1 Preliminary

Definition: A functional basis of a vector and tensor system is a set of their isotropic invariants such that every invariant can uniquely be expressed in terms the basis. Since Itskov’s [1] paper only deals with two symmetric tensors, \(\varvec{A}_1\) and \(\varvec{A}_2\), any scalar-valued isotropic function (invariant) of these tensors should satisfy the condition

$$\begin{aligned} f(\varvec{A}_1,\varvec{A}_2) = f(\varvec{Q}\varvec{A}_1\varvec{Q}^T,\varvec{Q}\varvec{A}_2\varvec{Q}^T) \,, { }\forall \varvec{Q}\in Orth^3 \,, \end{aligned}$$
(1)

where \(Orth^3\) denotes a group of all orthogonal transformations within the three-dimensional Euclidean space and \(\varvec{Q}\varvec{Q}^T = \varvec{Q}^T\varvec{Q}= \varvec{I}\) (the identity tensor). See also the Appendix.

The “classical" functional basis for tensors \(\varvec{A}_1\) and \(\varvec{A}_2\) contains, for example, the isotropic invariants [2]

$$\begin{aligned} \text {tr}\varvec{A}_1\,, \text {tr}\varvec{A}_1^2 \,, \text {tr}\varvec{A}_1^3 \,, \text {tr}\varvec{A}_2\,, \text {tr}\varvec{A}_2^2 \,, \text {tr}\varvec{A}_2^3 \,, \text {tr}(\varvec{A}_1\varvec{A}_2) \,, \text {tr}(\varvec{A}_1^2\varvec{A}_2) \,, \text {tr}(\varvec{A}_1\varvec{A}_2^2) \,, \text {tr}(\varvec{A}_1^2,\varvec{A}_2^2) \,, \end{aligned}$$
(2)

where \(\text {tr}\) denotes the trace of a tensor.

The spectral functional basis of these two tensors [7] contains the isotropic invariants

$$\begin{aligned} \lambda _i \,, { } {A}^{(2)}_{ij} = \text {tr}[\varvec{A}_2(\varvec{v}_i\otimes \varvec{v}_j ] = \varvec{v}_i\cdot \varvec{A}_2\varvec{v}_j \,, \end{aligned}$$
(3)

where \(\lambda _i\) and \(\varvec{v}_i\) are, respectively, the eigenvalues and orthonormal eigenvectors of \(\varvec{A}_1\), i.e.,

$$\begin{aligned} \varvec{A}_1 = \sum _{i=1} \lambda _i\varvec{v}_i\otimes \varvec{v}_i \, \end{aligned}$$
(4)

and we can express

$$\begin{aligned} \varvec{A}_2 = \sum _{i,j=1}^3 {A}^{(2)}_{ij} \varvec{v}_i\otimes \varvec{v}_j \,. \end{aligned}$$
(5)

It is shown in Shariff [7] all "classical" invariants such as those given in Eq. 2 can be explicitly expressed in terms of Shariff [7] spectral invariants. For example, the classical invariant

$$\begin{aligned} \text {tr}\varvec{A}_2^2 = \sum _{i,j=1}^3 {A}^{(2)}_{ij} {A}^{(2)}_{ji} \,, { }\text {tr}\varvec{A}_1\varvec{A}_2 = \sum _{i=1}^3 \lambda _i {A}^{(2)}_{ii} \,, \end{aligned}$$
(6)

Hence, \(\lambda _i, {A}^{(2)}_{ij}\) are spectral elements of a functional basis for the tensor set \(\{\varvec{A}_1,\varvec{A}_2\}\).

In Shariff [6], he showed that spectral invariants (such as \(\lambda _i, {A}^{(2)}_{ij}\)) can be expressed in terms of "classical invariants" (such as those given in Eq. 2).

Itskov [1] claimed, via a counterexample, that the scalars \( {A}^{(2)}_{ij}\) are not isotropic invariants. In this paper, we debunk Itskov’s claim, but before we invalidate Itskov’s [1] counterexample, a few simple basic examples are given below to facilitate our discussion and for clarity.

2 Examples

2.1 Example 1

Let \(\varvec{B}\) and \(\varvec{C}\) be tensors. The scalar function

$$\begin{aligned} f(\varvec{B},\varvec{C}) = \text {tr}(\varvec{B}\varvec{C}) \end{aligned}$$
(7)

is a scalar-valued isotropic function since

$$f(\varvec{Q}\varvec{B}\varvec{Q}^T,\varvec{Q}\varvec{C}\varvec{Q}^T) = \text {tr}\varvec{Q}\varvec{B}\varvec{Q}^T\varvec{Q}\varvec{C}\varvec{Q}^T = \text {tr}\varvec{Q}\varvec{B}\varvec{C}\varvec{Q}^T = \text {tr}\varvec{B}\varvec{C}\varvec{Q}^T\varvec{Q}= \text {tr}\varvec{B}\varvec{C}= f(\varvec{B},\varvec{C}) \,, { }\forall \varvec{Q}\in Orth^3 \,. $$

