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I am currently working on a subject that involves a lot of 4th order tensors computations including double dot product and inverse of fourth order tensors.

First the definitions so that we are on the same page. What I call the double dot product is :

$$ (A:B)_{ijkl} = A_{ijmn}B_{mnkl} $$

and for the double dot product between a fourth order tensor and a second order tensor :

$$ (A:s)_{ij} = A_{ijkl}s_{kl}$$

Using the convention of sommation over repeating indices.

What I call the identity of the fourth order tensors is the only tensor such that :

$$ A:I = I:A = I $$

it is defined by $ I = \delta_{ik}\delta_{jl} e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l} $.

What I call the inverse of a fourth order tensor is the inverse with respect to the double dot product, that is, the inverse of $A$ is the only tensor $B$ such that $AB = BA = I$.

The double dot product is easy to compute if you don't think about the efficiency of the code, just create an array and loop over the four indices. Computing the inverse is something else. Every tensor I use has the minor symmetries $A_{ijkl} = A_{jikl} = A_{ijlk}$ so I thought I would use the Mandel representation for second order and fourth order tensors mentioned on Wikipedia. The fourth order tensor can be put into a $6 \times6$ matrix with the following components :

$$ [C] = \begin{bmatrix} C_{1111} & C_{1122} & C_{1133} & \sqrt{2}C_{1123} & \sqrt{2}C_{1131} & \sqrt{2}C_{1112}\\ C_{2211} & C_{2222} & C_{2233} & \sqrt{2}C_{2223} & \sqrt{2}C_{2231} & \sqrt{2}C_{2212}\\ C_{3311} & C_{3322} & C_{3333} & \sqrt{2}C_{3323} & \sqrt{2}C_{3331} & \sqrt{2}C_{3312}\\ \sqrt{2}C_{2311} & \sqrt{2}C_{2322} & \sqrt{2}C_{2333} & 2C_{2323} & 2C_{2331} & 2C_{2312}\\ \sqrt{2}C_{3111} & \sqrt{2}C_{3122} & \sqrt{2}C_{3133} & 2C_{3123} & 2C_{3131} & 2C_{3112}\\ \sqrt{2}C_{1211} & \sqrt{2}C_{1222} & \sqrt{2}C_{1233} & 2C_{1223} &2C_{1231} & 2C_{1212} \end{bmatrix} $$

$C$ is a fourth order tensor with minor symmetries and $[C]$ is its Mandel representation. The reason why Mandel's representation exists according to different sources is such that the matrix-matrix and matrix-vector usual products coincide with the fourth order tensors double dot product and the inverse in each respective space (fourth order tensors and $6\times 6$ matrices) coincides as well, that is

$$ [A:B] = [A].[B] \qquad \qquad (1) $$ and $$ [A^{-1}] = [A]^{-1} \qquad \qquad (2) $$

where $.$ is the usual matrix-matrix product. But it doesn't work or at least there must be something I don't understand. If I put the identity 4th order tensor defined above into Mandel's notation, I get the following matrix :

$$ [I] = \begin{bmatrix} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&2&0&0\\ 0&0&0&0&2&0\\ 0&0&0&0&0&2 \end{bmatrix} $$

which is obviously different from the identity of $6 \times 6$ matrices so if I compute $[C].[I]$ using the usual matrix-matrix product I won't get the same $[C]$. I also wrote a little script to check relations (1) and (2) but wasn't able to find this result with random $4^{th}$ order tensors possessing minor symmetries.

What am I missing here ?

