[r-t] Conjugate permutations

Richard Smith richard at ex-parrot.com
Wed Aug 25 10:27:20 UTC 2010

Richard Smith wrote:

> Bizarrely, this function, c(r), turns out to have some very interesting 
> properties.  [...]
> However, what is particularly useful is that it provides an efficient 
> algorithm for constructing r from a and b when solving the equation:
>  b = r^-1 a r.
> [...] which gives the solution:
>  r = c(a) c(b)^-1.

Some people use exponentiation notation to denote 
conjugation -- in other words, they define

   a^r = r^-1 a r

One inspiration being

  (a^r)^s = s^-1 r^-1 a r s = (r s)^-1 a (r s) = a^(rs)

exactly as for real exponents.

Now, I'm a little uncomfortable with this notation because 
it seems to invite further extension.  For example, if n is 
an integer and a and r permutations, then (a^n)^r = (a^r)^n 
and it's tempting to write this a^(nr).  But what is nr?

But reservations aside, the equation b = r^-1 a r would now 
be written:

   b = a^r.

If a, b and r were real numbers, we would use logarithms to 
solve this for r:

   log b = r log a   =>   r = log b / log a.

(I've written log rather than ln as the choice of base is 

For permutations, the function c(r)^-1 serves the same role:

   r = c(a) c(b)^-1  =>   c(b)^-1 = c(a)^-1 r.

If you're going to use exponentiation to represent 
conjugation, is it unreasonable to define

   log r = c(r)^-1 ?

I'm not sure how much more mileage there is in this, though.


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