Let $(M,g)$ be a Riemannian manifold and let $f \in C^{\infty}(M)$. Let $X$ be a smooth vector field on $M$. In smooth local coordinates $(x^i)$ on $M$, we can write $g = g_{ij} dx^i \otimes dx^j$ as well as $X = X^i \partial_{x^i}$. Now I have computed that $\langle\operatorname{grad} f, X \rangle_g = X^i\frac{\partial f}{\partial x^i}$ as follows. If we let $g^{ij}$ be the inverse matrix of $g_{ij}$ We have
$$\begin{eqnarray*} \langle\operatorname{grad} f, X \rangle_g &=& \left\langle g^{ij} \frac{\partial f}{\partial x^i}\partial_{x^j} , X^k\partial_{x^k} \right\rangle_g\\
&=& g^{ij} \frac{\partial f}{\partial x^i}X^k\left\langle \partial_{x^j},\partial_{x^k} \right\rangle_g\\
&=& g_{jk}g^{ij} \frac{\partial f}{\partial x^i}X^k \\
&=& g_{kj}g^{ji}\frac{\partial f}{\partial x^i}X^k \hspace{1cm} (\text{by using that $g_{ij}$ is a symmetric matrix})\\
&=& X^i\frac{\partial f}{\partial x^i}. \end{eqnarray*}$$
My question is: Is there any geometric reason as to why $\langle \operatorname{grad} f, X \rangle_g$ should be independent of the metric $g$?
Best Answer
Yes, the reason is simple. By definition, $\operatorname{grad}f$ is the vector field (metric-dependent) so that $\langle\operatorname{grad}f,X\rangle_g = df(X)$, and the right-hand side is independent of the metric. But the vector field itself does depend on the metric. :)