Thanks a lot for your help and the discussions to come :)

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  • $\begingroup$ Are you sure about the presence of the $2$ coefficients? For instance, in Voigt notation note how the stress tensor is not given such coefficients, whereas the strain tensor is. Look especially at the stiffness tensor here. $\endgroup$ – user3658307 May 10 '17 at 19:01
  • $\begingroup$ Well, in any website, book or literature I found, the Mandel notation is exactly as I wrote. The Voigt notation you're suggesting doesnt have any factor but according to every source, it doesn't preserve the double dot product or the inverse when using the usual matrix-matrix product $\endgroup$ – Experience111 May 10 '17 at 20:22
  • $\begingroup$ There's definitely an issue with your identity tensor since $[I][I]\ne[I]$. These representations may depend on covariance/contravariance ... check out this review by Helnwein. $\endgroup$ – user3658307 May 10 '17 at 22:53
  • $\begingroup$ @user3658307 What I know for sure is that $I = \delta_{ik}\delta_{jl} e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l}$ is the proper identity tensor for the double dot product as I checked it with several random 4-th order tensors. However, you're right the representation as I gave it is not the right representation so there must be an issue. I already came accross the review you suggested but didn't read it extensively, I'll check it out again. $\endgroup$ – Experience111 May 11 '17 at 6:46
  • $\begingroup$ What I'm sure of so far is that the Mandel representation is correct i.e. $[A:B] = [A].[B]$ because I checked it by hand. The problem is definitely in my identity tensor. I'm not so sure anymore about the relation $I = \delta_{ik}\delta_{jl} e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l}$ But when I use $\frac{1}{2}( \delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk})e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l}$ my computation for the double product fails $A:I \neq A$ $\endgroup$ – Experience111 May 11 '17 at 7:38
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I'll answer my own question since I was able to find the solution to my problem with the help of one commentator. The definition of the identity tensor $I = \delta_{ik}\delta_{jl} e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l}$ is correct but it does not lead to a tensor with minor symmetries.

My mistake was in using this definition of the identity and applying the Mandel transformation to it. The Mandel transformation preserves the double dot product and the inverse if and only if the transformed tensors have the minor symmetries. As suggested in the review by Helnwein mentioned by @user3658307 whom I thank for his help, the definition one should use for the identity tensor in the case of minor symmetries is :

$$ I = \frac{1}{2}( \delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk})e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l} $$

Which has the minor symmetries and can thus be put into the Mandel representation that yields :

$$ [I] = \begin{bmatrix} 1 & 0&0&0&0&0 \\ 0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\end{bmatrix} $$

If you are interested in a more rigorous presentation of this subject (matrix representation of tensors), I highly recommend reading the aforementioned review. In particular, one should be really careful of the covariance and contravariance of the tensors he or she is handling as in some particular cases, their matrix representation can differ even if $A^{ij}_{kl} = A_{ij}^{kl}$.

Note on the inverse :

My original goal was to find an easy way to inverse fourth order tensors with minor symmetries using usual inversion algorithms for matrices. It is not always possible in the general case since the matrix representation of a general fourth order tensor possessing only minor symmetries is not always invertible in the space of $6\times 6$ matrices. However it is known that given a random $n \times n$ matrix with real coordinates, the 'probability' of it being invertible is $1$ in the sense of the Lebesgue measure. In linear elasticity or physics in general, one should thus be able to compute an inverse in every case.

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  • $\begingroup$ The link to the paper is no longer working. I think the referenced paper is: Helnwein, P. (2001). Some remarks on the compressed matrix representation of symmetric second-order and fourth-order tensors. Computer methods in applied mechanics and engineering, 190(22-23), 2753-2770. $\endgroup$ – Philipp Jan 19 at 22:24
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In this modern computer age, I'm not sure why anyone is enamored by these old compressed $\,6\times 6\,$ formats anymore.

