Subject: Re: A question and an answer Date: Sat, 7 Apr 2001 00:09:34 +0100 From: Tom Goodwillie To: Don Davis >Let G be a semi-simple, simply connected, compact Lie group, and H a >closed subgroup. What are the conditions on G and H so that G/H is a >spin manifold? One thing you can say is that it depends only on H. Thm: Let H be a closed subgroup of the Lie group G. If the representation Ad_H of H admits a spin structure, then the manifold G/H admits a spin structure. The converse is valid if G is (connected and) simply-connected. Pf: Consider the fibration sequence i j G/H ---> BH ---> BG A representation of H determines a vector bundle on BH and, by i*, a vector bundle on G/H. The tangent bundle of G/H arises in this way from a representation V defined by the exact sequence 0 ---> Ad_H ---> (Ad_G)|H ---> V ---> 0. The vector bundle on G/H determined in this way by (Ad_G)|H is trivial, for example because j composed with i is constant. Therefore, by the Cartan formula for Stiefel-Whitney classes, the vanishing of w_1 and w_2 for the tangent bundle of G/H is equivalent to the vanishing of w_1 and w_2 for i*(Ad_H). This in turn will follow from the vanishing of w_1 and w_2 for Ad_H itself. The argument is reversible if i* is a monomorphism in first and second cohomology, which will be true if G is simply-connected. Tom Goodwillie