2.2 Example 2

Let \(\varvec{B}\) be a symmetric tensor and \(\varvec{C}=\varvec{a}\otimes \varvec{b}\), where \(\otimes \) is the dyadic product, and \(\varvec{a}\) and \(\varvec{b}\) are arbitrary vectors. Consider the function

$$\begin{aligned} f(\varvec{B},\varvec{a},\varvec{b}) = \text {tr}[\varvec{B}(\varvec{a}\otimes \varvec{b})] = \varvec{a}\cdot \varvec{B}\varvec{b}= \varvec{b}\cdot \varvec{B}\varvec{a}\,. \end{aligned}$$
(8)

It is clear \(f(\varvec{B},\varvec{a},\varvec{b})\) is a scalar-valued isotropic function since

$$\begin{aligned} f(\varvec{Q}\varvec{B}\varvec{Q}^T,\varvec{Q}\varvec{a},\varvec{Q}\varvec{b}) = \varvec{Q}\varvec{a}\cdot \varvec{Q}\varvec{B}\varvec{Q}^T\varvec{Q}\varvec{b}= \varvec{a}\cdot \varvec{B}\varvec{b}= f(\varvec{B},\varvec{a},\varvec{b}) \,, { }\forall \varvec{Q}\in Orth^3\,. \end{aligned}$$
(9)

2.3 Example 3

In view of Eq. 4, the eigenvalues

$$\begin{aligned} \lambda _i(\varvec{A}_1) = f(\varvec{A}_1,\varvec{v}_i) = \text {tr}[\varvec{A}_1(\varvec{v}_i\otimes \varvec{v}_i)] = \varvec{v}_i\cdot \varvec{A}_1\varvec{v}_i \,, { }i \, \, \text {not summed} \,. \end{aligned}$$
(10)

The eigenvalues are isotropic function since

$$\begin{aligned} f(\varvec{Q}\varvec{A}_1\varvec{Q}^T,\varvec{Q}\varvec{v}_i) = \varvec{Q}\varvec{v}_i\cdot \varvec{Q}\varvec{A}_1\varvec{Q}^T\varvec{Q}\varvec{v}_i = \varvec{v}_i\cdot \varvec{A}_1\varvec{v}_i = f(\varvec{A}_1,\varvec{v}_i) \,, { }\forall \varvec{Q}\in Orth^3 \,. \end{aligned}$$
(11)

Equation 11 is true even if the eigenvectors of \(\varvec{A}_1\) are not unique.

2.4 Example 4

In view of Eq. 3\(_2\), we have

$$\begin{aligned} {A}^{(2)}_{ij}= f(\varvec{A}_2,\varvec{v}_i) = \text {tr}[\varvec{A}_2(\varvec{v}_i\otimes \varvec{v}_j)]= \varvec{v}_i\cdot \varvec{A}_2\varvec{v}_j \,. \end{aligned}$$
(12)

\( {A}^{(2)}_{ij}\) are isotropic invariants, since

$$\begin{aligned} f(\varvec{Q}\varvec{A}_2\varvec{Q}^T,\varvec{Q}\varvec{v}_i) = \varvec{Q}\varvec{v}_i\cdot \varvec{Q}\varvec{A}_2\varvec{Q}^T\varvec{Q}\varvec{v}_j = \varvec{v}_i\cdot \varvec{A}_2\varvec{v}_j = f(\varvec{A}_2,\varvec{v}_i) \,, { }\forall \varvec{Q}\in Orth^3 \,. \end{aligned}$$
(13)

Equation 13 is true even if the eigenvectors of \(\varvec{A}_1\) are not unique. This example clearly proves that \( {A}^{(2)}_{ij}\) are isotropic invariants; hence, no counterexample, including Itskov’s counterexample, can prove that they are not isotropic invariants. For example, Pythagoras theorem has been proven to be correct; no counterexample can prove that the theorem is incorrect. This clearly contradicts Itskov’s [1] claim that \( {A}^{(2)}_{ij}\) are not isotropic invariants.

2.4.1 Remark

We emphasize that in Examples 3 and 4, \(\lambda _i\) and the scalar \( {A}^{(2)}_{ij}\) are traces of two tensors, i.e.

$$\begin{aligned} \lambda _i = \text {tr}[\varvec{A}_1(\varvec{v}_i\otimes \varvec{v}_i)] \,, { } {A}^{(2)}_{ij} = \text {tr}[\varvec{A}_2(\varvec{v}_i\otimes \varvec{v}_j)] \,.\end{aligned}$$
(14)

The fact that traces of tensors are isotropic invariants simply proves that \(\lambda _i\) and the scalar \( {A}^{(2)}_{ij}\) are isotropic invariants according to the definition (1).

2.5 Example 5

Consider the orthogonally transformed tensors

$$\begin{aligned} \varvec{A}_1'= & {} \varvec{Q}\varvec{A}_1 \varvec{Q}^T = \varvec{Q}(\sum _{i=1}^3 \lambda _i \varvec{v}_i\otimes \varvec{v}_i ) \varvec{Q}^T = \sum _{i=1}^3 \lambda _i \varvec{Q}\varvec{v}_i\otimes \varvec{Q}\varvec{v}_i \,, \nonumber \\ \varvec{A}_2'= & {} \varvec{Q}\varvec{A}_2 \varvec{Q}^T = \varvec{Q}(\sum _{i,j=1}^3 {A}^{(2)}_{ij} \varvec{v}_i\otimes \varvec{v}_j ) \varvec{Q}^T = \sum _{i,j=1}^3 {A}^{(2)}_{ij} \varvec{Q}\varvec{v}_i\otimes \varvec{Q}\varvec{v}_j \,, { }\forall \varvec{Q}\in Orth^3 \,. \end{aligned}$$
(15)