Column-stacking will transform any $\,3\times 3\,$ matrix into a $\,9\times 1\,$ vector. $$ [S] = {\rm vec}(S) = \begin{bmatrix} S_{11} \\S_{21}\\S_{31}\\S_{12}\\S_{22}\\S_{32}\\S_{13}\\S_{23} \\S_{33} \end{bmatrix} $$ The index $(\alpha)$ of the vector elements can be calculated from the index-pair $(i,j)$ of the matrix elements with simple mod-3 arithmetic, and vice versa. $$\eqalign{ &\alpha = 1+(i-1)\,\%\,3 + (j-1)\times 3 \\ &i = 1+(\alpha-1)\,\%\,3,\quad j = 1+(\alpha-1)\,//\,3 \\ &\alpha\in[1\ldots 9],\quad\quad i,j\in[1\ldots 3] }$$ Extending this index mapping to an additional index/index-pair $\big(\beta,\, (k,l)\big)$ allows a fourth-order tensor $C_{ijkl}$ to be flattened into a matrix $[C]_{\alpha\beta}\,$ in a perfectly reversible way. $$ [C] = \begin{bmatrix} C_{1111} &C_{1121}&C_{1131}&C_{1112}&C_{1122}&C_{1132}&C_{1113}&C_{1123} &C_{1133}\\ C_{2111} &C_{2121}&C_{2131}&C_{2112}&C_{2122}&C_{2132}&C_{2113}&C_{2123} &C_{2133}\\ C_{3111} &C_{3121}&C_{3131}&C_{3112}&C_{3122}&C_{3132}&C_{3113}&C_{3123} &C_{3133}\\ C_{1211} &C_{1221}&C_{1231}&C_{1212}&C_{1222}&C_{1232}&C_{1213}&C_{1223} &C_{1233}\\ C_{2211} &C_{2221}&C_{2231}&C_{2212}&C_{2222}&C_{2232}&C_{2213}&C_{2223} &C_{2233}\\ C_{3211} &C_{3221}&C_{3231}&C_{3212}&C_{3222}&C_{3232}&C_{3213}&C_{3223} &C_{3233}\\ C_{1311} &C_{1321}&C_{1331}&C_{1312}&C_{1322}&C_{1332}&C_{1313}&C_{1323} &C_{1333}\\ C_{2311} &C_{2321}&C_{2331}&C_{2312}&C_{2322}&C_{2332}&C_{2313}&C_{2323} &C_{2333}\\ C_{3311} &C_{3321}&C_{3331}&C_{3312}&C_{3322}&C_{3332}&C_{3313}&C_{3323} &C_{3333} \end{bmatrix} $$

This is a very clean and easy to code transformation, since it doesn't require goofy scale factors like $(2,{\sqrt 2})$ on three-quarters of the terms.

A double-dot product between two tensors becomes a single-dot product in the flattened matrix representation, i.e. $$\eqalign{ C &= A:B &\implies C_{ijmn} = A_{ijkl}B_{klmn} \\ [C] &= [A]\cdot[B] &\implies [C]_{\alpha\lambda} = [A]_{\alpha\beta}\,[B]_{\beta\lambda} }$$

Finally, the inverse of a $\,9\times 9\,$ matrix $[C]^{-1}$ (should it exist) can be reconstituted into a fourth-order tensor. Not in some probabilistic sense, but in every single case.

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  • $\begingroup$ While you can indeed represent a fourth order tensor using a 9x9 matrix, I think the issue is when you want to preserve the operations. You will probably be able to inverse the 9x9 matrix representation of the fourth order tensor but I'm not sure it will match with the representation of the inverse of the tensor. I would be interested if you had any source about this :) $\endgroup$ – Experience111 Dec 10 '18 at 7:52
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    $\begingroup$ The equivalence of the operations $$C:S \Longleftrightarrow [C]\cdot[S]$$ is preserved by the above mapping. Look at the ordering of the first index pair in the columns of $C$ and $S$. It is the same ordering as the last index pair in the rows of $C$. The dot product is the sum over the index pair; the double-dot product sums over the same pair of indices. $\endgroup$ – greg Dec 10 '18 at 11:31
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    $\begingroup$ Also note that the fourth-order identity tensor in your original question gets mapped to the standard $9\times 9$ identity matrix -- because there is only one identity. The other fourth-order tensors mentioned in your answer are isotropic but they're not identity tensors. $\endgroup$ – greg Dec 10 '18 at 11:38

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