The Levi-Chivita Connection $\nabla$ on a Reimannian manifold $(M,g)$, satisfies the equation $$ X(g(Y,Z))=g(\nabla_X Y,Z)+g(Y,\nabla_X Z) $$ where $X,Y,Z$ are vector fields on $M$.
It might be a ridiculous question, but I do not understand how we say that $\nabla g=0$, if the arguments of $\nabla$ are vector fields, and g can be viewed either as a 2-tensor or as a real-valued function $g_{X,Y}(p)=g|_p(X(p),Y(p))$.
Best Answer
Once you have defined $\nabla$ on scalars (just the usual differential) and vector fields (via the Levi-Civita axioms), there is a unique extension to all tensors that satisfies the product rule $$\nabla(a \otimes b) = \nabla a \otimes b + a \otimes \nabla b$$ and commutes with contractions; and this extension is by definition the derivative operator we are using when we say $\nabla g = 0.$ To see that this equation is the same thing as the metric-compatibility equation you stated, we can apply the product rule to $\nabla_Z(g(X,Y))$ to find $$\nabla_X (g(Y,Z))=(\nabla_Xg)(Y,Z)+g(\nabla_XY,Z)+g(Y,\nabla_XZ).$$
Since $\nabla_X f= Xf$ for scalars $f$, this can be rearranged to $$(\nabla g)(X,Y,Z) = (\nabla_X g)(Y,Z) = X(g(Y,Z)) - g(\nabla_XY,Z)-g(Y,\nabla_XZ);$$ so $\nabla g= 0$ is equivalent to the RHS vanishing for all $X,Y,Z.$