Subject: top question
From: Wolfgang Ziller <Wolfgang@sas.upenn.edu>
Date: Mon, 18 Dec 2006 23:10:02 -0500

Dear topology friends,

I have an algebraic topology question which has applications in geometry
(nonnegative curvature)
It is a very concrete and special situation I am in.

I have a family of SO(k) principal bundles P over CP2 , k>4 ,
with w_2 \ne 0
P arises from a  concrete construction.
All I know about P though is that SO(k) acts freely and I know the 
cohomology
ring of P. From this knowledge can I can determine the bundle?
By Dold Whitney's classification of bundles over simply connected 
4-manifolds,
such bundles are classified by w_2 , w_4 and p_1.
p_1 can be interpreted as an integer by evaluating on a fundamental cycle.

The structure group of course reduces to 4 dim bundles where w_4 is euler 
class
mod 2.
But my bundles have k>4 and from their construction there is no easy way 
to
reduce the structure group explicitly.
>From this though one can determine the restriction that need to be 
satisfied for
these numbers since
p_1=2k+2l e=k-l if w_2 = 0  and p_1 = 2k+2l+1 , e= k-l if w_2 \ne 0 where 
k,l
are arbitrary integers.

It is not hard to recognize w_2 and  | p_1 | in the top of P :
w_2\ne 0 iff P is simply connected and if p_1 \ne 0 then H4(P,Z) is a 
finite
group of order |p_1|.

If w_2=0 I am ok since then p_1= 2 mod 4 implies w_4 \ne 0 and p_1 = 0 mod 
4
implies w_4 = 0

But when I have w_2 \ne 0 it only  implies that
if p_1 = 1 mod 4 then w_4 \ne 0 and if p_1 = 3 mod 4 then w_4=0.

So my problem is that I cannot recognize the sign of p_1 from the coh ring 
.
Is there some other information in the coh ring that will tell me what w_4 
is?
I assume that this is somehow encoded in the spectral sequence for the
universal  bundle  (maybe with Z_2 coefficients).
This is how one sees the above facts about w_2 and |p_1| for example

Another question would be if you know two such bundles  where p_1 differs 
by a
sign concretely enough so I can see what to expect from their cohomology 
ring?

Any advice would be appreciated.


Thanks!

Wolfgang