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$\mathbb{S}^4$, the sphere of dimension $d=4$ admits lie group structure?

I know that every Lie group has trivial vector bundle, so I have thought about calculating the vector bundle of $\mathbb{S}^4$ and if it is not trivial (as I suspect) then $\mathbb{S}^4$ will not admit a Lie group structure. What is the vector bundle of $\mathbb{S}^4$ and how to calculate it? Thank you.

Edit: I have never dealt with vector bundle, so this is new to me, I would appreciate any explanation.

Edit 2: I would like to specifically address this problem with basic tools of differential geometry, so that the vector bundle can easily be calculated and become non-trivial.

juan diego rojas
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user424241
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  • There they do not answer how to calculate vector fiber and do not show that it is non-trivial, so it is basically different. – user424241 Nov 15 '19 at 01:07
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    You mean trivial vector bundle. Just having one nowhere-zero section, in fact, tells you that the Euler characteristic $\chi$ must be zero, but $\chi(S^4) = 2$. – Ted Shifrin Nov 15 '19 at 01:13

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