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Symmetry of the splitting map forces $X_{j} = 0$ for every even $j$ , so the first possible modified-energy correction is of order $h^{2}$

Statement

Let $d \geq 1$ , let $(U, U_{ρ}, \tilde{A}, \tilde{B}, M, K, δ, Ψ_{h})$ be analytic splitting data (Analytic splitting data with uniform holomorphic neighbourhood, and the exact symmetric splitting map $Ψ_{h}$ ), and use the notation of The truncated binomial family $T_{h}^{s, N}$ , the truncated logarithm $L_{N} (h)$ , its coefficient fields $X_{j}$ and the truncations $X^{[N]}$ and Each coefficient field $X_{j}$ of the modified logarithm is a holomorphic symplectic field with a global holomorphic primitive $H_{j}$ on the star-shaped domain, and $X_{1} = X_{\tilde{A} + \tilde{B}}$ : the coefficient fields $X_{j}$ ( $j \geq 1$ , with $X_{j}$ acting on holomorphic functions as the coefficient operator $D_{j} = [h^{j}] L (h)$ , and $X_{j} = D_{j}$ in the notation of that lemma) and the normalized primitives $H_{j}$ with $d H_{j} = ι_{X_{j}} ω_{C}$ , $H_{j} (0) = 0$ , on the domains $D_{j}$ .

Reversibility. For the local flow germs at $h = 0$ , and their mutually reverse continuation branches wherever defined, $Ψ_{- h} = Ψ_{h}^{- 1}$ .
Parity. $X_{j} = 0$ for every even $j \geq 2$ .
Modified energy. Consequently $H_{2} = 0$ . For $N \geq 2$ define the finite complex modified Hamiltonian on $D_{N}$ by $H_{h} : = \tilde{H} + \sum_{j = 2}^{N} h^{j - 1} H_{j} .$ Then $H_{h} = \tilde{H} + \sum_{j = 3}^{N} h^{j - 1} H_{j}$ on $D_{N}$ , so the first possible correction is of order $h^{2}$ . For real $h$ , on the real part of $D_{N}$ , $H_{h}$ restricts to the corresponding real function with baseline $H = A + B$ .

Facts & Assumptions

Given: The analytic splitting data and the objects named in the Statement; for every holomorphic $f$ on $D_{0}$ the two families $T_{h}^{s} f : = \sum_{k \geq 0} (\binom{s}{k}) Δ_{h}^{k} f, L (h) f : = \sum_{k \geq 1} \frac{(- 1)^{k + 1}}{k} Δ_{h}^{k} f,$ read as formal power series in $h$ whose $h$ -degree- $j$ coefficients are finite sums (because $Δ_{h} = O (h)$ ); the concrete truncations $T_{h}^{s, N}$ , $L_{N} (h)$ of The truncated binomial family $T_{h}^{s, N}$ , the truncated logarithm $L_{N} (h)$ , its coefficient fields $X_{j}$ and the truncations $X^{[N]}$ agree with them through order $N$ .

[L1]

Inverse and composition laws: $φ_{F}^{- t} \circ φ_{F}^{t} = i d$ , and $(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1}$ where the inverses exist (The flow of a bounded holomorphic field on a polydisc: existence in complex time, displacement bound, holomorphy, and the composition law clause 3, Calculus of holomorphic maps: chain rule, Wirtinger derivatives, coefficient extraction in one variable, and locally uniform limits).

[L2]

Integer case: for every integer $m \geq 0$ , $T_{h}^{m} f = f \circ Ψ_{h}^{m}$ as formal series in $h$ (the sum is finite at each $h$ -degree and equals the finite binomial sum of The binomial family of the splitting pullback is multiplicative and bracket-preserving, and its logarithm is a derivation clause 1 for any truncation order $\geq m$ ). Consequently $T_{h}^{m} f$ is well defined for all $m \geq 0$ and all holomorphic $f$ on $D_{0}$ .

