REMARKS ON JORDAN ALGEBRAS (DIM 9, DEG 3), CUBIC SURFACES, AND DEL PEZZO SURFACES (DEG 6)

MARKUS ROST

The purpose of these notes is to record some formulas and remarks.
Everything deserves a check.

1. A construction of Jordan algebras (deg 3, dim 9)

The base field F has char ≠ 3.

Let us ask for a functorial construction

𝔹: (L, K) ↦ B(L, K)

which associates to an ordered pair of separable degree 3 extensions a 9-dimensional Jordan algebra of degree 3. Consider the split cases L = K = F ⊕ F ⊕ F and let B̃ = B(L, K). The functoriality of 𝔹 then yields a homomorphism

Ψ_{𝔹}: S_{3} × S_{3} = Aut(L) × Aut(K) → Aut(B̃).

Clearly 𝔹 is determined by B̃ and Ψ_{𝔹}.

Here is an example: Let

Z = F [x]/(x^{2} + x + 1), σ: Z → Z, σ(x) = x^{2}

and let

A = M_{3}(Z), τ: A → A, τ(a) = σ(a)^{t}.

Then (A, τ) is an algebra with involution of second kind. Put

B̃ = A^{τ} .

There are the (split) subalgebras

L = (\begin{matrix}F \\ & F \\ & & F \end{matrix} )

and

K = \frac{1}{3}(1 + β + β^{2})F ⊕ \frac{1}{3}(1 + xβ + x^{2}β^{2})F ⊕ \frac{1}{3}(1 + x^{2}β + xβ^{2})F

where

β = (\begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix})

Let G be the subgroup of Aut(B̃) (as Jordan algebra) leaving the subalgebras L and K invariant. The natural homomorphism G → Aut(L) × Aut(K) turns out to be an isomorphism.

Date: December 17, 1996, minor changes on July 29, 2003.

Lemma 1. Let 𝔹 be as above and suppose that Ψ_{𝔹} is injective. Then 𝔹 is isomorphic to the functor given by the example. □

Lemma 2. The last lemma can be extended to unordered pairs of cubic extensions, that is to say to a cubic extension over a quadratic extension. The underlying group is then (S_{3} × S_{3}) ⋊ ℤ/2.

One would like to see a rational description of B(L, K) for arbitrary K and L. Here it is for char F ≠ 2, 3:

Put

B = B(L, K) = L ⊗ K.

Let L_{0} ⊂ L, K_{0} ⊂ K be the subspaces of trace 0 elements. Define a Jordan product on B by the following formulas with α ∈ L_{0} and β ∈ K_{0}.

(1 ⊗ 1)^{2} = 1 ⊗ 1
(α ⊗ 1)^{2} = α^{2} ⊗ 1
(1 ⊗ β)^{2} = 1 ⊗ β^{2}
(α ⊗ β) (α ⊗ 1) = \frac{1}{4}(trace(α^{2}) − 2α^{2}) ⊗ β)
(α ⊗ β) (1 ⊗ β)^{2} = \frac{1}{4}α ⊗ (trace(β^{2}) − 2β^{2})
(α ⊗ β)^{2} = −\frac{1}{2}α^{2} ⊗ β^{2} + \frac{1}{8}(trace(α^{2}) ⊗ β^{2} + α^{2} ⊗ trace(β^{2}))

(One could clean these formulas a bit, by using the adjoint α^{#} = α − \frac{1}{2} trace(α^{2}).)

If L is cyclic and K is a Kummer extension, then B(L, K) = A^{+} where A is the usual crossed product.

From this it is not difficult to see that the H^{2}-mod3 invariant of B is the cup product of the H^{1}-mod3 invariants of L and K (all of these invariants are defined only up to sign).

Also concerning the “mod2-part” of B(L, K) there is a sort of product.

Lemma 3.

trace_{B(L,K)/F} ≃ 3 trace_{L/F} ⊗ trace_{K/F}

Proof. This follows from the above formulas. There might be a better proof. □

Note that the trace form of a cubic extension with discriminant δ is ⟨1, 2, 2δ⟩. The associated 2-fold Pfister form is ⟨⟨−2, −δ⟩⟩.

Lemma 4. Let δ_{L}, δ_{K} be the discriminants of L and K, respectively. Then the H^{3}(ℤ/2)-invariant of B is

(−2, −δ_{L}, −δ_{K}) ∈ H^{3}(F, ℤ/2).

