I read the following results on covariant differentiation (summation convention applied): If $X=X^i e_i$ and $Y=Y^je_j$, then

$$\nabla_XY = X^i\nabla_{e_i}(Y^je_j) = X^ie_i(Y^j)e_j + X^iY^j\Gamma_{ij}^ke_k = [X(Y^k)+X^iY^j\Gamma_{ij}^k]e_k.$$

I do not understand the usage of the third index $k$ here. Where does the third index come from, i.e. the $k$ in $\Gamma_{ij}^k$, please? Thank you! Consider the case where $n=2$. Could anyone explain how to fill the question marks, please? I think my question is where the third index come from in the definition of Christoffel symbols, i.e. $\nabla_{e_j}e_k =\Gamma_{jk}^i e_i$. Why do we need a third index $k$ for it?

$$\begin{align}\nabla_XY &= \nabla_{X^1e_1+X^2e_2}(Y^1e_1+Y^2e_2) \\ &= X^1\nabla_{e_1}(Y^1e_1) + X^1\nabla_{e_1}(Y^2e_2) + X^2\nabla_{e_2}(Y^1e_1) + X^2\nabla_{e_2}(Y^2e_2) \\ & = X^1e_1(Y^1)e_1 + X^1e_1(Y^2)e_2 + X^2e_2(Y^1)e_1 + X^2e_2(Y^2)e_2 \\ &+ X^1Y^1\nabla_{e_1}e_1 + X^1Y^2\nabla_{e_1}e_2 + X^2Y^1\nabla_{e_2}e_1 + X^2Y^2\nabla_{e_2}e_2 \\ &= \sum_{i=1}^2 X(Y^i)e_i + \sum_{i,j=1}^2 X^iY^j\nabla_{e_i}e_j\\&= \cdots??? \\ &= [X(Y^k)+X^iY^j\Gamma_{ij}^k]e_k\end{align}$$

  • $\begingroup$ there has to be a $k$ because the covariant derivative of a vector field is once more a vector field. Vector fields have components, hence the $k$. The Christoffel symbol quantifies how the basis vectors are rotating in the direction of the basis vector. In particular $\Gamma_{ij}^k$ tells how $e_j$ rotates into the $k$-direction as we move in direction of $e_i$. $\endgroup$ – James S. Cook May 18 '14 at 3:51
  • $\begingroup$ @JamesS.Cook Thanks. I think I got the point. It is like what we do in studying curves where we used the Frenet-Serret Formulae. Right? $\endgroup$ – LaTeXFan May 18 '14 at 3:54
  • $\begingroup$ The $k$ is a dummy index that's just used for summation. If you quit using the Einstein convention, and write out all the summations explicitly, I think everything will be clear (but longer). $\endgroup$ – bubba May 18 '14 at 4:16

By definition, $\nabla_{e_i}e_j = \Gamma_{ij}^k e_k $, again using the summation convention. The left hand side is a vector and you're just expanding it in the basis vectors, with the Christoffel symbols being the coefficients.

Maybe it would help you to not omit the summation: $\nabla_{e_i} e_j = \sum_k \Gamma_{ij}^k e_k$. Let's say instead of having some compound expression on the left side, you had some vector $v$. For simplicity, let's take the dimension to be $2$, as in your question. You want to write $v$ in terms of the basis vectors, $e_1,e_2$. Then $v = v_1e_1+v_2e_2$. In general, you write $v = v_1e_1+v_2e_2 + \dots v_ne_n$, or in short, $v = \sum_{k=1}^n v_ke_k$.

To be even briefer, one omits the summation and just says $v = v_ke_k$. Do you understand why we needed an extra index $k$ on the right hand side here, even though there weren't any on the left? Now let's go back to the original expression. Instead of having a fixed vector $v$, you have a bunch of vectors $v_{ij}$, given by $\nabla_{e_i}e_j$. To each of them you can apply the same procedure and expand in terms of $e_k$. Except now the coefficients will depend not only on $k$ but also $i$ and $j$. So this number which depends on these three indices is what you call $\Gamma_{ij}^k$.

  • $\begingroup$ Thank you for the comments. I think I asked the wrong question and I made update to it. Could you explain why we need the third index in the definition of Chirstoffel symbols in the first place, please? It seems to me that two indices are sufficient for the summation. $\endgroup$ – LaTeXFan May 18 '14 at 2:13
  • $\begingroup$ @20824 Does the edit answer your question? $\endgroup$ – ronno May 18 '14 at 2:21
  • $\begingroup$ Oh, I think I know now. Basically, we try to express $\nabla_{e_i}e^j$ in terms of the local coordinates, namely, $e_1, \cdots, e_n$ and the Christoffel symbols $\Gamma_{ij}^k$ can be treated as the coordinates/coefficients of $e_k$. Right? $\endgroup$ – LaTeXFan May 18 '14 at 2:29
  • $\begingroup$ It is not that «they can be treated as coefficients» of the e_k: they are! $\endgroup$ – Mariano Suárez-Álvarez May 18 '14 at 2:49
  • $\begingroup$ Thanks for the clarification. $\endgroup$ – LaTeXFan May 18 '14 at 3:55

The Christoffel symbols are defined so that for all $ i $ and $ j $ $$ \nabla_{e_i} e_j = \Gamma_{ij}^{k} e_k $$ This term comes up because of the product rule: $$ \nabla_{e_i} (Y^j e_j) = (\nabla_{e_i} Y^j)e_j + Y^j (\nabla_{e_i} e_j ) $$


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