For instance, I know that $\chi^{2}$ materials need to lack inversion symmetry, but what about $\chi^{3}$ materials?
1 Answer
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, with eigenvalues=1, 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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$\begingroup$ One more thing. Representation of the identity will always be the identity matrix, so its trace will be nonzero. Representation of group over the real-valued vectors will be real-valued (all entries in $M(a)$ will be real numbers). In case of $\chi^{(3)}$ one therefore will end up having a sum of real-valued things to 4-th power, with at least one of them being non-zero, so you will always end up with $n_{allowed}>0$ $\endgroup$– CryoCommented Oct 31, 2022 at 16:23