# Lines through four lines

(Reposting this from an earlier blog which I gave up on, but liked the
post so I added some pictures – all images produced using the amazing
free alternative to Maple/Mathematica, Sage).

I was reading Fulton-Pandharipande ("Notes on stable maps and quantum
cohomology") the other day and came across the classical result that
there are exactly two lines passing through a generic quadruple of
lines in \(\mathbf{CP}^3\). I encourage people to whom this fact is
unfamiliar to convince themselves of it. It was unfamiliar to me and I
found it hard to visualise, so I sat down and drew some pictures until
I understood it.

Suppose for starters that three of the lines (call them \(A\), \(B\) and
\(C\)) lie on a quadric surface. Recall that a quadric surface is ruled
in two different ways by families of lines – call these lines alpha-
and beta-curves. In the picture on the left they're yellow and red
respectively.

The alpha-curves are all mutually disjoint, the beta-curves are all
mutually disjoint and any alpha-curve intersects any beta-curve in
exactly one point transversely. Lines in general position in
\(\mathbf{CP}^3\) are disjoint, so our three lines are either all alpha-
or all beta-curves. Let’s say they’re all alpha-curves for
definiteness’s sake.

Our fourth line (\(D\), blue in the picture on the right) intersects the
quadric surface in precisely two points. But each intersection point
has a unique beta-curve passing through it. These two (red)
beta-curves intersect \(A\), \(B\), \(C\) (the yellow lines) and \(D\) (the
blue line) and these are the two lines in question. QED

Now how do we see that any three general lines lie on a quadric
surface? Fix a point on line \(A\) and draw all the lines from this
point which hit line \(B\). Precisely one of these will hit line
\(C\). This means that for each point \(p\) of \(A\) there is a unique line
\(L_p\) through that point which hits both \(B\) and \(C\). These lines
trace out a quadric surface containing the three lines as you vary the
point on \(A\). Note that the \(L_p\) are precisely the beta-curves of the
resulting surface.

Cute.

## Comments

On the original post, commenter Vijay Sharma asked:

Hello Jonny,

I too have been reading a lot on Gromov Witten Invariants
and trying to gain a little more grip at it.

And I has this fact only through GW invariants.
Thanks for the post for more intuitive explanations
with drawings. May I ask you drew these pictures in Sage?

Regards,
Vijay

In answer, here is the Sage code I used to produce the pictures. It's
probably highly suboptimal, and I haven't run it for many years now to
check it.

var('x,y,z')
p = implicit_plot3d(x^2+y^2-z^2==1, (x, -2, 2), (y, -2,2), (z, -2,2),opacity=0.8,color='grey')
u = var(‘u')
q = parametric_plot3d((u, 1, u), (u, -2, 2),color='yellow',thickness=5)
r = parametric_plot3d((u/sqrt(2), u/sqrt(2)+sqrt(2), u+1), (u, -3, 1),color='yellow',thickness=5)
s = parametric_plot3d((u/sqrt(10), -3*u/sqrt(10)+sqrt(10), u-3), (u, 0, 5),color='yellow',thickness=5)
t = parametric_plot3d((u, 0.7, 0.5*u+0.1),(u, -2, 2),color='blue',thickness=5)
n = (0.1-sqrt(0.1^2-4*(1-0.5^2)*(0.7^2-0.1^2-1)))/(2*(1-0.5^2))
m = 0.5*n+0.1
k = (0.1+sqrt(0.1^2-4*(1-0.5^2)*(0.7^2-0.1^2-1)))/(2*(1-0.5^2))
f = 2
g = -5
a = parametric_plot3d((-u/sqrt(1+f^2), -f*u/sqrt(1+f^2)+sqrt(1+f^2), u-f), (u, -1, 4),color='red',thickness=5)
b = parametric_plot3d((-u/sqrt(1+g^2), -g*u/sqrt(1+g^2)+sqrt(1+g^2), u-g), (u, -7, -3),color='red',thickness=5)
c = parametric_plot3d((n,0.7,m),(u,-2,2),color='red',thickness=15)
d = parametric_plot3d((k,0.7,0.5*k+0.1),(u,-2,2),color='red',thickness=15)
show(p+q+r+s+t+b+a+c+d,frame=False)

Commenter Grisvard asked:

"These lines trace out a quadric surface containing
the three lines as you vary the point on A" Why a quadric?

My answer was as follows. I think I'm right in saying that
\(PGL(4,\mathbf{C})\) acts transitively on generic triples of lines in
\(\mathbf{CP}^3\): at least this is heuristically justified by a
dimension count – the group has complex dimension 15 and the space of
triples of lines has dimension 12. This could be made rigorous if you
check that the infinitesimal action is surjective at your favourite
configuration of lines (provided your favourite triple is generic!)
because it would give you a Zariski open \(PGL(4,\mathbf{C})\)-orbit
in the space of line-triples (which you now take as your definition of
"generic"). It also acts transitively on smooth quadric surfaces
(because any two nondegenerate quadratic forms are equivalent over
\(\mathbf{C}\)). Therefore it suffices to fix your favourite
configuration of three lines and check that it's contained in a
quadric surface.