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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?

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2 Answers 2

The components that the formula $g^{ij} \partial_j f$ refers to are taken with respect to the natural tangent space basis induced by the coordinate system; these vectors are often denoted by $(\partial/\partial r, \partial/\partial \theta, \partial/\partial \phi)$, and they differ from the orthonormal frame $(\hat{r}, \hat{\theta}, \hat{\phi})$ by the usual normalization factors $1$, $r$, $r \sin\theta$, respectively.

EDIT: Let's think in terms of vector calculus. In that case, your manifold could be a surface in $\mathbf{R}^3$, or the whole space $\mathbf{R}^3$ (but described in a curvilinear coordinate system). The position vector of a point in the manifold is written as $\mathbf{r}(s,t)$ (for a parametrized surface) or $\mathbf{r}(r,\theta,\phi)$ (for the whole space in spherical coordinates). The tangent vectors to the surface are $\partial\mathbf{r}/\partial s$ and $\partial\mathbf{r}/\partial t$. For the whole space, you have a frame of vector fields $(\partial\mathbf{r}/\partial r, \partial\mathbf{r}/\partial \theta, \partial\mathbf{r}/\partial \phi)$ which are orthogonal at each point (this is what it means when we say that spherical coordinates are an orthogonal coordinate system), but they are not normalized. It is these un-normalized vectors that in differential geometry are referred to as $(\partial/\partial r, \partial/\partial \theta, \partial/\partial \phi)$. (In the abstract setting, the manifold is not embedded in a Euclidean space, so it doesn't make sense to talk about a position vector, and thus the $\mathbf{r}$ is omitted from the notation. Also, as you've probably seen, vectors are often defined as first order differential operators, and this notation conforms to that way of thinking.) You get $\hat{r}$ etc. by normalizing these vectors.

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Thanks for the response, could you elaborate on the connection between the basis vectors $\partial_i$ and $\hat{x}_i$? –  Christopher Reilly Nov 13 '11 at 21:37
@Christopher: I've updated my answer. –  Hans Lundmark Nov 14 '11 at 7:37

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!

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Could you please take a look at the question math.stackexchange.com/questions/81405/anti-curl-operator (where I have run out of patience)? –  Will Jagy Nov 13 '11 at 21:51
Welcome to Math.SE, professor Fecko! –  Willie Wong Nov 14 '11 at 12:09

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