Note that \(\varvec{A}_1\) in Eq. 4 and \(\varvec{A}_2\) in Eq. 5 are both expressed using the basis \(S = \{\varvec{v}_1,\varvec{v}_2,\varvec{v}_3\}\) (reference basis), and \(\varvec{A}_1'\) and \(\varvec{A}_2'\) in Eq. 15 are expressed using the basis \(S' = \{\varvec{Q}\varvec{v}_1,\varvec{Q}\varvec{v}_2,\varvec{Q}\varvec{v}_3\}\) (rotated basis). It is clear from above (and we strongly emphasize) that the S-invariant components \(\lambda _i\) and \( {A}^{(2)}_{ij}\) of \(\varvec{A}_1\) and \(\varvec{A}_2\), respectively, are the same as the \(S'\)-invariant components of \(\varvec{A}_1'\) and \(\varvec{A}_2'\), respectively. Furthermore, we can express

$$\begin{aligned} \varvec{A}_2' = \sum _{i,j=1}^3 {A'}^{(2)}_{ij} \varvec{Q}\varvec{v}_i\otimes \varvec{Q}\varvec{v}_j = \sum _{i,j=1}^3 {\bar{A'}}^{(2)}_{ij} \varvec{v}_i\otimes \varvec{v}_j \,, \end{aligned}$$
(16)

where

$$\begin{aligned} {A'}^{(2)}_{ij} = \varvec{Q}\varvec{v}_i \cdot \varvec{A}_2'\varvec{Q}\varvec{v}_j = \varvec{Q}\varvec{v}_i \cdot (\varvec{Q}\varvec{A}_2 \varvec{Q}^T) \varvec{Q}\varvec{v}_j = \varvec{v}_i \cdot \varvec{A}_2\varvec{v}_j = {A}^{(2)}_{ij} \, \end{aligned}$$
(17)

and

$$\begin{aligned} {\bar{A'}}^{(2)}_{ij} = \varvec{v}_i \cdot \varvec{A}_2'\varvec{v}_j = \varvec{v}_i \cdot \varvec{Q}\varvec{A}_2\varvec{Q}^T\varvec{v}_j \ne {A}^{(2)}_{ij} \,. \end{aligned}$$
(18)

It is obvious in Eq. 18 that the S components \( {\bar{A'}}^{(2)}_{ij}\) of \(\varvec{A}_2'\) are not isotropic functions and their values change with respect to different values of \(\varvec{Q}\); this fact is important because Itskov’s counterexample uses the reference basis S (instead of the rotated basis \(S'\)) to describe the transformed tensor \(\varvec{A}_2'\) and hence obtained the tensor components \( {\bar{A'}}^{(2)}_{ij}\) that are not isotropic functions.

2.6 Example 6

Consider the special case

$$\begin{aligned} \varvec{A}_2 = \sum _{i=1}^3 \bar{\lambda }_i \, \varvec{v}_i\otimes \varvec{v}_i = \bar{\lambda }_1 \, \varvec{v}_1\otimes \varvec{v}_1 + \bar{\lambda }_2 \, \varvec{v}_2\otimes \varvec{v}_2 + \bar{\lambda }_3 \, \varvec{v}_3\otimes \varvec{v}_3 \,. \end{aligned}$$
(19)

We then have

$$\begin{aligned} \varvec{A}_2' = \varvec{Q}\varvec{A}_2 \varvec{Q}^T =\sum _{k=1}^3 \bar{\lambda }_k \, \varvec{Q}\varvec{v}_k\otimes \varvec{Q}\varvec{v}_k \,. \end{aligned}$$
(20)

It is clear from Eqs. 19 and 20 that the S-invariant components \(\bar{\lambda }_i\) of \(\varvec{A}_2\) and the \(S'\)-invariant components of \(\varvec{A}'_2\) are the same. This is as expected, since \(\bar{\lambda }_i\) are eigenvalues of \(\varvec{A}_2\), they are isotropic functions and their values should not change. However if we express \(\varvec{A}_2'\) in using the reference basis S, we obtain

$$\begin{aligned} \varvec{A}_2' = \sum _{i,j=1}^3 (\varvec{v}_i\cdot \varvec{A}_2'\varvec{v}_j) \, \varvec{v}_i\otimes \varvec{v}_j \,, \end{aligned}$$
(21)

where on using Eq. 20

$$\begin{aligned} \varvec{v}_i\cdot \varvec{A}_2'\varvec{v}_j = \sum _{k=1}^3 \bar{\lambda }_k (\varvec{v}_i\cdot \varvec{Q}\varvec{v}_k)(\varvec{v}_j\cdot \varvec{Q}\varvec{v}_k) \,. \end{aligned}$$
(22)

It is clear from Eq. 22 that the tensor components \(\varvec{v}_i\cdot \varvec{A}_2'\varvec{v}_j\) are not isotropic functions. For example, consider the special case for \(\varvec{Q}\) given in Itskov’s [1] Eq. 7, we have

$$\begin{aligned} \varvec{Q}\varvec{v}_1 = -\varvec{v}_2 \,, { }\varvec{Q}\varvec{v}_2 = \varvec{v}_1 \,, { }\varvec{Q}\varvec{v}_3 = \varvec{v}_3 \,, \end{aligned}$$
(23)

we have from Eqs. 22 and 23

$$\begin{aligned} \varvec{A}_2' = \bar{\lambda }_2 \, \varvec{v}_1\otimes \varvec{v}_1 + \bar{\lambda }_1 \, \varvec{v}_2\otimes \varvec{v}_2 + \bar{\lambda }_3 \, \varvec{v}_3\otimes \varvec{v}_3 \,, \end{aligned}$$
(24)

where the tensor components of \(\varvec{A}_2\) in Eq. 19 are not the same as those of \(\varvec{A}_2'\) in Eq. 24.

