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} $


1 Answer 1


Bundle the indices into pairs $$ 11\to 1\\ 12\to 2\\ 13\to 3\\ 21\to 4\\ 22\to 5\\ \vdots $$ and then your rank four object becomes a rank-two matrix on a larger space. This will be 9 dimensional in the case $i$ or $j=$ 1,2,3 I was writing above. Such packaging is what one does for linear operators on tensor products of representations when one is making a Clebsh-Gordan decomposition, for example.

  • $\begingroup$ Often termed Voight notation. $\endgroup$ Commented Apr 23, 2021 at 16:06
  • $\begingroup$ I am familiar with this, and have also done this for the tensor in question. But it would be immensely useful for me to have an explicit expression in terms of the original indices. The fact that such a procedure is possible, does that not imply that there should exist such an expression? $\endgroup$ Commented Apr 23, 2021 at 18:46

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