Gradient of a function and inverse of metric

Question

Having some knowledge in differential geometry on $\mathbb{R}^n$, I'm reading a book on Information geometry by Amari. Let $S=\{p_{\theta}\}$, $\theta=(\theta_1,\dots,\theta_n)$ be an $n$-dimensional manifold parameterized by $\theta$. At some point the author writes, for a (differentiable) funtion $f:S\to \mathbb{R}$, $$(\text{grad}~f)_p=\sum_{i,j}\left(\frac{\partial f}{\partial \theta_i}\right)_pg^{i,j}\left(\frac{\partial}{\partial \theta_i}\right)_p,$$ where $[g^{i,j}]$ is the inverse of the Riemannian metric $[g_{i,j}]=[\langle \frac{\partial}{\partial \theta_i},\frac{\partial}{\partial \theta_j}\rangle]$. Can someone explain how does one arrive at this expression? Thank you.

Answer 1

As you mentioned, the gradient of a function $f$ is defined as its dual vector of $df$, we can immediately write $$(\mathrm{grad} f(p))^i=g^{ij}(df(p))_j$$

Proof: $$g_{ij}X^j=\left<\partial_i,\partial_j\right>X^j=\left<\partial_i,X^j\partial_j\right>=\left<\partial_i,X\right>=X_i$$

And we know $df(p)_j=\partial_j f(p)$, so it follows $$\mathrm{grad} f(p)=(\mathrm{grad} f(p))^i\partial_i=g^{ij}\partial_jf(p)\partial_i$$