# Non-tangent lines to lines in $\mathbb{P}^3(\mathbb{C})$.

Let $\pi:\mathbb{C}^4\setminus\{0\}\to\mathbb{P}^3(\mathbb{C})$ be the quotient map. Let $Q\subset\mathbb{P}^3(\mathbb{C})$ be a smooth quadric, let $q$ be the quadratic form such that $Q=\pi[\{v\in\mathbb{C}^4\,|\,|q(v)=0\}]$, and let $\sigma$ be the unique symmetric bilinear form for which $q(v)=\sigma(v,v)$ holds for all $v\in\mathbb{C}^4$.

Let $L\subset\mathbb{P}^3(\mathbb{C})$ be a line which is not tangent to $Q$, and let $p_1,p_2\in\mathbb{P}^3(\mathbb{C})$ be the two points of intersection. Then $L=\overline{p_1p_2}\not\subset Q$ and thus $$\sigma(p_1,p_1)\neq0\,\vee\,\sigma(p_2,p_2)\neq0\,\vee\,\sigma(p_1,p_2)\neq0,$$ but since $p_1,p_2\in Q$ it follows that $\sigma(p_1,p_2)\neq0$, so $p_1\in T_{p_1}Q$ and $p_1\notin T_{p_2}Q$, so $T_{p_1}Q\neq T_{p_2}Q$. Since $T_{p_1}Q$ and $T_{p_2}Q$ are not equal they intersect in a line $\mathcal{L}^{p_1}_{p_2}:=T_{p_1}Q\cap T_{p_2}Q$, so we get a map $\alpha$ from the set of non-tangent lines to the set of lines in $\mathbb{P}^3(\mathbb{C})$ by sending $L$ to $\mathcal{L}^{p_1}_{p_2}$.

I first thought that $L\cap\alpha(L)=\emptyset$ for all non-tangent lines, since $$\begin{split} L\cap(T_{p_1}Q\cap T_{p_2}Q)&=(L\cap T_{p_1}Q)\cap T_{p_2}Q \\ &=\{p_1\}\cap T_{p_2}Q \\ &=\emptyset, \end{split}$$ but now I'm not so sure anymore, so I'm wondering if anyone can shed light on this, or show why it is not true.

EDIT

It is clear that $p_1,p_2\notin L\cap\mathcal{L}^{p_1}_{p_2}$. Since $\mathcal{L}^{p_1}_{p_2}\subset T_{p_1}Q$ and $\mathcal{L}^{p_1}_{p_2}Q\subset T_{p_2}Q$, it follows that $\mathcal{L}^{p_1}_{p_2}\subset Q$, so indeed $L\cap\mathcal{L}^{p_1}_{p_2}=\emptyset$.

Questions: so $\alpha$ is invertible since $\alpha\circ\alpha=\mathrm{id}$, but how would you in general define what the image of a tangent line should be? How can you decide which points to use for the construction similar to $\alpha$?

Yes, $\alpha$ is invertible; in fact, it's an involution.
Let $\lambda_{p}$ and $\rho_{p}$ (for "left" and "right") denote the rulings of the quadric $Q$ passing through a point $p$ of $Q$. The tangent plane $T_{p}Q$ is the unique plane containing $\lambda_{p}$ and $\rho_{p}$.
Every "left" ruling (blue) intersects each "right" ruling (green) in exactly one point, so if $L = \overline{p_{1}p_{2}}$, then $\alpha(L)$ is the line through $p_{1}' := \lambda_{p_{1}} \cap \rho_{p_{2}}$ and $p_{2}' := \rho_{p_{1}} \cap \lambda_{p_{2}}$.
As the one-sheeted hyperboloid makes clear, \begin{align*} \lambda_{p_{1}'} &= \lambda_{p_{1}}, & \rho_{p_{1}'} &= \rho_{p_{2}}, \\ \lambda_{p_{2}'} &= \lambda_{p_{2}}, & \rho_{p_{2}'} &= \rho_{p_{1}}, \end{align*} and $\alpha(\overline{p_{1}' p_{2}'}) = \overline{p_{1} p_{2}}$.
• Does it not matter that $\alpha(\overline{p_1p_2})$ is contained in $Q$? How would you define the image of a general non-tangent line? Here it is straightforward how you define it, but how would you do it in general? – B. Pasternak May 26 '16 at 16:19
• The line $L = \overline{p_{1} p_{2}}$ and its image $\alpha(L) = \overline{p_{1}' p_{2}'}$ contain the two "excluded" edges of the tetrahedron with vertices $p_{i}$ and $p_{i}'$, so in fact neither is contained in $Q$. Since every non-tangent line hits $Q$ in a pair of distinct points, the diagram is general. :) – Andrew D. Hwang May 26 '16 at 22:38