## The Twenty-Seven Lines on the Cubic Surface

Ok, I just came across this proof, and it’s so cool I have to blog about it. This is not related to my ongoing series, nor is this anything to do with the conference I’m at (I won’t be blogging about this conference…though probably will for next year’s conferences).

Previously, I’d seen two proofs of the fact that a cubic surface has twenty-seven lines on it. The first and more elementary one involved using resultants to write down a polynomial that has a root for every line on a surface, and showed that it had degree 27 and no multiple roots. The second involved the hard route of showing that a cubic surface is the projective plane blown up at six points not lying on a conic, and then showing that the six conics, six exceptional divisors, and fifteen lines between pairs of blownup points all give lines, and nothing else does.

This proof is better than those, though it requires a little bit of background, but not so much that I can’t turn it into a seminar talk at some point in the future…

Take a cubic surface $S$. That’s a degree three equation $F(x,y,z,w)=0$. Now, a line lies on the surface if and only if $F$ vanishes along the line. We need a way to count these lines. Now, as a cubic surface, it sits in $\mathbb{P}^3$. There’s a nice space parameterizing all of the lines in $\mathbb{P}^3$, we call it the Grassmannian $Gr(2,4)$. Strictly, it consists of all two dimensional subspaces of $\mathbb{C}^4$. So the problem is to identify how many points of $Gr(2,4)$ lie on $S$.

Now, we investigate this Grassmannian a bit more closely. It naturally has a rank 2 vector bundle on it $Q$ which assigns to each point the two dimensional subspace of $\mathbb{C}^4$ that represents it. Now, it turns out that what we want to get at is its Chern classes. Or not exactly its Chern classes, but its third symmetric power. The reason for this is that the third symmetric power consists of degree three polynomials in two variables. Now, $F$ is a section of this vector bundle, and the section will vanish exactly at the points which represent lines where $F$ restricts to zero, that is, lines in $S$.

Now, the top Chern class, when integrated, gives how many points this is. So all we need to do is compute the top Chern class of $Sym^3(Q)$ on the Grassmannian, and then integrate it. To do this, we first need to work out the total Chern character of $Q$.

To compute the second Chern class of $Q$, we need a section. Taking a linear polynomial works, so over each point, we give it the value of the linear polynomial restricted to the line. When is this zero on $Gr(2,4)$? Precisely on the lines where the linear form vanished, that is, the ones in the hyperplane. We’ll denote this set by $\sigma_{1,1}$. The first Chern class is harder to work out, but it turns out to be $\sigma_1$ as defined below.

So now we need one more thing: the multiplicative structure on the cohomology of $Gr(2,4)$. This is an example of Schubert Calculus, which can be used to solve many problems. It’s a well known fact (meaning that I don’t want to explain it right now) that $Gr(2,4)$ has cohomology of rank 1 in degrees 0,2,6 and 8, rank 2 in degree 4, and zero otherwise, and is generated by Schubert Classes, which we will now give explicit descriptions of. The rank of a Schubert class $\sigma_{a,b}$ is going to just be $2(a+b)$. Now, the degree zero generator is $\sigma_0=1$, which is the cohomology class of the whole Grassmannian. All the others will in fact be classes of subvarieties called Schubert varieties.

To define the degree 2 generator, we fix a line $L$. Then $\sigma_1(L)=\{\ell\in Gr(2,4)|\ell\cap L\neq \emptyset\}$, so it’s the collection of lines intersecting a given line. For cohomology purposes, which line we chose doesn’t matter. The same with points and planes later, so we’ll just call this $\sigma_1$. For the degree 4 generators, $\sigma_2$ is the locus of lines passing through a specified point, whereas $\sigma_{1,1}$ are the lines contained in a given plane. For degree 6, we have $\sigma_{2,1}$ which are the lines in a plane containing a given point, and in degree 8 we have $\sigma_{2,2}$, which is the class of a point. This may seem a bit arbitrary, but it can be explained quite handily with flags and partitions of integers, and I WILL discuss this when I get around to doing Schubert calculus thoroughly.

So now the point remaining is to work out multiplication, which will just be intersection with different points, lines and planes chosen to define the classes being multiplied. There’s a combinatorial way to do this, but for this example we don’t need to go into the full multiplication rule, but for now I will mention the Pieri rule: if we have $\sigma_a\sigma_{b,c}$, then the product is just the sum of the classes $\sigma_{d,e}$ with $d\geq b$, $e\geq c$, $d\geq e$ and $d+e=a+b+c$. And finally, we note that the integral of $\sigma_{2,2}$ is just one.

Now all that’s left is the splitting principle. It lets us pretend that $Q$ is a product of line bundles, and so we write $(1+\alpha)(1+\beta)=1+\sigma_1+\sigma_{1,1}$ for the Chern character. Then $\alpha+\beta=\sigma_1$ and $\alpha\beta=\sigma_{1,1}$. And as we’re working on $Sym^3(Q)$ we pretend it’s $Sym^3(V_1\oplus V_2)$, and so we get $V_1^3\oplus V_2^2\otimes V_1\oplus V_1\otimes V_2^2\oplus V_2^3$.

