What is simple definition of the covariant derivative that looks like the definition of the derivative of a function in calculus?

definition of the derivative of a function in calculus is:

$$\frac {df(x)}{dx}=\lim_{\Delta x\to o}\frac {f(x+\Delta x)-f(x)}{\Delta x}$$

what about Covariant derivative, what is definition of the Covariant derivative of a function in calculus? $$\frac {{\mathcal D}f(x)}{dx}$$

Remark: ok lets say that:

$$\frac {{\mathcal D}f(x)}{dx}=\frac {df(x)} {dx} +\delta f(x)$$ where the covariant derivative is broken into two parts, the extrinsic normal component and the intrinsic covariant derivative component . Now, how do we define a simple definition for the intrinsic covariant derivative component $\delta f(x)$ (this small addition is the result of parallel translation)

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    $\begingroup$ If you would like a definition of the covariant derivative that looks like the definition of the derivative of a function in calculus, then you may want to refer to the Wikipedia article on parallel transport: en.wikipedia.org/wiki/Parallel_transport. $\endgroup$ Jan 2, 2013 at 8:27
  • $\begingroup$ @Haskell Curry I'm writing a scientific paper in differential geometry, so I need a simple definition to begin with $\endgroup$
    – Neo
    Jan 2, 2013 at 8:32
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    $\begingroup$ @Neo, if you want a definition of the precise form you seem to want, the parallel transport definition mentioned by Haskell Curry really, truly is the only show in town. $\endgroup$ Jan 2, 2013 at 17:39
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    $\begingroup$ @Neo, if you're unhappy with that definition, your only alternatives are basically Koszul's definition (en.wikipedia.org/wiki/Connection_(vector_bundle)) and Ehresmann's definition (en.wikipedia.org/wiki/Ehresmann_connection), which are no more elementary than the parallel transport definition, and which really take you far from anything resembling the definition of the derivative in freshman calculus. Your hands really are tied here... $\endgroup$ Jan 2, 2013 at 19:04
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    $\begingroup$ @Neo If you're dealing with a scalar-valued function, directional differentiation is all you have, so for your question even to make sense, you need to at least be dealing with vector-valued functions (if not vector fields or sections of some general vector bundle) for your question to even make sense. But then, if you're dealing with vector fields, the "intrinsic covariant derivative component" would be coming from the connection $1$-form... $\endgroup$ Jan 3, 2013 at 12:02

4 Answers 4


There is a very intuitive way to understand the covariant derivative (for the Levi-Civita connection) for the special case of isometrically embedded submanifolds in $\mathbb R^n$. Roughly speaking, first take the usual derivative in $\mathbb R^n$, and then project the answer onto the tangent plane of the submanifold. The image of the projection is then the covariant derivative.

In fact, this is how the concept of covariant derivative if often introduced to undergraduates in their first differential geometry course. For instance, if you consider the sphere $S^2 \subset \mathbb R^3$, then it is easy (both analytically and visually) to prove that the tangent vector field to a great circle has zero covariant derivative and therefore that great circles must be geodesics.

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    $\begingroup$ I like this answer, but I think the OP might be looking for a concrete formula. I'll upvote if you edit your answer to include one. $\endgroup$ Jan 3, 2013 at 7:30
  • $\begingroup$ I don't care too much about the points but I will provide more details if you like, probably tomorrow. $\endgroup$
    – treble
    Jan 3, 2013 at 9:14
  • $\begingroup$ "Pull back" is more accurate than "project". $\endgroup$ Jan 3, 2013 at 12:09
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    $\begingroup$ I think "project" is just fine. The projection (via the literal projection map $\pi\colon T_pM \to T_pS$) of the ambient Levi-Civita connection gives the intrinsic Levi-Civita connection, no? $\endgroup$ Jan 3, 2013 at 20:49
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    $\begingroup$ Comparing to the accepted answer; without giving any equation, this answer conveys the main idea, and that was what I needed. Thanks a lot. $\endgroup$
    – Our
    Apr 13, 2019 at 10:25

One can motivate the covariant differentiation using only vector calculus. It works for an oversimplified case though (but since the OP doesn't accept either the definition via Ehresmann connection nor the vector bundle definition, may be it's justified.)

