How Does the Sign of a Permutation Product Relate to Its Components?

  • Context: Graduate 
  • Thread starter Thread starter bomba923
  • Start date Start date
  • Tags Tags
    Permutations Proof
Click For Summary
SUMMARY

The discussion focuses on the relationship between the sign of a permutation product and its components within the symmetric group \( S_N \). It establishes that for permutations \( \sigma \) and \( \pi \) in \( S_N \), the equation \( \text{sgn}(\sigma \circ \pi) = \text{sgn}(\sigma) \cdot \text{sgn}(\pi) \) holds true. The proof involves analyzing the behavior of the determinant function \( \Delta \) under permutations and demonstrating that if \( \text{sgn}(\sigma \circ \pi) = 1 \), then both \( \text{sgn}(\sigma) \) and \( \text{sgn}(\pi) \) must also equal 1, and vice versa for -1.

PREREQUISITES
  • Understanding of symmetric groups, specifically \( S_N \)
  • Familiarity with the concept of permutations and bijections
  • Knowledge of determinants and the determinant function \( \Delta \)
  • Basic grasp of algebraic properties of signs in mathematical functions
NEXT STEPS
  • Study the properties of symmetric groups \( S_N \) in greater detail
  • Learn about the determinant function and its applications in linear algebra
  • Explore the concept of even and odd permutations and their implications
  • Investigate other algebraic structures that utilize signs, such as alternating groups
USEFUL FOR

Mathematicians, students studying abstract algebra, and anyone interested in combinatorial properties of permutations and their applications in linear algebra.

bomba923
Messages
759
Reaction score
0
Consider
[tex]S_N = \left\{ {\left. {\sigma :\left\{ {1, \ldots ,N} \right\} \to \left\{ {1, \ldots ,N} \right\}} \right|\sigma {\text{ is a bijection}}} \right\}[/tex]
i.e., the set of all permutations on 'N' values.

Define
[tex]\Delta \left( {x_1 , \ldots ,x_N } \right) = \prod\limits_{i < j} {\left( {x_i - x_j } \right)}[/tex]
and, for [tex]\sigma \in S_N[/tex],
[tex]\sigma \left( \Delta \right)\left( {x_1 , \ldots ,x_N } \right) = \prod\limits_{i < j} {\left( {x_{\sigma \left( i \right)} - x_{\sigma \left( j \right)} } \right)}[/tex]

Also, define [tex]{\mathop{\rm sgn}} : S_N \to \left\{ {\pm 1} \right\}[/tex] as
[tex]{\mathop{\rm sgn}} \left( \sigma \right) = \left\{ \begin{array}{l}<br /> 1,\;\sigma \left( \Delta \right) = \Delta \\ <br /> - 1,\;\sigma \left( \Delta \right) = - \Delta \\ <br /> \end{array} \right.[/tex]

How do I prove that, for [tex]\sigma ,\pi \in S_N[/tex],
[tex]{\mathop{\rm sgn}} \left( {\sigma \circ \pi } \right) = {\mathop{\rm sgn}} \left( \sigma \right){\mathop{\rm sgn}} \left( \pi \right) \; ?[/tex]
 
Last edited:
Physics news on Phys.org
sgn(sigma o pi)=1
then sigma(pi(delta))=delta
if pi(delta)=delta then sigma(delta)=delta so both sgn are 1.
if pi(delta)=-delta sigma(-delta)=delta, which is the same as sigma(delta)=-delta, you can see it by obserivng that sigma(delta)+sigma(-delta)=0.
the same goes when sgn (sigma(pi))=-1.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 16 ·
Replies
16
Views
4K