Clarifying the representation of isotropic symmetric tensor-valued functions of two symmetric tensors

Abstract

We reconsider the question of the representation of an isotropic symmetric second-order tensor-valued function of two symmetric second-order tensors. This question goes back roughly 70 years and represents the attempt to duplicate the stunning result possible when the function depends only on one tensor input. Over the years, there have been numerous attempts to derive a general representation theorem for such functions, and the literature is replete with conflicting statements as to the appropriate form. By focusing on the Cayley–Hamilton theorem for three-by-three matrices, we are able to arrive at a general result applicable to isotropic functions using nine tensor-valued basis functions. With the addition of an argument exploiting the Cayley–Hamilton theorem for two-by-two matrices, we are able to show that the complete function basis reduces to only involve eight tensor-valued basis functions. Our result clarifies which historically advocated representations are complete and non-redundant without need for complex qualifying cases based on the eigen-structure of the input tensors. The arguments are straightforward and only involve basic algebraic considerations with an intentional focus on Cayley–Hamilton-based reductions.

1. Revisiting isotropic representations

In this contribution, we reconsider the representation formulas for isotropic symmetric second-order tensor-valued functions of two symmetric second-order tensors. The topic has been studied for over 70 years, beginning with works by Rivlin and Ericksen [1] and Smith [2], followed by Wang [3], further refinements by Smith [4], and the works of Zheng [5], among quite a few others. In the case of symmetric tensor-valued functions of one symmetric tensor, the derivation of isotropic representations can be “cleanly” achieved through usage of the Cayley–Hamilton theorem, which guarantees a tensor A satisfies its own characteristic polynomial (see, e.g., Bowen and Wang [6, Theorem 26.1] or Gurtin et al. [7, §2.16]). Thus, powers of A higher than 3 can be represented as a combination of powers of A of order 2 or less. This results in the astonishing representation of all isotropic symmetric tensor-valued functions of a single symmetric tensor argument being expressible as:

$$\mathbf{B}=\widehat{\mathbf{B}}(\mathbf{A})={\phi }_0({\mathcal{I}}_{\mathbf{A}})\mathbf{1}+{\phi }_1({\mathcal{I}}_{\mathbf{A}})\mathbf{A}+{\phi }_2({\mathcal{I}}_{\mathbf{A}}){\mathbf{A}}^2,$$

(1)

where ${\phi }_0,{\phi }_1,{\phi }_2$ are scalar-valued functions of the principal invariants, ${\mathcal{I}}_{\mathbf{A}}=\{I_1(\mathbf{A}),I_2(\mathbf{A}),I_3(\mathbf{A})\}\equiv \{\mathrm{tr}(\mathbf{A}),\frac{1}{2}({(\mathrm{tr}(\mathbf{A}))}^2-\mathrm{tr}({\mathbf{A}}^2)),\det (\mathbf{A})\}$ , of A (see, e.g., Gurtin [8], §37)—astonishing, since there is no requirement that B be polynomial in A (see Appendix 1).

For functions of two or more tensors, however, the representations obtained since the 1950s have not relied (solely) on implications of the Cayley–Hamilton theorem. Rather, they have been surmised using algebraic considerations that are sometimes daunting to follow and have resulted in proposed forms that have required amendment over time. While it is certainly true that the necessity of a term within a representation can be shown by example (see Pennisi and Trovato [9]), it is difficult to deduce the span; that is, whether the proposed representation can be proven to generate all isotropic functions or at least some broad subset.

In this brief note, we take up this foundational issue and first look at how general of a result one can arrive at on the basis of applications of the 3D Cayley-Hamilton theorem alone. We will see that the best one can do is the reduction to the form:

$$\begin{array}{cccc}\hfill \mathbf{C}=\widehat{\mathbf{C}}(\mathbf{A},\mathbf{B})=c_{00}\mathbf{1}& +c_{10}\mathbf{A}+c_{11}\mathbf{B}+c_{20}{\mathbf{A}}^2+c_{21}{\mathbf{B}}^2+c_{22}(\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A})\hfill & \hfill & \hfill \\ \hfill & +c_{32}({\mathbf{A}}^2\mathbf{B}+\mathbf{B}{\mathbf{A}}^2)+c_{33}({\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2)+c_{42}({\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2),\hfill & \hfill \end{array}$$

(2)

where $\widehat{\mathbf{C}}$ is an isotropic symmetric second-order tensor-valued function of symmetric second-order tensors A and B, and the coefficients $c_{\mathrm{ij}}$ are any functions of the joint invariants of A and B. Thus, the tensorial function basis has nine elements, no more and no less, when only utilizing the 3D Cayley–Hamilton theorem. We will, however, show through an additional argument based on the 2D Cayley–Hamilton theorem and a lemma from Rivlin and Ericksen [1], that the fourth-order term, the last term in equation (2), is unneeded, leaving us with the final representation:

$$\begin{array}{cccc}\hfill \mathbf{C}=\widehat{\mathbf{C}}(\mathbf{A},\mathbf{B})=c_{00}\mathbf{1}& +c_{10}\mathbf{A}+c_{11}\mathbf{B}+c_{20}{\mathbf{A}}^2+c_{21}{\mathbf{B}}^2+c_{22}(\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A})\hfill & \hfill & \hfill \\ \hfill & +c_{32}({\mathbf{A}}^2\mathbf{B}+\mathbf{B}{\mathbf{A}}^2)+c_{33}({\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2).\hfill & \hfill \end{array}$$

(3)

There can be no fewer than the number of terms in equation (3) as can be shown by direct example [9]; thus, no more reduction is possible and the correct dimension of the tensor function basis is eight. This result only relies on the assumed existence of a tensorial expansion of $\widehat{\mathbf{C}}$ (see equation (4)) in terms of its arguments, irrespective of the eigen-structure of A and B, with coefficients that can be arbitrary functions of the scalar invariants. Our reasoning is detailed in what follows and differs somewhat from what is available in the current literature. Beyond adding clarity to the form of the representation, we also aim herein to deliver a clear step-wise derivation showing how it is obtained.

