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We have that De Rham cohomology of $SO(3) \simeq \mathbb{R}P^{3}$ is $\mathbb{R}$ in degree $0$ and $3$ and $0$ in degree $1$ and $2$. But I saw that $H^{*}(SO(3)) \simeq \mathbb{Z}_{2} $ in degree 2. (Probably I have to use the universal coefficient theorem to armonize these two results...). I have to calculate the De Rham cohomology of $SO(4)$ using the fiber bundle $SO(3) \rightarrow SO(4) \rightarrow S^{3}$. How can I do it using spectral sequence? In this situation can I say that $H^{2}(SO(3)) \simeq 0$?

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Your title does not match your question! – Juan S Feb 4 '13 at 23:54
up vote 3 down vote accepted

Your work seems slightly misleading. The De Rham cohomology $H_{dr}^2(SO(3)) \simeq 0$, whilst the ordinary cohomlogy $H^2(SO(3);\mathbb{Z}) \simeq \mathbb{Z}/2\mathbb{Z}$. It's good to use different notation for the two, so it is clear what you are trying to do.

Recall that, for smooth manifolds, $H_{dr}^*(X) \simeq H^*(X;\mathbb{R})$. Thus in your case the spectral sequence runs $$H^p(S^3;H^q(SO(3);\mathbb{R})) \Rightarrow H^{p+q}(SO(4);\mathbb{R})$$ and so you can indeed use $H^2(SO(3);\mathbb{R}) = 0$

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