Suppose that $P,N,M$ are smooth manifolds and that there exist two maps $\pi:P\rightarrow N$ and $\pi':P\rightarrow M$ such that $\pi$ and $\pi'$ are principal bundles with structure groups $G$ and $H$, respectively, where $H$ is a closed subgroup of $G$, with $G$ compact. Moreover, there exists a map $\pi'':N\rightarrow M$ which is a fiber bundle such that $\pi=\pi''\circ\pi'$.
Since $G$ is compact, there exists an $\operatorname{Ad}(G)$-invariant inner product $<,>$ on the Lie algebra $\mathfrak{g}$ of $G$. The Lie algebra $\mathfrak{h}$ is a subalgebra of $\mathfrak{g}$, therefore we can decompose $\mathfrak{g}$ into a direct sum
$$\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{t},$$
where $\mathfrak{t}$ is the orthogonal complement of $\mathfrak{h}$ in $\mathfrak{g}$ with respect to $<,>$. Suppose that $\omega$ is a connection form on the principal bundle defined by $\pi$: since $\omega$ takes values in $\mathfrak{g}$, we have that $\omega=\omega_{\mathfrak{h}}+\omega_{\mathfrak{t}}\in\mathfrak{h}\oplus\mathfrak{t}$.
If $V$ and $V'$ denote the vertical bundles of $\pi$ and $\pi'$, respectively, is it true that $\omega_{\mathfrak{t}}$ vanishes identically on $V'$? My idea was to prove that $\omega$ takes values in $\mathfrak{h}$ if evaluated on any vertical vector in $V'$, but I am not sure if it holds.
Best Answer
Let $p:P\rightarrow M$ be a principal $G$-bundle. Every element $A$ of the Lie algebra ${\cal G}$ of $G$ defines a vector field $X_A$ such that for every $x\in P$, $X_A(x)={d\over{dt}}_{t=0}x.exp(tA)$. For every $u\in T_xP$ such that $dp_x(u)=0$, there exists $A\in{\cal G}$ such that $X_A(x)=u$.
A connection defined on a principal $G$-bundle $p:P\rightarrow M$ is a $1$-form $\omega$ defined on $P$ which takes its values in ${\cal G}$ the Lie algebra of $G$ such that for every $x\in P$, $\omega_x(X_A(x))=A$ and $R_g^*\omega=Ad(g^{-1})\omega$.
Let $v'\in V'_x$ there exists $A\in \mathfrak{h}$ such that $X_A(x)=v'$, we have $\omega(v')=\omega_{\mathfrak{h}}(v')+\omega_{\mathfrak{t}}(v')=A'$ implies that $\omega_{\mathfrak{t}}(v')=0$ since it is an element of $\mathfrak{t}$.