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Let $P$ be a principal bundle over a manifold $M$ with structure group the Lie group $G$. Assume that $P$ admits to distinct topological reductions, say $Q_{1}$ and $Q_{2}$, where $Q_{a}$, $a=1,2$ are principal $H_{a}$-bundles over $M$. We have then:

$P = Q_{a}\times_{\rho_{a}} G$

where $\rho_{a}\colon H_{a}\to G$ is the homomorphism defining the corresponding reduction. My questions are:

1) Are there any "natural" compatibility conditions that one can impose in $Q_{1}$ and $Q_{2}$?. Let me provide an example. Let us take $M$ to be orientable and $2n$-dimensional and let us take $P$ to be its $Gl(2n,\mathbb{R})$-frame bundle. Let us further assume that $M$ it admits almost-complex structures. Then $P$ admits several different reductions:

a) $P$ admits a reduction to a $Sl(2n,\mathbb{R})$-bundle since $M$ it is orientable.

b) $P$ admits a reduction to a $O(2n,\mathbb{R})$-bundle just by choosing a Riemannian metric.

c) $P$ admits a reduction to a $Gl(n,\mathbb{C})$-bundle since it admits an almost-complex structure.

d) $P$ admits a reduction to a $Sp(2n, \mathbb{R})$-bundle since it admits a non-degenerate two-form.

At the end of the day $P$ admits a reduction to a $U(n)$-bundle, namely to the intersection of all the previous groups. Is this a generic situation (see question 2)), or this happens because there is certain compatibility condition between the different reductions a)-d) (for example, the metric $g$, the almost-complex structure $J$ and the non-degenerate two-form $\omega$ are required to satisfy $g(-,-) = \omega(J-,-)$). How can we phrase this compatibility from the point of view of principal bundles exclusively?

2) If $P$ admits topological reductions $Q_{a}$ is it true that then it also admits a topological reduction $Q_{3}$, where $Q_{3}$ is a principal bundle over $M$ with structure group $\rho_{1}(H_{1})\cap \rho_{2}(H_{2})\subset G$?

Thanks.

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    $\begingroup$ To answer (2), in general it is not true that if you have two reductions, then you can further reduce to a mutual subgroup. However, I proved a compatibility condition where this is indeed the case in my PhD thesis. I'll post a link later once I submit. $\endgroup$
    – ಠ_ಠ
    Mar 19 at 7:47
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    $\begingroup$ Excellent, I am looking forward to it! $\endgroup$
    – Bilateral
    Apr 3 at 20:08

1 Answer 1

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The bundle $p$ has a $Q_a$ reduction if and only if the $G/Q_a$ bundle $p_a:P_a\rightarrow M$ induced by $p$ has a global section. The $G/Q_1\cap Q_2$ bundle $p_{12}$ is the coimage of $(u_1,u_2):P\rightarrow P/Q_1\times P/Q_2$ where $u_a:P\rightarrow P/Q_a$ is the canonical quotient map.

Suppose that $Q_a$ is normal in $G$, then $p_a$ is a principal bundle and the existence of a global section on $P_a$ is equivalent to say that $P_a$ is trivial, in this case $p_{12}$ is trivial. In general (even if $Q_a$ is not normal) the fact that $p_a$ is trivial implies that $p_{12}$ is also trivial.

Thus $(2)$ is true if $Q_a$ is a normal subgroup or more generally if $p_a$ is trivial.

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  • $\begingroup$ Thanks for the answer. However, I am not sure I understand. The notation that you use certainly doesn't help (what is $p_{12}$ for example?). What $G/Q_{a}$ means? I think I know what you mean, but it is better to be precise. Could you pleas elaborate on this: The $G/Q_{1}\cap Q_{2} $ bundle $p_{12}$ is the pullback of $p_{1}$ and $p_{2}$. $\endgroup$
    – Bilateral
    Nov 25, 2016 at 18:35

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