Commutator calculus and groups of homotopy classes by Hans Joachim Baues

By Hans Joachim Baues

A basic challenge of algebraic topology is the category of homotopy kinds and homotopy sessions of maps. during this paintings the writer extends result of rational homotopy thought to a subring of the reason. The tools of facts hire classical commutator calculus of nilpotent team and Lie algebra thought and depend on an in depth and systematic research of the algebraic homes of the classical homotopy operations (composition and addition of maps, spoil items, Whitehead items and better order James-Hopi invariants). The account is basically self-contained and may be available to non-specialists and graduate scholars with a few history in algebraic topology and homotopy idea.

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EA] the equation Il a + ... + lka = (i1+... +ik)o +' ' n>2 cn(ii' ... , ik) o yn(a) where cn(i1 , lk) deDk 1'1' ... , ']0(d) 0 TT(d) n Proof of (2. 8). Let R : FQY - Y be the evaluation map with R (t, a) = Q(t). For the adjoint f : X - R Y of f : EX - Y we have f = R ° (Ef). (1) We consider the diagram 44 II v n>1 i1g + i2g where by use of (1) we have (2) (i19+i 2 i) - (Ea) = ila + i 2 a Since G is a homotopy equivalence, there exist mappings cn making the diagram homotopy commutative. We have to show (3) cn = c ri(i l i ), , 2 c =i +i , 1 1 2 as defined in (2.

Thus cn(a, (3) is needed only if I a l = 1,61 is even. In this case, we evaluate c n (a, p) in the graded Lie algebra 7r. Similar remarks apply to Rm n(a' p). , It is easily seen that tp in (3. 6) satisfies the relations in (3. 8) [x®a, y 0 0] = (x u y) 0 [a, 0] for the Lie bracket in (3. 4). For this, it is important that Cn and Rm are in fact homogeneous terms. This is the advantage of Rm n over Qn in (2. 2). Theorem (3. 7) can be proved along the same lines as (5. 9) in chapter II. ¢4. n The general type of Zassenhaus terms and its characterization modulo a prime We first generalize the Zassenhaus formula (1.

4) T(ul) 0 T(u2) = H*(J(Sn) x J(Sm), Q). The embedding 0 is defined to be the Lie algebra homomorphism with $(x) = ill 0 U1, )(Y) = {a2 9) u2 where fit, f12 : T -+ Q map u1 9) 1 to 1 and 1 0 u2 to and map all other elements un g) um to zero. 1 respectively Looking at the images of basic commutators we see that 0 and are actually embeddings. We remark that (4. 5) L(u1, u2) = n*(S2(ESn , ESm)) 9) Q For each N we have the mapping (4. 6) IrN : PN = (Sn)N - J(Sn) which is the restriction of the identification map n in (2.

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