International
Tables for
Crystallography
Volume D
Physical properties of crystals
Edited by A. Authier

International Tables for Crystallography (2013). Vol. D, ch. 1.1, pp. 5-7

Section 1.1.2. Basic properties of vector spaces

A. Authiera*

aInstitut de Minéralogie et de Physique des Milieux Condensés, 4 Place Jussieu, 75005 Paris, France
Correspondence e-mail: aauthier@wanadoo.fr

1.1.2. Basic properties of vector spaces

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[The reader may also refer to Section 1.1.4[link] of Volume B of International Tables for Crystallography (2008)[link].]

1.1.2.1. Change of basis

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Let us consider a vector space spanned by the set of n basis vectors [{\bf e}_{1}], [{\bf e}_{2}], [{\bf e}_{3},\ldots, {\bf e}_{n}]. The decomposition of a vector using this basis is written [{\bf x} = x^{i}{\bf e}_{i} \eqno(1.1.2.1)]using the Einstein convention. The interpretation of the position of the indices is given below. For the present, we shall use the simple rules:

  • (i) the index is a subscript when attached to basis vectors;

  • (ii) the index is a superscript when attached to the components. The components are numerical coordinates and are therefore dimensionless numbers.

Let us now consider a second basis, [{\bf e}'_{j}]. The vector x is independent of the choice of basis and it can be decomposed also in the second basis: [{\bf x} = x'^{i}{\bf e}'_{i}. \eqno(1.1.2.2)]

If [A\hskip1pt_{i}^{j}] and [B_{j}^{i}] are the transformation matrices between the bases [{\bf e}_{i}] and [{\bf e}'_{j}], the following relations hold between the two bases: [\left.\matrix{{\bf e}_{i} = A\hskip1pt_{i}^{j}{\bf e}'_{j}\semi\hfill &{\bf e}'_{j} = B_{j}^{i}{\bf e}_{i}\hfill \cr x^{i} = B_{j}^{i} x'^{j}\semi &x'^{j} = A\hskip1pt_{i}^{j} x^{i}\cr}\right\} \eqno(1.1.2.3)](summations over j and i, respectively). The matrices [A\hskip1pt_{i}^{j}] and [B_{j}^{i}] are inverse matrices: [A\hskip1pt_{i}^{j} B_{j}^{k} = \delta_{i}^{k} \eqno(1.1.2.4)](Kronecker symbol: [\delta_{i}^{k} = 0] if [i \neq k, = 1] if [i = k]).

Important Remark. The behaviour of the basis vectors and of the components of the vectors in a transformation are different. The roles of the matrices [A\hskip1pt_{i}^{j}] and [B_{j}^{i}] are opposite in each case. The components are said to be contravariant. Everything that transforms like a basis vector is covariant and is characterized by an inferior index. Everything that transforms like a component is contravariant and is characterized by a superior index. The property describing the way a mathematical body transforms under a change of basis is called variance.

1.1.2.2. Metric tensor

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We shall limit ourselves to a Euclidean space for which we have defined the scalar product. The analytical expression of the scalar product of two vectors [{\bf x} = x^{i}{\bf e}_{i}] and [{\bf y} = y\hskip 2pt^{j}{\bf e}_{j}] is [{\bf x} \cdot {\bf y} = x^{i}{\bf e}_{i} \cdot y\hskip 2pt^{j}{\bf e}_{j}.]Let us put [{\bf e}_{i} \cdot {\bf e}_{j} = g_{ij}. \eqno(1.1.2.5)]The nine components [g_{ij}] are called the components of the metric tensor. Its tensor nature will be shown in Section 1.1.3.6.1[link]. Owing to the commutativity of the scalar product, we have [g_{ij}= {\bf e}_{i} \cdot {\bf e}_{j} = {\bf e}_{j} \cdot {\bf e}_{i} = g_{ji}.]

The table of the components [g_{ij}] is therefore symmetrical. One of the definition properties of the scalar product is that if [{\bf x} \cdot {\bf y} = 0] for all x, then [{\bf y} = {\bf 0}]. This is translated as[x^{i}y\hskip 2pt^{j}g_{ij}= 0 \quad \forall x^{i}\ \Longrightarrow\ y\hskip 2pt^{j}g_{ij} = 0.]

In order that only the trivial solution [(y\hskip 2pt^{j} = 0)] exists, it is necessary that the determinant constructed from the [g_{ij}]'s is different from zero: [\Delta (g_{ij}) \neq 0.]This important property will be used in Section 1.1.2.4.1[link].

1.1.2.3. Orthonormal frames of coordinates – rotation matrix

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An orthonormal coordinate frame is characterized by the fact that [g_{ij} = \delta_{ij} \quad (= 0 \quad \hbox{if} \ \ i \neq j \ \hbox{ and } \ = 1 \ \hbox{ if } \ i = j). \eqno(1.1.2.6)]One deduces from this that the scalar product is written simply as [{\bf x} \cdot {\bf y} = x^{i}y\hskip 2pt^{j}g_{ij} = x^{i}y^{i}.]

