Subject: SU(3)/SO(3) From: Date: Sun, 19 Nov 2006 12:52:00 +0000 Dear Professor: Just now I found the question of P.Landweber aboutthe characteristic class of coset SU(3)/SO(3)(he posted this question on the discussion group) >From: Peter Landweber >Date: Mon, 7 Aug 2006 19:01:27 -0400 > >Here's a small question for the discussion group. > >I believe it is known that the 5-manifold SU(3)/SO(3), which is simply >connected, represents a generator of the 5-dimensional unoriented >cobordism ring. In fact, I believe there is some reference for this fact, >which should include the further fact that the product of Stiefel-Whitney >classes w_2w_3 is nonzero. Having forgotten the reference, I hope >someone can remind me of such a reference, or give an argument for these >results. I think we can consider this problem in the following sense: At first, one easily sees the fact that due to the fibration sequence SO(3)--->SU(3)---->M \pi_1(M)=0,\pi_2(M)=Z_2, so due to Hurewicz theorem, we have H_1(M)=0,H_2(M)=Z_2, then by using the Poincare duality, (here we use the Z_2 coefficient,certainly) H1(M)=0 H2(M)=Z_2 H3(M)=Z_2 H4(M)=0 H5(M)=Z_2 Let us analyze them carefully For the H2, let x_2 denote the non-trivial element,if we consider the classifying map of the fibration mentioned at the beginning of the email: f: M---->BSO(3) let w_2,w_3 denote the S-W class of BSO(3), then we see that f^*(w_2)=x2, this is a standard result of the cohomology structure of SU(n)/SO(n), which can be found in the book : Topology of Lie groups, I and II, by Mamoru Mimura, Hirosi Toda,AMS. We can simply state it as the cohomology ring of H^*(SU(n)/SO(n)) are determined by the pull-back of universal S-W class, furthermore it is a polynomial ring. Another fact is that Sq1(x_2)=x_3, this is because Sq3=Sq^1Sq2 and the definition of S-W class(see Milnor&Stasheff),then f^*(w_3)=x_3. So the next job will be to ensure that x_2 ,x_3 are just what we want. By Wu formula, we have x_2{\cup}x_3=Sq2(x_3), however, consider another kind of Wu formula with the fact M is simply connected, we have Sq2(x_3)=w_2(M){\cup}x_3. One immediatedly know that x_2=w_2(M), even x_3=w_3(M). I hope this is helpful more or less. _________________________________________________________________ ??????????,??? MSN Messenger: http://messenger.msn.com/cn