Let $\pi:P\to M$ be a principal $G$-bundle and $K$ a maximal compact subgroup of $G$. Then, $G/K$ is homeomorphic to a vector space $V$, so $P/K\to M$ is a fibre bundle whose fibres are homeomorphic to $V$. Is there a natural vector bundle structure on $P/K\to M$?
Vector bundle obtained by a maximal compact subgroup
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general-topology
differential-geometry
vector-bundles
topological-groups
principal-bundles
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0Does it have a section, usually? – 2017-01-22
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0@MarianoSuárez-Álvarez I don't know. Does it? – 2017-01-22
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0@MarianoSuárez-Álvarez For example, $\mathrm{GL}(n,\mathbb{C})$-bundles can always be reduced to $U(n)$ by choosing a metric, but I am not sure if this is valid in this more general context. – 2017-01-22
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1I don't think so. Consider the case where $M$ is a point, $G = \operatorname{GL}(n)$ and $K = \operatorname{O}(n)$. The quotient $G/K$ is the space of inner products on $\mathbb{R}^n$ which is homeomorphic to the collection of positive definite matrices. As far as I know, there isn't any meaningful way to attach a vector space structure the the space of inner products / positive definite matrices and I don't see any reason for it to be possible globally. – 2017-01-22
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1@levap: What about putting a vector space structure on the space of positive definite matrices by using the exponential map? Since exp is a bijection from symmetric matrices to positive definite ones, you could define an addition $\boxplus$ on the positive definite matrices by requiring that $\exp(A+B) = (\exp A)\boxplus (\exp B)$. I haven't thought this through very deeply, but one guess is that the resulting vector bundle is just the bundle of symmetric $2$-tensors. – 2017-01-23
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0@JackLee: Hmm, yeah, this seems to work but I'm not sure it will honor the natural action involved. Namely, we can use $\exp$ to identify $G/K$ with the vector space $S(n)$ of symmetric matrices and then translate the natural action of $G$ on the quotient $G/K$ to an action of $G$ on $S(n)$ and look at the associated fiber bundle. However I don't think this action will be linear. Of course one can choose a different action of $G$ on $S(n)$ such as conjugation but then I wouldn't call the resulting vector bundle natural. – 2017-01-23