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The type-A rational R-matrix satisfies Yang–Baxter and has exact singular parameters

Statement

Let $N \geq 2$ and let $R_{N} (x) = x I + i P_{N}$ be as in The type-A RTT Yangian in the rational i-normalization. For all $a, b, c \in C$ , with $x = a - b$ , $y = b - c$ , and $z = a - c = x + y$ ,

$R_{12} (x) R_{13} (z) R_{23} (y) = R_{23} (y) R_{13} (z) R_{12} (x)$

on $V_{N}^{\otimes 3}$ . Moreover, $R_{N} (x)^{- 1} = (x I - i P_{N}) / (x^{2} + 1)$ when $x \notin {i, - i}$ . At $x = i$ the kernel is $⋀^{2} V_{N}$ and the rank is $N (N + 1) / 2$ ; at $x = - i$ the kernel is ${Sym}^{2} V_{N}$ and the rank is $N (N - 1) / 2$ .

Facts & Assumptions

Given: $N \geq 2$ , the standard basis of $V_{N}$ , and arbitrary $a, b, c \in C$ .

[F1]

The type-A RTT Yangian in the rational i-normalization: $P_{N} (e_{j} \otimes e_{k}) = e_{k} \otimes e_{j}$ , $R_{N} (x) = x I + i P_{N}$ , and $P_{i j}$ swaps the indicated tensor positions.

[F2]

The rational spin-1/2 spectral R-matrix: on the ordered basis of $C^{2}$ , $P (e_{j} \otimes e_{k}) = e_{k} \otimes e_{j}$ and $R (x) = x I + i P$ .

[F3]

The rational spin-1/2 spectral R-matrix satisfies Yang-Baxter: for all complex spectral points, the corresponding un-checked three-leg operators satisfy the parameterized Yang–Baxter identity.

Proof

1.1givenF1F2F3

For $N = 2$ , the ordered basis and flip in [F1] are exactly those in [F2], so $R_{2} (x) = R (x)$ on the same $C^{2} \otimes C^{2}$ . The un-checked Yang–Baxter identity for this case is [F3].

1.2givenF1algebra

For $N \geq 3$ , put $s = P_{12}$ , $t = P_{13}$ , and $r = P_{23}$ . Acting on any basis tensor $e_{p} \otimes e_{q} \otimes e_{ℓ}$ , the products $s t, t r, r s$ all send $(p, q, ℓ)$ to $(q, ℓ, p)$ , while $s r, r t, t s$ all send it to $(ℓ, p, q)$ ; also $s t r = r t s$ as both send $(p, q, ℓ)$ to $(ℓ, q, p)$ . Hence, as operators, with $A = s t = t r = r s$ and $B = s r = r t = t s$ , those permutation identities hold on the whole basis.

1.3givenF1algebra

The flip satisfies $P_{N}^{2} = I$ , so direct multiplication gives $(x I + i P_{N}) (x I - i P_{N}) = (x^{2} + 1) I = (x I - i P_{N}) (x I + i P_{N})$ . Over $C$ , $x^{2} + 1 \neq 0$ exactly when $x \notin {i, - i}$ ; division by this scalar gives the stated inverse.

2.1givenF1step 1.1step 1.2algebra

Expanding the left side gives $x y z I + i (y z s + x y t + x z r) - (y s t + z s r + x t r) - i s t r$ ; the right side gives $x y z I + i (y z s + x y t + x z r) - (x r t + z r s + y t s) - i r t s$ . By step 1.2, the quadratic parts are respectively $- ((x + y) A + z B)$ and $- (z A + (x + y) B)$ , which agree since $x + y = z$ ; the scalar and linear parts match termwise and the cubic terms match by $s t r = r t s$ . This proves the Yang–Baxter identity for all $N \geq 3$ , and step 1.1 covers $N = 2$ .

3.1givenF1step 1.3algebra∎

The operators $P_{+} = (I + P_{N}) / 2$ and $P_{-} = (I - P_{N}) / 2$ are complementary projectors because $P_{N}^{2} = I$ . Their images have bases $e_{j} \otimes e_{j}$ , $e_{j} \otimes e_{k} + e_{k} \otimes e_{j}$ and $e_{j} \otimes e_{k} - e_{k} \otimes e_{j}$ for $j < k$ , respectively, so their dimensions are $N (N + 1) / 2$ and $N (N - 1) / 2$ . On these subspaces $R_{N} (x)$ acts by $x + i$ and $x - i$ ; therefore at $x = i$ it has kernel $im P_{-} = ⋀^{2} V_{N}$ , and at $x = - i$ it has kernel $im P_{+} = {Sym}^{2} V_{N}$ , with the ranks stated.

Diagram

Each arrow is a composite endomorphism of $V_{N}^{\otimes 3}$ , with its rightmost factor applied first. The diagram records the two sides of the equation just proved; the equality follows from the basis-permutation calculation in the proof, not from the diagram.

Depends on

Used by

Dependency tree · two levels

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Sources

A. I. Molev, M. L. Nazarov, and G. I. Olshanski, Yangians and classical Lie algebras (standard reference, not scraped)
F. Loebbert, Lectures on Yangian Symmetry (standard reference, not scraped)
A. P. Isaev, Lectures on Quantum Groups and Yang-Baxter Equations (standard reference, not scraped)