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Worked Fibonacci and Ising braid matrices

Example

For the Fibonacci three- $τ$ , total- $τ$ sector and the Ising three- $σ$ , total- $σ$ sector, use the F/R data of Fibonacci fusion data and chiral convention and Ising fusion data and chiral convention. In each sector the first generator is $B_{1} = diag (R_{1}, R_{2})$ and the second is $B_{2} = F B_{1} F$ . We verify both are unitary, satisfy the $B_{3}$ Artin relation, and have projective orders $10$ (Fibonacci) and $4$ (Ising).

Facts & Assumptions

Given: the two fixed two-dimensional fusion sectors and their orthonormal bases, F matrices, and R eigenvalues.

[F1]

Fibonacci fusion data and chiral convention: in the Fibonacci sector, $F = (\begin{array}{cc} u & v \\ v & - u \end{array})$ , $u = ϕ^{- 1}$ , $v = ϕ^{- 1 / 2}$ , $R_{1} = e^{- 4 π i / 5}$ , and $R_{2} = e^{3 π i / 5}$ .

[F2]

Ising fusion data and chiral convention: in the Ising sector, $F = 2^{- 1 / 2} (\begin{array}{cc} 1 & 1 \\ 1 & - 1 \end{array})$ , $R_{1} = e^{- i π / 8}$ , and $R_{2} = e^{3 i π / 8}$ .

[F3]

The braid group by Artin presentation: for three strands the defining adjacent-generator relation is $B_{1} B_{2} B_{1} = B_{2} B_{1} B_{2}$ .

[F4]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces: a unitary operator preserves the inner-product norm; in an orthonormal basis its matrix satisfies $B^{†} B = B B^{†} = I$ .

Verification

Given: the data in [F1] and [F2].

1.1F1F2F4algebra

In either sector write $B_{1} = diag (x, y)$ and $B_{2} = F B_{1} F$ . The F matrices in [F1] and [F2] are unitary by their displayed entries and the identities proved for the respective examples. Since $x, y$ are unit complex numbers, $B_{1}$ is unitary; $B_{2}$ is a product of unitary matrices and is therefore unitary as well.

1.2F1F2algebra

A projective class of $diag (x, y)$ has the same order as $r = y / x$ . In the Fibonacci case $r = e^{2 π i (- 3 / 10)}$ has exact order $10$ because $\gcd (3, 10) = 1$ . In the Ising case $r = i$ has exact order $4$ . These orders ignore only the common scalar $x$ .

2.1F1F3step 1.1algebra

In the Fibonacci case put $r = y / x = e^{- 3 π i / 5}$ and $D = diag (1, r)$ . Then $B_{1} = x D$ and $B_{2} = x C$ , where $C = F D F$ . Put $s = r^{2} + u r + 1$ . Direct multiplication gives $D C D - C D C = (\begin{array}{cc} u^{3} (1 - r) s & - u^{3} v (1 - r) s \\ - u^{3} v (1 - r) s & u^{3} (r - 1) s \end{array}) .$ Here $u^{2} + u = 1$ , $v^{2} = u$ , and $r + r^{- 1} = 2 \cos (3 π / 5) = - u$ , so $s = 0$ . Thus $D C D = C D C$ and $B_{1} B_{2} B_{1} = x^{3} D C D = x^{3} C D C = B_{2} B_{1} B_{2}$ .

2.2F2F3step 1.1algebra

In the Ising case put $r = y / x = i$ and $D = diag (1, r)$ . Then $B_{1} = x D$ and $C = F D F = \frac{1}{2} (\begin{array}{cc} 1 + r & 1 - r \\ 1 - r & 1 + r \end{array}), B_{2} = x C .$ Multiplication yields $D C D - C D C = \frac{1 + r^{2}}{4} (\begin{array}{cc} 1 - r & - (1 - r) \\ - (1 - r) & - (1 - r) \end{array}) = 0,$ because $r^{2} = - 1$ . Therefore $B_{1} B_{2} B_{1} = x^{3} D C D = x^{3} C D C = B_{2} B_{1} B_{2}$ in the Ising sector as well.

3.1F1F2F3F4step 1.1step 2.1step 2.2step 1.2∎

Steps 1.1–1.2 give the explicit unitary braid matrices, verify their Artin relation, and compute their projective generator orders in both stated sectors. No statement about other Ising sectors is used.

Depends on

Used by

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Dependency tree · two levels

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Sources

Chetan Nayak et al., Non-Abelian Anyons and Topological Quantum Computation (standard reference, not scraped)