[L3]

Formal identities in one element. Let $A$ be an element of the ring of formal series in $h$ with operator coefficients, with $A = O (h)$ . Define $\exp (A) : = \sum_{m \geq 0} A^{m} / m!$ and $\log (1 + U) : = \sum_{k \geq 1} (- 1)^{k + 1} U^{k} / k$ , both well defined coefficient-wise because $A = O (h)$ and $U = O (h)$ . Then: (i) $\exp (A) \exp (- A) = 1$ and $\exp (- A) \exp (A) = 1$ , because all powers of $A$ commute and the identity holds in the commutative $Q$ -algebra $Q [A]$ ; (ii) $\log (\exp (A)) = A$ , because both sides are universal rational-coefficient expressions in the single element $A$ and the identity holds in the commutative ring $Q ⟦ x ⟧$ (differentiate $t \mapsto \log (\exp (t x))$ and use $(\log)^{'} (1 + u) = (1 + u)^{- 1}$ coefficient-wise); (iii) hence $\exp$ is injective on ${A : A = O (h)}$ : if $\exp (A) = \exp (B)$ then $A = \log (\exp (A)) = \log (\exp (B)) = B$ .

[L4]

Formal binomial group law. For $B_{s} (x) = \sum_{k \geq 0} (\binom{s}{k}) x^{k}$ , the falling-factorial recurrence gives $(1 + x) \partial_{x} B_{s} = s B_{s}$ and $B_{s} (0) = 1$ . Comparing coefficients determines the series uniquely. Thus $B_{s} B_{t}$ satisfies the same equation as $B_{s + t}$ and the same constant term, so $B_{s} B_{t} = B_{s + t}$ . Differentiate in $t$ at zero: $\partial_{s} B_{s} = B_{s} \sum_{k \geq 1} (- 1)^{k + 1} x^{k} / k$ , by the derivative formula for $(\binom{t}{k})$ at zero. Substituting the single operator $x = Δ_{h} = O (h)$ makes all sums finite at any fixed $h$ -degree and gives $\partial_{s} T_{h}^{s} = L (h) T_{h}^{s}$ , with $T_{h}^{0} = i d$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , The binomial family of the splitting pullback is multiplicative and bracket-preserving, and its logarithm is a derivation).

[L5]

Uniqueness of formal solutions: if $s \mapsto F (s)$ and $s \mapsto G (s)$ are families whose $h$ -degree- $j$ coefficients are polynomials in $s$ with $F^{'} (s) = L (h) F (s)$ , $G^{'} (s) = L (h) G (s)$ and $F (0) = G (0)$ , then $F = G$ : comparing coefficients of $s^{n}$ determines each coefficient recursively from the preceding one, since $L (h)$ acts coefficient-wise in $h$ .

[L6]

Primitives and domains (Each coefficient field $X_{j}$ of the modified logarithm is a holomorphic symplectic field with a global holomorphic primitive $H_{j}$ on the star-shaped domain, and $X_{1} = X_{\tilde{A} + \tilde{B}}$ ): $d H_{j} = ι_{X_{j}} ω_{C}$ , $H_{j} (0) = 0$ , $D_{j}$ connected with $0 \in D_{j}$ ; and $X_{j} = D_{j}$ as operators on holomorphic functions, with the bounds of Bounds for the truncated logarithm: $∣ L_{N} (h) z_{i} / h ∣ \leq 8 M / 3$ and $∣ X_{j} ∣ \leq (8 M / 3) (C_{0} M j / ρ)^{j - 1}$ .

Proof

technique · direct; identify $\Psi_{-h}$ with the inverse, use the formal exponential of the logarithm to compare $L(h)$ and $L(-h)$, and differentiate in the binomial parameter

1.1givenL1algebra

Reversibility. Use the local flow germs, or continue each inverse factor along the reverse of the branch of its forward factor, as required by [L1]. For $h$ in their common domain, $Ψ_{h}^{- 1} = (φ_{\tilde{A}}^{h / 2} \circ φ_{\tilde{B}}^{h} \circ φ_{\tilde{A}}^{h / 2})^{- 1} = φ_{\tilde{A}}^{- h / 2} \circ φ_{\tilde{B}}^{- h} \circ φ_{\tilde{A}}^{- h / 2} = Ψ_{- h}$ by [L1]. This is clause 1.

