TITLE: What do "pivotal" and "spherical" mean for (unitary) fusion categories on the level of the $F$-symbols?
QUESTION [7 upvotes]: For me, a fusion category (over $\mathbb{C}$) is just a tensor $F$ (the associator, with $6$ simple-object labels and $4$ fusion space indices) and a tensor $d$ (the quantum dimensions, with one simple-object label), such that the $F$-tensor fulfils the pentagon equation. (This seems to be the most useful language for physics.)
If furthermore the fusion category is unitary, this means that (one can find a basis for the fusion spaces such that) the $F$-tensor is unitary when interpreted as a linear map between two groups of its indices.
What extra structure/conditions do I have to add in this tensor language to make the (unitary) fusion category pivotal/spherical?
For example, it seems that a unitary fusion category is spherical if the $F$-tensor is symmetric under the full tetrahedral symmetry group (after normalizing with the square roots of the quantum dimensions of four of its simple-object labels). This seems not to be the most general case, however.

REPLY [7 votes]: This is really an answer to the question in the comment of Noah's answer.
Let $F$ be a solution to the pentagon equations for some fusion category $\mathcal C$. Pivotal structures on $\mathcal C$ should be in $1-1$ correspondence with solutions to the polynomial equations
\begin{equation}
\epsilon_c^{-1} \epsilon_b \epsilon_a = \sum_{s=1}^{N_{b c^*}^{a^*}}\sum_{t=1}^{N_{c^* a}^{b^*}} 
F_{abc^*}^{\mathbf 1}\lbrack\begin{smallmatrix}i & c & 1 \\
1 & a^* & s\end{smallmatrix}\rbrack
F_{bc^*a}^{\mathbf 1}\lbrack\begin{smallmatrix}s & a^* & 1 \\
1 & b^* & t\end{smallmatrix}\rbrack
F_{c^* ab}^{\mathbf 1}\lbrack\begin{smallmatrix}t & b^* & 1 \\
1 & c & i\end{smallmatrix}\rbrack
\end{equation}
for all $i \in \lbrace 1,\ldots, N_{ab}^c\rbrace$, where $N_{ab}^c$ is the dimension of $hom(a \otimes b,c)$. The $\epsilon_a$ are referred to as pivotal coefficients. There are various ways in which the pivotal coefficients and quantum dimensions for a given solution can be played around with, given our initial solution $F$ to the pentagon equations. For ease, we'll go ahead and assume that our $F$ matrices use basis choices which agree with those in 1305.2229 (which are pretty standard choices), in which case the left and right quantum dimensions of an object $a$ are related to $(F,\epsilon)$ via
\begin{align}
q_l(a) &= \epsilon_a\left (
F_{a^*a a^*}^{a^*}\lbrack\begin{smallmatrix}1 & 1 & 1 \\
1 & 1 & 1\end{smallmatrix}\rbrack\right)^{-1}, &
q_r(a) &= \left (\epsilon_a
F_{a a^* a}^{a}\lbrack\begin{smallmatrix}1 & 1 & 1 \\
1 & 1 & 1\end{smallmatrix}\rbrack\right)^{-1}.
\end{align}