COMPLEX ALGEBRAIC SURFACES CLASS 14

RAVI VAKIL

Contents

1. Combinatorial aspects 1 2. del Pezzo surfaces 2 3. Cubic surfaces 3 3.1. Automorphisms of the intersection graph 5 3.2. Every smooth cubic has 27 lines, and is the blow-up of ℙ^{2} at 6 points 5

Last day we looked at a lot of rational surfaces. We made use of:

Useful proposition. Consider the blow-up of ℙ^{2} at n general points, giving exceptional divisors E_{1}, …, E_{n}. Then the intersection ring on ℙ^{2} is given by

ℤ[H, E_{1}, …, E_{n}]/H^{2} = 1, HE_{i} = 0, E_{i} · E_{j} = 0, E_{i}^{2} = −1.

We can understand divisors and sections of divisors in terms of divisors on ℙ^{2} with certain multiplicities at the E_{j}. More precisely: the vector space of sections of aH − b_{1}E_{1} − ⋯ − b_{n}E_{n} is naturally isomorphic to the vector space of degree a polynomials in ℙ^{2} vanishing with multiplicity at least b_{i} on E_{i}.

We also proved the following. Suppose S is a surface, and −K_{S} gives a map to projective space. Then (−1)-curves map isomorphically onto lines. Conversely, if S is a surface, and −K_{S} gives a map to projective space, then any curve mapping isomorphically onto a line is a (−1)-curve.

Today: Cubic surfaces. But first, some interesting combinatorial remarks, due to Tyler, Diane and others.

1. Combinatorial aspects

Proposition. The automorphism group of the Peterson graph is S_{5}. (I haven’t drawn the Peterson graph in these notes, sorry!)

Proof. Two (equivalent) proofs. Diane’s: To each vertex put a size 2 subset of {1, …, 5}. Join them by an edge if the don’t intersect.

Date: Friday, November 15.

Tyler’s: Label the edges as follows (3 have label 1, etc.). Observe that as soon as you’ve labeled one edge with a number, the other two are determined: they are those that are distance 3 from it. Hence if G is the automorphism group of the graph, we have a morphism G → S_{5}. It is surjective: you can get a 5-cycle (rotate) and a 2-cycle (do it). It is injective: suppose you have an automorphism fixing the colors. We’ll show that the three 1-labeled edges are fixed. Look at the two pairs of edges each 1-edged meets; note that this gives a partition of {2, 3, 4, 5} into two couples. There are 3 ways to do this, and they correspond to the 3 edges.

Connection between Diane’s and Tyler’s: in Diane’s construction: label an edge with the number missing in its vertex labels.

If you remember, I described a pattern of automorphism groups of the intersection graphs of the blow-up of ℙ^{2} at n up to 9 points. The answer was:

\begin{matrix}n = & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ dim Aut & [S_{3} × S_{2}] & W (A_{4}) = S_{5} & W (D_{5}) & W (E_{6}) & W (E_{7}) & W (E_{8}) & W (E_{9}) \end{matrix} .

I expressed disappointment that S_{3} × S_{2} was not isomorphic to D_{12}, the symmetries of hexagon, because that’s what the correct answer was. But in fact:

Proposition (Tyler). S_{3} × S_{2} ≅ D_{12}.

Proof. x = (12)(12) and y = (12)(123) generate S_{3} × S_{2}. They also satisfy the relations defining D_{12}: x^{2} = y^{6} = e, xyx = y^{−1}.

Remark. The map D_{12} → S_{3} corresponds to the induced permutation of the 3 diagonals. The map D_{12} → S_{2} corresponds to the permutation of the two inscribed equilateral triangles.

2. del Pezzo surfaces

Another remark about these surfaces.

Definition. A surface S is a del Pezzo surface if −K_{S} is ample.

This is yet another ancient idea that remains important. Most recently, they have come up in physics.

Examples from last day: ℙ^{1} × ℙ^{1}, and the blow-up of ℙ^{2} at up to 8 general points. Let’s get rid of that nasty word “general”, by making more precise which points you throw out.

Proposition. The blow up of ℙ^{2} at up to 8 distinct points, no 3 on a line and no 6 on a conic, is a del Pezzo surface.

