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The Symmetries of Octupolar Tensors

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Abstract

Octupolar tensors are third order, completely symmetric and traceless tensors. Whereas in 2D an octupolar tensor has the same symmetries as an equilateral triangle and can ultimately be identified with a vector in the plane, the symmetries that it enjoys in 3D are quite different, and only exceptionally reduce to those of a regular tetrahedron. By use of the octupolar potential, that is, the cubic form associated on the unit sphere with an octupolar tensor, we shall classify all inequivalent octupolar symmetries. This is a mathematical study which also reviews and incorporates some previous, less systematic attempts.

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Notes

  1. In the Introduction to [30] (p. IX), we read:

    About 25 years ago I started to write notes for a course for seniors and beginning graduate students at Carnegie Institute of Technology (renamed Carnegie-Mellon University in 1968). At first, the course was entitled “Tensor Analysis”. […] The notes were rewritten several times. They were widely distributed and they served as the basis for appendices to the books [9] and [44].

  2. The superscript \(^{(2)}\) reminds us that this tensor expresses the field induced by polarization as a quadratic function of the external field, whereas the ordinary susceptibility establishes a linear relationship between the two fields.

  3. More generally, we might consider potentials with contributions up to third order; thus we would have the sum of a scalar part, a vector one, another part described by a second order tensor, and finally the one described by the third order one. Here we focus on this last contribution, as the study of theories with scalar, vector, or second order tensor order parameters is standard (in principle; obviously concrete applications can present endless complications).

  4. As a general convention, we will denote the potentials in Cartesian coordinates by \(\varPhi \) (with several suffixes) and those in spherical coordinates—which we always consider only for \(r=1\)—by \(\varPsi \) (again with corresponding suffixes).

  5. It may be worth mentioning that (in particular, if we are satisfied with studying \(\varPhi \) on one hemisphere, which is justified by (3.2)) a third option is present, i.e., setting \(z = \pm \sqrt{1 - x^{2} - y^{2}}\) and considering \(\varPhi \) as a function of \(x\) and \(y\); these take value in the unit disk. This will be used in Sect. 5.3.

  6. In fact, as pointed out by Walcher [51], this kind of results follow ultimately from the work of Bezout on intersection theory dating back to the 18th century. See his paper [52] for details.

  7. It should be noted that the “disappearance” of real critical points—w.r.t. the generic situation described by Röhrl’s theorem—is related, at least in our model, to the appearance of a “monkey saddle” [14], i.e., of a critical point with a non-generic index; see below for detail.

  8. In fact, if \(\mathbf{v}\) is an eigenvector of \(M\) with eigenvalue \(\lambda \), then for any number \(\alpha \neq 0\) also \(\mathbf{w} = \alpha {\mathbf{v}}\) is an eigenvector with the same eigenvalue \(\lambda \).

  9. In this paper, the adjective “generic” is given the meaning common in algebraic geometry, that is, it designates a property valid away from the roots of a polynomial in parameter space [5].

  10. This means that we can rule out the possibility to have \(\alpha _{3} = 0\). In fact, even in the case this is a local maximum at height zero, we can always—see Remark 6—choose the North Pole to be an absolute maximum, and this is necessarily positive.

  11. In order to know the value for the corresponding \(\lambda \), one needs to express the solution in Cartesian coordinates and go back to (5.22); this is due to the fact that our change of coordinates was performed imposing \(r=1\) and thus the constraint term, which represents the dynamical origin of \(\lambda \), is absent in the angular coordinates.

  12. To compare the expressions worked out in this paper for the Hessian matrix of the octupolar potential with those featuring in [14], the reader should heed that these differ by a scaling factor: the Hessian matrix here is three times the Hessian matrix there.

  13. It should be noted that in our previous work [14] we have used a slightly different reparametrization, with \(\rho \) instead of \(\rho /2\). This accounts for the differences in many of the forthcoming formulas.

  14. There are also maps acting on \(\rho \) by changing its sign and leaving the potential invariant; these are not admitted as we have required \(\rho \in [0,2]\).

