2
$\begingroup$

enter image description here

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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

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

as

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

(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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.