The gradient in spherical coordinates is given by $$\left(\partial_r f, \frac{1}{r} \partial_\theta f, \frac{1}{r \sin \phi}\partial_\phi f\right)$$

However, I get a wrong answer if I try to compute it a different way, by lowering the index of the differential using the metric in spherical coordinates. The metric in spherical coordinates is $$g = \begin{pmatrix} 1 & 0 & 0\\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2 \phi \end{pmatrix}$$ So if I take $g^{-1} (df) = g^{-1} (\partial_r f \; dr + \cdots)$, then I get $$\left(\partial_r f, \frac{1}{r^2} \partial_\theta f, \frac{1}{r^2 \sin^2 \phi}\partial_\phi f\right)$$

What's going wrong here?

• I'm sorry, I don't understand what you mean. The basis is $\partial_r, \partial_\theta, \partial_\phi$ for the tangent space, and $dr, d\theta, d\phi$ for the cotangent space? To get from cotangent to tangent I just multiply by the inverse of the metric right? Commented Jun 28, 2014 at 6:16
• Aha, ok. So one basis is normalized and the others aren't. To normalize, we multiply by $1/r, 1/r \sin \theta$, and then we see that these vectors have length 1 under this metric. For future reference, there is another thread that explains something similar here math.stackexchange.com/questions/261830/… Commented Jun 28, 2014 at 6:43