Given is the metric $\gamma_{jk}$ for the surface of a Sphere $S^2$ with $\gamma_{22}=1,\gamma_{23}=\gamma_{32}=0$ and $\gamma_{33}=\sin^2(\theta)$. The coordinates are $x=$($t,r,\theta,\phi$) and $j$ and $k$ run over $x^2=\theta$ and $x^3=\phi$.
Therefore I get the Christoffels $\Gamma^2_{33}=-\sin(\theta)\cos(\theta)$ and $\Gamma^3_{23}=\Gamma^3_{32}=\frac{\cos(\theta)}{\sin(\theta)}$. All other entries of the Christoffels are zero.
Now I want to calculate the following second covariant derivatives with respect to this metric: \begin{equation} \nabla_{2}\nabla_{2} Y^l_m = \nabla_{2}\left(\partial_\theta Y^l_m + \Gamma^3_{23}Y^3_m - \Gamma^3_{23}Y^l_3\right)\\ \nabla_{3}\nabla_{3} Y^l_m = ...\\ \nabla_{3}\nabla_{2} Y^l_m = ..., \end{equation} where the $Y^l_m$ denotes the spherical harmonic functions. In the first line I only used the definition for the covariant derivative and cancel out all Christoffels, which are zero.
The solutions should be: \begin{equation} \nabla_{2}\nabla_{2} Y^l_m = \frac{\partial^2}{\partial\theta^2}Y^l_m\\ \nabla_{3}\nabla_{3} Y^l_m = \left(\frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial\phi^2}+\frac{\cos(\theta)}{\sin(\theta)}\frac{\partial}{\partial\theta}\right)Y^l_m\\ \nabla_{3}\nabla_{2} Y^l_m = \left(\frac{\partial}{\partial\theta}\frac{\partial}{\partial\phi}-\frac{\cos(\theta)}{\sin(\theta)}\frac{\partial}{\partial\phi}\right)Y^l_m. \end{equation} Does anyone know, how do get these expressions?


1 Answer 1


Since $f = f(\theta,\phi) = Y^l_m(\theta,\phi)$ is a scalar function, $\nabla_a f$ is a covariant vector, and so it's covariant derivative is that of the covariant derivative of a covariant vector, thus we have \begin{align} \nabla^3 \nabla_3 f &= g^{33} \nabla_3 \nabla_3 f \\ &= g^{33} (\partial_3 \nabla_3 f - \Gamma_{33}^a \nabla_a f) = g^{33} (\partial_3 \partial_3 f - \Gamma_{33}^2 \nabla_2 f) \\ &= g^{33} (\partial_3 \partial_3 f - \Gamma_{33}^2 \partial_2 f) = \frac{1}{\sin^2 \theta} \Bigl[\partial_3 \partial_3 f - (- \sin \theta \cos \theta) \partial_2 f\Bigr] \\ &= \left[\frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} + \frac{\cos \theta}{\sin \theta} \frac{\partial}{\partial \theta}\right] f \end{align} and \begin{align} \nabla_3 \nabla_2 f &= \partial_3 \nabla_2 f - \Gamma^a_{32} \nabla_a f = \partial_3 \partial_2 f - \Gamma^3_{32} \partial_3 f = \Biggl[\frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta} - \frac{\cos \theta}{\sin \theta} \frac{\partial}{\partial \phi}\Biggr] f \end{align} and $$ \nabla_2 \nabla_2 f = \partial_2 \nabla_2 f - \Gamma_{22}^a \nabla_a f = \frac{\partial^2}{\partial \theta^2} f.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.