# Cohomological invariants: exceptional groups and spin groups by Skip Garibaldi By Skip Garibaldi

This quantity matters invariants of G-torsors with values in mod p Galois cohomology - within the experience of Serre's lectures within the booklet Cohomological invariants in Galois cohomology - for varied easy algebraic teams G and primes p. the writer determines the invariants for the outstanding teams F4 mod three, easily attached E6 mod three, E7 mod three, and E8 mod five. He additionally determines the invariants of Spinn mod 2 for n

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5. Strongly orthogonal roots in G. The Weyl group of G acts transitively on the roots in Φ1 of the same length [ABS, §2, Lemma 1], so we may assume that µ1 equals π. For j = 2, 3, 4, we set: γj := α − π − µj . 36 II. INVARIANTS OF E6 , E7 , AND E8 Lemma. γ2 , γ3 , γ4 are pairwise strongly orthogonal long roots of G. For various x, y, the value of x, y is given by the table: y x α π µj γj α 2 1 1 0 π 1 2 0 −1 µj γj 1 0 0 −1 2 −1 −1 2 Proof. The top row of the table is the π-coordinate of y, and we know these already.

Fix η ∈ Z 1 (K, H) and let A be the central simple algebra of degree 5 representing π1 (η) ∈ H 1 (K, PGL5 ). 6, there is an extension L/K of 42 II. , equals c∗ (γ) for some γ ∈ H 1 (L, C). We have a commutative diagram with exact rows: 1 −−−−→ µ5 −−−−→ C × µ5 −−−−→ ⏐ ⏐ C ⏐ ⏐ t 1 −−−−→ µ5 −−−−→ H −−−−→ 1 π ×π 2 −−1−−→ PGL5 × PGL5 −−−−→ 1. 3, γ and η have the same image in H 1 (k, PGL5 × PGL5 ), namely the class of (π1 (η), π2 (η)). It follows that η and t∗ (γ) are in the same H 1 (k, µ5 )-orbit.

That is, we have identiﬁed H with the quotient of SL5 × SL5 by the subgroup generated by (ζ, ζ 2 ) for ζ ∈ µ5 . For i = 1, 2, write πi : H → PGL5 for the projection on the i-th factor. 3. Lemma. For η ∈ H 1 (k, H), write Ai for the central simple k-algebra of degree 5 deﬁned by πi (η). Then 2[A1 ] = [A2 ] in the Brauer group of k. The twisted group Hη is isomorphic to (SL(A1 ) × SL(A2 ))/µ5 . Proof. Consider the diagram with exact rows 1 −−−−→ µ5 × µ5 −−−−→ SL5 × SL5 −−−−→ PGL5 × PGL5 −−−−→ 1 ⏐ ⏐ ⏐ ⏐ q 1 −−−−→ µ5 −−−−→ H π ×π 2 −−1−−→ PGL5 × PGL5 −−−−→ 1 where q is given by (x, y) → y/x2 .