2
$\begingroup$

I cannot quite understand the fusion rules in conformal field theories: \begin{eqnarray} \phi_i\times\phi_j=N_{ij}^{k}\phi_k, \end{eqnarray} where the coefficient $N_{ij}^{k}$ according to textbooks can take general nonnegative integers. I cannot quite understand why that $N_{ij}^{k}$ is larger than 1 is meaningful. Since $N_{ij}^{k}$ is nonzero as long as \begin{eqnarray} \langle\phi_k\phi_i\phi_j\rangle\neq0, \end{eqnarray} the specific value $N_{ij}^{k}$ seems meaningless.

$\endgroup$
3
  • 1
    $\begingroup$ You should provide additional context on what the $\phi_k$ are, what fusion rules are etc so that your question is a little more self-contained. $\endgroup$ Sep 3, 2017 at 15:11
  • $\begingroup$ Hi Smart Yao: Is your real question why the fusion structure constants $N_{ij}^{k}$ (after appropriate normalization) is integer-valued? $\endgroup$
    – Qmechanic
    Sep 3, 2017 at 17:35
  • $\begingroup$ @Qmechanic I was simply confused the mathematical meaning of larger-than-1 $N_{ij}^k$. $\endgroup$
    – Yuan Yao
    Sep 4, 2017 at 14:04

1 Answer 1

2
$\begingroup$

In general, the decomposition of a tensor product of two representation $\Gamma_1\otimes \Gamma_2$ (take them to be irreducible for simplicity) will result in a sum of irreducible representations $$ \Gamma_1\otimes \Gamma_2 = \sum_k N_{12}^k \Gamma_k\, . \tag{1} $$ Eq.(1) is called the Clebsch-Gordan series. A particular irrep $\Gamma_k$ can generically occur more than once. How many times an irrep occurs depends on $\Gamma_1$ and $\Gamma_2$.

As an example, suppose $\Gamma_1$ and $\Gamma_2$ are both the 2-dimensional $\{2,1\}$ irrep of $S_3$. Then $\Gamma_1\otimes \Gamma_2$ is a 4-dimensional reducible representation which can, by elementary use of character orthogonality, be written as $$ \{2,1\}\otimes\{2,1\}=\{3\}\oplus\{2,1\}\oplus\{1,1,1\} \tag{2} $$ so in this specific example the various $N^{\{\lambda\}}_{\{2,1\}\{2,1\}}$s are all $1$.

Looking instead the coupling $\{3,1,1\}\otimes\{3,1,1\}$ of $S_5$ irreps, one finds \begin{align} \{3,1,1\}\otimes\{3,1,1\}& = \{5\}\oplus\{4,1\}\oplus 2\{3,2\}\oplus \{3,1,1\}\\ &\quad \oplus 2\{2^21\}\oplus\{21^3\}\oplus \{1^5\}\tag{3} \end{align} and you can see that $\{3,2\}$ and $\{2^2,1\}$ actually occur twice in the decomposition.

The same can occur for continuous groups. Consider the tensor product $(1,1)\otimes(\sigma,\sigma)$ of $SU(3)$ irrep. One can show using again elementary methods that \begin{align} (1,1)\otimes(\sigma,\sigma)&=(\sigma+1,\sigma+1) \oplus 2(\sigma,\sigma)\\ &\quad \oplus (\sigma-1,\sigma-1)\oplus \hbox{other irreps}\, . \tag{4} \end{align} In this specific example, the irrep $(\sigma,\sigma)$ occurs twice in the decomposition so this particular coefficient $N^{(\sigma,\sigma)}_{(1,1)(\sigma,\sigma)}=2$.

Some observations are in order. If you think of decomposing the product of two irreps of $SU(2)$, then $N^J_{j_1,j_2}$ is either $0$ or $1$. For other $SU(n)$, as illustrated in my second example, it may be that some coefficients are greater than $1$. In general, $N_{ij}^k >1$ will occur when the representations $\Gamma_i$ and $\Gamma_j$ have weight multiplicities. The problem of constructing the highest weight for irrep $k$ has multiple solutions due to the weight multiplicities. (In $SU(n)$, both irreps must have weight multiplicities, but I'm not sure about other cases.)

In your specific examples, your fields carry a representation of some algebra. Quoting this paper by J. Fuchs:

Fusion rule algebras are certain associative algebras over the complex numbers which arise in various areas of physics and mathematics, where they describe the possible couplings among three objects out of some given class. As examples let me mention:

....

  • Truncated tensor products of unitary representations of quantum groups with deformation parameter a root of unity...

[the emphasis is mine] so one can think of the fusion rules are a generalized form of the Clebsch-Gordan series of Eq.(1), as opposed to Eqs.(2), (3) and(4) which are examples of the standard series.

$\endgroup$
3
  • $\begingroup$ Pretty nice explanation although a little (but trivial) suggestion on the missed square on l.h.s. of Eq. (2). I have a further question for my specific example of fusion rules in CFT. If $N_{ij}^k>1$, does it mean that it is necessary to have at least $N_{ij}^k$ different primary fields with the same weight $h_k$? In this sense, does it imply nondegeneracy of highest weights in minimal models? $\endgroup$
    – Yuan Yao
    Sep 4, 2017 at 14:01
  • $\begingroup$ @SmartYao I'm not an expert in CFT, but the game is counting highest weights. From a representation theory perspective $N_{ij}^k>1$ means multiple distinct highest weights, irrespective of the degeneracy of weight $h_k$ (this weight might occur multiple times in other irreps where it is not dominant). Thanks for the heads up on the typo, which is now fixed. $\endgroup$ Sep 4, 2017 at 14:11
  • $\begingroup$ I now understand the source of this larger-than-one $N_{ij}^k$, but I was wrong in my last comment in that $N_{ij}^k\leq1$ does not mean the nondegeneracy in highest weight. $\endgroup$
    – Yuan Yao
    Sep 4, 2017 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.