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 $X \cap H_2 = L_2 \cap L_4$Now $L_3$ 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.

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Categories: algebraic geometry
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