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I have an application where I need to compute the covariant derivative of a vector field in spherical coordinates. I am decomposing the vector field into vector spherical harmonics to try to make it simpler, and I need to compute covariant derivatives of all the VSH. I am getting bogged down with one in particular. Consider the VSH defined as, \begin{align} \mathbf{Y}_l^m &= Y_l^m \hat{r} \\ \boldsymbol{\Psi}_l^m &= r \nabla Y_l^m \\ \boldsymbol{\Phi}_l^m &= \mathbf{r} \times \nabla Y_l^m \end{align} (as defined by Barrera et al, 1985). Then I want to compute: $$ \nabla_{\boldsymbol{\Psi}_l^m} \boldsymbol{\Psi}_{l'}^{m'} $$ My approach has been to decompose $\boldsymbol{\Psi}_l^m$ into the $e_i$ basis, e.g. $e_1 = \hat{r}$, $e_2 = r \hat{\theta}$, $e_3 = r \sin{\theta} \hat{\phi}$, and then use the usual formula for the covariant derivative to grind out the result. I am getting bogged down in pages of algebra, and I am hoping someone might know of a clever trick to calculate this in a more simple (and less error prone) way.

For reference, the above VSH can be represented as, $$ \boldsymbol{\Psi}_l^m = \frac{1}{r} \partial_{\theta} Y_l^m e_2 + \frac{1}{r \sin^2{\theta}} \partial_{\phi} Y_l^m e_3 $$ I have been using the usual formula for covariant derivative. So if $\mathbf{W}_l^m = W_l^{m,i} e_i$ and $\mathbf{V}_l^m = V_l^{m,i} e_i$ are VSH, then $$ \nabla_{\mathbf{W}_l^m} \mathbf{V}_{l'}^{m'} = W_l^{m,i} \left[ \frac{\partial V_{l'}^{m',k}}{\partial u^i} + V_{l'}^{m',j} \Gamma_{ij}^k \right] e_k $$ Maybe someone knows a nice trick to calculate these, taking advantage of the definitions of the VSH above? I was thinking that because the VSH themselves are a basis for $\mathbb{R}^3$, perhaps it is not necessary to use the $\hat{r},\hat{\theta},\hat{\phi}$ basis?

