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*Index symmetries:*

A stiffness tensor $ C $ is a fourth order tensor with components $ c_ {ijkl} $ which maps symmetrical second order tensors into symmetrical second order tensors, i.e. $ sigma_ {ij} = c_ {ijkl} varepsilon_ {kl} $ (linear elastic law), $ sigma $ (Stress) and $ varepsilon $ (Elongation) are any symmetrical second order tensors. Because of the symmetry of the second order tensors $ C $ may be slightly symmetrical, i.e. $ c_ {ijkl} = c_ {jikl} = c_ {ijlk} $. The not insignificant symmetrical part of $ C $ is irrelevant to the elastic law and is dropped. When the stress $ sigma $ is related to an elastic energy potential $ W $ (referred to as hyperelastic behavior), i.e. $ sigma = partial W / partial varepsilon $then, due to Black's theorem, the stiffness tensor $ c_ {ijkl} = partially ^ 2 W / partially varepsilon_ {ij} partially varepsilon_ {kl} $ must have the main symmetry, i.e. $ c_ {ijkl} = c_ {klij} $.

*Material symmetry:*

A material with stiffness $ C $ should have the material symmetry group $ G $ (e.g. triclinic, orthotropic, transverse isotropic, …) if

begin {equation}

C = Q star C qquad Q in G.

end {equation}

stops where $ Q $ are second order tensors, which are called symmetry transformations of $ C $. The product $ has {C} = Q star C $ (referred to here as the Rayleigh product) is defined in components as

begin {equation}

hat {c} _ {ijkl}

= Q_ {im} Q_ {jn} Q_ {ko} Q_ {lp} c_ {mnop}

end {equation}

For solids, $ G $ is a subset of the orthogonal group. In solid-state mechanics, it is sufficient to take rotation matrices into account $ Q $ from the rotation group $ SO (3) $. If $ G = {I } $, $ I $ so be the identity matrix $ C $ should be triclinic. If $ G $ has more than the identity transformation, then different material classes can be defined (different anisotropy types). If $ G = SO (3) $, the $ C $ should be isotropic (no directional dependence).

**I want to use Mathematica to get the number of independent parameters that the fourth order tensor needs $ C = Q asterisk C (Q in SO (3)) $ under the rotation of the group $ SO (3) $.**

Currently, I can only get 30 independent variables using the following method:

```
SymmetrizedIndependentComponents({3, 3, 3, 3},
Symmetric({1, 2, 3})) // Length
```

I still can't use the group's rotation $ S0 (3) $ to further reduce the number of independent variables. What should I do?