Home > algebraic geometry > Understanding the birationality of the projective plane and a nonsingular quadric surface

Understanding the birationality of the projective plane and a nonsingular quadric surface

Today I’m going to write up a beautiful and very classical little piece of projective geometry, which is quite a departure from the sort of algebraic geometry I usually post about. Here is my abstract motivation: the projective plane \mathbb{P}^2 and the surface \mathbb{P}^1 \times \mathbb{P}^1 both contain a copy of the affine plane \mathbb{A}^2 (the former as a standard affine open and the latter as \mathbb{A}^1 \times \mathbb{A}^1 \subset \mathbb{P}^1 \times \mathbb{P}^1). Thus these two varieties are birational, and it is well-known that a birational map of surfaces can be represented as a sequence of blow-ups and blow-downs. We will see, in a totally geometrical way, that \mathbb{P}^1 \times \mathbb{P}^1 is obtained by first blowing up two points of \mathbb{P}^2 and then blowing down the line which connects them. Inversely, one could first blow up a point of \mathbb{P}^1 \times \mathbb{P}^1 and then blow down the two lines passing through that point to get \mathbb{P}^2.

Think of \mathbb{P}^1 \times \mathbb{P}^1 as the quadric surface X = \{ xy = zw \} \subset \mathbb{P}^3 via the Segre embedding, fix x \in X, and realize \mathbb{P}^2 as the space of lines through x in \mathbb{P}^3. Write L_1,L_2 \in \mathbb{P}^2 for the two lines contained in X, and consider the morphism \mathbb{P}^2 \setminus \overline{L_1L_2} \to X defined by L \mapsto L \cap X (by \overline{L_1L_2} we mean the line in \mathbb{P}^2 connecting L_1 and L_2, or in our setup the projectivized tangent space \mathbb{P}(T_xX)). It has an inverse X \setminus (L_1 \cup L_2) \to \mathbb{P}^2 given by the formula y \mapsto \overline{xy}. Thus we have found a birational map \mathbb{P}^2 \dashrightarrow X which identifies copies of \mathbb{A}^2, and we claim this lifts to an isomorphism \text{Bl}_{L_1,L_2} \mathbb{P}^2 \cong \text{Bl}_x X.

Now \text{Bl}_{L_1,L_2} \mathbb{P}^2 can be thought of as a subvariety of \mathbb{P}^2 \times \mathbb{P}^1 \times \mathbb{P}^1, where the first and second copies of \mathbb{P}^1 consist of lines in \mathbb{P}^2 through L_1 and L_2 respectively, or equivalently planes in \mathbb{P}^3 containing L_1 and L_2 respectively. Namely, \text{Bl}_{L_1,L_2} \mathbb{P}^2 consists of triples (L,H_1,H_2) satisfying L \subset H_1 \cap H_2. On the other hand, \text{Bl}_x X is the subvariety of X \times \mathbb{P}^2 consisting of pairs (y,L) such that y \in L and if y = x then L is tangent to X at x.

First we exhibit a canonical isomorphism between the ambient spaces of the two blow-ups, so we must show how to identify X with the space of pairs (H_1,H_2), where H_1,H_2 \subset \mathbb{P}^3 are planes with L_1 \subset H_1, \ L_2 \subset H_2. For i = 1,2 the quadratic curve X \cap H_i \subset H_i \cong \mathbb{P}^2 contains the line L_i, so it must be the union of two lines: write X \cap H_1 = L_1 \cup L_3 and X \cap H_2 = L_2 \cap L_4Now L_3 and L_4 must intersect in a point y \in X (they are members of the two distinct rulings of X), and in fact (H_1,H_2) \mapsto y is the desired isomorphism. For the inverse, fix y \in Y and write L_3,L_4 \subset X for the two lines which pass through y. Then there are unique planes H_1,H_2 \subset \mathbb{P}^3 such that X \cap H_1 = L_1 \cup L_3 and X \cap H_2 = L_2 \cap L_4, provided one orders H_1 and H_2 correctly.

To finish, let us show that the conditions which cut out the blow-ups coincide under our identifications of their ambient spaces. Let (L,H_1,H_2) \in \text{Bl}_{L_1,L_2} \mathbb{P}^2 and write X \cap H_1 = L_1 \cup L_3 and X \cap H_2 = L_2 \cap L_4, so if L_3 \cap L_4 = \{ y \} then this triple corresponds to (y,L) \in X \times \mathbb{P}^2. If y \neq x it suffices to observe that y \in L_3 \cap L_4 \subset H_1 \cap H_2 = L. In the case y = x we have H_1 = H_2 = T_xX, whence L \subset T_xX as needed. Conversely, given a pair (y,L) \in \text{Bl}_x X, let (L,H_1,H_2) be the corresponding triple, and note that when y \neq x the lines L and H_1 \cap H_2 contain the same pair of distinct points x,y and hence must be identical. If y = x, then L \subset T_xX = H_1 = H_2, which completes the argument.

  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: