Background
----------


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
$$F^{(n)}F^{(n)} \equiv F^{(n)} \quad \textrm{and} \quad C\_iF^{(n)} \equiv 0 \equiv F^{(n)}C\_i \quad (\operatorname{mod} \mathcal{I})$$


Question
--------



> 
> 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 $(2)$.



Do Jones-Wenzl idempotents lift to anything interesting in the Hecke algebra?
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
$$u\_i^2=[2]u\_i$$
$$u\_iu\_j=u\_ju\_i\qquad\text{for $|i-j|>1$}$$
$$u\_iu\_{i+1}u\_i-u\_i=u\_{i+1}u\_iu\_{i+1}-u\_{i+1}$$


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.


