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 $I\_2(p \cdot p \cdot p) = 1$?



Computing 3 points Gromov-Witten invariants of the Grassmannian
Reading more carefully your question, I think that in the hint, the ruled surface is meant to only contain the lines that are part of the ruling, not the spurious ones that may come when you look at the scroll in P^3. Thus in the case of degree 1, you only get the lines through the point, in the case of degree two, you only get the lines in one ruling. I thus suspect that the three lines will be in the same ruling (otherwise they would intersect, which is not very generic) and hence the unique quadric containing them will be the 1 you need. You still need to make sure it is a reduced point.


(Si, sono il damiano che conosci!)