  15. Which thus are presumably more complicated than anticipated.

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Acknowledgements

E.G. Virga acknowledges the kind hospitality of the Oxford Centre for Nonlinear PDE, where part of this work was done while he was visiting the Mathematical Institute at the University of Oxford.

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  1. Dipartimento di Matematica, Università degli Studi di Milano, via Saldini 50, 20133, Milano, Italy

    Giuseppe Gaeta

  2. Dipartimento di Matematica, Università di Pavia, via Ferrata 5, 27100, Pavia, Italy

    Epifanio G. Virga

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  1. Giuseppe Gaeta
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Appendix: The Tetrahedron Group

Appendix: The Tetrahedron Group

In this Appendix we give further detail—beyond those mentioned in Sect. 6—on the tetrahedron group. We will work on a concrete realization of it in three-dimensional space; the points identifying the tetrahedron will be

$$ \biggl( 0 , \frac{2 \sqrt{2}}{3} , -\frac{1}{3} \biggr) , \ \biggl( \sqrt{ \frac{2}{3}} , -\frac{\sqrt{2}}{3} , -\frac{1}{3} \biggr) , \ \biggl( -\sqrt{\frac{2}{3}} , -\frac{\sqrt{2}}{3} , - \frac{1}{3} \biggr) , \ ( 0 , 0 , 1 ). $$

In angular coordinates \((\vartheta _{1},\vartheta _{2})\), these are

$$ \biggl( \theta _{0} , - \frac{\pi }{2} \biggr) ,\ \biggl(\theta _{0} , - \frac{\pi }{6} \biggr) , \ \biggl(\theta _{0} , \frac{5 \pi }{6} \biggr) , \ \biggl(\frac{\pi }{2} , * \biggr) , $$

where the symbol ∗ means that in this case \(\vartheta _{2}\) is not determined, and

$$ \theta _{0} =- \arcsin \biggl( \frac{1}{3} \biggr) \doteq -0.34. $$

The tetrahedron group \(T_{d} \subset O(3)\) is made of 12 matrices of determinant \(+1\), associated to rotations of an angle \(2 \pi / 3\) and multiples around each of the four axes of the tetrahedron, denoted as \(T_{d}^{+} \subset \mathit{SO}(3)\); and other 12 matrices of determinant −1, collectively denoted as \(T_{d}^{-}\), among which are those associated to reflections through planes containing axes of the tetrahedron.

We now give the twelve matrices in \(T_{d}^{+}\); these are:

$$\begin{aligned} M_{1} =& \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} , \quad \quad M_{2} = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} , \\ M_{3} =& \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2}&0 \\ \frac{\sqrt{3}}{2} & - \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} , \quad \quad M_{4} = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\ \frac{1}{2 \sqrt{3}} & -\frac{1}{6} & \frac{2 \sqrt{2}}{3} \\ \sqrt{\frac{2}{3}} & - \frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \\ M_{5} =& \begin{pmatrix} \frac{1}{2} & \frac{1}{2 \sqrt{3}} & \sqrt{\frac{2}{3} } \\ \frac{\sqrt{3}}{2} & - \frac{1}{6} & -\frac{\sqrt{2}}{3} \\ 0 & \frac{2 \sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \quad \quad M_{6} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ -\frac{1}{2 \sqrt{3}} & -\frac{1}{6} & \frac{2 \sqrt{2}}{3} \\ -\sqrt{\frac{2}{3}}& -\frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \\ M_{7} =& \begin{pmatrix} \frac{1}{2} & - \frac{1}{2 \sqrt{3}} & -\sqrt{\frac{2}{3}} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{6} & -\frac{\sqrt{2}}{3} \\ 0 & \frac{2 \sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \quad \quad M_{8} = \begin{pmatrix} - \frac{1}{2} & \frac{1}{2 \sqrt{3}} & \sqrt{\frac{2}{3}} \\ -\frac{1}{2 \sqrt{3}} & \frac{5}{6} & -\frac{\sqrt{2}}{3} \\ - \sqrt{\frac{2}{3}} & - \frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \\ M_{9} =& \begin{pmatrix} -\frac{1}{2} & -\frac{1}{2 \sqrt{3}} & -\sqrt{\frac{2}{3}} \\ \frac{1}{2 \sqrt{3}} & \frac{5}{6} & -\frac{\sqrt{2}}{3} \\ \sqrt{\frac{2}{3}} & - \frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \quad \quad M_{10} = \begin{pmatrix} 0 & \frac{1}{\sqrt{3}} & -\sqrt{\frac{2}{3}} \\ \frac{1}{\sqrt{3}} & - \frac{2}{3} & -\frac{\sqrt{2}}{3} \\ -\sqrt{\frac{2}{3}} & -\frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \\ M_{11} =& \begin{pmatrix} 0 & -\frac{1}{\sqrt{3}} & \sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{3}} & -\frac{2}{3} & -\frac{\sqrt{2}}{3} \\ \sqrt{\frac{2}{3}} & - \frac{\sqrt{2}}{3} & - \frac{1}{3} \end{pmatrix} , \quad \quad M_{12} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & \frac{2 \sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} . \end{aligned}$$

