Subject: BSU(n) and job From: Tom Goodwillie Date: Thu, 8 Dec 2005 10:12:35 -0500 Looks like we've more or less beaten this one to death. To sum up: (1) BSU ---> BU ---> BU(1) splits because of the section BU(1) ---> BU and the H-space structure on BU. (2) BSU(n) ---> BU(n) ---> BU(1) splits locally at any prime not dividing n, because an isomorphism of U(1) with the center of U(n) provides a diagram SU(n) ---> SU(n)xU(1) ---> U(1) I 1 I I p v v v SU(n) ---> U(n) ---> U(1) which after taking classifying spaces has vertical maps that are p-locally equivalences. (3) BSU(n) ---> BU(n) ---> BU(1) does not split if 1 BSU(n), then it would give a rank n complex bundle E on BU(n) such that (i) its determinant (nth exterior power) is the trivial line bundle, and (ii) its restriction to BSU(n) is the same as that of the canonical rank n bundle on BU(n). Let L_1, ... , L_n be the obvious generating line bundles on BT(n). The restriction of E to BT(n) is a sum of n line bundles, each of which is a monomial (negative exponents allowed) in the L's. Because E lived on BU(n), this polynomial expression in the L's is symmetric under permutation. Now you do a little bit of combinatorics (figuring out, for the symmetric group acting on the monomials, what are the orbits with at most n elements) and a little bit of algebra (using (i) and (ii)) to arrive at a contradiction. Tom Goodwillie P.S. The possible postdoc position at Brown that I mentioned the other day is now a definite postdoc position.