0

This is a follow up to a comment made on one of my previous questions that I feel makes more sense as a separate question. The comment was to the effect that the real solutions to a system of polynomials may not be parameterizable unless the system is "nice". When would a solution not be parameterizable? Could I not just choose arbitrarily some of the variables to be parameters, then solve the remaining system? Or is that too naive?

Michael Stachowsky
  • 2,943
  • As an example, how would you parametrize the real solutions to the fairly simple equation $$x^3+yx+y^3=0.$$ Plugging in an arbitrary value $y_0$ for $y$ yields a cubic in $x$, which has one or three real solutions depending on whether the discriminant $$-4y_0^3-27y_0^6=-y_0^3(4+27y_0^3),$$ is negative or not. So we need to distinguish cases for whether $y\in(-\tfrac{\sqrt[3]{4}}{3},0)$ or not, ruling out a 'nice' parametrization. In general, the higher the degree of the equations, the more cases we need to distinguish, and the harder it is to tell whether (real) solutions exist at all. – Servaes Jun 13 '16 at 14:28
  • Wow, I'd forgotten about this question, but this is really helpful. Your comment leads to a question: didn't you just parameterize the equation? You chose $y$ as the parameter and formed a cubic in $x$. You end up getting a number of cases, I agree, but doesn't that just mean the parameterization will be ugly? In other words, could I not just arbitrarily choose the first $n-1$ coordinates as parameters and leave the last one to be a function of those, and then have a parameterized system? – Michael Stachowsky Jun 13 '16 at 14:31
  • You could certainly do that. But the more coordinates you have and the higher the degree, the uglier it becomes. Which leads me to the point that a 'parametrization' means different things to different people (and mathematicians). What would you consider a parametrization of a solution set? – Servaes Jun 13 '16 at 14:35
  • I would define it loosely as: a set of parameters that can vary freely in some domain so that, for an arbitrary choice of parameters from that domain, all other unknowns in the system are determined and the system of equations is satisfied. Thinking about it now, the problem then must reduce to finding the domain for the parameters on which the system is satisfied, and I think that's what you meant - if we knew the domain then we'd know the solution set anyway and wouldn't really need the parameterization, right? – Michael Stachowsky Jun 13 '16 at 14:38
  • 1
    Roughly speaking, yes. Some systems of equations allow parametrizations with domain $\Bbb{R}^n$, which is a very nice domain. But if a lot of cases need to be distinguished the domain could become a horrible thing. In fact, with such a loose definition every solution set is parametrized by itself, which is of course not a useful parametrization at all. Often some requirements are made for the domain to be 'nice enough'. This in turn prevents some (often many!) systems of equations of having a (nice enough) parametrization. – Servaes Jun 13 '16 at 14:42
  • By the way, I just realized the equation I described does have a nice parametrization, which has nothing to do with the approach I just took. Take $$\varphi:\ \Bbb{R}-{-1}\ \longrightarrow\ \Bbb{R}^2:\ t\ \longmapsto\ \left(-\frac{t}{1+t^3},-\frac{t^2}{1+t^3}\right).$$ That just goes to show that it is in general difficult to tell whether systems have parametrizable solution sets. – Servaes Jun 13 '16 at 14:45
  • Out of curiosity, how did you do that? Your previous solution to one of my other questions seems to imply that there is a method (or something) that might help me. What should I look for? – Michael Stachowsky Jun 13 '16 at 14:46
  • In this case I noticed the curve is parametrizable because its genus is zero. The solution set to a system of equations is a curve when the number of equations is one less than the number of variables (and the equations have no common factors). Computing the genus of a curve (or higher-dimensional solution space!) in general is hard, but for an irreducible plane curve (given by a single equation like the one above) there is the genus-degree formula. – Servaes Jun 13 '16 at 15:14
  • 2
    Generally, solution sets to equations of degree $2$ (in any number of variables) are parametrizable by first finding one solution(if it exists), and then looking at all lines through this point. Every line contains precisely one solution other than the one already found, if you are working over an algebraically closed field (e.g. the complex numbers). This parametrizes the solution set with domain $\Bbb{C}^{n-1}$, where $n$ is the number of variables. The same thing works if you have an equation of higher degree $d$ with a singular point of multiplicity $d-2$ by taking lines through this point – Servaes Jun 13 '16 at 15:20
  • question in favorites. +1 for comment from Servaes. – mick May 21 '23 at 16:41

0 Answers0