Representations of finite groups

Janssen, T.

doi:10.1107/97809553602060000629

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

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International Tables for Crystallography (2006). Vol. D. ch. 1.2, pp. 36-37

Section 1.2.2.2. Representations of finite groups

T. Janssen^a ^*

^a Institute for Theoretical Physics, University of Nijmegen, 6524 ED Nijmegen, The Netherlands
Correspondence e-mail: ted@sci.kun.nl

1.2.2.2. Representations of finite groups

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As stated in Section 1.2.1, elements of point groups act on physical properties (like tensorial properties) and on wave functions as linear operators. These linear operators therefore generally act in a different space than the three-dimensional configuration space. We denote this new space by V and consider a mapping D from the point group K to the group of nonsingular linear operators in V that satisfies $[D(R)D(R')=D(RR')\quad \forall \,\,R,R'\in K. \eqno (1.2.2.3)]$ In other words D is a homomorphism from K to the group of nonsingular linear transformations [GL(V)] on the vector space V. Such a homomorphism is called a representation of K in V. Here we only consider finite-dimensional representations.

With respect to a basis $[{\bf e}_{i}]$ ( $[i=1,2,\ldots n]$ ) the linear transformations are given by matrices $[\Gamma (R)]$ . The mapping $[\Gamma]$ from K to the group of nonsingular $[n\times n]$ matrices GL( [n,R] ) (for a real vector space V) or GL( [n,C] ) (if V is complex) is called an n-dimensional matrix representation of K.

If one chooses another basis for V connected to the former one by a nonsingular matrix S, the same group of operators [D(K)] is represented by another matrix group $[\Gamma '(K)]$ , which is related to $[\Gamma (K)]$ by S according to $[\Gamma '(R)=S^{-1}\Gamma (R)S]$ ( $[\forall \,\,R\in K]$ ). Two such matrix representations are called equivalent. On the other hand, two such equivalent matrix representations can be considered to describe two different groups of linear operators [ [D(K)] and [D'(K)] ] on the same basis. Then there is a nonsingular linear operator T such that [D(R)T=TD'(R)] ( $[\forall \,\,R\in K]$ ). In this case, the representations [D(K)] and [D'(K)] are also called equivalent.

It may happen that a representation [D(K)] in V leaves a subspace W of V invariant. This means that for every vector $[v\in W]$ and every element $[R\in K]$ one has $[D(R)v\in W]$ . Suppose that this subspace is of dimension $[m \,\lt\, n]$ . Then one can choose m basis vectors for V inside the invariant subspace. With respect to this basis, the corresponding matrix representation has elements $[\Gamma (R) = \pmatrix{ \Gamma_{1}(R) & \Gamma_{3}(R) \cr 0 & \Gamma_{2}(R)}, \eqno (1.2.2.4)]$ where the matrices $[\Gamma_{1}(R)]$ form an m-dimensional matrix representation of K. In this situation, the representations [D(K)] and $[\Gamma (K)]$ are called reducible. If there is no proper invariant subspace the representation is irreducible. If the representation is a direct sum of subspaces, each carrying an irreducible representation, the representation is called fully reducible or decomposable. In the latter case, a basis in V can be chosen such that the matrices $[\Gamma (R)]$ are direct sums of matrices $[\Gamma_i (R)]$ such that the $[\Gamma_i (R)]$ form an irreducible matrix representation. If $[\Gamma_3 (R)]$ in (1.2.2.4) is zero and $[\Gamma_1]$ and $[\Gamma_2]$ form irreducible matrix representations, $[\Gamma]$ is fully reducible. For finite groups, each reducible representation is fully reducible. That means that if $[\Gamma (K)]$ is reducible, there is a matrix S such that $[\eqalignno{\Gamma (R) &= S\left [\Gamma_{1}(R)\oplus \ldots \oplus \Gamma_{n}(R)\right] S^{-1} &\cr&= S\pmatrix{ \Gamma_{1}(R) &0& \ldots & 0 \cr 0&\Gamma_{2}(R)&\ldots & 0\cr \vdots &\vdots & \ddots & \vdots \cr 0&0&\ldots &\Gamma_{n}(R)} S^{-1}.&\cr &&(1.2.2.5)}]$ In this way one may proceed until all matrix representations $[\Gamma_{i}(K)]$ are irreducible, i.e. do not have invariant subspaces. Then each representation $[\Gamma (K)]$ can be written as a direct sum $[\Gamma (R) = S\left [m_{1}\Gamma_{1}(R)\oplus \ldots \oplus m_{s}\Gamma_{s}(R) \right] S^{-1}, \eqno (1.2.2.6)]$ where the representations $[\Gamma_{1}\ldots \Gamma_{s}]$ are all nonequivalent and the multiplicities $[m_{i}]$ are the numbers of times each irreducible representation occurs. The nonequivalent irreducible representations $[\Gamma_{i}]$ for which the multiplicity is not zero are the irreducible components of $[\Gamma (K)]$ .

