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

International Tables for Crystallography (2006). Vol. D, ch. 1.2, pp. 51-53

Section 1.2.4. Tensors

T. Janssena*

aInstitute for Theoretical Physics, University of Nijmegen, 6524 ED Nijmegen, The Netherlands
Correspondence e-mail:

1.2.4. Tensors

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A vector is an element of an N-dimensional vector space that transforms under an orthogonal transformation, an element of [O(n]), as [x = \textstyle\sum\limits_{i=1}^{n}\xi_{i}{\bf a}_{i} \rightarrow x' = \textstyle\sum\limits_{i=1}^{n}\xi_{i}'{\bf a}_{i} = \textstyle\sum\limits_{ij}R_{ij}\xi_{j}{\bf a}_{i},\quad \{R_{ij}\}\in O(n).]A tensor of rank r under [O(n)] is an object with components [T_{i_{1}\ldots i_{r}}] ([i_{j}=1,2,\ldots, n]) that transforms as (see Section[link] )[T_{i_{1}\ldots i_{r}} \rightarrow T_{i_{1}\ldots i_{r}}' = \textstyle\sum\limits_{j_{1}=1}^{n}\ldots \textstyle\sum\limits_{j_{r}=1}^{n} R_{i_{1}j_{1}}\ldots R_{i_{r}j_{r}} T_{j_{1}\ldots j_{r}}.]A rank-zero tensor is a scalar, which is invariant under [O(n)]. A pseudovector (or axial vector) has components [x_{i}] and transforms according to [x_{i} \rightarrow x_{i}' = {\rm Det}(R)\textstyle\sum\limits_{j}R_{ij}\xi_{j}]and analogously for pseudotensors (or axial tensors – see Section[link] ).

A vector field is a vector-valued function in n-dimensional space. Under an orthogonal transformation it transforms according to [F_{i}({\bf r})' = \textstyle\sum\limits_{j=1}^{n}R_{ij} F_{j}(R^{-1}{\bf r}). \eqno (]Under a Euclidean transformation, the function transforms according to [F_{i}({\bf r})' = \textstyle\sum\limits_{j=1}^{n}R_{ij} F_{j}(R^{-1}({\bf r}-{\bf a})),\quad \{R|{\bf a}\}\in E(n). \eqno (]In a similar way, one has (pseudo)tensor functions under the orthogonal group or the Euclidean group. So it is important to specify under what group an object is a tensor, unless no confusion is possible.

The n-dimensional vectors form a vector space that carries a representation of the group O(n). Moreover, it is an irreducible representation space. To stress this fact, one could speak of irreducible tensors and vectors. Vectors are here just rank-one tensors. The three-dimensional Euclidean vector space carries in this way an irreducible representation of O(3). Such representations are characterized by an integer l and are [(2l+1)]-dimensional. The usual three-dimensional space is therefore an irreducible [l=1] space for O(3).

Since point groups are subgroups of the orthogonal group and space groups are subgroups of the Euclidean group, tensors inherit their transformation properties from their supergroups. As we have seen in Sections[link] and[link], one can also define tensors in a quite abstract way. Irreducible tensors under a group are then elements of a vector space that carries an irreducible representation of that group. Generally, tensors are elements of a vector space that carries a tensor product representation and (anti)symmetric tensors belong to a space with an (anti)symmetrized tensor product representation.

Because the point groups one usually considers in physics are subgroups of O(2) or O(3), it is useful to consider the irreducible representations of these groups. They are not finite, but they are compact, and for compact groups most of the theorems for finite groups are still valid if one replaces sums over group elements by integration over the group.

The group O(3) is the direct product [SO(3)\times C_{2}]. Therefore, there are even and odd representations. They have the property [D^{\pm}(R)=\Delta (R),\quad D^{\pm}(-R)=\pm \Delta (R),\quad R\in SO(3).]The irreducible representations are labelled by non-negative integers [\ell] and have character [\chi_{\ell}(R) = {{\sin (\ell + {{1}\over{2}})\varphi}\over{\sin {{1}\over{2}} \varphi}}\eqno (]if R is a rotation with rotation angle [\varphi]. From the character it follows that the dimension of the representation [D_{\ell}] is equal to [(2\ell +1)].

