Definition of sign

Question

The following definition is in my notes with no explanation:

$$\operatorname{sgn}(\sigma)=\begin{cases}1,&\text{if }\sigma(p)(x_1,\ldots,x_n)=p(x_1,\ldots,x_n)\\-1,&\text{if }\sigma(p)(x_1,\ldots,x_n)=-p(x_1,\ldots,x_n)\end{cases}$$

I'm a little confused by what it actually means in practice and what sigma represents. Any explanation or clarification would be great.

Answer 1

A permutation $\sigma$ on a set $X = \{1,\dotsc,n\}$ is a bijection of the set $X$ on itself. Meaning that $\sigma$ maps any element of $X$ to another unique element of $X$ (maybe the same). For example, a permutation could look like this:

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

Meaning that $\sigma(1)=2$, $\sigma(2)=5$, $\sigma(3)=4$ and so on. Permutations have applications in various fields. In linear algebra, they are extremely useful for defining the determinant as you can see for example here. The determinant is an example where you also need the sign of the permutation.

For $\sigma \in S_n$ (Permutation Group of the numbers $\{1,\dotsc,n\}$), let $P(x_1,\dotsc,x_n) = \prod_{i<n}(x_i-x_j)$, then we can define

$$\operatorname{sgn}(\sigma) = \frac{P(x_{\sigma(1)},\dotsc,x_{\sigma(n)})}{P(x_1,\dotsc,x_n)}$$

Since the polynomial $P(x_{\sigma(1)},\dotsc,x_{\sigma(n)})$ has the same factors as the polynoial $P(x_1,\dotsc,x_n)$ we have that either $\operatorname{sgn}(\sigma) = 1$ or $\operatorname{sgn}(\sigma) = -1$.

This is equivalent to

$$ P(x_{\sigma(1)},\dotsc,x_{\sigma(n)}) = \operatorname{sgn}(\sigma)P(x_1,\dotsc,x_n)$$

If you let $\operatorname{sgn}(\sigma) = 1$ or $\operatorname{sgn}(\sigma) = -1$ respectively, you get you definition.

Also possible and more common is the definitions of the sign as $\operatorname{sgn}(\sigma) = (-1)^n$ where $n$ is the number of transposition in the decomposition of $\sigma$. A transposition is a permutation where you only swap two elements and leave the rest untouched. So the decomposition is the product of transpositions $\tau_1,\cdots,\tau_n$ such that $\tau_1 \circ \cdots \circ \tau_n = \sigma$. This is well-defined as it can be proven that even though the decomposition of a permutation is not unique, the number of transpositions always stays the same.

Moreover, this is equivalent to $\operatorname{sgn}(\sigma) = (-1)^n$ where $n$ is the number of inversions of $\sigma$. An inversion is a pair $i,j \in \{1,\dotsc,n\}$ such that $i < j$ but $\sigma(i) > \sigma(j)$. If we look at the permutation above, we have for example the inversion $1 < 5$ but $2 = \sigma(1) > \sigma(5) = 1$.

Answer 2

The sign or signum function is defined as $$\operatorname{sgn}(x)=\begin{cases} -1&\text{if}\ x\lt 0,\\ 0&\text{if}\ x=0,\\ 1&\text{if}\ x\gt 0.\\ \end{cases}$$ It simply returns the sign of $x$ in the form of $-1$ or $1$. If $x$ is $0$, then $0$ is returned. Here are some examples, $$ \operatorname{sgn}(-5)=-1 $$ $$ \operatorname{sgn}\left(\frac{1}{2}\right)=1 $$ $$ \operatorname{sgn}(6-6)=0 $$

Since no context was provided for the functions $\sigma$ and $p$, I cannot provide a meaningful interpretation.

My best guess is that the $\sigma(p)$ function changes the sign of $p$ and outputs the modified $p$. In which $\operatorname{sgn}(\sigma(p))$ outputs $-1$ or $1$, assuming that $p$ and $\sigma(p)$ is not zero.