How to prove Riemann curvature identities

Question

Let $R_{abcd}$ be the Riemann curvature with all indices lowered. It is said that from the antisymmetric property $R_{abcd}=-R_{bacd}$ and the first Bianchi Identity $R_{abcd}+R_{acdb}+R_{adbc}=0$ it is possible to prove the symmetric property $R_{abcd}=R_{cdab}$. But I cannot find a way to prove this. Could anyone help me?

Answer 1

$\require{cancel}$

I will call BI: Bianchi Identity, and SS: Skew Symmetry, so that

\begin{eqnarray} R_{abcd} - R_{cdab} &\stackrel{\rm BI}{=}& R_{abcd} - (-R_{cabd} - R_{cbda})\\ &\stackrel{\rm SS}{=}& R_{abcd} - R_{acbd} - R_{bcda} \\ &\stackrel{\rm BI}{=}& R_{abcd} - (-R_{abdc} - R_{adcb}) - (- R_{bacd} - R_{bdac}) \\ &=& R_{abcd} + R_{abdc} + R_{adcb} + R_{bacd} + R_{bdac} \\ &\stackrel{\rm SS}{=}& \cancel{R_{abcd}}- \cancel{R_{abcd}} + R_{adcb} - R_{abcd} + R_{bdac} \\ &=& -R_{abcd} + R_{adcb} + R_{bdac} \\ &\stackrel{\rm SS}{=}& -(R_{abcd} - R_{adbc}) + R_{bdac} \\ &\stackrel{\rm BI}{=}& R_{acdb} + R_{bdac} \\ &\stackrel{\rm SS}{=}& -R_{acbd} + R_{bdac} \tag{1} \end{eqnarray}

Now, I'm going to change the name of the indices $a\leftrightarrow b$ and $c\leftrightarrow d$

\begin{eqnarray} R_{badc} - R_{dcba} &=& -R_{bdac} + R_{acbd} \\ R_{abcd} - R_{cdab} &\stackrel{\rm SS}{=}& -R_{bdac} + R_{acbd}\\ &\stackrel{(1)}{=}&-R_{acbd}+R_{bdac} \tag{2} \end{eqnarray}

From the last identity we have then

$$ -R_{bdac} + R_{acbd} = -R_{acbd}+R_{bdac} \tag{3} $$

which leads to

$$ R_{bdac} = R_{acbd} $$

... and that's it

Answer 2

The symmetry $R_{abcd}=R_{cdab}$ does not follow from $R_{abcd}=-R_{bacd}$ and the Bianchi Identity. You also need to assume antisymmetry in the last two indices: $R_{abcd}=-R_{abdc}$. With that additional assumption, the proof by @caverac works.

The additional symmetry, $R_{abcd}=-R_{abdc}$, depends on the fact that the Levi-Civita connection is compatible with the metric; without this assumption, the symmetry $R_{abcd}=-R_{abdc}$ might not hold.

It's interesting to note that the three symmetries reflect different aspects of the connection:

• $R_{abcd}=-R_{bacd}$ is true for the curvature of any connection on the tangent bundle (after lowering an index with respect to any metric at all).
• $R_{abcd}=-R_{abdc}$ follows from compatibility of $\nabla$ with the metric, as noted above.
• The Bianchi identity $R_{abcd} + R_{bcad} + R_{cabd}=0$ follows from the fact that the connection is torsion-free.
• $R_{abcd}=R_{cdab}$ follows from the three symmetries above.

(This Bianchi identity is slightly different from the one you wrote, because it involves the two indices that are independently known to be antisymmetric.) See pages 121-123 in my Riemannian Manifolds for more detail.