- #1

- 37

- 0

I read in Edmond's 'Angular momentum in Quantum Mechanics' that the symmetry group of the 9j symbol is isomorphic to the group [itex]S_3 \times S_3 \times S_2[/itex].

Why is this? Can anyone shed some light on this?

- Thread starter Yoran91
- Start date

- #1

- 37

- 0

I read in Edmond's 'Angular momentum in Quantum Mechanics' that the symmetry group of the 9j symbol is isomorphic to the group [itex]S_3 \times S_3 \times S_2[/itex].

Why is this? Can anyone shed some light on this?

- #2

Bill_K

Science Advisor

- 4,155

- 195

Likewise, the Wikipedia page on 9-j symbols describes it this way.we may permute the rows or columns in the matrix forming the 9-j symbol, or transpose the matrix itself...The symmetry group [is] the product of the three permutation groups of three, three and two objects respectively.

- #3

- 37

- 0

Labelling the rows [itex]r_1,r_2,r_3[/itex] and the columns [itex]c_1,c_2,c_3[/itex], it's easy to show that the subgroups of the row and column operations are both isomorphic to [itex]S_3[/itex]. Since any row permutation does not affect the order of the [itex]c_i[/itex], its an element of [itex]S_3 \times e[/itex] and any column permutation is in [itex]e \times S_3[/itex] in the same way.

So the subgroup of all symmetry operations not containing a transposition of the array is [itex]S_3 \times S_3[/itex]. But how do you take the transpositions into account?

I see that relevant subgroup is [itex]S_2[/itex], but I don't see exactly how you go from [itex]S_3 \times S_3[/itex] to [itex]S_3 \times S_3 \times S_2[/itex] by taking the transpositions into account.

- Last Post

- Replies
- 4

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 3K

- Replies
- 4

- Views
- 677

- Replies
- 29

- Views
- 5K

- Replies
- 5

- Views
- 5K

- Replies
- 5

- Views
- 2K

- Replies
- 11

- Views
- 3K

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 2K