2.7 Important remark

To avoid confusion, especially when two or more eigenvalues have the same value, we write

$$\begin{aligned} \varvec{A}_1(\lambda _1,\lambda _2,\lambda _3, \varvec{v}_1,\varvec{v}_2,\varvec{v}_3) = \lambda _1 \varvec{v}_1\otimes \varvec{v}_1 + \lambda _2 \varvec{v}_1\otimes \varvec{v}_2 +\lambda _3 \varvec{v}_3\otimes \varvec{v}_3 \,, \end{aligned}$$
(25)

with the following symmetrical properties,

$$\begin{aligned} \varvec{A}_1(\lambda _1,\lambda _2,\lambda _3, \varvec{v}_1,\varvec{v}_2,\varvec{v}_3) =\varvec{A}_1(\lambda _2,\lambda _1,\lambda _3, \varvec{v}_2,\varvec{v}_1,\varvec{v}_3) =\varvec{A}_1(\lambda _3,\lambda _2,\lambda _1, \varvec{v}_3,\varvec{v}_2,\varvec{v}_1) = \text {etc} \,. \end{aligned}$$
(26)

and

$$\begin{aligned} \varvec{Q}\varvec{A}_1\varvec{Q}^T= \lambda _1 \varvec{Q}\varvec{v}_1\otimes \varvec{Q}\varvec{v}_1 + \lambda _2 \varvec{Q}\varvec{v}_1\otimes \varvec{Q}\varvec{v}_2 +\lambda _3 \varvec{Q}\varvec{v}_3\otimes \varvec{Q}\varvec{v}_3 = \varvec{A}_1(\lambda _1,\lambda _2,\lambda _3, \varvec{Q}\varvec{v}_1,\varvec{Q}\varvec{v}_2,\varvec{Q}\varvec{v}_3)\,. \end{aligned}$$
(27)

For example, the identity tensor, where the eigenvalues have the same value,

$$\begin{aligned} \varvec{I}= \varvec{A}_1(\lambda _1=1,\lambda _2=1,\lambda _3=1, \varvec{v}_1,\varvec{v}_2,\varvec{v}_3) \end{aligned}$$
(28)

and

$$\begin{aligned} \varvec{Q}\varvec{I}\varvec{Q}^T = \varvec{A}_1(\lambda _1=1,\lambda _2=1,\lambda _3=1, \varvec{Q}\varvec{v}_1,\varvec{Q}\varvec{v}_2,\varvec{Q}\varvec{v}_3)\,. \end{aligned}$$
(29)

Consider the case when \(\varvec{Q}\varvec{v}_i\) take the values given in Eq. 23. We then have

$$\begin{aligned} \varvec{Q}\varvec{I}\varvec{Q}^T = \varvec{A}_1(\lambda _1=1,\lambda _2=1,\lambda _3=1, -\varvec{v}_2,\varvec{v}_1,\varvec{v}_3) = \varvec{A}_1(\lambda _1=1,\lambda _2=1,\lambda _3=1, \varvec{v}_1,\varvec{v}_2,\varvec{v}_3) \,.\end{aligned}$$
(30)

In Itskov’s counterexample, discussed below, he used the incorrect relation \(\varvec{A}_1(\lambda _1=1,\lambda _2=1,\lambda _3=1, \varvec{v}_1,\varvec{v}_2,\varvec{v}_3)\) for \(\varvec{Q}\varvec{I}\varvec{Q}^T\) and hence, used the incorrect eigenvectors \(\varvec{v}_i\) to evaluate the components of \(\varvec{Q}\varvec{A}_2 \varvec{Q}^T\) instead of the correct eigenvectors \(\varvec{Q}\varvec{v}_i\).

We are now in position to address Itskov’s [1] counterexample.

3 Itskov’s counterexample

In Itskov’s paper, [1] he uses

$$\begin{aligned} \varvec{A}_1 = \varvec{I}\,, \end{aligned}$$
(31)

In view of the repeated eigenvalues \(\lambda _1=\lambda _2 =\lambda _3=1\), the eigenvectors of \(\varvec{A}_1\) are not unique. \(\varvec{A}_1\) can be represented by infinitely many different bases associated with different orthonormal eigenvectors, i.e.,

$$\begin{aligned} \varvec{A}_1 = \varvec{I}= \sum _{i=1}^3 \varvec{v}_i\otimes \varvec{v}_i = \sum _{i=1}^3 \bar{\varvec{v}}_i\otimes \bar{\varvec{v}}_i = \sum _{i=1}^3 \hat{\varvec{v}}_i\otimes \hat{\varvec{v}}_i = \text {etc.} \,, \end{aligned}$$
(32)

where the bases \(\{\varvec{v}_1,\varvec{v}_2,\varvec{v}_3 \}\), \(\{\bar{\varvec{v}}_1,\bar{\varvec{v}}_2,\bar{\varvec{v}}_3 \}\), \(\{\hat{\varvec{v}}_1,\hat{\varvec{v}}_2,\hat{\varvec{v}}_3 \}\) etc. are different.

