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The Bravais lattice is a face-centered cubic lattice. Four inequivalent representatives of its lattice points in the conventional cubic unit cell, in fractional crystal coordinates, areBoundary copies in the drawing do not constitute additional independent lattice points. Attach the two-atom atomic basisto each lattice point. The resulting aluminium phosphide crystal structure is the zinc blende structure. For example, the four phosphorus positions reduced into the conventional unit cell are $(1/4,1/4,1/4)$, $(1/4,3/4,3/4)$, $(3/4,1/4,3/4)$, and $(3/4,3/4,1/4)$, reproducing the heights in the source drawing. There are four aluminium and four phosphorus chemical atoms per conventional unit cell, but only two chemical atoms per primitive unit cell.

This is a C-centered orthorhombic lattice. The translation $(1/2,1/2,0)$ takes each A molecule to another A molecule and each B molecule to another B molecule. Thus the conventional cell has two lattice points.

A two-molecule crystal motif can be chosen as A at $(0,0,0)$ and B at $(1/2,0,1/2)$. Attaching that same motif to both C-centered lattice points produces all four molecules, hence eight iodine atoms per conventional cell.

The centers alone appear face-centered, but $(1/2,0,1/2)$ exchanges A and B orientations. It is not a translation symmetry of the oriented molecular solid. An F-centered assignment would discard essential motif information.

A Bravais lattice is a set of equivalent points generated by integer combinations of three independent translation vectors. A crystal motif is the arrangement of atoms attached identically to every lattice point. Together they specify the periodic crystal structure: lattice translations repeat the motif without changing its internal arrangement.

A unit cell is a region whose translations tile the structure. A primitive unit cell contains one lattice point after sharing boundary points with neighboring cells. A conventional cell can contain more than one lattice point to display the crystal's symmetry with simple axes and angles; for example, a conventional face-centered cube has four lattice points but a less visually symmetric primitive cell has one.

Rotational symmetry compatible with a periodic three-dimensional lattice has order $1,2,3,4$ or $6$, as expressed by the crystallographic restriction theorem. A crystallographic fivefold axis is excluded from an ordinary periodic crystal. This statement concerns exact lattice symmetry; it does not exclude local fivefold atomic arrangements or nonperiodic ordered structures.

A motif can have fivefold symmetry. For example, repeat an identically oriented regular pentagonal molecular motif at every point of a square translation lattice. Each isolated motif has fivefold rotational symmetry, while the crystal structure as a whole does not: rotating about one motif fails to map the other motif centres onto lattice points. The symmetry of an atomic basis of a crystal need not be a symmetry of its Bravais lattice.

The Minkowski convex body theorem states that a centrally symmetric convex measurable set in ${\mathbb{R}}^d$ of volume greater than $2^d\mathrm{covol}(\Lambda )$ contains a nonzero point of the full lattice $\Lambda$. For a compact convex body, the weak inequality also suffices: apply the strict version to $(1+\epsilon )C$, let $\epsilon \downarrow 0$, and choose a constant subsequence of the resulting lattice points in the bounded set $2C$. Its nonzero value belongs to $C$ by closedness.

For $Q>1$, apply this borderline version with $\Lambda ={\mathbb{Z}}^2$ toThe linear transformation $(u,v)\mapsto (u,xu-v)$ has determinant of absolute value one, so this parallelogram has area four. A nonzero lattice point $(n,m)$ satisfies $|n|\le Q$ and $|nx-m|\le Q^{-1}$. If $n=0$, then $|m|<1$, giving the forbidden zero point; hence $n\ne 0$. It follows thatIf $x$ is irrational, these pairs cannot remain in a finite set as $Q\to \infty$: the positive errors $|nx-m|$ of any such finite set have a positive minimum, contradicting their bound $Q^{-1}$. If $x=p/q$ is rational in lowest terms with $q>0$, the pairs $(n,m)=(jq,jp)$ for $j\ge 1$ have zero error and are infinitely many distinct pairs. This proves the assertion for every real $x$.

Write $\tau =x+iy\in \mathfrak{h}$. Forthe imaginary part of the Möbius transformation isAmong the primitive integer pairs $(c,d)$, choose one minimizing the nonzero quantity $|c\tau +d|$. Such a minimum exists because only finitely many lattice points lie in a bounded region. Complete $(c,d)$ to a matrix $\gamma \in S{L}_2(\mathbb{Z})$. Then $\gamma \tau$ has maximal imaginary part in its modular group orbit.

Applying an integral translation does not change that imaginary part, so arrangeIf $\mathrm{Im}(\gamma \tau )<\sqrt{3}/2$, then $|\gamma \tau |<1$. The modular inversion $S:z\mapsto -1/z$ would givecontradicting maximality. Hence every orbit meets the stated region. This is the reduction to the standard modular region argument.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.

For a simple polygon whose vertices are lattice points in ${\mathbb{Z}}^2$, its area is $A=I+B/2-1$, where $I$ counts interior lattice points and $B$ counts boundary lattice points.

A member of a set of linearly independent vectors whose integer combinations generate every lattice point of a Bravais lattice. Such basis vectors specify translations; an atomic basis of a crystal specifies the chemical motif repeated under them.