# Finding the Drinfeld centre of a category

I have the following unitary monoidal spherical category C:

Simple objects: $\{1,x,y\}$.

Non-trivial Fusion Rules:

$$x\otimes y=x=y\otimes x$$ $$x\otimes x=1 \oplus 2x \oplus y$$

I would like to find the Drinfeld centre for this category. I am aware that the Drinfeld centre has elements of the form $(x,e_x)$ where $x\in C$ and $e_x=\{e_x(y)\in Hom(xy,yx),y\in C\}$ has to satisfy

(i) $f\otimes id_x o e_x(y)=e_x(z) o id_x \otimes f \forall f:y\rightarrow z$

(ii) $e_x(y\otimes z)=id_y\otimes e_z(z) o e_X(y) \otimes id_z \forall y,z \in C$

(iii) $e_x(1)=id_x$

I am trying to find explicit expressions for the maps $e_x(y)$.

One can consider a basis $(v^{yx}_{x_k})_io(v^{x_k}_{xy})_j$ where $x_k=1,x,y$ and $1\leq i,j \leq dim(Hom(xy,x_k))$ for the spaces Hom(xy,yx).

For example: $e_y(y)= \alpha (v^{yy}_1 o v^1_{yy})$ where $\alpha$ is some constant which we have to determine using the conditions above. If we use the first condition and choose z=1 then we get,

$$f \otimes id_y o e_y(y)=id_y \otimes f$$ where $f:y\rightarrow 1$. I am not sure how to proceed from here. Is this information enough to determine $\alpha$? I think I am missing something here and any help would be appreciated.

Edit: Since the category is spherical, one could maybe make use of the traces and quantum dimensions of the simple objects. But, I am not sure how to proceed.

• That is not enough information to determine the category or its Drinfeld center; you haven't told us anything about the associator, for example. – Qiaochu Yuan Jul 27 '18 at 20:40
• @Qiaochu Yuan. The associativity relations can be derived from the fusion rules by solving the Pentagon equation and the explicit form of all the association matrices are known. – Rajath Krishna R Jul 28 '18 at 1:02
• @QiaochuYuan Can you kindly elaborate on your comment? I am still struggling to find the answer. The appendix of arxiv.org/abs/0710.5761 contains all the half-braidings for the above category. But, I am not sure how to obtain them. – Rajath Krishna R Aug 4 '18 at 16:04