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I'm currently studying multilinear algebra and I felt the need to see just the basics of permutations since it's used to study symmetric and antisymmetric tensors. My doubt was: given some permutation how to find it's sign? I mean, how to find the number of inversions?

Searching this on google I've found Wikipedia's article on permutations and so I've read there about "disjoint cycles". I've found this method of finding the sign of a permutation very straightforward, however I'm still confused on how to work with it.

For instance, if we have:

$$\sigma=\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 2 & 1\end{pmatrix}$$

Then $\sigma(1)=3$, $\sigma(3)=5$ and $\sigma(5)=1$ while $\sigma(2)=4$ and $\sigma(4)=2$. So I understand pretty easily that this is equivalent to writting

$$\sigma=\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 2 & 1\end{pmatrix} = \begin{pmatrix}1 & 3 & 5\end{pmatrix}\begin{pmatrix}2 & 4\end{pmatrix}$$

However, on Wikpedia they further reduce this to the following:

$$\sigma=\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 2 & 1\end{pmatrix} = \begin{pmatrix}1 & 3 & 5\end{pmatrix}\begin{pmatrix}2 & 4\end{pmatrix} =\begin{pmatrix}1 & 5\end{pmatrix}\begin{pmatrix}1 & 3\end{pmatrix}\begin{pmatrix}2 & 4\end{pmatrix}$$

Now, why is that? What allows this last step? What's behind all of this things? There it says that this is a way to decompose any permutation into composition of other permutations, but how is that? I'm simply not understanding the reasoning behind this.

I know that the "correct" was to first study group theory and so on, but I'm really just trying to get the basics for now to finish Kostrkin's book.

If someone can give some help or some reference it'll be of great assistance. Thanks very much in advance.

Gold
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  • Any 3-cycle can be broken up into two two cycles as follows. $(a b c) = (a b)(a c)$. To see why it works, calculate where each of $a$, $b$ and $c$ go working from left to right. – John Douma Apr 22 '13 at 19:26
  • I think it's confusing that you do composition on the left, while in the question the OP does composition on the right. – davidlowryduda Apr 22 '13 at 19:26

2 Answers2

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The idea behind what they said is that any permutation $(a b c d e)$ can be written as $(a e ) (a d ) (a c ) (a b )$ (or the natural extension to any number of symbols). I encourage you to check this and justify why it's true to yourself.

Thus $(1 3 5) = (1 5 ) (1 3 )$.

So once you've written your permutation as disjoint cycles, you can find the sign by counting the number of transpositions by writing it as above. You might notice that you don't actually need to write it out, as you know how many transpositions are necessary from the length of the cycle. Any 3-cycle will be a product of $2$ transpositions. Any 4-cycle will be a product of $3$, and so on.

davidlowryduda
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We know that every permutation has a decomposition on product of transpositions and the signature of a permutation $\sigma$ is $$\epsilon(\sigma)=(-1)^k$$ where $k$ is the number of the transpositions, but also we know that the signature of a cycle $(a_1,\ldots,a_p)$ is $(-1)^{p-1}$ so we can find the signature of a permutation from the decompostion into product of cycles.