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I've recently met with the Temperley-Lieb algebra in my work. I'm in no way a specialist, and it's seems like a pretty simple question, but nevertheless. I'm interested in the subalgebra generated by two neighbour generator $ U_i, \, U_{i+1}$. This subalgebra is 5 dimensional with a basis: $$ \mathbb{1}, \, U_i, \, U_{i+1}, \, U_i U_{i+1}, \, U_{i+1} U_i. $$ It seems very simple, and it'll be very helpful to me if the subalgebra is isomorphic to a well-known matrix algebra, for example. I've looked throught references, but mostly they talk about representation-theoretical questions, and simple algebraic ones like mine are not discussed.

Therefore, my question is the following. Is the Temperley-Lieb subalgebra generated by two neighbor generators isomorphic to some well-known algebra? Or is there any way I can check it myself?

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    $\begingroup$ In the linked Wikipedia page, there's a parameter $\delta$. What is it here? It seems the subalgebra is $TL_3(\delta)$ isn't it. It"s described nowhere? To be complete you might have considered writing down the missing relations: $U_i^2=\delta U_i$, $U_iU_{i+1}U_i=U_i$, $U_{i+1}U_iU_{i+1}=U_{i+1}$ (and $U_iU_j=U_jU_i$ for $|i-j|\ge 2$, not relevant here). $\endgroup$
    – YCor
    Commented Mar 30, 2021 at 13:01
  • $\begingroup$ I quite quickly computed that the underlying Lie algebra is solvable iff $\delta^4=1$ (this should be double checked). The information that the underlying Lie algebra or not is solvable is relevant since it gives significant restrictions on the Wedderburn-Malcev decomposition. $\endgroup$
    – YCor
    Commented Mar 30, 2021 at 16:54

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As YCor mentions in the comments, this is (isomorphic to) the algebra $TL_3$, so I will call its generators $U_1,U_2$. I will henceforth write $R := TL_3$. I do not know what your ground ring is or what your parameter $\delta$ is. If $1-\delta^2$ is invertible, then $R$ contains a special idempotent $p$ named after Jones and Wenzl. The Jones–Wenzl idempotent is very special: $U_1 p = p U_1 = U_2 p = p U_2 = 0$. In particular it is central. It follows that your ring splits as $pR \times (1-p)R$. A formula is $$ p = 1 - \frac{\delta}{\delta^2 - 1} ( U_1 + U_2) + \frac{1}{\delta^2-1} (U_1 U_2 + U_2 U_1). $$

Since $U_ip = 0$, $pR$ is definitely 1-dimensional, and so $R' := (1-p)R$ is a noncommutative 4-dimensional ring. I claim it is a matrix ring. Note that $R'$ is spanned for example by the elements $1-p$, $U_1$, $U_2$, and $U_1U_2$. Consider the left $R'$-ideal $R'U_1 \subset R'$. It is 2-dimensional: the basis vectors map to $U_1, \delta U_1, U_2 U_1$, and $U_1$ respectively, and so we can take the basis $\{U_1, U_2 U_1\}$ of this ideal. In this basis, the action of $R'$ on the ideal is $$ 1-p \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad U_1 \mapsto \begin{pmatrix} \delta & 1 \\ 0 & 0 \end{pmatrix}, \quad U_2 \mapsto \begin{pmatrix} 0 & 0 \\ 1 & \delta \end{pmatrix}, \quad U_1 U_2 \mapsto \begin{pmatrix} 1 & \delta \\ 0 & 0 \end{pmatrix}. $$ Since I insisted that $1-\delta^2$ should be invertible, these four matrices are a basis for the $2\times 2$ matrix ring, proving the claim.

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  • $\begingroup$ I should emphasize that the case $\delta = \pm 1$ is also very interesting (and YCor's calculation in the comments makes me worry that I have a sign error somewhere). Note that $U_i \mapsto -U_i$ switches the sign of $\delta$, and so we might as well assume $\delta = 1$. Then the element that would be called "$(1-\delta^2)p$", i.e. $x := (U_1 + U_2) - (U_1 U_2 + U_2 U_1)$, solves $xU_1 = U_1 x = xU_2 = U_2 x = 0$. In particular, it is a central element solving $x^2 = 0$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Mar 31, 2021 at 13:52
  • $\begingroup$ Write "$k$" for whatever ground ring you are working over (perhaps $k = \mathbb{C}$). Then $R = TL_3(\delta=1)$ is an algebra over $k[x]/(x^2)$. The quotient $R' = R / (x)$ is 4-dimensional. Again we can look at the 2-dimensional ideal $R'U_1 \subset R'$, but now the action map $R' \to \mathrm{Mat}(2)$ is not injective. Rather, its kernel is spanned by $y := U_1 - U_1U_2 = U_2U_1 - U_2$. In other words, I am saying that $R'$ contains a 1-dimensional 2-sided ideal spanned by $y$. In particular, it is not a matrix ring. So not only $R$ but also $R'$ is not a sum of matrix rings. $\endgroup$
    – Theo Johnson-Freyd
    Commented Mar 31, 2021 at 14:08

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