§ The similarity between labellings and representations

a:1 -- b:2
c:2 -- d:1
are isomorphic since I can send a -> d and b -> c.
a:1-b:2-c:3
d:1-e:3-f:2
is not isomorphic (though they would be if we forget the numbering), since the center vertices b and e have different labels.
  • Let's think of the equation l(f(v))=l(v)l'(f(v)) = l(v). Since ff is a bijection, we have V=V|V| = |V'|, so ll and ll' are both bijections to the same set [V]=[V][|V|] = [|V'|]. So we can invert the equation to write f(v)=l1(l(v))f(v) = l'^{-1}(l(v)). This tells us that ff is determined by the labellings!
  • The point of having a labelling is that it forces upon us a unique isomorphism (if it exists), given by the equation f(v)l1(l(v))f(v) \equiv l^{-1}(l(v)).
  • This collapses hom sets to either empty, or a unique isomorphism, which is far tamer than having many possible graph isomorphisms that we must search for / enumerate !
  • In analogy to representation theory, if we consider two irreducible representations of a group GG, say α:GGL(V)\alpha: G \rightarrow GL(V) and β:GGL(W)\beta: G \rightarrow GL(W), Schur's lemma tells us that the Hom-set between the two representations (an intertwining map) is either the zero map (which is like having no isos) or a scaling of the identity map (which is like having a uniquely determined iso).
  • In this sense, we can think of an irrep as a "labelling" of group elements in a particularly nice way, since it constrains the potential isomorphisms of the "labelled objects"!