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Truncated sl2 fusion data give the sine Hopf-link matrix

Statement

Let $ℓ \geq 3$ , $t = π / ℓ$ , and suppose a spherical ribbon fusion category has simple labels $a = 0, \dots, ℓ - 2$ with

$N_{a b}^{c} = 1 for c = ∣ a - b ∣, ∣ a - b ∣ + 2, \dots, \min (a + b, 2 ℓ - 4 - a - b),$

and all other $N_{a b}^{c} = 0$ , with $d_{a} = \sin ((a + 1) t) / \sin t$ and $θ_{a} = e^{i t a (a + 2) / 2}$ . Then

$s_{a b} = \frac{\sin ((a + 1) (b + 1) t)}{\sin t}, D^{2} = \frac{ℓ}{2 \sin^{2} t}, S_{a b} = \sqrt{\frac{2}{ℓ}} \sin ((a + 1) (b + 1) t) .$

The sine matrix is invertible and $S^{2} = I$ . This is a conditional calculation from the displayed category data; it does not assert that a particular tilting quotient has those data.

Facts & Assumptions

Given: A spherical ribbon fusion category with the labels, fusion multiplicities, dimensions, and twists in the Statement.

[A1]

The complete truncated fusion rule, dimension function, and twist convention displayed in the Statement are hypotheses of this conditional lemma, not an excerpt from an unavailable supplier.

[F1]

The balancing equation and the fact that $h_{a} (b) = s_{a b} / d_{a}$ is a fusion-ring character are proved in Balancing and the Hopf-link character identity.

[F2]

The finite sine-transform orthogonality identity is proved in Finite sine-transform orthogonality at level ell.

Proof

Given: The hypotheses in [A1].

1.1A1F1algebra

For each fixed $a$ , put $h_{a} (b) = s_{a b} / d_{a}$ . By [F1], $h_{a} (0) = 1$ and the truncated products $b \otimes 1 = (b - 1) \oplus (b + 1)$ for $1 \leq b \leq ℓ - 3$ give $h_{a} (b + 1) = h_{a} (1) h_{a} (b) - h_{a} (b - 1)$ . The endpoint product $(ℓ - 2) \otimes 1 = ℓ - 3$ gives the corresponding terminal recurrence.

2.1A1F1step 1.1algebra

For $1 \leq a \leq ℓ - 3$ , balancing gives $s_{a 1} = θ_{a}^{- 1} θ_{1}^{- 1} (θ_{a - 1} d_{a - 1} + θ_{a + 1} d_{a + 1}) = q^{- (a + 2)} d_{a - 1} + q^{a} d_{a + 1}$ , where $q = e^{i t}$ . Substituting the sine dimensions and using $e^{- i (a + 2) t} \sin (a t) + e^{i a t} \sin ((a + 2) t) = \sin (2 (a + 1) t)$ yields $h_{a} (1) = 2 \cos ((a + 1) t)$ . For $a = 0$ , the single summand in $0 \otimes 1$ gives $h_{0} (1) = d_{1} = 2 \cos t$ . For $a = ℓ - 2$ , the single summand $ℓ - 3$ gives $s_{a 1} = q^{- ℓ} d_{ℓ - 3} = - 2 \cos t = 2 \cos ((ℓ - 1) t) d_{ℓ - 2}$ , so the same formula holds at the upper endpoint.

3.1A1step 1.1step 2.1algebra

The sequence $g_{b} = \sin ((a + 1) (b + 1) t) / \sin ((a + 1) t)$ has $g_{0} = 1$ , $g_{1} = 2 \cos ((a + 1) t)$ and satisfies $g_{b + 1} = 2 \cos ((a + 1) t) g_{b} - g_{b - 1}$ by the sine addition identity. Steps 1.1 and 2.1 therefore give $h_{a} (b) = g_{b}$ for all $0 \leq b \leq ℓ - 2$ by induction. Multiplying by $d_{a}$ yields the displayed formula for $s_{a b}$ .

4.1A1F2step 3.1algebra∎

By [F2] with its two sine indices equal to $1$ , $\sum_{a = 0}^{ℓ - 2} d_{a}^{2} = ℓ / (2 \sin^{2} t)$ ; choose the positive square root $D = \sqrt{ℓ / 2} / \sin t$ , which is permitted because $0 < t < π$ and each $d_{a} > 0$ . Dividing $s$ by $D$ gives the displayed $S$ . Applying [F2] to rows of $S$ proves $S^{2} = I$ ; hence $S$ and $s$ are invertible.

Depends on

Used by

A genus-two RT dimension from factorization Example
The ell=3 framing anomaly and Kirby blow-up Example
The sl2 modular data at ell=3 Example
The sl2 modular data at ell=4 Example
The standard root-of-unity sl2 tilting ribbon quotient is modular Theorem

Dependency tree · two levels

15 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

Bakalov and Kirillov, Lectures on Tensor Categories and Modular Functor, §3.3 (standard reference, not scraped)
Stephen F. Sawin, Quantum Groups at Roots of Unity and Modularity, §3.3 (standard reference, not scraped)