The tensor product of two irreducible representations of SO(3) is generally reducible: [D_{\ell}\otimes D_{m} = {\bigoplus\limits_{j=|\ell -m|}^{\ell +m}}D_{j} \eqno (] and the symmetrized and antisymmetrized tensor products are [\eqalignno{(D_{m}\otimes D_{m})_{s} &= {\bigoplus\limits_{j=0}^{m}}D_{2j}, &(\cr (D_{m}\otimes D_{m})_{a} &= {\bigoplus\limits_{j=1}^{m}}D_{2j-1}. & (}%fd1.2.4.6]

If the components of the tensor [T_{i_{1}\ldots i_{r}}] are taken with respect to an orthonormal basis, the tensor is called a Cartesian tensor. The orthogonal transformation R then is represented by an orthogonal matrix [R_{ij}]. Cartesian tensors of higher rank than one are generally no longer irreducible for the group O(n). For example, the rank-two tensors in three dimensions have nine components [T_{ij}]. Under SO(3), they transform according to the tensor product of two [\ell =1] representations. Because [D_{1} \otimes D_{1} = D_{0} \oplus D_{1} \oplus D_{2},]the space of rank 2 Cartesian tensors is the direct sum of three invariant subspaces. This corresponds to the fact that a general rank 2 tensor can be written as the sum of a diagonal tensor, an antisymmetric tensor and a symmetric tensor with trace zero. These three tensors are irreducible tensors, in this case also called spherical tensors, i.e. irreducible tensors for the orthogonal group.

An irreducible tensor with respect to the group [O(3)] transforms, in general, according to some reducible representation of a point group [K\in O(3)]. If the group K is a symmetry of the physical system, the tensor should be invariant under K, i.e. it should transform according to the identity representation of K.

Consider, for example, a symmetric second-rank tensor under [O(3)]. This means that it belongs to the space that transforms according to the representation [D_{0} \oplus D_{2}][see ([link]]. If the symmetry group of the system is the point group [K = 432], the representation [D_{0}(K) \oplus D_{2}(K)]has character[\matrix{R\colon\phantom{=C_{3}}\quad\varepsilon\phantom{=C_{3}}\quad\beta =C_{3}\quad\alpha^{2}=C_{4z}^{2}\quad \alpha =C_{4z}\quad\alpha \beta =C_{2}\cr \hrulefill\cr\chi(R)\colon\phantom{C_{3}}\quad6\phantom{1=C_{3}}\quad 0\phantom{1=C_{3}}\quad 2\phantom{1=C_{3}}\quad 0\phantom{1=C_{3}}\quad 2\phantom{1=C_{3}}}]and is equivalent to the direct sum[\Gamma_{1} \oplus \Gamma_{3} \oplus \Gamma_{5}.]The multiplicity of [\Gamma_{1}] is one. Therefore, the space of tensors invariant under K is one-dimensional. Consequently, there is only one parameter left to describe such a symmetric second-rank tensor invariant under the cubic group [K=432]. Noninvariant symmetric second-rank tensors are sums of tensors which transform according to the [\Gamma_3] and [\Gamma_5] representations. Here we are especially interested in invariant tensors. Invariants

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The dimension of the space of tensors of a certain type which are invariant under a point group K is equal to the number of free parameters in such a tensor. This number can be found as the multiplicity of the identity representation in the tensor space. For the 32 three-dimensional point groups this number is given in Table[link] for general second-rank tensors, symmetric second-rank tensors and a number of higher-rank tensors.

Invariant tensors, i.e. tensors of a certain type left invariant by a given group, may be constructed in several ways. The first way is a direct calculation. Take as an example again a second-rank symmetric tensor invariant under the cubic group 432. This means that [Rf = f\quad \forall\,\, R\in K,]which is a concise notation for [(Rf)_{ij} = \textstyle\sum\limits_{kl}R_{ik}R_{jl}f_{kl} = f_{ij}.]The group has two generators. Because each element of K is the product of generators, a tensor is left invariant under a group if it is left invariant by the generators. Therefore, one has in this case for [f = \pmatrix{ a_{1} & a_{2} & a_{3}\cr a_{2} & a_{4} & a_{5}\cr a_{3} & a_{5} & a_{6} }]the equation [\eqalign{&\pmatrix{ 0 & -1 & 0\cr 1&0&0\cr 0&0&1 } f \pmatrix{ 0&1&0\cr -1&0&0\cr 0&0&1 } \cr&\quad= \pmatrix{ 0&1&0\cr 0&0&1\cr 1&0&0 } f \pmatrix{ 0&0&1\cr 1&0&0\cr 0&1&0 } = f.}]These equations form a system of 12 linear algebraic equations for the coefficients of f with the solution [a_{1} = a_{4} = a_{6};\quad a_{2} = a_{3} = a_{5} =0.]Up to a factor there is only one such tensor: [f = \pmatrix{ a&0&0\cr 0&a&0\cr 0&0&a }, ]in agreement with the finding that the space of invariant second-rank symmetric tensors is one-dimensional. An overview of these relations for the 32 point groups can be found in Section 1.1.4[link] in this volume.

