Existence (or non existence) of principal bundle charts compatible with an $f$-reduction

Question

I asked this question on math stack exchange here, but I wonder if it would be better received here.

Let $\pi:P\rightarrow M$ and $\pi':P'\rightarrow M$ be principal $G$ and $H$ bundles respectively, and $f:G\rightarrow H$ a Lie group homomorphism. Recall that an $f$-reduction of principal bundles is the principal bundle $P$ together with a smooth bundle homomorphism $F:P\rightarrow P'$ that is $f$ equivariant. In other words we have the following two properties: $$\pi'\circ F=\pi$$ and: $$F(p\cdot g)=F(p)\cdot f(g)$$ for all $p\in P$, and $g\in G$.

My question is then this: let $U\subset M$ be an open neighborhood of $x\in M$, and $(U,\phi)$ be a principal bundle chart for $P'$, that is an $H$ equivariant diffeomorphism: $$\phi:\pi'^{-1}(U)\longrightarrow U\times H$$ Does there then exist a principal bundle chart $(U,\psi)$ for $P$ that is compatible with the $f$ reduction, in the sense that: $$\phi\circ F=(\text{Id}_U\times f)\circ \psi$$ In the case where $f$ is a Lie group isomorphism I believe this trivial. If $f$ is an embedding, then $F$ is a $G$ reduction of $H'$, and so $P$ is a principal subbundle of $P'$. In this case, let: $$\phi(p')=(\pi'(p'),h(p'))$$ and: $$\psi(p)=(\pi(p),g(p))$$ for some smooth equivariant maps $g:P_U\rightarrow G$ and $h:P_U\rightarrow H$, then our requirement reduces to: $$\phi(F(p))=(\pi'\circ F(p),h(F(p)))=(\pi(p),f(g(p)))=(\text{Id}_U\times f)\circ \psi(p)$$ since $F$ is a bundle homomorphism we have that: $$h(F(p))=f(g(p))$$ Which I believe is always the case, due to the equivariance of $F$ and $h$, but I can't rigorously demonstrate why. If I am mistaken please let me know.

The case I am more interested is when $f$ is a surjective Lie group homomorphism. In particular, this implies that $G/\ker f\cong H$, as Lie groups, and that $f$ is a smooth submersion. In this case, I strongly suspect that it is not true, as I suspect it would imply that global sections of $f$ exist, which is false in general.

My motivation for this problem, is to specifically apply it when $M$ is an oriented Riemannian spin manifold, and $P=\text{Spin}(M)$, and $P'=SO(M)$, i.e. $P$ is a spin structure associated to the bundle of oriented orthonormal frames.

If such charts existed, I would be easily able to prove that for every local section of $SO(M)$, there exist precisely two local sections of $\text{Spin}(M)$ that map to original local section of $SO(M)$ under $F$.

Any help would be greatly appreciated, as I have not been able to find anything regarding questions such as this in the literature.

Edit: At least in the case of $\text{Spin}(M)$ and $SO(M)$, I believe the existence of such charts to be true actually. This is because th existence of such sections of $\text{Spin}(M)$ implies the existence of such charts.

Answer 1

The answer is yes, in full generality, after possibly restricting to a smaller open set where $P$ admits a principal bundle chart. The restriction to a smaller open set can be seen to be essential by considering the case where $P'$ is the (unique) trivial-group bundle over $M$.

This is straightforward to see using the perspective that the existence of a section $\sigma$ of a principal bundle $P$ on an open set $U$ is equivalent to $P$ admitting a principal bundle chart on $U$. (Given a section $\sigma$ for $x \in U$, every point in the fiber over $x$ can be written uniquely as $\sigma(x)g$ for some $g \in G$. The chart is $\sigma(x)g \mapsto (x,g)$. Conversely, take the identity section of the product chart).

So let $U \subset M$ be a neighborhood so $P$ has a local section $\sigma$ on $U$. Then, in your notation, $F \circ \sigma$ is a section of $P'$ on $U$, and the $f$-compatibility condition $F(\sigma(x)g) = F(\sigma(x)) f(g)$ gives exactly what you want in the associated charts for $\sigma$ and $F \circ \sigma$ as above.

EDIT: If you want a chart for $P$ compatible with the specific chart $\phi$ on $U$, this can be arranged for surjective homomorphisms $f$ by using the fact that Lie group homomorphisms have constant rank, and so surjections admit local models based on standard projections $\mathbb{R}^n \to \mathbb{R}^k$ in a neighborhood of $e$. In particular, if we denote by $V_G, V_H$ neighborhoods of $e$ in $G,H$ so $f|_{V_G}$ has a standard model in charts for $V_G, V_H$, there is a smooth section $\nu: V_H \to V_G$ of $f$, i.e. one has $f \circ \nu = \text{id}$.

To use this, let $\sigma'$ be the section corresponding to $\phi$. Given any two sections $\sigma_1, \sigma_2$ of $P'$ on $U$, they differ by a map $\alpha: U \to H$ (i.e. so $\sigma_1(x) = \sigma_2(x)\alpha(x))$. So let $\alpha$ be so $\sigma'(x) = F(\sigma(x))\alpha(x)$ for $x \in U$. Pick $p \in U$. Modifying $\sigma$ by some $g \in G$, we can arrange for $F(\sigma(p)) = \sigma'(p)$. In a neighborhood $U''$ of $p$, in the chart for $P'$ induced by $F \circ \sigma$, the section $\sigma'$ is valued in $U \times V_H$, and so $\nu: V_H \to V_G$ induces a section $\eta: U \to P$ by specifying $\eta(x) = \sigma(x)\nu(\alpha(x))$ for $x \in U''$. Then use $\eta$ in place of $\sigma$ in the above construction to get the desired chart. By construction, $F(\eta(x)) = F(\sigma(x))f(\nu(\alpha(x))) = \sigma'(x)$, so one obtains a compatible chart for $P$.