2. Brief historical recap

The literature contains a number of conflicting and difficult to follow developments by some of the seminal figures in continuum mechanics. Without in any way attempting to be comprehensive in citation of all works, we provide a brief road map to some of the most cited developments that took place with respect to the special representation question addressed in this note, viz. representations for isotropic symmetric second-order-tensor-valued functions of two symmetric second-order tensors. The developments are shown in Table 1 in a chronological order. Some useful definitions and facts for understanding the papers are as follows:

• $\widehat{\mathbf{C}}$ is a polynomial isotropic tensor function if it is a sum of joint powers of A and B with scalar coefficients. We will denote this case simply as polynomial. This does not imply that, as a series expansion in terms of A and B, the scalar coefficients are constants—a restriction that would be too severe for practical work [10, §7].

• The 10 basic invariants are ${\mathcal{I}}_{\mathbf{A},\mathbf{B}}=(\mathrm{tr}\mathbf{A},\mathrm{tr}{\mathbf{A}}^2,\mathrm{tr}{\mathbf{A}}^3,\mathrm{tr}\mathbf{B},\mathrm{tr}{\mathbf{B}}^2,\mathrm{tr}{\mathbf{B}}^3,\mathrm{tr}\mathbf{A}\mathbf{B},\mathrm{tr}\mathbf{A}{\mathbf{B}}^2,\mathrm{tr}\mathbf{B}{\mathbf{A}}^2,\mathrm{tr}{\mathbf{A}}^2{\mathbf{B}}^2)$ . The coefficients of the final representation are often determined or simply stated to be either rational, polynomial, or general functions in the 10 basic invariants (or at times of simply the components of A and B).

• The methods of proof usually differ in their approach. If a paper utilizes an approach invoking Cayley–Hamilton arguments, we denote it by CH. If a paper utilizes solvability conditions based on non-zero determinants of linear equations, we denote it as S. If a paper utilizes a co-set invariance argument, we denote it as Co. If the method of analysis is unclear to us, we denote it as unc.

It can be seen in the table that the ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ term in the representation appears and disappears over the last seven decades, even over common sets of assumptions. It is interesting to note that this fourth-order term comes and goes in papers with common authors without any comment, and similarly, when the term disappeared in the 1970s, it did so also without any explicit comments on why the prior analyses were in error. One of our goals in this paper is to show that this term is indeed not necessary.

Table 1.

Brief historical summary of the fourth-order term and the chronological developments.

Publication	Assumptions on $\widehat{\mathbf{C}}$	${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ included?	Coefficients	Method
Rivlin and Ericksen [1, §27,40]	No	No	Rational	S
Rivlin [11, eqn (1.7)]	Polynomial	Yes	Polynomial	CH
Spencer and Rivlin [12, eqn (4.4) & Thm 5]	Polynomial	Yes	Polynomial	CH
Wang [13, eqn (1.14)]	No	Yes	General	Co
Smith [14, eqn (2.7)]	No	Yes	General	unc a
Wang [3, p. 215]	No	Yes	General	Co
Smith [4, eqns (4.4) & (4.5)]	No	No	General	Co b
Pennisi and Trovato [9, eqn (2.3)]	No	No	General	unc c
Zheng [5, eqn (3.23)]	No	No	General	S+Co

a

The main point of the paper is to provide counter examples to Wang [13] for the case of more than two arguments; proofs of the basic results are not given.

b

Smith [4] states that he follows the method of Wang [3], while sharpening it, but does not provide any proofs or details of where Wang [3] can be sharpened; he only points out where he feels [3] is incomplete.

c

Pennisi and Trovato [9] adopt the results of Smith [4] without proof but declare them to be correct. Pennisi and Trovato [9] further go on to show that the Smith [4] results are irreducible via use of the Rouché–Capelli theorem.

3. The two tensor argument case

We take a moment to attempt a derivation which appears different from what we see in the existing literature, focused on the case of symmetric tensor-valued functions of two symmetric tensors. ¹ The representation derived below relies on reductions from the Cayley–Hamilton considerations and is (i) guaranteed to span a particular isotropic function space and (ii) never encounters “corner cases” where certain choices of A and B make one or more coefficients in the representation indeterminate or infinite. ² Letting A and B be two symmetric tensors, observe that any general polynomial series in A and B with symmetric output:

$$\begin{array}{cccc}\hfill \mathbf{C}=\widehat{\mathbf{C}}(\mathbf{A},\mathbf{B})=c_{00}\mathbf{1}& +c_{10}\mathbf{A}+c_{11}\mathbf{B}+c_{20}{\mathbf{A}}^2+c_{21}{\mathbf{B}}^2+c_{22}(\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A})+c_{30}{\mathbf{A}}^3+c_{31}{\mathbf{B}}^3\hfill & \hfill & \hfill \\ \hfill & +c_{32}({\mathbf{A}}^2\mathbf{B}+\mathbf{B}{\mathbf{A}}^2)+c_{33}({\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2)+c_{34}\mathbf{A}\mathbf{B}\mathbf{A}+c_{35}\mathbf{B}\mathbf{A}\mathbf{B}+c_{40}{\mathbf{A}}^4\hfill & \hfill & \hfill \\ \hfill & +c_{41}{\mathbf{B}}^4+c_{42}({\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2)+c_{43}(\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B})+c_{44}\mathbf{A}{\mathbf{B}}^2\mathbf{A}+\dots \hfill & \hfill \end{array}$$

(4)

is an isotropic symmetric-tensor-valued function. ³ The scalars $c_{\mathrm{ij}}={ĉ}_{\mathrm{ij}}({\mathcal{I}}_{\mathbf{A},\mathbf{B}})$ are arbitrary scalar-valued functions of ${\mathcal{I}}_{\mathbf{A},\mathbf{B}}$ , the joint invariants of A and B. Note the convention that the first index i represents the order of the term $c_{\mathrm{ij}}$ multiplies, and j counts through all the terms of the same order.

