Sign Function

Def

Given two groups, $(S_k,\circ )$ and $(\{-1,1\},\cdot )$, we can define a group homomorphism, denoted as the sign function as $\begin{array}{rl}sgn:S_k& \to \{-1,1\}\\ \sigma & \mapsto (-1{)}^{\text{number of permutations of}\sigma }\end{array}$

• Recall, every $\sigma \in S_k$ can be decomposed into k transpositions. the number of transpositions = number of permutations as it is the number of times we swap two numbers!

Where does it come from?

suppose we have an $f\in F[x_1,...,x_k]$ where we can define the following group action for function of the following form : f = \prod_{i < j} (x_i - x_j)$$$$\sigma \cdot f = \prod_{i < j} (x_{\sigma(i)} - x_{\sigma(j)}) lets do a simple example, $S_3$, how does $\sigma \cdot f$ look like for all $\sigma \in S_k$?

$f=(x_1-x_2)(x_1-x_3)(x_2-x_3)$ $e\cdot f=f$ $(12)\cdot f=(x_2-x_1)(x_2-x_3)(x_1-x_3)=-(x_1-x_2)(x_1-x_3)(x_2-x_3)$ $(13)\cdot f=(x_3-x_2)(x_3-x_1)(x_2-x_1)=-(x_2-x_3)\cdot -(x_1-x_3)\cdot -(x_1-x_2)=-(x_1-x_2)(x_1-x_3)(x_2-x_3)$ $(23)\cdot f=(x_1-x_3)(x_1-x_2)(x_3-x_2)=-(x_1-x_2)(x_1-x_3)(x_2-x_3)$

$(123)\cdot f=(x_2-x_3)(x_2-x_1)(x_3-x_1)=(x_1-x_2)(x_1-x_3)(x_2-x_3)$ $(132)\cdot f=(x_3-x_2)(x_3-x_2)(x_2-x_3)=(x_1-x_2)(x_1-x_3)(x_2-x_3)$

Notice how the functions with odd number of permutations have a negative sign, while the ones that have an even number is the original function? That is our goal that we want to encode into our function!

Hence we can actually define the above as a group action! $S_k\times f\mapsto f,\sigma \cdot f=sign(\sigma )\cdot f$

Prove it is a group Homomorphism

• Note : permutations add up!

$sgn(\sigma \tau )=(-1{)}^{\#\sigma perm+\#\tau Perm}=(-1{)}^{\#\sigma perm}\cdot (-1{)}^{\#\tau perm}=sgn(\sigma )sgn(\tau )$

anagrams

$\{e,a\},\{e,b\}$

$a^2=a$

e, ae, ea, eb, be, ab

1, 2, 2, 2, 2, 4

e, a, b, ab

1, 2, 2, 4

$Z/2[X,Y]=\{1,x,y,xy\}$

$Z/3[X,Y]=\{1,x,x^2,y,y^2,x{y}^2,x^{2}y,x^2{y}^2\}$ ab = cd

Fundamental properties :

$2n,n\cdot n$

trihedral group : $D_{3n}=\{a^3=e,b^3=e,ab=ba\},order=9$

$a^3=a$, $ab=ba$

baabb = a aabb = ba

(1, 2, 3)(1, 2, 3) $\{e,a,aa\},\{e,b,bb\}$

e, ea, ae, eb, be, eaa, aae, ebb, bbe, abb, bba, aabb, bbaa, ba, ab

a, b, c, d, e, f, g, h, i e, a, b, aa, bb, abb, bba, aabb, bbaa

1, 2, 2, 3, 3, 4, 4, 5, 5

$e\cdot bb\cdot bbaa+a\cdot abb\cdot bba+b\cdot aa\cdot aabb$

$1\cdot 3\cdot 16+2\cdot 6\cdot 6+2\cdot 3\cdot 16$

$aa\cdot a\cdot bbaa$ $2\cdot 1\cdot 16$

1 x 4 +

$\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$

det : aei - afh + bdi - bfg + cdh - ceh

aei - bfg + cdh = afh - bdi + ceh