Temperley-Lieb Diagrams and Representations of U_q(sl_2)

A Temperley-Lieb diagram is a crossingless matching of $$2n$$ points. We think of this matching as living in a rectangle, with $$n$$ points on top and the other $$n$$ on the bottom. To $$n$$ points we can assign $$V^{\otimes n}$$, where $$V$$ is a special $$2$$-dimensional representation of $$U_q(sl_2)$$. To every Temperley-Lieb diagram, one assigns a map $$V^{\otimes n} \to V^{\otimes n}$$. This map is built from the "cups" and "caps" in the diagram, which correspond to evaluation and coevaluation maps. I think this is essentially where WRT invariants come from.

My question is as follows: suppose we have a $$\mathbb{Z}[q,q^{-1}]$$-linear combination of Temperley-Lieb diagrams, such that the induced endomorphism of $$V^{\otimes n}$$ is an isomorphism. Does it follow that the original linear combination is the scaled identity diagram, which consists of $$n$$ vertical strands? In other words, are the isomorphisms of $$V^{\otimes n}$$ which are induced by Temperley-Lieb diagrams exactly those which are scaled identity?

Absolutely not. The representation $$V^{\otimes n}$$ is not irreducible and the Temperley-Lieb algebra is $$\mathrm{End}_{U_q(\mathfrak{sl}_2)}(V^{\otimes n})$$. So, at the very least, one can rescale the irreducible factors independently.
• @Ross $U_q(\mathfrak{sl}_2)$-invariant endomorphisms, I think – Jules Lamers Apr 18 at 21:12