Algebraic Combinatorics -- Injective symmetry classes

Injective symmetry classes

Injective symmetry classes

Such an f is injective if and only if the mapping described in the first item of corollary is injective and corresponding cyclic factors of bar (g) and bar (h) have the same length. The number of such mappings is

Õ_j[a_j( bar (h)) choose a_j( bar (g))]a_j( bar (g))!,

while the second item of corollary says that we have to multiply this number by P_jj^{a_j( bar (g))} in order to get the number | Y^X_i,(h,g) | of fixed points of (h,g) on Y_i^X. Thus we have proved

Corollary: The number of fixed points of (h,g) on Y_i^X is
| Y^X_i,(h,g) | = Õ_j[a_j( bar (h)) choose a_j( bar (g))] j^{a_j( bar (g))}a_j( bar (g))!,
and hence, by restriction, the numbers of fixed points of g and of h are:
| Y^X_i,g | = [ | Y | choose | X | ] | X | ! if bar (g)=1

| Y^X_i,g | = 0 otherwise,
and
| Y^X_i,h | = [a₁( bar (h)) choose | X | ] | X | !.

An application of the Cauchy-Frobenius Lemma yields the desired number of injective symmetry classes:

Theorem: The number of injective H ´G-classes is
| (H ´G) \\Y_i^X | =(1)/( | H | | G | ) å_(h,g) Õ_j[a_j( bar (h)) choose a_j( bar (g))]j^{a_j
( bar (g))}a_j( bar (g))!,
so that we obtain by restriction the number of injective G-classes
| G \\Y_i^X | =( | X | !)/( | bar (G) | )[ | Y | choose | X | ]= [ | Y | choose | X | ] | S_X/ bar (G) | ,
and the number of injective H-classes
| H \\Y_i^X | =( | X | !)/( | H | ) å_{k= | X |}^{| Y |} | {h ÎH | a₁( bar (h))=k } | [k choose | X | ].

Try to compute the number of injective symmetry classes for various group actions.

harald.fripertinger@kfunigraz.ac.at,
last changed: August 28, 2001

Injective symmetry classes