Symmetry adapted bases |

This means that we have to find anV=V_{1}Å...ÅV_{r}.

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

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

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

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

Ordering this basis in a proper way:D(p)b^{a}_{ij}=å_{k=1}^{fa}d^{a}_{kj}(p)b^{a}_{ik}.

the matrix corresponding to the (linear) action ofB={...,b^{a}_{11},...,b^{a}_{1,fa}, b^{a}_{21}, ...,b^{a}_{2,fa}, ...,b^{a}_{m(a),1}, ...,b^{a}_{m(a),fa}, ...},

*First of all we note that such bases do in fact exist*, this is
obvious from
linear algebra.
In order to construct such a basis from given tabulated matrix representations
* D^{a} *, i.e. from given numbers

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

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

where the last equation is obtained from . This shows that the following is true, for=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},

This yields, for the product of two such operators:P^{a}_{ji}e^{b}_{kl}=e^{a}_{kj}if a=b,i=l,

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

Another important consequence isP^{g}_{rs}P^{a}_{ji}=d_{ga}d_{sj}P^{a}_{ri}.

We are now in a position to construct a symmetry adapted basis. For this purpose we pick, for eachCorollary: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) >> .

(Recall from the corollary above that the rank ofb^{a}_{1}:=P^{a}_{11}(v) not =0.

and we claim that these vectors form a symmetry adapted basis ofb^{a}_{ij}:=P^{a}_{ji}(b^{a}_{1}),

Using this linear combination we obtain:b^{a}_{ij}=å_{k=1}^{m(a)}z_{k}e^{a}_{kj}.

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}

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

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

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

shows that thedet(xI-D((1...n)))=x^{n}-1

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

A nice application is the following derivation of Cardano's formula for the solutions of cubic equations: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 more important application of symmetry adapted bases is the evaluation of the irreducible polynomial representationsExample: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

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).áañ:=id [a],

harald.fripertinger@kfunigraz.ac.at,

last changed: August 28, 2001

Symmetry adapted bases |