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Harmonic Decomposition, Irreducible Basis Tensors, and Minimal Representations of Material Tensors and Pseudotensors

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Abstract

We propose a general and efficient method to derive various minimal representations of material tensors or pseudotensors for crystals. By a minimal representation we mean one that pertains to a specific Cartesian coordinate system under which the number of independent components in the representation is the smallest possible. The proposed method is based on the harmonic and Cartan decompositions and, in particular, on the introduction of orthonormal irreducible basis tensors in the chosen harmonic decomposition. For crystals with non-trivial point group symmetry, we demonstrate by examples how deriving restrictions imposed by symmetry groups (e.g., \(C_{2}\), \(C_{s}\), \(C_{3}\), etc.) whose symmetry elements do not completely specify a coordinate system could possibly miss the minimal representations, and how the Cartan decomposition of SO(3)-invariant irreducible tensor spaces could lead to coordinate systems under which the representations are minimal. For triclinic materials, and for material tensors and pseudotensors which observe a sufficient condition given herein, we describe a procedure to obtain a coordinate system under which the explicit minimal representation has its number of independent components reduced by three as compared with the representation with respect to an arbitrary coordinate system.

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Change history

  • 05 August 2023

    The original online version of this article was revised: Equation (50) as published in the original article, published online date 11 April 2023, has been updated.

  • 04 August 2023

    A Correction to this paper has been published: https://doi.org/10.1007/s10659-023-10030-z

Notes

  1. Here we follow Sirotin and Shaskolskaya [4, p. 47] in calling the 32 geometric crystal classes (as defined by the point groups of crystals) “symmetry classes”. Later we shall also refer to “symmetry classes” of elements in material tensor or pseudotensor space, which carry a completely different meaning; see Sect. 4.2.

  2. As pointed out by Forte and Vianello [7], Schouten [8, p. 162] in a book published in 1951 writes down two assertions in a footnote with no further elaboration: “The number of constants [for the triclinic and monoclinic “sub-systems”] could be reduced to 18 and 12 respectively. No reduction is possible in the other systems.” The second assertion is incorrect.

  3. Harmonic tensors are defined in Sect. 2.3. There is a theorem which asserts that any \(r\)-th order tensor \(\boldsymbol {A}\) can be expressed in terms of harmonic tensors of orders not higher than \(r\), the second-order identity tensor \(\boldsymbol {I}\), and the third-order alternating tensor \(\boldsymbol {\varepsilon }\). Each such expression is called a harmonic decomposition of tensor \(\boldsymbol {A}\). See [9, Section 17.4] and [10] for details.

  4. See [9] and the references therein for details.

  5. In exhibiting the matrix \([D^{l}_{mn}(\boldsymbol {R})]\) we follow the standard practice in the physics literature (see, e.g., Biedenharn and Louck [14, p. 46]). We let the indices \(m\) and \(n\) run from \(l\) to \(-l\) when reading the matrix from top to bottom and when reading from left to right, respectively.

  6. The adjective “harmonic” comes from the fact that there is a linear isomorphism between \(\mathscr{H}^{r}\) and the space \(\mathcal {H}^{r}\) of complex-valued homogeneous polynomials of degree \(r\) that satisfy the Laplace equation, whose solutions are called harmonic functions. The elements of \(\mathcal {H}^{r}\) are called homogeneous harmonic polynomials of degree \(r\). See, e.g., [9, Chap. 17] and the references therein for details.

  7. This assertion will be justified in Sect. 3.4, where the relationship between (30) with restriction (31) and the classical harmonic decomposition of tensor spaces will be laid bare.

  8. See Weyl [26, pp. 52–56], Gurevich [27, Section 16].

  9. Sirotin and Shaskolskaya [4] call the independent components in a minimal representation “independent invariants”, a term that we shall also use.

  10. By (21) the inversion \(\boldsymbol {\mathcal {I}}\) is a symmetry operation for all even-order material tensors. Hence a crystal with \(G_{\mathrm{cr}} = C_{2h}\) or \(C_{s}\) has the same minimal representations as those of a crystal with \(G_{\mathrm{cr}} = C_{2}\). In fact, the Hall-effect tensors of \(C_{2}\), \(C_{2h}\), and \(C_{s}\) crystals all belong to the symmetry class \([C_{2h}]\).

