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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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  • $\begingroup$ The commutator gives the alternating group, i.e.,$[S_n,S_n]=A_n$. So one could think of the commutator as permutations of signum $1$. $\endgroup$ Mar 5 '19 at 13:16
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    $\begingroup$ The commutator $[x,y]$ is the difference between the compositions $xy$ and $yx$. $\endgroup$ Mar 5 '19 at 13:35
  • $\begingroup$ In my opinion as a geometric group theorist, commutators are intrinsically less geometric and more algebraic than conjugates. One has to stretch further in order to tell geometric stories about commutators, and they tend to be shaggy dog stories. I could tell a geometric story about commutators in free groups, or surface groups, but I don't know such a story for permutation groups. Alternatively, I could tell geometric stories about the commutator subgroup of a group, if not about individual commutators. $\endgroup$
    – Lee Mosher
    Mar 5 '19 at 14:39
  • $\begingroup$ @LeeMosher I would be very interested in a geometric story for all those things (in free groups, surface groups, and the commutator subgroup), as a substitute for my question about permutation groups. I would be very grateful if you could explain those in an answer, and perhaps an explanation why commutators are intrinsically less geometric (if there is such an explanation). I have been thinking about this for a while, and your insight would be greatly appreciated (In fact, I promise to give you a 100 points bounty on an answer addressing those things). $\endgroup$
    – user56834
    Mar 5 '19 at 15:41
  • $\begingroup$ Do you happen to own a Rubik's cube? Start with a solved cube and try interleaving two short sequences of moves and their inverses. $\endgroup$
    – j.p.
    Mar 7 '19 at 7:45
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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.

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  • $\begingroup$ This is the ring definition of commutator. My question is about the group definition, though I would find it nice to have a similar intuitive understanding of the ring commutator. Nevertheless, I don’t really find your explanation very intuitive. I would like to have an understanding that allows me to geometrically immediately see, given two operations, what their commutator is. I can’t do that if I just know the symbolic definition $\endgroup$
    – user56834
    Mar 5 '19 at 13:57
  • $\begingroup$ I used the ring definition in my explicit answer because I think it is intuitively easier; I have added my interpretation of the group commutator :) $\endgroup$ Mar 5 '19 at 14:37

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