TITLE: Gradient in differential geometry
QUESTION [9 upvotes]: I am a graduate student in physics trying to learn differential geometry on my own, out of a book written by Fecko.
He defines the gradient of a function as:
$
\nabla f = \sharp_g df = g^{-1}(df, \cdot )
$
This makes enough sense to me.  However, when I try to calculate the gradient of a function in spherical coordinates:
$
g^{-1} (df, \cdot) = g^{ij} \partial_i(df) \otimes \partial_j = g^{ij} \partial_i f  \partial_j
$
So the $j^{th}$ component of the gradient of f is:
$
g^{ij} \partial_if
$
The coefficients of the metric tensor are:
$
g = 
\begin{pmatrix} 
1 & 0 & 0 \\ 
0 & r^2 & 0 \\ 
0 & 0 & r^2 \sin^2{\theta} 
\end{pmatrix} 
$
So the inverse of a diagonal matrix ($g^{-1}$) is just a diagonal matrix whose entries are the reciprocals of the original matrix:
$
g^{-1} = 
\begin{pmatrix} 
1 & 0 & 0 \\
0 & r^{-2} & 0 \\ 
0 & 0 & r^{-2} \csc^2{\theta} 
\end{pmatrix} 
$
So it seems our expression doesn't match the vector calculus definition of the gradient in spherical coordinates.  For instance, differential geometry gives us a  $\hat{\theta}$ component of $ r^{-2} \partial_\theta f$ but vector calculus tells us this is $ r^{-1} \partial_\theta f$.
Where is my mistake?

REPLY [7 votes]: In vector calculus one usually computes in terms of orthonormal (rather than coordinate) components of vector fields.
This stuff is discussed in detail a bit later in the book :-) 
See pages 183-4 (Section 8.5.)
You find explicit expressions in both components, there.
Good luck!