I am trying to derive the expression for gradient of a vector using tools from tensor calculus and I am not able to get the correct expression. I have written down the derivation down here : Notation : $Z^i$ are contravariant vector components i.e. $r,\theta$. $\mathbf{Z}_i$ are covariant vectors. $\mathbf{g}$ is metric tensor such that $g_{ij}=\mathbf{Z}_i . \mathbf{Z}_j$
For a scalar function $S(r, \theta)$, the definition gradient is $$ \nabla S = \frac{\partial S}{\partial Z^i} g^{ij} \mathbf{Z}_j =\frac{\partial S}{\partial Z^i}\mathbf{Z}^i = \frac{\partial S}{\partial r} \hat{r} + \frac{1}{r}\frac{\partial S}{\partial \theta} \hat{\theta} $$ In the above expression $\mathbf{Z}^2 = \hat{\theta}/r$. Now using above formula, I tried to write gradient of a vector, which gives $$\nabla \mathbf{V} = \frac{\partial \mathbf{V}}{\partial Z^i}\mathbf{Z^i} = \frac{ \partial \big( V^k \mathbf{Z}_k \big)}{\partial Z^i}\mathbf{Z^i} =g^{ij} \Big( \frac{\partial V^k}{\partial Z^i} + \Gamma^k_{im}V^m \Big) Z_k Z_j = T^{kj} Z_k Z_j.$$
For Polar co-ordinates $$ \mathbf{g} = \begin{bmatrix} 1 & 0 \\ 0 & r^2 \end{bmatrix} \hspace{20pt} \mathbf{g}^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & r^{-2} \end{bmatrix} $$ and $$\Gamma_{22}^{1}=-r \quad \Gamma_{12}^{2}=\frac{1}{r} \quad \Gamma_{21}^{2}=\frac{1}{r} $$ and rest of the value of christoffel tensor are zero.
Now using this information, we can write $$T^{11} = g^{i1} \Big( \frac{\partial V^1}{\partial Z^i} + \Gamma^1_{im} V^m\Big) $$ $$ = g^{11} \Big( \frac{\partial V^1}{\partial Z^1} + \Gamma^1_{11} V^1 + \Gamma^1_{12} V^2\Big) = \frac{\partial V^1}{\partial r} $$ Which is an expected answer. But other values that I calculate are wrong. For example, $$T^{12} = g^{i2} \Big( \frac{\partial V^1}{\partial Z^i} + \Gamma^1_{im} V^m\Big) $$ $$ = g^{22} \Big( \frac{\partial V^1}{\partial Z^2} + \Gamma^1_{21} V^1 + \Gamma^1_{22} V^2\Big) $$ $$ = \frac{1}{ r^2 } \Big( \frac{\partial V^1}{\partial Z^2} + \frac{V^1}{r} - r V^2 \Big)$$ Above calculation implies that $$T^{12} = \frac{\partial V_r}{\partial \theta} + \frac{V_r}{r^2} - \frac{V_\theta}{r} $$ but this is not the correct answer. The correct answer is $$T^{12}= \frac{\partial V_r}{\partial \theta} - \frac{V_\theta}{r} $$ I am not sure, why am I getting extra $V_r/r^2$ term. Can you please let me know what am I doing wrong. I will then complete the calculation and upload the final answer with all calculations here.