Does this implication hold in a ribbon fusion category? Let $\mathcal{C}$ be a ribbon fusion category in the sense of e.g. Bakalov-Kirillov Lecture Notes on tensor categories and modular functors.
The notion of the $S$-matrix is of course well-known.
Letting $Irr$ denote a set of representatives of isomorphism classes of simple objects, $|Irr| = n$, the $n \times n$ matrix is defined as having as $S_{U, V}$-entry the trace of the double braiding of the simple object $U$ with the simple object $V$.
And object of $\mathcal{C}$ is called transparent if its double-braiding with any other object is trivial.
I would like to know whether

$V \in Irr$ not transparent  $\implies$ $\sum_{U \in Irr} d_U S_{U, V} = 0$

is true, where $d_U = tr_U(id_U)$ is the quantum dimension of $U$.
I know that it's true if $\mathcal{C}$ is modular - i.e. $S$ is invertible, or equivalently, the monoidal unit 1 is the only transparent object (and of course direct sums of 1 with itself)
For simple examples, e.g. the Fibonacci categories, this can easily be verified.
Note also that if $V$ is transparent, $\sum_{U \in Irr} d_U S_{U, V} = \sum_{U \in Irr} d_U^2 d_V = d_V \cdot dim(\mathcal{C})$, where $dim(\mathcal{C})$ is the dimension of the category, which need in general not be non-zero, so the converse of the above is not true.
 A: Yes, this is true.  The argument below uses the version of the $S$-matrix that utilizes the canonical spherical structure of $\mathcal C$ coming from the fact that $\mathcal C$ is ribbon.  Some authors use another normalization $\tilde{S}_{U,V}=\theta_U^{-1}\theta_V^{-1}S_{U,V}$, so please be careful translating this argument to your preferred notation.
Following the arguments of proposition 8.13.11 in EGNO's Tensor Categories, for each $X\in\text{Irr}$, we can construct characters of the Grothendieck ring $h_X:K_0(\mathcal C)\to\mathbb K$ by the formula
$$h_X:Y\mapsto\frac{S_{X,Y}}{d_X}\,$$
for any $Y\in\text{Irr}$, then extending by linearity.  This formula makes sense over any algebraically closed base field $\mathbb K$ where $\mathsf{char}(\mathbb K)=0$, since $d_X\neq0$ for any simple object (c.f. Prop 4.8.4 in EGNO).
By lemma 8.14.1, for any $X,Y\in\text{Irr}$ if $h_X\neq h_Y$, then the following formula holds
$$0\;=\;\sum_{U\in\text{Irr}}h_X(U^*)h_Y(U)\,.$$
Now let $X=\mathbf 1$, and let $Y=V$ be your non-transparent object.  The statement that $V$ is not transparent is equivalent to the statement that $h_{\mathbf 1}\neq h_V$ as characters.  Thus we can apply the above equation to find that
$$0\;=\;\sum_{U\in\text{Irr}}\frac{S_{\mathbf 1,U^*}}{d_{\mathbf 1}}\frac{S_{V,U}}{d_V}\,=\;\frac{1}{d_V}\cdot\sum_{U\in\text{Irr}}d_U\cdot S_{V,U}\,.$$
This is equivalent to your desired conclusion, since $S_{V,U}=S_{U,V}$.