Let us consider a change of basis between two orthonormal systems of coordinates: [{\bf e}_{i} = A\hskip1pt_{i}^{j}{\bf e}'_{j}.]Multiplying the two sides of this relation by [{\bf e}'_{j}], it follows that [{\bf e}_{i} \cdot {\bf e}'_{j} = A\hskip1pt_{i}^{j} {\bf e}'_{k} \cdot e'_{j} = A\hskip1pt_{i}^{j} g'_{kj} = A\hskip1pt_{i}^{j} \delta_{kj} \ \hbox{ (written correctly),}]which can also be written, if one notes that variance is not apparent in an orthonormal frame of coordinates and that the position of indices is therefore not important, as [{\bf e}_{i} \cdot {\bf e}'_{j} = A\hskip1pt_{i}^{j} \ \hbox{ (written incorrectly)}.]

The matrix coefficients, [A\hskip1pt_{i}^{j}], are the direction cosines of [{\bf e}'_{j}] with respect to the [{\bf e}_{i}] basis vectors. Similarly, we have [B_{j}^{i} = {\bf e}_{i} \cdot {\bf e}'_{j}]so that [A\hskip1pt_{i}^{j} = B_{j}^{i} \quad \hbox{or} \quad A = B^{T},]where T indicates transpose. It follows that [A = B^{T} \hbox{ and } A = B^{-1}]so that [\left. \matrix{A^{T} \ = A^{-1}\quad \Rightarrow\quad A^{T}A = I\cr B^{T} \ = B^{-1}\quad \Rightarrow\quad B^{T}B = I.\cr}\right\} \eqno(1.1.2.7)]The matrices A and B are unitary matrices or matrices of rotation and [\Delta (A)^{2} = \Delta (B)^{2} = 1 \Rightarrow \Delta (A) = \pm 1. \eqno(1.1.2.8)]

  • If [\Delta (A) = 1] the senses of the axes are not changed – proper rotation.

  • If [\Delta (A) = - 1] the senses of the axes are changed – improper rotation. (The right hand is transformed into a left hand.)

One can write for the coefficients [A\hskip1pt_{i}^{j}][A\hskip1pt_{i}^{j}B_{j}^{k} = \delta_{i}^{k};\quad A\hskip1pt_{i}^{j}A_{j}^{k} = \delta_{i}^{k},]giving six relations between the nine coefficients [A\hskip1pt_{i}^{j}]. There are thus three independent coefficients of the [3 \times 3] matrix A.

1.1.2.4. Covariant coordinates – dual or reciprocal space

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1.1.2.4.1. Covariant coordinates

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Using the developments (1.1.2.1)[link] and (1.1.2.5)[link], the scalar products of a vector x and of the basis vectors [{\bf e}_{i}] can be written [x_{i} = {\bf x} \cdot {\bf e}_{i} = x\hskip 1pt^{j}{\bf e}_{j} \cdot {\bf e}_{i} = x\hskip 1pt^{j}g_{ij}. \eqno(1.1.2.9)]The n quantities [x_{i}] are called covariant components, and we shall see the reason for this a little later. The relations (1.1.2.9)[link] can be considered as a system of equations of which the components [x\hskip 1pt^{j}] are the unknowns. One can solve it since [\Delta (g_{ij}) \neq 0] (see the end of Section 1.1.2.2[link]). It follows that [x\hskip 1pt^{j} = x_{i}g^{ij} \eqno(1.1.2.10)]with [g^{ij}g_{jk} = \delta_{k}^{i}. \eqno(1.1.2.11)]

The table of the [g^{ij}]'s is the inverse of the table of the [g_{ij}]'s. Let us now take up the development of x with respect to the basis [{\bf e}_{i}]: [{\bf x} = x^{i}{\bf e}_{i}.]

Let us replace [x^{i}] by the expression (1.1.2.10)[link]: [{\bf x} = x_{j}g^{ij}{\bf e}_{i}, \eqno(1.1.2.12)]and let us introduce the set of n vectors [{\bf e}\hskip 1pt^{j} = g^{ij}{\bf e}_{i} \eqno(1.1.2.13)]which span the space [E^{n}\,\,(j = 1, \ldots, n)]. This set of n vectors forms a basis since (1.1.2.12)[link] can be written with the aid of (1.1.2.13)[link] as [{\bf x} = x_{j}{\bf e}\hskip 1pt^{j}. \eqno(1.1.2.14)]

The [x_{j}]'s are the components of x in the basis [{\bf e}\hskip 1pt^{j}]. This basis is called the dual basis. By using (1.1.2.11)[link] and (1.1.2.13)[link], one can show in the same way that [{\bf e}_{j} = g_{ij}{\bf e}\hskip 1pt^{j}. \eqno(1.1.2.15)]

