Bravais types of lattices and Bravais classes

Souvignier, B.

doi:10.1107/97809553602060000921

International
Tables for
Crystallography
Volume A
Space-group symmetry
Edited by M. I. Aroyo

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International Tables for Crystallography (2016). Vol. A. ch. 1.3, pp. 34-37

Section 1.3.4.3. Bravais types of lattices and Bravais classes

B. Souvignier^a ^*

^a Radboud University Nijmegen, Faculty of Science, Mathematics and Computing Science, Institute for Mathematics, Astrophysics and Particle Physics, Postbus 9010, 6500 GL Nijmegen, The Netherlands
Correspondence e-mail: souvi@math.ru.nl

1.3.4.3. Bravais types of lattices and Bravais classes

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In the classification of space groups into geometric crystal classes, only the point-group part is considered and the translation lattice is ignored. It is natural that the converse point of view is also adopted, where space groups are grouped together according to their translation lattices, irrespective of what the point groups are.

We have already seen that a lattice can be characterized by its metric tensor, containing the scalar products of a primitive basis. If a point group $[{\cal P}]$ acts on a lattice $[{\bf L}]$ , it fixes the metric tensor $[{\bi G}]$ of $[{\bf L}]$ , i.e. $[{\bi W}^{\rm T} \cdot {\bi G} \cdot {\bi W} = {\bi G} ]$ for all $[{\bi W}]$ in $[{\cal P}]$ and is thus a subgroup of the Bravais group $[Aut({\bf L})]$ of $[{\bf L}]$ . Also, a matrix group $[{\cal B}]$ is called a Bravais group if it is the Bravais group $[Aut({\bf L})]$ for some lattice $[{\bf L}]$ . The Bravais groups govern the classification of lattices.

Definition

Two lattices $[{\bf L}]$ and $[{\bf L}']$ belong to the same Bravais type of lattices if their Bravais groups $[Aut({\bf L})]$ and $[Aut({\bf L}')]$ are the same matrix group when written with respect to suitable primitive bases of $[{\bf L}]$ and $[{\bf L}']$ .

Note that in order to have the same Bravais group, the metric tensors of the two lattices $[{\bf L}]$ and $[{\bf L}']$ do not have to be the same or scalings of each other.

Example

The mineral rutile (TiO₂) has a space group of type [P4_2/mnm] (136) with a primitive tetragonal cell with cell parameters a = b = 4.594 Å and c = 2.959 Å. The metric tensor of the translation lattice L is therefore $[{\bi G} = \pmatrix{ 4.594^2 & 0 & 0 \cr 0 & 4.594^2 & 0 \cr 0 & 0 & 2.959^2 } ]$ and the Bravais group of the lattice is generated by the fourfold rotation $[\pmatrix{ 0 & -1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 }]$ around the z axis, the reflection $[\pmatrix{ -1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 }]$ in the plane x = 0 and the reflection $[\pmatrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & -1 } ]$ in the plane z = 0.

The silicate mineral cristobalite also has (at low temperatures) a primitive tetragonal cell with a = b = 4.971 Å and c = 6.928 Å, and the space-group type is [P4_12_12] (92). In this case the metric tensor of the translation lattice $[{\bf L}' ]$ is $[{\bi G}' = \pmatrix{ 4.971^2 & 0 & 0 \cr 0 & 4.971^2 & 0 \cr 0 & 0 & 6.928^2 } ]$ and one checks that the Bravais group of $[{\bf L}']$ is precisely the same as that of L. Therefore, the translation lattices L for rutile and $[{\bf L}' ]$ for cristobalite belong to the same Bravais type of lattices.

The different Bravais types of lattices, their cell parameters and metric tensors are displayed in Tables 3.1.2.1 (dimension 2) and 3.1.2.2 (dimension 3): in dimension 2 there are 5 Bravais types and in dimension 3 there are 14 Bravais types of lattices.

It is crucial for the classification of lattices via their Bravais groups that one works with primitive bases, because a primitive and a body-centred cubic lattice have the same automorphisms when written with respect to the conventional cubic basis, but are clearly different types of lattices.