The multiplication table for these matrices is the following:

$$ P_{11} =\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 2 & 3 & 1 & 11 & 7 & 8 & 12 & 10 & 4 & 6 & 9 & 5 \\ 3 & 1 & 2 & 9 & 12 & 10 & 5 & 6 & 11 & 8 & 4 & 7 \\ 4 & 12 & 6 & 5 & 1 & 11 & 9 & 2 & 10 & 7 & 3 & 8 \\ 5 & 8 & 11 & 1 & 4 & 3 & 10 & 12 & 7 & 9 & 6 & 2 \\ 6 & 4 & 12 & 10 & 8 & 7 & 1 & 11 & 3 & 2 & 5 & 9 \\ 7 & 10 & 9 & 2 & 11 & 1 & 6 & 5 & 12 & 4 & 8 & 3 \\ 8 & 11 & 5 & 6 & 10 & 12 & 2 & 9 & 1 & 3 & 7 & 4 \\ 9 & 7 & 10 & 12 & 3 & 4 & 11 & 1 & 8 & 5 & 2 & 6 \\ 10 & 9 & 7 & 8 & 6 & 5 & 3 & 4 & 2 & 1 & 12 & 11 \\ 11 & 5 & 8 & 7 & 2 & 9 & 4 & 3 & 6 & 12 & 1 & 10 \\ 12 & 6 & 4 & 3 & 9 & 2 & 8 & 7 & 5 & 11 & 10 & 1 \end{array}\displaystyle \right ). $$

These generate several subgroups; in particular—apart from the trivial ones consisting of \(M_{1}\) alone and of the full group \(T_{d}^{+}\)—we have four subgroups of order three,

$$\begin{aligned} &G_{1} = \{ M_{1}, M_{2}, M_{3} \} , \quad \quad G_{2} = \{M_{1}, M_{4} , M_{5} \} , \quad \quad G_{3} = \{ M_{1} , M_{6} , M_{7} \} , \\ & G_{4} = \{ M_{1}, M _{8} , M_{9} \} ; \end{aligned}$$

three groups of order two,

$$ G_{5} = \{ M_{1} , M_{10} \} , \quad \quad G_{6} = \{ M_{1} , M_{11} \} , \quad \quad G _{7} = \{ M_{1} , M_{12} \} ; $$

and one group of order four,

$$ G_{8} = \{ M_{1} , M_{10} , M_{11} , M_{12} \} . $$

The latter is the only nontrivial normal subgroup, and also the only one acting freely.

