q$, the complex curve corresponding to $C_1$ has a Puiseux expansion of the form $y = \sum_i a_i \zeta^{jp_i}x^{r_i}$, where $p_i = qr_i$. In particular, taking $j =q/2$ (so that $\zeta^j = -1$), we see that the complex curve corresponding to $C_1$ has an expansion of the form $$y = \sum_i a_i (-1)^{p_i}x^{r_i}. \quad\quad\quad (2)$$ It follows by the minimality of assumption on $q$ that there is $i$ such that $a_i\neq 0$ and $p_i$ is odd , and consequently, $(1)$ and $(2)$ give different real curves, and it follows that $C_1$ does not end abruptly.

PS: The above arguments only show that $C_1$ has at least two end points on the boundary of a small enough disk centered at $O$. But it can not have more than two, because for all $j \not\in \lbrace q/2, q\rbrace$, $\zeta^j$ is non-real, so the corresponding parametrization does not give any real points.

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