Example

The silicate mineral zircon (ZrSiO₄) has a body-centred tetragonal cell with cell parameters a = b = 6.607 Å and c = 5.982 Å. The body-centred translation lattice $[{\bf L}']$ is spanned by the primitive tetragonal lattice $[{\bf L}]$ with basis $[{\bf a}, {\bf b}, {\bf c} ]$ with $[\alpha = \beta = \gamma = 90^\circ]$ and the centring vector $[{\bf v} = \textstyle{{1}\over{2}} ({\bf a} + {\bf b} + {\bf c})]$ . A primitive basis of $[{\bf L}']$ is obtained as $[({\bf a}', {\bf b}', {\bf c}') = ({\bf a}, {\bf b}, {\bf c}) {\bi P} ]$ with $[{\bi P} = {{1}\over{2}} \pmatrix{ -1 & 1 & 1 \cr 1 & -1 & 1 \cr 1 & 1 & -1 } ,]$ i.e. $[{\bf a}' = \textstyle{{1}\over{2}} (-{\bf a} + {\bf b} + {\bf c}) = -{\bf a} + {\bf v} ]$ , $[{\bf b}' = \textstyle{{1}\over{2}} ({\bf a} - {\bf b} + {\bf c}) = -{\bf b} + {\bf v} ]$ , $[{\bf c}' = \textstyle{{1}\over{2}} ({\bf a} + {\bf b} - {\bf c}) = -{\bf c} + {\bf v} ]$ and the metric tensor $[{\bi G}']$ of $[{\bf L}']$ with respect to the primitive basis $[{\bf a}', {\bf b}', {\bf c}']$ is $[ \eqalign{{\bi G}' &= {\bi P}^{\rm T} \pmatrix{ 6.607^2 & 0 & 0 \cr 0 & 6.607^2 & 0 \cr 0 & 0 & 5.982^2 } {\bi P} \cr&= \pmatrix{ 5.547^2 & -12.880 & -8.946 \cr -12.880 & 5.547^2 & -8.946 \cr -8.946 & -8.946 & 5.547^2 }. }]$ The Bravais group of the primitive tetragonal lattice $[{\bf L}]$ is generated (as in the previous example) by $[\displaylines{{\bi W}_1 = \pmatrix{ 0 & -1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 },\quad {\bi W}_2 = \pmatrix{ -1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 } \cr {\rm and} \ {\bi W}_3 = \pmatrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & -1 } ,}]$ and these matrices also generate the Bravais group of the body-centred tetragonal lattice $[{\bf L}']$ , but written with respect to the primitive basis $[{\bf a}', {\bf b}', {\bf c}']$ these matrices are transformed to $[\eqalign{{\bi W}_1' &= {\bi P}^{-1} {\bi W}_1 {\bi P} = \pmatrix{ 0 & 1 & 0 \cr 0 & 1 & -1 \cr -1 & 1 & 0 } ,\cr {\bi W}_2' &= {\bi P}^{-1} {\bi W}_2 {\bi P} = \pmatrix{ 1 & 0 & 0 \cr 1 & 0 & -1 \cr 1 & -1 & 0 }\ {\rm and}\cr {\bi W}_3' &= {\bi P}^{-1} {\bi W}_3 {\bi P} = \pmatrix{ 0 & -1 & 1 \cr -1 & 0 & 1 \cr 0 & 0 & 1 } .}]$

That the primitive and the body-centred tetragonal lattices have different types ultimately follows from the fact that the body-centred lattice $[{\bf L}']$ does not have a primitive basis consisting of vectors $[{\bf a}'', {\bf b}'', {\bf c}'' ]$ which are pairwise perpendicular and such that $[{\bf a}'']$ and $[{\bf b}'']$ have the same length. This would be required to have the matrices $[{\bi W}_1]$ , $[{\bi W}_2]$ and $[{\bi W}_3]$ in the Bravais group of $[{\bf L}']$ .

