7 Applications 431thenVkis anmk-dimensional subspace ofRn(k= 1 ,...,p)andRnis the direct sum
ofV 1 ,...,Vp. Moreover each eigenspaceVkis invariant underA(s)for everys∈G.
Hence, by Maschke’s theorem (which holds also for representations in a real vector
space),Vkis a direct sum of real-irreducible invariant subspaces. It follows that there
exists a real non-singular matrixTsuch that, for everys∈G,T−^1 A(s)T=⎛
⎜
⎜
⎝
A 1 (s) 0 ··· 0
0 A 2 (s) ··· 0
··· ··· ··· ···
00 ··· Aq(s)⎞
⎟
⎟
⎠,
wheres →Ak(s)is a real-irreducible representation ofG,ofdegreenksay(k =
1 ,...,q),andT−^1 B−^1 CT=
⎛
⎜
⎜
⎝
λ 1 In 1 0 ··· 0
0 λ 2 In 2 ··· 0
··· ··· ··· ···
00 ··· λqInq⎞
⎟
⎟
⎠.
If the real-irreducible representationss→Ak(s)(k = 1 ,...,q)are also complex-
irreducible, then their degrees and multiplicities can be found by character theory.
Thus by decomposing the representationρofGinto its irreducible components we
can determine the degeneracy of the normal frequencies.
We will not consider here the modifications needed when some real-irreducible
component is not also complex-irreducible. Also, it should be noted that it may
happen ‘accidentally’ thatλj=λkfor somej=k.
As a simple illustration of the preceding discussion we consider the ammonia
moleculeNH 3. Its internal configuration may be described by the six internal coor-
dinatesr 1 ,r 2 ,r 3 andα 23 ,α 31 ,α 12 ,whererjis the change from its equilibrium value
of the distance from the nitrogen atom to thej-th hydrogen atom, andαjkis the change
from its equilibrium value of the angle between the rays joining the nitrogen atom to
thej-th andk-th hydrogen atoms.
We will determine the characterχof the corresponding representationρof the
symmetry groupS 3. In the notation of the character table previously given forS 3 ,
there is an elements∈C 3 for which the symmetry operationA(s)cyclically permutes
r 1 ,r 2 ,r 3 andα 23 ,α 31 ,α 12. Consequentlyχ(s)=0ifs∈C 3. Also, there is an element
t∈C 2 for which the symmetry operationA(t)interchangesr 1 withr 2 andα 23 with
α 31 ,butfixesr 3 andα 12. Consequentlyχ(t)=2ift∈C 2. Since it is obvious that
χ(e)=6, this determinesχand we adjoin it to the character table ofS 3 :
|C| 13 2
CC 1 C 2 C 3
χ 1 111
χ 2 1 − 11
χ 3 20 − 1
χ 620