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We know that the matrices $A$ and $B$ don't commute even if $e^{A}$ and $e^{B}$ commute.

However, if the problem now is that $e^{xA}$ and $e^{yB}$ commute $\forall x,y\in \mathbb{R}$, then do A and B commute?

The following is my proof:

$\frac{\partial }{\partial y}\frac{\partial }{\partial x}(e^{xA}e^{yB})=(\frac{\mathrm{d} }{\mathrm{d} x}e^{xA})(\frac{\mathrm{d} }{\mathrm{d} y}e^{yB})=Ae^{xA}Be^{yB}$

$\frac{\partial }{\partial y}\frac{\partial }{\partial x}(e^{yB}e^{xA})=(\frac{\mathrm{d} }{\mathrm{d} y}e^{yB})(\frac{\mathrm{d} }{\mathrm{d} x}e^{xA})=Be^{yB}Ae^{xA}$

$\because0=\frac{\partial }{\partial y}\frac{\partial }{\partial x}[e^{xA},e^{yB}]=Ae^{xA}Be^{yB}-Be^{yB}Ae^{xA}$

Then set $x=y=0$, we have $[A,B]=0$.

Is my proof correct?

William Huang
  • 437

1 Answers1

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We can do better.

Proposition. Assume that there is a complex convergent sequence $(t_n)_n$ s.t. if $m\not= n$, then $t_m\not= t_n$ and, for every $n$, $e^{t_nA}e^{t_nB}=e^{t_nB}e^{t_nA}$; then $AB=BA$.

Proof. Definition. We say that a complex finite subset $((a_i))$ is $2i\pi$ congruence-free (denoted $2i\pi$ CF) iff there are no distinct $a_i,a_j$ s.t. $a_i-a_j\in 2i\pi\mathbb{Z}^*$.

Let $spectrum(A)=(\lambda_i)$, $spectrum(B)=(\mu_i)$. Assume that $AB\not= BA$. Then, for every $n$, the spectra of $t_nA,t_nB$, that is $(t_n\lambda_i),(t_n\mu_i)$ are not both $2i\pi$ CF; cf. http://www.ams.org/journals/proc/1997-125-06/S0002-9939-97-03643-5/S0002-9939-97-03643-5.pdf

$spectrum(t_nA)$ (for example) is not $2i\pi$ CF for an infinity of values of $n$. It is easy to see that there are integers $k_1,k_2$, s.t., for an infinity of $n$, $t_n\lambda_{k_1}-t_n\lambda_{k_2}\in 2i\pi\mathbb{Z}^*$. Let $n_0$ be s.t., for every distinct $p,q\geq n_0$, $|t_p-t_q|<2\pi/|\lambda_{k_1}-\lambda_{k_2}|$.

On the other hand, there are $p_0,q_0\geq n_0$ s.t. $(t_{p_0}-t_{q_0})(\lambda_{k_1}-\lambda_{k_2})\in 2i\pi\mathbb{Z}^*$, that is contradictory.