I'm reading up on quantum groups and their applications and I've come across a question I just can't find an answer to. I know about the basic representation theory of $U\_q(\mathfrak{sl}\_2)$ and I know that when $q$ is a primitive root of unity (say, not equal to 1 or -1) then there are two irreducible representations in each dimension strictly smaller than the degree of $q$ and an infinite family of them in the dimension equal to this degree. Now, I've often heard people say that there is however some sort of finiteness to its category of representations because all the rep's of dimension $deg(q)$ are "of quantum dimension zero". The only notion of "quantum dimension" I know of is in pivotal tensor categories (which the aforementioned are) by taking the trace of identity morphisms (or the pivotal structure maps if you will) but I don't see why this dimension should vanish for these objects. Am I interpreting this right but just am too blind to see it? 


Edit:
Suppose $d=min(n\in \mathbb{N}:q^n=1)$ then by $deg(q)$ I mean the number $d$ if $d$ is odd and $d/2$ if $d$ is even. An example of the context I'm talking about would be the paper [Topological invariants from non-restricted quantum groups](http://arxiv.org/pdf/1009.4120.pdf) as I'm trying to deal with these Turaev-Viro-type models which should be formulated over fusion categories (so in particular categories with finitely many simple objects) so that the sum over dimensions in their definition does not diverge. However, this is unfortunately all hear-say my advisors have thrown at me making it rather difficult for me to actually find reliable sources. 



$U_q(\mathfrak{sl}_2)$ representations of "quantum dimension" zero
Yes, you are interpreting it right. For $U\_q(\mathfrak{sl}\_2)$, there's a very concrete interpretation of the quantum dimension: it's the trace (in the usual sense) of the element $K$. So, on the Weyl module of highest weight $n$ (the representation generated by $v$ with the relation $Kv=q^nv$ and $F^{(n+k)}v=E^{(k)}v=0$ for all $k>0$), the quantum dimension is $\frac{q^{n+1}-q^{-n-1}}{q-q^{-1}}=q^n+q^{n-2}+\cdots+q^{-n}$. So, if $q^{2n+2}=1$, then this representation has quantum dimension 0. 


