I am following the proof of Ambrose-Singer theorem for Riemannian homogeneous spaces in this book https://www.cambridge.org/core/books/homogeneous-structures-on-riemannian-manifolds/EA6AACEAF54C4AED688DF0C014289250.
Let $(M,g)$ be a connected, simply connected, and complete Riemannian manifold and $O(M,g)$ be the orthonormal frame bundle. We denote by $B_1, \ldots, B_n$ the standard horizontal vector fields with respect to the metric connection $\tilde{\nabla}$ and the standard basis of $\mathbb{R}^n$.
My question is that why is the holonomy bundle $\tilde{J_u} = \{v \in O(M,g) \mid u\sim v\},$ where $u\sim v$ means that there exists a horizontal curve $\gamma \colon [0,1] \to O(M,g)$ from $u$ to $v$, parallelizable?
There is a following sentence "The restrictions of $B_1, \ldots, B_n, A^\#_1, \ldots, A^\#_r$ determine an absolute parallelism on $\tilde{J_u}$" in this book on page 27.
I know that the fundamental vector fields $A^\#_1, \ldots, A^\#_r \subset \Gamma(T\tilde{J_u})$ of $A_1, \ldots, A_r \subset \mathfrak{hol}(n)$ defines a global trivialization of vertical subbundle $V \subset T\tilde{J_u}.$ However, I still be wondering whether the standard horizontal vector fields $B_1, \ldots, B_n$ consist of a global frame of the horizontal subbundle $H \subset T\tilde{J_u}$ or not.