• Our goal is to collapse this space of tensor-valued functions into a representation with the fewest terms allowable using repeated application of Cayley–Hamilton-based reduction.

The intentionally broad form of equation (4) is chosen to be able to represent general isotropic tensor functions, noting that it contains infinitely more terms than any of the previously suggested representations. However, we cannot rule out at this stage the possibility that there may exist an isotropic function that is inexpressible via equation (4). But we further recognize that equation (4) is sufficiently general for most applications. As such, we restrict our attention to relations of this form.

3.1. Cayley–Hamilton reducers

First, we determine which joint powers of A and B can be expressed in terms of lower-order powers using the Cayley–Hamilton theorem. The Cayley–Hamilton theorem ensures that any (not necessarily symmetric) tensor M satisfies its own characteristic polynomial. In three dimensions, this implies:

$${\mathbf{M}}^3=\mathrm{tr}(\mathbf{M}){\mathbf{M}}^2-I_2(\mathbf{M})\mathbf{M}+\det (\mathbf{M})\mathbf{1}.$$

(5)

Let us apply this rule to progressively higher-order polynomials in A and B. Starting with an order one polynomial, we have, for arbitrary scalars $s_0$ and $s_1$ ,

$$\begin{array}{cccc}\hfill {(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B})}^3=\mathrm{tr}(\mathbf{1}& +s_0\mathbf{A}+s_1\mathbf{B}){(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B})}^2-I_2(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B})(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B})\hfill & \hfill & \hfill \\ \hfill & +\det (\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B})\mathbf{1}.\hfill & \hfill \end{array}$$

(6)

We introduce the shorthand $o(\mathbf{i},j)$ to denote a well-defined ith-order tensor polynomial in A and B whose scalar coefficients are ( ${\mathcal{I}}_{\mathbf{A},\mathbf{B}}$ -dependent) jth-order or lower polynomials in the s variables. For example, the right-hand side of equation (6) is $o(\mathbf{2},3)$ because no tensorial monomial exceeds order 2, but the scalar coefficients that multiply the tensorial monomials can depend on $s_0$ and $s_1$ up to order 3. We shall use $o(\mathbf{i})\equiv o(\mathbf{i},0)$ for tensor polynomials with no dependence on s variables. Expanding the left-hand side of equation (6) gives:

$${(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B})}^3=s_0^3{\mathbf{A}}^3+s_0^2{s}_1({\mathbf{A}}^2\mathbf{B}+\mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{B}{\mathbf{A}}^2)+s_0{s}_1^2({\mathbf{B}}^2\mathbf{A}+\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{A}{\mathbf{B}}^2)+s_1^3{\mathbf{B}}^3.$$

(7)

Since right-hand side of equation (6) is $o(\mathbf{2},3)$ , and hence has no tensorially third-order terms, when we gather terms with like powers of the s variables in equation (6), after applying the expansion in equation (7), we obtain:

$$\begin{array}{cccc}\hfill s_0^3[{\mathbf{A}}^3+o(\mathbf{2})]+s_0^2{s}_1[{\mathbf{A}}^2\mathbf{B}+& \mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{B}{\mathbf{A}}^2+o(\mathbf{2})]\hfill & \hfill & \hfill \\ \hfill & +s_0{s}_1^2[{\mathbf{B}}^2\mathbf{A}+\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{A}{\mathbf{B}}^2+o(\mathbf{2})]+s_1^3[{\mathbf{B}}^3+o(\mathbf{2})]+o(\mathbf{2},2)=\mathbf{0}.\hfill & \hfill & \hfill \end{array}$$

The above is a polynomial in $s_0$ and $s_1$ equal to the zero function. Thus, each term in brackets must be 0, yielding the following three third-order Cayley–Hamilton reductions:

$$\begin{array}{ccc}\hfill {\mathbf{A}}^3=o(\mathbf{2}),{\mathbf{A}}^2\mathbf{B}+\mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{B}{\mathbf{A}}^2=o(\mathbf{2}),{\mathbf{B}}^2\mathbf{A}+\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{A}{\mathbf{B}}^2=o(\mathbf{2}),{\mathbf{B}}^3=o(\mathbf{2}).& \hfill & \hfill \end{array}$$

(8)

The first and last reductions in the above are well known from the one-input case; e.g., that ${\mathbf{A}}^3$ can be expressed in terms of ${\mathbf{A}}^2$ , A, and 1, the usual Cayley–Hamilton result for a tensor. But the middle two arise only when considering a second input tensor. Heretofore, we use the term “n-reducer” to refer to a polynomial comprised of only nth degree powers in A and B, which can be rewritten in terms of lower-order powers, i.e., equation (8) shows four 3-reducers.

The same technique can be applied to determine reductions at higher orders. For example, if we use a second-order tensor polynomial on the left-hand side of equation (5), we get:

$$\begin{array}{cccc}\hfill & {(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B}+s_2{\mathbf{A}}^2+s_3{\mathbf{B}}^2+s_4\mathbf{A}\mathbf{B}+s_5\mathbf{B}\mathbf{A})}^3\hfill & \hfill & \hfill \\ \hfill & =\mathrm{tr}(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B}+s_2{\mathbf{A}}^2+s_3{\mathbf{B}}^2+s_4\mathbf{A}\mathbf{B}+s_5\mathbf{B}\mathbf{A}){(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B}+s_2{\mathbf{A}}^2+s_3{\mathbf{B}}^2+s_4\mathbf{A}\mathbf{B}+s_5\mathbf{B}\mathbf{A})}^2\hfill & \hfill & \hfill \\ \hfill & -I_2(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B}+s_2{\mathbf{A}}^2+s_3{\mathbf{B}}^2+s_4\mathbf{A}\mathbf{B}+s_5\mathbf{B}\mathbf{A})(\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B}+s_2{\mathbf{A}}^2+s_3{\mathbf{B}}^2+s_4\mathbf{A}\mathbf{B}+s_5\mathbf{B}\mathbf{A})\hfill & \hfill & \hfill \\ \hfill & +\det (\mathbf{1}+s_0\mathbf{A}+s_1\mathbf{B}+s_2{\mathbf{A}}^2+s_3{\mathbf{B}}^2+s_4\mathbf{A}\mathbf{B}+s_5\mathbf{B}\mathbf{A})\mathbf{1}.\hfill & \hfill \end{array}$$

(9)

Expanding both sides of the above gives very long expressions. However, all terms of order five or higher in A and B on the left side have no terms on the right to cancel them. With these fifth- and sixth-order terms, as before, we can separate and match terms by their corresponding powers of the s variables. These groupings, in turn, give rise to a set of twenty 5-reducers all equal to $o(\mathbf{4})$ :

$$\begin{array}{cccccccc}\hfill & 2{\mathbf{A}}^3{\mathbf{B}}^2+{\mathbf{A}}^2{\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2{\mathbf{A}}^2+2{\mathbf{B}}^2{\mathbf{A}}^3=o(\mathbf{4}),\hfill & \hfill & \mathbf{A}\mathbf{B}\mathbf{A}{\mathbf{B}}^2+\mathbf{A}{\mathbf{B}}^2\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & 2{\mathbf{A}}^3\mathbf{B}\mathbf{A}+{\mathbf{A}}^2\mathbf{B}{\mathbf{A}}^2+\mathbf{A}\mathbf{B}{\mathbf{A}}^3+2\mathbf{B}{\mathbf{A}}^4=o(\mathbf{4}),\hfill & \hfill & \mathbf{A}{\mathbf{B}}^4+{\mathbf{B}}^2\mathbf{A}{\mathbf{B}}^2+{\mathbf{B}}^4\mathbf{A}=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^2\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{A}\mathbf{B}{\mathbf{A}}^2\mathbf{B}+\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}=o(\mathbf{4}),\hfill & \hfill & \mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A}{\mathbf{B}}^2\mathbf{A}+{\mathbf{B}}^2\mathbf{A}\mathbf{B}\mathbf{A}=o(\mathbf{4})\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{B}{\mathbf{A}}^2\mathbf{B}\mathbf{A}+\mathbf{B}\mathbf{A}\mathbf{B}{\mathbf{A}}^2=o(\mathbf{4}),\hfill & \hfill & {\mathbf{A}}^4\mathbf{B}+{\mathbf{A}}^2\mathbf{B}{\mathbf{A}}^2+\mathbf{B}{\mathbf{A}}^4=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & 2{\mathbf{A}}^2{\mathbf{B}}^3+\mathbf{B}{\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2\mathbf{B}+2{\mathbf{B}}^3{\mathbf{A}}^2=o(\mathbf{4}),\hfill & \hfill & 2{\mathbf{A}}^4\mathbf{B}+{\mathbf{A}}^3\mathbf{B}\mathbf{A}+{\mathbf{A}}^2\mathbf{B}{\mathbf{A}}^2+2\mathbf{A}\mathbf{B}{\mathbf{A}}^3=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & 2\mathbf{A}{\mathbf{B}}^4+\mathbf{B}\mathbf{A}{\mathbf{B}}^3+{\mathbf{B}}^2\mathbf{A}{\mathbf{B}}^2+2{\mathbf{B}}^3\mathbf{A}\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & 2\mathbf{B}\mathbf{A}{\mathbf{B}}^3+{\mathbf{B}}^2\mathbf{A}{\mathbf{B}}^2+{\mathbf{B}}^3\mathbf{A}\mathbf{B}+2{\mathbf{B}}^4\mathbf{A}=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{cccc}\hfill & {\mathbf{A}}^2\mathbf{B}\mathbf{A}\mathbf{B}+{\mathbf{A}}^2{\mathbf{B}}^2\mathbf{A}+\mathbf{B}{\mathbf{A}}^3\mathbf{B}+\mathbf{B}{\mathbf{A}}^2\mathbf{B}\mathbf{A}+\mathbf{B}\mathbf{A}\mathbf{B}{\mathbf{A}}^2+{\mathbf{B}}^2{\mathbf{A}}^3=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & \mathbf{A}{\mathbf{B}}^2\mathbf{A}\mathbf{B}+\mathbf{A}{\mathbf{B}}^3\mathbf{A}+\mathbf{B}{\mathbf{A}}^2{\mathbf{B}}^2+\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A}{\mathbf{B}}^2\mathbf{A}+{\mathbf{B}}^2{\mathbf{A}}^2\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^2{\mathbf{B}}^3+\mathbf{A}\mathbf{B}\mathbf{A}{\mathbf{B}}^2+\mathbf{A}{\mathbf{B}}^2\mathbf{A}\mathbf{B}+\mathbf{A}{\mathbf{B}}^3\mathbf{A}+{\mathbf{B}}^2{\mathbf{A}}^2\mathbf{B}+{\mathbf{B}}^2\mathbf{A}\mathbf{B}\mathbf{A}=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & \mathbf{A}\mathbf{B}\mathbf{A}{\mathbf{B}}^2+\mathbf{A}{\mathbf{B}}^3\mathbf{A}+\mathbf{B}{\mathbf{A}}^2{\mathbf{B}}^2+\mathbf{B}\mathbf{A}{\mathbf{B}}^2\mathbf{A}+{\mathbf{B}}^2\mathbf{A}\mathbf{B}\mathbf{A}+{\mathbf{B}}^3{\mathbf{A}}^2=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^2{\mathbf{B}}^2\mathbf{A}+\mathbf{A}\mathbf{B}{\mathbf{A}}^2\mathbf{B}+\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{A}{\mathbf{B}}^2{\mathbf{A}}^2+\mathbf{B}{\mathbf{A}}^3\mathbf{B}+\mathbf{B}{\mathbf{A}}^2\mathbf{B}\mathbf{A}=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^3{\mathbf{B}}^2+{\mathbf{A}}^2\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{A}\mathbf{B}{\mathbf{A}}^2\mathbf{B}+\mathbf{A}{\mathbf{B}}^2{\mathbf{A}}^2+\mathbf{B}{\mathbf{A}}^3\mathbf{B}+\mathbf{B}\mathbf{A}\mathbf{B}{\mathbf{A}}^2=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & {\mathbf{B}}^5=o(\mathbf{4}),{\mathbf{A}}^5=o(\mathbf{4}),\hfill & \hfill & \hfill \end{array}$$

