# Dielectric permittivity and Kronecker symbol

I tried to express the dielectric permittivity tensor $$\varepsilon_{ij}$$ as: $$\begin{equation} \varepsilon_{ij} = \varepsilon_{r} \varepsilon_{0} = \left(\chi_{ij} +1\right) \varepsilon_{0}\text{,} \end{equation}$$ where $$\varepsilon_{0}$$ is the electrical permittivity of free space, $$\varepsilon_{r}$$ is the relative permittivity or dielectric constant of a material and $$\chi_{ij}$$ is the dielectric susceptibility tensor.

However, I was told a Kronecker symbol was missing. If I understand properly, it is because the central part of the equation ($$\varepsilon_{r} \varepsilon_{0}$$) must be a tensor to match the left and right hand sides. Would it be correct to write the equation as: $$\begin{equation} \varepsilon_{ij} = \varepsilon_{r} \varepsilon_{0}\delta_{ij} = \left(\chi_{ij} +1\right) \varepsilon_{0} \end{equation}$$ or do I misunderstand something?

Your equation, $$\varepsilon_{ij} \stackrel{?}{=} \varepsilon_{r} \varepsilon_{0}\delta_{ij} \stackrel{?}{=} \left(\chi_{ij} +1\right) \varepsilon_{0},$$ is incorrect.
To see why, consider the case where the material is not present, and you're just using your (now over-complicated) theory to describe a vacuum. In this case, your equation says that $$\varepsilon_{ij} = \varepsilon_0$$, and it says that this is true for every combination of $$i$$ and $$j$$, which can't be right.
Instead, the correct version is $$\varepsilon_{ij} = \left(\chi_{ij} +\delta_{ij}\right) \varepsilon_{0},$$ which uses the correct lifting of the scalar operator $$1$$ to the identity matrix. If you must have a relative permittivity $$\varepsilon_r$$, then this must also be a tensor, and you can write $$\varepsilon_{ij} = \left(\chi_{ij} +\delta_{ij}\right) \varepsilon_{0} = \varepsilon_{r,ij}\varepsilon_{0}$$.