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 n_d 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 n_d is its dimension.

The Molien series is then by definition the formal power series

M(t) =	∑	ndtd.
	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 n_d 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 t^d in this series is the dimension n_d 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 S₃ 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 S₃; so det(I − te) = (1 − t)³,det(I − tσ₂) = (1 − t)(1 − t²) and det(1 − tσ₃) = (1 − t³) where σ₂ = (1,2) and σ₃ = (1,2,3).

Then

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

References

• David A. Cox, John B. Little, Donal O'Shea (2005), Using Algebraic Geometry, pp. 295–8
• Molien, Th. (1897). "Uber die Invarianten der linearen Substitutionsgruppen.". Sitzungber. Konig. Preuss. Akad. Wiss. (J. Berl. Ber.) 52: 1152–1156. JFM 28.0115.01. http://books.google.com/?id=EIxK-opAmJYC&pg=PA1152. 
• Mukai, S. (2002). An introduction to invariants and moduli. Cambridge Studies in Advanced Mathematics. 81. ISBN 9780521809061. http://www.cambridge.org/catalogue/catalogue.asp?isbn=0521809061.