Diffgeometer1

Mathematician working mainly in differential geometry with an interest in physics.

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Diffgeometer1_at_mathstodon.xyz@mostr.pub

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npub14mnqckz48cqng6rjfnvywwcy73u9ewuff8khnfk0vs7zege096fscctm7l

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Last Notes

2024-10-04 01:57:24

Let \(\pi: SO(n+1)\rightarrow S^n\) be the map given by\[\pi(P)=Px\]where \(x\in S^n\) is some fixed element of \(S^n\). Then \(\pi: SO(n+1)\rightarrow S^n\) is a principal \(SO(n)\)-bundle. Assume \(n>2\). However, my intuition is that this cannot be a trivial principal bundle since it would imply\[SO(n+1)\simeq S^n\times SO(n)\]which I don't think is right. I don't know what the de Rham cohomology of \(SO(n)\) is. If I can show that the left and right sides have different de Rham cohomology groups that would prove that \(\pi: SO(n+1)\rightarrow S^n\) cannot be a trivial principal bundle. Does anyone have a simple way of showing that \(SO(n+1)\) cannot be homeomorphic to \(SO(n)\times S^n\)? Also I want to avoid the use of characteristic classes or \(K\)-theory. I want to limit things to homology, cohomology, and fundamental groups only.