Proof. Check using our useful proposition that −K or −2K is very ample. If n ≤ 6, −K is ample, i.e. cubics vanishing at the n points gives something that separates points and tangent vectors. If n = 7 or 8, −K isn’t very ample, but −2K is: the linear system corresponding to sextics vanishing at the n points separate points and tangent vectors.

Proposition. The blow up of ℙ^{2} at up to 8 distinct points, with 3 on a line or 6 on a conic, is not a del Pezzo surface.

Proof. First, note that a del Pezzo surface can’t have a (-2)-curve, i.e. a genus 0 curve C with self-intersection ≤ −2. Reason: genus formula gives

−2 = 2g − 2 = C · (K + C) = C · K + C · C ≤ C · K − 2

so 0 ≤ C · K. But −K is ample, so K · C < 0 for all C.

If at least there are at least 3 points on a line, then there is a genus 0 curve of selfintersection at most −2.

Similarly, if there are at least 6 points on a conic, then there is a genus 0 curve of selfintersection at most −2.

Remark: Slight extension. You can allow “infinitely near” points: blow up p_{1}, and blow up a point p_{2} on the exceptional divisor E_{1}; somewhat archaic (but still-used) terminology is that p_{2} is an “infinitely near point” to E_{2}.

But as soon as you blow up a point on an exceptional divisor, you have a genus 0 curve of self-intersection at most -2, and hence it can’t be a del Pezzo surface.

Theorem. The only del Pezzo surfaces are the above blow-ups of ℙ^{2} (up to 8 points, no 3 on a line, no 6 on a conic), and ℙ^{1} × ℙ^{1}.

Proof: later in the course.

3. Cubic surfaces

Blow up ℙ^{2} at six points. 27 (-1)-curves. Reminder of where they are: 6 exceptional divisors E_{i}. \binom{6}{2} lines L_{ij}. 6 conics C_{ijklm}.

Anticanonical map gives embedding in to ℙ^{3}. (Remember: (−1)-curves ↔ lines.)

Get smooth cubic surface in ℙ^{3} with lines. We’ll see: (i) there are no more lines on this surface, (ii) that almost all smooth cubic surfaces are ℙ^{2} blown up at six points and hence have 27 lines, (iii) all smooth cubic surfaces have 27 lines, and (iv) all smooth cubic surfaces are ℙ^{2} blown up at six points.

Proposition. We have found all the lines.

Proof. Diophantine equations. Suppose D = aH − ∑ b_{i}E_{i} is a line, i.e. a (−1)-curve.

First case: some b_{i} < 0. Then D · E_{i} = −b_{i} < 0. But D is the class of a line, so D · E_{i} ≥ 0 unless the line and E_{i} have a component in common. Thus they must be equal, i.e. D = E_{i}. Hence we’ve found six lines: E_{1}, …, E_{6}.

Second case: all b_{i} ≥ 0. Also note that a > 0.

Then degree = 1 means (3H − ∑ E_{i}) · D = 1, from which 3a − ∑ b_{i} = 1.

D^{2} = −1 implies a^{2} − ∑ b_{i}^{2} = −1.

Recall (or re-prove) the quadratic mean - arithmetic mean inequality:

\sqrt{(\frac{∑ b_{i}^{2}}{6} )} ≥ \frac{∑ b_{i}}{6}

from which ∑ b_{i}^{2} ≥ (∑ b_{i})^{2}/6. Thus

a^{2} = ∑ b_{i}^{2} − 1 ≥ (∑ b_{i})^{2}/6 − 1 = (3a − 1)^{2}/6 − 1

from which 6a^{2} + 6 ≥ 9a^{2} − 6a + 1 ⇒ 0 ≥ 3a^{2} − 6a − 5 ⇒ 8 ≥ 3(a − 1)^{2}. Thus a = 1 or 2.

If a = 1, then we are considering the class of lines, then we get ∑ b_{i} = 2 and ∑ b_{i}^{2} = 2, from which two of the b_{i}’s are 1 and the rest are 0.

If a = 2, we get ∑ b_{i} = 5 and ∑ b_{i}^{2} = 5. The five b_{i}’s must be equal to 1, and the last equal to 0.

