3
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 '16 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.