Let $P_{H}$ be a principal bundle over a manifold $M$ with fiber the Lie group $H$ and let $P_{G}$ be a principal bundle with fiber the Lie group $G$ over the same manifold $M$. Let $h_{ab}\colon U_{ab}\to H$ and $g_{ab}\colon U_{ab}\to G$ be the corresponding $\check{\mathrm{C}}$ech cocycles (rather, some representatives), where $U_{ab}$ denotes the non-empty intersection of open sets in a common trivializing open cover $\mathcal{U}$ of $M$. Let $\tau\colon H \to Aut(G)$ be an homomorphism, so we can consider the semidirect product $G\rtimes_{\tau} H$. I was trying to build the "semidirect product bundle" of $P_{H}$ and $P_{G}$ by taking the "pointwise" semidirect product of $g_{ab}$ and $h_{ab}$. In other words, I take $\mathcal{U}$ and define a new bundle $P_{G\rtimes H}$ by the following transition functions:

$\tilde{g}_{ab} = (g_{ab},h_{ab})\colon U_{ab} \to G\rtimes_{\tau} H$.

However, $\tilde{g}_{ab}$ does not satisfy the cocycle condition. Is there any sensible way to define the semidirect product of two principal bundles?



1 Answer 1


I don't think so unless $M$ has the homotopy type of a sphere (or more generally, the suspension of a based space): there is a fibration of classifying spaces $$ BG \to B(G\rtimes H) \to BH\, . $$ This comes from the (split) short exact sequence $1\to G\to G\rtimes H \to H\to 1$.

One of your bundles is classified by a map $M \to BG$ and your other one by a map $M \to BH$. Pushforward the first map and apply the section to the second one to get a pair of maps $f,g: M \to B(G\rtimes H)$.

If we could add these maps then you would have a model for the semi-direct product bundle. If $M$ is a homotopy sphere this is clear since homotopy classes maps out of a sphere form a group.

  • 1
    $\begingroup$ Thanks. In proposition 4 of math.muni.cz/~kolar/KOLAR26.pdf they seem to construct a principal bundle with group $G\rtimes_{\tau} H$ from a $P_{H}$ bundle and a $P_{G}$ bundle. $\endgroup$
    – Bilateral
    Mar 23, 2016 at 0:58
  • 4
    $\begingroup$ I looked at your link: it seems that they are doing something different from that: they have a principal $G$-bundle $P \to M$ and "group object in the category of bundles" $Q\to M$ that acts fiberwise on the right: $P \times_M Q \to P$, Then they construct a fiberwise semi-direct product of that over $M$. This is different from the question you are asking (it seems). $\endgroup$
    – John Klein
    Mar 23, 2016 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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