# $sgn(\alpha)=(-1)^k=(-1)^k$ help proof

For $n\geqslant 2$, if $\alpha\in S_n$ and $\alpha=\tau_1...\tau_k=\lambda_1...\lambda_m$ are factorizations of $\alpha$ as the product of transpositions in $S_n$, so $m$ and $k$ have the same parity, this means,$(-1)^k=(-1)^m$.

The sign of $\alpha$ is denoted $sgn(\alpha)=(-1)^k$.

Following some notes I did not understand the following proof:

If $\alpha$ is a transposition:

$\prod_{j>i}(\alpha(j)-\alpha(i))=-a$

$i,j\in\{1,...,n\}$

So if now $\alpha=\tau_1...\tau_k=\lambda_1...\lambda_m$

$\prod_{j>I}(\alpha(j)-\alpha(i))=(-1)^k\times a=(-1)^m\times a$.

Questions:

1)I think this proof is incomplete. How can I complete it?

2)How can $\prod_{j>i}(\alpha(j)-\alpha(i))$ measure the number of transpositions? As far as I understand it measuring the number of transformations that is not the number of transpositions that compose $\alpha$.

• 1) Why do you think it is incomplete? 2) What do you mean "number of transformations"? – Trevor Gunn Oct 1 '17 at 18:40
• Number of transformations. Consider the permutation $(1324)\in S_4$, the number of transformations would be 1to3, 3to2, 2to4and 4to 1, that equals 4.I think it is incomplete because I doubt the notes are complete. – Pedro Gomes Oct 1 '17 at 18:45
• "I think it is incomplete because I doubt the notes are complete." - This remark is not as informative as perhaps you'd like it to be. – Matt Oct 1 '17 at 19:02

Let me address your second concern first. The action of $\alpha$ on $\prod_{j > i}(j - i)$ isn't counting the number of transformations, it is counting the number of inversions. Or rather, the parity (even- or odd-ness) of this number.

Definition: An inversion of a permutation $\alpha$ is a pair $\{i,j\}$ with $i < j$ and $\alpha(i) > \alpha(j)$.

For example the permutation $(12)(34)$ has two inversions: $\{1,2\}$ and $\{3,4\}$. Notice that if $\{i,j\}$ is an inversion of $\alpha$ and $(j - i)$ is positive, then $(\alpha(j) - \alpha(i))$ is negative. The sign of a permutation can alternatively be written as $(-1)^n$ where $n$ is the number of inversions.

When we apply $(12)(34)$ to the product $$(4 - 1)(4 - 2)(4 - 3)(3 - 1)(3 - 2)(2 - 1)$$ we get

$$(3 - 2)(3 - 1)\color{red}{(3 - 4)}(4 - 2)(4 - 1)\color{red}{(1 - 2)}$$

The red terms are where the inversions are and also where we pick up a factor of $-1$.

On the other hand, the permutation $(1324)$ has 5 inversions:

$$\color{red}{(1 - 3)}\color{red}{(1 - 4)}\color{red}{(1 - 2)}\color{red}{(2-3)}\color{red}{(2 - 4)}(4 - 3).$$

Therefore $(12)(34)$ is an even permutation ($\operatorname{sgn}((12)(34)) = (-1)^2 = 1$) and $(1324)$ is an odd permutation ($\operatorname{sgn}((1324)) = (-1)^5 = -1$).

Caution: It is tempting to say that the number of inversions of $\sigma \pi$ is the sum of the number of inversions of $\sigma$ plus the number of inversions of $\pi$. However, this is incorrect as simply taking $\sigma = \pi = (12)$ will show you. Both $\sigma$ and $\pi$ have 1 inversion and $\sigma \pi = \iota$ has zero inversions.

Now with this in mind, we can prove that the sign of a permutation is well-defined as follows. Note that the concept of inversions is only used in the first step.

1. Show that if $(ij)$ is a transposition, then $(ij)$ has only one inversion, namely $\{i,j\}$. Therefore $$(ij)\cdot\prod_{j > i}(j - i) = -\prod_{j > i}(j - i).$$

2. It follows from (1) that given a product $\alpha = \tau_1 \cdots \tau_k$ of $k$ permutations then $$\alpha \cdot\prod_{j > i}(j - i) = (-1)^k \prod_{j > i}(j - i).$$

3. Therefore if $\alpha$ is a product of $k$ transpositions and of $m$ transpositions then $(-1)^k = (-1)^m$.

• Thank you! The answer is marvellous! – Pedro Gomes Oct 1 '17 at 19:54