When is a principal bundle with group $G \times H$ the product of a principal bundle with group $G$ and one with group $H$?

Question

In general, it is not true that a fiber bundle with a product fiber is the product of two fiber bundles with the factors as fibers (think of vector bundles). However, I read somewhere that every 2-torus bundle is the product of two circle bundles, and I do not know why. So, my question is both why is this true in the torus case and what can we say in the more general case of the product of two Lie groups?

Answer 1

I think a principal bundle with structure group $G\times H$ is always a product in the sense given in the comments.

Specifically, we have the following

Proposition: Suppose $P\xrightarrow{\pi} M$ is a principal $G\times H$ bundle. Then there is a principal $G$ bundle $P_G\rightarrow M$ and a principal $H$-bundle $P_H\rightarrow M$ with the property that $P$ is isomorphic to the pullback of the $G\times H$ bundle $P_G\times P_H\rightarrow M\times M$ under the diagonal map $M\rightarrow M\times M$.

Before proving this, let me just mention some notation I'll be using. I will identity $G$ with $G\times \{e\}$ and $H$ with $\{e\}\times H$ (where $e$ denotes the identity of either group.). Given $p\in P$, I'll write multiplication on the left: for $(g,h)\in G\times H$, I'll write $(g,h)\ast p = ghp$, where I am thinking of $g$ as $(g,1)$ and $h$ as $(1,h)$. The $G$, $H$, and $(G\times H)$ orbits through $p$ will be denoted by $pG$, $pH$, and $p(G\times H)$ respectively. The letters $g$ and $h$ will always indicate members of $G$ and $H$ respectively.

Proof: Consider first the manifold $P_G:= P/H$. This is a smooth manifold upon which $G$ natural acts via $g\ast (pH) = gpH$. This action is again free, and $(P_G)/G$ is naturally diffeomorphic to $M$. In particular, the quotient map $P_G\rightarrow (P_G)/H\cong M$ is a principal $G$-bundle. Similarly, one can consider the manifold $P_H:=P/G$.

Form the principal $(G\times H)$-bundle $P_G\times P_H$ over $M\times M$, and let $i:M\rightarrow M\times M$ denote the diagonal map $i(m) = (m,m)$.

We claim that $i^\ast(P_G\times P_H) \cong P$.

Then one can easily verify that the map $\phi:P\rightarrow i^\ast(P_G\times P_H)$ given by $\phi(p) = (\pi(p), pH, pG)$ is a $(G\times H)$ equivariant map covering the identity.

Proving $\phi$ is a bijection is a bit more difficult, so I'll provide the proof.

Injectivity: Assume $\phi(p) = \phi(p')$, so $pH = p'H$ and $pG = p' G$. This means that there is an $h\in H$ and a $g\in G$ with $p = hp'$ and $p = gp' $. Thus, $hp' = gp'$, so $(g^{-1}h)p' = p'$. This tells us that $g^{-1}h = (e,e)$, so $g = h$. However, $g\in G$ and $H\in H$, and $G\cap H = \{(e,e)\}$, so we must have $g = h = e$. Thus, $p = p'$, as desired.

Surjectivity: Suppose $(m, p_1 H, p_2 G)\in i^\ast(P_G\times P_H)$. Because $p_1(G\times H) = m= p_2(G\times H)$, there is a $(g,h)\in G\times H$ with $gp_1 = hp_2$. We can rewrite this as $h^{-1} p_1 = g^{-1}p_2$.

Consider the point $g^{-1} p_2 \in P$. We claim that $\phi(g^{-1}p_2) = (m, p_1 H, p_2 G)$. Since $\phi(g^{-1}p_2) = (\pi(g^{-1} p_2), g^{-1} p_2 H, g^{-1}p_2 G)$, we will verify the equality $$(\pi(g^{-1} p_2), g^{-1} p_2 H, g^{-1}p_2 G) = (m, p_1 H, p_2 G)$$ one coordinate at a time.

For the first coordinate, we find $\pi(g^{-1} p_2) = \pi(p_2) = m$, as desired. For the third coordinate, we have $g^{-1} p_2 G = p_2 G$ as desired.

The second coordinate works as follows: $g^{-1} p_2 H = h^{-1} p_1 H = p_1 H.$ $\square$