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For instance, I know that $\chi^{2}$ materials need to lack inversion symmetry, but what about $\chi^{3}$ materials?

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Long answer:

Basically, $\chi^{(3)}$ is a rank four tensor with 81 components. You then apply the Neuman principle saying that symmetries of the crystal that supports $\chi^{(3)}$ should also be the symmetries of the tensor. Then you look for independent components of this tensor, which boils down to looking for trivial irreducible sub-representations in the representation of the symmetry group of the crystal over that rank four tensor.

So lets say the representation of your crystal symmetry group ($G$) over vectors is given by: $M\left(a\right): \mathbb{R}^3\to \mathbb{R}^3$, where $a\in G$

Then the induced representation for $\chi^{(3)}$ is $K\left(a\right)=M\left(a\right)\otimes M\left(a\right) \otimes M\left(a\right) \otimes M\left(a\right)$. If you want to know the independent components of $\chi^{(3)}$, simply define the projection operator, to project into trivial irreps:

$P=\frac{1}{\#G}\sum_{a\in G} K\left(a\right)$

$P:\mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3\to\mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3 $

The eigenvectors of this projection operator will be the allowed components of $\chi^{(3)}$, if you only want to know the number of allowed components $n_{allowed}$, it is given by the trace of the projection operator:

$n_{allowed}=Tr\left(P\right)=\frac{1}{\#G}\sum_{a\in G} Tr\left(K\left(a\right)\right)=\frac{1}{\#G}\sum_{a\in G} Tr\left(M\left(a\right)\right)^4$


Short answer:

Have a look in a suitable nonlinear optics book, e.g. Popov, Svirko, Zheludev "Susceptibility Tensors for Nonlinear Optics", Appendix F. They list all the allowed components of $\chi^{(3)}$ for all crystallographic classes. There is no class where $\chi^{(3)}$ is not allowed, but in some cases the number of allowed components can be as low as 2 (Cubic system, class 432) or even 1 (isotropic medium).

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