Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $H^{2}(SO(3)) \simeq 0$?

share|improve this question
1  
Your title does not match your question! –  Juan S Feb 4 '13 at 23:54
add comment

1 Answer 1

up vote 2 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 $H^2(SO(3);\mathbb{R}) = 0$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.