Why is $\frac{d}{dt}|\xi(t)|^2=2<\xi, \nabla \xi>$, when $\nabla$ is the covariant derivative?

Question

I understand that:

$(1) \frac{d}{dt}|\xi(t)|^2=2<\xi, \nabla \xi>$

if this is happening in $\mathbb{R}^n$. But my question is in the following setting: $\nabla \xi:=K_{c^*\pi} \circ T \xi.\partial t$, where $K_{c^*\pi} $ is the connection induced by the induced bundle of a function $c:S \rightarrow M$, where $S$ is the sphere and $M$ is the manifold (which I'm supposing has a connection on a bundle $\pi : E \rightarrow M$, and $\partial t$ is the canonical tangent vector of the sphere - This is according to Klingenberg's Lectures on Closed Geodesics).

I'm having trouble understanding why $(1)$ holds in this case.

Answer 1

In Riemannian manifold $(M,g)$, $X(f)$ implies that if $c(t)$ is a curve with $c'(0)=X$ then $$ X(f)=\frac{d}{dt}\bigg|_{t=0} f(c(t)) $$

So $$\frac{d}{dt} |\xi|^2= X |\xi|^2=X g(\xi,\xi) $$

In $\mathbb{R}^3$, $M$ is a surface. Then covariant derivative is defined as : $$ \nabla_X \xi = Proj_{T_{c(0)} M} \frac{d}{dt}\xi $$

where $Proj$ is a projection. So if $g$ on $M$ is a metric induced from $\mathbb{R}^3$, then $$ Xg(\xi,\zeta) = g( \frac{d}{dt}\xi,\zeta) + g( \xi, \frac{d}{dt}\zeta) =g(\nabla_X\xi,\zeta) + g( \xi,\nabla_X\zeta) \ \ast$$

In general case, $\nabla$ is defined to satisfy $\ast$ (cf. Levi-Civita connection)