qa.quantum algebra - Is there a source for a diagrammatic description of the induction functor C->Z(C)?

I suggest trying this paper of Mueger: http://arxiv.org/abs/math/0111205. In section 3.3 he defines a map $Hom_{C}(X,Y) rightarrow Hom_{Z(C)}((X,e_X),(Y,e_Y))$ (actually a trace preserving conditional expectation) which, as far as I understand, is what you are looking for. It implies $R(I(V)) = oplus_X X otimes V otimes X^*$ almost tautologically, I would say.

If I remember well, Mueger assumed $dim C$ to be non-zero, but this turned out to be always true (proven in the paper of Etingof-Nykshych-Ostrik).

Edit I read your question again, and now I see which half-braiding you mean. Proof and diagrams are almost the same as the ones used to show that $oplus_X X boxtimes X^{op}$ is a Frobenius algebra in $C boxtimes C^{op}$. I doubt it has been written in detail anywhere.

Concerning the theorem, I don't know if the following is helpful.
Denote $Q = oplus_X X boxtimes X^{op}$. $Z(C)$ is equivalent to the tensor category of $Q-Q$-bimodules. For $V in C$, $I(V)$ corresponds to $Q otimes (V boxtimes 1) otimes Q$. Notice that in general $(V^* boxtimes 1) otimes Q sim (1 boxtimes V^{op}) otimes Q$ and there is a canonical choice for this isomorphism.

For $(Y,e_Y) in Z(C)$, $(Y boxtimes 1) otimes Q$ has the structure of a $Q-Q$-bimodule: let $Q$ act on the right the obvious way, and use the half braiding $e_y$ to let $Q$ act from the left as well. All $Q-Q$ bimodules are of this form (up to isomorphism). Thus the restriction functor sends $(Y boxtimes 1) otimes Q$ to $Y$.

As you mentioned, $oplus_X X otimes V otimes X^*$ has a natural choice of half braiding: choose a basis for each $Hom_C(X otimes Y, Z)$, where $X,Y,Z$ span a complete set of irreducibles, and use them (together with the rigidity structure) to build an isomorphism $(oplus_X X otimes V otimes X^*) otimes Y sim Y otimes (oplus_Z Z otimes V otimes Z^*)$ satisfying the necessary conditions.

Or define it through this sequence of isomorphims (so maybe you can avoid diagrams, in the end):

$$(Y boxtimes 1) otimes ((oplus_X X otimes V otimes X^*) boxtimes 1) otimes Q sim (Y boxtimes 1) otimes (oplus_X (X otimes V ) boxtimes X^{op}) otimes Q$$

$$sim (Y boxtimes 1) otimes Q otimes ( V boxtimes 1) otimes Q sim (1 boxtimes Y^{op *}) otimes ((oplus_X X otimes V otimes X^*) boxtimes 1) otimes Q$$

$$sim ((oplus_X X otimes V otimes X^*) boxtimes 1) otimes ( 1 boxtimes Y^{op *}) otimes Q sim ((oplus_X X otimes V otimes X^*) boxtimes 1) otimes (Y boxtimes 1) otimes Q.$$

Thus $((oplus_X X otimes V otimes X^*) boxtimes 1) otimes Q sim Q otimes (V boxtimes 1) otimes Q$ not only as objects in $C boxtimes C^{op}$, but also as $Q-Q$-bimodules (and $(R(I(V)) sim oplus_X X otimes V otimes X^*$).

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