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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?

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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.

I highly doubt that the representation is multiplicity-free, so the TL-algebra is probably quite a bit larger than that.

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  • $\begingroup$ Interesting, thank you. I didn't know that TL is the full endomorphism ring. $\endgroup$ – Ross Apr 18 at 13:31
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    $\begingroup$ @Ross $U_q(\mathfrak{sl}_2)$-invariant endomorphisms, I think $\endgroup$ – Jules Lamers Apr 18 at 21:12

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