As we have seen, the metric tensors of lattices belonging to the same Bravais type need not be the same, but if they are written with respect to suitable bases they are found to have the same structure, differing only in the specific values for certain free parameters.

Definition

Let $[{\bf L}]$ be a lattice with metric tensor $[{\bi G}]$ with respect to a primitive basis and let $[{\cal B}]$ = $[Aut({\bf L})]$ = $[ \{ {\bi W} \in {\rm GL}_3({\bb Z}) \mid {\bi W}^{\rm T} \cdot {\bi G} \cdot {\bi W} = {\bi G} \} ]$ be the Bravais group of $[{\bf L}]$ . Then $[\displaylines{ {\bf M}({\cal B}): = \{ {\bi G}' \ {\rm symmetric\ 3 \times 3\ matrix} \mid\hfill\cr\hfill {\bi W}^{\rm T} \cdot {\bi G}' \cdot {\bi W} = {\bi G}' \ {\rm for\ all }\ {\bi W} \in {\cal B} \} }]$ is called the space of metric tensors of $[{\cal B}]$ . The dimension of $[{\bf M}({\cal B})]$ is called the number of free parameters of the lattice $[{\bf L}]$ .

Analogously, for an arbitrary integral matrix group $[{\cal P}]$ , $[\displaylines{\quad{\bf M}({\cal P}): = \{ {\bi G}'\ {\rm symmetric\ 3 \times 3\ matrix} \mid \hfill\cr\hfill{\bi W}^{\rm T} \cdot {\bi G}' \cdot {\bi W} = {\bi G}' \ {\rm for\ all }\ {\bi W} \in {\cal P} \} }]$ is called the space of metric tensors of $[{\cal P}]$ . If $[\dim {\bf M}({\cal P}')]$ = $[\dim {\bf M}({\cal P})]$ for a subgroup $[{\cal P}']$ of $[{\cal P}]$ , the spaces of metric tensors are the same for both groups and one says that $[{\cal P}']$ does not act on a more general lattice than $[{\cal P}]$ does.

It is clear that $[{\bf M}({\cal B})]$ contains in particular the metric tensor $[{\bi G}]$ of the lattice $[{\bf L}]$ of which $[{\cal B}]$ is the Bravais group. Moreover, $[{\cal B}]$ is a subgroup of the Bravais group of every lattice with metric tensor in $[{\bf M}({\cal B}) ]$ .

Example

Let $[{\bf L}]$ be a lattice with metric tensor $[\pmatrix{ 17 & 0 & 0 \cr 0 & 17 & 0 \cr 0 & 0 & 42 } ,]$ then $[{\bf L}]$ is a tetragonal lattice with Bravais group $[{\cal B}]$ of type 4/mmm generated by the fourfold rotation $[{\bi W}_1 = \pmatrix{ 0 & -1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 }]$ and the reflections $[{\bi W}_2 = \pmatrix{ -1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 } \ {\rm and}\ {\bi W}_3 = \pmatrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & -1 }.]$ The space of metric tensors of $[{\cal B}]$ is $[ {\bf M}({\cal B}) = \left\{ \pmatrix{ g_{11} & 0 & 0 \cr 0 & g_{11} & 0 \cr 0 & 0 & g_{33} } \mid g_{11},g_{33} \in {\bb R} \right\} ]$ and the number of free parameters of $[{\bf L}]$ is 2.

For every lattice $[{\bf L}']$ with metric tensor $[{\bi G}']$ in $[{\bf M}({\cal B})]$ such that $[g_{11} \neq g_{33}]$ , one can check that the Bravais group of $[{\bf L}']$ is equal to $[{\cal B} ]$ , hence these lattices belong to the same Bravais type of lattices as $[{\bf L}]$ . On the other hand, if it happens that $[g_{11} = g_{33}]$ in the metric tensor $[{\bi G}']$ of a lattice $[{\bf L}']$ , then the Bravais group of $[{\bf L}']$ is the full cubic point group of type $[m\bar{3}m]$ and $[{\cal B}]$ is a proper subgroup of the Bravais group of $[{\bf L}']$ . In this case the lattice $[{\bf L}']$ is of a different Bravais type to $[{\bf L}]$ , namely cubic.

