$$\mathbf{A} = A^{1}\mathbf{e_{1}}+A^{2}\mathbf{e_{2}}$$ $$A= \sum_{i=1}^{n}A^{i}\mathbf{e_{i}}$$

Taking the derivative wrt the tangent basis vector and dropping the summation by Einstein convention: $$\left ( \frac{\partial A^{i}}{\partial x^{j}}+\Gamma _{jk}^{i}A^{k}\right )\mathbf{e_{i}}$$

The covariant derivative of this contravector is

$$\nabla_{j}A^{i}\equiv \frac{\partial A^{i}}{\partial x^{j}}+\Gamma _{jk}^{i} A^{k}$$

Now, I would like to determine the covariant derivative of a covariant vector but ran into some problem. Namely, with the red highlighted parts in bold which does not appear in my sketch.

$$\nabla_{j}A_{i}\equiv\frac{\partial A_{i}}{\partial x^{j}}{\color{Red} -}\Gamma_{{\color{Red} ij}}^{{\color{Red} k}}A_{k}$$ which the above is a covariant derivative of a covariant vector.

Any help is appreciated.

  • $\begingroup$ Your formula seems incomplete. Should there not be a contraction of the $A_i$ with the Christoffel symbols? $\endgroup$
    – Muphrid
    Commented Oct 27, 2015 at 3:22
  • $\begingroup$ ah yes! Thanks for the heads up. Haven't had my cup of coffee. $\endgroup$ Commented Oct 27, 2015 at 3:26
  • $\begingroup$ Your indices on $A$ should be down, not up in that formula. Notice that means it's impossible to contract $A$ with the index $i$ in the Christoffel symbol. $\endgroup$
    – Muphrid
    Commented Oct 27, 2015 at 3:28
  • $\begingroup$ Edit has been done. $\endgroup$ Commented Oct 27, 2015 at 3:40
  • $\begingroup$ Okay, do you still have a question? Your formula is now correct, save for the index still being up in the partial derivative. $\endgroup$
    – Muphrid
    Commented Oct 27, 2015 at 3:42

1 Answer 1


I will proceed with Einstein summation as you have.

Let $\partial_i$ denote $\dfrac{\partial}{\partial x^i}$. Let $u^m$ be a contravector and $\mu_m$ a covector.

First note that by the product rule, \begin{align} \partial_i(\mu_m u^m) &= (\partial_i \mu_m)u^m + \mu_m(\partial_i u^m)\\ \mathrm{So\ by\ rearranging,}\quad (\partial_i \mu_m)u^m &= \partial_i(\mu_m u^m) - \mu_m(\partial_i u^m) \end{align} We now proceed by finding how $\nabla_i\mu_m$ acts on an arbitrary vector to show how it acts on all vectors. (The covariant derivative of a covector is itself a covector, i.e. a map that acts on vectors, so this is precisely what we must find out.) \begin{align} (\nabla_i\mu_m)(u^m) &= \partial_i (\mu_m u^m) - \mu_m(\nabla_i u^m)\\ &= \partial_i (\mu_m u^m) - \mu_m(\partial_iu^m + \Gamma_{ij}^m u^j) \mathrm{\quad(by\ the\ formula\ given\ in\ your\ question)}\\ &= (\partial_i\mu_m)u^m - \Gamma_{ij}^m \mu_m u^j \mathrm{\quad(by\ the\ product\ rule\ above)} \end{align} By the summation convention, we have that $(\partial_i\mu_m)u^m - \Gamma_{ij}^m \mu_m u^j = (\partial_i\mu_m)u^m - \Gamma_{ij}^p \mu_p u^j$. (As the two terms in the equation are linked by subtraction, we do not change anything by re-labelling the second $m$ index to another letter.) In the same way, we relabel the $j$ to $m$. $$(\nabla_i\mu_m)(u^m) = (\partial_i\mu_m)u^m - \Gamma_{im}^p \mu_p u^m$$ As the contravector was arbitrary, this shows that $$\nabla_i\mu_m = \partial_i\mu_m- \Gamma_{im}^p \mu_p$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .