How high can the order of Strang splitting go?

15 Sep, 2026

This is an older question I had (from my PhD years) but which I didn't think would be useful to pursue. Now with smart LLMs, it becomes feasible to just let them look into it for me, with some guidance.

First, some notation: assume $B$ and $C$ are continuous-time Markov generators, i.e., $B, C \in ℝ^{n \times n}$ , $B 1 = C 1 = 0$ , and their off-diagonal entries are non-negative. We will also define $A = B + C$ . You don't really need to know what a Markov generator is for the remainder.

We are interested in approximating $e^{A h}$ , which is the matrix exponential of $A h$ , $h \geq 0$ . There are various splitting methods that can help with this.

Lie-Trotter Splitting

A typical way is

e^{A h} \approx e^{B h} e^{C h},

which is called the Lie-Trotter splitting. We do this approximation because it is assumed that $e^{A h}$ is too expensive to compute directly.

Using Taylor expansions, we can compute the local error of this approximation:

- \frac{[B, C] h^{2}}{2} + O (h^{3}) .

Here

[B, C] : = B C - C B

is the commutator of $B$ and $C$ .

Question: Can we pick $B$ and $C$ in some clever way so as to get

e^{A h} - e^{B h} e^{C h} = O (h^{3}) ?

That would require $[B, C] = 0$ . However, it is a property of this splitting that if $[B, C] = 0$ , then

e^{A h} = e^{B h} e^{C h};

that is, the splitting is exact. So with the Lie-Trotter splitting, the local error is either $O (h^{2})$ or the error vanishes completely.

For the sake of discussion, let's assume that splittings for which $[B, C] = 0$ are uninteresting and continue on to another splitting.

Strang Splitting

The Strang splitting is defined as

e^{B h / 2} e^{C h} e^{B h / 2},

and we can again compute the local error as before:

- \frac{h^{3}}{24} [B + 2 C, [B, C]] + O (h^{4}) .

Here the answer is not as obvious. Of course, we can pick our splitting so that $[B, C] = 0$ (for example, $B = a A$ and $C = (1 - a) A$ for some $a \in (0, 1)$ ), in which case the error vanishes (and is an uninteresting splitting).

Alternatively, we can ask that

[B, C] \neq 0 and [B + 2 C, [B, C]] = 0 .

This cancels the cubic term without making the splitting exact. Since the first nonzero local-error order of the symmetric Strang splitting must be odd, this actually gives us a local error of at least order $h^{5}$ .

Question: By picking $B$ and $C$ , we may be able to increase the local-error order without making the splitting exact. How high can we go?

I gave this question to GPT-6 Astra, probing it to give me upper and lower bounds instead of trying to get an exact count. First things first, the result:

Bounds (still conjectured, need to check the proof)

Let $M_{n}$ be the largest finite local order attained by a nonexact Strang splitting of two $n \times n$ Markov generators.

One can show that

2 ⌊ \frac{n}{2} ⌋ + 1 \leq M_{n} \leq n (n - 1) + 1 .

The lower bound comes from an explicit family of noncommuting examples. These examples can even be chosen so that every off-diagonal rate is strictly positive.

Commentary

The proof is not that involved, but has a few distinct steps. I'm in the process of checking and simplifying them, while Codex is formalizing the argument in Lean.

Separately, another Codex agent has been working to calculate the first values of $M_{n}$ using exact symbolic computation. It's not an easy task. So far, the exact calculations give the following comparison:

$n$	Lower bound $2 ⌊ n / 2 ⌋ + 1$	Known value of $M_{n}$	Upper bound $n (n - 1) + 1$
2	3	3	3
3	3	3	7
4	5	5	13
5	5	5	21
6	7	7	31
7	7	9	43

The lower bound is sharp for $n = 2, \dots, 6$ . The first surprise occurs at $n = 7$ , where an explicit nonexact splitting of local order $9$ exists.

The next unknown case is $M_{8}$ . The general bounds currently give

9 \leq M_{8} \leq 57,

although additional finite-dimensional arguments can reduce the upper bound substantially. Determining the exact value appears to be considerably harder.