Where is the "covariance" in a covariant derivative I have the following definition of a covariant derivative. Consider a general fibre bundle $E \rightarrow M$ with a connection given by a parallel transport, i.e. along a path $\gamma$ in $M$ we have a transport $\Gamma(\gamma)^t_s : E_{\gamma(s)} \rightarrow E_{\gamma(t)}$ with a covariant derivative $\nabla_{\dot{\gamma}(0)} \sigma(x) := \frac{d}{d t}\mid_{t=0}(\Gamma(\gamma)^t_0)^{-1} \circ \sigma \circ \gamma(t)$.
My question is, what does "covariant" refer to in the name covariant derivatrive? I have two main guesses:

*

*It reflects the fact that local forms of the covariant derivative "commute" with the transition maps of the bundle - but it is pretty obvious as the derivative is defined globally and its local expressions are defined so that it makes sens.

*It is purely historical and stems from the fact that the above covariant derivative is a generalisation of the covariant derivative of a metric connection which "vector field component" changes covariantly.

 A: I think that I found a post on mathoverflow that sorts it out:
https://mathoverflow.net/questions/85171/terminology-of-covariant-derivative-and-various-connections
According to the above link, the name "covariant" refers to both being "independant of coordinate choice" and the metric, or more generally Koszul, connection being a map $ \nabla: \Gamma (E) \rightarrow \Gamma(T^* M \otimes E)$ where the "component" $T^* M$ changes covariantly.
(in the linked question only the proposition of the covariant derivative in the form $\nabla \sigma := T\sigma$ seems to be wrong)
A: Covariant derivative extends the notions of directional derivative of multivariate calculus.

*

*$\nabla_v f(x) = f'_v(x) = D_v f(x) = Df(x)(v) = \partial_v f(x) = v \cdot \nabla f(x) = v \cdot \frac{\partial f(x)}{\partial x} $
The type of definition of derivative depends on the problem you are trying to solve and the geometric object you were looking at.

*

*$\displaystyle (\nabla_v f(x))_p = (f \circ \phi)'(0) = \lim_{t \to 0} \frac{f(\phi(t)) - f(p)}{t} $
where $\phi(t): [-1,1]\to M$ is a "path" of some kind.  This seems to be called the Lie derivative since we're differentiating a scalar function along a vector field.
These definitions had to be "covariant" with respect to change of coordinates.  The physics didn't change just because you used a different ruler or camera.
A: You may first learn the concept of covariant and contravariant vectors through Charles Francis's answer in the following post: https://physics.stackexchange.com/questions/541822/covariant-vs-contravariant-vectors.
To summarize, there are two ways to represent a vector $v$ in $\mathbb{R} ^n$.
First, let $\{e_i\}_{i=1}^n$ be a basis. Then $v=v^ie_i$.
We can also consider $v$ as a covector where $v(u)= \left\langle u,v \right\rangle $ for $u \in \mathbb{R} ^n$. Let $\left\{ e^i \right\}_{i=1}^n $ be the dual basis of $\left\{ e_i \right\}_{i=1}^n $. Then $v=v_ie^i$.
Now we have two sets of parameters, $(v^i)_{i=1}^n$ and $(v_i)_{i=1}^n$, both representing the same vector $v$ based on the basis $\left\{ e_i \right\}_{i=1}^n $ we choose.
However, the two representations change in different ways if we halve the basis vector and consider the basis $\left\{ \tilde{e}_i=\mathbf{\frac{1}{2}}e_i \right\}_{i=1}^n $.
On the one hand, $v=\tilde{v}^i \tilde{e}_i$ where $\tilde{v}^i=\mathbf{2}v^i$. The parameters $v^i$ change their length "contrarily" to the change of basis vectors. The parameters $v^i$ are therefore called "contra"variant component of $v$.
On the other hand, $v=\tilde{v}_i \tilde{e}^i$ where $\tilde{v}_i=\mathbf{\frac{1}{2} }v_i$. The parameters $v_i$ change their length "along" with the change of basis vectors. The parameters $v_i$ are therefore called "co"variant component of $v$.
Now come back to your question about covariant derivative. From my understanding, the name "covariant" comes from the fact that given a section $\sigma $ of the bundle $E$, the covariant derivative $\nabla \sigma $ takes tangent vectors as input and behaves like a covector; thus regarded as "covariant".
