From Do Carmo 《Riemannian Geometry》
Definition:
A connection is said to be compatible with the metric $\langle ,\rangle$, when for any smooth curve $c$ and any pair of parallel vector fields $P$ and $P'$ along $c$, we have $\langle P,P'\rangle =constant$.
it is equivalent to the proposition
Proposition:
Let $M$ be a Riemannian manifold. A connection $\nabla$ on $M$ is compatible with a metric if and only if for any vector fields $V$ and $W$ along the differentiable curve $c:I \to M$ we have
$\frac{d}{dt}\langle V,W\rangle =\langle\frac{DV}{dt},W\rangle+\langle V,\frac{DW}{dt}\rangle$,$ t \in I$
and here is the proof in the textbook
$proof$.It is obvious that the proposition implies that $\nabla$ is compatible with $\langle ,\rangle$. Therefore, let us prove the converse. Choose an orthonormal basis {$P_1(t_0),...,P_n(t_0)$} of $T_{x(t_0)}(M)$,$t_0 \in I$. We can extend the vectors $P_i(t_o)$ along $c$ by parallel translation. Because $\nabla$ is compatible with the metric, {$P_1(t),...,P_n(t)$} is an orthonormal basis of $T_{c(t)}(M)$, for any $t\in I$. Therefore, we can write
$V=\sum\limits_{i}v^iP_i,$ $W=\sum\limits_{i}w^iP_i$
where $v^i$ and $w^i$ are differentiable functions on $I$. It follows that
$\frac{DV}{dt}=\sum\limits_{i}\frac{dv^i}{dt}P_i,\frac{DW}{dt}=\sum\limits_{i}\frac{dw^i}{dt}P_i$
Therefore,
$\langle\frac{DV}{dt},W\rangle+\langle V,\frac{DW}{dt}\rangle=\sum\limits_{i}\left\{\frac{dv^i}{dt}w_i+\frac{dw^i}{dt}v_i\right\}=\frac{d}{dt}\left\{\sum\limits_{i} v^iw^i\right\}=\frac{d}{dt}\langle V,W\rangle$.
My question is:
In the proof, I can't see the necessity of the compatible condition, what I mean is that I think without the compatible condition the proposition can still be induced, so can someone help me to figure out: in the process of proving, at where the compatible condition is used and why it is necessary?
Thanks!
It is there explicitly, in the sentence "Because $\nabla$ is compatible with the metric ...". The fact that the scalar product does not change when the basis is subject to parallel transport is true because of this assumption.