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.