For a rank 2 tensor ${A^a}_b$ in 3 dimensions, an explicit expression for its inverse in index notation is given by,
$ {{(A^{-1})^a}_{b}} = 3\frac{ \epsilon^{a \beta_1 \beta_2} \epsilon_{b \alpha_1 \alpha_2} {A^{\alpha_1}}_{\beta_1} {A^{\alpha_2}}_{\beta_2} }{ \epsilon^{\beta_3 \beta_4 \beta_5} \epsilon_{\alpha_3 \alpha_4 \alpha_5} {A^{\alpha_3}}_{\beta_3} {A^{\alpha_4}}_{\beta_4} {A^{\alpha_5}}_{\beta_5} } $
such that
$ {{(A^{-1})^a}_{k}} {A^k}_b = {\delta^a}_b. $
I have searched, but not yet found a generalization of this formula for rank 4 tensors ${M^{ab}}_{kl}$, such that
$ {(M^{-1})^{ab}}_{kl} {M^{kl}}_{ij} = {\delta^a}_i {\delta^b}_j $
Naively, one might try a generalization like
$ {{(M^{-1})^{ab}}_{cd}} \overset{?}{=} 3\frac{ \epsilon^{a \gamma_1 \gamma_2} \epsilon^{b \delta_1 \delta_2} \epsilon_{c \alpha_1 \alpha_2} \epsilon_{d \beta_1 \beta_2} {M^{\alpha_1 \beta_1 }}_{\gamma_1 \delta_1} {M^{\alpha_2 \beta_2 }}_{\gamma_2 \delta_2} }{ \epsilon^{\gamma_3 \gamma_4 \gamma_5} \epsilon^{\delta_3 \delta_4 \delta_5} \epsilon_{\alpha_3 \alpha_4 \alpha_5} \epsilon_{\beta_3 \beta_4 \beta_5} {M^{\alpha_3 \beta_3 }}_{\gamma_3 \delta_3} {M^{\alpha_4 \beta_4 }}_{\gamma_4 \delta_4} {M^{\alpha_5 \beta_5 }}_{\gamma_5 \delta_5} } $
But trying this out in Mathematica it doesn't seem like this works.
Does anyone know of such a generalization that works?
As a direct application, in linear elasticity, the stress tensor is given in terms of the strain tensor by
$ \sigma^{ij} = {C^{ij}}_{kl} e^{kl} $
and a closed form for the inverse of $C^{ij}_{kl}$ would yield the inverse relation
$ e^{ij} = {(C^{-1})^{ij}}_{kl} \sigma^{kl} $
