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Regular point and exchange logarithmic derivative

Statement

Let $t (u)$ and $U$ be the transfer matrix and cyclic shift defined by the ordered monodromy construction. Use the periodic convention $P_{N, N + 1} : = P_{N, 1}$ . Then $t (0) = i^{N} U, i t^{'} (0) t (0)^{- 1} = \sum_{j = 1}^{N} P_{j, j + 1} .$ In particular, $t (0)$ is invertible. The derivative is with respect to the dimensionless complex parameter $u$ .

Statement references

Faddeev, §3, equations (49)–(65)

Classical and Quantum Integrable Systems, Lecture 15, §15.2

Facts & Assumptions

Given: An integer $N \geq 3$ and the finite-dimensional spaces and operators of Ordered auxiliary-space monodromy.

[F1]

Ordered auxiliary-space monodromy: $T_{a} (u) = R_{a N} (u) \dots R_{a 1} (u)$ , $t (u) = {tr}_{a} T_{a} (u)$ , $R_{a j} (u) = u I + i P_{a j}$ , and $U$ sends $v_{1} \otimes \dots \otimes v_{N}$ to $v_{N} \otimes v_{1} \otimes \dots \otimes v_{N - 1}$ .

Proof

1.1givenF1algebra

At $u = 0$ , every factor is $R_{a j} (0) = i P_{a j}$ , so $t (0) = i^{N} {tr}_{a} (P_{a N} \dots P_{a 1})$ . On an elementary chain tensor $e_{j_{1}} \otimes \dots \otimes e_{j_{N}}$ , start the auxiliary factor at each basis vector $e_{a}$ in turn. The swaps pass $e_{j_{1}}, \dots, e_{j_{N}}$ successively to the auxiliary factor and shift the preceding entries one site right; the auxiliary trace retains exactly the term with $e_{a} = e_{j_{N}}$ . The output is therefore $e_{j_{N}} \otimes e_{j_{1}} \otimes \dots \otimes e_{j_{N - 1}}$ . These tensors form a basis, hence ${tr}_{a} (P_{a N} \dots P_{a 1}) = U$ . As $U$ permutes a basis, it is invertible, proving the first identity and invertibility of $t (0)$ .

2.1givenF1step 1.1algebra

Differentiate the finite product defining $T_{a} (u)$ and use $R_{a j}^{'} (0) = I_{a j}$ . For each $m$ , let $D_{m} = {tr}_{a} (P_{a N} \dots P_{a, m + 1} P_{a, m - 1} \dots P_{a 1})$ , with the factor $P_{a m}$ omitted. Following the auxiliary basis vector through the remaining swaps and taking its trace gives, for $2 \leq m < N$ , $D_{m} (e_{j_{1}} \otimes \dots \otimes e_{j_{N}}) = e_{j_{N}} \otimes (⨂_{r = 1}^{m - 2} e_{j_{r}}) \otimes e_{j_{m}} \otimes e_{j_{m - 1}} \otimes (⨂_{r = m + 1}^{N - 1} e_{j_{r}}) .$ An empty tensor block in this display is omitted. For $m = N$ the same calculation gives $e_{j_{N - 1}} \otimes e_{j_{1}} \otimes \dots \otimes e_{j_{N - 2}} \otimes e_{j_{N}}$ ; for $m = 1$ it gives $e_{j_{1}} \otimes e_{j_{N}} \otimes e_{j_{2}} \otimes \dots \otimes e_{j_{N - 1}}$ . In each case this is exactly $U P_{m - 1, m}$ applied to the input, with $P_{0, 1} : = P_{N, 1}$ . Since the elementary tensors form a basis, $D_{m} = U P_{m - 1, m}$ as operators. Thus $t^{'} (0) = i^{N - 1} \sum_{m = 1}^{N} U P_{m - 1, m}$ .

3.1givenF1step 1.1step 2.1algebra∎

From step 1.1, $t (0)^{- 1} = i^{- N} U^{- 1}$ . The definition of $U$ shows $U P_{m - 1, m} U^{- 1} = P_{m, m + 1}$ for $2 \leq m \leq N$ , while $U P_{N, 1} U^{- 1} = P_{1, 2}$ . Therefore $t^{'} (0) t (0)^{- 1} = i^{- 1} \sum_{m = 1}^{N} P_{m, m + 1} = - i \sum_{m = 1}^{N} P_{m, m + 1}$ . Multiplication by $i$ proves the claimed logarithmic-derivative identity.

Remarks

The statement uses the ordered monodromy $R_{a N} \dots R_{a 1}$ and the stated orientation of $U$ . Reversing either convention can change the direction of the cyclic shift and must be accompanied by a corresponding reindexing of the periodic bonds.

Depends on

Ordered auxiliary-space monodromy