Quadric and cubic surfaces

Often when you hear the term “quadric surface” or “cubic surface” this refers to a hypersurface S : F = 0 in P^3 defined by a homogeneous polynomial F ∈ K[X_0, X_1, X_2, X_3] of degree 2 or 3. As usual we say that S : F = 0 and S' : F' = 0 define the same surface if F = λ F' for some nonzero λ ∈ K. We say S is irreducible if F is irreducible. We say that S is smooth or nonsingular if the system of equations

F = F_0 = F_1 = F_2 = F_3 = 0

in X_0, X_1, X_2, X_3 has no nonzero solution over the algebraic closure of K. Here F_i is the partial derivative of F with respect to X_i.

I'm going to discuss a way to construct rational curves on S : F = 0, i.e., morphisms

φ = (G_0, G_1, G_2, G_3) : P^1 —→ P^3

which map into S, i.e., such that F(G_0, G_1, G_2, G_3) = 0 as a polynomial. My discussion of this will not be exhaustive and it will not even be a good way of doing this (I think).

Given a quadric or cubic surface S : F = 0 then we can consider C : F(Y_0, Y_1, Y_2, 0) = 0 which is a conic or a cubic in the P^2 with coordinates Y_0, Y_1, Y_2. We can more generally take 4 linear forms L_0, L_1, L_2, L_3 in Y_0, Y_1, Y_2 and consider the conic or cubic

C : F(L_0, L_1, L_2, L_3) = 0

in P^2. If L_0, L_1, L_2, L_3 span the degree 1 part of K[Y_0, Y_1, Y_2] we say that this is a plane section of S. Namely, the condition on the span means exactly that

ψ = (L_0, L_1, L_2, L_3) : P^2 —> P^3

is a morphism. We have seen in our earlier work that if S is a quadric surface, then the curve C can be parametrized or contains a line, i.e., there is a nonconstant map

χ = (H_0, H_1, H_2) : P^1 —> P^2

into C where H_0, H_1, H_2 ∈ K[Z_0, Z_1] are homogeneous of degree 2. Combining this with ψ we obtain

ψ o χ : P^1 —> P^3

whose image is contained in S. What does this mean? It just means that

ψ o χ = (L_0(H_0, H_1, H_2), …)

in other words, you substitute H_i for Y_i. If S is a cubic surface, then you can try to do the same thing by finding L_i such that F(L_0, L_1, L_2, L_3) is singular and using the parametrization of singular cubic curves.

Exercise 24: Use the ideas above to prove (from scratch) that a cubic surface over an algebraically closed field has at least one rational curve on it.

It turns out that both quadrics and cubics always have lines over the algebraic closure. A line on S is just a morphism

φ = (G_0, G_1, G_2, G_3) : P^1 —→ P^3

of degree 1 which map into S. It is a marvelous classical fact that a nonsingular cubic has exactly 27 lines over an algebraically closed field.

Exercise 25: Work as a group to give a guess as to the possible numbers of lines nonsingular quadric surface over Z/pZ has. (It turns out not every quadric surface has the same number of lines.)

Exercise 26: (Too hard probably/just a curiosity.) Can you find a cubic surface over Z/pZ which has 27 lines over Z/pZ?

Exercise 27: The methods above give low degree rational curves. Can you think of methods for finding higher degree rational curves on the cubic surface S : X_0^3 + X_1^3 + X_2^3 + X_3^3 = 0 over Z/2Z? This is relevant for our project as we later need to do the same thing for the degree 5 Fermat equation in P^5. Thus I suggest you really try this: first think of algorithms! Don't implement algorithms that take a lot of steps.

Continue reading why rational curves?. Back to the start page.

 
surfaces.txt · Last modified: 2012/06/06 10:13 by johan
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