Consider, for example, the case \(\bar{\varvec{v}}_i = \varvec{Q}\varvec{v}_i\). For \(\varvec{Q}\ne \varvec{I}\), some of the \(\bar{\varvec{v}}_i \ne \varvec{v}_i\) and the bases \(S = \{\varvec{v}_1,\varvec{v}_2,\varvec{v}_3\}\) and \(S' = \{\bar{\varvec{v}}_1,\bar{\varvec{v}}_2,\bar{\varvec{v}}_3\}\) are different. In Itskov’s paper, he lets \(\varvec{v}_i=\varvec{e}_i\) and \(\varvec{Q}\) is given in his Eq. 7. We then obtain

$$\begin{aligned} \bar{\varvec{v}}_1= -\varvec{e}_2 \,, { }\bar{\varvec{v}}_2 = \varvec{e}_1 \,, { }\bar{\varvec{v}}_3 = \varvec{e}_3 \,, \end{aligned}$$
(33)

which gives the basis \(S' = \{-\varvec{e}_2,\varvec{e}_1,\varvec{e}_3\}\) and it is not the same as the basis \( S=\{\varvec{e}_1,\varvec{e}_2,\varvec{e}_3\} \). Itskov expresses

$$\begin{aligned} \varvec{A}_1 = \varvec{I}= \sum _{i=1}^3 \varvec{e}_i\otimes \varvec{e}_i \,. \end{aligned}$$
(34)

We then have

$$ \varvec{A}_1' = \varvec{Q}\varvec{A}_1\varvec{Q}^T = \sum _{i=1}^3 \varvec{Q}\varvec{e}_i\otimes \varvec{Q}\varvec{e}_i = \sum _{i=1}^3 \bar{\varvec{v}}_i\otimes \bar{\varvec{v}}_i = \varvec{A}_1(\lambda _1=1,\lambda _2=1,\lambda _3=1, -\varvec{e}_2,\varvec{e}_1,\varvec{e}_3) $$
$$\begin{aligned} = \varvec{A}_1(\lambda _1=1,\lambda _2=1,\lambda _3=1, \varvec{e}_1,\varvec{e}_2,\varvec{e}_3) \, \end{aligned}$$
(35)

(see Section 2.7).

In view of Eq. 32, we have \(\varvec{A}_1'=\varvec{A}_1\). However, it is important to note that although \(\varvec{A}_1'=\varvec{A}_1\), their bases are not unique. This contradicts Itskov’s statement ST: “Due to the fact that \(\varvec{A}_1'=\varvec{A}_1\) the eigenvectors (5) remain unchanged."

Hence, Itskov’s statement ST is false and he contradicts his own statement in the last paragraph of his paper stating that

“Alternatively, the students can use arbitrary orthonormal bases since every of them represents a set of eigenvectors of the identity tensor \(\varvec{A}_1\)."

Alternatively, without using basis, we can show that \(\varvec{A}_1'=\varvec{A}_1\), via the relations \(\varvec{A}'_1 = \varvec{Q}\varvec{A}_1\varvec{Q}^T=\varvec{Q}\varvec{I}\varvec{Q}^T =\varvec{I}=\varvec{A}_1\). In Itskov’s paper,

$$\begin{aligned} \varvec{A}_2= \sum _{i=1}^3 {A}^{(2)}_{ij} \varvec{e}_i\otimes \varvec{e}_j \,, { } {A}^{(2)}_{ij} = \varvec{e}_i\cdot \varvec{A}_2\varvec{e}_j \,. \end{aligned}$$
(36)

For \(\varvec{A}_2'\), we have

$$\begin{aligned} \varvec{A}_2' =\varvec{Q}\varvec{A}_2\varvec{Q}^T= \sum _{i=1}^3 {A}^{(2)}_{ij} \bar{\varvec{v}}_i\otimes \bar{\varvec{v}}_j \,. \end{aligned}$$
(37)

It is clear from Eqs. 36\(_2\) and 37 that \( {A}^{(2)}_{ij}\) remain unchanged under the orthogonal transformation given in Eq. 7 of Itskov’s paper. In fact Section 2.4 proves that the values of \( {A}^{(2)}_{ij}\) remain unchanged under any orthogonal transformation, provided that \(\varvec{A}_2'\) is described by the basis \(S' = \{\varvec{Q}\varvec{v}_1,\varvec{Q}\varvec{v}_2,\varvec{Q}\varvec{v}_3\}\). However, we strongly make a point that Itskov incorrectly uses the basis S (instead of correctly using the basis \(S'\) for \(\varvec{A}_1'\)—see Section 2.7) to represent \(\varvec{A}_2'\), i.e.,

$$\begin{aligned} \varvec{A}_2' = \varvec{Q}\varvec{A}_2\varvec{Q}^T = \sum _{i=1}^3 B_{ij} \varvec{e}_i\otimes \varvec{e}_j \,, { }B_{ij} = \varvec{e}_i\cdot (\varvec{Q}\varvec{A}_2\varvec{Q}^T)\varvec{e}_j \,. \end{aligned}$$
(38)

