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I am reading these notes on differential geometry from a course at MIT. I have been verifying the computations for myself and I have a concern about the expression for $de^k$ on the final line. When I evaluate $de^k(e_i, e_j)$ according to the lemma, I obtain $-{\Gamma^k_{ij}} - -{\Gamma^k_{ji}}$ which is zero because the Christoffel symbols are symmetric in the lower indices. This would imply that $de^k = 0$ but that is not what seems to be written. Is there something that I am not understanding?


1 Answer 1


It is, I think, a bit misleading to call that a Christoffel symbols: in general if we have a local coordinate $(x^1, \cdots, x^n)$, then it gives you a coordinate frame

$$\frac{\partial}{\partial x^1}, \cdots, \frac{\partial}{\partial x^n}$$

and the Christoffel symbols is defined as (write $\partial_i$ for the coordinate vector for simplicity)

$$\nabla_{\partial_i} \partial_j = \Gamma_{ij}^k \partial_k.$$

In this case, $\Gamma_{ij}^k = \Gamma_{ji}^k$ as by the torsion free condition,

$$\nabla_{\partial_i} \partial_j - \nabla_{\partial_j} \partial_i = [\partial_i , \partial_j] = 0.$$

However, if $e_1, \cdots, e_n$ is an arbitrary basis, and $\Gamma_{ij}^k$ is defined as

$$\nabla_{e_i} e_j = \Gamma_{ij}^k e_k,$$

then $\Gamma_{ij}^k\neq \Gamma_{ji}^k$ (as $[e_i, e_j]\neq 0$).

Going back to your calculation, you have

$$de^k(e_i, e_j) = -(\Gamma_{ij}^k - \Gamma_{ji}^k),\ \ \ \forall i, j, k.$$

This implies

$$de^k = -\sum_{a, b} \Gamma_{ab}^k e^a\wedge e^b$$


$$-\sum_{a, b} \Gamma_{ab}^k e^a\wedge e^b \ (e_i, e_j) = -\Gamma_{ij}^k- (- \Gamma_{ji}^k)$$

(as in the summation there are both $e^i\wedge e^j$ and $e^j\wedge e^i$.)

Remark: It is my habit that I write $\Theta_{ij}^k$ instead of $\Gamma_{ij}^k$ if $e_1, \cdots, e_n$ are not coordinate basis.

  • $\begingroup$ So in essence the torsion-free connection only guarantees that the symbols are symmetric in the lower indices for the coordinate symbols? $\endgroup$
    – Memeozuki
    Sep 17, 2015 at 3:41
  • $\begingroup$ Yes. @Memeozuki ${}{}{}{}$ $\endgroup$
    – user99914
    Sep 17, 2015 at 3:43

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