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:

\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


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.

  • $\begingroup$ 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. $\endgroup$ – Jerry Aug 21 '18 at 10:43
  • $\begingroup$ 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. $\endgroup$ – Alejandro Menaya Aug 21 '18 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.