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$S U (2)$ level $k$ has $k + 1$ torus states

Physical scope

Pure mathematics; no physical premises or physical claim.

Example

For an integer level $k \geq 1$ , let $ℓ = k + 2$ and take the actual positive-trace ribbon semisimplification of the finite-dimensional integrable type-1 Lusztig divided-power $U_{q} ({s l}_{2})$ tilting modules at $q = e^{π i / ℓ}$ . Its simple labels are $a = 0, 1, \dots, k$ , and its normalized matrix is

$S_{a b} = \sqrt{\frac{2}{k + 2}} \sin ⁣ (\frac{(a + 1) (b + 1) π}{k + 2}), 0 \leq a, b \leq k .$

Its unmarked torus state space has dimension $k + 1$ . At $k = 1$ , its closed genus- $g$ state-space dimension is $2^{g}$ for every integer $g \geq 0$ . These are mathematical consequences for the specified tilting quotient, not claims about a physical realization.

Example references

The standard root-of-unity sl2 tilting ribbon quotient is modular

Modular dimensions follow from exact sewing

Facts & Assumptions

Given: An integer level $k \geq 1$ , the convention $ℓ = k + 2$ and $q = e^{π i / ℓ}$ , and the weighted RT state spaces of the actual root-of-unity quotient.

[A1]

Full AC and its inherited countable-choice consequences are retained from the modularity and weighted RT suppliers (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The standard root-of-unity sl2 tilting ribbon quotient is modular establishes the actual positive-trace tilting quotient for every $ℓ \geq 3$ , with simple labels $a = 0, \dots, ℓ - 2$ and normalized matrix $S_{a b} = \sqrt{2 / ℓ} \sin ((a + 1) (b + 1) π / ℓ)$ .

[F2]

Modular dimensions follow from exact sewing gives $\dim V (Σ_{g}) = \sum_{a \in I} S_{0 a}^{2 - 2 g}$ for all closed unmarked surfaces and $\dim V (T^{2}) = ∣ I ∣$ .

Verification

1.1A1F1givenalgebra

Since $k \geq 1$ , $ℓ = k + 2 \geq 3$ . By [F1], the actual quotient is modular and its label set is $I = {0, \dots, ℓ - 2} = {0, \dots, k}$ , with the displayed $S$ matrix under the positive-trace convention. For each $a \in I$ , $S_{0 a} = \sqrt{2 / ℓ} \sin ((a + 1) π / ℓ) \neq 0$ because $0 < (a + 1) π / ℓ < π$ .

2.1F1F2step 1.1algebra

The genus-one formula in [F2] has exponent zero and therefore contributes one for each of the $k + 1$ nonzero entries just identified. Thus $\dim V (T^{2}) = ∣ I ∣ = k + 1$ .

3.1F1F2step 1.1algebra∎

For $k = 1$ , $ℓ = 3$ and $S_{00} = S_{01} = \sqrt{2 / 3} \sin (π / 3) = 1 / \sqrt{2}$ . Applying [F2] gives, for every $g \geq 0$ , $\dim V (Σ_{g}) = 2 {(\frac{1}{\sqrt{2}})}^{2 - 2 g} = 2 \cdot 2^{g - 1} = 2^{g} .$ The exponent formula is valid also at $g = 0$ , and no limiting or path-integral argument is used.

Depends on

Dependency tree · 2 levels

Sources

Stephen F. Sawin, Quantum Groups at Roots of Unity and Modularity, arXiv:math/0308281v2

SU(2)SU(2) level kk has k+1k+1 torus states