Spencer's quantum-characteristic criterion for Jones–Wenzl existence

Let $K$ be a field, $\delta \in K$, let $\mathrm{\ell }$ be the quantum characteristic from Temperley–Lieb quantum integers and quantum characteristic, and let $p=\mathrm{char}K$ (write $p=\mathrm{\infty }$ in characteristic zero). A normalized idempotent $e\in {\mathrm{T}\mathrm{L}}_n(K,\delta )$ satisfying $\epsilon (e)=1$ and $d_{i}e=0$ for every $1\le i<n$ exists exactly in the following cases:

1. if $\mathrm{\ell }=\mathrm{\infty }$, every finite $n$ satisfies the criterion $n<\mathrm{\ell }$;
2. if $\mathrm{\ell }<\mathrm{\infty }$ and $p$ is finite, then $n<\mathrm{\ell }$ or $n=ap(i)-1$ for some $i\ge 1$ and integer $1\le a<p$;
3. if $\mathrm{\ell }<\mathrm{\infty }$ and $p=\mathrm{\infty }$, then $n<\mathrm{\ell }$ or $n=a\mathrm{\ell }-1$ for some positive integer $a$.

The criterion is over fields. It does not assert descent to an arbitrary coefficient ring.

The proof uses an integral diagram representation and an explicit quantum-binomial zero-pattern argument. It does not infer arbitrary-field behavior from the ordinary generic quantum-group centralizer theorem, and does not claim the full Spencer composition-factor classification. The latter is a substantially richer theorem unnecessary for this particular existence criterion. A failed rank-by-rank Jones–Wenzl recursion is not a nonexistence test; for example the characteristic-zero finite-$\mathrm{\ell }$ branch includes $n=a\mathrm{\ell }-1$ even when an earlier recursion denominator vanished.