Line through two points in $\mathbb{P}^3$; Segre embedding

Question

Let $\Psi:\mathbb{P}^1\times \mathbb{P}^1\longrightarrow \mathbb{P}^3$ be the Segre embedding. If I take points $(A_1,A_2)$, $(B_1, B_2)$ in $\mathbb{P}^1\times \mathbb{P}^1$, I want to know how does the line that goes through $\Psi(A_1,A_2)$ and $\Psi(B_1,B_2)$ look like. Can someone help me with this?

Answer 1

In coordinates, the Segre embedding of $\mathbb{P}^1 \times \mathbb{P}^1$ into $\mathbb{P}^3$ sends a closed point $([x_0 : x_1], [y_0 : y_1]) \in \mathbb{P}^1 \times \mathbb{P}^1$ to the point $[x_0y_0 : x_0y_1 : x_1y_0 : x_1y_1] \in \mathbb{P^3}$. This is the n = m = 1 case of the more general equation given at https://en.wikipedia.org/wiki/Segre_embedding.

Meanwhile, the line connecting two points C and D in $\mathbb{P}^n$ is described set theoretically as $\{sC + tD \: | \: [s:t] \in \mathbb{P}^1\}$.

Careful use of these two equations, which appear in, say, Harris's Algebraic Geometry: A First Course, should give you your answer.