elements are written in terms of generators S and T
1 , S, T, ST, TS, T2, STS, ,TST, T2ST, TST2, T2S, ST2
How do I multiply S with T2ST from left
How do I multiply S with TST2 from left
For the second example, I can use the first relation TST2= S T2ST and then left multiply with S, then S TST2 = S2 T2ST = T2ST
@Martin,
Now while multiplying T2 S T with T S T2 we get, T2 S T2 S T2 = T -1 S -1 T -1 S -1 T -1 then I left multiply with SS -1 then I get S S -1 T -1 S -1 T -1 S -1 T -1 = S (ST) -3 =S, do you agree?
In most cases, and certainly for A4, the definition of a group in terms of generators is the most tedious and less helpful one. Define it by permutations and their natural cycle decomposition than you can obtain the products of arbitrary pairs by a simple algorithm.
Ok now I have all elements of 12x12 matrix in th S T generator notation. Next question is how to rewrite then in terms of permutations. For example S means (12)(34) then how to write other elements such as
ST or STS
My STS is agreeing with your (142) but my ST is (134) please let me know your ST once again. Thanks
OK, If I work from right to left, ST is (243), STS is the same either way!
It may help to view A4 as a subgroup of S4, the permutation group on 4 elements. S4 has 4!=24 elements. A4 sits (embeds) in S4 naturally as the group of even permutations.
Now I have got correspondence between elements in S, T notation and the cycle notation
1 --> E
S --> (12)(34)
ST --> (243)
STS --> (142)
T --> (123)
TS --> (134)
T2 --> (132)
ST2 --> (143)
T2 S --> (234)
TST --> (124)
T S T2 --> (14)(23)
T2 S T --> (13)(24)
@ Martin, using the multiplication tables we see that 1, S, TST2, and T2 ST forms a subgroup, now if we consider closure property, then a subgroup is also an invariant class, do you agree with it?
No, I meant conjugacy, but I see that there is a theorem that the identity is always a conjugacy class by itself (e e e-1=e). Then S, TST2, T2ST forms a different conjugacy class. That is identity is not a conjugate element of any of them. Then three elements (which are also three other elements of the said subgroup apart from identity) are conjugates of each other as can be checked from group multiplication tables.
All conjugacy classes can now be calculated using 12x12 multiplication table. I first took the element S, and found all 12 conjugates of S defined by the relation g S g-1, it turns out that the result is always contained in the set. {S,TST2, T2ST}. Next, I take another element of the group which is not contained in the first set. We choose T and find all 12 conjugates of T, the result is now contained in the set {T,ST,TS,STS}. In the next step of this algorithm, we take an element which is not contained in the first two sets. This time I choose, T2, and find all 12 conjugates of it. We find that the result is contained in the set {T2, TST, T2S, ST2}. Now we have covered the whole group in 4 disjoint sets. Members of each set is a conjugate element of every other element within the group. That is we get 4 conjugacy classes
E, {S,TST2, T2ST}, {T,ST,TS,STS}, {T2, TST, T2S, ST2}.
Because number of irreducible representations should be equal to number of conjugacy classes, we get that there are exactly 4 representations. Then we write 12= 12+12+12+32, that is sum of squares of exactly 4 integers. We thus get, a triplet and three singlets of A4
@Martin
Using orthogonality relations we get following characters for four classes listed in the post above.
But using orthogonality of characters there is no way to fix which is one prime and which is one prime prime. Singlet and triplet on the other hand can be uniquely fixed.
Is there a standard convention which fixes one prime and one prime prime?
When we multiply two representations, characters have to be balanced for each conjugacy class, following this rule, we get branching rules from character tables,
1 x 1'= 1', 1 x 1' ' = 1' ', 1' x 1' ' = 1
3 x 1 = 3, 3 x 1'= 3, 3 x 1' ' = 3, 3 x 3 = 1 + 1' + 1' ' + 3 + 3
Note that 3x3 branching rule is allowed by comparing characters of LHS and RHS, but there are other possibilities also, for example 3 1s plus 3 1' plus 3 1' '
Many thanks for your answer. Also, symmetric and anti symmetric triplets cannot be distinguished using character tables. Do you agree ?
I was looking at Tinkham's book. In the appendix it discusses the group C3 which has more than one singlets. There is an extra column in it's character table at the extreme left, which can distinguish between two or more singlet representations. They may be related to a basis choice.
@ Martin,
We can write any square matrix M as, M=(M+MT)/2 + (M - MT)/2, that is a symmetric part and antisymmetric part, M=Msym + Mantisym. This relation preserves trace or character of representation meaning trace of M is the sum of the traces of symmetric and antisymmetric parts, or in other words from character tables we can write,
3x3 = 1 + 1' + 1 ' ' + 2 3S + 2 3A
But this multiplicity of 2 is not explicitly written in papers. I think they absorb a factor of two inside their triplet representation.
Another point is confusing me. Suppose a set of matrices Mi forms a representation of a group G and I separate each matrix into Ms and MA then will Msi and MAi separately form two different representations of G or not.
OK then following your line of thought, if I proceed I get the following. In the left hand side there is a tensor product of two matrices. What it means is that there exits four tensor indices which are to be contracted. If all four are contracted we get singlets and if two out of four are contracted we get matrices. Now we can contract two out of four indices in two possible ways, namely symmetric (using metric tensor) and antisymmetric (using levi-civita tensor) contractions. Consequently in the RHS we get two triplet representations namely 3S and 3A
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