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Recall that the gradient operator in curvilinear coordinates is, $$ \nabla = e^i \frac{\partial}{\partial u^i} = g^{ij} e_j \partial_i $$ where $e^i = \nabla u^i$ and $e_i = \partial \mathbf{r} / \partial u^i$. In spherical coordinates, we have $(u^1,u^2,u^3) = (r,\theta,\phi)$. Now, for an arbitrary vector $v$ and scalar function $f$, \begin{align} \nabla_{r \nabla f} (v) &= \nabla_{r g^{ij} e_j \partial_i f} (v) \\ &= r g^{ij} \partial_i f \nabla_{e_j} (v) \\ &= r g^{ij} \partial_i f \nabla_j (v) \tag 1 \end{align} Now consider, \begin{align} \nabla_j \left( r \nabla f \right) &= \left( \nabla_j r \right) \nabla f + r \nabla_j \left( \nabla f \right) \\ &= ( \partial_j r ) \nabla f + r \nabla_j \left( \nabla f \right) \\ &= \delta_{j1} \nabla f + r \nabla_j \left( \partial_k f e^k \right) \\ &= \delta_{j1} \nabla f + r \left[ \partial_j \partial_k f e^k + (\partial_k f) \nabla_j e^k \right] \tag 2 \end{align} Now, \begin{align} \nabla_j e^k &= \nabla_j \left( g^{ku} e_u \right) \\ &= (\nabla_j g^{ku}) e_u + g^{ku} \nabla_j e_u \\ &= \partial_j g^{ku} e_u + g^{ku} \Gamma_{ju}^p e_p \\ &= \left( \partial_j g^{kp} + g^{ku} \Gamma_{ju}^p \right) e_p \tag 3 \end{align} Noting that $\boldsymbol{\Psi}_l^m = r \nabla Y_l^m$ and putting these results together, we have, \begin{align} \nabla_{\boldsymbol{\Psi}_l^m} \boldsymbol{\Psi}_{l'}^{m'} &= r g^{ij} \partial_i Y_l^m \nabla_j \left( \boldsymbol{\Psi}_{l'}^{m'} \right) \\ &= r g^{ij} \partial_i Y_l^m \left\{ \delta_{j1} \nabla Y_{l'}^{m'} + r \left[ \partial_j \partial_k Y_{l'}^{m'} e^k + (\partial_k Y_{l'}^{m'}) \nabla_j e^k \right] \right\} \\ &= r g^{i1} \partial_i Y_l^m \nabla Y_{l'}^{m'} + r^2 g^{ij} \partial_i Y_l^m ( \partial_j \partial_k Y_{l'}^{m'} ) e^k + r^2 g^{ij} \partial_i Y_l^m \partial_k Y_{l'}^{m'} \nabla_j e^k \tag 4 \end{align} The first term is zero because $g^{ij}$ is diagonal in spherical coordinates, and $\partial_r Y_l^m = 0$, so we get, \begin{align} \nabla_{\boldsymbol{\Psi}_l^m} \boldsymbol{\Psi}_{l'}^{m'} &= r^2 g^{ij} \partial_i Y_l^m ( \partial_j \partial_k Y_{l'}^{m'} ) e^k + \\ & r^2 g^{ij} \partial_i Y_l^m \partial_k Y_{l'}^{m'} \left( \partial_j g^{kp} + g^{ku} \Gamma_{ju}^p \right) e_p \end{align} In spherical coordinates, $g^{ij} = \frac{1}{h_i^2} \delta^{ij}$ where the $h$ factors are $h_r = 1$, $h_{\theta} = r$, $h_{\phi} = r \sin{\theta}$. So the above expression can be further simplified to, \begin{align} \nabla_{\boldsymbol{\Psi}_l^m} \boldsymbol{\Psi}_{l'}^{m'} &= \frac{r^2}{h_j^2} \partial_j Y_l^m ( \partial_j \partial_k Y_{l'}^{m'} ) e^k + \\ & \frac{r^2}{h_j^2} \partial_j Y_l^m \partial_p Y_{l'}^{m'} \left( \partial_j \left( \frac{1}{h_p^2} \right) e_p + \frac{1}{h_p^2} \Gamma_{jp}^q e_q \right) \end{align} It might be possible to simplify further by examining the Christoffel symbols in spherical coordinates. It would also be nice if this could be expressed in terms of the VSH basis vectors $\mathbf{Y}$, $\boldsymbol{\Psi}$, $\boldsymbol{\Phi}$, but I don't see an easy way to do that at the moment.

EDIT

I had forgotten that for (3), the covariant derivative of the dual basis vector is in fact related to the negative of the Christoffel symbols, $$ \nabla_j e^k = -\Gamma_{jp}^k e^p \tag {3b} $$ Substituting this into (4) now yields (again neglecting the first term and using the diagonal metric property), \begin{align} \nabla_{\boldsymbol{\Psi}_l^m} \boldsymbol{\Psi}_{l'}^{m'} &= \frac{r^2}{h_j^2} \partial_j Y_l^m \left( \partial_j \partial_p Y_{l'}^{m'} - \partial_k Y_{l'}^{m'} \Gamma_{jp}^k \right) e^p \tag 5 \end{align} EDIT 2 By taking dot products of (5) with $e_i$, I was able to show, \begin{align} \nabla_{\boldsymbol{\Psi}_l^m} \boldsymbol{\Psi}_{l'}^{m'} \cdot e_1 &= -\frac{1}{r} \boldsymbol{\Psi}_l^m \cdot \boldsymbol{\Psi}_{l'}^{m'} \\ \nabla_{\boldsymbol{\Psi}_l^m} \boldsymbol{\Psi}_{l'}^{m'} \cdot e_2 &= \boldsymbol{\Psi}_l^m \cdot \partial_{\theta} \boldsymbol{\Psi}_{l'}^{m'} \\ \nabla_{\boldsymbol{\Psi}_l^m} \boldsymbol{\Psi}_{l'}^{m'} \cdot e_3 &= \boldsymbol{\Psi}_l^m \cdot \partial_{\phi} \boldsymbol{\Psi}_{l'}^{m'} - \cos{\theta} \boldsymbol{\Phi}_l^m \cdot \boldsymbol{\Psi}_{l'}^{m'} \end{align}

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