The matrix given in the author’s Eq. 8 is \(B_{ij}\) (there is a typo in Itskov’s Eq. 8, \(\varvec{A}_2' = \varvec{Q}\varvec{A}_1\varvec{Q}^T\) should be \(\varvec{A}_2' = \varvec{Q}\varvec{A}_2\varvec{Q}^T\)). Note that the “tensor" components \( {A}^{(2)}_{ij}\) are isotropic invariants, and it is clear from Eq. 38\(_2\) that \(B_{ij}\) are not isotropic invariants, \(B_{ij} \ne {A}^{(2)}_{ij}\) and, clearly, the \(B_{ij}\) values depend on the values of \(\varvec{Q}\), as exemplified in Itskov’s Eq. 8.

4 Proof by contradiction

4.1 Proof 1

In this section, we use

$$\begin{aligned} \varvec{A}_1 = \sum _{i=1}^3 \varvec{e}_i\otimes \varvec{e}_i \,, { }\varvec{A}_2 = \sum _{i=1}^3 \bar{\lambda }_i \varvec{e}_i\otimes \varvec{e}_i \,, \end{aligned}$$
(39)

instead of \(\varvec{A}_2\) given in Itskov’s Eq. 4 to invalidate his counterexample by proving (via contradiction) that if Itskov’s counterexample is correct then the eigenvalues \(\bar{\lambda }_i\) are not isotropic functions. For simplicity, we let the eigenvalues \(\bar{\lambda }_i\) to take the values

$$\begin{aligned} \bar{\lambda }_1=1\,, { }\bar{\lambda }_2=2 \,, { }\bar{\lambda }_3 = 3 \, \end{aligned}$$
(40)

We note that

$$\begin{aligned} \bar{\lambda }_i = {A}^{(2)}_{ii} \,, { } {A}^{(2)}_{ij} = 0 \,, { }i \ne j. \end{aligned}$$
(41)

Since \(\bar{\lambda }_i\) are isotropic invariants, \( {A}^{(2)}_{ii}\) are also isotropic invariants (another example showing that \( {A}^{(2)}_{ij}\) are isotropic scalar functions). With respect to the basis \(S' = \{\varvec{Q}\varvec{e}_1,\varvec{Q}\varvec{e}_2,\varvec{Q}\varvec{e}_3\}\), we have, for the transformed tensor,

$$\begin{aligned} \varvec{A}_2' = \varvec{Q}\varvec{A}_2\varvec{Q}^T = \sum _{i=1}^3 \bar{\lambda }_i \varvec{Q}\varvec{e}_i\otimes \varvec{Q}\varvec{e}_i = 1 \varvec{Q}\varvec{e}_1\otimes \varvec{Q}\varvec{e}_1 + 2 \varvec{Q}\varvec{e}_2\otimes \varvec{Q}\varvec{e}_2 + 3 \varvec{Q}\varvec{e}_3\otimes \varvec{Q}\varvec{e}_3 \,, \end{aligned}$$
(42)

for \(\varvec{Q}\) given in Itskov’s (7). Since \(\bar{\lambda }_i\) are isotropic invariants, their values are not expected to change in the transformed tensor, as shown in Eq. 42. However, Itskov uses the basis \(S = \{\varvec{e}_1,\varvec{e}_2,\varvec{e}_3\}\) for \(\varvec{A}_2'\). Following the work given in Section 2.6, we obtain

$$\begin{aligned} \varvec{A}_2' = \varvec{Q}\varvec{A}_2 \varvec{Q}^T = 2 \varvec{e}_1\otimes \varvec{e}_1 + 1 \varvec{e}_2\otimes \varvec{e}_2 + 3 \varvec{e}_3\otimes \varvec{e}_3 \,, \end{aligned}$$
(43)

which implies

$$\begin{aligned} \bar{\lambda }_1 =2 \,, { }\bar{\lambda }_2=1 \,, { }\bar{\lambda }_3 = 3 \,. \end{aligned}$$
(44)

Hence, if the basis S is assumed to be the corret basis for the transformed tensor, then the different values of the eigenvalues given in Eqs. 40 and 44 show that the eigenvalues \(\bar{\lambda }_i\) are not isotropic invariants: This false conclusion further invalidate Itskov’s counterexample.

4.2 Proof 2

In this section, we prove that an existence of a non-isotropic function \( {A}^{(2)}_{ij}\) leads to a self-refuting result.

Proof. Assuming, as claimed by Itskov, there exist a set of eigenvectors \(\{\varvec{g}_1,\varvec{g}_2,\varvec{g}_3\}\) that indicates that \( {A}^{(2)}_{ij}(\varvec{g}_i\otimes \varvec{g}_j,\varvec{A}_2)\) are not isotropic functions. Since

$$ {A}^{(2)}_{ij}(\varvec{g}_i\otimes \varvec{g}_j,\varvec{A}_2) = \varvec{g}_i\cdot \varvec{A}_2\varvec{g}_j = \varvec{Q}\varvec{g}_i\cdot \varvec{Q}\varvec{A}_2\varvec{Q}^T\varvec{Q}\varvec{g}_j = {A}^{(2)}_{ij}(\varvec{Q}\varvec{g}_i\otimes \varvec{g}_j\varvec{Q}^T,\varvec{Q}\varvec{A}_2\varvec{Q}^T) \, $$

which proves that \( {A}^{(2)}_{ij}\) are isotropic invariants.

