Homotopical Nilpotence of S3

In [l] Berstein and Ganea define the nilpotence of an ü-space to be the least integer » such that the »-commutator is nullhomotopic. We prove that S3 with the usual multiplication is 4 nilpotent. Let X be an ii-space. The 2-commutator c2: XXX—>X is defined by c2(x, y) =xyx~1y~1 where the multiplication and inverses are given by the ü-space structure of X. The »-commutator cn: X"-+X is defined inductively by c„ = c2(cn_iXl). Let T?(X) denote the subset of Xn consisting of those «-tuples (xi, • • • , x„) such that x{ = * (the base point) for at least one i. It is well known that c„\ T?(X)~*. Thus a map „: Xn/Tl(X)-^X may be defined such that the homotopy class of „ be the homotopy class of „ = 0. The usual multiplication for S3 is that obtained by considering S3 to be the set of unit quaternions. With this multiplication QK, infinite quaternionic projective space, is a classifying space for S3. Let T: 7r(2S"-\ X)-nr(S»-\ OX) be defined by (Tf(s))(t) =/(/, s), where sÇS"-1,/Gir^S"-1, X), and tEIFis an isomorphism of the homotopy groups. Samelson [3] has shown that if j: Si-J>Q„ is inclusion, Tj: S3 —>QQK is an ZFhomomorphism which is also a homotopy equivalence. He uses this to show that T[[j, j], j] = (Tj)*$z where the product on the left is the 3-fold iterated Whitehead product. Since T is an isomorphism, to show that S3 is 4 nilpotent it suffices to show that the four-fold iterated Whitehead product of j is zero and the threefold product is nonzero. Let Î4 be the identity map on S4. Hilton [2] has shown that 0^[[î4, it], n]ETrio(S4) is the image of an element in 7r9(53) under the suspension homomorphism. He uses this fact to prove [[[ii, Ù], Ù], ii] = 0. Lemma 1. [[[j,j],j],j] = 0.