Does anyone knows what is the general form of an $8$th order isotropic tensor? $2$th order is $\delta_{ij}$, $4$th order it is $\lambda \delta_{ij} \delta_{kl}+\mu(\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk})$. what is the form of an $8$th order isotropic tensor? Thanks.
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Note that the expression you have for the fourth order is obviously from an elasticity text, since it has only two coefficients. In general, there are three independent coefficients for isotropic order 4 tensors. You have to impose additional conditions to get it down to two. – user_of_math Jul 09 '14 at 18:36
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Constructing the complete set of isotropic 8th order tensors is messy - there are 105(!) fundamental tensors, of which only 91 are linearly independent.
There is a paper "Linearly Independent Sets of Isotropic Cartesian Tensors of Ranks up to Eight" which gives a procedure to construct them (free download from NIST) and also contains a "minimal" list.
The original reference for representations of isotropic tensors is Hermann Weyl's "The Classical Groups", 1939, Princeton Press.
Also see the responses to this question; someone had suggested a paper by Harold Jeffreys.
May I ask why you are interested in these representations (if it is not a problem)?
user_of_math
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I need such tensor to calculate certain inequalities on 4th order tensors. I found a paper "Inversion of higher order isotropic tensors with minor symmetries and solution of higher order heterogeneity problems" which explicitly provide such tensors with minor symmetries. – Gal Jul 10 '14 at 21:29
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This is kind of very late to the party, but if anyone needs this stuff again, in this question
Finding basis of isotropic tensors of rank $n$
bases for isotropic tensors up to 8th-order have been computed using Mathematica, such that you can use them for explicit computations.
