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We know that the partial derivative of a tensor is not a tensor. But how is this problem fixed by adding to the partial derivatives, a field of Christoffel symbols? Christoffel symbols are a completely arbitrary field because you can define any arbitrary connection you want. How is it that adding a completely arbitrary field fixes the problem, and that the resultant covariant derivative transforms as a tensor?

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Christoffel symbols are a completely arbitrary field because you can define any arbitrary connection you want.

No, the Christoffel symbols are not arbitrary. They are defined (see Christoffel symbols - Definition in Euclidean space) by how the base vectors $\mathbf{e}_i$ depend on the coordinates $x^j$. $$\frac{\partial\mathbf{e}_i}{\partial x^j} = \Gamma^k_{ij}\ \mathbf{e}_k$$ or equivalently $$d \mathbf{e}_i = \Gamma^k_{ij}\ \mathbf{e}_k\ dx^j \tag{1}$$

It is this definition, from which you can derive that for a tensor field $A^i$ the expressions $$\frac{\partial A^i}{\partial x^j}+\Gamma^i_{jk}\ A^k$$ are components of a tensor, while the partial derivatives $$\frac{\partial A^i}{\partial x^j}$$ are not.

You can derive this in a straight-forward way by starting with the invariant differential $d\mathbf{A}$ of a vector field between two positions in space.

$$\begin{align} d\mathbf{A} &= d(A^i\ \mathbf{e}_i) \\ &= dA^i\ \mathbf{e}_i + A^i\ d\mathbf{e}_i & \text{use definition (1)} \\ &= \frac{\partial A^i}{\partial x^j} dx^j\ \mathbf{e}_i + A^i\ \Gamma^k_{ij}\mathbf{e}_k\ dx^j & \text{in the second term swap indices $i$ and $k$} \\ &= \frac{\partial A^i}{\partial x^j} dx^j\ \mathbf{e}_i + A^k\ \Gamma^i_{kj}\mathbf{e}_i\ dx^j \\ &= \left( \frac{\partial A^i}{\partial x^j}+A^k\ \Gamma^i_{kj} \right) \mathbf{e}_i\ dx^j \end{align}$$ or equivalently $$\frac{\partial\mathbf{A}}{\partial x^j} = \left( \frac{\partial A^i}{\partial x^j}+A^k\ \Gamma^i_{kj} \right) \mathbf{e}_i$$

You see, the covariant derivative emerged as the components of $\partial\mathbf{A}/\partial x^j$ in a quite natural way.

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  • $\begingroup$ Christoffel symbols can be choosen to one's wills.. the choice shows about how the space he is working on is curved. And, could you give a source for this idea of the 'invariant' differential, this is the first time I've ever seen it. $\endgroup$ Apr 15, 2022 at 14:08
  • $\begingroup$ And secondly that wiki section you linked clearly states that the symbols discussed for euclidean space only $\endgroup$ Apr 15, 2022 at 14:09
  • $\begingroup$ @Buraian I don't remember where I saw this approach long time ago. The closest thing I could find now is video The Covariant Derivative (Component Definition) which is part of a larger video series about Tensor Calculus. $\endgroup$ Apr 15, 2022 at 15:16

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