1.2givenL2L4L5algebra

The exponential of the logarithm. By [L4] the family $s \mapsto T_{h}^{s} f$ solves the formal initial value problem of [L5] with initial value $f$ ; the family $s \mapsto \log$ -exponential $\exp (s L (h)) f$ solves the same problem, because $\frac{d}{d s} \exp (s L (h)) = \sum_{m \geq 1} L (h)^{m} s^{m - 1} / (m - 1)! = L (h) \exp (s L (h))$ by commutation of $L (h)$ with $s L (h)$ . By [L5], $T_{h}^{s} f = \exp (s L (h)) f$ as formal series in $s$ ; taking $s = 1$ and using [L2] with $m = 1$ , $Ψ_{h}^{*} f = f \circ Ψ_{h} = T_{h}^{1} f = \exp (L (h)) f .$

2.1givenstep 1.2L2L3

Parity of the logarithm. Fix an integer $m \geq 0$ . By clause 1 and [L2], $T_{- h}^{m} f = f \circ (Ψ_{- h})^{m} = f \circ (Ψ_{h}^{- 1})^{m}$ ; by step 1.2 applied to the data at step $- h$ this equals $\exp (m L (- h)) f$ . The same expression, read with the data at step $h$ , equals $(Ψ_{h}^{*})^{- m} f = \exp (L (h))^{- m} f = \exp (- m L (h)) f$ by [L3]. Hence $\exp (m L (- h)) f = \exp (m (- L (h))) f$ as formal series in $h$ , for every integer $m \geq 0$ . At fixed $h$ -degree both sides are polynomials in $m$ (all defining operations are polynomial at each degree), so the equality at all $m \geq 0$ extends to all complex $m$ ; taking $m = 1$ gives $\exp (L (- h)) f = \exp (- L (h)) f$ for every holomorphic $f$ on $D_{0}$ , and injectivity of $\exp$ from [L3] gives $L (- h) = - L (h) .$

3.1givenstep 2.1L6algebra

Vanishing of the even coefficients. Write $L (h) = \sum_{j \geq 0} h^{j} D_{j}$ (so $D_{0} = 0$ ). Then $L (- h) = \sum_{j} (- 1)^{j} h^{j} D_{j}$ and $- L (h) = \sum_{j} - h^{j} D_{j}$ ; comparing $h$ -degree- $j$ coefficients in the identity of step 2.1 gives $(- 1)^{j} D_{j} = - D_{j}$ , hence $D_{j} = 0$ for even $j$ . Since $X_{j}$ acts as $D_{j}$ by [L6], $X_{j} = 0$ for every even $j \geq 2$ . This is clause 2.

4.1givenstep 3.1L6∎

Modified energy. For $j = 2$ , $X_{2} = 0$ gives $ι_{X_{2}} ω_{C} = 0$ , so $d H_{2} = 0$ ; since $D_{2}$ is connected and $H_{2} (0) = 0$ , $H_{2} = 0$ by [L6]. The definition of $H_{h}$ in clause 3 therefore gives $H_{h} = \tilde{H} + \sum_{j = 3}^{N} h^{j - 1} H_{j}$ on $D_{N}$ , and, for real $h$ , its restriction to the real part of $D_{N}$ is real-valued: the baseline $\tilde{H}$ and each normalized primitive $H_{j}$ are real there by the primitive supplier. On $U$ the baseline is $H = A + B$ . No nonzero $H_{3}$ is asserted; for example, an exact one-part splitting can have every correction zero. This is clause 3.

Remarks

The whole argument is formal in the step variable: only finitely many terms contribute to each $h$ -degree, and no convergence of the infinite logarithmic series is used. The concrete truncated functions $L_{N} (h) f$ of The truncated binomial family $T_{h}^{s, N}$ , the truncated logarithm $L_{N} (h)$ , its coefficient fields $X_{j}$ and the truncations $X^{[N]}$ have coefficients agreeing with $L (h) f$ through order $N$ , by Bounds for the truncated logarithm: $∣ L_{N} (h) z_{i} / h ∣ \leq 8 M / 3$ and $∣ X_{j} ∣ \leq (8 M / 3) (C_{0} M j / ρ)^{j - 1}$ clause 4.
Parity is a property of the symmetric Strang-type splitting: it uses $Ψ_{- h} = Ψ_{h}^{- 1}$ and fails for the asymmetric first-order splittings $φ_{\tilde{A}}^{h} \circ φ_{\tilde{B}}^{h}$ . No such claim is made for them.

Symmetry of the splitting map forces $X_{j} = 0$ for every even $j$ , so the first possible modified-energy correction is of order $h^{2}$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Symmetry of the splitting map forces Xj=0Xj​=0 for every even jj, so the first possible modified-energy correction is of order h2h2

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Symmetry of the splitting map forces $X_{j} = 0$ for every even $j$ , so the first possible modified-energy correction is of order $h^{2}$