We can also easily determine the subspaces \(F_{k}\) admitting each of these \(G_{k}\) as symmetry subgroups; in particular,

$$\begin{aligned} &F_{1} = (0,0,z) , \quad \quad F_{2} = (\sqrt{6} z , \sqrt{2} z , z ) , \quad \quad F _{3} = (- \sqrt{6} z , \sqrt{2} z , z ) , \\ & F_{4} = (0 , - 2 \sqrt{2} z , z ) ; \end{aligned}$$

the subgroups \(G_{k}\), \(k=1,2,3,4\) act as rotations (by an angle \(2 \pi / 3\)) around these axes, which are just the axes of the tetrahedron. Moreover,

$$ F_{5} = \biggl( - \sqrt{\frac{3}{2}} z , - \sqrt{ \frac{1}{2}} z , z \biggr) , \quad \quad F_{6} = \biggl( \sqrt{ \frac{3}{2}} z , - \sqrt{ \frac{1}{2}} z , z \biggr) , \quad \quad F_{7} = ( 0 , \sqrt{2} z , z ) ; $$

these subgroups \(G_{k}\), \(k=5,6,7\), represent rotations by \(\pi \) around the given axes \(F_{k}\). Note that \(F_{8} = \{ (0,0,0) \}\), and correspondingly \(G_{8}\) represents combined \(\pi \) rotations around the \(F_{5},F_{6},F_{7}\) axes.

We can give as well the twelve matrices in \(T_{d}^{-}\); these are:

$$\begin{aligned} M_{1}^{-} =& \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} , \quad \quad M_{2}^{-} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2},0 \\ 0 & 0 & 1 \end{pmatrix} , \\ M_{3}^{-} =& \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & -\frac{1}{2},0 \\ 0 & 0 & 1 \end{pmatrix} , \quad \quad M_{4}^{-} = \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{1}{2 \sqrt{3}} & -\frac{1}{6} & \frac{2 \sqrt{2}}{3} \\ \sqrt{\frac{2}{3}} & -\frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \\ M_{5}^{-} =& \begin{pmatrix} -\frac{1}{2} & -\frac{1}{2 \sqrt{3}} & -\sqrt{\frac{2}{3}} \\ \frac{\sqrt{3}}{2} & -\frac{1}{6} & -\frac{\sqrt{2}}{3} \\ 0 & \frac{2\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \quad \quad M_{6}^{-} = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\ -\frac{1}{2 \sqrt{3}} & -\frac{1}{6} & \frac{2 \sqrt{2}}{3} \\ -\sqrt{\frac{2}{3}} & -\frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \\ M_{7}^{-} =& \begin{pmatrix} -\frac{1}{2}&\frac{1}{2\sqrt{3}} & \sqrt{\frac{2}{3}} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{6} & -\frac{\sqrt{2}}{3} \\ 0,\frac{2 \sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \quad \quad M_{8}^{-} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2 \sqrt{3}} & -\sqrt{\frac{2}{3}} \\ -\frac{1}{2 \sqrt{3}} & \frac{5}{6} & -\frac{\sqrt{2}}{3} \\ -\sqrt{\frac{2}{3}} & -\frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \\ M_{9}^{-} =& \begin{pmatrix} \frac{1}{2} & \frac{1}{2 \sqrt{3}} & \sqrt{\frac{2}{3}} \\ \frac{1}{2 \sqrt{3}} & \frac{5}{6} & -\frac{\sqrt{2}}{3} \\ \sqrt{\frac{2}{3}} & -\frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \quad \quad M_{10}^{-} = \begin{pmatrix} 0 & -\frac{1}{\sqrt{3}} & \sqrt{\frac{2}{3} } \\ \frac{1}{\sqrt{3}} & -\frac{2}{3} & -\frac{\sqrt{2}}{3} \\ -\sqrt{\frac{2}{3}} & -\frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \\ M_{11}^{-} =& \begin{pmatrix} 0 & \frac{1}{\sqrt{3}} & -\sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{3}} & -\frac{2}{3} & -\frac{\sqrt{2}}{3} \\ \sqrt{\frac{2}{3}} & -\frac{\sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} , \quad \quad M_{12}^{-} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2\sqrt{2}}{3} \\ 0 & \frac{2 \sqrt{2}}{3} & -\frac{1}{3} \end{pmatrix} . \end{aligned}$$

Once \(M_{1}^{-}\) has been defined, they are built by

$$ M_{k}^{-} =M_{1}^{-} M_{k} . $$

It is obvious that \(M_{1}^{-}\) represents a reflection (across the \((y,z)\) plane), so that the matrices \(M_{k}^{-}\) represent the combination of rotations and reflections.

