By Skip Garibaldi

This quantity issues invariants of G-torsors with values in mod p Galois cohomology - within the feel of Serre's lectures within the booklet Cohomological invariants in Galois cohomology - for varied uncomplicated algebraic teams G and primes p. the writer determines the invariants for the phenomenal 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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G. ) Similarly, we obtain w(t)xαe−π (u) = xπ (c t−3 u)w(t) for some c ∈ {±1}. 8) give h(−1)xπ (u) = xπ (−u)h(−1) and since h(−1) = w(−1)2 , we have: xπ (−u)h(−1) = w(−1)2 xπ (u) = xπ (cc u)h(−1). So c = −c. 9. Remark. We can describe this copy of SL2 concretely in the notation of Dynkin [Dy 57b, Ch. III]. For simplicity, we consider the cases where G is simply laced, so we may identify roots and coroots by deﬁning all roots to have length 2 with respect to the Weyl-invariant inner product ( , ).

6]. [A] for uniquely determined λ1 , λ2 ∈ R3 (k0 ). But the algebra J(A, 1) is also split for every A. It follows that λ1 is zero. This proves that g3 spans Invnorm (F4 , Z/3Z). 5, we have found just three interesting invariants of F4 , namely g3 , f3 , and f5 . 7. Open problem. 4], [PeR 94, Q. 1, p. 205]) Is the map g3 × f3 × f5 : H 1 (∗, F4 ) → H 3 (∗, Z/3Z) × H 3 (∗, Z/2Z) × H 5 (∗, Z/2Z) injective? That is, is an Albert algebra J determined up to isomorphism by its invariants g3 (J), f3 (J), and f5 (J)?

Namely, we assume that π is long (equivalently, G is not of type C), the rank of G is at least 4, and the characteristic is = 2. , for every root β, the integer α, β is the coordinate of π in β. For example, the π-coordinate of α is α, α = 2. For w0 the longest element of the Weyl group of G, clearly w0 (α) = −α, hence w0 (π) = −π. 11, so that Q is generated over ksep by the Uα where α has positive π-coordinate. We do this both to agree with R¨ ohrle’s notation and for the convenience of working with positive roots.