# Contraction of the metric tensor

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.

Here

$$R_{abcd}\equiv K(x)(g_{bd}g_{ac}-g_{ad}g_{bc})$$ 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?

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.
• 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
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