Suppose that $S \subset \mathbb{P}^3$ is a smooth quadric surface. Let $p \in S$. Can more than two lines $l \subset S$ pass through $p$? If not, why?
2026-04-03 05:17:04.1775193424
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Number of lines passing through a fixed point in a quadric surface
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The tangent space $T_p S$ can be thought as a hyperplane $H\subset \mathbb{P}^3$ and any line $l\subset S$ passing through $p$ is contained in this hyperplane. On the other hand, as $S$ is smooth, it is irreducible, so the intersection $H\cap S$ is of dimension $1$ and has the same degree as $S$, namely $2$. Each line in $H\cap S$ adds one to this degree so there are at most two lines. The same argument works for a surface of degree $d$, showing that the number of lines is at most $d$.
An alternative approach to this problem is to use the isomorphism from $\mathbb{P}^1 \times \mathbb{P}^1$ to the smooth quadric surface $S \subset \mathbb{P}^3$. In coordinates, the map $\mathbb{P}^1 \times \mathbb{P}^1 \rightarrow \mathbb{P}^3$ is given by $$ (X_0;X_1), (Y_0;Y_1) \mapsto (X_0 Y_0; X_0 Y_1; X_1 Y_0; X_1 Y_1), $$ with the image being $S = \{ (X;Y;Z;W) : XW = YZ \}$. One can show that the map is an isomorphism from $\mathbb{P}^1 \times \mathbb{P}^1$ to $S$. And a curve of degree $(d,e)$ in $\mathbb{P}^1 \times \mathbb{P}^1$ maps to a curve of degree $d + e$ in $\mathbb{P}^3$. In particular, lines on $S$ correspond to curves in $\mathbb{P}^1 \times \mathbb{P}^1$ of degree $(1,0)$ or $(0,1)$. Thus, for a point $(P,Q) \in \mathbb{P}^1 \times \mathbb{P}^1$, there are two lines through $(P,Q)$, namely $\{P\} \times \mathbb{P}^1$ and $\mathbb{P}^1 \times \{Q\}$.