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The Temperley Lieb algebra $TL_n$ at roots of unity is not semisimple. The standard representations $V_{n,p}$ are indecomposable but, in general, not irreducible. If $K_{n,p}$ is the sub-representation of $V_{n,p}$ given by the kernel of the usual bilinear form, the quotients $V_{n,p}/K_{n,p}$ are irreducible. For one dimensional representations, we have the trivial representation, and additionally, we sometimes have certain special one dimensional representations at roots of unity - for example at $\delta = 1$ ($q = \zeta_{12}$), we have two additional one dimensional representations - If we let $u_1,..,u_{n-1}$ be the generators of $TL_n$, we could send every $u_i$ to either $+1$ or $-1$, and this would define a representation. My question is: are there any others? Could there by some other indecomposable representation which is not listed above?

Thank you in advance.

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  • $\begingroup$ David Ridout kindly informed me that the answer is in: arxiv.org/pdf/1605.05159.pdf. This is Theorem 3.10. The notation of that theorem is as follows: $S_k$ is the standard representation, $P_k$ is the projective cover of the irreducible obtained by quotienting by the radical (the kernel of the bilinear form). $B^l_k$ and $T^{l'}_k$ were new to me, and are defined in Section 3.1 (page 20). $\endgroup$ Commented Jun 24, 2020 at 7:36

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