# An operator exponential/commutator question

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 $$XY=YX.$$