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The rational R-matrix intertwines two fundamental evaluation orders

Statement

Let $N \geq 2$ , $a, b \in C$ , and let $W_{a} = W_{b} = V_{N}^{\lor}$ carry the evaluation actions with parameters $a$ and $b$ from Evaluation modules of the type-A RTT Yangian. On the same ordered space $W_{a} \otimes W_{b}$ , write $ρ_{Δ}$ for the coproduct action and $ρ_{Δ^{o p}}$ for the action obtained by using $Δ^{o p} = σ \circ Δ$ , where $σ (x \otimes y) = y \otimes x$ . Their auxiliary generating matrices are respectively $T_{0 a} (u) T_{0 b} (u)$ and $T_{0 b} (u) T_{0 a} (u)$ . Identify the indexed dual bases with the standard basis when defining $R_{a b} = R_{N} (a - b) = (a - b) I + i P_{N}$ on this space. For every generating-series coefficient,

$ρ_{Δ} (t_{i j} (u)) R_{a b} = R_{a b} ρ_{Δ^{o p}} (t_{i j} (u)) .$

Thus $R_{a b}$ is an intertwining linear map; it is an isomorphism when $a - b \notin {i, - i}$ . At the two exceptional differences the same intertwining identity holds, but the map is singular as described in The type-A rational R-matrix satisfies Yang–Baxter and has exact singular parameters.

Facts & Assumptions

Given: The Hopf coproduct, the two evaluation modules, and the rational spectral $R$ -matrix with its spectral Yang–Baxter equation.

[F1]

The RTT Yangian has its standard Hopf structure: $Δ (T (u)) = T^{[1]} (u) T^{[2]} (u)$ is a coassociative coproduct; applying the tensor flip to $Δ$ reverses the two matrix factors.

[F2]

The gl_N evaluation formula satisfies RTT: for any ${g l}_{N}$ -module and parameter $c$ , $t_{i j} (u) = δ_{i j} - i ρ (E_{i j}) / (u - c)$ defines a Yangian action; on $V_{N}^{\lor}$ its auxiliary Lax matrix is $R_{N} (u - c)$ .

[F3]

The type-A rational R-matrix satisfies Yang–Baxter and has exact singular parameters: $R_{12} (x) R_{13} (x + y) R_{23} (y) = R_{23} (y) R_{13} (x + y) R_{12} (x)$ , and $R_{N} (d)$ is invertible exactly when $d \notin {i, - i}$ .

[F4]

Evaluation modules of the type-A RTT Yangian: $V_{N}^{\lor} (a)$ has the dual-natural action and its auxiliary series is $I + i P_{N} / (u - a)$ , whose polynomial Lax matrix is $R_{N} (u - a)$ .

Proof

1.1givenF1F2F4algebra

On the same ordered space $W_{a} \otimes W_{b}$ , the coproduct action has auxiliary matrix $T_{0} (u) = T_{0 a} (u) T_{0 b} (u)$ by [F1]. The opposite coproduct reverses those two matrix factors, giving $T_{0 b} (u) T_{0 a} (u)$ . Multiplying both by the scalar $(u - a) (u - b)$ gives the polynomial products $L_{0 a} (u - a) L_{0 b} (u - b)$ and $L_{0 b} (u - b) L_{0 a} (u - a)$ , since [F2] and [F4] give $L_{0 c} (u - c) = R_{0 c} (u - c)$ for $c = a, b$ .

2.1givenF2F3step 1.1algebra

Apply the spectral Yang–Baxter identity [F3] on auxiliary space $0$ and quantum spaces $a, b$ , using their indexed-basis identifications and spectral points $u, a, b$ . It gives $L_{0 a} (u - a) L_{0 b} (u - b) R_{a b} (a - b) = R_{a b} (a - b) L_{0 b} (u - b) L_{0 a} (u - a)$ . Since $(u - a) (u - b)$ is nonzero and invertible in the scalar Laurent-series field at infinity, cancel it from the two polynomial Lax products in step 1.1. The resulting identity is the claimed intertwining relation for $T_{0} (u)$ , and equality of its matrix entries gives it for each $t_{i j} (u)$ coefficient.

3.1givenF3step 2.1algebra∎

If $a - b \notin {i, - i}$ , [F3] makes $R_{a b}$ invertible, so the intertwiner is an isomorphism. At $a - b = i$ its kernel is antisymmetric, and at $a - b = - i$ its kernel is symmetric, by [F3]; the identity in step 2.1 still holds because the Yang–Baxter identity is polynomial and no inverse was used.

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Sources

A. I. Molev, M. L. Nazarov, and G. I. Olshanski, Yangians and classical Lie algebras (standard reference, not scraped)
L. D. Faddeev, How Algebraic Bethe Ansatz Works for Integrable Model (standard reference, not scraped)
A. P. Isaev, Lectures on Quantum Groups and Yang-Baxter Equations (standard reference, not scraped)