notes-091029

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

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Unformatted text preview: MATH 681 Notes Combinatorics and Graph Theory I 1 Equivalence classes of symmetries This definition is probably familiar, but it’s 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,π ( a )) ∈ R G for all a ∈ S and π ∈ 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 π ∈ G such that b = π ( a ). By the definition of a symmetry group, π- 1 is also in G , and since it is the inverse of π , it follows that a = π- 1 b . Since π- 1 ∈ G and b ∈ S , bR G π- 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 π ∈ G such that b = π ( a ), and σ ∈ G such that c = σ ( b ). Thus c = σ ( π ( a )) = ( σπ )( a ). By closure of the permutation group, σπ ∈ G , and since a ∈ S , it follows that aR G ( σπ )( 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...
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notes-091029 - MATH 681 Notes Combinatorics and Graph...

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