In the previous post, we introduced the Fano scheme of a subscheme of projective space, as the Hilbert scheme of planes of a certain dimension on that subscheme. In this post, I’d like to work out an explicit example, of the 27 lines on a smooth cubic surface in $\mathbb{P}^3$; as we’ll see, the Fano scheme is 27 reduced points, and the count can be made with a little calculation on the Grassmannian. Although the calculation is elementary, I found it worthwhile to work carefully through it, not only for its intrinsic interest but also as motivation for the study of intersection theory on moduli spaces in general. Once again, most of this material is from Eisenbud-Harris’s draft book 3264 and All That.

1. The normal bundle as self-intersection

Suppose ${X = S}$ is a smooth surface, imbedded in some projective space, and consider the scheme ${F_1 S}$ of lines in ${S}$.

Fix a line ${L}$ in $S$. In this case, the normal sheaf ${N_{S/L}}$ is actually a vector bundle of normal vector fields, given by the adjunction formula

$\displaystyle N_{S/L} = \left(\mathcal{I}_L/\mathcal{I}_L^2\right)^{\vee} = \left(\mathcal{O}_S(-L)/\mathcal{O}_S(-2L)\right)^{\vee} = \mathcal{O}_L(L).$

In particular, ${N_{S/L}}$ is a line bundle on ${L \simeq \mathbb{P}^1}$ and has a well-defined degree. This degree is in fact the self-intersection ${L.L}$ of ${L}$, considered as a divisor on the smooth surface ${S}$.

To see this, let’s recall the definition of the intersection multiplicity on a smooth surface: to find ${L.L}$, one needs to compute the Euler characteristic

$\displaystyle L.L = \chi( \mathcal{O}_L \stackrel{\mathbb{L}}{\otimes_{\mathcal{O}_S}} \mathcal{O}_L ),$

where the tensor product is taken in the derived sense. In other words, the “derived tensor product” ${\mathcal{O}_L \stackrel{\mathbb{L}}{\otimes_{\mathcal{O}_S}}\mathcal{O}_S}$ accounts for the fact that transversality fails. To compute this, we can use the resolution on ${S}$,

$\displaystyle 0 \rightarrow \mathcal{O}_S(-L) \rightarrow \mathcal{O}_S \rightarrow \mathcal{O}_L \rightarrow 0,$

and tensor with ${\mathcal{O}_L}$ to get that the derived tensor product is represented by the two-term complex

$\displaystyle \mathcal{O}_L(-L) \rightarrow \mathcal{O}_L .$

It follows that the Euler characteristic is given by

$\displaystyle L.L = \chi(\mathcal{O}_L) - \chi(N_{S/L}^{\vee}) = \deg N_{S/L},$

by Riemann-Roch. (This is not specific to lines in ${S}$.)

Geometrically, the degree of the normal bundle on ${L}$ is a measure of its “positivity:” a greater degree indicates more sections, which in turn indicates that ${L}$ can be (at least infinitesimally) deformed to a greater degree. This in turn should correspond to the positivity of the intersection multiplicity: the statement ${L.L < 0}$ implies that ${L.L}$ cannot be deformed into general position.

In general, we have one more piece of information about the self-intersection ${L.L}$ if we know the surface ${S}$. Namely, we have the adjunction formula

$\displaystyle K_L \simeq \mathcal{O}_{\mathbb{P}^1}(-2) = K_S|_L \otimes \mathcal{O}_L(L),$

and, taking degrees, this implies that

$\displaystyle -2 = K_S . L + L.L,$

where ${K_S}$ is the divisor of the canonical line bundle on ${S}$.

Suppose that ${S \subset \mathbb{P}^3}$ is a surface of degree ${d}$, so that we can use adjunction again to conclude that ${K_S = (d - 4) H}$ for ${H = \mathcal{O}_S(1)}$ the hyperplane class. In this case, since ${H.L = d}$, we get

$\displaystyle -2 = (d-4) + L.L,$

so that ${L.L = 2 - d}$. As ${d \rightarrow \infty}$, this suggests that the surface ${S}$ is less and less likely to contain lines, or at least that they will be extremely “rigid.”

Another interpretation of this is that, once ${d > 3}$, the Hilbert scheme of curves on ${S}$ is smooth at ${L}$, and is a (reduced) point near ${L}$: that is, more generally, the Fano scheme ${F_1 S}$ consists of reduced points. In fact, the negativity of the normal bundle (${H^0( N_{S/L}) = 0}$) implies that there are no first-order deformations of ${L}$, so that the tangent space of ${F_1 S}$ vanishes at ${L}$.

