The Minkowski metric tensor have the relation $\eta_{ij} \eta^{jk}=\delta_i {^k}$. That is the inverse of the Minkowski matrix is the matrix itself.
Analogously, is it true that $g_{ij} g^{jk}=\delta_i {^k}$, where $g_{ij}$ is the metric tensor in a curved space? If yes how to prove this? I came up with the confusion while finding the chrischoffel symbol I came up with a equation $$2 \Gamma^{\gamma}_{\alpha j} g_{i \gamma}=g_{ij,\alpha}+g_{\alpha i,j}-g_{j\alpha,i} .$$ To eliminate $g_{i\gamma}$ I have to find the inverse of the metric tensor in tensor notation. Can anyone suggest how would I solve this problem?


The matrix $(g^{-1})^{\mu\nu}$ of components of the inverse metric tensor field$^1$ is not necessarily equal to the matrix $g_{\mu\nu}$ of components of the metric tensor field, if that's what you're asking.

However, there is a standard notational shorthand convention to write $(g^{-1})^{\mu\nu}$ as $g^{\mu\nu}$ because the upper indices already indicate that we're talking about the inverse metric.


$^1$ The metric tensor field is a covariant symmetric $(0,2)$ tensor field, while the inverse metric tensor field is a contravariant symmetric $(2,0)$ tensor field.

  • $\begingroup$ One should not really think of it as a matrix either as matrices typically are of the form $m^{\mu} _{\nu}$ with the first (upper) index indicating cotravariance and the second (lower) index representing covariance. $\endgroup$ – Paul Childs Mar 22 '19 at 3:02
  • $\begingroup$ 1. $g^{j m}$ in the equation $\Gamma$: $\Gamma^k_{li}=\frac{1}{2}g^{jm}(\frac{\partial g_{ij}}{\partial q^l}+\frac{\partial g_{lj}}{\partial q^i}-\frac{\partial g_{il}}{\partial q^j})$ of Christoffel symbol is the component of the inverse of the matrix representing the metric tensor. Is it right? 2. But it is it a covariant metric tensor too? $\endgroup$ – walber97 Mar 23 '19 at 14:14
  • 2
    $\begingroup$ 1. Yes. 2. No, it is instead contravariant. $\endgroup$ – Qmechanic Mar 23 '19 at 17:03

Even in flat spacetime, the metric does not have to have the form $\operatorname{diag}(1,-1,-1,-1)$. For example, if you want to do physics in SI units rather than natural relativistic units where $c=1$, then you want something like $g=\operatorname{diag}(c,-1,-1,-1)$. This is not the same as its own inverse.


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.