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The rational spin-1/2 spectral R-matrix satisfies Yang-Baxter

Statement

Let $V$ , $R (u)$ , $R_{i j} (u)$ , and ${\overset{ˇ}{R}}_{j} (u)$ be as in The rational spin-1/2 spectral R-matrix. For arbitrary $x, y, z \in C$ , set $u = x - y$ , $v = y - z$ , and $w = x - z = u + v$ . In ${End}_{C} (V^{\otimes 3})$ the un-checked operators satisfy

$R_{12} (u) R_{13} (w) R_{23} (v) = R_{23} (v) R_{13} (w) R_{12} (u) .$

The checked operators also satisfy the braid-form spectral identity

${\overset{ˇ}{R}}_{1} (u) {\overset{ˇ}{R}}_{2} (w) {\overset{ˇ}{R}}_{1} (v) = {\overset{ˇ}{R}}_{2} (v) {\overset{ˇ}{R}}_{1} (w) {\overset{ˇ}{R}}_{2} (u) .$

Facts & Assumptions

Given: The fixed space $V = C^{2}$ and the endomorphisms defined in The rational spin-1/2 spectral R-matrix, together with arbitrary $x, y, z \in C$ .

[F1]

The rational spin-1/2 spectral R-matrix: $P_{12}$ , $P_{13}$ , and $P_{23}$ act on the standard tensor basis by the corresponding transpositions; $R_{i j} (t) = t I + i P_{i j}$ and ${\overset{ˇ}{R}}_{1} (t) = t P_{12} + i I$ , ${\overset{ˇ}{R}}_{2} (t) = t P_{23} + i I$ .

Proof

1.1givenF1algebra

Write $s = P_{12}$ , $t = P_{13}$ , and $r = P_{23}$ . On each basis tensor $e_{a} \otimes e_{b} \otimes e_{c}$ , the products $s t$ , $t r$ , and $r s$ all give $e_{b} \otimes e_{c} \otimes e_{a}$ ; the products $s r$ , $r t$ , and $t s$ all give $e_{c} \otimes e_{a} \otimes e_{b}$ ; and $s t r$ and $r t s$ both give $e_{c} \otimes e_{b} \otimes e_{a}$ . Also $s r s$ and $r s r$ both give $e_{c} \otimes e_{b} \otimes e_{a}$ , while $s^{2} = r^{2} = I$ . Since the eight elementary tensors form a basis, these are equalities of endomorphisms.

2.1givenF1step 1.1algebra

Expand the left side of the un-checked identity as $u w v I + i (w v s + u v t + u w r) - (v s t + w s r + u t r) - i s t r$ . Expand the right side as $u w v I + i (w v s + u v t + u w r) - (u r t + w r s + v t s) - i r t s$ . By step 1.1, put $A = s t = t r = r s$ and $B = s r = r t = t s$ ; then the left quadratic term is $- (u + v) A - w B = - w (A + B)$ , and the right quadratic term is $- w A - (u + v) B = - w (A + B)$ . The scalar, linear, and cubic terms agree as well, using $u + v = w$ and $s t r = r t s$ . This proves the un-checked identity.

3.1givenF1step 1.1algebra∎

The checked left side expands to $i (u v - 1) I - w (s + r) + i w (u s r + v r s) + u v w s r s$ . The checked right side expands to $i (u v - 1) I - w (s + r) + i w (v r s + u s r) + u v w r s r$ . These expressions agree by step 1.1 and $w = u + v$ , proving the checked identity.

Diagram

Read each arrow label as a composite of endomorphisms, with the rightmost factor applied first. The two arrows have the same source and target; this is a notation diagram for the checked identity above, not part of its proof.

Depends on

The rational spin-1/2 spectral R-matrix

Used by

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)