# Is the twist of the direct sum of simple objects the direct sum of the twists?

$\DeclareMathOperator{\id}{id} \DeclareMathOperator{\tr}{tr}$I am reading Bakalov and Kirillov's Lectures on tensor categories and modular functors, and based on notation from section $2.4$, and regarding equation $(3.1.2)$.

The setting is a semisimple abelian ribbon category $\mathcal{C}$ over $\mathbb{C}$ with simple tensor unit. The twist is $\theta$, and let $I$ be the set of isomorphism classes of objects with $\{V_i\}_{i\in I}$ some representatives of those classes.

Given fusion rules $$V_i\otimes V_j \cong \bigoplus_k N^k_{ij}V_k$$ I want to understand why \begin{align*}\tag{$\star$} \mathrm{tr}\ \theta_{V_i\otimes V_j} = \sum_k N^k_{ij} \theta_k d_k, \end{align*} where the numbers $\theta_k,\ d_k$ are defined by \begin{align*} \theta_{V_k} = \theta_k \mathrm{id}_{V_k},\qquad d_k = \mathrm{tr}\ \mathrm{id}_{V_k}. \end{align*} While this is easily verifiable in the category of finite-dimensional vector-spaces, I don't see it generally.

Let us for simplicity and wlog assume that $V_i\otimes V_j\cong V_k\oplus V_k$ for some $k$. Surely then $$\tr\theta_{V_i\otimes V_j} = \tr \theta_{V_k \oplus V_k},$$ and it is also easy that $(\star)$ would certainly follow from $\theta_{V_k \oplus V_k} = \begin{pmatrix}\theta_{V_k} & 0 \\ 0 &\theta_{V_k} \end{pmatrix}$.

This in turn would follow from $\theta_{V_k\oplus V_k} \circ i_j = i_j \circ \theta_{V_k}\$, using the properties of the biproduct.

How to see this? Or am I on a completely wrong track here?

$\DeclareMathOperator{\id}{id} \DeclareMathOperator{\tr}{tr}$Well, I once again forgot to make sure that I was aware of all the things I'm working with. This time, I forgot that the twist is a natural isomorphism $$\theta: 1_{\mathcal C} \Rightarrow 1_{\mathcal C}.$$

Then $$\theta_{V_k\oplus V_k} \circ i_j = i_j \circ \theta_{V_k}$$ is just the naturality of $\theta$. With this we conclude \begin{align*} \pi^i\circ \theta_{V_k\oplus V_k} \circ i_j &= \pi^i\circ i_j \circ \theta_{V_k} = \delta^{i}_j\ \theta_{V_k}, \end{align*} so indeed $\theta_{V_k\oplus V_k} = \theta_k\cdot\id_{V_k\oplus V_k}$. Then \begin{align*} \tr \theta_{V_i\otimes V_j} &= \tr\theta_{V_k\oplus V_k} \\ &= \theta_k\tr\id_{V_k\oplus V_k} \\ &= 2 \theta_k d_k\\ &= \sum_k N^k_{ij}\ \theta_k\ d_k\ . \end{align*}