and a set of twenty 6-reducers also equal to $o(\mathbf{4})$ :

$$\begin{array}{cccccccc}\hfill & {\mathbf{A}}^4{\mathbf{B}}^2+{\mathbf{A}}^2{\mathbf{B}}^2{\mathbf{A}}^2+{\mathbf{B}}^2{\mathbf{A}}^4=o(\mathbf{4}),\hfill & \hfill & {\mathbf{A}}^5\mathbf{B}+{\mathbf{A}}^3\mathbf{B}{\mathbf{A}}^2+\mathbf{A}\mathbf{B}{\mathbf{A}}^4=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^4\mathbf{B}\mathbf{A}+{\mathbf{A}}^2\mathbf{B}{\mathbf{A}}^3+\mathbf{B}{\mathbf{A}}^5=o(\mathbf{4}),\hfill & \hfill & {\mathbf{A}}^2{\mathbf{B}}^4+{\mathbf{B}}^2{\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^4{\mathbf{A}}^2=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^3\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{A}\mathbf{B}{\mathbf{A}}^3\mathbf{B}+\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}{\mathbf{A}}^2=o(\mathbf{4}),\hfill & \hfill & {\mathbf{A}}^2\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{B}{\mathbf{A}}^3\mathbf{B}\mathbf{A}+\mathbf{B}\mathbf{A}\mathbf{B}{\mathbf{A}}^3=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}{\mathbf{B}}^2+\mathbf{B}\mathbf{A}{\mathbf{B}}^3\mathbf{A}+{\mathbf{B}}^3\mathbf{A}\mathbf{B}\mathbf{A}=o(\mathbf{4}),\hfill & \hfill & \mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \mathbf{A}\mathbf{B}\mathbf{A}{\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2{\mathbf{A}}^2\mathbf{B}+\mathbf{B}{\mathbf{A}}^2\mathbf{B}\mathbf{A}\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & \mathbf{A}{\mathbf{B}}^2\mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{B}{\mathbf{A}}^2{\mathbf{B}}^2\mathbf{A}+\mathbf{B}\mathbf{A}\mathbf{B}{\mathbf{A}}^2\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}=o(\mathbf{4})\hfill & \hfill & \mathbf{A}{\mathbf{B}}^5+{\mathbf{B}}^2\mathbf{A}{\mathbf{B}}^3+{\mathbf{B}}^4\mathbf{A}\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \mathbf{B}\mathbf{A}{\mathbf{B}}^4+{\mathbf{B}}^3\mathbf{A}{\mathbf{B}}^2+{\mathbf{B}}^5\mathbf{A}=o(\mathbf{4}),\hfill & \hfill & \mathbf{A}\mathbf{B}\mathbf{A}{\mathbf{B}}^3+\mathbf{A}{\mathbf{B}}^3\mathbf{A}\mathbf{B}+{\mathbf{B}}^2\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & \hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{cccc}\hfill & {\mathbf{A}}^3{\mathbf{B}}^2\mathbf{A}+{\mathbf{A}}^2\mathbf{B}{\mathbf{A}}^2\mathbf{B}+\mathbf{A}\mathbf{B}{\mathbf{A}}^2\mathbf{B}\mathbf{A}+\mathbf{A}{\mathbf{B}}^2{\mathbf{A}}^3+\mathbf{B}{\mathbf{A}}^4\mathbf{B}+\mathbf{B}{\mathbf{A}}^2\mathbf{B}{\mathbf{A}}^2=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & \mathbf{A}{\mathbf{B}}^2\mathbf{A}{\mathbf{B}}^2+\mathbf{A}{\mathbf{B}}^4\mathbf{A}+\mathbf{B}{\mathbf{A}}^2{\mathbf{B}}^3+\mathbf{B}\mathbf{A}{\mathbf{B}}^2\mathbf{A}\mathbf{B}+{\mathbf{B}}^2\mathbf{A}{\mathbf{B}}^2\mathbf{A}+{\mathbf{B}}^3{\mathbf{A}}^2\mathbf{B}=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^3{\mathbf{B}}^3+{\mathbf{A}}^2{\mathbf{B}}^2\mathbf{A}\mathbf{B}+\mathbf{A}\mathbf{B}{\mathbf{A}}^2{\mathbf{B}}^2+\mathbf{A}{\mathbf{B}}^3{\mathbf{A}}^2+{\mathbf{B}}^2{\mathbf{A}}^3\mathbf{B}+{\mathbf{B}}^2\mathbf{A}\mathbf{B}{\mathbf{A}}^2=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^2\mathbf{B}\mathbf{A}{\mathbf{B}}^2+{\mathbf{A}}^2{\mathbf{B}}^3\mathbf{A}+\mathbf{B}{\mathbf{A}}^3{\mathbf{B}}^2+\mathbf{B}\mathbf{A}{\mathbf{B}}^2{\mathbf{A}}^2+{\mathbf{B}}^2{\mathbf{A}}^2\mathbf{B}\mathbf{A}+{\mathbf{B}}^3{\mathbf{A}}^3=o(\mathbf{4}),\hfill & \hfill & \hfill \\ \hfill & {\mathbf{B}}^6=o(\mathbf{4}),{\mathbf{A}}^6=o(\mathbf{4}).\hfill & \hfill & \hfill \end{array}$$

