Let $M$, $P$ and $Q$ be C$^\infty$-manifolds.
Let $\pi_E:E\rightarrow M$ be a $\mathbb{C}$-vector bundle of rank r.
Let $\pi_P:P\rightarrow M$ be the frame bundle of $E$. Here, $P$ is a principal GL($\mathbb{C}^r)$-bundle and for $x\in M$, $\pi_P^{-1} (x)=\{\zeta:\mathbb{C}^r\rightarrow \pi_E^{-1}(x)$|$\zeta$ is an isomorphism as a vector space$\}$.
Let $h$ be a fiber metric on $E$ and $(\ ,\ )_{\text{std}}$ be the standard metric of $\mathbb{C^r}$.
Let $\pi_Q:Q\rightarrow M$ be the frame bundle of $(E,\ h)$. Here, $Q$ is a principal $U(r)$-bundle and for $x\in M$, $\pi_Q^{-1} (x)=\{\zeta\in\pi_P^{-1}(x)|\zeta$ preserves the metrics$\}$.
Let $\theta$ be a connection form on $P$ (i.e. a one-form on $P$ having values on End$(\mathbb{C}^r)$ and metting some conditions).
Let $\nabla^\theta$ be the connection induced by $\theta$.
I will ask why "$\nabla^\theta$ preserves $h$" and "$\theta$ is reducible to $Q$" are equivalent.
Here, "$\theta$ is reducible to $Q$" means that $Q$ is a subbundle of $P$ and for any $\zeta\in Q$, Ker $\theta_\zeta\subset T_\zeta Q$. I have proved that $Q$ is a subbundle of $P$ by myself.
I know that when we define $A_\alpha$ as the represent matrix of $\theta$, "$\nabla^\theta$ preserves $h$" means that the sum of $A_\alpha$ and the conjugate of $A_\alpha$ is equal to $0$.
Also, I know that the Lie algebra of $U(r)$ is the set of matrices meeting the condition the sum of those and the cojugate of those is equal to $0$.
So, If the latter condition of "$\theta$ is reducible to $Q$" were "$\theta_\zeta\subset T_\zeta Q(=$ the Lie algebra of $U(r)$)", I would have no problem.
However, the condition is not "$\theta_\zeta\subset T_\zeta Q"$ but "Ker $\theta_\zeta\subset T_\zeta Q$".
Please advise me.