a curve in $\mathbb P^5$

Question

I am wondering if there is a nice description of the set of points $[0,0,x,y,z,t] \in \mathbb P^5$ satisfying the equation $t^2+4xy-z^2=0$. What is its complement ? Thanks.

Answer 1

If the projective coordinates are $[v:w:x:y:z:t]$ then by de-projectivizing this equation with respect to $z$ we get $T^2 + 4 XY = 1$ in the affine space $\Bbb A^5$ of coordinates $(V,W,X,Y,T)$, where $V = \dfrac v z$, $W = \dfrac w z$, $X = \dfrac x z$, $Y = \dfrac y z$, $T = \dfrac t z$.

In order to recognize what quadric this is, make the change of variables $X = \dfrac {R + S} 2$ and $Y = \dfrac {R - \Bbb i S} 2$, whihc will turn your equation into $T^2 + R^2 + S^2 = 1$. In the subspace $\{(0,0)\} \times \Bbb A^3$ given by $V=W=0$, this is precisely the equation of a sphere of center $(0,0,0)$ and radius $1$, call it $\Sigma$.

We conclude that the de-projectivized equation represents the closed set $\Bbb A^2 \times \Sigma$, and the original projective equation is just the projective closure of this.

The complementary does not have any special name, it's just the complementary.