This follows from Lemma 3. Before I was aware of Lemma 3 I used the following arguments.

Proof. To check this one looks at our split example B̃ (which has H^{3}(ℤ/2)-invariant (−3, −1, −1)) and restricts to a ℤ/2 × ℤ/2 subgroup of G.

To be specific, introduce the following coordinates

B̃ = \{ (\begin{matrix}a & ū & w̄ \\ u & b & v̄ \\ w & v & c \end{matrix}) | a, b, c ∈ F, u, v, w ∈ Z \}

in B̃B (with ¯ = σ). Moreover let

α = (\begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix})

and

ϵ, θ: B → B, ϵ = ad_{α}, θ = ^{t}.

Then ϵθ is an element of order 2 in Aut(L) ⊂ G and θ is an element of order 2 in Aut(K) ⊂ G.

The trace form of B has the diagonal form

trace_{B} = ⟨1, 1⟩ ⊥ ⟨1⟩ ⊥ 2⟨1, 3⟩ ⊥ 2⟨1, 3⟩ ⊗ ⟨1, 1⟩

with respect to the coordinates

((a, b), c, u, (v, w)) ∈ F^{2} ⊕ F ⊕ Z ⊕ Z^{2} and Z = F ⊕ \sqrt{−3}F.

After twisting one has

trace_{B} = 2⟨1, δ_{L}⟩ ⊥ ⟨1⟩ ⊥ 2⟨1, 3δ_{K}⟩ ⊥ 2⟨1, 3δ_{L}δ_{K}⟩ ⊗ 2⟨1, δ_{L}⟩.

This gives

trace_{B} = ⟨1, 1, 1⟩ ⊥ 2⟨⟨−3δ_{K}δ_{L}⟩⟩ ⊗ ⟨2, 2δ_{L}, δ_{L}⟩
= ⟨1, 1, 1⟩ ⊥ 2⟨⟨−3δ_{K}δ_{L}⟩⟩ ⊗ ⟨⟨−2, −δ_{L}⟩⟩′

Finally note ⟨⟨−3δ_{K}δ_{L}, −2, −δ_{L}⟩⟩ = ⟨⟨−2, −δ_{L}, −δ_{K}⟩⟩. □

2. Twisting sums of four cubes

Consider the cubic form

Φ: L_{0} ⊗ K_{0} → F, Φ = (N_{L/F} |L_{0}) ⊗ (N_{K/F} |K_{0}).

It turns out that Φ is also given by the norm form of B(L, K):

Φ = N_{B(L,K)/F} |(L_{0} ⊗ K_{0}).

Let

C = C(L, K) = {Φ = 0} ⊂ ℙ(L_{0} ⊗ K_{0})

be the associated cubic surface.

Suppose that K = F [x]/(x^{3} − b). Then

Φ(α ⊗ x + α′ ⊗ x^{2}) = N_{L/F} (α)b + N_{L/F} (α′)b^{2}

In particular, for L = F [x]/(x^{3} − a) this gives the diagonal cubic form

Φ = ab⟨1, a, b, ab⟩

If b = 1 and L = F ⊕ F ⊕ F, then Φ has the form

uv(u + v) + st(s + t)

Lemma 5. The surface C(L, K) has a rational point if and only if the H^{2}-mod3-invariant of B(L, K) is trivial (i.e., B(L, K) has zero divisors).

Proof. A cubic form is isotropic if and only it is isotropic over a quadratic extension. We may therefore assume that L = F [x]/(x^{3} − a) and K = F [x]/(x^{3} − b). In this case the algebras is (L, b) and the cubic form is

Φ = ab⟨1, a, b, ab⟩

which is isotropic if and only if the equation

b = N_{L/F} (\frac{u + vx}{w + tx})

has a solution. But any element in L = F [x]/(x^{3} − a) is of the form

\frac{u + vx}{w + tx}.

So the cubic form is isotropic if and only if b ∈ N_{L/F} (L^{×}), i.e., the algebra has zero divisors. □

Lemma 6. The surface C(L, K) has a rational point if and only if it is rational.

Proof. If L is split, L = e_{1}F ⊕ e_{2}F ⊕ e_{3}F, then (e_{i} − e_{j}) ⊗ K_{0} give 3 disjoint lines in the cubic surface. Hence in general there is a set of three lines in C defined over F. As Colliot-Thelene informed me, in this case C is rational if and only if C has a rational point. The reference is:

[1] Swinnerton-Dyer, H. P. F. , The birationality of cubic surfaces over a given field. Michigan Math. J. 17 1970 289–295.

