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To better understand spectral sequences, I tried to apply it for the following problem. Let's consider an $SO(3)$ bundle, $E$, over $S^2$. I know that there are two inequivalent $SO(3)$ bundles over $S^2$; one is the trivial bundle $S^2\times SO(3)$ and the other is a nontrivial bundle. Let's consider the nontrivial case. In order to calculate the cohomology of $E$ we may use the spectral sequence with $E_2^{p,q}=H^p(S^2, H^q(SO(3);Z)); d_2:E_2^{p,q}\to E_2^{p+2,q-2+1}$. We have the following table for $E_2$:

$$ \begin{array}{l|c|c|c|} E_2^{0,3}=Z &0 & E_2^{2,3}=Z& 0 &0 \\ \hline E_2^{0,2}=Z/2Z& 0 &Z/2Z &0& 0\\ \hline E_2^{0,1}=0 &0 &0 &0 &0 \\ \hline E_2^{0,0}=Z& 0 & Z & 0 & 0 \end{array} $$ Now, obviously $d_3: E_3^{p,q}\to E_3^{p+3,q-2}$ is the zero map in this case, implying, $E_3=E_\infty=H(E)$. Our goal is to find $E_3$. We notice that $d_2=0$, except on $E_2^{0,3}$. So, we get $E_3=$ $$ \begin{array}{l|c|c|c|} E_3^{0,3}=M &0 & E_3^{2,3}=Z& 0 &0 \\ \hline E_3^{0,2}=Z/2Z& 0 &coker(d_2) &0& 0\\ \hline E_3^{0,1}=0 &0 &0 &0 &0 \\ \hline E_3^{0,0}=Z& 0 & Z & 0 & 0 \end{array} $$ where, $M=ker(d_2)$. My question is what is $M$? I am interested in the question because $M$ seems to be the only cohomological "parameter" which might distingush the nontrivial and the trivial $SO(3)$ bundle (over $S^2$).

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  • $\begingroup$ Please give a proper table, or a picture, or some other description of the $E_2$ page; as you have it right now it's basically unreadable. Also you should probably mention that you're talking about the Serre spectral sequence. $\endgroup$ Apr 14 at 19:22
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    $\begingroup$ @BenSteffan The table appeared in the preview but did not appear in the posted question. Anyway I have edited the question with a properly formatted table. $\endgroup$ Apr 14 at 19:30

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That potentially nonzero differential lands in the $\mathbb{Z}/2\mathbb{Z}$, so if it's nontrivial, the kernel will still be $\mathbb{Z}$. This suggests the vanishing of $H^4$ of your total space can be used to detect whether they are different bundles. Running the Serre spectral sequence in homology makes this clearer though, it's the same thing, with the vanishing of $H_1$, and in this instance this is the abelianised fundamental group (which is abelian, and at most $\mathbb{Z}/2\mathbb{Z}$ by LES in homotopy groups. So the thing distinguishing these bundles is the fundamental group.

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  • $\begingroup$ In th second line, do you mean ""vanishing of $H^4$"? $\endgroup$ Apr 15 at 6:11
  • $\begingroup$ The cohomology of the total space of the bundle is (up to extensions) given by the pieces on slopes of $-1$ in the final page of the spectral sequence. So we see in the case of a trivial bundle, we have $H^4(E_{triv})=\mathbb{Z}/2\mathbb{Z}$, and in the nontrivial bundle, this group is zero. I accidentally counted slopes wrong, your right it should be $H^4$, fixed. $\endgroup$
    – Chris H
    Apr 15 at 7:45

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