There is "an important lemma" related to the Baker-Campbell-Haussdorff theorem which says that

$$ e^XYe^{-X} = Y + [X,Y] + \frac{1}{2!}[X,[X,Y]]+\ldots $$

Clearly if $[X,Y]=0$ we get (noting that $e^{-X}e^{X} = \mathbf{1}$) that

$$ e^X Y-Ye^X = [e^X,Y]=0 $$

Thus $[X,Y]=0$ implies $[e^X,Y]=0$. Is the converse of this true? If $[e^X,Y]=0$ then does that imply $[X,Y]=0$? By the above formula this would imply that if

$$ [X,Y] + \frac{1}{2!}[X,[X,Y]]+\ldots =0 $$

then $[X,Y]=0$. It is not immediately obvious to me why this might be the case.


If $[X,Y]=0$ all terms in your sum are zero, so it's no surprise that the sum is zero.

I will use the holomorphic functional calculus in a Banach algebra; this allows us to define $f(A)$ for any function $f$ holomorphic on an open domain that contains that contains the spectrum of the matrix $A$. There is probably a less technical way of doing this.

For $s\in[0,1]$, consider the function $f_s(z)=z^s$ (this requires choosing a branch for the logarithm, but that's no issue). Since the spectrum $\sigma(e^X)$ of $e^X$ is discrete, we can choose an domain $D$ with smooth boundary, $0\not\in D$, and $\sigma(e^X)\subset D$. Note that $$ f_s(e^X)=(f_s\circ\exp)(X)=e^{sX}, $$ since we have $f_s\circ\exp (z)=e^{sz}$. Then, since products are continuous, and integrals are limits of sums, $$ f_s(e^X)Y=\left(\frac1{2\pi i}\int_{\Gamma} f(z)\,(zI-e^X)^{-1}\,dz\right)Y =\left(\frac1{2\pi i}\int_{\Gamma} f(z)\,(zI-e^X)^{-1}Y\,dz\right) =Y\left(\frac1{2\pi i}\int_{\Gamma} f(z)\,(zI-e^X)^{-1}\,dz\right)=Yf_s(e^X). $$ (note that $(zI -e^X)Y=Y(z-e^X)$, so $(zI -e^X)^{-1}Y=Y(z-e^X)^{-1}$). So we have that $$ e^{sX}Y=Ye^{sX},\ \ s\in[0,1]. $$ Now differentiate both sides and evaluate at $s=0$, to get $$ XY=YX. $$


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