Formula for product of matrix exponential, commutator, converging sequence.

Question

Let $x$, $y \in \text{M}_k(\mathbb{R})$. For $k \gg 0$, do we have$$e^{{1\over k}x} e^{{1\over k} y} e^{-{1\over k} (x + y)} = e^{{1\over{k^2}}\left({1\over2}[x, y] + z_k\right)},$$where $[x, y] = xy - yx$ and $z_k \in \text{M}_k(\mathbb{R})$ is a sequence such that $\lim_{k \to \infty} |z_k| = 0$?

Answer 1

Yes.

It is straightforward to derive this from the Baker–Campbell–Hausdorff formula:

$$e^{tx} \, e^{ty} = e^{t(x+y) + \frac{t^2}{2} [x,y] + o(t^2) }$$

For notational comfort I have written $t = 1/k$. So we can write

\begin{align*} e^{tx} \,e^{ty}\, e^{- t(x+y)} & = e^{t(x+y) + \frac{t^2}{2} [x,y] + o(t^2) }\,e^{- t(x+y)} \end{align*}

Using the Baker–Campbell–Hausdorff formula again: \begin{align*} e^{tx} \,e^{ty}\, e^{- t(x+y)} &= e^{\left(\frac{t^2}{2} [x,y] + o(t^2) \right) + \frac{1}{2}\left(\left[t(x+y) + \frac{t^2}{2} [x,y] + o(t^2), - t(x+y)\right]\right) + o(t^2)}\\ & = e^{\frac{t^2}{2} [x,y] + o(t^2)} \end{align*}

As you probably know, the notation "$o(t^2)$" is a conventional shortcurt for "$t^2 \,z_t$ where $z_t$ is some sequence such that $\lim z_t = 0$" (but this sequence can be different every time I write "$o(t^2)$")