The subgroup $[{\cal P}]$ of $[{\cal B}]$ generated only by the fourfold rotation $[{\bi W}_1]$ has the same space of metric tensors as $[{\cal B}]$ , thus this subgroup acts on the same types of lattices as $[{\cal B}]$ (i.e. tetragonal lattices). On the other hand, for the subgroup $[{\cal P}']$ of $[{\cal B}]$ generated by the reflections $[{\bi W}_2 ]$ and $[{\bi W}_3]$ , the space of metric tensors is $[ {\bf M}({\cal P}') = \left\{ \pmatrix{ g_{11} & 0 & 0 \cr 0 & g_{22} & 0 \cr 0 & 0 & g_{33} } \mid g_{11}, g_{22}, g_{33} \in {\bb R} \right\} ]$ and is thus of dimension 3. This shows that the subgroup $[{\cal P}']$ acts on more general lattices than $[{\cal B}]$ , namely on orthorhombic lattices.

Remark : The metric tensor of a lattice basis is a positive definite² matrix. It is clear that not all matrices in $[{\bf M}({\cal B})]$ are positive definite [if $[{\bi G} \in {\bf M}({\cal B})]$ is positive definite, then $[-{\bi G}]$ is certainly not positive definite], but the different geometries of lattices on which $[{\cal B}]$ acts are represented precisely by the positive definite metric tensors in $[{\bf M}({\cal B})]$ .

The space of metric tensors obtained from a lattice can be interpreted as an expression of the metric tensor with general entries, i.e. as a generic metric tensor describing the different lattices within the same Bravais type. Special choices for the entries may lead to lattices with accidental higher symmetry, which is in fact a common phenomenon in phase transitions caused by changes of temperature or pressure.

One says that the translation lattice $[{\bf L}]$ of a space group $[{\cal G}]$ with point group $[{\cal P}]$ has a specialized metric if the dimension of the space of metric tensors of $[{\cal B} = Aut({\bf L}) ]$ is smaller than the dimension of the space of metric tensors of $[{\cal P}]$ . Viewed from a slightly different angle, a specialized metric occurs if the location of the atoms within the unit cell reduces the symmetry of the translation lattice to that of a different lattice type.

Example

A space group $[{\cal G}]$ of type P2/m (10) with cell parameters a = 4.4, b = 5.5, c = 6.6 Å, $[\alpha = \beta = \gamma = 90^\circ]$ has a specialized metric, because the point group $[{\cal P}]$ of type 2/m is generated by $[{\bi W} = \pmatrix{ -1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & -1 }]$ and $[{\bi -I}]$ , and has $[{\bf M}({\cal P}) = \left\{ \pmatrix{ g_{11} & 0 & g_{13} \cr 0 & g_{22} & 0 \cr g_{13} & 0 & g_{33} } \mid g_{11}, g_{22}, g_{33}, g_{13} \in {\bb R} \right\} ]$ as its space of metric tensors, which is of dimension 4. The lattice $[{\bf L}]$ with the given cell parameters, however, is orthorhombic, since the free parameter $[g_{13}]$ is specialized to $[g_{13} = 0 ]$ . The automorphism group $[Aut({\bf L})]$ is of type mmm and has a space of metric tensors of dimension 3, namely $[\left\{ \pmatrix{ g_{11} & 0 & 0 \cr 0 & g_{22} & 0 \cr 0 & 0 & g_{33} } \mid g_{11}, g_{22}, g_{33} \in {\bb R} \right\} .]$

The higher symmetry of the translation lattice would, for example, be destroyed by an atomic configuration compatible with the lattice and represented by only two atoms in the unit cell located at 0.17, 1/2, 0.42 and 0.83, 1/2, 0.58. The two atoms are related by a twofold rotation around the b axis, which indicates the invariance of the configuration under twofold rotations with axes parallel to b, but in contrast to the lattice L, the atomic configuration is not compatible with rotations around the a or the c axes.

