More than One Character Table? A Warning on the Use of the Rules of the Irreducible Representations and Their Characters Carlos ~ontreras-0rlega1and Leonel Vera Universidad Catolica del Norte, Casilla 1260, Antofagasta, Chile Eduardo Quiroz-Reyes Universidad Austral de Chile, Casilla 567, Valdivia, Chile into account the possible values for the unknowns. Thus, I t is well known that there is one and only one character table for each svmmetrv uoint n o u ~Also. . that all tables we find: must obey the &own 41;s den'ved krom the Great Orthox2=l ~ = 1 zn=l @=-1 w = -1 gonality Theorem (GOT) about the irreducible reprec3=1 m=-1 sentations and their characters. Based on this, textbooks x3=l I n=l on the field show. a s a method. how to construct character x4=l v4=-I I v4=-1 wn=l tables from these rules and so,'it has became a general beHowever, the above solution can he found under several lief among most chemists that all tables can he derived different forms; a s many a s permutations are possible from them. among the several rows of values given above while mainFollowing the above belief, we worked out the character taining the names of the variables fixed. Each of these table for the group D4h. Surprisingly enough, a detailed forms of the solution implies a different character table. and rigorous handling of the equations involved in our We conclude this a s follows: any permutation we make problem led us to a very striking result. We found for the among the several rows of the values above will change the group D 4 h not only one, hut six character tables holding right half of a t least two representations while leaving unperfectly well all the known rules. How is that possible? changed the others halves. Because the left halves of rz, Where is the failure in the accepted procedure? Which is r3,and r4are all different and the same is true for the the correct method to obtain the correct table? right halves, there is no chance of repeating all the repreIn an article published in this Journal (11, we put forward sentations after each permutation, thus giving rise to a s without any proof those six tables and showed which is the many character tables as permutations are possible (3!). correct one, hut we did not answer any of the above questions. Six, in total (1).The six tables hold perfectly well all the The purpose of the present work is to answer those questions. known rules, a s the reader can check. If this fact is a proof The elements of group D4h listed by classes are shown a t of the correctness of a character table, we are then forced the heading row of the table we present here. Because the to say that the six tables are equally valid a s character number of classes is 10, the number of irreducible repretables for the group D4h, and that there is not any valid sentations is also 10. I t is then easy to show (I)that the reason, upon this fact, to prefer one character table over characters a t the upper left and lower left quadrants of the another one. tahle must he those of the group 0 4 . Also, because any group has a one-dimensional representation whose characters are all equal to one, we arbitrarily take this repreConclusions seutation a s rl.The other characters and the relationships We have shown that the accepted procedure of constructbetween the unknowns a t the upper right and lower right ing character tables by making use of the known rules of quadrants are found by applying the rules of the irreducthe irreducible representations and their characters can ihle representations and their characters as they are prelead to mistakes. This will always occur in groups containsented elsewhere (2). This work, though tedious and not ing a subgroup and the operation i or the operation a h ; simple, does not present any theoretical difficulty. Theregroups D,h and C,h ( I ) . fore, we shall go straightforward to the point where the problems arise. The application of the mentioned rules, Character Table for the Symmetry Point Group D4h allows us to establish that: E 2C4 C2 2C2 2C'2 D4h i 2% oh 20" 20.3 Xi = fl,y ; = +I,2; = fl,"i = f1,wj = fl, 1
for i=2, 3, 4. Also the following equation:
r,. r, = o = x2+zyz +z2+2u2+2w2
(I'
and the other similar products:
r, - r, = r, r, = I-, .r, = r,. r, = r ,. r, = o
1 1
r2
1
r3
1
-1
1 2
-1
I 1 1 1
I
1-4
rs
rs The solution to the system formed by eqs r7 1-2 is found on inspection of them, taking r, (2)
I-9
'Author to whom correspondence should be ad- I-lo dressed.
2
o 1 -1 -1
1 I 1 1 -2 1 1
1 1 0
-2
1
-I I -1
1 -1 -1 1
1
1
1
1
1
x2
~2
z2
VL
m
X x4
~
M zq
o
-2 -I
o
o
1
1
-1
-I
- 2
1
-x3 -x4 -2
-E -p -y4 0
-1 1
-1 0
-1 -1 0
Z
y4
2
~
v4
Y
w4
o -1
o -1
-a -a
-a
-m
-c3
-en
-24
-v4 0
-w4 0
2
Volume 72 Number 9 September 1995
821
~
U
The fact above arises because the character table for groups as the one mentioned, contains a part which is independent from the rest of the character table-the cluadrants below the subgroupand another one, the otherAhalf, whose characters are found from homogeneous equations for which the values of their solutions can be permuted. This set of possibilities that for homogeneous ecluations forming a single system actually is just one solution, is not so for the table as a whole because the several possible permutations must be combined with a part whose rows of values are all different and remain fixed. The GOT states that every symmetry point group has one and only one character table. This is an absolute result. The failure in the known procedure lies in the fact that, a t least for the groups D,h and Cnh, it gives only necessarv conditions for a table to be valid. but not the suficientones. Thus, in this case, the rules derived from the GOT do not imdv what the GOT states. This is not s t r a n ~ e from a mathekatical point of view. A theorem can imGy certain rules (corollaries, other theorems, etc.), but the opposite is not necessarily true. What it is named a counterexample is enough to show it. This is exactly what we have done with the example of the six tables for the group D4h. In such case. the accom~lishmentof the rules bv the tables derived is oily a necesBary condition for themto be valid, but not a suff~cientone. The condition of sufficieucy must be obtained from symmetry criteria for each particular case studied (1).Probably this insufficiency in the method had not been observed so far, because the ;se of the rules had not been forced to a such extreme case as is the D4h group. Anyway, the given example (a counterexample) invalidates the known procedure as a general one for obtaining character tables from the rulesof the irreducible representations and their characters. We need to recall here that we are not questioning the fact that all character tables must adhere to the rules, but that all character tables can unequivocally be derived from them. The possibility of redundant solutions, for a general case, was announced a rather long time ago (3).However, its demonstration obli-
822
Journal of Chemical Education
gates to go deeply into the mathematics of group theory, which might be beyond the interests of users. Also and though the author gives some examples on how to work out some tables, no examples showing the commented amhiguitv are provided. 1 n a previous article (1)we developed a method to get the correct character table for the m o u D4h. ~ with no claim to the GOT rules, except the one
