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The algebraic Verlinde formula

Statement

Let $C$ be modular with $S = s / D$ as in Modular ribbon fusion category and Hopf-link data. For simple labels $i, j, k$ ,

$N_{i j}^{k} = \sum_{x \in I} \frac{S_{i x} S_{j x} S_{k^{*} x}}{S_{0 x}} .$

For simultaneous diagonalization, define the fusion matrix $N_{i}$ with rows indexed by the second input $j$ , columns indexed by the output $k$ , and entries $(N_{i})_{j, k} = N_{i j}^{k}$ . Then

$N_{i} S = S diag ⁣ {(\frac{S_{i x}}{S_{0 x}})}_{x \in I} .$

This is the algebraic formula; no complex conjugation is used.

Facts & Assumptions

Given: A complex modular ribbon fusion category, its Hopf-link matrix $s$ , chosen square root $D$ , normalized matrix $S = s / D$ , and fusion coefficients $N_{i j}^{k}$ .

[F1]

The Hopf-link character identity is $s_{x i} s_{x j} = d_{x} \sum_{k} N_{i j}^{k} s_{x k}$ , with $h_{x} (i) = s_{x i} / d_{x}$ a unital character (Balancing and the Hopf-link character identity).

[F2]

The matrix identity is $s^{2} = D^{2} C$ , where $C_{i j} = δ_{i, j^{*}}$ , and therefore $\sum_{x} s_{x ℓ} s_{x, k^{*}} = D^{2} δ_{ℓ, k}$ (The modular Hopf-link matrix squares to D² times duality).

[F3]

$S$ is symmetric, $S_{0 x} = d_{x} / D$ , and $S$ is invertible (Modular ribbon fusion category and Hopf-link data, Balancing and the Hopf-link character identity, The modular Hopf-link matrix squares to D² times duality).

Proof

Given: The data in the Statement.

1.1F1F2givenalgebra

Divide the character identity in [F1] by $d_{x}$ to get $s_{x i} s_{x j} / d_{x} = \sum_{ℓ} N_{i j}^{ℓ} s_{x ℓ}$ . Multiply by $s_{x, k^{*}}$ and sum over $x$ . By [F2], the right side is $\sum_{ℓ} N_{i j}^{ℓ} D^{2} δ_{ℓ, k} = D^{2} N_{i j}^{k}$ . Thus $\sum_{x} s_{x i} s_{x j} s_{x, k^{*}} / d_{x} = D^{2} N_{i j}^{k}$ .

2.1F2F3step 1.1algebra

Since $S_{0 x} = d_{x} / D$ and $S = s / D$ , dividing step 1.1 by $D^{2}$ rewrites it as $N_{i j}^{k} = \sum_{x} S_{i x} S_{j x} S_{k^{*} x} / S_{0 x}$ , using symmetry of $S$ to exchange $S_{x i}$ with $S_{i x}$ .

3.1F1F3givenalgebra∎

For the stated matrix convention, $(N_{i} S)_{j, x} = \sum_{k} N_{i j}^{k} S_{k x}$ . The character equation in [F1] gives $\sum_{k} N_{i j}^{k} S_{k x} = (S_{i x} / S_{0 x}) S_{j x}$ . Hence $N_{i} S = S diag (S_{i x} / S_{0 x})$ . Since $S$ is invertible, it simultaneously diagonalizes these explicitly indexed fusion matrices.

Depends on

Used by

Modular Hom state spaces have exact sewing inverses and the dimension formula Lemma

Dependency tree · two levels

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Sources

Etingof, Gelaki, Nikshych, and Ostrik, Tensor Categories, Corollaries 8.14.4–8.14.5 (standard reference, not scraped)
Bakalov and Kirillov, Lectures on Tensor Categories and Modular Functor, §3.1 (standard reference, not scraped)