# Trace of the Riemann Curvature Tensor

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)

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:

$$R_{abcd} = -R_{bacd} = R_{badc}$$

so that taking a contraction (trace) gives:

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

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:

$$S^{ab} A_{ab} = -S^{ba}A_{ba} = -S^{ab} A_{ab} \Rightarrow S^{ab} A_{ab} = 0$$

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

For your first question, there is not a general consensus. Trace of a two rank tensor $\mathbf{A}$ is uniquely defined $Tr(\mathbf{A})=A_{\mu\nu}g^{\mu\nu}=A^{\mu\nu}g_{\mu\nu}={A_\mu}^\mu$ with $g_{\mu\nu}$ the metric tensor, but for a general tensor, the inidices which will be contracted should be explicit.

However, and answering the second question, Riemann tensor enjoys a lot of symmetries. This symmetries make all the contractions proportional or vanish. Take the Riemann tensor ${R_{abc}}^d$ and the antisymmetry in the first two indices ${R_{abc}}^d=-{R_{bac}}^d$. The contraction with the first index will be proportional with the contraction with the second index with proportionality factor $-1$.

If instead we try to contract the third and the fourth index:

$${R_{abc}}^c={R_{abcd}}g^{cd}={R_{abcd}}g^{(cd)}={R_{ab(cd)}}g^{cd}$$

And the last expression must vanish since the tensor is antisymmetric in those two indices while the metric tensor is symmetric.It is a good exercise to give a proof that the rest of the contractions also vanish or are proportional to the one you have given.

• In the book, the author stated "By the antisymmetry properties (1) and (3), the trace of the Riemann tensor over its first two or last two indices vanishes. " I don't really get what the meaning of "the trace of the Riemann tensor over its first two or last two indices vanishes. " is. – Jerry Aug 21 '18 at 10:43
• The trace of the first two indices $R_{abcd}g^{ab}=0$, the trace of the last two indices $R_{abcd}g^{cd}=0$. Trace of the first and third $R_{abcd}g^{ac}=0$. Trace over two indices is contract those two indices. – Alejandro Menaya Aug 21 '18 at 10:48