On the order three Brauer classes for cubic surfaces

is bijective.

Proof. This proof has a computer part. Using gap, we made **the** observation that

rk Pic(S ) H = 1 **for** every H ⊂ W(E 6 ) such that H 1 (H, Pic(S )) ∼ =/3.

To verify **the** assertion, we will show injectivity. According to Lemma 3.17.i), it

suffices to test this on **the** 3-Sylow subgroups of H and U t . But **the** 3-Sylow subgroup

∼ =/3×/3×/3is abelian. Thus, Lemma 3.17.ii) immediately implies

**the** assertion.

□

U (3)

t

3.19. Corollary. –––– Let H ′ ⊆ H ⊆ U t be arbitrary. Then, **for** **the** restriction

map res: H 1 (H, Pic(S )) −→ H 1 (H ′ , Pic(S )), **the**re are **the** following limitations.

i) If H 1 (H, Pic(S )) = 0 **the**n H 1 (H ′ , Pic(S )) = 0.

ii) If H 1 (H, Pic(S )) ∼ =/3and H 1 (H ′ , Pic(S )) ≠ 0 **the**n res is an injection.

iii) If H 1 (H, Pic(S )) ∼ =/3×/3**the**n H 1 (H ′ , Pic(S )) ∼ =/3×/3or 0.

In **the** **for**mer case, H ′ = H. In **the** latter case, H ′ = 0.

Proof. We know from Remark 3.16.i) that both groups may be only 0,/3, or

/3×/3.

i) If H 1 (H ′ , Pic(S )) were isomorphic to/3or/3×/3**the**n **the** restriction

from U t to H ′ would be **the** zero map.

ii) is immediate from **the** computations above.

iii) This assertion is obvious as H is of **order** **three**.

□

4 Computing **the** **Brauer**-Manin obstruction

A splitting field **for** **the** **Brauer** class.

4.1. –––– The **the**ory developed above shows that **the** non-trivial **Brauer** **classes**

are, in a certain sense, always **the** same, as long as **the**y are of **order** **three**. This may

certainly be used **for** explicit computations. The **Brauer** class remains unchanged

under suitable restriction maps. These correspond to extensions of **the** base field.

We will, however, present a different method here. Its advantage is that it avoids

large base fields.

4.2. Lemma. –––– Let S be a non-singular **cubic** surface and let T 1 and T 2 be

two pairs of Steiner trihedra that define disjoint sets of lines.

Then, **the** 18 lines defined by T 1 and T 2 contain exactly **three** double-sixes.

These **for**m a triple of azygetic double-sixes.

10