Relation between gradient of a function with respect two linear connection

Question

Let $\bar{\nabla}$ and $\nabla$ be two linear connection over $(M,\bar{g})$ and $(M,g)$ resp. and suppose that this two connection are related by the following equality: $$\bar{\nabla}_XY=\nabla_XY+\psi(X,Y),\qquad \forall\, X,Y\in \mathfrak{X}(M),$$ where $\psi$ is a given $(2,1)$-tensor field. Let $f$ be a smooth function on $M$. Then How $\bar{\nabla}f$ and $\nabla f$ are related (and then their Hessians)?

My attempt: $$\bar{\nabla} (fY)=(\bar{\nabla} f)Y+f\bar{\nabla}Y=(\bar{\nabla} f)Y+f\nabla Y+f\psi(.,Y),$$ $$\bar{\nabla}(f Y)=\nabla(fY)+\psi(., fY)=(\nabla f)Y+f\nabla Y+f\psi(.,Y),$$ So $(\nabla f)Y=(\bar{\nabla} f)Y$ for all $Y$. Thus $\bar{\nabla}f=\nabla f$. Is this correct?

Answer 1

Yes, it's correct. For any connection $\nabla$, the covariant derivative of a scalar function satisfies $\nabla f = df$, so it doesn't depend on the connection. The connection only makes a difference when you take covariant derivatives of vector or tensor fields.

Answer 2

According to definition of Hessian: $${\rm Hess}\, f(X,Y)=X.(df(Y))-df(\nabla_XY),$$ we have \begin{align}\overline{{\rm Hess}}\, f(X,Y)=X.(df(Y))-df(\bar{\nabla}_XY)&=X.(df(Y))-df(\nabla_XY+\psi(X,Y))\\ &={\rm Hess}\, f(X,Y)-df(\psi(X,Y)). \end{align}