(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?


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Ben
    Aug 1, 2013 at 19:21
  • $\begingroup$ 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. $\endgroup$
    – Ben
    Aug 1, 2013 at 19:57
  • 1
    $\begingroup$ 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?" $\endgroup$
    – Ben
    Aug 1, 2013 at 20:03
  • $\begingroup$ 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. $\endgroup$
    – Ben Webster
    Aug 1, 2013 at 21:09
  • $\begingroup$ Sorry for being lame, but in what category is your $\operatorname{End}$ defined? (Endomorphisms of what-modules?) $\endgroup$ Sep 16, 2014 at 1:28

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.