In flat space, the metric tensor is (in one of the two conventions)

$$\eta^\mu{} ^\nu = \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} = \eta_\mu{}_\nu$$ What is $\eta^\mu{}_\nu$ or $\eta_\mu{}^\nu$? I read here it was the same! But if we use the metric tensor to convert covariant components and contravariant components, it seems like the answer should be the identity matrix.

If we use the definition of the metric as the cofficients of an infitesimal arc length element $ds^2$, I think this also shows that the answer should be the identity matrix.

Can someone explain (in a simple way for someone who doesn't know much about differential geometry) how to obtain the correct answer, whatever it is? Preferably with an answer that doesn't just rely on the memorized rules of einstein index notation.

  • 1
    $\begingroup$ Yup, your reasoning is perfectly right, and I have no idea what that source is trying to say. $\endgroup$
    – knzhou
    Commented May 1, 2017 at 21:36
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/119126/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented May 1, 2017 at 21:52

1 Answer 1


For any metric (either on flat or curved spacetime), $g^\mu_{\ \ \nu} = g^{\ \ \nu}_\mu = \delta^\mu_\nu$. See here for the explanation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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