Why do the symmetries of a simple cubic lattice not include a 4-fold rotational axis through the lattice points?

When I read about classifying lattices based on symmetry, for the simple cubic lattice (or, as Wikipedia calls it, a primitive cubic lattice), there are only three 4-fold axes of rotational symmetry, which go through each face centre $$-$$ but the classification does not include 4-fold rotations with axes that go through the lattice points, i.e., such that the edges of the cube are rotational-symmetry axes.

Could someone explain why these symmetries are not included?

• I think you are assuming too much domain knowledge on the part of the reader. I would suggest adding more info on what you mean by simple cubic lattice. Is it the same as in Wiki? What is the point group for this system? Once you know this, it should be sufficient to look at conjugacy classes (of Tetrahedral group, I guess). – Cryo Jan 2 at 20:29
• As it stands, you expect the reader to do much more work than you put into it - this is not polite. Add more work/info please – Cryo Jan 2 at 20:31
• What is a “4-fold axis”? – G. Smith Jan 2 at 20:51
• Oh I'm sorry about it I didn't write more cus I'm only starting to learn the introductory course on material science now. – 123 Jan 2 at 20:55

the cubic lattice is defined by the four 3-fold rotation axes along the main diagonals of the cube, not by the presence of 4-fold ones. This is the reason why you have primitive, face-centred and body centred cubic lattices (but not a base centred cubic lattice as in the orthorhombic system that would break the 3-fold rotation symmetry).

there are several cubic space groups where no 4-fold rotation symmetry is present; for example the cubic space group P23 (n° 195) has only 2- and 3-fold rotation axes, but 4-fold.

A simple cubic lattice is a Bravais lattice, i.e., it can be thought of as originating from the set of (infinite) translation of a cube of side $$a$$ along three orthogonal axes parallel to the cube edges, according to the formula $${\bf R} = n_1 a {\bf \hat x} + n_2 a {\bf \hat y}+ n_3 a {\bf \hat z},~~~~~~~~~~~~(n_1,n_2,n_3) \in {\mathbb Z}^3.$$ It is probably more useful for physical applications to consider the crystal structure made by a monoatomic base, i.e., the cubic cell containing exactly one atom.

It is important to understand that the vectors $${\bf R}$$ represent displacements of the cubic cell whatever is its location and wherever the atomic base has been located. In particular, null displacement doesn't need to coincide with the atomic position or with a vertex or any other high-symmetry point of the cubic unitary cell.

However, in addition to the translational symmetry represented by the Bravais lattice, atomic positions in crystal structures may also have other symmetries. In particular, point symmetries correspond to space transformations leaving at least one fixed point. In the general case, it is also possible to have symmetry transformations made by special compositions of non-Bravais translations and point symmetries (skew axes or glide planes).

Of course, point transformations must be compatible with the Bravais lattice of the translations. This is the reason why some of the possible rotational symmetries are absent (for example, only 2-, 3-, 4-, and 6-fold axes are allowed in a crystal). To make evident this compatibility, it is handy to localize the atoms at points where the full point symmetry is evident in the case of a monoatomic structure. Even better, one can build a special (generally non-cubic) unit cell, the Wigner-Seitz (WS) cell.

In the case of a simple cubic crystal structure, the WS cell is a cube equal to the unit cell, making explicit that the point symmetries of that lattice are the symmetries of the cube.

Therefore there are three 4-fold axes, and it is equivalent to consider each of these axes as passing through an edge of one cell of the lattice or the center of one cell (aligned with the edges). The basic reason is that there is only one point symmetry of that kind.