0
$\begingroup$

Let $(M,g)$ be a pseudo-riemann manifold and $(U,\psi=(x^1,\ldots,x^n))$ a local chart around some point $p$ in $M$. It is easy to show that if $\partial g_{ij}/\partial x^k=0$ in $p$ for all $i,j,k$ then $\Gamma_{ij}^k(p)=0$ for all $i,j,k$ because: $$\Gamma_{ij}^k =\frac{1}{2}\sum_{h=1}^n \Big(\frac{\partial g_{ih}}{\partial x^j}+\frac{\partial g_{jh}}{\partial x^i}-\frac{\partial g_{ij}}{\partial x^h}\Big) g^{kh}$$ is the converse also true?

$\endgroup$
4
  • $\begingroup$ are you asking whether $g_{ij,k}=0$? $\endgroup$
    – Thomas
    Jun 19, 2016 at 19:07
  • $\begingroup$ My question is: if $\Gamma_{ij}^k(p)=0$ then $g_{ij,k}(p)=0$? $\endgroup$
    – FUUNK1000
    Jun 19, 2016 at 19:09
  • 1
    $\begingroup$ This is not what you've written in the headline $\endgroup$
    – Thomas
    Jun 19, 2016 at 19:09
  • 1
    $\begingroup$ Is M specifically a 4-manifold, or should the first line say $\psi = (x^1, \dots, x^n)$? $\endgroup$ Jun 19, 2016 at 20:54

1 Answer 1

2
$\begingroup$

The defining equation of the metric connection is $\nabla g = 0$, or in coordinates

$$ \nabla_i g_{jk} = \partial_i g_{jk} - \Gamma_{ij}^l g_{lk} - \Gamma_{ik}^l g_{jl} = 0.$$

Evaluating this at $p$ where $\Gamma = 0$ gives $\partial g=0$ at $p$.

Alternatively, if you want to work with nothing but the coordinate expression for $\Gamma$ in terms of $\partial g$, it's easier to work with $$\gamma_{ijk} = g_{kl} \Gamma^l_{ij}=\frac12\left(-\partial_k g_{ij} + \partial_i g_{jk} + \partial_jg_{ki}\right).$$ Note that $$\gamma_{ijk} + \gamma_{kji} = \partial_jg_{ik}.$$

Since $\Gamma = 0$ at $p$ we must have $\gamma = 0$ there also; so we get $\partial_j g_{ik} = 0$.

$\endgroup$

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.