If we write \(M_{k+12} = M_{k}^{-}\), the full multiplication table is given in block form by

$$ P = \begin{pmatrix} P_{11} & P_{12} \\ P_{21} & P_{22} \end{pmatrix} , $$

where \(P_{11}\) has been given above, and the other blocks are:

$$\begin{aligned} P_{12} =&\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\ 15 & 13 & 14 & 21 & 24 & 22 & 17 & 18 & 23 & 20 & 16 & 19 \\ 14 & 15 & 13 & 23 & 19 & 20 & 24 & 22 & 16 & 18 & 21 & 17 \\ 18 & 16 & 24 & 22 & 20 & 19 & 13 & 23 & 15 & 14 & 17 & 21 \\ 19 & 22 & 21 & 14 & 23 & 13 & 18 & 17 & 24 & 16 & 20 & 15 \\ 16 & 24 & 18 & 17 & 13 & 23 & 21 & 14 & 22 & 19 & 15 & 20 \\ 17 & 20 & 23 & 13 & 16 & 15 & 22 & 24 & 19 & 21 & 18 & 14 \\ 21 & 19 & 22 & 24 & 15 & 16 & 23 & 13 & 20 & 17 & 14 & 18 \\ 20 & 23 & 17 & 18 & 22 & 24 & 14 & 21 & 13 & 15 & 19 & 16 \\ 23 & 17 & 20 & 19 & 14 & 21 & 16 & 15 & 18 & 24 & 13 & 22 \\ 22 & 21 & 19 & 20 & 18 & 17 & 15 & 16 & 14 & 13 & 24 & 23 \\ 24 & 18 & 16 & 15 & 21 & 14 & 20 & 19 & 17 & 23 & 22 & 13 \end{array}\displaystyle \right ), \\ P_{21} =&\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\ 14 & 15 & 13 & 23 & 19 & 20 & 24 & 22 & 16 & 18 & 21 & 17 \\ 15 & 13 & 14 & 21 & 24 & 22 & 17 & 18 & 23 & 20 & 16 & 19 \\ 16 & 24 & 18 & 17 & 13 & 23 & 21 & 14 & 22 & 19 & 15 & 20 \\ 17 & 20 & 23 & 13 & 16 & 15 & 22 & 24 & 19 & 21 & 18 & 14 \\ 18 & 16 & 24 & 22 & 20 & 19 & 13 & 23 & 15 & 14 & 17 & 21 \\ 19 & 22 & 21 & 14 & 23 & 13 & 18 & 17 & 24 & 16 & 20 & 15 \\ 20 & 23 & 17 & 18 & 22 & 24 & 14 & 21 & 13 & 15 & 19 & 16 \\ 21 & 19 & 22 & 24 & 15 & 16 & 23 & 13 & 20 & 17 & 14 & 18 \\ 22 & 21 & 19 & 20 & 18 & 17 & 15 & 16 & 14 & 13 & 24 & 23 \\ 23 & 17 & 20 & 19 & 14 & 21 & 16 & 15 & 18 & 24 & 13 & 22 \\ 24 & 18 & 16 & 15 & 21 & 14 & 20 & 19 & 17 & 23 & 22 & 13 \end{array}\displaystyle \right ), \\ P_{22} =&\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 3 & 1 & 2 & 9 & 12 & 10 & 5 & 6 & 11 & 8 & 4 & 7 \\ 2 & 3 & 1 & 11 & 7 & 8 & 12 & 10 & 4 & 6 & 9 & 5 \\ 6 & 4 & 12 & 10 & 8 & 7 & 1 & 11 & 3 & 2 & 5 & 9 \\ 7 & 10 & 9 & 2 & 11 & 1 & 6 & 5 & 12 & 4 & 8 & 3 \\ 4 & 12 & 6 & 5 & 1 & 11 & 9 & 2 & 10 & 7 & 3 & 8 \\ 5 & 8 & 11 & 1 & 4 & 3 & 10 & 12 & 7 & 9 & 6 & 2 \\ 9 & 7 & 10 & 12 & 3 & 4 & 11 & 1 & 8 & 5 & 2 & 6 \\ 8 & 11 & 5 & 6 & 10 & 12 & 2 & 9 & 1 & 3 & 7 & 4 \\ 11 & 5 & 8 & 7 & 2 & 9 & 4 & 3 & 6 & 12 & 1 & 10 \\ 10 & 9 & 7 & 8 & 6 & 5 & 3 & 4 & 2 & 1 & 12 & 11 \\ 12 & 6 & 4 & 3 & 9 & 2 & 8 & 7 & 5 & 11 & 10 & 1 \end{array}\displaystyle \right ). \end{aligned}$$