It can be shown that the basis vectors [{\bf e}\hskip 1pt^{j}] transform in a change of basis like the components [x\hskip 1pt^{j}] of the physical space. They are therefore contravariant. In a similar way, the components [x_{j}] of a vector x with respect to the basis [{\bf e}\hskip 1pt^{j}] transform in a change of basis like the basis vectors in direct space, [{\bf e}_{j}]; they are therefore covariant: [\left. \matrix{{\bf e}\hskip 1pt^{j} = B\hskip1pt_{k}^{j} {\bf e}'^{k}\semi &{\bf e}'^{k} = A_{j}^{k} {\bf e}\hskip 1pt^{j}\cr x_{i} = A\hskip1pt_{i}^{j} x'_{j}\semi &x'_{j} = B_{j}^{i}x_{i}.\cr}\right\} \eqno(1.1.2.16)]

1.1.2.4.2. Reciprocal space

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Let us take the scalar products of a covariant vector [{\bf e}_i] and a contravariant vector [{\bf e}\hskip 1pt^{j}]: [{\bf e}_{i} \cdot {\bf e}\hskip 1pt^{j} = {\bf e}_{i} \cdot g\hskip 1pt^{jk}{\bf e}_{k} = {\bf e}_{i} \cdot {\bf e}_{k} g\hskip 1pt^{jk} = g_{ik}g\hskip 1pt^{jk} = \delta\hskip1pt_{i}^{j}][using expressions (1.1.2.5)[link], (1.1.2.11)[link] and (1.1.2.13)[link]].

The relation we obtain, [{\bf e}_{i} \cdot {\bf e}\hskip1pt^{j} = \delta\hskip1pt_{i}^{j}], is identical to the relations defining the reciprocal lattice in crystallography; the reciprocal basis then is identical to the dual basis [{\bf e}^{i}].

1.1.2.4.3. Properties of the metric tensor

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In a change of basis, following (1.1.2.3)[link] and (1.1.2.5)[link], the [g_{ij}]'s transform according to [\left. \matrix{g_{ij} \ = A_{i}^{k} A_{j}^{m} g'_{km}\cr g'_{ij} \ = B_{i}^{k} B_{j}^{m} g_{km}.\cr}\right\} \eqno(1.1.2.17)]Let us now consider the scalar products, [{\bf e}^{i} \cdot {\bf e}\hskip 1pt^{j}], of two contravariant basis vectors. Using (1.1.2.11)[link] and (1.1.2.13)[link], it can be shown that [{\bf e}^{i} \cdot {\bf e}\hskip 1pt^{j} = g^{ij}. \eqno(1.1.2.18)]

In a change of basis, following (1.1.2.16)[link], the [g^{ij}]'s transform according to [\left. \matrix{g^{ij} \ = B\hskip1pt_{k}^{i} B\hskip1pt_{m}^{j} g'^{km}\cr g'^{ij} \ = A_{k}^{i} A\hskip1pt_{m}^{j} g^{km}.\cr}\right\} \eqno(1.1.2.19)]The volumes V ′ and V of the cells built on the basis vectors [{\bf e}'_{i}] and [{\bf e}_{i}], respectively, are given by the triple scalar products of these two sets of basis vectors and are related by [\eqalignno{V' &= ({\bf e}'_{1}, {\bf e}'_{2}, {\bf e}'_{3}) &\cr &= \Delta (B_{j}^{i}) ({\bf e}_{1}, {\bf e}_{2}, {\bf e}_{3}) &\cr & = \Delta (B_{j}^{i}) V, &(1.1.2.20)\cr}]where [\Delta (B_{j}^{i})] is the determinant associated with the transformation matrix between the two bases. From (1.1.2.17)[link] and (1.1.2.20)[link], we can write [\Delta (g'_{ij}) = \Delta (B_{i}^{k}) \Delta (B_{j}^{m}) \Delta(g_{km}).]

If the basis [{\bf e}_{i}] is orthonormal, [\Delta (g_{km})] and V are equal to one, [\Delta (B_{j})] is equal to the volume V ′ of the cell built on the basis vectors [{\bf e}'_{i}] and [\Delta (g'_{ij}) = V'^{2}.]This relation is actually general and one can remove the prime index: [\Delta (g_{ij}) = V^{2}. \eqno(1.1.2.21)]

In the same way, we have for the corresponding reciprocal basis[\Delta (g^{ij}) = V^{*2},]where [V^{*}] is the volume of the reciprocal cell. Since the tables of the [g_{ij}]'s and of the [g^{ij}]'s are inverse, so are their determinants, and therefore the volumes of the unit cells of the direct and reciprocal spaces are also inverse, which is a very well known result in crystallography.

References

International Tables for Crystallography (2008). Vol. B, Reciprocal Space, edited by U. Shmueli. Heidelberg: Springer.








































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