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Normalized right closure preserves the diagram trace radical

Statement

Let $n \geq 1$ and $δ \in C^{\times}$ . Define the trace-pairing radical

J_{n} : = {x \in {T L}_{n} (C, δ) : τ_{n} (y x) = 0 for every y \in {T L}_{n} (C, δ)} .

Then $J_{n}$ is a two-sided ideal, $ι_{n} (J_{n}) \subseteq J_{n + 1}$ , and $E_{n} (J_{n + 1}) \subseteq J_{n}$ . Consequently the inclusion and normalized right closure descend to the quotient tower ${T L}_{n} / J_{n}$ .

Facts & Assumptions

Given: $δ \in C^{\times}$ , the trace-pairing radicals, and the maps from Finite diagram tower inclusion and normalized right closure.

[F1]

The diagram tower maps $ι_{n}$ and $E_{n}$ are bimodular and trace-preserving. (Right closure is a trace-preserving diagram-tower expectation)

[F2]

$τ_{n}$ is the normalized closure trace on the finite diagram algebra, and stacking defines its algebra products. (Temperley–Lieb planar diagram data and endomorphism modules, Normalized Temperley–Lieb trace from diagram closure)

[F3]

The closure trace is cyclic: $τ_{n} (x y) = τ_{n} (y x)$ . (The normalized closure trace is cyclic and satisfies the Markov rules)

Proof

technique · direct

1.1F2F3givenalgebra

Let $x \in J_{n}$ and $a, b, y \in {T L}_{n}$ . The defining radical condition gives $τ_{n} ((y a) x) = 0$ , so $τ_{n} (y a x) = 0$ . By cyclicity, $τ_{n} (y x b) = τ_{n} ((b y) x) = 0$ , since $b y \in {T L}_{n}$ . Thus both $a x$ and $x b$ lie in $J_{n}$ , proving that $J_{n}$ is a two-sided ideal.

2.1F1step 1.1algebra

If $x \in J_{n}$ and $z \in {T L}_{n + 1}$ , trace preservation and right bimodularity yield $τ_{n + 1} (z ι_{n} (x)) = τ_{n} (E_{n} (z ι_{n} (x))) = τ_{n} (E_{n} (z) x) = 0$ , because $E_{n} (z) \in {T L}_{n}$ . Hence $ι_{n} (x) \in J_{n + 1}$ .

3.1F1step 1.1step 2.1algebra∎

If $x \in J_{n + 1}$ and $a \in {T L}_{n}$ , then $τ_{n} (a E_{n} (x)) = τ_{n} (E_{n} (ι_{n} (a) x))$ by left bimodularity, and trace preservation makes this $τ_{n + 1} (ι_{n} (a) x) = 0$ by the definition of $J_{n + 1}$ . Thus $E_{n} (x) \in J_{n}$ . These ideal inclusions make both quotient maps independent of representatives, so the tower and right-closure maps descend.

Depends on

Used by

The positive-root Jones quotient is the type-A path tower with positive right closure Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

David Penneys, Exercises in Quantum Algebra (standard reference, not scraped)