We first discuss two special representations. The simplest representation in one-dimensional space is obtained by assigning the number 1 to all elements of K. Obviously this is a representation, called the identity or trivial representation. Another is the regular representation. To obtain this, one numbers the elements of K from 1 to the order N of the group ( [|K|=N] ). For a given $[R\in K]$ there is a one-to-one mapping from K to itself defined by $[R_{i}\rightarrow R_{j} \equiv RR_{i}]$ . Consider the $[N\times N]$ matrix $[\Gamma (R)]$ , which has in the ith column zeros except on line j, where the entry is unity. The matrix $[\Gamma (R)]$ then has as only entries 0 or 1 and satisfies $[RR_{i} = \Gamma (R)_{ji}R_{j},\quad (i=1,2,\ldots, N). \eqno (1.2.2.7)]$ These matrices $[\Gamma (R)]$ form a representation, the regular representation of K of dimension N, as one sees from $[\eqalign{(R_{i}R_{j})R_{k} &= R_{i}\textstyle\sum\limits_{l=1}^{N}\Gamma (R_{j})_{lk}R_{l} = \textstyle\sum\limits_{l=1}^{N}\textstyle\sum\limits_{m=1}^{N}\Gamma (R_{j})_{lk}\Gamma (R_{i})_{ml}R_{m}\cr &= \textstyle\sum\limits_{m=1}^{N}\left [\Gamma (R_{i})\Gamma (R_{j})\right]_{mk}R_{m} = \textstyle\sum\limits_{m=1}^{N} \Gamma (R_{i}R_{j})_{mk}R_{m}. }]$

A representation in a real vector space that leaves a positive definite metric invariant can be considered on an orthonormal basis for that metric. Then the matrices satisfy $[\Gamma (R)\Gamma (R)^{T} = E]$ (^T denotes transposition of the matrix) and the representation is orthogonal. If V is a complex vector space with positive definite metric invariant under the representation, the latter gives on an orthonormal basis matrices satisfying $[\Gamma (R)\Gamma (R)^{\dagger} = E]$ ( $[^{\dagger}]$ denotes Hermitian conjugation) and the representation is unitary. A real representation of a finite group is always equivalent with an orthogonal one, a complex representation of a finite group is always equivalent with a unitary one. As a proof of the latter statement, consider the standard Hermitian metric on V: $[f(x,y)=\textstyle\sum_{i}x_{i}^{*}y_{i}]$ . Then the positive definite form $[F(x,y) = ({1}/{N})\textstyle \sum \limits_{R\in K} f\left(D(R)x,D(R)y\right) \eqno (1.2.2.8)]$ is invariant under the representation. To show this, take an arbitrary element [R'] . Then $[\eqalignno{F(D(R')x,D(R')y) &= (1/N)\textstyle\sum\limits_{R\in K} f(D(R'R)x,D(R'R)y)&\cr &= F(x,y). &(1.2.2.9)}]$ With respect to an orthonormal basis for this metric [F(x,y)] , the matrices corresponding to [D(R)] are unitary. The complex representation can be put into this unitary form by a basis transformation. For a real representation, the argument is fully analogous, and one obtains an orthogonal transformation.

From two representations, $[D_{1}(K)]$ in $[V_{1}]$ and $[D_{2}(K)]$ in $[V_{2}]$ , one can construct the sum and product representations. The sum representation acts in the direct sum space $[V_{1}\oplus V_{2}]$ , which has elements ( $[{\bf a},{\bf b}]$ ) with $[{\bf a}\in V_{1}]$ and $[{\bf b}\in V_{2}]$ . The representation $[D_{1}\oplus D_{2}]$ is defined by $[\left [\left(D_{1}\oplus D_{2} \right) (R)\right] ({\bf a},{\bf b}) = (D_{1}(R){\bf a},D_{2}(R){\bf b}). \eqno (1.2.2.10)]$ The matrices $[\Gamma_{1}\oplus\Gamma_{2}(R)]$ are of dimension $[n_{1}+n_{2}]$ .

