The final expression for the tensor product $\mathbf{F}{ij} \mathbf{F}{jk}$ in terms of the magnetic field components $B_i$ is: $\mathbf{F}{ij} \mathbf{F}{jk} = B^2 \delta_{ik} - B_i B_k$
Where:
The result is derived using the definition of the magnetic field tensor and the vector identity for the product of two Levi-Civita symbols.
Substitute the definition $\mathbf{F}{ij} = \epsilon{ijm} B_m$ (using $m$ as a dummy index for the first $\mathbf{F}$) and $\mathbf{F}{jk} = \epsilon{jkl} B_l$ (using $l$ for the second $\mathbf{F}$):
$$ \mathbf{F}{ij} \mathbf{F}{jk} = (\epsilon_{ijm} B_m) (\epsilon_{jkl} B_l) = B_m B_l (\epsilon_{ijm} \epsilon_{jkl}) $$
Rearrange the Levi-Civita symbols to group the repeated summation index $j$:
$$ \epsilon_{jkl} = -\epsilon_{kjl} $$
$$ \mathbf{F}{ij} \mathbf{F}{jk} = - B_m B_l (\epsilon_{ijm} \epsilon_{kjl}) $$
Use the $\epsilon-\delta$ identity: $\epsilon_{abc} \epsilon_{dbc} = \delta_{ad} \delta_{cb} \delta_{cb} - \delta_{ab} \delta_{dc} \delta_{cb}$. A simpler form for the expression $\epsilon_{ijm} \epsilon_{kjl}$ (with summation over $j$) is:
$$ \epsilon_{ijm} \epsilon_{kjl} = \delta_{ik} \delta_{ml} - \delta_{il} \delta_{mk} $$
Substitute the identity back:
$$ \mathbf{F}{ij} \mathbf{F}{jk} = - B_m B_l (\delta_{ik} \delta_{ml} - \delta_{il} \delta_{mk}) $$
$$ \mathbf{F}{ij} \mathbf{F}{jk} = - B_m B_l \delta_{ik} \delta_{ml} + B_m B_l \delta_{il} \delta_{mk} $$
Apply the Kronecker deltas (using the summation convention):
First term: $B_m B_l \delta_{ml} = B_m B_m = B^2$.
$$ B_m B_l \delta_{ik} \delta_{ml} = B_m B_m \delta_{ik} = B^2 \delta_{ik} $$
Second term: $\delta_{il}$ sets $l=i$, and $\delta_{mk}$ sets $m=k$.
$$ B_m B_l \delta_{il} \delta_{mk} = B_k B_i $$
Combine the terms to get the final expression:
$$ \mathbf{F}{ij} \mathbf{F}{jk} = - B^2 \delta_{ik} + B_i B_k $$
$$ \mathbf{F}{ij} \mathbf{F}{jk} = B_i B_k - B^2 \delta_{ik} $$