$\DeclareMathOperator\GL{GL}$Let $M$ be a smooth manifold, $G$ a Lie group and $P(M,G)$ a principal $G$-bundle and $\rho: G \to \GL(V)$ of $G$ a representation with $V$ finite-dimensional $\mathbb{F}$-vector space (where $\mathbb{F} = \mathbb{R},\mathbb{C}$) with inner product $\langle\cdot \, , \cdot \rangle_V$. Let $g$ be a metric in the associated vector bundle $E(M,V,G,P)$.

Now, in Kobayashi-Nomizu *"Foundations of differential geometry"* on page 117 it is stated the following:

>The set 
$$Q:= \{u \in P \, | \, g(u\xi,u\eta) = \langle\xi, \eta\rangle_V \ \forall \ \xi,\eta \in V \}$$
is a closed submanifold of $P$ and has a canonical structure as principal $H$-bundle over $M$, where 
$$H := \rho^{-1}(O(V, \langle\cdot\, , \cdot\rangle_V)) = \{a \in G \ | \ \langle \rho(a)\xi, \rho(a)\eta\rangle_V = \langle\xi, \eta \rangle_V \ \forall \ \xi,\eta \in V\}.$$

I want to proof this fact and did the following:

It is clear that 
$$Q\cdot h = Q$$ 
for all $h \in H$. Further it is clear that for $u,u' \in Q$ with $\pi(u) = \pi(u')$ and $u' = ug$ for some $g \in G$ it follows that $g \in H$.
Also $H$ is clearly a closed subgroup of $G$.

What is missing is that for all $x \in M$ there should be an open neighborhood $U \subseteq M$ of $x$ and a local section $\sigma: U \to P$ of $P$ with 
$$\sigma(U) \subseteq Q.$$
Why should this exist? Why is even $Q \neq \emptyset$? *Is this* statement above even *true*?
In Kobayashi-Nomizu it is stated that it is **"easy to verify"**.

I tried Gram-Schmidt but it didn't work:

If $G$ is extended by $\rho$ to the whole $\GL(V)$ then it should work by choosing an local orthonormal frame $(e_i)$ of $E$ and local section $\sigma$ of $P$ on $U$. Then $e_i(\pi(u)) = u\cdot f_i(u)$ for $u \in P$ for some $G$-equivariant  smooth $f_i: \pi^{-1}(U) \to V$ and $g(e_i,e_j) = \delta_{ij}$. Now choose a smooth function $b: U \to \GL(V)$ such that $b(x)^{-1} (f_1(\sigma(x)),...,f_n(\sigma(x)))$ is an orthonormal basis of $V$ for all $x \in U$. Set $\tilde\sigma := \sigma\cdot b$ and $\tilde f_i = b(\cdot)^{-1}f_i$. Then $e_i = \tilde \sigma \cdot \tilde f_i$ and thus $g(e_i,e_j) = \delta_{ij} = \langle\tilde f_i,\tilde f_j \rangle_V$.

The **problem**: $b$ has image in $\GL(V)$ and **not** in $G$. 

Also a simple example didn't work:

Take $M = P$, $G = 1$ and $V = \mathbb{R}$. Then $E = M \times \mathbb{R}$. Take the standard inner product in $\mathbb{R}$. Take some positive smooth function $f: M \to (0,\infty)$ with $f(x) \neq 1$ for all $x \in M$ and define $g_x((x,s),(x,t)) := f(x)st$ for $x \in M$ and $s,t \in \mathbb{R}$. Then $g$ is a metric in $E$ and $Q = \emptyset$. So the above is not true.

What is *wrong* here? What am I *missing*? *Is this* an *error* in Kobayashi-Nomizu?

(See also https://math.stackexchange.com/q/5011091).