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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?

2 Answers 2