By looking at the spaces of metric tensors, space groups can be classified according to the Bravais types of their translation lattices, without suffering from complications due to specialized metrics.

Definition

Let $[{\bf L}]$ be a lattice with metric tensor $[{\bi G}]$ and Bravais group $[{\cal B} = Aut({\bf L})]$ and let $[{\bf M}({\cal B}) ]$ be the space of metric tensors associated to $[{\bf L}]$ . Then those space groups $[{\cal G}]$ form the Bravais class corresponding to the Bravais type of $[{\bf L}]$ for which $[{\bf M}({\cal P}) = {\bf M}({\cal B})]$ when the point group $[{\cal P}]$ of $[{\cal G}]$ is written with respect to a suitable primitive basis of the translation lattice of $[\cal G]$ . The names for the Bravais classes are the same as those for the corresponding Bravais types of lattices.

The Bravais groups of lattices provide a link between lattices and point groups, the two building blocks of space groups. However, although the Bravais group of a lattice is simply a matrix group, the fact that it is expressed with respect to a primitive basis and fixes the metric tensor of the lattice preserves the necessary information about the lattice. When the Bravais group is regarded as a point group, the information about the lattice is lost, since point groups can be written with respect to an arbitrary basis. In order to distinguish Bravais groups of lattices at the level of point groups and geometric crystal classes, the concept of a holohedry is introduced.

Definition

The geometric crystal class of a point group $[{\cal P}]$ is called a holohedry (or lattice point group, cf. Chapters 3.1 and 3.3 ) if $[{\cal P}]$ is the Bravais group of some lattice $[{\bf L}]$ .

Example

Let $[{\cal P}]$ be the point group of type $[\bar{3}m]$ generated by the threefold rotoinversion $[{\bi W}_1 = \pmatrix{ 0 & 1 & 0 \cr -1 & 1 & 0 \cr 0 & 0 & -1 }]$ around the z axis and the twofold rotation $[{\bi W}_2 = \pmatrix{ 1 & -1 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & -1 },]$ expressed with respect to the conventional basis $[{\bf a}, {\bf b}, {\bf c} ]$ of a hexagonal lattice. The group $[{\cal P}]$ is not the Bravais group of the lattice $[{\bf L}]$ spanned by $[{\bf a}, {\bf b}, {\bf c} ]$ because this lattice also allows a sixfold rotation around the z axis, which is not contained in $[{\cal P}]$ . But $[{\cal P} ]$ also acts on the rhombohedrally centred lattice $[{\bf L}']$ with primitive basis $[{\bf a}' = \textstyle{{1}\over{3}} (2 {\bf a} + {\bf b} + {\bf c}) ]$ , $[{\bf b}' = \textstyle{{1}\over{3}} (-{\bf a} + {\bf b} + {\bf c})]$ , $[{\bf c}' = \textstyle{{1}\over{3}} (-{\bf a} - 2 {\bf b} + {\bf c})]$ . With respect to the basis $[{\bf a}', {\bf b}', {\bf c}']$ the rotoinversion and twofold rotation are transformed to $[{\bi W}_1' = \pmatrix{ 0 & 0 & -1 \cr -1 & 0 & 0 \cr 0 & -1 & 0 } \ {\rm and} \ {\bi W}_2' = \pmatrix{ 0 & -1 & 0 \cr -1 & 0 & 0 \cr 0 & 0 & -1 } ,]$ and these matrices indeed generate the Bravais group of $[{\bf L}' ]$ . The geometric crystal class with symbol $[\bar{3}m]$ is therefore a holohedry.

Note that in dimension 3 the above is actually the only example of a geometric crystal class in which the point groups are Bravais groups for some but not for all the lattices on which they act. In all other cases, each matrix group $[{\cal P}]$ corresponding to a holohedry is actually the Bravais group of the lattice spanned by the basis with respect to which $[{\cal P}]$ is written.

References

International Tables for Crystallography (2016). Vol. A. ch. 1.3, pp. 34-37