## 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 $(2)$.