Posts on related issues can be found from here or here.

*Index symmetries:*

A stiffness tensor $ C$ is a fourth-order tensor with components $ c_{ijkl}$ which maps symmetric second-order tensors into symmetric second-order tensors, i.e., $ \sigma_{ij} = c_{ijkl} \varepsilon_{kl}$ (linear elastic law), $ \sigma$ (stress) and $ \varepsilon$ (strain) being arbitrary symmetric second-order tensors. Due to the symmetry of the second-order tensors, $ C$ is allowed to be minor symmetric, i.e., $ c_{ijkl} = c_{jikl} = c_{ijlk}$ . The not minor symmetric part of $ C$ is irrelevant for the elastic law and is dropped. If the stress $ \sigma$ is to be related to an elastic energy potential $ W$ (referred to as hyperelastic behavior), i.e., $ \sigma = \partial W / \partial \varepsilon$ , then, due to Schwarz’s theorem, the stiffness tensor $ c_{ijkl} = \partial^2 W / \partial \varepsilon_{ij} \partial \varepsilon_{kl}$ has to possess the major symmetry, i.e., $ c_{ijkl} = c_{klij}$ .

*Material symmetry:*

A material with stiffness $ C$ is said to possess the material symmetry group $ G$ (e.g., triclinic, orthotropic, transversally isotropic, …) if

\begin{equation} C = Q \star C \qquad Q \in G \end{equation}

holds, where $ Q$ are second-order tensors, referred to as symmetry transformations of $ C$ . The product $ \hat{C} = Q \star C$ (referred to here as 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 mechanics, if suffices to consider rotation matrices $ Q$ from the rotational group $ SO(3)$ . If $ G = \{I\}$ , $ I$ being the identity matrix, then $ C$ is said to triclinic. If $ G$ possesses more than the identity transformation, then different material classes can be defined (different anisotropy types). If $ G = SO(3)$ , the $ C$ is said to be isotropic (no direction dependency).

**I want to use Mathematica to get the number of independent parameters needed for the fourth-order tensor to make $ C = Q \star C (Q \in SO(3))$ under the rotation of group $ SO(3)$ .**

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

`SymmetrizedIndependentComponents[{3, 3, 3, 3}, Symmetric[{1, 2, 3}]] // Length `

However, I still can’t use the rotation of group $ SO(3)$ to further reduce the number of independent variables. What should I do?