# The definition of unitary fusion category

I just come across a definition of the unitary fusion category:

A fusion category $\mathcal{C}$ over the complex number is said to be unitary if we have:

1. We have a Hilbert space structure on each Hom space, and we denote the inner product by $\langle \cdot, \cdot \rangle$.
2. We have a contravariant endofunctor * on $\mathcal{C}$ which is identity on objects.
3. We have $\|fg\| \leq \|f\|\|g\|$ and $\|f^*f\| = \|f\|^2$ for each $f \in Hom(Y,Z)$ and $g \in Hom(X,Y)$.
4. We have $(f \otimes g)^* = f^* \otimes g^*$ for any morphism $f$ and $g$.
5. All structure isomorphisms for simple objects are unitary.

I am confused about the notation in condition 4. Does $\|f\| = \sqrt{ \langle f, g\rangle}$ or $\|f\|$ means the operator norm? From my understanding, I guess $\|f\|$ in the definition should means the operator norm. Would anybody please clarify this for me? Thank you!

• The hom spaces are a priori just Banach spaces. On the other hand you want to ask positivity of the involution. It turns out that for every C* category there is a faithful representation and the norm coincides with the operator norm, see projecteuclid.org/euclid.pjm/1102703884 in the case of fusion categories it is simpler because all hom spaces are finite dim – Marcel Bischoff Oct 21 '15 at 2:37
• Marcel, thank you for replying. As you said that $Hom$ is just a Banach spaces. For a fusion category $Hom$ is a finite $C^*$-algebra, thus we can make it a Hilbert space by considering the inner product induced by the trace. I guess my point is that in the definition $\|f\|$ does not equals $tr(f^*f)^{1/2}$. I think that is also what you are trying to say. If I am wrong, please let me know. – heller Oct 21 '15 at 9:58
• The scalar product should be defined by the left inverse $(x,y)=\phi_\sigma(x^\ast y)=1/d_\sigma \mathrm{tr}_\sigma(x^\ast y)$ for $\mathrm{Hom}(\sigma, \tau)$. If $\sigma$ is simple/irreducible then this coincides with the norm, because all elements are multiple of isometries. – Marcel Bischoff Oct 21 '15 at 13:21
• Can you give a reference from where you got this definition. Usually the hom spaces are only asked to be Banach spaces and that there is a scalar product follows later. Condition 3 should say there is a family norm such that.... – Marcel Bischoff Oct 21 '15 at 14:47
• I first encounter the definition in the survey paper (Definition 2.18) http://arxiv.org/abs/1503.05675, and a similar definition in http://lanl.arxiv.org/pdf/1209.2022v2.pdf. But as you said, I guess the authors just mean the operator norm. – heller Oct 22 '15 at 8:47

Given a unitary fusion category, there is more than one natural norm on the (finite dimensional) hom-space $Hom(X,Y)$.

There is the operator norm:
That's the norm under which the unitary fusion category is a $C^*$-category. Let's denote that norm $\|\,\,\|_\infty$. That norm satisfies the $C^*$-identity $\|f^*f\|_\infty=\|f\|_\infty^2$.

And there's the Hilbert-Schmidt norm:
(That's the norm used by Baez et al in Higher-Dimensional Algebra II). Let's denote that norm by $\|\,\,\|_2$. With respect to that norm, a morphism $f:X\to Y$ which factors through an irreducible object will satisfy $\|f\|_2^2 = \|f\|_\infty^2$. Using the above formula, the Hilbert-Schmidt norm of an arbitrary morphism is then uniquely determined by means of the polarization identity.

Warning:
There are more than one way to normalize the Hilbert-Schmidt norm, and so some care is needed when dealing with that notion. Given a morphism $f:X\to Y$ that factors through an irreducible object $Z$, one could also decide to let $\|f\|_2^2 = d_Z\cdot\|f\|_\infty^2$, where $d_Z$ is, e.g., the quantum dimension of the object $Z$.

Once the normalization of Hilbert-Schmidt norm has been fixed, then there is also the associated
Trace-class norm $\|\,\,\|_1$:
It satisfies $\|f^*f\|_1=\|f\|_2^2$ for any morphism $f$. It also satisfies$\|pu\|_1=\|p\|_1$ for any positive morphism $p:Y\to Y$ and unitary morphism $u:X\to Y$.