The top Chern class of this will then be the product of the top Chern classes of the factors in the tensor product, and so we get $(3\alpha)(2\alpha+\beta)(\alpha+2\beta)(3\beta)$. On first multiplication and doing all the replacement possible, we get $9\sigma_{1,1}(\alpha+\sigma_1)(\beta+\sigma_1)$. This multiplies to $9\sigma_{1,1}(\alpha\beta+\alpha\sigma_1+\beta\sigma_1+\sigma_1^2)$, which is $9\sigma_{1,1}(\sigma_{1,1}+(\alpha+\beta)\sigma_1+\sigma_1^2)$. And so it reduces to $9\sigma_{1,1}(\sigma_{1,1}+2\sigma_1^2)=9\sigma_{1,1}^2+18\sigma_{1,1}\sigma_1^2$. Working with the Pieri rule and just doing intersection for $\sigma_{1,1}^2$ (which is just that there is a unique line contained in the intersection of two planes) we get $27\sigma_{2,2}$, which integrates to 27. So there are then 27 lines on a cubic surface.

Of course, the blowup method gives their configuration and a few other nice things, but still, I like this method, because coming up with the number 27 is an almost trivial calculation once you’ve set things up, whereas with blowing up it’s nontrivial to prove that there are ONLY 27, and with the resultant, it’s not so easy to prove that there actually are twenty-seven distinct lines. Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
This entry was posted in Algebraic Geometry, Algebraic Topology, Combinatorics, Enumerative Geometry. Bookmark the permalink.

### 13 Responses to The Twenty-Seven Lines on the Cubic Surface

1. D. Eppstein says:

Re your final line: so why not do the blowup to lower bound the number of lines, do the resultant to upper bound the number of lines, and observe that the two bounds are equal?

2. Charles says:

That’s another way to get the number twenty seven, but that argument is for that precise fact, and isn’t useful in other places (at least, not that I’ve seen).

I like this way better partly because it generalizes. I adapted it from the method of proving that there are 2875 lines on a quintic three-fold, but more generally, this method can be adapted to count the number of k-planes on a degree d hypersurface of projective space without much trouble (so long as the number is finite).

3. Max says:

Reminds me of the paper “The Moduli Space of Curves and Its Tautological Ring” by Ravi Vakil in the Notices. In a small part of it he explains how the question “How many lines in C^3 meet four given lines L_1,…,L_4” is really a question about the cohomology of the Gr(2,4) (more precisely, computing \sigma_1^4) – an even simpler example in the same spirit.

4. Charles says:

I haven’t seen that paper (though now I’m definitely going to look it up), but that’s precisely the same thing, it’s Schubert calculus on Gr(2,4), which has been something people have worked on since the 1800s. Enumerative geometry is very classical, and thanks to string theory is coming back into fashion. And in fact, as far as I know, the number of lines intersecting four given lines is the simplest example that you actually need Grassmannians for.

5. Matt says:

Your proof shows that the subvariety of the Grassmannian consisting of lines on a cubic hypersurface has degree 27, (when the variety is 0-dimensional, which it always is for a cubic surface) — not that there are 27 distinct lines. You mention the quintic threefold case; already in this case there are smooth quintic threefolds with infinitely many lines even though the computation tells you 2875.

6. Charles says:

Well, true. The missing bit is that a generic cubic surface will have 27 lines, and a generic quintic threefold has 2875. I don’t know as much about the threefold as I should…do you have an example of a smooth quintic threefold with infinitely many lines? And if its coefficients are perturbed, does it go to one with finitely many?

7. Matt says:

Yes! For example consider the Fermat quintic:

x_0^5 + x_1^5 + x_2^5 + x_3^5 + x_4^5 = 0

This is smooth. Suppose there’s a point in \PP^2 [a,b,c] such that a^5 + b^5 + c^5 = 0.

Then the line [s,t] \rightarrow [s, -s, t*a, t*b, t*c] will be contained on the Fermat quintic. As the collection of [a,b,c] satisfying the equation x^5 +y^5 + z^5 = 0, is one dimensional, you will get a one dimensional family of lines on the Fermat. In fact, you will have many components of one dimensional families of lines (all described in a similar way).

As the moduli space of degree 5 hypersurfaces in \PP^4 is connected, the coefficients can be perturbed to one with finitely many lines.

8. Charles says:

Thanks Matt. Sorry it took so long to respond, have been running around. And now, of course, I feel stupid, because when my class did the basics of mirror symmetry last semester, we discussed the Fermat quintic in detail, so I should have known that. Oh well, one of the advantages of being a student is that my mistakes (except perhaps on Orals this Fall) aren’t that big a deal, so long as I get the point eventually.

9. Seth Neel says:

where can i find a proof using the resultant?

• Charles Siegel says:

The reference where I learned it was, if I recall correctly, Hulek’s “Elementary Algebraic Geometry.”

10. Anonymous says:

I just counted the edges of a cube. There are 12.