The quadric surface $xy = zw$ in $\PP^3$ and its two rulings

Let $k$ be algebraically closed. Consider the surface $Q$ in $\PP^3$ defined by $xy - zw = 0$; a surface is a variety of dimension $2$.

Show that $Q$ is the Segre embedding of $\PP^1 \times \PP^1$ in $\PP^3$, for a suitable choice of coordinates.
Show that $Q$ contains two families of lines $\ts{L_t}$ and $\ts{M_t}$, each parametrized by $t \in \PP^1$, with these properties: if $L_t \neq L_u$ then $L_t \intersect L_u = \emptyset$; if $M_t \neq M_u$ then $M_t \intersect M_u = \emptyset$; and $L_t \intersect M_u$ is a single point for all $t, u$. A line is a linear variety of dimension $1$.
Show that $Q$ contains curves other than these lines, and deduce that the Zariski topology on $Q$ is not carried by $\psi$ to the product topology on $\PP^1 \times \PP^1$, where each factor has its Zariski topology.

The quadric surface Q \subseteq \PP^3 with the lines L_0 and M_0 of its two families. — The quadric surface $Q \subseteq \PP^3$ with the lines $L_0$ and $M_0$ of its two families.

Solution

Use the coordinate order $[x:y:z:w]$ on $\PP^3$ and the reordered Segre map \[ \psi([a:b],[c:d])=[ac:bd:ad:bc]. \] The topology in (c) is the topology on the classical sets of projective points over $k$.

1. The image of $\psi$ is exactly $Q$.

1.This proves (a).

2. The ruling lines in (b) are\[ \boxed{L_{[a:b]}=\{[ac:bd:ad:bc]:[c:d]\in\PP^1\},\qquad M_{[c:d]}=\{[ac:bd:ad:bc]:[a:b]\in\PP^1\}.} \]

For fixed $[a:b]$, the vectors in the first parametrization have the form \[ c(a,0,0,b)+d(0,b,a,0). \] The two displayed vectors are linearly independent, since $(a,b)\ne(0,0)$ and their nonzero coordinates lie in disjoint pairs of positions. Their projectivized span is consequently a line by P-AGH211LINEAR. Similarly, for fixed $[c:d]$, the second parametrization is the projectivized span of the independent vectors $(c,0,d,0)$ and $(0,d,0,c)$. All these lines lie in $Q$ by step 1.

Under the bijection $\psi$, the first family is the image of the subsets $\{t\}\times\PP^1$, and the second is the image of $\PP^1\times\{u\}$. Two distinct members of either family are therefore disjoint. For every $t,u$, their intersection is precisely the one point $\psi(t,u)$. This also proves that each family is parametrized without repetition by $\PP^1$.

3. The closed curve\[ C=Q\cap Z(z-w)=\{[a^2:b^2:ab:ab]:[a:b]\in\PP^1\} \]is not a line.

In the coordinates of $\psi$, the extra equation $z=w$ is $ad=bc$. For nonzero two-dimensional vectors, this determinant equation is equivalent to $[a:b]=[c:d]$. Thus $C$ is precisely the image under $\psi$ of the diagonal of the set $\PP^1\times\PP^1$. Its displayed parametrization is the quadratic Veronese embedding of $\PP^1$ in the plane $z=w$, by P-AGH212DUPLE. It is therefore a closed irreducible curve. It contains the points \[ [1:0:0:0],\qquad[0:1:0:0],\qquad[1:1:1:1], \] whose representative vectors are linearly independent. A projective line is the projectivization of a two-dimensional vector space, so it cannot contain these three points. Hence $C$ is a curve other than either ruling line, in every characteristic.

4. The product topology in (c) does not make $\psi$ a homeomorphism onto $Q$.

Every proper Zariski-closed subset of $\PP_k^1$ is finite. Indeed, it is contained in the zero set of a nonzero homogeneous polynomial, whose zeros on either affine chart are zeros of a nonzero one-variable polynomial, apart from a possible missing endpoint. Conversely, every finite subset is closed because points are closed. The field $k$ is infinite, so the set of projective points is infinite and this topology is the cofinite topology.

Let $\Delta=\{(t,t):t\in\PP^1\}$. Every nonempty basic open rectangle $A\times B$ in the product topology meets $\Delta$: both $A$ and $B$ have finite complement, so $A\cap B\ne\varnothing$. Thus $\Delta$ is dense in the product topology. It is a proper subset, since there are distinct projective points, and hence is not closed.

But step 3 gives $\psi^{-1}(C)=\Delta$, whereas $C$ is Zariski-closed in $Q$. Therefore $\psi$ from the product-topological space to $Q$ is not even continuous, and in particular is not a homeomorphism. This proves (c).

5. Q.E.D.

Step 1 proves (a), step 2 proves (b), and steps <1>3–<1>4 prove both assertions in (c).

Source Collections

Hartshorne, Algebraic Geometry, I.2.15

Dependencies

P-AGH214SEGREThe Segre embedding of $\PP^r \times \PP^s$uses

Backlinks

P-AGH45QUADSURFBIRThe quadric surface $xy = zw$ is birational but not isomorphic to $\PP^2$uses