This method can always be used for groups with a finite number of generators. Another method for determining invariant tensors is using projection operators.

If a group, for example a point group, acts in some linear vector space, for example the space of tensors of a certain type, this space carries a representation. Then it is possible to construct a basis such that the representation corresponds to a choice of matrix representation. In particular, if the representation is reducible, it is possible to construct a basis such that the matrix representation is in reduced form. This can be achieved with projection operators.

Suppose the element [R\in K] acts in a space as an operator [D(R)] such that the representation [D(K)] is equivalent with a matrix representation [\Gamma (K)] which has irreducible components [\Gamma_{\alpha}(K)]. Then choose a vector v in the representation space and construct the [d_{\alpha}] vectors[v_{i} = {({1}/{N})}\textstyle\sum\limits_{R\in K}\Gamma_{\alpha}(R)_{ji}^{*}D(R)v\eqno (]with j fixed. If v does not have a component in the invariant space of the irreducible representation [D_{\alpha}], these vectors are all zero, but for a sufficiently general vector the [d_{\alpha}] vectors form a basis for the irreducible representation. This property follows from the orthogonality relations.

Using this relation one can write for an invariant symmetric second-rank tensor [f = {({1}/{N})}\textstyle\sum\limits_{R\in K} D(R)f' = {({1}/{N})}\textstyle\sum\limits_{R\in K} \Gamma (R) f'\Gamma (R)^{T}]for an arbitrary symmetric second-rank tensor [f']. For the group [K=432 ] this would give a tensor with components [f_{ij}=a\delta_{ij}]. Of course, this is a rather impractical method if the order of the group is large. A simple example for a very small group is the construction of the symmetrical and antisymmetrical components of a function: [f_{\pm}(x)=[f(x)\pm f(-x)]/2]. Clebsch–Gordan coefficients

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The tensor product of two irreducible representations of a group K is, in general, reducible. If [{\bf a}_{i}] is a basis for the irreducible representation [\Gamma_{\alpha}] ([i=1,\ldots, d_{\alpha}]) and [{\bf b}_{j}] one for [\Gamma_{\beta}] ([j=1,\ldots, d_{\beta}]), a basis for the tensor product space is given by[{\bf e}_{ij} = {\bf a}_{i}\otimes{\bf b}_{j}.]On this basis, the matrix representation is, in general, not in reduced form, even if the product representation is reducible. Suppose that [\Gamma_{\alpha}\otimes\Gamma_{\beta} \sim \textstyle\sum\limits_{\gamma} ^{}{}^{\oplus} m_{\gamma}\Gamma_{\gamma}.]This means that there is a basis [\psi_{\gamma\ell k} \quad (\ell =1,\ldots, m_{\gamma};\,k=1,\ldots, d_{\gamma}),]on which the representation is in reduced form. The multiplicity [m_{\gamma}] gives the number of times the irreducible component [\Gamma_{\gamma}] occurs in the tensor product. The basis transformation is given by[\psi_{\gamma \ell k}= \sum\limits_{ij}\left.\left(\matrix{ \alpha & \beta\cr i & j}\right|\matrix{\gamma & \cr k & \ell}\right){\bf a}_{i}\otimes{\bf b}_{j}. \eqno (]The basis transformation is unitary if one starts with orthonormal bases and has coefficients [\left.\left(\matrix{ \alpha & \beta\cr i & j}\right|\matrix{\gamma & \cr k & \ell}\right)\eqno (]called Clebsch–Gordan coefficients. For the group O(3) they are the original Clebsch–Gordan coefficients; for bases [|\ell m\rangle] and [|\ell' m'\rangle] of the ([2\ell+1])- and ([2\ell'+1])-dimensional representations [D_{\ell}] and [D_{\ell'}], respectively, of O(3) one has [\eqalignno{&|JM\rangle = \sum\limits_{m\,m'}\left.\left(\matrix{ \ell & \ell'\cr m & m'}\right|\matrix{J\cr M}\right) |\ell m\rangle\otimes |\ell'm'\rangle,&\cr&\quad(J=|\ell-\ell'|,\ldots, \ell+\ell').&(}]The multiplicity here is always zero or unity, which is the reason why one leaves out the number [\ell] in the notation.

