Two linear conections on a Riemannian manifold and one ... !!!

Question

Let $(M,g)$ be a Riemannian manifold and $\nabla$ and ${\nabla}^*$ be two linear connections. If $f$ is a real valued function defined on $M$ and $X$ is a vector field,
my question is: Does the value of $Xf$ change depending on the connection?
If yes, what is the deference?
If no, can I get the value of $X(g(Y,Z))$ for a linear connection using a Riemannian one?

Answer 1

There is some (intentional) ambiguity in the term "connection" and the symbol $\nabla$.

A "linear" connection $\nabla$ is defined to be a map $\nabla \colon \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)$ which is $C^{\infty}(M)$-linear in the first slot and satisfies a Lebniz-like rule $$ \nabla_X (f Y) = \mathrm{d}f(X) Y + f \nabla_X Y $$ where $\nabla_X Y$ is a shortcut for $\nabla(X,Y)$, and $f \in C^{\infty}(M)$. Thus, initially $\nabla$ is defined to act only on vector fields.

By insisting on the Leibniz rule we extend the action of $\nabla$ to all tensor fields, that is sections of finite tensor products of some copies of $T M$ and $T^* M$.

The Leibniz rule for this extended connection is stated as $$ \nabla_X (T \otimes S) = (\nabla_X T) \otimes S + T \otimes \nabla_X S $$ where $T,S$ are some tensor fields.

Comparing the two forms of the Leibniz rule we arrive to a convention that smooth functions can be treated as tensors of zero rank, and by definition $f \otimes T : = f T$.

To complete the picture we need to extend the action of $\nabla$ to smooth functions, and if we want the entire construction to be natural, the only possibility is to set $\nabla_X f := \mathrm{d}f(X)$.

In the nutshell, all connections agree on smooth functions.

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To compute the value of $X(g(Y,Z))$ we just use the rules and the information available. Suppose, you have a linear connection $\nabla$ as above, and the Riemannian metric on $M$, so that you also have the Levi-Civita connection $\nabla^{LC}$. Recall that $\nabla^{LC} g = 0$, and $\nabla^{LC}$ is torsion free.

Let $S(X,Y)$ be the difference tensor of the two connections $\nabla$ and $\nabla^{LC}$, that is $$ \nabla_X Y = \nabla^{LC}_X Y + S(X,Y) $$

Now, $$ X(g(Y,Z)) = \nabla_X g (Y,Z) = \nabla g (X,Y,Z) $$

To proceed, we need to understand how the connections differ on covariant tensors.

Claim. $\nabla_X \omega (Y) = \nabla^{LC}_{X} \omega (Y) - \omega(S(X,Y))$, where $\omega$ is a 1-form on $M$.

Using the multilinearity, we can find that $$ \nabla_X g (Y,Z) = \nabla^{LC}_X g (Y,Z) - g(S(X,Y),Z) - g(Y,S(X,Z)) $$ but $\nabla^{LC}_X g (Y,Z) = 0$, so finally $$ \nabla_X g (Y,Z) = - g(S(X,Y),Z) - g(Y,S(X,Z)) $$