Proposition. Almost all smooth cubic surfaces are ℙ^{2} blown up at six points.

Proof. By dimension count. Dimension of space of cubic surfaces: count cubic equations in 4 variables (20). Subtract 1 to projectivize, to get 19. There is a ℙ^{19} parametrizing all smooth cubic surfaces. An open set S corresponds to the smooth ones.

Now how many ways can we get a cubic surface by blowing up six points in the plane? Choose six points in the plane (12 dimensions), except mod out by the automorphisms of the plane dim P GL(3) = 8. Then map to projective space using four linearly independent sections of 3H − ∑ E_{i}. There is a four-dimensional space of sections, so we have 16 dimensions of choice of 4 sections. But any multiple of such a 4-tuple gives the same embedding, so −1. Get:

12 − 8 + 16 − 1 = 19.

Thus we have a 19-dimensional family!

It’s not an open subset yet; there may be, and in fact are, many ways of representing the surface as a blow-up of six points. But there are only a finite number: any such description corresponds to six lines on the surface, no 2 intersecting. So the image of this 19-dimensional family lies in S, and must be dense in S.

We’ll see soon: Every smooth cubic has 27 lines, and is the blow-up of ℙ^{2} at 6 points 3.1. Automorphisms of the intersection graph. Again, we can make a graph of the combinatorial structure of the 27 lines. It again is highly symmetric. Theorem. Its automorphism group is W (E_{6}), a finite group of order 51840 = 2^{7} × 3^{4} × 5. Sketch of proof. Recall the definition of W (E_{6}). 6 skew lines gives full structure. Get map to ℙ^{2}. See S_{6} in it. Last one: Cremona transformation.

Interpretation of this group in two ways:

Geometrically: as the monodromy of the 27 lines as you make loops in S.

Algebraically: Let M be the variety parametrizing (surface, line). There is a morphism from M to S, of degree 27. M turns out to be irreducible. Hence we have a field extension of degree 27 k(S) ⊂ k(M). If we take the Galois closure of the this field extension, we get a field extension of order 51840, and its Galois group is W (E_{6}).

Next day we’ll show that:

3.2. Every smooth cubic has 27 lines, and is the blow-up of ℙ^{2} at 6 points. Let S be a smooth cubic surface.

Earlier today (in response to a question of Eric’s) I said: Proposition. S is anticanonically embedded.

Proof: Adjunction formula. S has degree 3, so K_{S} = O_{ℙ^{3}}(−4 + 3)|_{S}.

Hence lines correspond to (−1)-curves.

The strategy of proof that every smooth cubic is a blow-up of ℙ^{2} at 6 points is as follows. Proposition. Every cubic surface (even singular ones) contains a line. Proposition. Given any line l in S, there are exactly 10 other lines in S meeting l (distinct from l). These fall into 5 disjoint pairs of concurrent lines. Corollary. Any smooth cubic contains two skew lines. Finally: Theorem. S is the blow-up of ℙ^{2} at 6 points.

We’ll see instead that S is the blow-up of ℙ^{1} × ℙ^{1} at 5 points. We’d seen earlier that ℙ^{2} blown up at 2 points is ℙ^{1} blown up at one point, so we’ll be done. Proof. Let l and l′ be disjoint lines. We define rational maps ϕ : l × l′ −−→ S and ψ : S −−→ l × L′ as follows: if (p, p′) is a generic point of l × l′, the line ⟨p, p′⟩ meets S in a third point p′′. Define ϕ(p, p′) = p′′. For s ∈ S − l − l′, set p = l ∩ ⟨s, l′⟩, p′ = l′ ∩ ⟨s, l⟩ and put ϕ(s) = (p, p′). It is clear that ϕ and ψ are inverses. Moreover, ϕ is a morphism: we can define at a points of l (or l) by replacing the plane ⟨s, l⟩ by the tangent plane to S at s (checking that this gives a morphism, explain). Thus ψ is a birational morphism, and is a composite of blow-ups. Which curves are blown-down? Those lines meeting both l and l′.

We’ll see why there are precisely 5 of them, one of each the pairs described above.