  11. Both Khatkevich and Federov adopt the passive view of rotations.

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Acknowledgements

We thank two anonymous reviewers for their comments, which helped us improve the presentation of our paper.

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The original online version of this article was revised: Equation (50) as published in the original article, published online date 11 April 2023, has been updated.

Appendix: Lists of Irreducible Basis Tensors

Appendix: Lists of Irreducible Basis Tensors

In what follows we assume that in the physical space a right-handed orthonormal triad \(\boldsymbol {e}_{1}\), \(\boldsymbol {e}_{2}\), and \(\boldsymbol {e}_{3}\) has been chosen, which defines a Cartesian coordinate system. The components of every tensor \(\boldsymbol {H}= H_{i_{1} i_{2} \cdots i_{r}} \boldsymbol {e}_{i_{1}} \otimes \boldsymbol {e}_{i_{2}} \otimes \cdots \otimes \boldsymbol {e}_{i_{r}} \in V^{\otimes r}_{c}\), where the summation convention is in force, are referred to this coordinate system.

For each set of irreducible basis tensors, we show only those \(\boldsymbol {H}^{k,s}_{m}\) with \(m \geq 0\). Each \(\boldsymbol {H}^{k,a}_{\bar{m}}\) can be written down easily by applying (29) to \(\boldsymbol {H}^{k,s}_{m}\).

1.1 A.1 Irreducible Basis Tensors for the Second-Order Hall-Effect Tensor

The tensor space in question is \(V^{\otimes 2}_{c}\). Each tensor \(\boldsymbol {\rho }^{(H)}\) in this space is represented by a \(3 \times 3\) matrix as usual.

$$\begin{gathered} \boldsymbol {H}^{0}_{0} = \frac{1}{\sqrt{3}}\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\displaystyle \right ), \qquad \boldsymbol {H}^{1}_{0} = \frac{1}{\sqrt{2}}\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\displaystyle \right ), \qquad \boldsymbol {H}^{1}_{1} = \frac{1}{2}\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & -i \\ 0 & 0 & -1 \\ i & 1 & 0 \end{array}\displaystyle \right ), \\ \boldsymbol {H}^{2}_{0} = \frac{1}{\sqrt{6}}\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{array}\displaystyle \right ), \qquad \boldsymbol {H}^{2}_{1} = \frac{1}{2}\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 1 \\ 0 & 0 & -i \\ 1 & -i & 0 \end{array}\displaystyle \right ), \qquad \boldsymbol {H}^{2}_{2} = \frac{1}{2}\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} -1 & i & 0 \\ i & 1 & 0 \\ 0 & 0 & 0 \end{array}\displaystyle \right ). \end{gathered}$$
(122)

1.2 A.2 Irreducible Basis Tensors for Piezoelectricity Tensor

The tensor space in question is \(V^{\otimes}_{c} \otimes [V^{\otimes 2}_{c}]\). Each tensor \(\boldsymbol {d}\) in this space is represented by a \(3 \times 6\) matrix (see, e.g., Sirotin and Shaskolskaya [4, p. 606], Newnham [2, p. 88]):

$$ \boldsymbol {d}= \left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} d_{11} & d_{12} & d_{13} & d_{14} & d_{15} & d_{16} \\ d_{21} & d_{22} & d_{23} & d_{24} & d_{25} & d_{26} \\ d_{31} & d_{32} & d_{33} & d_{34} & d_{35} & d_{36} \end{array}\displaystyle \right ) := \left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} d_{111} & d_{122} & d_{133} & 2d_{123} & 2d_{113} & 2d_{112} \\ d_{211} & d_{222} & d_{233} & 2d_{223} & 2d_{213} & 2d_{212} \\ d_{311} & d_{322} & d_{333} & 2d_{323} & 2d_{313} & 2d_{312} \end{array}\displaystyle \right ). $$
(123)

