I often heard that each smooth cubic surface contains even $27$ straight lines. I cannot prove it today, but I'll do my best to do it soon. However how to prove that each cubic surface contains a straight line?
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1Dear Nikita, I'm not sure that it's much easier to show the existence of one line than to show the existence of all $27$. One way I can think of is to intersect the cubic with a plane, to obtain a plane cubic curve, and then vary the plane, and try to argue that you can take a limit where the plane cubic curve degenerates into the union of a line and a conic. But I don't see directly how to do that. Regards, – Matt E Jan 26 '14 at 15:43
2 Answers
Shafarevich Volume 1 has a pretty elementary argument: Chapter 1.6, Theorem 10. It's a standard incidence correspondence proof: one considers the variety $\mathscr I$ of pairs $(X,L)$ where $X$ is a cubic surface, and $L$ is a line in $\mathbf{P}^3$ such that $L \subset X$. One calculates the dimension of $\mathscr I$ by considering the projection to $G(2,4)$ (the parameter space of lines) and finds that it is 19, which is the same as the dimension of the space $\mathbf{P}^{19}$ of cubic surfaces. Finally, one writes down a single cubic that contains only finitely many lines, proving that the map $\mathscr I \rightarrow \mathbf{P}^{19}$ is generically finite onto its image, hence (since the dimensions are equal) surjective.
As Matt E comments, once you know there is one line, it's not that hard to find all 27: Shafarevich goes on to do this in Chapter 4.
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3A bit off-topic for the original question, but let me mention that my favourite proof is to view the cubic as defining a section of a certain vector bundle on $G(2,4)$; calculations with Chern classes then show that the number of zeroes of this section (= number of lines on the cubic) equals 27. I like it because exactly the same method shows that there are 2875 lines on the quintic threefold. See Joe Harris, "Galois groups of enumerative problems", Duke. – Jan 27 '14 at 13:51
You should look at Section 7 of Miles Reid's wonderful book.
If you're more advanced, you can without too much difficulty prove that if you blow up $6$ general points in $\Bbb P^2$, the surface embeds as a cubic surface in $\Bbb P^3$. You get $6$ lines from the $6$ exceptional divisors, $6$ more lines from the proper transforms of the $6$ conics through five of the six points, and $15$ more lines from the $\binom 62$ proper transforms of the lines through pairs of the points.
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Dear Ted: yes, but then you also have to prove that every cubic surface in $\mathbb P^3$ is obtained by your blowing-up procedure. – Georges Elencwajg Jan 26 '14 at 22:48
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Right, @GeorgesElencwajg: This argument is given, not in immediate fashion, in Griffiths/Harris. – Ted Shifrin Jan 26 '14 at 22:52
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