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:

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*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.


Covariant derivative extends the notions of directional derivative of multivariate calculus.

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*$\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.

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*$\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.