This paper is not available to me till now. □

Corollary 7. The stable birational equivalence class of C(L, K) depends only on the H^{2}-mod3-invariant of B(L, K) (defined up to sign).

Proof. Indeed, if C is rational over F (C′), and vice versa, then C × C′ is stable birational to C and C′. □

Question 1. What about birational equivalence?

3. A construction of (all) del Pezzo surface of degree 6

A del Pezzo surface of degree 6 is a form of ℙ^{2} blown up in 3 points in general position. They may be constructed as follows. Let B be a Jordan algebra (of the type as above) and let L ⊂ B be a separable associative subalgebra of degree 3. Define

Y (B, L) = { [b] ∈ ℙ(B) | {b, L_{0}, b} = 0 }

Here {b, λ, b} denotes the Jordan triple product.

In the split situation B = M_{3}^{+} and L = Δ (diagonal matrices) this gives

Y (M_{3}, Δ) = { [X] ∈ ℙ(M_{3}) | XΔ_{0}X = 0 }

Any matrix X with [X] ∈ Y (M_{3}, Δ) has rank 1, so that X = v · w^{t} for some 3-vectors v, w. This gives an identification

Y (M_{3}, Δ) = { ([v], [w]) ∈ ℙ^{2} × ℙ^{2} | v_{1}w_{1} = v_{2}w_{2} = v_{3}w_{3} }

In other words, Y (M_{3}, Δ) is the “quadratic correspondence of ℙ^{2}”, as described in Hartshorne’s book.

Let’s discuss the corresponding automorphism groups.

On M_{3} the group PGL_{3} ⋊ℤ/2 acts by conjugation and transposition. The subgroup leaving Δ_{0} invariant is of the form

H = T^{2} ⋊ (S_{3} ⋊ Z/2)

with T^{2} a 2-dimensional torus (=projective diagonal matrices).

So H acts on Y = Y (M_{3}, Δ), and one finds that H = Aut(Y), since on such a del Pezzo surface the hexagon consisting of the 6 exceptional lines is left invariant under all automorphisms of Y —and so Aut(Y) consists of the automorphisms of the toric structure on Y.

Corollary 8. There is a bijection between (isomorphism classes of ) pairs (B, L) and del Pezzo surfaces of degree 6.

Question 2. What about the (stable) birational classification of the Y ’s?

The stable question this is not difficult to answer by using the toric structure. There are classifying H^{2}-mod3 and H^{2}-mod2 invariants.

Let (A, τ) be an algebra with involution of second kind with center Z such that B = A^{τ}. Then the imbedding of Y to ℙ^{2} × ℙ^{2} twists to an embedding

Y (B, L) ⊂ R_{Z/F} (SB(A))

If Z is split, i.e., B = A′^{+} for a central simple algebra A′, then

Y (B, L) ⊂ SB(A′) × SB(A′^{op})

The projection to any of the factors is the blow down of 3 lines R_{L/F} (ℙ^{1}) ⊂ Y (B, L). Let still B = A^{τ} and let

S = { [b] ∈ ℙ(B) | rank b = 1, b^{2} = 0 }

If B = M_{3}^{+}, then

S = { ([v], [w]) ∈ ℙ^{2} × ℙ^{2} | v_{1}w_{1} + v_{2}w_{2} + v_{3}w_{3} = 0 }

The intersection S ∩ Y is exactly the hexagon on Y.

4. Blowing down the cubic surface

We return to the case L = F ⊕ F ⊕ F and K = F [x]/(x^{3} − 1). Note that then B(L, K) = M_{3}^{+}. Then the cubic surface is given by

C = {uv(u + v) + st(s + t) = 0}

Consider the map

C → ℙ^{2} × ℙ^{2}, (u, v, s, t) ↦ ([−vs, st, uv], [−ut, uv, st]).

If I am not mistaken, this map is everywhere defined, maps to Y and the map C → Y is a blow down of 3 lines. The map C → ℙ^{2} (given by any of the two projections) should be the blow up in the 6 points

[1, 0, 0], [0, 1, 0], [0, 0, 1], [1, 1, 1], [1, ζ, ζ^{2}], [1, ζ^{2}, ζ]

with ζ a primitive 3^{rd} of unity.

Question 3. How to describe these blow downs in the non split situation?