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Let $C$ be semisimple category with simple objects $X_1, \dots, X_r$.

Suppose we have a fusion relation $X_i\otimes X_j =\bigoplus_{l=1}^r N_{ij}^l X_l$.

Let $m\in \mathbb{N}$ and let $g:mX_j \to mX_j$ be a morphism given by $m$ by $m$ matrix over a field $k$.

Using the relation above, the algebra $\mathrm{End}((X_i\otimes mX_j)\otimes X_l)=\mathrm{End}(X_i \otimes (mX_j \otimes X_l))$ is equal to $$\otimes_{s=1}^r \mathrm{Mat}_{mn_{s}}(k),$$ where

$$n_s=\sum_{p=1}^rN_{ij}^p N_{pl}^s=\sum_{q=1}^r N_{iq}^s N_{jl}^q.$$

I want to show that in this algebra, we have

$$(\mathrm{id}_{X_i} \otimes g ) \otimes \mathrm{id}_{X_l}=\bigoplus_{p=1}^r \mathrm{id}_{N_{ij}^p} \otimes g \otimes \mathrm{id}_{N_{pl}^s}.$$

[That formula needs to be fixed; for example, $s$ is undefined.]

I don't understand how morphisms are changed by the fusion relation.

EDIT It is really hard to see but actually $s$ is defined. The subscript in $\otimes_{s=1}^r \mathrm{Mat}_{mn_{s}}(k)$ is $mn_{s}$.

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  • $\begingroup$ Yes, in $\otimes_{s=1}^r \mathrm{Mat}_{m n_s}(k)$, the variable $s$ is bound, and also $n_s$ is well-defined. But in $\otimes_{p=1}^r \mathrm{id}_{N_{i j}^p} \otimes \mathrm{id}_{N_{p l}^s}$, the variable $s$ is free, i.e. "undefined". $\endgroup$ – Daniel Gerigk Nov 13 '15 at 17:20
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Just some thoughts.

Let's consider a very simple fusion algebra, just some matrices. We can define the square matrix: $E_{ij} = e_i \otimes e_j$.

$$ \left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array} \right) = \left( \begin{array}{c} 1 & 0 & 0 \end{array} \right) \otimes \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right) $$

As algebras, we can always write any matrix in terms of matrix elements $E_{ij} = e_i \otimes e_j^\ast$ so that:

$$ Mat_{3 \times 3} = \oplus_{i=1}^9 \mathbf{1}$$

More complicated version of this appear in representation theory of groups. Your fusion relation says tensor products decompose into block-diagonal "matrices" whose elements are tensors of the objects you describe. In general $X \not \subseteq \mathbb{C}[G]$ these objects will not live in a group ring (these are semi-simple right?), they will be something more exotic.


This is no substitute for a more serious discussion on "monoidal categories" but more often than not they are just very fancy words for matrices.

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