We now have subgroups involving elements of both \(T_{d}^{+}\) and \(T_{d}^{-}\); in particular, using again the 1 through 24 numeration and denoting \(M_{k}\) directly by \(k\), we have the following subgroups; those of types \(\mathcal{G}_{a}\) through \(\mathcal{G}_{c}\) extending those seen above, while those of type \(\mathcal{G}_{d}\) involve no other element of \(T_{d}^{+}\) but the identity:

$$\begin{aligned} \mathcal{G}_{a}: & (1,2,3,13,14,15) , (1,2,3,16,21,23) , (1,2,3,17,19,24) , (1,2,3,18,20,22) ; \\ & (1,4,5,13,18,19) , (1,4,5,14,16,22) , (1,4,5,15,21,24) , (1,4,5,17,20,23) ; \\ & (1,8,9,13,20,21) , (1,8,9,14,19,23) , (1,8,9,15,17,22) , (1,8,9,16,18,24) ; \\ \mathcal{G}_{b}: & (1,10,13,23) , (1,10,14,17) , (1,10,15,20) , (1,10,16,19) , (1,10,18,21) ,\\ & (1,10,22,24) ; \\ & (1,11,13,22) , (1,11,14,21) , (1,11,15,19) , (1,11,16,20) , (1,11,17,18) ,\\ & (1,11,23,24) ; \\ & (1,12,13,24) , (1,12,14,18) , (1,12,15,16) , (1,12,17,21) , (1,12,19,20) ,\\ & (1,12,22,23) ; \\ \mathcal{G}_{c}: & (1,10,11,12,13,22,23,24) , (1,10,11,12,14,17,18,21) ,\\ & (1,10,11,12,15,16,19,20) ; \\ \mathcal{G}_{d} : & (1,13) , (1,14) , (1,15) , (1,20) , (1,21) , (1,24). \end{aligned}$$

Groups of type \(\mathcal{G}_{a}\) contain rotations by \(2 \pi /3\) around an axis and reflections through a plane containing that axis, hence they are of type \(D_{3h}\), and only the first one leaves the distinguished point at the North Pole untouched; those of type \(\mathcal{G}_{b}\) contain rotations by \(\pi \) around an axis and reflections through a plane containing that axis, hence are of type \(D_{2h}\), but none of these leaves the distinguished point at the North Pole untouched; those of type \(\mathcal{G}_{c}\) combine rotations and reflections through different axes, and none of them preserves the North Pole; and those of type \(\mathcal{G}_{d}\) consist just of reflections in a plane, hence are of type \(D_{h}\); the first three preserve the North Pole.

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Gaeta, G., Virga, E.G. The Symmetries of Octupolar Tensors. J Elast 135, 295–350 (2019). https://doi.org/10.1007/s10659-018-09722-8

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