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**Extra resources for Algebraization of Hamiltonian systems on orbits of Lie groups**

**Example text**

19. Let K /k be a finite Galois extension such that all lines are defined over K . Set G = Gal(K /k). Let L be any line on SK . The sum over all conjugates σ ∈G σ (L) is invariant under G by construction, hence it is a divisor defined over k. ) If the Picard number of Sk is one then, because the hyperplane class generates Pic(Sk ), these orbit sums are all multiples of the hyperplane class −K S . This establishes that (1) implies (2). Next suppose that {L 1 , . . , L t } is an orbit consisting of non-intersecting lines.

In particular, a smooth cubic surface is rational over k if it contains two skew ¯ lines defined over k (of the twenty-seven lines on the surface defined over k). Proof. Let X ⊂ P2n+1 be the cubic hypersurface, and let L 1 and L 2 be the two linear spaces on X . Consider the map φ : L1 × L2 X (P, Q) → third intersection point X ∩ P Q. This defines a birational map from L 1 × L 2 to X . The map is well defined because a typical line intersects X in exactly three points (counting multiplicities).

The point x is in the image of φ precisely when at least one of these lines intersects S in a point other than s. In particular, the projection from x, πx : S → S ⊂ P4 can not be one-to-one. Indeed, x has a unique preimage under φ precisely when the projection πx collapses exactly two points of S to a single point. On the other hand, if πx : S → S is not of degree one, then x has infinitely many preimages under φ. Thus φ is finite and dominant if and only if the generic projection of S from a point x ∈ X is one-to-one except on a finite set.

### Algebraization of Hamiltonian systems on orbits of Lie groups

by Anthony

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