# What is the intuition or the derivation of covariant derivative?

I asked this question in mathematics but the answer I got was a bit too abstract for me so I hope that my fellow physicists can give me more of an intuition or an easier explaination of my question. This is my original post:  So the derivation in my textbook for the covariant derivative of a vector field $$\vec{u}$$ in curvilinear coordinates $$\xi^k$$ is the following: $$\frac{\partial \vec{u}}{\partial\xi^j}=\frac{\partial (u^i\vec{a}_i)}{\partial\xi^j}=\frac{\partial u^i}{\partial\xi^j}\vec{a}_i+\frac{\partial \vec{a}_i}{\partial\xi^j}u^i=\vec{a}_i\left(\frac{\partial u^i}{\partial\xi^j}+u^k\Gamma_{jk}^i\right)$$ So the covariant derivative is defined as: $$\nabla_ku^i=\left(\frac{\partial u^i}{\partial\xi^k}+u^j\Gamma_{jk}^i\right)$$ And the same way is derived for covector field: $$\nabla_ku_i=\left(\frac{\partial u_i}{\partial\xi^k}-u_j\Gamma_{ik}^j\right)$$ And then out of nowhere it says that for higher rank tensors it must be defined as: $$\nabla_k\Phi^{i_1 i_2 ...i_p}_{j_1j_2...j_q}=\frac{\partial \Phi^{i_1 i_2 ...i_p}_{j_1j_2...j_q}}{\partial\xi^k}+\Phi^{s i_2 ...i_p}_{j_1j_2...j_q}\Gamma_{sk}^{i_1}+...+\Phi^{i_1 i_2 ...s}_{j_1j_2...j_q}\Gamma_{sk}^{i_p}-\Phi^{i_1 i_2 ...i_p}_{sj_2...j_q}\Gamma_{j_1k}^{s}-...-\Phi^{i_1 i_2 ...i_p}_{j_1j_2...s}\Gamma_{j_qk}^{s}$$ And I just can't understand what makes us think that it must be the way the higher rank tensor field covariant derivatives should look like that without any intuition of proof that follows from the first two examples. Can you please help me understand it?

You need to write out the tensor field with in some basis and then repeat the argument that you already verified. You need to apply the Leibniz rule multiple times now \begin{align*} \frac{\partial}{\partial x^k}\Phi &= \frac{\partial}{\partial x^k} (\Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \partial_{i_1}\dots\partial_{i_p}\mathrm{d}x^{j_1}\dots\mathrm{d}x^{j_q}) \\ &= \frac{\partial \Phi^{i_1 \dots i_p}_{j_1 \dots j_q}}{\partial x^k}\partial_{i_1}\dots\partial_{i_p}\mathrm{d}x^{j_1}\dots\mathrm{d}x^{j_q} \\ &\qquad + \Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \frac{\partial}{\partial x^k}(\partial_{i_1}) \partial_{i_2}\dots \partial_{i_p} \mathrm{d}x^{j_1}\dots\mathrm{d}x^{j_q} + \Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \partial_{i_1}\frac{\partial}{\partial x^k}(\partial_{i_2}) \partial_{i_3}\dots \partial_{i_p} \mathrm{d}x^{j_1}\dots\mathrm{d}x^{j_q} + \dots \\ &\qquad + \Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \partial_{i_1}\dots \partial_{i_p} \frac{\partial}{\partial x^k}(\mathrm{d}x^{j_1})\mathrm{d}x^{j_2}\dots \mathrm{d}x^{j_q} + \Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \partial_{i_1}\dots \partial_{i_p} \mathrm{d}x^{j_1}\frac{\partial}{\partial x^k}(\mathrm{d}x^{j_2})\mathrm{d}x^{j_3}\dots \mathrm{d}x^{j_q} + \dots \end{align*} Replacing the derivative with the usual Christoffel symbols then leads to \begin{align*} &= \frac{\partial \Phi^{i_1 \dots i_p}_{j_1 \dots j_q}}{\partial x^k}\partial_{i_1}\dots\partial_{i_p}\mathrm{d}x^{j_1}\dots\mathrm{d}x^{j_q} \\ &\qquad + \Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \Gamma_{ki_1}^l \partial_l \partial_{i_2}\dots \partial_{i_p} \mathrm{d}x^{j_1}\dots\mathrm{d}x^{j_q} + \Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \partial_{i_1}\Gamma_{ki_2}^l \partial_l \partial_{i_3}\dots \partial_{i_p} \mathrm{d}x^{j_1}\dots\mathrm{d}x^{j_q} + \dots \\ &\qquad - \Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \partial_{i_1}\dots \partial_{i_p} \Gamma_{kl}^{j_1}\mathrm{d}x^l\mathrm{d}x^{j_2}\dots \mathrm{d}x^{j_q} - \Phi^{i_1 \dots i_p}_{j_1 \dots j_q} \partial_{i_1}\dots \partial_{i_p} \mathrm{d}x^{j_1}\Gamma_{kl}^{j_2}\mathrm{d}x^l\mathrm{d}x^{j_3}\dots \mathrm{d}x^{j_q} - \dots \\ &= \left[\frac{\partial \Phi^{i_1 \dots i_p}_{j_1 \dots j_q}}{\partial x^k} + \Phi^{li_2 \dots i_p}_{j_1 \dots j_q}\Gamma^{i_1}_{kl} + \Phi^{i_1 l i_3 \dots i_p}_{j_1 \dots j_q}\Gamma^{i_2}_{kl} + \dots - \Phi^{i_1 \dots i_p}_{l j_2 \dots j_q}\Gamma^l_{k j_1} - \Phi^{i_1 \dots i_p}_{j_1 l j_3 \dots j_q}\Gamma^l_{k j_2} - \dots \right] \partial_{i_1} \dots \partial_{i_p} \mathrm{d}x^{j_1}\dots\mathrm{d}x^{j_q} \end{align*}
• Sorry, I used the standard notation from differential geometry. It boils down to vectors living in the tangent space $T_pM$ and a natural basis for the tangent space is given by the coordinate derivatives $\partial_\mu$. Similarly, the cotangent space has a canonical basis given by $\mathrm{d}x^\mu$. You can just replace all $\partial_i$ by $\vec{a}_i$ and similarly for the 1-form basis. Oct 19, 2023 at 11:58