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(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?

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up vote 7 down vote accepted

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 E^{m}v,E^mv\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 $m$.

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Thanks Ben, that's a great answer. It's certainly believable that this could be the answer over $\mathbb{Z}$ too. I don't think this representation theory argument immediately implies this though - usually arguments involving cellular intersection forms require the base ring to be local in order to guarantee that, for instance, idempotents in the associated graded will lift to true idempotents. –  Ben Aug 1 '13 at 19:21
    
For what its worth, when $[m]=0$, one can observe that $\binom{m-1}{k}_q$ is invertible for all $k$. So, as desired, $JW_{m-1}$ appears to be defined. –  Ben Aug 1 '13 at 19:57
    
Finally, do you know of any reasonable paper where I could quote this result, or something similar? Is this fact well-known enough to be "folklorish?" –  Ben Aug 1 '13 at 20:03
    
I certainly don't know of a reference; knowing the answer, it might not be too hard to check that in the JW is defined in this case. There are formulas for the coefficients in a paper of Morrison though I don't see instantly why they don't blow up if the qbc is non-zero. This would have the additional advantage of working over the integers. –  Ben Webster Aug 1 '13 at 21:09
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