Subalgebras of the Temperley-Lieb algebra 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?
 A: 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.
