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The rational spectral R-matrix has singular parameter values

Example

Use $V = C^{2}$ and the operators of The rational spin-1/2 spectral R-matrix. In the ordered basis $(e_{1} \otimes e_{1}, e_{1} \otimes e_{2}, e_{2} \otimes e_{1}, e_{2} \otimes e_{2})$ , the matrix of $R (u) = u I + i P$ is

$(\begin{array}{cccc} u + i & 0 & 0 & 0 \\ 0 & u & i & 0 \\ 0 & i & u & 0 \\ 0 & 0 & 0 & u + i \end{array}) .$

The symmetric subspace has eigenvalue $u + i$ and dimension three, and the antisymmetric subspace has eigenvalue $u - i$ and dimension one. For $\overset{ˇ}{R} (u) = P R (u)$ , the corresponding eigenvalues are $u + i$ and $i - u$ . Hence both operators are invertible exactly when $u \notin {i, - i}$ . The parameterized Yang-Baxter identities hold for all parameter triples, including triples containing these singular parameter values, but a singular operator cannot be assigned to a braid-group generator.

Facts & Assumptions

Given: The matrices $R (u)$ and $\overset{ˇ}{R} (u)$ of The rational spin-1/2 spectral R-matrix, the identities of The rational spin-1/2 spectral R-matrix satisfies Yang-Baxter, and the braid-group representation interface.

[F1]

The rational spin-1/2 spectral R-matrix: $R (u) = u I + i P$ and $\overset{ˇ}{R} (u) = P R (u)$ .

[F2]

The rational spin-1/2 spectral R-matrix satisfies Yang-Baxter: for all $x, y, z \in C$ , both the un-checked and checked spectral equations hold with parameters $x - y$ , $x - z$ , and $y - z$ .

[F3]

Yang–Baxter operators on an object: a Yang-Baxter operator is an invertible endomorphism satisfying the braid-form equation; invertibility is part of the definition.

[F4]

A Yang–Baxter operator gives braid-group representations: an invertible Yang-Baxter operator gives a homomorphism from each Artin braid group B_n into the automorphism group of the corresponding tensor power.

Verification

1.1givenF1algebra

The flip $P$ has eigenvalue $1$ on the three-dimensional symmetric subspace, with basis $e_{1} \otimes e_{1}$ , $e_{1} \otimes e_{2} + e_{2} \otimes e_{1}$ , $e_{2} \otimes e_{2}$ , and eigenvalue $- 1$ on the one-dimensional antisymmetric subspace spanned by $e_{1} \otimes e_{2} - e_{2} \otimes e_{1}$ . Therefore $R (u) = u I + i P$ has eigenvalues $u + i$ and $u - i$ on these two subspaces, while $\overset{ˇ}{R} (u) = u P + i I$ has eigenvalues $u + i$ and $i - u$ .

2.1givenF1step 1.1algebra

At $u = i$ , $R (u)$ vanishes on the antisymmetric line and has rank three; at $u = - i$ , it vanishes on the symmetric subspace and has rank one. Thus both $R (u)$ and $\overset{ˇ}{R} (u)$ are invertible exactly away from $u = i$ and $u = - i$ . At $u = 0$ , $R (0) = i P$ and $\overset{ˇ}{R} (0) = i I$ are invertible.

3.1givenF2F3step 2.1algebra

Set $x = i$ , $y = 0$ , $z = 0$ . Then the checked identity in [F2] has the factors ${\overset{ˇ}{R}}_{1} (i)$ , ${\overset{ˇ}{R}}_{2} (i)$ , and ${\overset{ˇ}{R}}_{1} (0)$ , and both ${\overset{ˇ}{R}}_{1} (i)$ and ${\overset{ˇ}{R}}_{2} (i)$ are singular by step 2.1. Thus these maps cannot be images of braid-group generators in an automorphism group: the required invertibility in [F3] fails.

4.1givenF1F3F4step 2.1algebra∎

At $u = 0$ the checked operator is $i I$ . Its local operators on every tensor power are scalar $i I$ and satisfy the Artin braid and distant-commutativity relations; they are invertible because $i$ is nonzero. Thus the assignment $σ_{j} \mapsto i I$ does define a braid-group representation, consistent with [F4].

Depends on

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Sources

L. D. Faddeev, How Algebraic Bethe Ansatz Works for Integrable Model (standard reference, not scraped)
Pavel Etingof and Mykola Semenyakin, A Brief Introduction to Quantum Groups (standard reference, not scraped)