Why is the Lattice of a crystal required to have at least as much symmetry as its motif?

Question

I know that a crystal structure is formed by the addition of a motif to a lattice (crystal structure = lattice + motif). I also know that an arbitrary lattice will in general exhibit certain symmetries (It must by definition have translation symmetry but it may also have rotational symmetries, reflections etc as well). An arbitrary motif on the other hand may also exhibit certain symmetries (i.e reflection, rotation etc) however it may only exhibit point symmetries and not translational symmetries. Additionally, the crystal structure itself will then exhibit certain symmetry (translational is required but it may exhibit further point symmetries as well).

What I don't understand is why the symmetries of these two entities (a lattice and a basis) have to be compatible with one another in order to form a valid crystal structure. In the first link provided below, it is stated that "If a motif has certain symmetry, the lattice must have at least that much symmetry." This statement is labelled as an all-important second law of crystallography. Yet it is left entirely unjustified. To illustrate this law, the link shows as an example a square lattice (with its characteristic 4 fold rotational symmetry) coupled with a motif that has 3-fold rotational symmetry. The combination of these two is shown below

The link states that this is an "impossible combination of a lattice and a motif". But what makes this combination impossible? It is clear that the combination of these two results in a structure that exhibits the same translational symmetry as its underlying lattice. The structure is periodic and even has a horizontal mirror line as well. So why is it deemed an invalid or impossible crystal structure ? It is also clear that the lattice has 4-fold rotational symmetry while the motif has 3-fold rotational yet the resulting structure has neither 4-fold nor 3-fold rotational symmetry. The combination of these two evidently destroys some of the symmetry. But so what? Why is nature not okay with the resulting structure?

I can extend my issue further by instead populating a square lattice with a motif consisting of a circle. In this case, the motif has an infinite-fold continuous rotational symmetry while the square-lattice certain does not. So according to the crystallographic law in question this is an invalid combination. If this is so, why can we have simple cubic crystals like $\alpha$ polonium? In this case, the motif is a polonium atom which is presumably spherically symmetric. Hence it should be incompatible with a cubic lattice according to the law in question. But we know that this is not the case. Clearly, something is amiss here. So why is it requirement that the lattice of a crystal have at least as much symmetry as its motif?

EDIT I think most of my issue can be resolved if the following can be answered:

1. what is the plane group of the structure in fig 11.29 (if it has a plane group)

2. If it doesn't have a plane group, why does it's symmetry group (its mirror plus square translational symmetry) not count as one of the 17 crystallographic plane groups

3. Does nature ever allow for the existance of crystals that do not have the symmetry of one of the 230 space groups (3d crystals) or 17 plane groups? If so, what is the use of the 230 space groups/17 plane groups?

link: https://geo.libretexts.org/Bookshelves/Geology/Mineralogy_(Perkins_et_al.)/11%3A_Crystallography/11.03%3A_New_Page/11.3.3%3A_How_May_Motifs_and_Lattices_Combine%3F

Answer 1

After inspection of the source you mention, I am not entirely sure about what they meant by impossible but here is my take. The structure they describe should be possible. However, because of the mismatch between the symmetries of the lattice and those of the motif, the correct way to describe it would be by differentiate each site by a different label. Say the blue site at the center is A and the three ones around are B, C and D. Doing so, four bravais lattices are superimposed. B,C and D are now treated as inequivalent sites and the rotational symmetries between B sites are presents and the same are those of the lattice (same argument for the C sites and the D sites). A good example to visualise this would be to examine the Lieb lattice or the Kagome lattice:

In the Lieb lattice above (from Phys. Rev. B 96, 054305), Bravais lattice is a square lattice and the motif in the unit cell is made of a Black site, one nearest neighbouring blue one and one nearest neighbouring blue site. The motif does not have the same symmetries as the Bravais lattice and as a result the correct description needs to be the superposition of three Bravais lattice (that is three inequivalent sites per unit cell).

In summary, the second law of crystallography tells that in the motif you mentioned, the three blue sites are not equivalent.

Answer 2

I think the source you cited made a slightly more complex story too short, and your example of Polonium structure clearly shows that something is missing.

From the point of view of the classification of the 2D isometries of the plane, the impossible combination of the square lattice and the equilateral triangular motif shown in the figure corresponds to one of the $17$ wallpaper groups, the 2D analogous of the $230$ space groups in 3D.

The criterion to classify the wallpaper or space groups is not based on the cited second law of crystallography the way it was stated in the link. It is instead based on the identification of symmetries left after combining the possible translational, point, and glide symmetries of the crystal structure (obtained by combining the Bravais lattice and the motif).

Looking at the figure, and with reference to the identification table in the Wikipedia page, one can recognize that

1. there is a reflection plane;
2. there is no rotational symmetry of the 2D structure (although the motif has a 3-fold rotational axis);
3. there is no glide axis off mirrors.

Therefore the wallpaper group is $pm$.

Why is the statement in the linked page "If a motif has a certain symmetry, the lattice must have at least that much symmetry?" If we are dealing with real drawings on the wallpaper, there is no compelling reason for the lattice point symmetry to include the point group of the motif. However, suppose we are dealing with groups of atoms. In that case, there will be an effect of the interaction with the neighbor groups, and such an interaction has the local symmetry induced by the Bravais lattice. For example, in the case of the impossible combination, it is unlike the threefold rotational axis of the motif would be preserved by the interaction. Nevertheless, in the presence of extremely strong bonds, we cannot exclude it. In every case, the symmetry group would remain $pm$.