In fact, a very general surface of degree ${d \geq 4}$ in ${\mathbb{P}^3}$ contains only divisors of degrees dividing ${d}$: the Picard group is generated by the hyperplane class ${\mathcal{O}(1)}$, by a theorem of Noether and Lefschetz. (In higher dimensions, Grothendieck’s version of the Lefschetz hyperplane theorem implies that the Picard group of a smooth hypersurface is always generated by ${\mathcal{O}(1)}$, but in dimension ${2}$, one needs ${d \geq 4}$ and “very general.”)

3. Counting

Let ${S}$ be a smooth cubic surface, so that ${S}$ is the zero locus in ${\mathbb{P}^3}$ of a section ${s \in H^0( \mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(3))}$. Our goal in this section is to analyze the scheme ${F_1 S}$ of lines on ${S}$. In the previous section, we saw that ${F_1 S}$ is always reduced and finite: in fact, by the analysis there, any line ${L \subset S}$ has self-intersection ${-1}$.

In the previous post, we saw another computationally useful expression for ${F_1 S}$ as a subscheme of the Grassmannian ${\mathbb{G}(1, 3)}$ of lines in ${\mathbb{P}^3}$: ${F_1 S}$ is the zero locus in ${\mathbb{G}(1, 3)}$ of a certain section ${\sigma}$ of a certain four-dimensional vector bundle ${\mathcal{V}}$ on ${\mathbb{G}(1, 3)}$. The vector bundle in question assigned to each line ${L \subset \mathbb{P}^3}$ the global sections

$\displaystyle L \mapsto H^0( L, \mathcal{O}_L(3));$

that is, it assigned to ${L}$ the restriction of all the cubic polynomials in ${\mathbb{P}^3}$ to ${L}$. (As we saw, this vector bundle was well-defined and could be defined as a direct image.) Since ${s}$ is a global section of ${\mathcal{O}_{\mathbb{P}^3}(3)}$, it naturally defines a section ${\sigma}$ of ${\mathcal{V}}$.

The zero-locus, both set-theoretically and scheme-theoretically, of ${\sigma}$ defines precisely the scheme ${F_1 S}$ of lines in ${S}$. Now, the statement that ${F_1 S}$ is reduced amounts precisely to saying that the section ${\sigma}$ of ${\mathcal{V} \rightarrow \mathbb{G}(1, 3)}$ is transverse to the zero section: in other words, the number of points in the zero locus is precisely the top Chern class (Euler class) of ${\mathcal{V}}$, integrated over ${\mathbb{G}(1, 3)}$. So, to count the number of lines on ${S}$, we need to compute ${c_4( \mathcal{V})}$! In particular, the answer we’ll get is independent of the smooth surface ${S}$, and it’ll require a calculation on the Grassmannian.

4. The Grassmannian

The Grassmannian ${\mathbb{G}(1, 3)}$ is a four-dimensional smooth variety (it is a quadric hypersurface in ${\mathbb{P}^5}$), and its cohomology or Chow ring has concrete generators given by the Schubert cycles. Fix a point ${p \in \mathbb{P}^3}$, a line ${\ell \subset \mathbb{P}^3}$, and a 2-plane ${\Lambda \subset \mathbb{P}^3}$ which are “general.”

Then one has a natural hypersurface in the Grassmannian given by

$\displaystyle \Sigma_{\ell} = \left\{L: L \cap \ell \neq \emptyset\right\},$

consisting of lines meeting ${\ell}$. (In fact, it is the intersection of the Grassmannian with a hyperplane under the Plücker embedding ${\mathbb{G}(1, 3) \subset \mathbb{P}^5}$.) There are natural codimension two loci

$\displaystyle \Sigma_{p} = \left\{L: p \in L\right\} , \quad \Sigma_H = \left\{L: L \subset H\right\},$

and a codimension three subvariety

$\displaystyle \Sigma_{p, H} = \left\{L: p \in L \text{ and } L \subset H\right\} .$

It is a basic fact that the Chow ring (or cohomology ring) of the Grassmannian is the free module on these four classes, together with the fundamental class and ${1}$. In other words