It is clear that some of these reductions could have been inferred from the 3-reducers in equation (8). For example, ${\mathbf{A}}^6$ clearly reduces down since we known ${\mathbf{A}}^3$ does. However, many of the above reductions are independent of the ones derived previously at third order, such as the 6-reducer $\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}$ . Our goal at the moment is to be exhaustive rather than concise, to make sure we include every reducible grouping that follows from 3D Cayley–Hamilton, even if there is some duplication.

With a careful eye, it can be seen that equation (9) also discloses a set of 4-reducers. The left and right sides of equation (9) each produce tensorially fourth-order terms. While many of these cancel each other, some do not. For example, fourth-order terms multiplying $s_0^2{s}_5$ will exist on the left side but not the right. The surviving fourth-order terms generate a set of 4-reducers, listed below:

$$\begin{array}{cccccccc}\hfill & {\mathbf{A}}^3\mathbf{B}+{\mathbf{A}}^2\mathbf{B}\mathbf{A}+\mathbf{A}\mathbf{B}{\mathbf{A}}^2=o(\mathbf{3})\hfill & \hfill & {\mathbf{A}}^2\mathbf{B}\mathbf{A}+\mathbf{A}\mathbf{B}{\mathbf{A}}^2+\mathbf{B}{\mathbf{A}}^3=o(\mathbf{3})\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^2{\mathbf{B}}^2+\mathbf{A}{\mathbf{B}}^2\mathbf{A}+{\mathbf{B}}^2{\mathbf{A}}^2=o(\mathbf{3})\hfill & \hfill & 2{\mathbf{A}}^3\mathbf{B}+{\mathbf{A}}^2\mathbf{B}\mathbf{A}+\mathbf{A}\mathbf{B}{\mathbf{A}}^2+2\mathbf{B}{\mathbf{A}}^3=o(\mathbf{3})\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & {\mathbf{A}}^2{\mathbf{B}}^2+2\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{A}{\mathbf{B}}^2\mathbf{A}+\mathbf{B}{\mathbf{A}}^2\mathbf{B}+\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}=o(\mathbf{3})\hfill & \hfill & 2\mathbf{A}{\mathbf{B}}^3+\mathbf{B}\mathbf{A}{\mathbf{B}}^2+{\mathbf{B}}^2\mathbf{A}\mathbf{B}+2{\mathbf{B}}^3\mathbf{A}=o(\mathbf{3})\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{A}{\mathbf{B}}^2\mathbf{A}+\mathbf{B}{\mathbf{A}}^2\mathbf{B}+2\mathbf{B}\mathbf{A}\mathbf{B}\mathbf{A}+{\mathbf{B}}^2{\mathbf{A}}^2=o(\mathbf{3})\hfill & \hfill & {\mathbf{A}}^2{\mathbf{B}}^2+\mathbf{B}{\mathbf{A}}^2\mathbf{B}+{\mathbf{B}}^2{\mathbf{A}}^2=o(\mathbf{3})\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \mathbf{A}{\mathbf{B}}^3+\mathbf{B}\mathbf{A}{\mathbf{B}}^2+{\mathbf{B}}^2\mathbf{A}\mathbf{B}=o(\mathbf{3})\hfill & \hfill & \mathbf{B}\mathbf{A}{\mathbf{B}}^2+{\mathbf{B}}^2\mathbf{A}\mathbf{B}+{\mathbf{B}}^3\mathbf{A}=o(\mathbf{3})\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & 3{\mathbf{B}}^4=o(\mathbf{3})\hfill & \hfill & 3{\mathbf{A}}^4=o(\mathbf{3}).\hfill & \hfill & \hfill & \hfill & \hfill \end{array}$$

Aside from the primary reductions listed so far, reductions of a secondary nature must also be considered. That is, any reduction shown on one of the above lists can be multiplied on the left or right by A or B to produce a reduction at the next order higher. For example, the 3-reducer ${\mathbf{A}}^2\mathbf{B}+\mathbf{A}\mathbf{B}\mathbf{A}+\mathbf{B}{\mathbf{A}}^2$ implies that ${\mathbf{A}}^2{\mathbf{B}}^2+\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}+\mathbf{B}{\mathbf{A}}^2\mathbf{B}$ is a 4-reducer. By repeatedly multiplying all the Cayley–Hamilton reductions on the left/right by A or B, we can construct all secondary reductions up to some desired order—for reasons that will become apparent, one need only do this up to order six in order to obtain the final result. We neglect writing these all down for brevity. The accumulated set of all primary and secondary n-reducers is denoted by $R(n)$ .

