Home > Higher Algebra, Linear Algebra, Problem Solving > Commutators and the identity matrix – Miklos Schweitzer 2012 Problem 6

## Commutators and the identity matrix – Miklos Schweitzer 2012 Problem 6

Suppose ${A,B,C}$ are complex ${n\times n}$ matrices such that ${[A,B]=C,\ [B,C]=A}$ and ${[C,A]=B}$, where ${[X,Y]=XY-YX}$ is the usual commutator. Prove that ${e^{4\pi A}}$ is the identity matrix.

Miklos Schweitzer 2012 Problem 6

Proof: Let’s try to use the Hadamard formula because it involves both the commutators and the exponential:

$\displaystyle e^{S}T e^{-S} = T+\left[S,T\right]+\frac{1}{2!}[S,[S,T]]+\frac{1}{3!}[S,[S,[S,T]]]+\cdots$

If we replace ${X}$ by ${\lambda X}$ we get

$\displaystyle e^{\lambda S}T e^{-\lambda S} =T+\lambda\left[S,T\right]+\frac{\lambda^2}{2!}[S,[S,T]]+\frac{\lambda^3}{3!}[S,[S,[S,T]]]+\cdots$

Take ${S=A, T=B}$ in the above formula. We get

$\displaystyle e^{\lambda A}Be^{-\lambda A}=(1-\frac{\lambda^2}{2}+\frac{\lambda^4}{4!}+...)B+(\lambda -\frac{\lambda^3}{3!}+\frac{\lambda^5}{5!}... )C=$

$\displaystyle = \frac{e^{i\lambda}+e^{-i\lambda}}{2}B +\frac{e^{i\lambda}-e^{-i\lambda}}{2i}C$

If we choose ${\lambda=\pi/2}$ we get ${e^{\frac{\pi}{2}A}B=Ce^{\frac{\pi}{2}A}}$ from where we get ${e^{\frac{\pi}{2}A}e^{\frac{\pi}{2}B}=e^{\frac{\pi}{2}C}e^{\frac{\pi}{2}A}}$

If we denote ${X=e^{\pi/2A},Y=e^{\pi/2B},Z=e^{\pi/2C}}$, the last relation translates to ${XY=ZX}$ and because we can obtain the same results by making a cyclic permutation we have ${XY=YZ=ZX}$. (these won’t be of any use in the sequel)

For ${\lambda=\pi}$ we obtain ${ e^{\pi A}B=-Be^{\pi A}}$ which implies that ${e^{\pi A} e^{\frac{\pi}{2} B}=e^{-\frac{\pi}{2} B}e^{\pi A}}$, namely ${X^2Y=Y^{-1}X^2}$ which means that ${YX^2Y=X^2}$.

If we apply the Hadamard like formula to ${S=B, T=A}$ and ${\lambda=\pi}$ we obtain in the same way that ${Y^2X=X^{-1}Y^2}$ which is equivalent to ${XY^2X=Y^2}$.

Now ${(X^2Y^2)^2=X \cdot XY^2X \cdot XY^2=XY^2XY^2=Y^4}$. In the same way we have ${(X^2Y^2)^2= X^2Y\cdot YX^2Y\cdot Y=X^2YX^2Y=X^4}$.

So we have reached ${X^4=Y^4=(X^2Y^2)^2}$ (from here the problem is solved using only simple group theory). In particular we have ${X^2=Y^2X^2Y^2=Y^2(Y^2X^2Y^2)Y^2=X^{10}}$. Since ${X}$ is an exponential matrix, it is invertible, and therefore ${X^8=I}$, which translates to ${e^{4\pi A}=I}$ and finishes the problem.

Note: It can be proved that if ${ a,b\in G }$, where ${ G }$ is a group and ${ a^2=b^2=(ab)^2 }$ then ${ a^4=b^4=e }$ where ${ e }$ is the identity element of the group ${ G }$.