I'm reading "Elements of Noncommutative Geometry" by Garcia-Bondía. I have a questions regarding the the contraction of the Levi-Civita connection $\nabla^g$ with a vector field $Z$. First I will give you some definitions.
For a vector bundle $E\to M$ the graded $C^\infty$-module of "$E$-valued differential forms on $M$, $\Omega^\bullet (M,E)$, is defined by $\Omega^k(M,E)=\Gamma^\infty(M,E)\otimes_{C^\infty(M)}\Omega^k(M)\cong\Gamma^\infty(M,\bigwedge^kT^*M\otimes E)$.
A connection on $E$ is a linear map $\nabla:\Gamma^{\infty}\to\Omega^\bullet (M,E)$ satisfying the Leibniz rule $\nabla(s\otimes\omega)=(\nabla s)\otimes\omega+s\otimes\mathrm{d}\omega \quad \mathrm{for}\,s\in\Gamma^{\infty}(M,E), \omega\in \Omega^\bullet(M).$
Define the operator $\nabla_X:=\iota_X\circ\nabla+\nabla\circ\iota_X$, where $\iota_X$ is given by $\iota_X(s\otimes\omega)=s\otimes\iota_X\omega$ with $\iota_X\omega$ the contraction of forms.
Let $(M,g)$ be a Riemannian manifold. Then the Levi-Civita connection $\nabla^g$ on $TM$ is the unique affine, torsion-free connection satisfying: $\quad g(\nabla^gX,Y)+g(X,\nabla^gY)=\mathrm{d}(g(X,Y)) \quad \forall\,\,X,Y\in\mathfrak{X}(M,\mathbb{R}).$
It was mentioned that it holds that $g(\nabla^g_ZX,Y)+g(X,\nabla^g_ZY)=Z(g(X,Y))$. I have no idea how to show this.
Thanks for your help.
The equation you want to prove is exactly the equation written in 4., just using the notation $d(g(X,Y))$ as the covectorfield associated to $g(X,Y)$. This covectorfield applied to $Z$ yields $Z(g(X,Y))$. The left hand side of the equation in 4. is just waiting for the $Z$ to be inserted.