I was doing some calculations, but now I am stuck at a silly point. Suppose $A\in $GL$(n,\mathbb{R})$ and suppose $X$, $Y$ and $Z$ are vector fields on a Riemannian manifold $M$. Consider the Levi-Civita connection $\Delta$.
I am interested in calculating, for example, what is $X(g(AY,AZ))$, where $g$ is the Riemannian metric.
I know that $Xg(AY,AZ) = g(\Delta_XA Y, AZ)+g(A\Delta_XY, AZ)+g(AY, \Delta_XAZ)+g(AY, A\Delta_XZ)$.
But, I don't know that how can I simply it further. In particular, can I do anything further with the first two terms on the RHS of the above equation ? How can I make sense of the covariant derivative of the matrix $A$ along the vector field $X$ ?
Thanks
An $n \times n$ matrix $A$ does not determine an endomorphism of tangent spaces; that is, "$AX$" is not defined unless you fix a local trivialization of the tangent bundle.
Your Leibniz rule identity treats $A$ as a field of endomorphisms of the tangent bundle, in which case the covariant derivative $\nabla_{X}A$ has its usual meaning. (Presumably you meant to write $$ Xg(AY,AZ) = g\bigl((\nabla_{X}A)Y, AZ\bigr) + g\bigl(A(\nabla_{X}Y), AZ\bigr) + g\bigl(AY, (\nabla_{X}A)Z\bigr) + g\bigl(AY, A(\nabla_{X}Z)\bigr), $$ however, with the covariant derivative of an endomorphism field acting on a vector field rather than vice versa.)