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Consider the real $n \times n$-matrices $A$ and $B$.

Can $A, B$ fail to commute if $e^A=e^B=e^{A+B}=id$ ?

Shaun
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Siggi
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2 Answers2

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Yes, $A$ and $B$ can fail to commute. Consider $$ A = \left[\begin{array}{@{}rrr@{}} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right],\qquad B = \left[\begin{array}{@{}ccc@{}} 0 & -\tfrac{1}{2} & \tfrac{\sqrt{3}}{2} \\ \tfrac{1}{2} & 0 & 0 \\ -\tfrac{\sqrt{3}}{2} & 0 & 0 \end{array}\right], $$ so that $$ A + B = \left[\begin{array}{@{}ccc@{}} 0 & \tfrac{1}{2} & \tfrac{\sqrt{3}}{2} \\ -\tfrac{1}{2} & 0 & 0 \\ -\tfrac{\sqrt{3}}{2} & 0 & 0 \end{array}\right]. $$ Since each of $A$, $B$, and $A + B$ has Frobenius norm equal to $\sqrt{2}$, we have $$ \exp(2\pi A) = \exp(2\pi B) = \exp\bigl(2\pi (A + B)\bigr) = I. $$ However, it's easy to check $AB$ is not symmetric, so $$ BA = (-B^{T})(-A^{T}) = B^{T}A^{T} = (AB)^{T} \neq AB. $$ It follows, of course, that $2\pi A$ and $2\pi B$ don't commute, either.

Andrew D. Hwang
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    Nice example. It would be interesting to determine if there is any $2 \times 2$ example. – Hans Engler Jun 08 '14 at 15:51
  • If a matrix $M$ has Frobenius norm equal to $\sqrt{2}$, then $\exp(2 \pi M)=I$ ? Ok nice. But can you give me some keywords to find further readings to this, please? – Siggi Jun 08 '14 at 16:39
  • @HansEngler: There are no smaller examples because the space of skew-symmetric $2\times 2$ matrices is one-dimensional (so the bracket is trivial). :) – Andrew D. Hwang Jun 08 '14 at 16:42
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    @Siggi: I neglected to emphasize that the matrices in question are skew-symmetric. The example rests on the fact that if $A$ is a real, skew-symmetric matrix of Frobenius norm $\sqrt{2}$, then $\exp(2\pi A) = I$. For further reading, you might start here. – Andrew D. Hwang Jun 08 '14 at 16:49
  • Thanks a lot for your help. – Siggi Jun 08 '14 at 16:58
  • @Siggi: You're very welcome. – Andrew D. Hwang Jun 08 '14 at 17:33
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This is wrong. If $e^{A+B}=I$, then $A+B=2n\pi i$, for all $n\in\mathbf{Z}$, i.e., $2n\pi i-A=B$. Consider $[A,B]=AB-BA=A(2n\pi i-A)-(2n\pi i-A)A=-AA+AA=0$. Hence, if $e^{A+B}=I$, then $A$ and $B$ commute.