Molien's formula

Derivation

Let \chi_1, \dots, \chi_r denote the irreducible characters of a finite group G and V, R as above. Then the character \chi_{R_n} of R_n can be written as: :\chi_{R_n} = \sum_{i=1}^r a_{i, n} \chi_i. Here, each a_{i, n} is given by the inner product: :a_{i, n} = \langle \chi_{R_n}, \chi_i \rangle = (# G)^{-1} \sum_{g \in G} \overline{\chi_i}(g) \chi_{R_n}(g) = (# G)^{-1} \sum_{g \in G} \overline{\chi_i}(g) \sum_{|\alpha| = n} \lambda(g)^{\alpha} where \lambda(g)^{\alpha} = \prod_{i=1}^m \lambda_i(g)^{\alpha_i} and \lambda_1(g), \dots, \lambda_m(g) are the possibly repeated eigenvalues of g : V^* \to V^. Now, we compute the series: : \begin{align} \sum_{n = 0}^{\infty} a_{i, n} t^n &= (# G)^{-1} \sum_{g \in G} \overline{\chi_i}(g) \sum_{\alpha} (\lambda_1(g)t)^{\alpha_1} \cdots (\lambda_m(g)t)^{\alpha_m} \ &= (# G)^{-1} \sum_{g \in G} \overline{\chi_i}(g) (1 - \lambda_1(g)t)^{-1} \cdots (1 - \lambda_m(g)t)^{-1} \ &= (# G)^{-1} \sum_{g \in G} \overline{\chi_i}(g) \det(1 - tg|V^)^{-1}. \end{align} Taking \chi_i to be the trivial character yields Molien's formula.

Example

Consider the symmetric group S_3 acting on R3 by permuting the coordinates. We add up the sum by group elements, as follows. Starting with the identity, we have

: \det \begin{pmatrix} 1-t & 0 & 0 \ 0 & 1-t & 0 \ 0 & 0 & 1-t \end{pmatrix} = (1-t)^3 .

There is a three-element conjugacy class of S_3, consisting of swaps of two coordinates. This gives three terms of the form

: \det \begin{pmatrix} 1 & -t & 0 \ -t & 1 & 0 \ 0 & 0 & 1-t \end{pmatrix} = (1-t)(1-t^2) .

There is a two-element conjugacy class of cyclic permutations, yielding two terms of the form

: \det \begin{pmatrix} 1 & -t & 0 \ 0 & 1 & -t \ -t & 0 & 1 \end{pmatrix} = \left(1 - t^3 \right) .

Notice that different elements of the same conjugacy class yield the same determinant. Thus, the Molien series is

: M(t) = \frac 1 6 \left(\frac{1}{(1-t)^3} + \frac 3 {(1-t)(1-t^2)} + \frac{2}{1-t^3}\right) = \frac{1}{(1-t)(1-t^2)(1-t^3)}.

On the other hand, we can expand the geometric series and multiply out to get : M(t) = (1 + t + t^2 + t^3 + \cdots)(1+ t^2 + t^4 + \cdots)(1 + t^3 + t^6 + \cdots) = 1 + t + 2t^2 + 3t^3 + 4t^4 + 5t^5 + 7t^6 + 8t^7 + 10 t^8 + 12 t^9 + \cdots

The coefficients of the series tell us the number of linearly independent homogeneous polynomials in three variables which are invariant under permutations of the three variables, i.e. the number of independent symmetric polynomials in three variables. In fact, if we consider the elementary symmetric polynomials : \sigma_1 = x + y + z : \sigma_2 = xy + xz + yz
: \sigma_3 = xyz

we can see for example that in degree 5 there is a basis consisting of \sigma_3 \sigma_2, \sigma_3 \sigma_1^2, \sigma_2^2 \sigma_1, \sigma_1^3 \sigma_2, and \sigma_1^5 .

(In fact, if you multiply the series out by hand, you can see that the t^k term comes from combinations of t, t^2, and t^3 exactly corresponding to combinations of \sigma_1, \sigma_2, and \sigma_3, also corresponding to partitions of k with 1, 2, and 3 as parts. See also Partition (number theory) and Representation theory of the symmetric group.)

References

• David A. Cox, John B. Little, Donal O'Shea (2005), Using Algebraic Geometry, pp. 295–8
• {{cite journal | author1-link= Theodor Molien
• {{cite book
• {{cite journal |doi-access=free

References

1. The formula is also true over an algebraically closed field of characteristic not dividing the order of ''G''.