Consider new coordinates $y^i$ on $\mathbb{R}^n$ (e.g. spherical). We require orthonormality on that coordinate system, $$\mathrm{d}s^2=(u_i)^2\mathrm{d}y^i.$$ One has $\mathbf{x}=x^i(y)e_i$ and defines $$e'_j=\frac{\partial\mathbf{x}(y) }{\partial y^j}.$$ Then the metric is given by $$g_{ij}=e_i'\cdot e_j'.$$ If one considers a vector field $X:\mathbb{R}^n\to \mathbb{R}^n$ one can write $X=X^ie_i'$. If we now wish to differentiate $X$, we have to take into account the change of the components $X^i$ and of the basis $e'_j$, which is no longer rigid. That is $$\frac{\partial X }{\partial y^j}=\frac{\partial (X^i e'_i)}{\partial y^j}=\frac{\partial X^i}{\partial y^j}e_i'+X^i\frac{\partial e'_i}{\partial y^j}.$$ One can write $\frac{\partial e'_i}{\partial y^j}$ as linear combination of the $e_i'$ s, i.e. for some functions $\Gamma_{ij}^k$ $$ \frac{\partial e'_i}{\partial y^j}=\Gamma_{ij}^k e_k'.$$ Upon taking inner product of this equation with $e_l'$, one sees that these coefficients are given by $$\Gamma_{ij}^k=g^{kl}e_l'\cdot \frac{\partial e'_i}{\partial y^j}.$$

Now, in index notation, the covariant derivative of $X^i$ is given by the

$$\nabla_jX^i=\frac{\partial X^i}{\partial y^j}+\Gamma_{jk}^iX^k.$$ This is of the form $\frac {{\mathcal D}f(x)}{dx}=\frac {df(x)} {dx} +\delta f(x)$, but $f$ must be a vector field (or higher rank tensor), otherwise the covariant and ordinary derivatives concide.

  • $\begingroup$ it seems $\Gamma_{ij}^k=g^{jl}e_l'\cdot \frac{\partial e'_i}{\partial y^j}$ was incorrect (index $k$ is missing). : I think it should be : $\Gamma_{ij}^k=g^{kl}e_l'\cdot \frac{\partial e'_i}{\partial y^j}$ $\endgroup$
    – user359328
    Nov 15, 2017 at 22:23
  • $\begingroup$ good catch, corrected. $\endgroup$
    – c.p.
    Nov 16, 2017 at 9:14

$${\mathcal D}_{X} V=\lim_{\Delta x \to o}\frac {\Gamma(\gamma)^o_{\Delta x}V_{(\gamma)\Delta x}-V_{(\gamma)o}}{\Delta x},$$ where the ${\mathcal D}_{X}=\nabla_{X}$

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    $\begingroup$ This equation is still complicated. But some can use it to write a simple equation. $\endgroup$
    – herbert
    Jan 3, 2013 at 9:10
  • $\begingroup$ I guess $\Gamma(\gamma)^o_{\Delta x}$ are the maps between the tangent space at o and $\Delta x$ $\endgroup$
    – MSIS
    Jan 28, 2022 at 22:00

Ok, you want to do the derivative quotient $ \frac {df}{dx} $, as you do it in $\mathbb R^n$ ( using norms, matrices, etc. when $n>1$.

The reason you need additional machinery is that , unlike the case of $\mathbb R^n$, vector spaces are not naturally-isomorphic. You use this natural isomorphism to declare a vector at $T_p \mathbb R^n$ is the same at each point of $ \mathbb R^n$. But in a manifold, Tangent Spaces , as vector spaces, are not naturally isomorphic. The connection provides a choice of isomorphism between tangent spaces at different points.

The old difference quotient then evaluates the difference of the function at a vector and that at it's (isomorphic) translate.


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