I want to write the Riemann tensor in two dimensions as


Now I know that the Ricci scalar is defined as $$R={R^{a}}_{a}=g^{ab}R_{ba}=g^{ab}R_{ab}$$ and the Ricci tensor is $$R_{ab}={R^m}_{amb}=g^{mn}R_{namb}.$$ So I found $$R=g^{ab}g^{mn}R_{namb}.$$ How does this return the identity I am looking for?


The purely covariant components of the Riemann-tensor satisfy the following symmetries: $$ R_{abcd}=R_{cdab} \\ R_{abcd}=-R_{bacd}. $$

It is of course skew-symmetric under the exchange of $c-d$ as well, but that follows from the first property.

So basically, $R_{(ab)(cd)}$ (the brackets are NOT symmetrization here!!!) is symmetric in $(ab)\leftrightarrow(cd)$ and skew-symmetric under the exchange of any two indices in the brackets.

In two dimensions, there is only one independent component of skew-symmetric second order tensors, as if $A_{ab}$ is skew, then if you know $A_{12}$, you know $A_{21}$ and you also know that $A_{11}=A_{22}=0$.

Which means that in two dimensions, a skew-symmetric index-pair $(ab)$ has only one effective value.

Obviously, then a symmetric (but, basically, any) array built out of such pairs also have only one value. Hence, the fact that $R$ has only one independent component in two dimensions follow from $R$'s symmetries alone, and not from the definition of $R$.

Which also means that the space of tensors which satisfy the symmetries of $R_{abcd}$ is one dimensional at every point.

It is easy to check that the tensor $$ X_{abcd}=g_{ac}g_{bd}-g_{ad}g_{bc} $$ satisfies the symmetries of the curvature tensor, and $X$ is manifestly not zero anywhere. So $X$ provides a basis for the one-dimensional space of tensors-that-have-the-symmetries-of-R at every point, therefore there exists a function $K$ such that $$ R_{abcd}=KX_{abcd}. $$

The function $K$ can be determined by contracting $R$: $$ R=R_{abcd}g^{ac}g^{bd}=KX_{abcd}g^{ac}g^{bd}=K(g_{ac}g_{bd}-g_{ad}g_{bc})g^{ac}g^{bd}=K(4-2)=2K, $$ from which $$ R_{abcd}=\frac{K}{2}(g_{ac}g_{bd}-g_{ad}g_{bc}) $$ follows.

  • $\begingroup$ Why $g_{ac}g_{bd}g^{ac}g^{bd}=4$? I know $g_{mn}g^{nk}=\delta^k_m$, but where does the $4$ come from? $\endgroup$ – Christian Willis Oct 8 '17 at 10:17
  • $\begingroup$ $g_{ac}g^{ac}=\delta^{a}_{a}=D$ in $D$ dimensions. So $g_{ac}g_{bd}g^{ac}g^{bd}=2\times 2=4$. $\endgroup$ – Demosthene Oct 8 '17 at 17:21

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