The product representation acts in the tensor space, which is the space spanned by the vectors $[{\bf e}_{i}\otimes {\bf e}_{j}]$ ( $[i=1,2,\ldots ,{\rm dim}V_{1}]$ ; $[j=1,2\ldots, {\rm dim}V_{2}]$ ). The dimension of the tensor space is the product of the dimensions of both spaces. The action is given by $[\left [\left(D_{1}\otimes D_{2} \right) (R)\right] {\bf a}\otimes{\bf b} = D_{1}(R){\bf a}\otimes D_{2}(R){\bf b}. \eqno (1.2.2.11)]$ For bases $[{\bf e}_{i}]$ ( $[i=1,2,\ldots, d_{1}]$ ) for $[V_{1}]$ and $[{\bf e'}_{j}]$ ( $[j=1,2,\ldots , d_{2}]$ ) for $[V_{2}]$ , a basis for the tensor product of spaces is given by $[{\bf e}_{i}\otimes {\bf e'}_{j},\quad i=1,\ldots, d_{1};\; j=1,2,\ldots, d_{2}, \eqno (1.2.2.12)]$ and with respect to this basis the representation of K is given by matrices $[\left(\Gamma_{1}\otimes \Gamma_{2} \right)(R)_{ik,jl} = \Gamma_{1}(R)_{ij}\Gamma_{2}(R)_{kl}. \eqno(1.2.2.13)]$ As an example of these operations, consider $[\eqalign{\pmatrix{ 1&0\cr 0&-1}\oplus \pmatrix{0&1\cr 1&0}&= \pmatrix{ 1&0&0&0\cr 0&-1&0&0\cr 0&0&0&1\cr 0&0&1&0}\semi\cr \pmatrix{ 1&0\cr 0&-1}\otimes \pmatrix{ 0&1\cr 1&0}&= \pmatrix{0&1&0&0\cr 1&0&0&0\cr 0&0&0&-1\cr 0&0&-1&0}.}]$

If two representations $[D_{1}(K)]$ and $[D_{2}(K)]$ are equivalent, there is an operator S such that $[SD_{1}(R) = D_{2}(R)S\quad \forall\,\, R \in K. ]$ This relation may also hold between sets of operators that are not necessarily representations. Such an operator S is called an intertwining operator. With this concept we can formulate a theorem that strictly speaking does not deal with representations but with intertwining operators: Schur's lemma.

Proposition. Let M and N be two sets of nonsingular linear transformations in spaces V (dimension n) and W (dimension m), respectively. Suppose that both sets are irreducible (the only invariant subspaces are the full space and the origin). Let S be a linear transformation from V to W such that [SM=NS] . Then either S is the null operator or S is nonsingular and $[SMS^{-1}=N]$ .

Proof: Consider the image of V under S: $[{\rm Im}_{S}V\subseteq W]$ . That means that $[S{\bf r}\in{\rm Im}_{S}V]$ for all $[{\bf r}\in V]$ . This implies that $[NS{\bf r}=]$ $[SM{\bf r}\in{\rm Im}_{S}V]$ . Therefore, $[{\rm Im}_{S}V]$ is an invariant subspace of W under N. Because N is irreducible, either $[{\rm Im}_{S}V=0]$ or $[{\rm Im}_{S}V=W]$ . In the first case, S is the null operator. In the second case, notice that the kernel of S, the subspace of V mapped on the null vector of W, is an invariant subspace of V under M: if $[S{\bf r}=0]$ then $[NS{\bf r}=0]$ . Again, because of the irreducibility, either $[{\rm Ker}_{S}]$ is the whole of V, and then S is again the null operator, or $[{\rm Ker}_{S}=0]$ . In the latter case, S is a one-to-one mapping and therefore nonsingular. Therefore, either S is the null operator or it is an isomorphism between the vector spaces V and W, which are then both of dimension n. With respect to bases in the two spaces, the operator S corresponds to a nonsingular matrix and $[M=S^{-1}NS]$ .

This is a very fundamental theorem. Consequences of the theorem are:

(1) If N and M are nonequivalent irreducible representations and , then .
(2) If a matrix S is singular and links two irreducible representations of the same dimension, then .
(3) A matrix S that commutes with all matrices of an irreducible complex representation is a multiple of the identity. Suppose that an $[n\times n]$ matrix S commutes with all matrices of a complex irreducible representation. S can be singular and is then the null matrix, or it is nonsingular. In the latter case it has an eigenvalue $[\lambda \neq 0]$ and $[S-\lambda E]$ commutes with all the matrices. However, $[S-\lambda E]$ is singular and therefore the null matrix: $[S=\lambda E]$ . This reasoning is only valid in a complex space, because, generally, the eigenvalues $[\lambda]$ are complex.

References

International Tables for Crystallography (2006). Vol. D. ch. 1.2, pp. 36-37