If the multiplicity [m_{\gamma}] is unity, the coefficients for given [\alpha, \beta, \gamma] are unique up to a common factor for all [i,j,k]. This is no longer the case if the multiplicity is larger, because then one can make linear combinations of the basis vectors belonging to [\Gamma_{\gamma}]. Anyway, one has to follow certain conventions. In the case of O(3), for example, there are the Condon–Shortley phase conventions. The degree of freedom of the Clebsch–Gordan coefficients for given matrix representations [\Gamma_{\alpha}] can be seen as follows. Suppose that there are two basis transformations, S and [S'], in the tensor product space which give the same reduced form: [S\left(D_{\alpha}\otimes D_{\beta} \right)S^{-1} = S'\left(D_{\alpha}\otimes D_{\beta} \right)S'^{-1} = D = \bigoplus m_{\gamma}D_{\gamma}. \eqno (]Then the matrix [S'S^{-1}] commutes with every matrix [D(R)] ([R\in K]). If all multiplicities are zero or unity, it follows from Schur's lemma that [S'S^{-1}] is the direct sum of unit matrices of dimension [d_{\gamma}]. If the multiplicities are larger, the matrix [S'S^{-1}] is a direct sum of blocks which are of the form [\pmatrix{ \lambda_{11}E & \lambda_{12}E & \ldots &\lambda_{1m_{\gamma}}E \cr \lambda_{21}E & \lambda_{22}E & \ldots &\lambda_{2m_{\gamma}}E \cr \vdots & \vdots & \ddots & \vdots \cr \lambda_{m_{\gamma}1}E & \ldots & \ldots & \lambda_{m_{\gamma}m_{\gamma}}E }, ]such that [{\rm Det}(\lambda_{ij})=1], and the E's are [d_{\gamma}]-dimensional unit matrices. This means that for multiplicity-free ([m_{\gamma} \,\leq \, 1]) cases, the Clebsch–Gordan coefficients are unique up to a common factor for all coefficients involving one value of [\gamma].

The Clebsch–Gordan coefficients satisfy the following rules:[\displaylines{\left.\left(\matrix{ \alpha & \beta\cr i & j}\right|\matrix{\gamma & \cr k & \ell}\right) = \left.\left(\matrix{ \beta & \alpha\cr j & i}\right|\matrix{\gamma & \cr k & \ell}\right)\cr\left.\left(\matrix{ \alpha & \beta\cr i & j}\right|\matrix{\gamma & \cr k & \ell}\right) = 0, \hbox{ if } D_\alpha\otimes D_\beta \hbox{ does not contain } D_\gamma \cr \sum\limits_{k\ell}\left.\left(\matrix{ \alpha & \beta\cr i & j}\right|\matrix{\gamma & \cr k & \ell}\right)^* \left.\left(\matrix{ \alpha & \beta\cr i' & j'}\right|\matrix{\gamma & \cr k & \ell}\right) = \delta_{ii'}\delta_{jj'} \cr \sum\limits_{ij}\left.\left(\matrix{ \alpha & \beta\cr i & j}\right|\matrix{\gamma & \cr k & \ell}\right)^* \left.\left(\matrix{ \alpha & \beta\cr i & j}\right|\matrix{\gamma & \cr k' & \ell'}\right) = \delta_{kk'}\delta_{\ell \ell '}.}]For the basis vectors of the invariant space belonging to the identity representation [\Gamma_{1}], one has [\gamma =d_{\gamma}=1]. Consequently, [\psi_{\ell} = \sum\limits_{ij}\left.\left(\matrix{ \alpha & \beta\cr i & j}\right|\matrix{1 & \cr 1 & \ell}\right) {\bf a}_{i}\otimes {\bf b}_{j}.]

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