When are Jones-Wenzl projectors defined?

Question

(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?

Answer 1

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.

Answer 2

If $\delta = q+ q^{-1}$ for $q$ a primitive $2m$-th root of unity and $p$ is the characteristic of $\mathbb{k}$ then $JW_n$ is defined over $\mathbb{k}$ iff $n < m$ or $n = am p^{r} - 1$ for $1\le a < p$.

This statement is equivalent to @ben-webster's condition that the quantum binomial coefficients do not vanish, but we can show it without reverting to $U_q(\mathfrak{sl}_2)$ if we like. The trick is to show that the trivial module is its own projective cover (or equivalently that no other cell module has a trivial factor), which is done here. Unfortunately the determinants of the cell modules are not easily related to quantum binomials (the algebra is almost never semi-simple) so it takes some work.

Suppose that $[n]=0$... Does $JW_{n−1 }$ exist? If not, what additional conditions are required for it to exist?

If $[n] = 0$ and $\mathbb{k}$ is a field, then $n$ is divisible by $m$. The condition above says that it must actually be a power of $p$ multiplied by $m$ to have $JW_{n-1}$ defined.