I read the Wikipedia page about the covariant derivative, my main problem is in this part:


Some of the formulas seem to lead to contradictions, I assume I'm making some mistakes.

Here are some formulas from that page.

They define the Covariant derivative in the direction $\mathbf e_j$, denoted $\nabla_{\mathbf e_j}$ or $\nabla_j$ so that:

$\nabla_{\mathbf e_j} \mathbf e _i = \nabla_j \mathbf e _i = \Gamma^k_{\ \ ij}\mathbf e_k$

And define it so it obeys Leibniz' rule.

They then go on to show that

Covariant Derivative

Where it seems they used

$\nabla_{\mathbf e_i} u^j = \frac {\partial u^j}{\partial x^i}$

But then later they define here: http://en.wikipedia.org/wiki/Covariant_derivative#Notation

$\nabla_{\mathbf e_i} u^j = \frac {\partial u^j}{\partial x^i} + u^k \Gamma^j_{\ \ ki}$

1) Is this a misunderstanding of mine or a problem in Wikipedia?

Also instead of the definition:

$\nabla_j \mathbf e _i = \Gamma^k_{\ \ ij}\mathbf e_k$

I saw in other places the Christoffel symbols defined so

$\partial_j \mathbf e _i = \Gamma^k_{\ \ ij}\mathbf e_k$

2) Is the covariant derivative of basis vectors the same as the regular derivative of a basis vector?or are these just two different definition of the Christoffel symbols?

Another contradiction I saw is that they write the following formula:

in the end of the section "Coordinate Description"

where you add here a Gamma for each upper index and subtract a Gamma for each lower index according to the rule written there.

According to this it seems to me that:

$\nabla_j \mathbf e _i = \partial_j \mathbf e _i - \Gamma^k_{\ \ ij}\mathbf e_k$

Which is also inconsistent with how they defined the covariant derivatove

3) Is this a contradiction or a confusion of mine?

Thank you very much, sorry it's so long

If it's a problem I can break the question up into two questions or something

  • 1
    $\begingroup$ Ordinarily we like to have only one question per post, but these are closely related enough that it's probably fine to leave them together. $\endgroup$
    – David Z
    Jul 31, 2012 at 8:04

1 Answer 1


1) The confusion comes from an omission of parentheses in these notations. In the first case we do indeed have $$\nabla_{\vec{e}_i}(u^j) = \frac{\partial u^j}{\partial x^i},$$ since $u^j$ is just one non-specific component of $\vec{u}$. In the second case, they mean to take the component after differentiating the tensor: $$u^j{}_{;i} = (\nabla_{\vec{e}_i} \vec{u})^j = \frac{\partial u^j}{\partial x^i} + u^k \Gamma^j_{ki}.$$ I am using arrows instead of Roman type to indicate vectors in order to emphasize which things are full vectors (which may have subscripts, e.g. $\vec{e}_i$ is the $i$-th vector in your basis) and which things are components.

2) There should only be one set of Christoffel symbols. In what context was this the definition?

Also, covariant derivatives reduce to partial derivatives on scalars.

3) The confusion here comes from the use of $i$ in $\vec{e}_i$ as a label on which basis vector is being used, rather than on which component of a given vector is in place. Think of $\vec{e}_i$ as one symbol, such as $\hat{x}$ or $\hat{y}$. (This is indicated by the Roman as opposed to Italic typeface in the question, which again I've switched to an arrow to draw attention to the vector nature of the symbol.) We use lower subscripts so they don't interfere with the upper superscripts that would label the components. That is, $\vec{e}_i$ has components $e_i^0$, $e_i^1$, etc. As an object whose components are indexed with upper indices, one uses a positive Christoffel term: $$(\nabla_j \vec{e}_i)^k = \partial_j e_i^k + \Gamma^k_{jl} e_i^l.$$ Note that $e_i^k = \delta_i^k$, which is a constant and therefore has vanishing partial derivative. Contracting the Christoffel symbol with the Kronecker delta in the second term leaves only $\Gamma^k_{ji}$, as expected.

  • $\begingroup$ about question 2, it was online on a video lecture, which I won't make you sit through unless you're interested, by I also saw it on a forum post on Physics forum while searching for an answer for this on google, here's the post: physicsforums.com/showpost.php?p=3991829&postcount=3 $\endgroup$
    – fiftyeight
    Jul 31, 2012 at 1:40
  • $\begingroup$ Another reason why that definition makes sense to me is that on the first derivation I displayed, it seems like the covariant derivative acts like the regular derivative when acting on $u^j$, just like you wrote in answer 1, I don't understand why this is, I don't understand what you mean by $u^j$ is a "non-specific component" $\endgroup$
    – fiftyeight
    Jul 31, 2012 at 1:44
  • $\begingroup$ @fiftyeight That forum post would be correct if all partials were replaced by covariant derivatives. The author is working with (nonstandard) notation in which derivatives of vectors are understood to be covariant, rather than the vector of partial derivatives of the components. Keep in mind though that we demand that the covariant derivative agree with the partial derivative when acting on scalars, so $\partial_\alpha (A^i) = \nabla_\alpha (A^i)$ always. $\endgroup$
    – user10851
    Jul 31, 2012 at 2:43
  • $\begingroup$ @fiftyeight To clarify regarding your second comment: Yes, the covariant and partial derivatives of $u^j$ are the same. By "non-specific component" I meant $u^j$ could be any one of the four scalar components of $\vec{u}$, and which one it is depends on which term in the implicit sum you're considering. $\endgroup$
    – user10851
    Jul 31, 2012 at 2:51
  • 1
    $\begingroup$ As for why the two derivatives are made to agree on scalars, my response is that this is a desired property of the covariant derivative that we enforce, just like linearity and obeying the Leibniz (product) rule. (Perhaps someone has a more enlightening reason.) These properties, together with a couple other properties relating to contractions and derivatives of the metric, uniquely define the covariant derivative we use in standard GR. $\endgroup$
    – user10851
    Jul 31, 2012 at 2:55

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