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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.

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Any unitary fusion category has a canonical spherical structure. See Example 2.12 of this paper and the references therein. It also follows from a more general result Prop 8.23 in this paper. So you don't need to do anything in the unitary case.

For a non-unitary fusion category, I think you're looking for Bruce Bartlett's explicit description of pivotal structures. There he shows that you can think of a pivotal structure as a choice of a number $\gamma_i$ attached to each simple object such that the condition of Theorem 5.4 holds: $$\gamma_j \gamma_k \mathrm{Id} = \gamma_i T_{jk}^i \text{ whenever $X_i$ is a summand of $X_j \otimes X_k$}.$$ Here $T_{jk}^i$ are the pivotal operators which are defined from the F-matrices by Thm 4.17. (There's a slightly tricky point there which I'm unclear on, which is that the formula in 4.17 only works for a "fair basis" which should translate into some simple condition on how you've gauaged your F-matrix, but it's not obvious to me what that condition is.) The pivotal structure is spherical if all the $\gamma_i$ are $\pm 1$.

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  • $\begingroup$ Ah, good to know. But does that mean that being "spherical" cannot even be formulated as an additional condition/structure on the level of $F$-tensors, or that those conditions/structures follow directly from the unitarity and pentagon equation? $\endgroup$ – Andi Bauer Feb 3 '18 at 17:28
  • $\begingroup$ Deleted some comments and incorporated and expanded their content as a new paragraph in the answer. $\endgroup$ – Noah Snyder Feb 3 '18 at 21:08
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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}

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  • $\begingroup$ Just to check that I understand, there’s an equation for every quadruple (a,b,c,i) where i indexes a basis for Hom(a b,c)? $\endgroup$ – Noah Snyder Feb 4 '18 at 14:37
  • $\begingroup$ Yes, and the answer has been updated. This is (partially) gone through in the proof of Proposition 3.12 of 1305.2229, particularly on pages 33-37 to go from a ribbon category $\mathcal C$ to the equations above, and then pages 41-43 to go from a set of coefficients back to a pivotal structure. $\endgroup$ – Matthew Titsworth Feb 4 '18 at 20:25
  • $\begingroup$ One more question, in this version it seems like the sphericality condition isn't exactly $\epsilon=\pm 1$ because the F-matrices in the formulas for the dimensions are not the same F-matrix. But presumably you can always gauge the F-matrices in order to make those particular two equal for every a? $\endgroup$ – Noah Snyder Feb 5 '18 at 14:19
  • $\begingroup$ Correct. Sphericality at this point is just that $\forall a, q_l(a)=q_r(a)$. The presence of the F-matrix entries here correspond to the specific choice of birth/death maps being used, as stated in (31). To obtain the condition mentioned on the pivotal coefficients, DHW choose new birth/death maps in (55) and (56) on page 43. Here (so no one has to go back and look) $u_a$ is the $F_{a a^* a}^a$ entry mentioned above. $\endgroup$ – Matthew Titsworth Feb 5 '18 at 23:08

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