Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is perhaps a simple tensor calculus problem -- but I just can't see why...

I have notes (in GR) that contains a proof of the statement

In space of constant sectional curvature, $K$ is independent of position.


$$R_{abcd}\equiv K(x)(g_{bd}g_{ac}-g_{ad}g_{bc})$$ where $R_{abcd}$ is the Riemann curvature tensor and $g_{ab}$ is the metric of the spacetime.

The proof goes like this:

Contract the defining equation with $g^{ac}$, giving $$R_{bd}=3Kg_{bd}.$$ and so on.

Problem is I don't understand why the contraction gives $$R_{bd}=3Kg_{bd}.$$ I can see the first term gives $$g^{ac}g_{bd}g_{ac}=4g_{bd}$$ since it's 4D spacetime. But as far as I can tell, the second term gives $g^{ac}g_{ad}g_{bc}=\delta_{bd}$ which is not necessarily $g_{bd}$.

Where have I gone wrong?

share|cite|improve this question
up vote 4 down vote accepted

As soon as you get something like $\delta_{bd}$, alarm bells should ring, as this is not a tensor.

The inverse metric $g^{ac}$ is defined by the identity $$ g^{ac}g_{cb} = \delta^a_b $$ If you plug this into your expression (and use the fact that $g$ is symmetric), you will obtain the correct equation.

share|cite|improve this answer
Thanks, Rhys. I was incredibly silly. Btw I didn't know that $\delta_{ab}$ is not a tensor... Why? – Clarice Apr 29 '13 at 15:50
With both indices 'downstairs', it is not invariant under coordinate changes. In other words, even if the components of a tensor are given by $\delta_{ab}$ in some coordinate system, they will take different values in other coordinate systems. On the other hand, $\delta^a_b$ is invariant, so can appear in valid tensor equations. – Rhys Apr 29 '13 at 15:54
Thanks again! :) – Clarice Apr 29 '13 at 17:14

You can simplify the equation as such: $$\begin{align} g^{ac}g_{ad}g_{bc}=&g^{ca}g_{ad}g_{bc} \\ g^{ac}g_{ad}g_{bc}=&\delta^{c}_{\phantom{.}d}g_{bc} \\ g^{ac}g_{ad}g_{bc}=&g_{bd} \end{align}$$ This is how you obtain the last metric tensor

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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