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Consider the three lines of $\mathbb{P}^3$ given by

$L: \, z_0 = z_1 = 0 \\ M: \, z_2=z_3 = 0 \\ N: \, z_0 = z_2, \, z_1 = z_3.$

It is claimed in Exercise 2.12 of Harris (a first course) that the union of all lines that meet all $L,M,N$ is projectively equivalent to the segre variety $\Sigma_{1,1} = \sigma(\mathbb{P}^1 \times \mathbb{P}^1)$, where $\sigma$ is the Segre embedding.

I have been thinking about this problem but i am still confused about some important things:

1) Conceptually, a line of $\mathbb{P}^3$ can be represented by a point of $\mathbb{P}^3 \times \mathbb{P}^3$. On the other hand $\Sigma_{1,1}$ lives inside $\mathbb{P}^3$. So i don't see how a point of $\Sigma_{1,1}$ may correspond to a line.

2) If i take a line that meets all three $L,M,N$, then the coefficients of the equations of the line satisfy three quadratic equations and that's precisely the space of lines that meet all three $L,M,N$. Now, these are three equations in $8$ unknowns and i can't see how they could be replaced by one equation in $4$ unknowns, which is the equation that describes $\Sigma_{1,1}$.

Any insights?

Manos
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  • I don't understand (1). Lines meeting $L, M, N$ are subsets of $\mathbb{P}^3$ and you want to take the union of all of these subsets. I don't think it's relevant for the problem but lines in $\mathbb{P}^3$ are parametrized by the Grassmannian $\mathbb{G}(1, 3)$, which I think is treated later in the book. Of course there should be a map from $\mathbb{P}^3 \times \mathbb{P}^3$ less the diagonal to $\mathbb{G}(1,3)$. – Hoot Apr 07 '15 at 21:21
  • Dear Manos, a line in $\mathbb P^3$ is really a line and not a point of $\mathbb{P}^3 \times \mathbb{P}^3$ (although there is a non-trivial relation between both concepts). The claim is that any line in $\mathbb P^3$ intersecting your three lines $L,M,N$ will lie completely in $\Sigma_{1,1}$. An example is the line $z_0=z_2=0$ which cuts your three lines in the points $(0:0:0:1),(0:1:0:0)(0:1:0:1)$. – Georges Elencwajg Apr 07 '15 at 21:31
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    Also: beware that Harris has nice descriptions of the scenery of classical algebraic geometry but calculates essentially nothing serious in his book, so that his exercises are essentially impossible to do by his claimed readership of beginners. I'm saying this to ENCOURAGE you ( slightly paradoxically!) – Georges Elencwajg Apr 07 '15 at 21:32

1 Answers1

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Let $X$ be the locus of lines $\ell\subset\mathbb P^3$ meeting $L,M,N$.

Hints:

  1. Up to an element of $PGL(4,\mathbb C)$, you can assume $L,M,N$ are contained in the quadric surface $Q\subset \mathbb P^3$ (belonging to one of the two families of lines). This should say $X\supseteq Q$.
  2. A generic line $\ell\subset\mathbb P^3$ meets $Q$ in only two points. This should say $X\subseteq Q$.
Brenin
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  • Dear Brenin, here is what i understand from your answer: $Q$, which i called $\Sigma_{1,1}$, is covered by two families of lines. One family is given by $(F1): , z_1= \lambda z_0, , z_3 = \lambda z_2$ and the other family is given by $(F2): , z_2= \lambda z_0, , z_3 = \lambda z_1$. Now, i can see that every line from $(F1)$ does meet all $L,M,N$. The issue is that every line from $(F2)$ does not meet all $L,M,N$, since it does not meet e.g. $L$. So i don't see how you conclude that $Q \subset X$. Could you please explain? Your hint 2 is clear. – Manos Apr 08 '15 at 02:57
  • So if your lines $L,M,N$ belong to the family $F2$, any line belonging to the family $F1$ will meet them all, as you say. This is all you care about, because any point $(p,q)\in Q$ belongs to exactly one line of the family $F1$. – Brenin Apr 08 '15 at 06:20
  • I see! I had not noticed that the family $(F1)$ alone covers $Q$. – Manos Apr 08 '15 at 13:43