The irreducible basis tensors are as follows:

$$\begin{gathered} \boldsymbol {H}_{0}^{1,1} =\frac{\sqrt{3}}{3}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \end{array}\displaystyle \right ), \quad \boldsymbol {H}_{1}^{1,1}=\frac{\sqrt{6}}{6}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}-1 & -1 & -1 & 0 & 0 & 0 \\ i & i & i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\displaystyle \right ), \\ \boldsymbol {H}_{0}^{1,2} =\frac{\sqrt{15}}{15}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 \\ -1 & -1 & 2 & 0 & 0 & 0 \end{array}\displaystyle \right ), \\ \boldsymbol {H}_{1}^{1,2}=\frac{\sqrt{30}}{30} \left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}-2 & 1 & 1 & 0 & 0 & 3i \\ -i & 2i & -i & 0 & 0 & -3 \\ 0 & 0 & 0 & 3i & -3 & 0 \end{array}\displaystyle \right ), \quad \boldsymbol {H}_{0}^{2} =\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\displaystyle \right ), \\ \boldsymbol {H}_{1}^{2}=\frac{\sqrt{6}}{6}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & -i & i & 0 & 0 & 1 \\ -1 & 0 & 1 & 0 & 0 & i \\ 0 & 0 & 0 & -1 & -i & 0 \end{array}\displaystyle \right ),\quad \boldsymbol {H}_{2}^{2}=\frac{\sqrt{6}}{6}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & -1 & -i & 0 \\ 0 & 0 & 0 & i & -1 & 0 \\ i & -i & 0 & 0 & 0 & 2 \end{array}\displaystyle \right ), \\ \boldsymbol {H}_{0}^{3} =\frac{\sqrt{10}}{10}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 1 & 1 & -2 & 0 & 0 & 0 \end{array}\displaystyle \right ), \quad \boldsymbol {H}_{1}^{3}=\frac{\sqrt{30}}{60}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}-3 & -1 & 4 & 0 & 0 & 2i \\ i & 3i & -4i & 0 & 0 & -2 \\ 0 & 0 & 0 & -8i & 8 & 0 \end{array}\displaystyle \right ), \\ \boldsymbol {H}_{2}^{3} =\frac{\sqrt{3}}{6}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & 2i & -2 & 0 \\ 0 & 0 & 0 & 2 & 2i & 0 \\ -1 & 1 & 0 & 0 & 0 & 2i \end{array}\displaystyle \right ), \quad \boldsymbol {H}_{3}^{3}=\frac{\sqrt{2}}{4}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}1 & -1 & 0 & 0 & 0 & -2i \\ -i & i & 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\displaystyle \right ). \end{gathered}$$
(124)

1.3 A.3 Irreducible Basis Tensors for Elasticity Tensor

The following irreducible basis tensors in \([V^{\otimes 2}_{c}]^{\otimes 2}\)] first appeared in [30]. They are given in the Kelvin notation (see, e.g., [30] and the references therein.

