Principal bundles where the the fibre is a product Lie group

Question

Let $P$ be a $G\times H$ principal bundle over a smooth manifold $M$. Then the transition functions $\phi_{ij}$ take value in $G\times H$, and if we set the projection $\pi_G:G\times H\rightarrow G$ it seems that we can obtain transition functions for a principal $G$ bundle by setting: $$g_{ij}=\pi_G\circ g_{ij}$$ Since $\pi_G$ is a Lie group homomorphism, I think these transition functions then determine a principal $G$ bundle, as they should easily satisfy the cocycle conditions.

Is this true? Is there a more constructive way of doing this by constructing a formal bundle atlas, and showing that the the bundle admits a smooth right $G$ action which is free and transitive on the fibres?

I feel like there should then also be a principal bundle homomorphism $P\rightarrow P_G$, but I am unsure of how to define because this construction seems to tell me very little about the actual structure of $P_G$. Any help would be greatly appreciated.

Answer 1

The bundle you need can be given in terms of its trivializations. Consider a local trivialization $\{U_j\}$ for the principal bundle $P_{G\times H} \to M$ with $(G\times H)$-equivariant diffeomorphisms $$ h_j : P|_{U_j} \longrightarrow U_j \times(G\times H) $$ then $P|_{U_j}$ has a free $H$-action which is also proper because $H$ naturally acts by right multiplication on $U_j \times (G \times H)$. We can then consider the quotient for this action obtaining $$ k_j : \frac{P|_{U_j}}{H} \longrightarrow \frac{U_j \times (G \times H)}{H} \simeq U_j \times G $$ which is also diffeomorphic since $h_j$ was.

Now we only need to check $k_j$'s define transitions fuctions of a principal $G$-bundle:

• They are $G$-equivariant: $$ k_j(p.g) = h_j(p.gH) = h_j(p).gH = k_j(p).g $$
• The action trivially preserves fibres
• The action is free and transitive on each fibre: Consider $x \in U_j$ and $p \in P/H|_x$, then $p = k_j(x, g_p)$, since $k_j$ is equivariant, if $p = p.g$, we find $$ k_j(x, g_p) = k_j(x, g_p).g = k_j(x, g_p \, g)$$ whence $g = 1$ since $k_j$ is injective. For transitivity notice that for any $q \in P/H|_x$ we have $$ q = k_j(x, g_q) = k_j(x, 1_G) g_q =:p_{0} . g_q $$ and it's easy to see that $g_q$ does not depend on the trivialization choice.