notes-091029 - MATH 681 Notes Combinatorics and Graph...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 681 Notes Combinatorics and Graph Theory I 1 Equivalence classes of symmetries This definition is probably familiar, but its useful for discussing classification under symmetry. Definition 1. A set R of ordered pairs from S S is called a relation on S ; two elements a and b are said to be related under R , written aRb , if ( a,b ) R . A relation R is said to be reflexive if for all a S , aRa . R is symmetric if for all a,b S , if aRb , then bRa . R is transitive if for all a,b,c in S such that aRb and bRc , it follows that aRc . A relation which is reflexive, symmetric, and transitive is known as an equivalence relation . This lends itself to our previous investigation of symmetry as such: Proposition 1. For S a set of n-element objects and G a permutation group consisting of permu- tations of length n , there is a natural relation R G induced by letting ( a, ( a )) R G for all a S and G . Then R G is an equivalence relation on S . Proof. Since G is a permutation group, the identity element e G . By the definition of R G , it is true that for all a S , aR G e ( a ), or, since e ( a ) = a , aR G a Thus R G is reflexive. Now, consider a and b from S such that aR G b . By the definition of R G , this is true because there is a permutation G such that b = ( a ). By the definition of a symmetry group, - 1 is also in G , and since it is the inverse of , it follows that a = - 1 b . Since - 1 G and b S , bR G - 1 b , or in other words, bR G a . Thus R G is symmetric. Lastly, consider a , b , and c in S such that aR G b and bR G c . By the definition of R G , there is G such that b = ( a ), and G such that c = ( b ). Thus c = ( ( a )) = ( )( a ). By closure of the permutation group, G , and since a S , it follows that aR G ( )( a ), or in other words, aR G c . Thus R G is transitive. This is very useful for our investigation of symmetry because, if G represents some collection of symmetries, then aR G b if and only if a and b represent different orientations of the same object. We thus want to collect elements of S into subsets based on their equivalencies. In fact doing so is an established way of identifying an equivalence relation: Proposition 2. The equivalence relations on S are in a one-to-one correspondence with the set- partitions of S . Proof. A set-partition of S necessariily induces an equivalence relation: let aRb iff a and b are in the same partition; it is fairly easy to show that this relation definition satisfies the equivalence conditions. Conversely, if we have an equivalence relation R , we may define a set-partition by letting a and b be in the same partition if and only if aRb . The equivalence conditions guarantee that this is a well-defined procedure: we would only run into inconsistencies if a were required not to be in the same partition as itself (forbidden by reflexivity) or if different established members of a single...
View Full Document

This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.

Page1 / 6

notes-091029 - MATH 681 Notes Combinatorics and Graph...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online