2-Divisible groups over Z by Abrashkin V.A.

By Abrashkin V.A.

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A Q-slgebra, We have thus established (Ar+l). But now (Br+I) follows from (Ar+I) and (Br) by an easy argument. The first equation in Lamina 3 just tells us that ~ is a map of R-algebras. All the maps occuring in the lest two equations of the lemma are homomorphism of algebras preserving identities. In each case it then suffices to verify that the images of the generators di of the algebra E(LF) coincide, and this follows from the explicit description given earlier on. PROOF o_~fTheorem i (D and w Prop.

2) . ~ M | ttM [I@ < <81 ("associative law" - here we identify (M @ M) @ M = M ~ (M @ 14)). M ~ RM (here t is the "twisting map", t(x @ y) = y 8 x ; "commutative law"). 6) / ROEM < -" M -- >M@ER A hi algebra is given by (i) a coalgebra {M, <, (ii) the structure of an associative (but not necessarily commutative) R-algebra on M with identity a, 8} , [Exercise : describe by diagre~ ] . Here m is to coincide with the algebra structure map R § M, and and 8 are to be homomorphisms of R-algebras. (x 2 @ y2 ) = xlx 2 @ yly2.

44 Each Lie algebra L has an enveloping algebra E(L). 1) Note: Homassoc(ECL),A ) > HomLieCL~CA)). All associative algebras have identities, and Homasso c is the set of homomorphisms preserving identities. 1), taking A = R we get from the n1111 a homomorphism of assoclative algebras : ECL) § R. As E(L) has an identity we also have a homomorphism o : R § E(L). Next if L 1 and L 2 are Lie algebras, then their cartesian set product L I x L 2 has again a Lie algebra structure, and E(Ti x L2) =~ ECLl) eR ECL2).

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