[Physics] Trace of the Riemann Curvature Tensor

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Referring to Wald's General Relativity, I have two questions.

Let ${R_{abc}}^d$ be the Riemann curvature tensor.

  1. The author has never defined what it means by "trace of a tensor" before page 40 of the book, but he used this term on this page. According to this link, in case we are taking trace on a vector slot and a dual vector slot, it is simply equivalent to taking contraction of the tensor. But what if the two slots are both vectors/dual vector? In this case, contraction may not be defined (since it may be coordinate dependent). So how is it defined? For instance, how do we define the trace of ${R_{abc}}^d$ on the first and second vector slots?
  2. Why is taking the trace on the first and third slots of ${R_{abc}}^d$ equivalent to doing the same thing on the second and fourth slots? (This is how the Ricci tensor defined)

Best Answer

  1. You can't define a trace on $2$ similar indices (both up or both down). Evaluating a trace implies contraction of indices in a coordinate invariant way (just what you said). Remember to follow Einstein's summation convention to avoid getting stuck- for a contraction, an index occurs at most twice, with one of them up and the other down: $A^{\cdots \mu \cdots} B_{\cdots \mu \cdots}$.

So, in your example of ${R_{abc}}^d$, trace on indices $a$ and $b$ in the sense of ${R_{aac}}^d$ is not the way to go. But ${{R^a}_{ac}}^d$ is fine.

  1. This comes from the antisymmetry property of indices of Riemann tensor:

\begin{equation} R_{abcd} = -R_{bacd} = R_{badc} \end{equation}

so that taking a contraction (trace) gives:

\begin{equation} g^{ac}R_{abcd} = {R^c}_{bcd} = R_{bd} \quad \text{(contraction on indices 1 and 3)}\\ g^{ac}R_{badc} = {{R_b}^c}_{dc} = R_{bd} \quad \text{(contraction on indices 2 and 4)} \end{equation}

Also, to clarify a comment made by the OP in another answer about why the trace of Riemann tensor over its first $2$ or last $2$ indices vanishes - this comes from the fact that contracting a symmetric rank $2$ tensor ($S^{ab}$) with an antisymmetric rank $2$ tensor ($A_{ab}$) vanishes:

\begin{equation} S^{ab} A_{ab} = -S^{ba}A_{ba} = -S^{ab} A_{ab} \Rightarrow S^{ab} A_{ab} = 0 \end{equation}

where $a$ and $b$ are dummy indices.

In our case, the metric tensor is the symmetric tensor, and the first (last) $2$ indices of Riemann tensor form an antisymmetric pair

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