Inside the Temperley-Lieb algebra $TL_n$ (with loop value $\delta=-[2]$ and standard generators $e_1,\ldots,e_{n-1}$), the Jones-Wenzl idempotent is the unique non-zero element $f^{(n)}$ satisfying $$ f^{(n)}f^{(n)} = f^{(n)} \quad \textrm{and} \quad e_i\;f^{(n)} = 0 = f^{(n)}e_i \quad \textrm{for each } i.$$ Consider the Iwahori-Hecke algebra $\mathcal{H}_n$, $n\ge3$, normalized so that $(T_i-q)(T_i+q^{-1})=0$, where $q$ is generic. Let $\mathcal{I}$ be the two-sided cellular ideal generated by canonical basis element $$C_{121} = T_1T_2T_1-qT_1T_2-qT_2T_1+q^2T_1+q^2T_2-q^3.$$ The assignment $\mathcal{H}_n \rightarrow TL_n$ given by $T_i \mapsto e_i + q$ is a surjective $\mathbb{C}(q)$-algebra homomorphism with kernel $\mathcal{I}$.

We can lift the generators $e_i$ in $TL_n$ to the Kazhdan-Lusztig elements $C_i=T_i-q \in \mathcal{H}\_n$. In fact, we have $C_{121} = C_1C_2C_1 - C_1$, hence the relation down below. Rescaling a bit, $E=-\frac{1}{[3]!}C_{121}$ is an idempotent, corresponding to the partition $(1,1,1)$. Actually, all of the primitive idempotents in $\mathcal{H}_n$ that correspond to Young diagrams with more than two rows live in the ideal $\mathcal{I}$.

Now, any preimage of $f^{(n)}$ in the Hecke algebra (call it $F^{(n)}$) satisfies $$F^{(n)}F^{(n)} \equiv F^{(n)} \quad \textrm{and} \quad C_iF^{(n)} \equiv 0 \equiv F^{(n)}C_i \quad (\operatorname{mod} \mathcal{I})$$


Can we choose $F^{(n)}$ to be an idempotent in $\mathcal{H}_n$?

When $n=2$, the map is an isomorphism and we have no choice. $$F^{(2)} = \frac{1}{[2]}(T_1+q^{-1}),$$ which projects onto the $q$-eigenspace for $T_1$. In other words, it is the idempotent corresponding to the partition $(2)$.


Yes. No problem. This is the $q$-analogue of the symmetriser. In terms of $T_i$ we have $$ \frac{1}{[n]!}\sum_{\pi\in S_n} q^{\ell(\pi)} T_\pi$$ where for $T_\pi$ we take a reduced word for $\pi$ and $\ell(\pi)$ is the length of a reduced word.

There is another presentation for the Hecke algebra which I am used to writing as generators $u_i$ and defining relations $$u_i=[2]u_i$$ $$u_iu_j=u_ju_i\qquad\text{for $|i-j|>1$}$$ $$u_iu_{i+1}u_i-u_i=u_{i+1}u_iu_{i+1}-u_{i+1}$$

Strictly speaking this is the subring of the Hecke ring which is invariant under the bar involution. This algebra is defined over $\mathbb{Z}[\delta]$. The Hecke algebra is the algebra over $\mathrm{Z}[q,q^{-1}]$ obtained by the specialisation $\delta\mapsto q+q^{-1}$.

Then we have $u_i=-C_i$.

The Temperley-Lieb algebra is the quotient by $u_iu_{i\pm 1}u_i=u_i$.

This has an involution given by $u_i\leftrightarrow \delta-u_i$.

To define the idempotents we need to divide by $[n]!$.

Define $R_i(k)=1-\frac{[k]}{[k+1]}u_i$. Then these satisfy the Yang-Baxter equation $$R_i(r)R_{i+1}(r+s)R_i(s)=R_{i+1}(s)R_i(r+s)R_{i+1}(r)$$

Using this you can write everything (well a lot, at least) explicitly. For example the idempotents can be written

$$1\qquad R_1(1)\qquad R_1(1)R_2(2)R_1(1)\qquad R_1(1)R_2(2)R_1(1)R_3(3)R_2(2)R_1(1)$$ and so on. You get another sequence of idempotents by applying the involution. Under the involution we have $R_i(k)\leftrightarrow R_i(-k)$. The three string idempotent is $$\left(\frac{u_1}{\delta}\right)\left(1-\delta u_2\right)\left(\frac{u_1}{\delta}\right)$$ which gives the relation for the Temperley-Lieb algebra.

  • $\begingroup$ Bump! Along those lines, does there exist some interesting 'lift' of the Temperley-Lieb-Jones categories (i.e. to some modular category where the objects are Hecke idempotents instead)? $\endgroup$ – S Valera Oct 15 '18 at 13:41
  • $\begingroup$ Don't you want a square in the first relation, as in: $u_i^2 = [2] u_i$ ? $\endgroup$ – Ines Institoris Nov 28 '20 at 1:23
  • 1
    $\begingroup$ @InesInstitoris Yes. I just tried to fix it but I couldn't edit. $\endgroup$ – BWW Nov 29 '20 at 8:56

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