Return to Answer

Partial answer

The second fusion ring admits no unitary categorification. It is proved in this paper, using Character Table and Fourier Analysis. Here are some explanations:

Let $\mathcal{F}$ be a commutative fusion ring. Then its fusion matrices $(M_i)$ are commuting and normal so simultaneously diagonalisable: there is an invertible matrix $P$ such that $P^{-1}M_iP = diag(\lambda_{i,j})$. Let us call $\Lambda=(\lambda_{i,j})$ the character table of $\mathcal{F}$. Note that if $G$ is a finite group and $\mathcal{F}$ the Grothendieck ring of $Rep(G)$ then $\Lambda$ is the usual character table of $G$.

If $\mathcal{F}$ is the Grothendieck ring of a unitary fusion category, and if we choose $\lambda_{i,1}$ to be $\Vert M_i \Vert$, then by Corollary 7.5 in this paper, for every triple $(j,k,\ell)$  $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$ Note that for $Rep(G)$ this identity admits a combinatorial/probabilistic interpretation (see here).

Now observe the character table of the second fusion ring above, in particular its last column, then $$ \frac{1^3}{1} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{1^3}{6} + \frac{(-3)^3}{7} + \frac{2^3}{7} = -\frac{65}{42}<0.$$ It follows that the second fusion ring above admits no unitary categorification.

Note that the identity holds for the first fusion ring.

The first fusion ring is excluded from complex categorification in [LPR,$\S$14.1] using triangular prism equations. Note that the proof involves only the multiplicity one part of the fusion data, which is shared with the second fusion ring; thus, the second fusion ring is excluded as well.

Previously, the second fusion ring (only) was excluded from unitary categorification in [LPW,$\S$8.4] using quantum Fourier analysis. Here are some explanations:

Let $\mathcal{F}$ be a commutative fusion ring. The fusion matrices $(M_i)$ of $\mathcal{F}$ are commuting and normal, hence simultaneously diagonalisable. There exists an invertible matrix $P$ such that $P^{-1}M_iP = \mathrm{diag}(\lambda_{i,j})$. Let us denote $\Lambda=(\lambda_{i,j})$ as the character table of $\mathcal{F}$. Note that if $G$ is a finite group and $\mathcal{F}$ is the Grothendieck ring of $\mathrm{Rep}(G)$, then $\Lambda$ corresponds to the usual character table of $G$.

If $\mathcal{F}$ is the Grothendieck ring of a unitary fusion category and we choose $\lambda_{i,1}$ to be $\Vert M_i \Vert$, then by Corollary 8.5 in [LPW], for every triple $(j,k,\ell)$, we have: $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$ Note that for $\mathrm{Rep}(G)$, this identity admits a combinatorial/probabilistic interpretation (see here).

Now, consider the character table of the second fusion ring mentioned above, specifically its last column: $$ \frac{1^3}{1} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{1^3}{6} + \frac{(-3)^3}{7} + \frac{2^3}{7} = -\frac{65}{42} < 0.$$ This inequality implies that the second fusion ring does not admit a unitary categorification.

References

[LPR] Liu, Zhengwei; Palcoux, Sebastien; Ren, Yunxiang. Triangular prism equations and categorification, arXiv:2203.06522, 33 pages
[LPW] Liu, Zhengwei; Palcoux, Sebastien; Wu, Jinsong. Fusion bialgebras and Fourier analysis: analytic obstructions for unitary categorification. Adv. Math. 390 (2021), Paper No. 107905, 63 pp.

Partial answer

The second fusion ring was excluded from unitary categorification in this paper, using its Character Table and Fourier Analysis. Here are some explanations:

Let $\mathcal{F}$ be a commutative fusion ring. Then its fusion matrices $(M_i)$ are commuting and normal so simultaneously diagonalisable: there is an invertible matrix $P$ such that $P^{-1}M_iP = diag(\lambda_{i,j})$. Let us call $\Lambda=(\lambda_{i,j})$ the character table of $\mathcal{F}$. Note that if $G$ is a finite group and $\mathcal{F}$ the Grothendieck ring of $Rep(G)$ then $\Lambda$ is the usual character table of $G$.

If $\mathcal{F}$ is the Grothendieck ring of a unitary fusion category, and if we choose $\lambda_{i,1}$ to be $\Vert M_i \Vert$, then by Corollary 7.5 in this paper, for every triple $(j,k,\ell)$ $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$ Note that for $Rep(G)$ this identity admits a combinatorial/probabilistic interpretation (see here).

Now observe the character table of the second fusion ring above, in particular its last column, then $$ \frac{1^3}{1} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{1^3}{6} + \frac{(-3)^3}{7} + \frac{2^3}{7} = -\frac{65}{42}<0.$$ It follows that the second fusion ring above admits no unitary categorification.

Note that the identity holds for the first fusion ring.

Partial answer

The second fusion ring admits no unitary categorification. It is proved in this paper, using Character Table and Fourier Analysis. Here are some explanations:

Let $\mathcal{F}$ be a commutative fusion ring. Then its fusion matrices $(M_i)$ are commuting and normal so simultaneously diagonalisable: there is an invertible matrix $P$ such that $P^{-1}M_iP = diag(\lambda_{i,j})$. Let us call $\Lambda=(\lambda_{i,j})$ the character table of $\mathcal{F}$. Note that if $G$ is a finite group and $\mathcal{F}$ the Grothendieck ring of $Rep(G)$ then $\Lambda$ is the usual character table of $G$.

If $\mathcal{F}$ is the Grothendieck ring of a unitary fusion category, and if we choose $\lambda_{i,1}$ to be $\Vert M_i \Vert$, then by Corollary 7.5 in this paper, for every triple $(j,k,\ell)$ $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$ Note that for $Rep(G)$ this identity admits a combinatorial/probabilistic interpretation (see here).

Now observe the character table of the second fusion ring above, in particular its last column, then $$ \frac{1^3}{1} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{1^3}{6} + \frac{(-3)^3}{7} + \frac{2^3}{7} = -\frac{65}{42}<0.$$ It follows that the second fusion ring above admits no unitary categorification.

Note that the identity holds for the first fusion ring.

Partial answer

The second fusion ring was excluded from unitary categorification in this paper, using Fourier Analysis.

Partial answer

The second fusion ring was excluded from unitary categorification in this paper, using its Character Table and Fourier Analysis. Here are some explanations:

Let $\mathcal{F}$ be a commutative fusion ring. Then its fusion matrices $(M_i)$ are commuting and normal so simultaneously diagonalisable: there is an invertible matrix $P$ such that $P^{-1}M_iP = diag(\lambda_{i,j})$. Let us call $\Lambda=(\lambda_{i,j})$ the character table of $\mathcal{F}$. Note that if $G$ is a finite group and $\mathcal{F}$ the Grothendieck ring of $Rep(G)$ then $\Lambda$ is the usual character table of $G$.

If $\mathcal{F}$ is the Grothendieck ring of a unitary fusion category, and if we choose $\lambda_{i,1}$ to be $\Vert M_i \Vert$, then by Corollary 7.5 in this paper, for every triple $(j,k,\ell)$ $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$ Note that for $Rep(G)$ this identity admits a combinatorial/probabilistic interpretation (see here).

Now observe the character table of the second fusion ring above, in particular its last column, then $$ \frac{1^3}{1} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{1^3}{6} + \frac{(-3)^3}{7} + \frac{2^3}{7} = -\frac{65}{42}<0.$$ It follows that the second fusion ring above admits no unitary categorification.

Note that the identity holds for the first fusion ring.