Symmetry adapted bases |

We discussed finite actions * _{G}X* and their decomposition into orbits. The
corresponding problem in linear representation theory is the
decomposition of a representation space

V=V_{1}Å...ÅV_{r}.

This means that we have to find an *adapted basis*

{v_{1},...,v_{f1},v_{f1+1},...,v_{f1+f2},...},

where *adapted* means that *{v _{1},...,v_{f1}}* is a basis of

gb_{k}:=D(g)b_{k}=å_{j=1}^{f1}d^{i}_{jk}(g)b_{j}, ifD^{i}(g)= (d^{i}_{jk}(g)).

A basis of *V* which has this property, is called a *symmetry adapted
basis* of
*V* with respect to the given system * D^{1},...,D^{s}* of irreducible
matrix representations. Such a symmetry adapted basis has great advantages,
since we know in advance, how each of the vectors

Suppose we are given a complete system of ordinary irreducible matrix
representations * D^{a} * of

B:={b^{a}_{ij}| a|¾n,1£i£m(a),1£j£f^{a}},

such that

D(p)b^{a}_{ij}=å_{k=1}^{fa}d^{a}_{kj}(p)b^{a}_{ik}.

Ordering this basis in a proper way:

B={...,b^{a}_{11},...,b^{a}_{1,fa}, b^{a}_{21}, ...,b^{a}_{2,fa}, ...,b^{a}_{m(a),1}, ...,b^{a}_{m(a),fa}, ...},

the matrix corresponding to the (linear) action of *pÎS _{n} * takes the
following form:
The square box containing the matrices

obvious from
linear algebra.
In order to construct such a basis from given tabulated matrix representations
* D^{a} *, i.e. from given numbers

P^{a}_{ji}:=(f^{a})/(n!)å_{pÎSn }d^{a}_{ij}(p^{-1})D(p).

Assume that the vectors *e ^{b}_{kl},1£k£m(b),1£l£f^{b}*
form asymmetry adapted basis. Then

P^{a}_{ji}e^{b}_{kl}=(f^{a})/(n!)å_{p}d^{a}_{ij}(p^{-1}) D(p)e^{b}_{kl}

=f^{a}å_{m=1}^{fa}((1)/(n!)å_{p}d^{a}_{ij}(p^{-1})d^{b}_{ml}(p)) e^{b}_{km}= d_{ab}d_{il}b^{a}_{kj},

where the last equation is obtained from . This shows that
the following is true, for *any symmetry adapted basis* of *V*:

P^{a}_{ji}e^{b}_{kl}=e^{a}_{kj}if a=b,i=l,

P^{a}_{ji}e^{b}_{kl}= 0 otherwise.

This yields, for the product of two such operators:

P^{g}_{rs}P^{a}_{ji}=d_{ga}d_{sj}P^{a}_{ri}.

Another important consequence is

Corollary:The linear operatorPintroduced above is a projection operator that maps^{a}_{ji}Vonto the subspaceP^{a}_{ji}(V)=W^{a}_{j}:=<< e^{a}_{kj}| 1£k£m(a) >> .

We are now in a position to construct a symmetry adapted basis. For this
purpose we pick, for each *a|¾n* which is contained in *D*,
*any vector* *vÎV* such that

b^{a}_{1}:=P^{a}_{11}(v) not =0.

(Recall from the corollary above that the rank of *P ^{a}_{11}* is nonzero if and
only if

b^{a}_{ij}:=P^{a}_{ji}(b^{a}_{1}),

and we claim that these vectors form a symmetry adapted basis of *V*.
In order to check this we use the last corollary which says that, for a fixed
symmetry adapted bases *{e ^{b}_{rs}}*, the vector

b^{a}_{ij}=å_{k=1}^{m(a)}z_{k}e^{a}_{kj}.

Using this linear combination we obtain:

D(p)b^{a}_{ij}=å_{k}D(p)e^{a}_{kj}=å_{k}z_{k}å_{l=1}^{fa}d^{a}_{lj}(p)e^{a}_{kl}=å_{l}d^{a}_{lj}(p) å_{k}z_{k}e^{a}_{kl}

=å_{l}d^{a}_{lj}(p)P^{a}_{lj}å_{k}z_{k}e^{a}_{kj}=å_{l}d^{a}_{lj}(p)P^{a}_{lj}b^{a}_{ij}=å_{l}d^{a}_{lj}(p) P^{a}_{lj}P^{a}_{ji}(b^{a}_{1}),

which is, by the formula for the product of two such operators,

=å_{l}d^{a}_{lj}(p)P^{a}_{li}(b_{1})=å_{l}d^{a}_{lj}(p) b^{a}_{ij},

as it is stated.

Some of the many interesting applications combine symmetry adapted bases with
eigenvalue considerations. For example, the natural representation of the
cyclic group *C _{n}=á(1...n)ñ* is generated by the matrix
Its characteristic polynomial

det(xI-D((1...n)))=x^{n}-1

shows that the *eigenvalues*
are exactly the powers of a primitive *n*-th root
of unity *w:=e ^{2pi/n}*:

w_{j}:=w^{j}=e^{2pij/n}.

This implies that the homogeneous components are onedimensional (which means
that the irreducible constituents of the natural representation of the cyclic
group are pairwise different, this representation is in fact a *model*
for the cyclic group). A symmetry adapted basis therefore consists of
eigenvectors corresponding to pairwise different eigenvalues, and so the
arguments used in the proof imply the following:

Corollary:Each matrix that commutes withbecomes diagonal if we change the basis into a symmetry adapted one. Hence in particular the eigenvectors ofD((1...n))are eigenvectors of the commuting matrix, too.D((1...n))

A nice application is the following derivation of Cardano's formula for the solutions of cubic equations:

Example:The following matrix clearly commutes with: The characteristic polynomial of this matrix isD((123))In order to derive its roots we can use corollary. Puttingx^{3}-3abx-a^{3}-b^{3}.w:=e, we obtain fromthis corollary the following eigenvectors of^{2pi/3}M: Hence the eigenvalues of the matrixMturn out to beWe can apply this now to a cubic equation of the forml_{1}=a+b,l_{2}=aw^{2}+bw,l_{3}=aw+bw^{2}.obtainingx^{3}+px+q=0,p=-3ab,q=-(a, from which we derive^{3}+b^{3})Cardano's formulae

A more important application of symmetry adapted bases is the evaluation
of the irreducible polynomial representations *áañ* of
the general linear group *GL _{m}(Q)*, which arise by symmetrizing the
identity map:

áañ:=id [a],

but the details of this form quite a long story, and so I refer to the corresponding literature (see the chapter with comments and references).

harald.fripertinger "at" uni-graz.at | http://www-ang.kfunigraz.ac.at/~fripert/ |

UNI-Graz | Institut für Mathematik |

UNI-Bayreuth | Lehrstuhl II für Mathematik |

Symmetry adapted bases |