5 Remark

Let, for example, a free energy function

$$\begin{aligned} W_e = \bar{W}(\varvec{A}_1,\varvec{A}_2) = \bar{W}(\varvec{Q}\varvec{A}_1\varvec{Q}^T,\varvec{Q}\varvec{A}_2\varvec{Q}^T) = W(\lambda _i, {A}^{(2)}_{ij}) \,. \end{aligned}$$
(45)

As explained in the literature [3, 4, 7, 8], it is important to note W must contain spectral invariants that satisfy the P-property, to deal with symmetry and coalescence of eigenvalues. For example, a general spectral invariant to describe W is of the form [5]

$$\begin{aligned} I = \sum _{i=1}^3\,g(\lambda _i) \varvec{v}_i \cdot \varvec{T}\varvec{v}_i \,, \end{aligned}$$
(46)

where \(\varvec{T}\) is a second-order tensor. For example, the invariants

$$\begin{aligned} \text {tr}\varvec{A}_2^2 = \sum _{i,j=1}^3 {A}^{(2)}_{ij} {A}^{(2)}_{ji} \,, { }g(\lambda _i) =1 \,, { }\varvec{T}= \varvec{A}_2^2 \,, \end{aligned}$$
(47)
$$\begin{aligned} \text {tr}(\varvec{A}_1\varvec{A}_2) = \sum _{i=1}^3 \lambda _i {A}^{(2)}_{ii} \,, { }g(\lambda _i) =\lambda _i \,, { }\varvec{T}= \varvec{A}_2\,, \end{aligned}$$
(48)
$$\begin{aligned} \text {tr}\varvec{A}_1^2 = \sum _{i=1}^3 \lambda _i^2 \,, { }g(\lambda _i) =\lambda _i^2 \,, { }\varvec{T}= \varvec{I}\,. \end{aligned}$$
(49)

In Shariff [8], it is shown that all “classical" invariants can be explicitly expressed in terms of the spectral invariants developed in [8].

6 Subjective student scenario

We personally believe that if one has given a robust and correct counterexample, then one need not give a subjective scenario, such as the student scenario given in [1], to further justify one’s counterexample. Since, the student scenario is subjective, we are reluctant to address it. However, for the readers’ benefit, we will discuss it in this section. We strongly emphasize that such a subjective scenario does not prove whether a scalar function is isotropic or not. For example, consider two students who do not know each other and are asked independently to construct the “classical" functional basis for \(\varvec{A}_1\) and \(\varvec{A}_2\). Since they do not know each other, there is a possibility that student 1 will construct the functional basis given in Eq. 2 and student 2 will construct the functional basis

$$\begin{aligned} I_1 \,, I_2 \,, I_3 \,, \text {tr}\varvec{A}_2^3 \,, \text {tr}(\varvec{A}_1\varvec{A}_2) \,, \text {tr}(\varvec{A}_1^2\varvec{A}_2) \,, \text {tr}(\varvec{A}_1\varvec{A}_2^2) \,, \text {tr}(\varvec{A}_1^2,\varvec{A}_2^2) \,, \end{aligned}$$
(50)

where

$$\begin{aligned} I_1 = \text {tr}\varvec{A}_1 \,, { }I_2 = \lambda _1\lambda _2 + \lambda _1\lambda _3 + \lambda _2\lambda _3 \,, { }I_3=\lambda _1\lambda _2\lambda _3 \,. \end{aligned}$$
(51)

They were then asked to evaluate the values of their invariants for \(\varvec{A}_1\) and \(\varvec{A}_2\) given in Itskov (4). Student 1 will obtain the values

$$\begin{aligned} \text {tr}\varvec{A}_1 = \text {tr}\varvec{A}_1^2 = \text {tr}\varvec{A}_1^3 = 3 \,, { }\text {etc.} \, \end{aligned}$$
(52)

and student 2 will obtain the values

$$\begin{aligned} I_1 = I_2=3 \,, { }I_3=1 \,, { }\text {etc.} \,. \end{aligned}$$
(53)

The different numerical values in Eqs. 52 and 53 may make the students think that elements of their functional basis are not isotropic functions. The above different values do not prove, at all, that the classical invariants are not isotropic functions. A similar but somewhat different situation occurs in Itskov’s student scenario.

Itskov constructs a scenario, where student 1 is asked to obtain the components of \(\varvec{A}_1\) and \(\varvec{A}_2\), and student 2 to obtain the components of \(\varvec{A}_1 (= \varvec{A}_1')\) and \(\varvec{A}_2'\). He has given the condition the students do not know each other and are asked independently to calculate the spectral invariants (3). We discuss this via the following two scenarios, taking note that \(\varvec{A}_1\) and \(\varvec{A}_2\) are those given in Itskov (4) and \(\varvec{Q}\) is given in Itskov (7).

