So, I have learned that the trace of a second order tensor $\textbf{Q}^{(2)}$ is given by:
$Tr(\textbf{Q}^{(2)})=Q_{11}+Q_{22}+Q_{33}$
Now, I want to calculate the trace of a third order tensor $\textbf{Q}^{(3)}$, is it true that:
$Tr(\textbf{Q}^{(3)})=Q_{111}+Q_{222}+Q_{333}$ ?
Or maybe this just doesn't make sense... In particular I want to calculate the trace of:
$Q_{ijk}=\int\bigg[r_{i}r_{j}r_{k}-\frac{1}{5}r^{2}(r_{i}\delta_{jk}+r_{j}\delta_{ik}+r_{k}\delta_{ij})\bigg]\rho(\textbf{r})d^{3}r^{\prime}$
where $\rho(\textbf{r})$ is a function of $r_{i}$, also $i,j,k=1,2,3$ (It is supposed to be a traceless tensor).
Any help with this?
Trace generalizes to tensor contraction which is defined for pairs of indices.
I suspect that by calling $Q$ a "traceless" tensor, the author/ your professor means that contracting $Q$ over any pair of indices yields $0$. Which certainly is the case here (try summing the part in brackets over any two of the indices and then notice that this is a totally symmetric tensor).