To see piezoelectricity in a crystalline material, you must have a crystal symmetry which is non-centrosymmetric (i.e. broken inversion symmetry). The reverse is almost always true (i.e. broken inversion implies piezoelectricity), except for the single exception of crystals in the point group 432 (space groups 207-214). Crystals with this symmetry have broken inversion symmetry, but cannot have piezoelectricity.

Can someone give an intuitive explanation about what is so special about point group 432 compared to all the others? Why doesn't it display piezoelectricity? Ideally, pictures should be included in the explanation.

For example, consider the structure of AsH$_3$, which falls under point group 432, does not exhibit piezoelectricity despite breaking inversion symmetry.

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3 Answers 3


A quote from "Piezoelectricity and Crystal Symmetry" by Kholkin et al.:"According to the definition of the piezoelectric effect, all components of the piezoelectric tensor should vanish in crystals possessing the center of symmetry. In the remaining 21 noncentrosymmetric crystallographic classes, the piezoelectricity may exist, except for the cubic class 432, where the piezoelectric charges developed along the <111> polar axes cancel each other". Intuitively, for a crystal to be piezoelectric it needs a polar axis, i.e. an axis with different structures at opposite sides. This rules out all centrosymmetric crystals. Among non-centrosymmetric crystals, only 432 does not have a polar axis.

  • $\begingroup$ Would you be able to draw how these piezoelectric charges "cancel out"? I accepted your answer but realized I am still having trouble visualizing this $\endgroup$
    – KF Gauss
    Jul 4, 2022 at 18:54
  • $\begingroup$ @KFGauss. 432 Is not centrosymmetric , but there are enough rotational symmetries to kill piezoelectricy. Think chirality. The key point, illustrated by the gyroid, is that absence of a polar axis does not imply central symmetry. $\endgroup$ Jul 5, 2022 at 7:34

I'm not sure that there is a nice intuitive explanation for this - if there is I don't know what it is.

Having said that, the calculations that show that the crystals belonging to point group 432 are not piezoelectric are discussed in chapter 7 of Nye's 'Physical Properties of Crystals' (I consulted the 2006 reprint of the 1985 edition).

Specifically, the exhaustive method for determining this is outlined in Chapter 7.4.2, starting on page 117 - the only complete example provided is for the class $\overline{4}2m$.

Table 8, on page 123, contains a summary of the results for all the crystal classes and shows that all the components of the piezoelecticity tensor vanish for the group 432 (note that instead of the third rank piezoelectricity tensor the results are presented in a simpler derived matrix form - this matrix is non-square, and therefore not a tensor).

If I have time I will try to write up the calculation, but I will have to remind myself how to read crystal stereograms first.


According to Neumann's principle the tensors representing a physical property of a crystal must be invariant under all symmetry transformations. A third rank tensor, such as the piezoelectric tensor, changes the sign under inversion symmetry, i.e., $d_{ijk}=-d_{ijk}=0$. Hence, we conclude that centrosymmetric crystals do not exhibit a piezoelectric effect.

In the case of non-centrosymmetric crystals, the effect is not prohibited by inversion symmetry. However, we still need to check the other crystal symmetries. The space groups 207-2014 have a high symmetry. They have 24 distinct symmetry operations. For example, the symmetry $C_{2x}$ requires all terms with an even number of $x$ to be zero. For example $d_{xxy}=-d_{xxy}=0$ or $d_{xxx}=-d_{xxx}=0$. $C_{2y}$ and $C_{2z}$ pose similar constraints. Now, we can still have terms of the form $d_{ijk}$ where $i\neq j \neq k$, but we also have many more symmetries that might prohibit these terms. Maybe you can find which symmetry operations prohibit these terms?

To summarize, the inversion symmetry operation itself does not prohibit the piezoelectric effect for these space groups. However, by considering all other symmetries of these large point groups, we still conclude that the piezoelectric effect vanishes.

Note that I have not double-checked your claim myself, but if it is correct, I believe this is the proper explanation.


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