With $R(n)$ in hand up to $n=6$ , we can begin to remove terms from equation (4). Since our ultimate goal is to represent symmetric-tensor-valued outputs, we first symmetrize by adding each element of $R(n)$ to its transpose to produce the set of symmetric-valued n-reducers, $S(n)$ . We can then compute the dimension, $D_S(n)$ , of the polynomial space spanned by $S(n)$ . For example, a brute-force computer calculation ⁴ of the dimension of the polynomial space spanned by $S(3)$ can be done by writing each element of $S(3)$ as a vector in terms of monomial basis elements, and computing the rank of the set, which gives $D_S(3)=4$ . We also define $D_{\mathrm{Tot}}(n)$ , which is the dimension of the space of all symmetric-valued combinations of nth-degree powers of A and B; this is also equal to the total number of nth degree terms showing up in equation (4). A simple counting problem reveals that $D_{\mathrm{Tot}}(n)=(2^n-2^{\lceil n∕2\rceil })∕2+2^{\lceil n∕2\rceil }$ , where $\lceil x\rceil$ is the ceiling function of x, the smallest integer greater than x. For example, this formula gives $D_{\mathrm{Tot}}(3)=6$ , and indeed, from equation (4), it can be seen that there are six independent third-order terms, i.e., ${\mathbf{A}}^3,{\mathbf{B}}^3,{\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2,{\mathbf{A}}^2\mathbf{B}+\mathbf{B}{\mathbf{A}}^2,\mathbf{A}\mathbf{B}\mathbf{A},\mathrm{and}\mathbf{B}\mathbf{A}\mathbf{B}$ . Altogether, we obtain $D_{\mathrm{Tot}}(3)-D_S(3)=2$ , so exactly two “irreducible” third-order terms from equation (4) will remain in our final representation.

The same method for determining the number of irreducible terms can be used at progressively higher orders without affecting results from previous orders. We find the number of irreducible terms at fourth-, fifth-, and sixth-order are, respectively, $D_{\mathrm{Tot}}(4)-D_S(4)=1$ , $D_{\mathrm{Tot}}(5)-D_S(5)=0$ , and $D_{\mathrm{Tot}}(6)-D_S(6)=0$ . Moreover, another calculation reveals that when n reaches 6, the space of $R(6)$ —the space generated by all not-necessarily-symmetric n-reducers—has dimension $D_R(6)=2^6$ . This is precisely the dimension of the entire space generated by sixth-order monomials. Likewise, all sixth-order polynomial terms are reducible. Since any seventh-order monomial can be constructed by multiplying a sixth-order monomial on the left or right by A or B, it follows that secondary reductions from $R(6)$ are sufficient to span the entire space of seventh-order polynomial terms. By continuing the same construction order-by-order, we conclude that all polynomial terms of order $n\ge 7$ must be reducible and can be eliminated from the representation. Altogether, we find any function of the form of equation (4) can be truncated after fourth-order. Note that generating higher-order Cayley–Hamilton reductions beyond those from equations (6) and (9) is unnecessary since these additional results would only affect orders $>6$ , which have already been shown to be fully reducible.

3.2. Greatest reduction via strictly 3D Cayley–Hamilton-based arguments

A particular representation can be found by identifying two third-order terms and one fourth-order term, which, when included within the sets $S(3)$ and $S(4)$ , respectively, increase the dimension of the spaces spanned by those sets to $D_{\mathrm{Tot}}(3)$ and $D_{\mathrm{Tot}}(4)$ . Multiple options exist. Below is one such solution:

$$\begin{array}{cccc}\hfill \mathbf{C}=\widehat{\mathbf{C}}(\mathbf{A},\mathbf{B})=c_{00}\mathbf{1}& +c_{10}\mathbf{A}+c_{11}\mathbf{B}+c_{20}{\mathbf{A}}^2+c_{21}{\mathbf{B}}^2+c_{22}(\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A})\hfill & \hfill & \hfill \\ \hfill & +c_{32}({\mathbf{A}}^2\mathbf{B}+\mathbf{B}{\mathbf{A}}^2)+c_{33}({\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2)+c_{42}({\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2).\hfill & \hfill \end{array}$$

(10)

This final result, which includes the fourth-order term, is identical to that of Wang [3], Rivlin [11], Spencer and Rivlin [12], Wang [13] and Smith [14] but differs from Rivlin and Ericksen [1], Smith [4], Zheng [5] and Pennisi and Trovato [9] whose representations are the same except without the fourth-order term. Any representation without the fourth-order term, as we have derived, would require reductive algebra beyond 3D Cayley–Hamilton considerations.

4. Showing the ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ term is unnecessary

The previous argument concluded that the representation for $\widehat{\mathbf{C}}$ needs no more than the nine terms in equation (10). We shall now show through a different kind of argument that the ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ term can always be expressed as a linear combination of the other eight terms regardless of the choice of A and B. Thus, it can always be removed from the representation.

First, Rivlin and Ericksen [1] (in §27) showed that the set of tensors:

$$K\equiv \{\mathbf{1},\mathbf{A},\mathbf{B},{\mathbf{A}}^2,{\mathbf{B}}^2,\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A},{\mathbf{A}}^2\mathbf{B}+\mathbf{B}{\mathbf{A}}^2,{\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2\},$$

is a basis for the space of all 3×3 symmetric tensors in the special case that (i) A and B do not both have repeated eigenvalues and (ii) A and B share no common eigenvectors. ⁵ These conditions exemplify the special cases one sees frequently in the existing literature on this topic. The result can be shown in a rather straightforward fashion by writing each element of K as a 6D vector using Voigt notation in the eigenbasis of A, and showing that the matrix of these vectors has non-zero determinant when the aforementioned conditions are met. Consequently, for any A and B fulfilling these conditions, ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ can be expressed as a linear combination of the members of elements of K, and thus, the function $\widehat{\mathbf{C}}$ can be written without the ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ term.

To show that the ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ term is in fact never needed, we must show that ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ is a linear combination of elements of K even when condition (i) or (ii) is not met.

