How can one prove that the Einstein tensor (as it is usually defined in the field equations) is the contraction of the double of the Riemann curvature tensor?

Specifically, I want to show

$$ R^\mu_\nu-G^\mu_\nu = \dfrac{1}{2}R\delta^{\mu}_{\nu} $$ where $G^\mu_\nu$ is the contraction of the double dual of the Riemann curvature tensor. I have proceeded by taking cases: $\mu=\nu$ and $\mu\neq\nu$. The second case went fine (I got zero on the righthand side), but the first case did not.

Here is my attempt in the first case. By the definition of the contraction of the double dual, I got $$R^{\mu}_{\mu}-G^{\mu}_{\mu}=R-\dfrac{1}{4}\epsilon^{\alpha\mu\sigma\omega}\epsilon_{\alpha\mu\zeta\xi}R^{\sigma\omega}_{\zeta\xi}$$ Using the formula $$\epsilon_{i_1\dots i_k\ell_{k+1}\dots\ell_n}\epsilon^{j_1\dots j_k\ell_{k+1}\dots\ell_n}=\det(g)(n-k)!\delta^{i_1\dots i_k}_{j_1\dots j_k}$$ (Here, $g$ is the metric tensor) I got

$$R^\mu_\mu-G^\mu_\mu = R+\dfrac{1}{2}\delta^{\zeta\xi}_{\sigma\omega}R^{\sigma\omega}_{\zeta\xi} =R+\dfrac{1}{2}(\delta^{\zeta}_{\sigma}\delta^{\xi}_{\omega}-\delta^{\xi}_{\sigma}\delta^{\zeta}_{\omega})R^{\sigma\omega}_{\zeta\xi} =R+\dfrac{1}{2}(R^{\sigma\omega}_{\sigma\omega}-R^{\sigma\omega}_{\omega\sigma}) =R+\dfrac{1}{2}(2R)=2R $$ which is not the desired result of $\dfrac{1}{2}R$.

  • 1
    $\begingroup$ But $\tfrac12R\delta^\mu_\mu=2R$ in $4$ dimensions. $\endgroup$
    – J.G.
    Oct 29 '20 at 7:01
  • $\begingroup$ Thanks! That's what I was missing! So obvious. $\endgroup$ Oct 29 '20 at 7:03

I may as well convert my comment to an answer. The $\mu=\nu$ calculation is actually a contraction over all cases, not just the case not satisfying $\mu\ne\nu$. Since the latter obtains $0$, $R^\mu_\nu-G^\mu_\nu=A\delta^\mu_\nu$ in general iff $R^\mu_\mu-G^\mu_\mu=4A$.

  • $\begingroup$ Good point. I did not need to take cases. $\endgroup$ Oct 29 '20 at 21:46
  • $\begingroup$ I appreciate the response. $\endgroup$ Oct 29 '20 at 21:46

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