2021-04-26_Cohomology_of_SOn

Basically, all of these computations are based on the fact that \(SO^n\) acts transitively on \(S^{n-1}\) with stabilizer \(SO^{n-1}\), producing fibrations of the form \(SO^{n-1} \to SO^n \to S^{n-1}\). The overall question: can this be used to inductively determine \(H^*(SO^n)\) for all \(n\)?

1) Cohomology of SO3

A priori, we can use the fact that \(SO^3\) is diffeomorphic to \({\mathbf{RP}}^3\) to obtain

\begin{align*}H^*(SO^3) = H^*({\mathbf{RP}}^3) = [{\mathbf{Z}}, 0, {\mathbf{Z}}_2, {\mathbf{Z}}, 0, 0, \cdots]\end{align*}

But supposing we didn’t know the cohomology of the total space, the Serre spectral sequence can be applied by using the fibration

\begin{align*} S^1 \to SO^3 \to S^2\end{align*}

and then the usual theorems give the formula (in cohomology)

\begin{align*} E_2^{p,q} = H^p(S^2) \otimes H^q(S^1).\end{align*}

This yields the following \(E_2\) and \(E_3\) pages respectively (indexing columns with \(p\) and rows with \(q\)):

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