Condition (i) not met—Repeated eigenvalues. We will prove the stronger result that ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ can be expressed as a linear combination of the members of K if A or B has a repeated eigenvalue. Suppose, without loss of generality, that A has a repeated eigenvalue. Let its (unordered) eigenvalues be $\{a_1,a_1,a_3\}$ . In this case, the following always holds:

$${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2=(a_1+a_3)(\mathbf{A}{\mathbf{B}}^2+{\mathbf{B}}^2\mathbf{A})-2a_1{a}_3{\mathbf{B}}^2,$$

(11)

and thus, ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ is a linear combination of elements of K. One way to obtain this result is to utilize the principal basis of A, in which we can write:

$$[\mathbf{A}]=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill a_1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill a_3\hfill \end{array}\right].$$

The repeated eigenvalue means that A satisfies the analogous 2D Cayley–Hamilton theorem, i.e.,

$${\mathbf{A}}^2=(a_1+a_3)\mathbf{A}-a_1{a}_3\mathbf{1},$$

where $a_1+a_3$ is the analogous trace and $a_1{a}_3$ is the analogous determinant. Equation (11) thus follows directly on multiplying this result on either side by ${\mathbf{B}}^2$ and adding the two results together.

Condition (ii) not met—common eigenvector. Suppose A and B share a common eigenvector. In the principal basis of A, the matrices of A and B can be expressed as:

$$[\mathbf{A}]=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill a_2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill a_3\hfill \end{array}\right],[\mathbf{B}]=\left[\begin{array}{ccc}\hfill b_1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill b_{22}\hfill & \hfill b_{23}\hfill \\ \hfill 0\hfill & \hfill b_{23}\hfill & \hfill b_{33}\hfill \end{array}\right].$$

Here, we obtain the result:

$${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2=(a_2+a_3)(\mathbf{A}{\mathbf{B}}^2+{\mathbf{B}}^2\mathbf{A})-2a_2{a}_3{\mathbf{B}}^2+2b_1^2\left({\mathbf{A}}^2-(a_2+a_3)\mathbf{A}+a_2{a}_3\mathbf{1}\right),$$

(12)

which shows, again, ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ is a linear combination of elements of K. This result can be verified through direct calculation and can be most easily deduced by again appealing to the 2D Cayley–Hamilton theorem.

Since we have now shown that ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ is always representable as a linear combination of elements of K for all choices of A and B, we have thus proven that:

$$\begin{array}{cccc}\hfill \mathbf{C}=\widehat{\mathbf{C}}(\mathbf{A},\mathbf{B})=c_{00}\mathbf{1}& +c_{10}\mathbf{A}+c_{11}\mathbf{B}+c_{20}{\mathbf{A}}^2+c_{21}{\mathbf{B}}^2+c_{22}(\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A})\hfill & \hfill & \hfill \\ \hfill & +c_{32}({\mathbf{A}}^2\mathbf{B}+\mathbf{B}{\mathbf{A}}^2)+c_{33}({\mathbf{B}}^2\mathbf{A}+\mathbf{A}{\mathbf{B}}^2).\hfill & \hfill \end{array}$$

(13)

Since, as shown by counterexample in Pennisi and Trovato [9], the representation for $\widehat{\mathbf{C}}$ cannot be reduced further than the terms included in equation (13), we conclude equation (13) is the minimal representation needed to represent any series-expandable (as in equation (4)) isotropic function $\widehat{\mathbf{C}}$ .

5. Summary

In this paper, we have revisited the topic of isotropic representations of tensor-valued functions of multiple tensors, a topic with a long and somewhat daunting literature of development over the last 70 years or so. Our emphasis here was twofold. Our first goal was to help resolve the question of how many terms the representation needs in the case of two symmetric inputs and one symmetric output. This case carries prime relevance in mechanics where one might want, for example, a function for stress in terms of a symmetric deformation variable (e.g., strain or strain-rate) and a symmetric tensorial variable representing the structural state of the material. The second goal of this paper was to carry out the derivation in a clear fashion relying primarily on consequences of the Cayley–Hamilton theorem in three dimensions and then in two dimensions.

The derivation described herein takes the following route to obtain the final result. First, we consider the broad class of functions in equation (4). We then apply the 3D Cayley–Hamilton theorem to first- and second-order combinations of monomials of A and B to obtain a set of Cayley–Hamilton-based reductions—i.e., polynomial expressions in A and B that are identical to lower-order polynomials. Second, we use computer-assisted linear algebra ⁶ to determine the dimensionality of the order-by-order space of tensorial monomials that can be built from the Cayley–Hamilton reducers (including secondary reductions) and hence removed from inclusion in the representation. This wipes away all terms in the function expansion above fourth order, and many lower-order terms as well, leaving a nine-term expression that includes the disputed fourth-order term ${\mathbf{A}}^2{\mathbf{B}}^2+{\mathbf{B}}^2{\mathbf{A}}^2$ . We then appeal to a theorem in Rivlin and Ericksen [1] whose proof is straightforward from algebra, which we can assert to see that there are only two conditions in which the fourth-order term might not be expressible in terms of the other eight. In each of those two cases, using the 2D Cayley–Hamilton theorem, we find explicit formulas for the fourth-order term in terms of the other eight, thereby finalizing the proof. The only explicit limitation of the proof offered here is the initial assumption that the tensor-valued function can be expanded into a non-constant coefficient tensorial power series, which is an assumption that need not be made in the classical and far-simpler one-input case shown in equation (1) (see Appendix 1 for such a proof).

Supplemental Material

sj-m-2-mms-10.1177_10812865251340424 – Supplemental material for Clarifying the representation of isotropic symmetric tensor-valued functions of two symmetric tensors

Supplemental material, sj-m-2-mms-10.1177_10812865251340424 for Clarifying the representation of isotropic symmetric tensor-valued functions of two symmetric tensors by Ken Kamrin and Sanjay Govindjee in Mathematics and Mechanics of Solids

Supplemental Material

sj-py-1-mms-10.1177_10812865251340424 – Supplemental material for Clarifying the representation of isotropic symmetric tensor-valued functions of two symmetric tensors

Supplemental material, sj-py-1-mms-10.1177_10812865251340424 for Clarifying the representation of isotropic symmetric tensor-valued functions of two symmetric tensors by Ken Kamrin and Sanjay Govindjee in Mathematics and Mechanics of Solids