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

January 8, 2013 Leave a comment Go to comments

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 }.

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