(I am hoping that someone well-versed in the literature of Temperley-Lieb algebras or of quantum groups at roots of unity can answer my question. Fingers crossed.)


Consider the Temperley-Lieb algebra on $n$ strands, $TL\_n$, viewed as an algebra over the base ring $\mathbb{Z}[\delta]$. The closed loop evaluates to $-\delta$. Thinking of $\delta$ as the quantum number $[2]$, one can view any quantum number $[m]$ as a polynomial in $\delta$. For any extension $\Bbbk$ of $\mathbb{Z}[\delta]$, we write $TL\_{n,\Bbbk}$ for the corresponding algebra after base change.


The Jones-Wenzl projector $JW\_n$ (if it exists) is the unique element of $TL\_{n,\Bbbk}$ which is orthogonal to every cup and cap, and whose coefficient of the identity diagram (in the basis of crossingless matchings) is $1$. My first (and hardest) question is:


For which $\Bbbk$ does $JW\_n$ exist?


This seems like a question for which the answer should be in the literature, but for the life of me I can not find it. I would like a precise answer! Is there an obvious representation-theoretic reason?


For example, the recursive formulae for the Jones-Wenzl projector imply that it will certainly exist when $[k]$ is invertible in $\Bbbk$, for all $k \le n$. Commonly in the literature, one is concerned with the case when $\Bbbk = \mathbb{C}$ and $\delta$ is specialized to $q+q^{-1}$ for a primitive $2m$-th root of unity $q$. In this case, $[k]$ is invertible for all $k < m$, and $[m]=0$. It is not hard to see that $JW\_m$ does not exist.


However, in this case, the literature does not seem to state which Jones-Wenzl projectors exist for $k>m$! For example, when $[2]=0$, $JW\_3$ is still well-defined (and some of its coefficients will vanish).


So my second question is:


When q is a primitive $2m$-th root of unity, which Jones-Wenzl projectors exist?


A more algebraic version of the question is:


Suppose that $[m]=0$. Note that it is entirely possible that $[k]=0$ (or is non-zero, but non-invertible) for $k<m$, when $k$ and $m$ are not relatively prime. Does $JW\_{m-1}$ exist? If not, what additional conditions are required for it to exist?



When are Jones-Wenzl projectors defined?
**tl;dr:** *The JW projector $JW\_n$ exists if and only if the q-binomial coefficient $\binom{n}{m}\_q$ (which is actually a polynomial in $\delta$!) is non-zero in your field for all $1< m<n$.*


Your question is, in essence, one about the decomposition of the tensor product $V^{\otimes n}$ over $U(\mathfrak{sl}\_2)$ at a $2m$th root of unity. You'd like to know if there is a summand which is a specialization of the $n+1$st dimensional representation at $q$ generic (since this is what the JW projector must project to). 


Since $V^{\otimes n}$ is tilting and has a 1-dimensional space of weight $n+1$, this will happen only if the tilting module with highest weight $n+1$ is simple/coincides with the Weyl module (otherwise, all the summands containing the highest weight space will have a dimension that is too large). This in turn will happen if and only if the Weyl module with highest weight $n+1$ is simple.


In order to check this, you have to see if the q-Shapovalov form stays non-degenerate (since the simple is the quotient of the Weyl module by the radical of this form. That is, if we let $v$ be the highest weight vector, we need to calculate $\langle F^{(m)}v,F^{(m)}v\rangle $; if this is 0 for any m, there is highest weight vector of weight $m$ in the Weyl module, and there is no Jones-Wenzl projector; if it's always non-zero for $m\leq n/2$, then there is a JW projector. 


If you work it out, what you'll get is the quantum binomial coefficient $\binom{n}{m}\_q$, so you want that to be non-zero for all $m$. 


**EDIT**: I've since realized there's a much better argument here. On the $n-2m$ weight space of the tensor power $V^{\otimes n}$, the operator $F^{(m)}E^{(m)}$ acts by $\binom{n}{m}\_q JW\_n$. Thus, if we work in the ring $A=\mathbb{Z}[\delta,\binom{n}{1}\_q^{-1},\binom{n}{2}\_q^{-1},\dots, \binom{n}{n-1}\_q^{-1}]$, we can write $JW\_m=\sum\_m \binom{n}{m}\_q^{-1}F^{(m)}E^{(m)} 1\_m$, the latter obviously being an endomorphism defined over $A$. We still have to give an argument that $TL\_n\otimes\_{\mathbb{Z}[\delta]} A=\mathrm{End}\_{U\_q(\mathfrak{sl}\_2)}(V^{\otimes n}\otimes\_{\mathbb{Z}[\delta]} A)$. It suffices to prove this after base change to a field, and then you can use the fact that $V^{\otimes n}$ is tilting to compute the dimension of this endomorphism ring, so the fact that $TL\_n$ acts faithfully is enough to prove it.


