2
$\begingroup$

If I have a permutation $\sigma$ on the set $A$ written by disjoint cycles. There are $n$ disjoint cycles can I then write the sign of the permutation as:

$\operatorname{sign}(\sigma) = (-1)^{|A|-n}$?

$\endgroup$
1
  • 1
    $\begingroup$ Does fixed points of permutation regarded as cycles of length one? $\endgroup$
    – Andrew
    Oct 11, 2011 at 19:13

2 Answers 2

8
$\begingroup$

You know a cycle of length $r$ can be written as $r-1$ transpositions (if you don't, it's good to convince yourself of why) so each disjoint cycle can be decomposed into transpositions.

If you can write $\sigma$ as $n$ disjoint cycles, each of length $r_i$, for $1\leq i\leq n$, you then have $$\operatorname{sgn}(\sigma) = \prod_{i=1}^n (-1)^{r_i-1} = (-1)^{\sum_{i=1}^n (r_i - 1)}$$ But notice $$\sum_{i=1}^n(r_i - 1) = \sum_{i=1}^n r_i - \sum_{i=1}^n 1 = \sum_{i=1}^n r_i - n.$$

So $$\operatorname{sgn}(\sigma) = (-1)^{\left(\sum_{i=1}^n r_i\right) - n}.$$ Now what's the sum of the lengths of all the disjoint cycles?

As mentioned in the comments, it turns out the sign of a permutation is really quite a nice function. Namely, it is a homomorphism. One way to see this is to show that for any permutation $\sigma$ and transposition $\tau$ (permutations with the same underlying set of course) we have $\operatorname{sgn}(\sigma\tau) = -\operatorname{sgn}(\sigma)$. Using this we can then use the decomposition of an $r$-cycle into $r-1$ transpositions again, and one has, for any $\sigma_1,\sigma_2$ acting on $A$, of length $r_1,r_2$ respectively, $$ \operatorname{sgn}(\sigma_1\sigma_2) = (-1)^{r_1 - 1}\operatorname{sgn}(\sigma_2).$$ But we can see $\operatorname{sgn}(\sigma_1) = (-1)^{r_1 - 1}$ and so $$ \operatorname{sgn}(\sigma_1\sigma_2) = \operatorname{sgn}(\sigma_1)\operatorname{sgn}(\sigma_2).$$

$\endgroup$
4
  • $\begingroup$ Your answer is much better then mine. Very organized! :-) Maybe you would like to add the observation that $\mathrm{sgn}$ is a homomorphism of groups. $\endgroup$ Oct 11, 2011 at 19:46
  • 1
    $\begingroup$ @AndréCaldas: Done, and thanks! I had the pleasure of going over a problem like this last night, so it was rather fresh in my mind :) $\endgroup$
    – Alex
    Oct 11, 2011 at 20:10
  • $\begingroup$ Nice way to show that sign is a homomorphism! Well, it depends on how one defines the "sign". $\endgroup$ Oct 11, 2011 at 20:40
  • $\begingroup$ @AndréCaldas Well it doesn't actually matter how you defined your sign function as long as you've shown that composition with a transposition negates the value. You remove $\sigma_1$ one transposition at a time to get the first equality, then you "add" each of those transposition to the identity permutation to get the second equality. I didn't actually realize this until you mentioned the definition though, but it's rather neat! $\endgroup$
    – Alex
    Oct 12, 2011 at 15:53
6
$\begingroup$

Use the fact that $\mathrm{sgn}$ is a homomorphism to reduce the problem to the cycles. Then, show that for a cycle of length $\ell$, the sign for the cycle is $(-1)^{\ell-1}$.

So, $$ \mathrm{sgn}(\sigma) = \prod_{j=1}^n (-1)^{\ell_j - 1} = (-1)^{\sum_{j=1}^n (\ell_j - 1)} = (-1)^{|A| - n}, $$ because $\sum \ell = |A|$ and $\sum 1 = n$.

$\endgroup$
2
  • $\begingroup$ Dear André: I like your answer very much! +1 $\endgroup$ Oct 11, 2011 at 19:52
  • $\begingroup$ @Pierre-YvesGaillard: Thanks! :-) I found Alexe's answer much better detailed... but I think it is important to emphasize the fact that $\mathrm{sgn}$ is a homomorphism of groups. That's the only advantage of my answer... ;-) $\endgroup$ Oct 11, 2011 at 20:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .