This is from an exercise in Koch, Vainsencher - An invitation to quamtum cohomology.

**Background**

The exercise asks to compute the 3-points Gromov-Witten invariants of the Grassmannian $G = \mathop{Gr}(1, \mathbb{P}^3) = \mathop{Gr}(2, 4)$ via the enumerative interepretation. In particular my problem is with computing the invariant $I_2(p \cdot p \cdot p)$, where $p$ is the class of a point on $G$. This is the number of rational curves of degree $2$ through $3$ generic points on $G$. Here we see $G$ as embedded by the Plucker map, and the degree is defined accordingly.

A rational curve $C \subset G$ of degree $d$ will sweep out a rational ruled surface $S$ of degree $d$ in $\mathbb{P}^3$; up to here I agree with the hints of the book. The problem is the following hint:

Show that the condition on $C$ of passing through a point $q \in G$ corresponds to the condition on $S$ of containing the line in $\mathbb{P}^3$ corresponding to $q$.

This seems to me plain false. Of course one implication is true, but is absolutely possible that $S$ contains a line without $C$ passing through the corresponding point.

For instance, when $d = 1$, $C$ is a line on the Grassmannian, and it is well-known that these have the form $\{ \ell \mid a \in \ell \subset A \}$, where $a$ is a point and $A$ a plane of $\mathbb{P}^3$. In this case $S$ is the plane $A$, so it contains many lines which do not pass through $a$, hence these lines are not parametrized by $C$.

Similarly, when $d = 2$, the surface $S$ can be a smooth quadric, which has two distinct rulings of lines; one will correspond to lines parametrized by $C$, but the other one will not. To see that a smooth quadric can actually arise, just invert the construction. Starting from a smooth quadric $S$ take any line $\ell \subset S$. There is a natural map $\ell \to G$ given by sending a point $q \in \ell$ to the unique line in the other ruling passing through $q$. The image of this map is a curve $C \subset G$, such that the associated surface is $S$ itself.

Given the hint, the book goes on to say

Show that $I_2(p \cdot p \cdot p) = 1$, by interpreting this number as a count of quadrics containing three lines.

Now I certainly agree that given three generic lines in $\mathbb{P}^3$ there is a unique quadric containing them. To see this, just choose $3$ points on each line: a quadric will contain the lines iff it contains the $9$ points, and it can be shown that these $9$ points give $9$ independent conditions.

Still I do no see how this implies the count $I_2(p \cdot p \cdot p) = 1$. What I guess happens is that generically we will have two lines in one ruling and one line in the other, so that the curve $C \subset G$ which sweeps $S$ will only pass through one or two of the assigned points.

**Question**

What is the right count? Is there something wrong in what I said above? Is it even true that $I_2(p \cdot p \cdot p) = 1$?