Scenario 1: Students 1 and 2, by chance (or cheated), both use the same eigenvectors \(\varvec{e}_i\) for \(\varvec{A}_1=\varvec{I}\).

Student 1, using the reference basis \(S=\{\varvec{e}_1,\varvec{e}_2,\varvec{e}_3\}\), will obtain the spectral invariants \(\lambda _i=1 \,, { } {A}^{(2)}_{ij}\), for, respectively, \(\varvec{A}_1\) and \(\varvec{A}_2\), and student 2, using the correct the rotated basis \(S'=\{\varvec{Q}\varvec{e}_1,\varvec{Q}\varvec{e}_2,\varvec{Q}\varvec{e}_3\}\), will also obtain the same spectral invariants \(\lambda _i=1\) and \( {A}^{(2)}_{ij}\) for, respectively, \(\varvec{A}_1'\) and \(\varvec{A}_2'\). However, if by mistake student 2 (thinking that the identity tensor \(\varvec{I}\) has a unique basis and assuming that the bases S and \(S'\) are the same) uses the unrotated basis S for \(\varvec{A}_1'\) and \(\varvec{A}_2'\), he/she will then obtain different tensor components \(B_{ij}= \varvec{e}_i\cdot (\varvec{Q}\varvec{A}_2\varvec{Q}^T)\varvec{e}_j \ne {A}^{(2)}_{ij}\) for \(\varvec{A}_2\). This scenario does not prove that \( {A}^{(2)}_{ij}\) are not isotropic invariants

Scenario 2: Since \(\varvec{A}_1=\varvec{I}\) has infinitely many bases, there is a high probability that students 1 and 2 will use different eigenvectors to describe \(\varvec{A}_1\). For example, student 1 will use

$$\begin{aligned} \varvec{A}_1 = \sum _{i=1}^3 \varvec{e}_i\otimes \varvec{e}_i \end{aligned}$$
(54)

and obtain the components \( {A}^{(2)}_{ij}\) for \(\varvec{A}_2\). Student 2 will use

$$\begin{aligned} \varvec{A}_1 = \sum _{i=1}^3 \bar{\varvec{e}}_i\otimes \bar{\varvec{e}} _i \,, { }\bar{\varvec{e}}_i \ne \varvec{e}_i \, \end{aligned}$$
(55)

and obtain \( {\bar{A}}^{(2)}_{ij} = \bar{\varvec{e}}_i\cdot \varvec{A}_2\bar{\varvec{e}_j}\) for \(\varvec{A}_2\). Student 2, using the correct rotated basis \(S' = \{\varvec{Q}\bar{\varvec{e}}_1,\varvec{Q}\bar{\varvec{e}}_2,\varvec{Q}\bar{\varvec{e}}_3\}\), will also obtain \( {\bar{A}}^{(2)}_{ij}\) for \(\varvec{A}_2'\). Although \( {\bar{A}}^{(2)}_{ij} \ne {A}^{(2)}_{ij}\), this scenario does not prove that invariants \( {\bar{A}}^{(2)}_{ij}\) and \( {A}^{(2)}_{ij}\) are not isotropic invariants: Both of them are isotropic invariants as exemplified in Section 2. It is important to note that, as explained in Section 5 (see also reference [5]), to satisfy the P-property the spectral invariants must be of the form Eq. 46. Consider the following examples (for \(\lambda _i=1\)):

$$\begin{aligned} I = \sum _{i=1}^3\,g(\lambda _i) {A}^{(2)}_{ii} = \sum _{i=1}^3\,g(\lambda _i) {\bar{A}}^{(2)}_{ii} = g(1) \text {tr}\varvec{A}_2 \,, \end{aligned}$$
$$\begin{aligned} \text {tr}(\varvec{A}_1\varvec{A}_2) = \sum _{i=1}^3 \lambda _i {A}^{(2)}_{ii} = \sum _{i=1}^3 \lambda _i {\bar{A}}^{(2)}_{ii} = \sum _{i=1}^3 {A}^{(2)}_{ii} = \sum _{i=1}^3 {\bar{A}}^{(2)}_{ii} = \text {tr}\varvec{A}_2 \,, \end{aligned}$$
(56)

where they have the same values even though \( {\bar{A}}^{(2)}_{ij} \ne {A}^{(2)}_{ij}\).

Important: Even though \( {A}^{(2)}_{ij}\) have different values for different coordinate systems (material frames), in view of the P-property [5], they will give the same values for the strain energy or stresses at the same strain. Spectral invariants, such as \( {A}^{(2)}_{ij}\), together with the P-property, have been used in the literature (see reference [8] and references within) to model mechanical behaviors in solid mechanics.

7 Conclusion

Itskov’s [1] objective is to prove, via a counterexample, that Shariff spectral invariants [7] are not isotropic scalar functions. In Section 3 (also in Section 2.4), we prove that his counterexample is wrong because, from his statement “Due to the fact that \(\varvec{A}_1'=\varvec{A}_1\) the eigenvectors (5) remain unchanged," he incorrectly assumes that the basis for the identity tensor \(\varvec{I}\) is unique; hence, he uses the reference basis (instead of the rotated basis) to evaluate the transformed tensors and obtains an incorrect result. In Section 2 and in the literature [6, 7], we have shown that Shariff spectral invariants are isotropic scalar functions. In Section 6, we address the subjective students’ scenario.