$$ \boldsymbol {H}_{0}^{0,1} =\frac{\sqrt{5}}{15}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}2 & -1 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 & 0 \\ -1 & -1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 \end{array}\displaystyle \right ), \qquad \boldsymbol {H}_{0}^{0,2} =\frac{1}{3}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\displaystyle \right ), $$
$$ \boldsymbol {H}_{0}^{2,1} =\frac{\sqrt{14}}{42}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}2 & -4 & 2 & 0 & 0 & 0 \\ -4 & 2 & 2 & 0 & 0 & 0 \\ 2 & 2 & -4 & 0 & 0 & 0 \\ 0 & 0 & 0 & -3 & 0 & 0 \\ 0 & 0 & 0 & 0 & -3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6 \end{array}\displaystyle \right ), $$
$$ \boldsymbol {H}_{1}^{2,1} =\frac{\sqrt{42}}{84}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & 4i & 2 & 0 \\ 0 & 0 & 0 & -2i & -4 & 0 \\ 0 & 0 & 0 & -2i & 2 & 0 \\ 4i & -2i & -2i & 0 & 0 & 3\sqrt{2} \\ 2 & -4 & 2 & 0 & 0 & -3\sqrt{2}i \\ 0 & 0 & 0 & 3\sqrt{2} & -3\sqrt{2}i & 0 \end{array}\displaystyle \right ), $$
$$\begin{aligned} \boldsymbol {H}_{2}^{2,1} & =\frac{\sqrt{21}}{42}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}-2 & 0 & 2 & 0 & 0 & \sqrt{2}i \\ 0 & 2 & -2 & 0 & 0 & \sqrt{2}i \\ 2 & -2 & 0 & 0 & 0 & -2\sqrt{2}i \\ 0 & 0 & 0 & 3 & 3i & 0 \\ 0 & 0 & 0 & 3i & -3 & 0 \\ \sqrt{2}i & \sqrt{2}i & -2\sqrt{2}i & 0 & 0 & 0 \end{array}\displaystyle \right ), \end{aligned}$$
$$ \boldsymbol {H}_{0}^{2,2} =\frac{1}{6}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}2 & 2 & -1 & 0 & 0 & 0 \\ 2 & 2 & -1 & 0 & 0 & 0 \\ -1 & -1 & -4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\displaystyle \right ), \qquad \boldsymbol {H}_{1}^{2,2}=\frac{\sqrt{3}}{6}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & -i & 1 & 0 \\ 0 & 0 & 0 & -i & 1 & 0 \\ 0 & 0 & 0 & -i & 1 & 0 \\ -i & -i & -i & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\displaystyle \right ), $$
$$ \boldsymbol {H}_{2}^{2,2}=\frac{\sqrt{6}}{12}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}-2 & 0 & -1 & 0 & 0 & \sqrt{2}i \\ 0 & 2 & 1 & 0 & 0 & \sqrt{2}i \\ -1 & 1 & 0 & 0 & 0 & \sqrt{2}i \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \sqrt{2}i & \sqrt{2}i & \sqrt{2}i & 0 & 0 & 0 \end{array}\displaystyle \right ), $$
$$ \boldsymbol {H}_{0}^{4} =\frac{\sqrt{70}}{140}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}3 & 1 & -4 & 0 & 0 & 0 \\ 1 & 3 & -4 & 0 & 0 & 0 \\ -4 & -4 & 8 & 0 & 0 & 0 \\ 0 & 0 & 0 & -8 & 0 & 0 \\ 0 & 0 & 0 & 0 & -8 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{array}\displaystyle \right ), $$
$$\begin{aligned} \boldsymbol {H}_{1}^{4} & =\frac{\sqrt{14}}{56}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & -\sqrt{2}i & 3\sqrt{2} & 0 \\ 0 & 0 & 0 & -3\sqrt{2}i & \sqrt{2} & 0 \\ 0 & 0 & 0 & 4\sqrt{2}i & -4\sqrt{2} & 0 \\ -\sqrt{2}i & -3\sqrt{2}i & 4\sqrt{2}i & 0 & 0 & 2 \\ 3\sqrt{2} & \sqrt{2} & -4\sqrt{2} & 0 & 0 & -2i \\ 0 & 0 & 0 & 2 & -2i & 0 \end{array}\displaystyle \right ), \\ \boldsymbol {H}_{2}^{4} & =\frac{\sqrt{7}}{28}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}-2 & 0 & 2 & 0 & 0 & \sqrt{2}i \\ 0 & 2 & -2 & 0 & 0 & \sqrt{2}i \\ 2 & -2 & 0 & 0 & 0 & -2\sqrt{2}i \\ 0 & 0 & 0 & -4 & -4i & 0 \\ 0 & 0 & 0 & -4i & 4 & 0 \\ \sqrt{2}i & \sqrt{2}i & -2\sqrt{2}i & 0 & 0 & 0 \end{array}\displaystyle \right ), \end{aligned}$$
$$ \boldsymbol {H}_{3}^{4} =\frac{\sqrt{2}}{8}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & \sqrt{2}i & -\sqrt{2} & 0 \\ 0 & 0 & 0 & -\sqrt{2}i & \sqrt{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \sqrt{2}i & -\sqrt{2}i & 0 & 0 & 0 & 2 \\ -\sqrt{2} & \sqrt{2} & 0 & 0 & 0 & 2i \\ 0 & 0 & 0 & 2 & 2i & 0 \end{array}\displaystyle \right ), $$
$$ \boldsymbol {H}_{4}^{4} =\frac{1}{4}\left ( \textstyle\begin{array} [c]{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}1 & -1 & 0 & 0 & 0 & -\sqrt{2}i \\ -1 & 1 & 0 & 0 & 0 & \sqrt{2}i \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ -\sqrt{2}i & \sqrt{2}i & 0 & 0 & 0 & -2 \end{array}\displaystyle \right ). $$
(125)

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Man, CS., Du, W. Harmonic Decomposition, Irreducible Basis Tensors, and Minimal Representations of Material Tensors and Pseudotensors. J Elast 154, 3–41 (2023). https://doi.org/10.1007/s10659-023-10010-3

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  • DOI: https://doi.org/10.1007/s10659-023-10010-3

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