Isomorphism symmetry group of 6j symbol

[Yoran91]

Hi everyone,

I read in 'Angular momentum in Quantum Mechanics' by A.R Edmonds that the symmetry group of the 6j symbol is isomorphic to the symmetry group of a regular tetahedron.

Is there an easy way of seeing this? I've tried working out what the symmetry relations of the 6j symbol do to the associated tetahedron, but I can't see the bigger picture. I have also tried constructing an isomorphism to S4 (which is isomorphic to the full symmetry group of a regular tetahedron), but no succes.

Can anyone help me out?

 

[DrDu]

I had a look at wikipedia:
http://en.wikipedia.org/wiki/6-j_symbol
So the 6j's are invariant under a permutation of the colums and is S3. I start from (j1 j2 j3 / j4 j5 j6), then every column is described by the upper j alone, which I take as the label of a vertex in the tetrahedron. So apparently this group is isomorphic to the interchange of any of the three vertices in one face of the tetrahedron.
But is also invariant under the exchange of the elements in one column. You can see that this is equivalent to the choice of another triangular face where e.g. J1 is replaced by j4.
So this yields all the symmetry operations of the tetrahedron.

More mathematically, the permutations of the columns span the normal sub-group S3 (C3v) of S4 (Td) and the permutation of upper/ lower elements in a column from the corresponding coset.

 

[Yoran91]

Ah, I see! Thank you for your help.

Can you explain that last part a bit more? I see that the subgroup generated by the column permutations is isomorphic to S_3, but I don't see why the subgroup generated by the permutation of the upper and lower arguments is isomorphic to the corresponding coset.

 

[DrDu]

Yes, you are right, I had second thoughts about this this morning, too.
Apparently the second class of symmetries specified in Wikipedia is not correct.
According to:
http://books.google.de/books?id=r4c...gBMAk#v=onepage&q=6j symbols symmetry&f=false

The 6j's are only invariant under a permutation of the upper and lower j's in two columns, which leaves only half of the permutations claimed in wikipedia. Then the isomorphism with the tetrahedral group becomes clear identifying the j's with the edges (not vertices) of a tetrahedron).

Interestingly, this are not all the symmetries of the 6j symbol. There are so-called Regge symmetries which were discovered later:
http://www.google.de/url?sa=t&rct=j...q4GICA&usg=AFQjCNFIFsfxJeOBEF8ZL5h49h1HwqbRfQ

 

[Vanadium 50]

The easiest way is probably to recognize that a 6j is non-zero if and only if the value of the representations' spins correspond to the edges of a tetrahedron. This causes the four triangle inequalities (which are now the faces of the tetrahedron) to be automatically fulfilled.