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I think there was a theorem, like

  • every cubic hypersurface in $\mathbb P^3$ has 27 lines on it.

What is the exact statement and details?

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As discussed below, the smooth cubics all have 27 lines. But if you are interested also in the singular cubic surfaces, you will find in Chapter 9 of Dolgachev's notes ( math.lsa.umich.edu/~idolga/topics.pdf ) a discussion of the number of lines on each of them, and much more... –  Vivek Shende Nov 18 '10 at 23:03

9 Answers 9

up vote 15 down vote accepted

The exact statement is that every smooth cubic surface in $\mathbb P^3$ (over an algebraically closed field) has exactly $27$ lines on it. Many books on algebraic geometry include a proof of this famous fact. The proof that I first learned comes from chapter V of Hartshorne, where cubic surfaces arise as the blowup of $\mathbb P^2$ at $6$ points, and where the formula $27=6+15+6$ is explained.

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Hi Daniel, I'm not sure that there is a proof in Hartshorne of the fact that every smooth cubic surface has 27 lines, but he has a reference (Remark 4.7.1) to proofs, by Manin (the cubic forms reference Pasha Zusmanovich gives below) and Nagata. He gives an informal dimension count in Remark 4.7.2 which shows that most cubic surfaces are of this form; this can certainly easily be made precise using the methods the reader will know at at that point in the book. –  Ravi Vakil Sep 4 '11 at 17:16
One can also find real cubic surfaces with 27 real lines, with the same construction. –  F. C. Feb 4 '14 at 20:39

Cayley, gaily, opines:

"On a cubic surface, there are 27 lines"

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+1. However, at least according to www-history.mcs.st-andrews.ac.uk/HistTopics/Cubic_surfaces.html Cayley showed rather that the number of lines was finite, and Salmon then proved it to be 27. That might be difficult to put into song, though. –  user5117 Nov 18 '10 at 8:24

To see the 27 lines, take 6 generic points on P(2) and consider the linear system of cubics through them. This linear system defines a rational map from P^2 to P^3 which is nothing more than the blow-up of P^2 on these 6 points.

The image is a cubic surface and the lines are:

  • 6 exceptional divisors;
  • 15 lines connecting two among the six points;
  • 6 conics through five of the six points.
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There is a completely elementary (i.e. no surface theory or Chern classes needed) way to carry out this computation. I give you a brief sketch omitting the details. First use an incidence correspondence to prove that every cubic (smooth or not) contains at least one line.

Once you have a line l consider planes H containing l. The intersection of H with the cubic is either (l + a smooth conic) or three lines. Moreover in the second case the three lines are distinct and do not meet in one point, otherwise the cubic would not be smooth there.

The planes containing l form a $\mathbb{P}^1$, and a simple computation tells you when the residual conic in the plane is smooth. Namely the vanishing of the determinant of the residual conic is an equation of degree 5. This equation has distinct roots, again because the cubic is smooth.

We conclude that for a given line l there are exactly 5 planes through l on which the cubic decomposes as the union of three lines. Said differently, every line meets exactly 10 other lines.

From the last statement it is a combinatioral exercise to prove that the total number of lines is 27.

I hope the sketch is clear enough; feel free to ask more details if some step is too obscure.

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Maybe I should't comment on something nearly a year old, but I've only just noticed this question. Anyway, "the three lines are distinct and do not meet in one point, otherwise the cubic would not be smooth there". The smooth cubic surface $x^3 + y^3 = z^3 + w^3$ meets the plane $x=z$ in three lines meeting at one point. Points on a cubic surface where three lines intersect like this are called Eckhardt points. This doesn't affect the rest of the argument at all, though. –  Martin Bright Nov 18 '10 at 10:46
You are right, there is a problem, and a small argument is needed. For the combinatorial count to work one needs to know that there is at least one plane $H$ such that $H \cap S$ consists of three distinct lines. (Every other line on $S$ will meet one of those three, yielding a total of $3 + 3 \cdot 8 = 27$ lines.) If there was no such plane, it is easy to see that all lines on $S$ would meet in a single point, which would then be singular. –  Andrea Ferretti Nov 19 '10 at 14:12

It's a nice application of dimension theory to show that the set of cubic surfaces containing exactly 27 lines corresponds to an open subset of projective 19-space; the remaining surfaces either contain an infinite number of lines or a nonzero finite number less than 27. The proof can be found in several places, including Shafarevich's book and my online algebraic geometry notes.

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I vaguely remember that Miles Reid gives a hands-on proof of this towards the end of his LMS Student Text on "Undergraduate Algebraic Geometry", but I don't know how common this is in libraries or if CUP have reprinted it.

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... or, also beautiful Manin's "Cubic Forms" (it has a chapter entitled "27 lines").

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Is it online somewhere? –  Ilya Nikokoshev Nov 1 '09 at 19:04
Yes. It is available in Kolkhoz (Russian edition for sure, about English edition I don't know), and I have seen the English edition floating around somewhere else (one may try duansci.com/e/action/ShowInfo.php?classid=7&id=50084 though I am not sure how useful it can be). –  Pasha Zusmanovich Nov 1 '09 at 19:18
Oh, that collection. Sure. –  Ilya Nikokoshev Nov 25 '09 at 20:43

In addition to the nice answers above, here is a QDOS computation you can do in your head. If you consider a family of form XYZ + tF = 0, where F is a general cubic, then there are infinitely many lines on the three planes XYZ=0, but in the given t direction, the only ones that are limits of lines on F=0 are those passing through pairs of points where F=0 meets the axes of the system of planes XYZ = 0. Since F meets each axis 3 times, there are in each plane 3x3 = 9 lines, for a total of 27 limiting lines, hence the general F had 27 lines.

Another: In a family QH+tF = 0 where Q is quadratic, H linear, the limiting lines pass through the 6 points where F meets the conic Q=H=0. Exactly 12 lines on Q and 15 on H do this. (These occur in the book by Beniamino Segre.)

Or project the cubic to the plane from a point on itself, after blowing up that point to gain a new exceptional line. The images in P^2 of the lines on the surface, are lines everywhere tangent to the branch curve, which is a quartic. Then the 28 bitangents to a plane quartic finish the computation.

There is also a nice computation of the degree of the map from the space of lines to the space of cubics, in Mumford's yellow book, AG I, Complex projective varieties, last chapter. Basically he reduces it to the case of the fermat surface X^3 + Y^3 + Z^3 + W^3 = 0, where one can solve for the lines by hand.

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What does "QDOS" mean? –  Ravi Vakil Sep 4 '11 at 16:56
early PC term for "quick and dirty operating system". –  roy smith Sep 4 '11 at 17:10
I'd like to put in a plug for Mumford's argument and exposition, which is (as is typical of Mumford's writing) fast and devastatingly elegant. Many proofs in the usual sources (unlike Mumford) don't do everything (e.g. some just show that every smooth cubic has 27 lines, others show that most are blow-ups of $\mathbb{P}^2$ at six points (in suitably general position), and that those have 27 lines. –  Ravi Vakil Sep 4 '11 at 23:43

See the beautiful book of Fuchs and Tabachnikov, Mathematical Omnibus, for an answer.

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