Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having trouble finding a parameterization for the following curve: $x^4 - 2x^2yz + y^2z^2 - y^3z = 0$ taken to be a curve in $\mathbb{C}\mathbb{P}^2$. I followed the example on Wikipedia where they parameterized a circle's equaton, ie., I demohogenized the polynomial at $z$, so I reduce to the curve $x^4 - 2x^2y + y^2 - y^3 = 0$. Then I observed that the curve contains the point $(0,1)$, so I considered the line through $(0,1)$ with slope $t$, ie. $y = tx + 1$ and I plugged this into my equation.

I ended up with an ugly quartic polynomial in $x$ with coefficients in $t$...Specifically, I got $x^4 -2tx^3 + (2+t^2)x^2 + (-t-2t)x + 2 = 0$ and hopefully I didn't make any mistakes. Solving for $x$ doesn't seem easy. Is this the right approach to doing this?

share|improve this question
    
This method of parametrizing a curve typically only works if it's a degree-2 curve; that's because you end up with a quadratic with one known root, so the other root is well defined. Here it's better if you dehomogenize by setting $x=1$, since the resulting curve is quadratic in $z$ - but it's still cubic in $y$, so this method still doesn't seem to apply. –  Greg Martin Apr 3 '12 at 6:32

1 Answer 1

up vote 1 down vote accepted

Let us de-homogenize the equation by setting $X=x/y$, $Z=z/y$ (which is the same as putting $y=1$): $$Z^2-(1+2X^2)Z+X^4=0.\qquad(+)$$ Let us project to the $X$-axis and see where the branch points are: those are the $X\in\mathbb{C}$ where the equation for $Z$ has a double root, i.e. where $1+4X^2=0$. We thus have a double cover of the $X$-sphere with the branch points $\pm i/2$, hence the curve is indeed rational. To get a rational parametrization, we just need to find a rational function $\mathbb{CP}^1\to\mathbb{CP}^1$, $T\mapsto X(T)$, which gives the same branched double cover of $\mathbb{CP}^1$. We can take e.g. $(X+i/2)/(X-i/2)=T^2$, i.e. $$X(T)=-\frac{i}{2}\frac{1+T^2}{1-T^2}.$$ We can now compute $Z(T)$ out of $(+)$ and get the following: $$Z(T)=\frac{T^8-6T^6+14T^4-8T^2+1+4iT(T^2+1)^3}{4(1-T^2)^4}$$ and we have a rational parametrization. If you prefer, in homogeneous coordinates, setting $T=t/s$, we get $$z=\frac{t^8-6t^6s^2+14t^4s^4-8t^2s^6+s^8+4its(t^2+s^2)^3}{4}$$ $$x=\frac{i}{2}(s^2+t^2)(s^2-t^2)^3$$ $$y=(s^2-t^2)^4.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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