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.
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