Subject: Re: two more on BSU(n) From: Clarence W Wilkerson Date: Mon, 5 Dec 2005 12:38:35 -0500 (EST) I disagree with the conclusion of the following post ( my reasons follow the included post): >>>>Subject: Re: two more on BSU(n) >>>>From: jnkf@math.mit.edu >>>>Date: 02 Dec 2005 18:24:32 -0500 >>>> >>>>Mike Hill commmented to me that, contrary to some of the >>>>posts, BU is NOT homotopy equivalent to BSU x BU(1) at ANY prime p. >>>>The action of the mod p reduced powers distinguishes them. To see >>>>this it suffices to look at P1 c_2, which involves c_1.* >>>> >>>>Therefore BSU(n) x BU(1) is not homotopy equivalent to BU(n) at p >>>>so long as n is large enough relative p. >>>> >>>>John >>>> >>>>* One computes this by pulling back to BT where P1 c_2 is a certain >>>>symmetric function. The problem from there, to rewrite that Newton >>>>polynomial in terms of elementary symmetric functions, is combinatorics. Counterclaim: Consider SU(n) as a subgroup of U(n). Its center Z(SU(n)) consists of powers of the scalar diagonal matrix with each entry \zeta, where \zeta is a primitive n-th root of unity. On the other hand, the center of U(n) consists of all scalar diagonal matrices with the scalar of absolute value 1. In particular Z(U(n)) \iso S1 as groups. For any subgroup H of U(n), the multiplication map \phi_H : Z(U(n)) x H \to U(n) by (z,h) \to zh=hz is a group homomorphism. If H=SU(n), then \phi_SU(n) is onto and its kernel is generated by (\zeta^-1,\zeta), i.e., it's a diagonal copy of Z/nZ in S1 x SU(n). So we have a short exact sequence 1 \to Z/nZ \to S1 x SU(n) \to U(n) \to 1, where the maps are group homomorphisms. This gives a fibration BZ/nZ \to BS1 x BSU(n) \to BU(n). >From the Serre SS, for any prime $p$ not dividing $n$, H^*(BZ/nZ,F_p) = 0 for * > 0 and H^*(BU(n),F_p) \to H^*(BS^1xBSU(n),F_p) is an isomorphism. Or for these primes, passing to p-completions gives BU(n)^_p = ( BS1 x BSU(n))^_p. Thus the "Therefore ....." line is incorrect. As for >>>>this it suffices to look at P1 c_2, which involves c_1: This is not the whole story. If BU(n)^_p = (BS1 x BSU(n))^_p then this only means that in H^*(BU(n),Z/pZ) = Z/pZ[c_1,c_2,\hdots c_n) one can choose another (different) set of polynomial generators for H^*(BU(n),Z/pZ) which behave under the Steenrod operations like those of BS1 x BSU(n). For example for n=2 and p=2, Sq1 == 0, Sq^2c_1 = c_12, Sq^2c_2 = c_1c_2, Sq4 c_2 = c_22 and no choice of another generator c_2' will rid Sq^2c_2' of the c_1 involvement. However for n=3 and p=2, we have Sq^2c_2 = c_3 +c_1_c_2, which looks bad, because one can't elimnate the c_1 terms by repicking a c_2'. However, by picking c_3' = c_3 +c_1c_2, we can avoid the problem. One can calculate that Sq^2c_3' = 0 , Sq^4c_3' = c_2c_3', and Sq^6c_3' = c_3'^2. The piece Z/2Z[c_2,c_3'] is iso over the Steenrod alg. to H^*(BSU(3),Z/2Z). Clarence