Molien series

In mathematics, a Molien series is a generating function attached to a linear representation ρ of a group G on a finite-dimensional vector space V. It counts the homogeneous polynomials of a given total degree d that are invariants for G. It is named for Theodor Molien.

Formulation

More formally, there is a vector space of such polynomials, for each given value of d = 0, 1, 2, ..., and we write nd for its vector space dimension, or in other words the number of linearly independent homogeneous invariants of a given degree. In more algebraic terms, take the d-th symmetric power of V, and the representation of G on it arising from ρ. The invariants form the subspace consisting of all vectors fixed by all elements of G, and nd is its dimension.

The Molien series is then by definition the formal power series

$M(t) = \sum_d n_d t^d.$

This can be looked at another way, by considering the representation of G on the symmetric algebra of V, and then the whole subalgebra R of G-invariants. Then nd is the dimension of the homogeneous part of R of dimension d, when we look at it as graded ring. In this way a Molien series is also a kind of Hilbert function. Without further hypotheses not a great deal can be said, but assuming some conditions of finiteness it is then possible to show that the Molien series is a rational function. The case of finite groups is most often studied.

Formula

Molien showed that

$M(t) = \frac{1}{|G|} \sum_{g\in G} \frac{1}{\det(I-tg)}$

This means that the coefficient of td in this series is the dimension nd defined above. It assumes that the characteristic of the field does not divide |G| (but even without this assumption, Molien's formula in the form $\left|G\right| \cdot M\left(t\right) = \sum_{g\in G} \frac{1}{\det(I-tg)}$ is valid, although it does not help with computing M(t)).

Example

Consider $S_3$ acting on R3 by permuting the coordinates. Note that $\det(I-tg)$ is constant on conjugacy classes, so it is enough to take one from each of the three classes in $S_3$; so $\det(I-te) = (1-t)^3,\det(I-t\sigma_2) = (1-t)(1-t^2)$ and $\det(1-t\sigma_3) = (1-t^3)$ where $\sigma_2=(1,2)$ and $\sigma_3 = (1,2,3)$.

Then

$M(t) = \frac16\left(\frac{1}{(1-t)^3} + \frac3{(1-t)(1-t^2)} + \frac{2}{1-t^3}\right) = \frac{1}{(1-t)(1-t^2)(1-t^3)}$