I have an expression given to me in tensor/index notation and I need help expanding out into something more readable. The expression is
$$\frac{b_{ij}b_{jk}b_{ki}}{6}$$
and $b$ is a symmetric second order tensor, and the indices range from 1 to 3.
Can somebody please expand this expression for me?
I'm trying to code this expression up and I keep getting complex numbers, so I know I'm doing something wrong.
This is my attempt: $(b_{11}b_{11}b_{11} + b_{11}b_{12}b_{21} + b_{11}b_{13}b_{31} + b_{12}b_{21}b_{11} + b_{12}b_{22}b_{21} + b_{12}b_{23}b_{31} + b_{13}b_{31}b_{11} + b_{13}b_{32}b_{21} + b_{13}b_{33}b_{31} + b_{21}b_{11}b_{12} + b_{21}b_{12}b_{22} + b_{21}b_{13}b_{32} + b_{22}b_{21}b_{12} + b_{22}b_{22}b_{22} + b_{22}b_{23}b_{32} + b_{23}b_{31}b_{12} + b_{23}b_{32}b_{22} + b_{23}b_{33}b_{32} + b_{31}b_{11}b_{13} + b_{31}b_{12}b_{23} + b_{31}b_{13}b_{33} + b_{32}b_{21}b_{13} + b_{32}b_{22}b_{23} + b_{32}b_{23}b_{33} + b_{33}b_{31}b_{13} + b_{33}b_{32}b_{23} + b_{33}b_{33}b_{33})/ 6$
EDIT: I was computing the tensor $b$ incorrectly
Let us define a second-order symmetric tensor $\boldsymbol{B} = [b_{i j}]$ given by its coordinates $b_{\bullet\bullet}$ over an orthonormal basis (the bullets could be any distinct symbols). Using the dot product and Einstein notation, one can write $$ [(b^3)_{i \ell}] = \boldsymbol{B}^3 = \boldsymbol{B}\cdot\boldsymbol{B}\cdot\boldsymbol{B} = [ b_{i j}b_{jk}b_{k\ell} ] \, , $$ where $(b^3)_{\bullet \bullet}$ denotes the coordinates of $\boldsymbol{B}^3$. Taking the trace, the result in @Chappers comment follows: $$ \frac{1}{6}(b^3)_{i i} = \frac{b_{i j}b_{jk}b_{k i}}{6} = \frac{1}{6}\text {tr} (\boldsymbol{B}^3)\, . $$ Note that the symmetry hypothesis was not used here.
Let us use the tensor's symmetry to obtain an alternative formula. We apply the spectral theorem and the Cayley–Hamilton theorem, so that $$ \boldsymbol{B}^3 - {B}_\text{I}\, \boldsymbol{B}^2 + {B}_\text{II}\, \boldsymbol{B}^1 - B_\text{III} \boldsymbol{B}^0 = \boldsymbol{0} \, . $$ The coefficients -- principal invariants of $\boldsymbol{B}$ -- in the characteristic polynomial above are computed separately: \begin{aligned} {B}_\text{I} &= \text{tr}(\boldsymbol{B}) &&= b_{ii}\, ,\\ {B}_\text{II} &= \tfrac{1}{2} \left(\text{tr}(\boldsymbol{B})^2 - \text{tr}(\boldsymbol{B}^2)\right) &&= \tfrac{1}{2} \left((b_{ii})^2 - b_{ij}b_{ij}\right) ,\\ {B}_\text{III} &= \det \boldsymbol{B} &&= \varepsilon_{ijk}b_{1i}b_{2j}b_{3k} \, , \end{aligned} where $\varepsilon_{ijk}$ is the Levi-Civita permutation symbol. Taking the trace, one obtains \begin{aligned} \text {tr} (\boldsymbol{B}^3) &= 3 {B}_\text{III} - {B}_\text{II}{B}_\text{I} + {{B}_\text{I}}\, \text {tr} (\boldsymbol{B}^2)\\ &= 3 {B}_\text{III} - 3 {B}_\text{II}{B}_\text{I} + {{B}_\text{I}}^3 \, . \end{aligned}
Note that the invariants are related to the eigenvalues $\lambda_1, \lambda_2, \lambda_3$ of $\boldsymbol{B}$ through \begin{aligned} {B}_\text{I} &= \lambda_1 + \lambda_2 + \lambda_3\, ,\\ {B}_\text{II} &= \lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \lambda_2 \lambda_3 \, ,\\ {B}_\text{III} &= \lambda_1 \lambda_2 \lambda_3 \, . \end{aligned}