How should we think of the commutator of two permutations?

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I have a fairly good intuitive concept of what the conjugate of $x$ with $y$ does: $yxy^{-1}$. It “applies the transformation $y$ to the transformation $x$” rather than to the underlying set. For example, if $y$ is 45 degree rotation of the plane, and $x$ is reflection along the vertical-axis, then the conjugate of $x$ with $y$ will give you a rotated reflection, i.e. a reflection along the diagonal. (Hence, the $y$-conjugate operation rotates the operation $x$ itself, rather than the plane on which $x$ is acting).

I do not have a similar intuitive understanding of what the commutator does in a permutation group. I went through a number of examples, and couldn’t find the common pattern. Is there a similarly intuitive concept of what the commutator does?

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The commutator of two transformations $A$ and $B$ represents the difference between applying first $A$ then $B$ and first $B$ then $A$.

In linear vector spaces it's pretty easy to follow. Imagine that you have a vector $v$ and two linear operators $A$ and $B$. Then, the commutator $[A,B]$ represents the operation you need to do on the vector $v$ such that applying $BA$ to the transformed vector is equivalent to applying $AB$ to the original vector; consider the transformed vectors obtained by first applying the commutator to the vector $v$ $$ \hat v = [A,B] v$$

Then, applying $BA$ to the transformed vector gives $$\begin{aligned} z& = BA \hat v \\ &= BA [A,B] v \\ &= BA (A^{-1} B^{-1}A B) v \\& = AB v. \end{aligned}$$

In other words, the commutator represents the operation you need to do on $AB$ to turn it into $BA$: $$BA = AB[A,B].$$

To paraphrase it, the commutator represents the transformation such that $xy$ on the transformed set acts like $yx$ on the underlying set.