bio | website | chemie.uni-hamburg.de/ac/AKs/… |
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location | Hamburg | |
age | 53 | |
visits | member for | 3 years, 10 months |
seen | Oct 23 at 13:23 | |
stats | profile views | 1,196 |
knot theory dilettant!
Oct 22 |
comment |
Classification properties of fusion rings
(contd) Assume you do their construction, but no crossings are allowed in the cubic graph. I think you still can get an invariant, with "Elliott" 6j symbols. If someone could tell me the smallest self-dual, no multiplicity fusion ring that is NOT ribbon, I could try my Clebsch-O-Matic on it and settle the matter. (A try with AA=1,AB=B,B*B=1+A+2B suggests you are right but don't know whether I programmed multiplicity correctly.) |
Oct 22 |
comment |
Classification properties of fusion rings
@Turion: Thus "style" - I don't know enough on the field. |
Oct 21 |
asked | Classification properties of fusion rings |
Oct 9 |
accepted | No basis change in a fusion ring allowed? |
Oct 8 |
asked | No basis change in a fusion ring allowed? |
Jul 14 |
revised |
Non-semisimple Lie algebra tensors
Reduced it to the relevant. May go into details with new question. |
Jul 11 |
comment |
Non-semisimple Lie algebra tensors
Also: If you start with the tensor $C^i_{jk}$ and you can NOT invert $g_{lm}$ to get "the indices up", how do you arrive at $g^{no}$ at all? (I'm always working with matrix generators, thus my "complex conjugate transpose"). |
Jul 11 |
comment |
Non-semisimple Lie algebra tensors
I'm not at all familiar with Lie algebras, but obviously e.g. $C_{ijk}$ is identically zero for aff1 (because of dimension 2). Do you include this degenerate possibility in your statement? (Do such Lie algebras have a special name?) |
Jul 10 |
revised |
Non-semisimple Lie algebra tensors
More definitions. Expanded. |
Jul 9 |
asked | Non-semisimple Lie algebra tensors |
Jul 2 |
awarded | Inquisitive |
Jul 2 |
awarded | Curious |
Jun 7 |
comment |
“Multiplying” Clebsch-Gordan series
Yup, those. And after perusing his opus magnum for three years or so I understand it a bit :-) But we have digressed so much that we better take this issue to PM... (reddmannatchemiedotuni-hamburgdotde) |
Jun 2 |
comment |
“Multiplying” Clebsch-Gordan series
You're right. Problem is I have no formal education in higher math and usually don't even know the correct terms. In any case, you can peek into "Birdtracks" by Cvitanovic (he's kind of a maverick too, but at least he knows the math :-) - on p. 232 of his "Group Theory" pdf I found the 8^2=1+9+27+27. The table header lists "Rep" so it's well possible that his vector spaces are built from several irreps. |
May 30 |
comment |
“Multiplying” Clebsch-Gordan series
The span (?) of the vector space of $A\bigotimes{B}$ doesn't come out right otherwise! OK, you have then 8 irreps of (2,2,2)*(2,2,2) (I have to trust you on that) but still, it's vector space needs only four basis vectors to build the R matrix. (Is the span 8 nevertheless?) In all dimension formulas I know, the "clustered" irreps appear. |
May 28 |
comment |
“Multiplying” Clebsch-Gordan series
Thus: 2A2(16)^2=(1,8+8,1)^2=(1,8)^2+(8,1)^2+2*(1,8)*(8,1)=(1,1)+(1,8)+(1,8)+(1,10)+(1,10)+(1,27)+(1,1)+(8,1)+(8,1)+(10,1)+(10,1)+(27,1)+(8,8)+(8,8). Those are the blocks, but which must be "clustered" to get the right dimensions? |
May 28 |
comment |
“Multiplying” Clebsch-Gordan series
By reading it somewhere :-) Myself, I of course would have used the naive distributive law. But here is an even more striking example from 3A1: Since it's on the Vogel plane and a member of the quarternion series, one just needs to look up the relevant (Westbury) paper to get adjoint^2: 9^2=1+9+2+15+27+27. Dissecting this: 9 is 113+131+311, 15 is 115+151+511, 1 is 111, 2 is the other two 111+111 (from 113*113), 27 is 133+313+331 (from 113*113), other 27 is 133+313+331 (from 113*131). The distributive law gives the building blocks, but "clustering" of irreps is completely wacky. |
May 27 |
comment |
“Multiplying” Clebsch-Gordan series
@Darij: THX. Here is an example the intuitive law can't work when G=g. SU2=A1 defining dimension already works. (I label by dimension, not J.) 2*2=1+3. (2,2)*(2,2)=1+3+3+9 (so far, so good). (2,2,2)*(2,2,2)=1+9+27+27. With simple distributive law, you would get eight terms! (But (1,1,3),(1,3,1),(3,1,1) somehow fall together, and so do (1,3,3) etc., which explains the dimensions. At least they do for the span of the clebsch - the R matrix fulfils a cubic equation and that's what relevant to me.) |
May 27 |
comment |
A funny property of E8-family Clebsch-Gordan series
Notabene: For any point on the Vogel plane (i.e. R being the adjoint) the property is obvious indeed (just plug in the known dimensions and check that the formula gives a square). For the rest you still need magic :P |
May 27 |
asked | “Multiplying” Clebsch-Gordan series |