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:

  1. 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.
  2. 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.
  • $\begingroup$ Can you write the statement of “covariance” you want to show? Even if using other notions? Is it a transformation law of the covariant derivative under frame changes? $\endgroup$
    – Keshav
    Jan 22 at 16:54
  • $\begingroup$ To be honest this precisely what I am asking about - I edited the question. $\endgroup$
    – szantag
    Jan 22 at 19:23

3 Answers 3


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)

  • 2
    $\begingroup$ This is a link only answer. Please include a summary of the answer, and why you think it answers the question. $\endgroup$
    – amWhy
    Jan 22 at 23:42

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.


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".


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