Gradient vector derived from the metric tensor According to Frankel's book "The Geometry of Physics", the components of a contravariant gradient vector can be obtained from the inverse of the metric tensor as follows (in section 2.1d, Page 73):
$$
(\nabla f)^i = \sum_j g^{ij} \frac{\partial f}{\partial x^j},
$$
while the metric sensor is:
$$
g_{ij} = \bigg\langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j} \bigg\rangle.
$$
Take the spherical coordinate as example,
\begin{align}
x &= r\sin\theta\cos\phi \\
y &= r\sin\theta\sin\phi \\
z &= r\cos\theta
\end{align}
and let $J$ be the Jacobi matrix
$$
J = \begin{pmatrix}
\sin\theta\cos\phi & -r\sin\phi\sin\theta & r\cos\phi\cos\theta \\
\sin\theta\sin\phi & r\cos\phi\sin\theta & r\sin\phi\cos\theta \\
\cos\theta & 0 & -r\sin\theta
\end{pmatrix},
$$
the metric tensor can be obtained as
$$
\left( g_{ij} \right) = J^TJ =
\begin{pmatrix}
1 & 0 & 0 \\
0 & r^2\sin^2\theta & 0 \\
0 & 0 & r^2
\end{pmatrix}.
$$
Finally, the contravariant gradient vector can be obtained as follows according to the first equation:
$$
\nabla f = \begin{pmatrix}\displaystyle{
\frac{\partial f}{\partial r} \\
\frac{1}{r^2\sin^2\theta} \frac{\partial f}{\partial \phi} \\
\frac{1}{r^2} \frac{\partial f}{\partial \theta}}
\end{pmatrix}.
$$
However, the correct answer is
$$
\nabla f = \begin{pmatrix}\displaystyle{
\frac{\partial f}{\partial r} \\
\frac{1}{r\sin\theta} \frac{\partial f}{\partial \phi} \\
\frac{1}{r} \frac{\partial f}{\partial \theta}}
\end{pmatrix}.
$$
I don't know why I cannot get the correct answer using metric tensor. Could you please help me figure it out? Thank you!
 A: In your first equation for $\nabla f$,
$$
  \nabla f = \Big( \frac{\partial f}{\partial r}, \frac{1}{r^2 \sin^2 \theta} \frac{\partial f}{\partial \phi}, \frac{1}{r^2} \frac{\partial f}{\partial \theta} \Big),
$$
what you've written down is an expression for $\nabla f$ with respect to the coordinate vectors $\tfrac{\partial}{\partial r}, \tfrac{\partial}{\partial \phi}, \tfrac{\partial}{\partial \theta}$. That is, your equation means
$$
   \nabla f = \frac{\partial f}{\partial r} \frac{\partial}{\partial r} + \frac{1}{r^2 \sin^2 \theta} \frac{\partial f}{\partial \phi} \frac{\partial}{\partial \phi} + \frac{1}{r^2} \frac{\partial f}{\partial \theta} \frac{\partial}{\partial \theta}.
$$
Note that these coordinate vectors are not unit vectors; if we use $\boldsymbol{\hat{r}}, \boldsymbol{\hat{\phi}}, \boldsymbol{\hat{\theta}}$ to denote the associated unit vectors, then
$$
  \boldsymbol{\hat{r}} = \frac{\partial}{\partial r}, \qquad \boldsymbol{\hat{\phi}} = \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi}, \qquad \boldsymbol{\hat{\theta}} = \frac{1}{r} \frac{\partial}{\partial \theta}.
$$
Your second equation (the standard one written in physics) is written in terms of these unit vectors; it states
$$
   \nabla f = \frac{\partial f}{\partial r} \boldsymbol{\hat{r}} + \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \boldsymbol{\hat{\phi}} + \frac{1}{r} \frac{\partial f}{\partial \theta} \boldsymbol{\hat{\theta}}.
$$
It should now be clear that there is no inconsistency between these equations.