$\displaystyle H^*(\mathbb{G}(1, 3); \mathbb{Z}) = \mathbb{Z}\left\{1, \Sigma_{\ell}, \Sigma_p, \Sigma_H, \Sigma_{p, H}, [\ast]\right\},$

where ${[\ast]}$ is the fundamental class (i.e., the class of a point). Moreover, one can compute the ring structure by intersecting cycles in general position: for instance, clearly

$\displaystyle \Sigma_p . \Sigma_H = \Sigma_{p. H}.$

Similarly,

$\displaystyle \Sigma_p^2 = [\ast], \quad \Sigma_H^2 = [\ast], \quad \Sigma_p . \Sigma_H = 0,$

because, for instance, the first intersection consists of lines passing through two general points ${p, p'}$. The third intersection is zero if ${p \notin H}$.

Less clearly,

$\displaystyle \Sigma_{\ell}^2 = \Sigma_p + \Sigma_H. \ \ \ \ \ (1)$

Here is an informal argument for this. To compute ${\Sigma_{\ell}^2}$, we take lines ${\ell, \ell'}$ in general position and compute the intersection of cycles ${\Sigma_{\ell} \cap \Sigma_{\ell'}}$, which consists of lines ${L}$ that meet two general lines ${\ell, \ell'}$. However, instead of taking ${\ell, \ell'}$ in “truly” general position, we take them simply distinct and meeting at a point ${p}$; then the intersection of cycles consists of lines that either pass through the intersection ${\ell \cap \ell'}$ or through the plane that ${\ell, \ell'}$ span.

More precisely, to show that ${\Sigma_{\ell}^2 = \Sigma_p + \Sigma_H}$, one can use Poincaré duality: it suffices to compute the intersection of both sides with ${\Sigma_p}$ and ${\Sigma_H}$. Now

$\displaystyle \Sigma_{\ell}^2 . \Sigma_p, \quad \Sigma_{\ell}^2 . \Sigma_H,$

both consist of single points by choosing two general lines and a general point or plane. For instance, ${\Sigma_{\ell}^2 . \Sigma_p}$ is represented by lines that pass through a point and through two general lines ${\ell, \ell'}$: that means the line has to be in the intersection of the planes spanned by ${\left\{p, \ell\right\}}$ and ${\left\{p, \ell'\right\}}$.

Example 1 This calculation implies that

$\displaystyle \Sigma_{\ell}^4 = (\Sigma_p + \Sigma_H)^2 = 2[\ast],$

or that there are two lines in ${\mathbb{P}^3}$ passing through four general lines.

Let’s now see how the Chern classes of the two-dimensional tautological bundle ${\mathcal{V}}$ on ${\mathbb{G}(1, 3)}$ given by ${L \mapsto H^0(L, \mathcal{O}_L(1))}$ look in this basis. By definition, a section of ${H^0( \mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(1))}$ gives a section of ${\mathcal{V}}$ whose zero locus is precisely the lines contained in a hyperplane: so

$\displaystyle c_2(\mathcal{V}) = \Sigma_H.$

Given two linearly independent elements of ${H^0( \mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(1))}$, defining two hyperplanes in ${\mathbb{P}^3}$, the degeneracy locus of the two induced sections of ${\mathcal{V}}$ consist of lines ${L \subset \mathbb{P}^3}$ on which the restrictions of the two hyperplanes intersect: that is, lines ${L}$ which meet the intersection of the two hyperplanes. So

$\displaystyle c_1(\mathcal{V}) = \Sigma_{\ell}.$

Using this, we can compute ${c_4( \mathrm{Sym}^3 \mathcal{V})}$ (which is the vector bundle ${L \mapsto H^0(L, \mathcal{O}_L(3))}$) using the splitting principle. Namely, if we write formally for the “Chern roots” of ${\mathcal{V}}$ the set ${\left\{t_1, t_2\right\}}$, then the Chern roots of the symmetric cube are ${3t_1 ,2t_1 + t_2 , t_1 + 2t_2, 3t_3}$, so the Euler class is

$\displaystyle 9 t_1 t_2 (t_1 + 2t_2)(2t_1 + t_2) = 9 c_2( 2c_1^2 + c_2) ,$

by expressing in terms of the elementary symmetric polynomials. In our case, this means that

$\displaystyle c_4(\mathrm{Sym}^3 \mathcal{V}) = 9 \Sigma_H ( 2 \Sigma_{\ell}^2 + \Sigma_H) = 27,$

by the previous formulas, and we get the twenty-seven lines on a cubic surface, as desired.