I totally reformulate my question to express precisely my problem. I reason on tensor $(1,1)$ for simplicity even if what I say could be more general.
A quantity $T^{\nu}_{\mu}$ i said to be a tensor if under a change of variable $x \rightarrow \widetilde{x}$, it follows the property:
$$T^{\nu}_{\mu}=\frac{\partial \widetilde{x}^{\alpha} }{\partial x^{\mu}} \frac{\partial x^{\nu}}{\partial \widetilde{x}^{\beta} } \widetilde{T}^{\beta}_{\alpha}$$
Consider the quantity $T_{\mu}^{\nu}=\partial_{\mu} A^{\nu}$
I know that we usually say it is not a tensor and I understand the reason we associate to it. If I consider a change of frame $x \rightarrow \widetilde{x}$, then I have:
$$\partial_{\mu} A^{\nu}=\frac{\partial \widetilde{x}^{\alpha} }{\partial x^{\mu}}\widetilde{\partial}_{\alpha} \left( \frac{\partial x^{\nu}}{\partial \widetilde{x}^{\beta} } \widetilde{A}^{\beta} \right)$$
$$\partial_{\mu} A^{\nu}=\frac{\partial \widetilde{x}^{\alpha} }{\partial x^{\mu}} \frac{\partial x^{\nu}}{\partial \widetilde{x}^{\beta} } \widetilde{\partial}_{\alpha} \widetilde{A}^{\beta} + \frac{\partial \widetilde{x}^{\alpha} }{\partial x^{\mu}} \frac{\partial^2 x^{\nu}}{\partial \widetilde{x}^{\alpha} \partial \widetilde{x}^{\beta} } \widetilde{A}^{\beta}$$
Because of this extra second derivative term, it doesn't follow the property I am asking. Then it cannot be a tensor. I think I understand this approach.
Here, I emphasize that the reason it is not a tensor is because we implicitly define $\widetilde{T}^{\beta}_{\alpha}$ as being $T^{\nu}_{\mu}$ where we replaced all vector and covector components appearing by the corresponding components in $\widetilde{x}$. I.e, what we expect to call $\widetilde{T}^{\beta}_{\alpha}$ would be $\widetilde{T}^{\beta}_{\alpha}=\widetilde{\partial}_{\alpha} \widetilde{A}^{\beta}$. I go back to it in the next part.
Now consider the following:
I choose a particular frame $x$ in which I say $T^{\nu}_{\mu}=\partial_{\mu} A^{\nu}$.
Then, I define the quantity $\widetilde{T}^{\beta}_{\alpha}=\frac{\partial \widetilde{x}^{\beta} }{\partial x^{\nu}} \frac{\partial x^{\mu}}{\partial \widetilde{x}^{\alpha} } T^{\nu}_{\mu}$
Thus, here we have $\widetilde{T}^{\beta}_{\alpha} \neq \widetilde{\partial}_{\alpha} \widetilde{A}^{\beta} $. The interpretation of $\widetilde{T}^{\beta}_{\alpha}$ is then different from what we wanted to have in the previous part: it has only the interpretation of derivative of a vector component in the initial frame $x$. It loses this interpretation in $\widetilde{x}$.
Why cannot I say that this quantity is a tensor ?
It follows the good law of transformation and it is well defined.
Indeed, consider any vector $a$ and covector $b$, if I compute $T(a,b)$ in the frame $x$ it gives me:
$$T(a,b)=\left( \partial_{\mu} A^{\nu} \right) a^{\mu} b_{\nu}$$
In the frame $\widetilde{x}$ it gives me:
$$T(a,b)=\left( \partial_{\mu} A^{\nu} \right) \frac{\partial \widetilde{x}^{\beta} }{\partial x^{\nu}} \frac{\partial x^{\mu}}{\partial \widetilde{x}^{\alpha} } \widetilde{a}^{\alpha} \widetilde{b}_{\beta}=\left( \partial_{\mu} A^{\nu} \right) a^{\mu} b_{\nu}$$
Thus the map $T$ is well defined (the map doesn't depend on the frame I use to describe it, it exists indepentantly of any frame).
My questions:
Various answer tell me that my way of definining the tensor is ill-defined (I guess in the sense my map cannot exist) ? Why ? For me it works I don't see any contradiction.
If there is indeed no contradiction, why isn't this way used to define a "valid" tensor $\partial_{\mu} A^{\nu}$. Is it for aesthetical reasons or there are good mathematical reasons that says that this quantity would be bad and would'nt represent the variation of a vector field ?
Best Answer
Your approach is completely fine, and it is subsumed into covariant derivatives. One thing that makes your approach less general is that it takes one particular coordinate system as special, and assumes that the Christoffel symbols vanish in that system (and hence in particular the resulting affine connection is flat). You will have trouble if, for example, your manifold cannot be covered by a single coordinate chart. The advantage of the Christoffel symbols is that you don't have any issues with nontrivial topologies, and you will be able to talk about non-flat affine connections.
Let us see how we can "hide" the special coordinate system and discover that the Christoffel symbols were lurking beneath this. In your notation, we can write the components of $T$ in the $\tilde x$-coordinates as $$ \tilde T^\beta_\alpha=\tilde{\partial}_{\alpha} \tilde{A}^{\beta} + \frac{\partial \tilde{x}^{\beta} }{\partial x^{\nu}} \frac{\partial^2 x^{\nu}}{\partial \tilde{x}^{\alpha} \partial \tilde{x}^{\gamma} } \tilde{A}^{\beta}, $$ that is, $$ \tilde T^\beta_\alpha=\tilde{\partial}_{\alpha} \tilde{A}^{\beta} + \tilde\Gamma^\beta_{\alpha\gamma} \tilde{A}^{\beta}, \qquad\textrm{where}\qquad \tilde\Gamma^\beta_{\alpha\gamma} = \frac{\partial \tilde{x}^{\beta} }{\partial x^{\nu}} \frac{\partial^2 x^{\nu}}{\partial \tilde{x}^{\alpha} \partial \tilde{x}^{\gamma} }. $$ In a new coordinate system we call $\bar x$, we would obviously have $$ \bar T^\beta_\alpha=\bar{\partial}_{\alpha} \bar{A}^{\beta} + \bar\Gamma^\beta_{\alpha\gamma} \bar{A}^{\beta}, \qquad\textrm{where}\qquad \bar\Gamma^\beta_{\alpha\gamma} = \frac{\partial \bar{x}^{\beta} }{\partial x^{\nu}} \frac{\partial^2 x^{\nu}}{\partial \bar{x}^{\alpha} \partial \bar{x}^{\gamma} }. $$ It is not difficult to derive the following transformation law for the $\Gamma$-coefficients: $$ \bar\Gamma^\beta_{\alpha\gamma} = \tilde\Gamma^\lambda_{\mu\nu} \frac{\partial \bar{x}^{\beta} }{\partial \tilde{x}^{\lambda}} \frac{\partial \tilde{x}^{\mu} }{\partial \bar{x}^{\alpha}} \frac{\partial \tilde{x}^{\nu} }{\partial \bar{x}^{\beta}} +\frac{\partial \bar{x}^{\beta} }{\partial \tilde{x}^{\nu}} \frac{\partial^2 \tilde{x}^{\nu}}{\partial \bar{x}^{\alpha} \partial \bar{x}^{\gamma} } . $$ Thus, once the $\Gamma$-coefficients are known in some coordinate system, its values can be computed in any other coordinate system. We never need to refer to the original "special" $x$-coordinate system. These $\Gamma$-coefficients are actually called the Christoffel symbols.