Algebra Some Current Trends by Avramov L.L. (ed.), Tchakerian K.B. (ed.)

By Avramov L.L. (ed.), Tchakerian K.B. (ed.)

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Then r ≤ r and there exists w ∈ W0 such that βi = w(βi ) for i ≤ r . Proof. 14 (d) we may suppose that β1 = β1 . Now the proof goes by induction: both β2 , . . , βr and β2 , . . , βr belong to ∆1 and, hence, there exists w ∈ W0 (the Weil group of l0 ) such that βi = w(βi ) for i ≤ r . It remains to notice that wβ1 = β1 . Now we can describe the orbit decomposition and the Pyasetskii pairing. 17 Let α1 , . . , αr ∈ ∆1 be any maximal sequence of pairwise orthogonal long roots. Set ek = eα1 + .

Then we have a basis vector e1 ∧ . . ∧ edim V0 ⊗ f1 ∧ . . ∧ fdim V1 ∈ Det V. 4) we get a number det(V, ∂, e) called the Cayley determinant of a based supervector space with an exact differential. If we fix other bases {˜ e1 , . . , e˜dim V0 } in V0 and {˜ e1 , . . 5) where (A0 , A1 ) ∈ GL(V0 ) × GL(V1 ) are transition matrices from bases e to bases e˜. One upshot of this is the fact that if bases e and e˜ are equivalent over some subfield k0 ⊂ k then the Cayley determinants with respect to these bases are equal up to a non-zero multiple from k0 .

D) W0 acts transitively on long roots in ∆1 . Proof. (a) Indeed, since l1 is an irreducible g-module, β is the unique lowest weight of the g-module l1 . e. to γ. Hence β is long. (b) Since l1 = U g · β, the β-height of any root in ∆1 is equal to 1. (c) Suppose that α, α ∈ ∆1 . Since [eα , eα ] = 0, it follows that α + α is not a root, therefore, (α, α ) ≥ 0. (d) Suppose that α0 ∈ ∆1 is a long root, α0 = β + α∈Π0 nα α. If we have (α0 , α) < 0 for some α ∈ Π0 then wα (α0 ) > α0 and we may finish by 36 2.

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