Algebraic determination of asymmetric unit (aka irreducible wedge) in Brillouin zone of lattice In Solid State physics the reciprocal space is of utmost importance to predict the band structure and thus most of the electrical transport parameters like effective mass, etc.
The First Brillouin Zone of a certain crystal lattice with its symmetries specified by its space and point groups, is defined as the volume whose points are closer to a lattice point in the reciprocal lattice than to all other reciprocal lattice points (thus the equivalent to the Wigner Seitz cell in real space).
In the International Tables for Crystallography and also on the excellent Bilbao Crystallographic Server (listed in the row "GP"), the asymmetric unit (aka irreducible wedge) is specified as the polyhedron which is the first Brillouin zone reduced by all of the symmetries in the point group of the lattice. The geometry of this zone can be calculated from the given equations by doing a vertex enumeration and can be described by the inequality relation
$$\textbf{m}.\vec{r}-\vec{b}\leq0$$
with $\textbf{m}$ being the row matrix of the normal vectors of the surface planes and $\vec{b}$ being the distance of the plane to the coordinate center.
A similiar set of inequality relations can be defined for the full first Brillouin zone.
Now for my question: One can easily show that by applying all point group operations of the related point group of the lattice to the asymmetric unit we can get back to the first Brillouin Zone of the lattice. However is there an algebraic way how to deduce the asymmetric unit through the space group operations applied to the first Brillouin zone (thus the inverse operation applied on the inequality relation describing the first Brillouin zone)?
 A: Yes, there is. Let me start with a very general notion from metric geometry. Let $(X,d)$ be a metric space, $\Gamma$ a discrete group of isometries of $X$ by which I mean that all elements of $\Gamma$ preserve the metric $d$ and if $\gamma_i\in \Gamma$ is any infinite sequence of distinct elements then for all $x\in X$
$$
\lim_{i\to\infty} d(x, \gamma_i x)=\infty. 
$$
Examples of this setup you should think about are:
a. "Point-groups" of a Euclidean space $E^n$, $(X,d)=E^n$.
b. Finite subgroups of the orthogonal group $O(n)$, where $(X,d)= S^{n-1}$ is the unit sphere with the standard angular metric.
Back to the general setting: For each $x\in X$ its orbit $Y=\Gamma x=\{\gamma x: \gamma\in \Gamma\}$ is a discrete closed subset of $X$, hence, one has the associated Voronoi tiling of $X$: Tiles $V_y$ "centered" at $y\in Y$ are given by the inequalities
$$
V_y= \{z\in X: d(z,y)\le d(z, y') ~~\forall y'\in Y\setminus \{y\}\}.  
$$
In the two examples above, each tile is a Euclidean/spherical polyhedron bounded by Euclidean (spherical) hyperplanes equidistance from some pairs $y, y'$. Such tiles are permuted transitively by the group $\Gamma$, hence, for every tile $V_y$ we have
$$
\Gamma V_y= X. 
$$
The action of $\Gamma$ on the tiles need not be simply-transitive: It is simply transitive if and only if the stabilizer $\Gamma_x$ (the "point-group" or the "isotropy group") of $x$ consists only of the identity transformation $1$,
$$
\Gamma_x= \{\gamma\in \Gamma: \gamma x=x\}.  
$$
For instance, if $\Gamma$ is a lattice in $E^n$ (consists only of translations) then $\Gamma_x=\{1\}$ for all $x\in E^n$ and, hence, the action of $\Gamma$ on the set of tiles is simply-transitive. The tile $V_x$ is also known as the Brillouin zone of $x$. A mathematician would call it a fundamental set of the action of $\Gamma$ on $X$.
Even if a group of Euclidean/spherical isometries $\Gamma$ contains nontrivial elements fixing some points in $X$, using discreteness of $\Gamma$ one can always find $x$ such that $\Gamma_x=\{1\}$, in which case $V_x$ is also an irreducible Brillouin zone of $x$, which means that for every closed subset $C\subsetneq V_x$, $\Gamma C\ne X$.
The question raised by OP is how to find a "subtile" $V_x'\subset V_x$ which would give an irreducible Brillouin zone of $x$ for general $x$, even one with nontrivial $\Gamma_x$.
A mathematician would call such $V_x'$ a fundamental domain of the action of $\Gamma$ on $X$:
Definition. Let $\Gamma$ be a discrete group of isometries of a metric space $(X,d)$.  A subset $D\subset X$ is called a fundamental domain for the action of $\Gamma$ on $X$ if the following conditions are met:

*

*$\Gamma D=X$.


*If $y$ is an interior point of $D$ then for all $\gamma\in \Gamma \setminus \{1\}$, $\gamma y\notin D$.


*Few more technical conditions which I will ignore.
These conditions imply that if $C\subsetneq D$ ($C$ is closed), then $\Gamma C\ne X$, which is what you want from an irreducible Brillouin zone.
Now, back to the discrete groups of Euclidean isometries. Assume now that $\Gamma_x< \Gamma$ is a nontrivial subgroup. Since $\Gamma$ is discrete, $\Gamma_x$ is a finite subgroup of the orthogonal group. Let $V_x$ be the Voronoi cell of $x$ as above. The subgroup $\Gamma_x$ fixes $x$ and, hence, preserves $V_x$.  I will now pick a point $z\in E^n$ not fixed by an element of $\Gamma_x$ (any generic choice of $z$ would work; for sure, you do not want $z=x$) and let $W_z$ denote its Voronoi tile with respect to the subgroup $\Gamma_z$:
$$
W_z=\{ w\in E^n: d(w, z)\le d(w, \gamma z)~~ \forall \gamma \in \Gamma_x \setminus \{1\} \}. 
$$
This domain is a convex polyhedral cone with tip at $x$. Now, take
$$
V'_x= V_x\cap W_z. 
$$
It is a nice exercise to see that $V'_x$ will be a fundamental domain (actually, a convex polyhedron) of $\Gamma$, equivalently, an irreducible Brillouin zone of $x$. The polyhedron $V'_x$ will have $x$ as one of its boundary points (frequently, but not always, $x$ is a vertex of $V'_x$).
Another way to define $W_z$ is to use spherical geometry: Take a generic unit vector $u$ (where I think of $x$ as $0$ in ${\mathbb R}^n$) and let $S_u$ be the Voronoi tile of $u$  in the unit sphere $S^{n-1}$ with respect to the subset $\Gamma_x u\subset S^{n-1}$. Then $S_u$ is a convex spherical polyhedron. It equals the intersection of the cone $W_u$ with  $S^{n-1}$. You can also recover $W_u$ from $S_u$ as the set of multiples $t v$, $t\ge 0, v\in S_u$.
A side remark: The definition of Voroinoi tile $V_x$ requires one to go through the entire list of elements of $\Gamma$ which could be infinite or finite but very large (hence, impractical). There is a trick which allows one (e.g. in Euclidean or spherical setting) to reduce (algorithmically) the construction of $V_x$ to finitely many elements of $\Gamma$ (or simply a smaller subset of $\Gamma$ if the latter is finite). But this is another story...
