Molien's formula

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

Precisely, it says: given a finite-dimensional complex representation V of G and Rn=[V]n=Symn(V*), the space of homogeneous polynomial functions on V of degree n (degree-one homogeneous polynomials are precisely linear functionals), if G is a finite group, the series (called Molien series) can be computed as:[1]

n=0dim(RnG)tn=(#G)1gGdet(1tg|V*)1.

Here, RnG is the subspace of Rn that consists of all vectors fixed by all elements of G; i.e., invariant forms of degree n. Thus, the dimension of it is the number of invariants of degree n. If G is a compact group, the similar formula holds in terms of Haar measure.

Derivation

Let χ1,,χr denote the irreducible characters of a finite group G and V, R as above. Then the character χRn of Rn can be written as:

χRn=i=1rai,nχi.

Here, each ai,n is given by the inner product:

ai,n=χRn,χi=(#G)1gGχi(g)χRn(g)=(#G)1gGχi(g)|α|=nλ(g)α

where λ(g)α=i=1mλi(g)αi and λ1(g),,λm(g) are the possibly repeated eigenvalues of g:V*V*. Now, we compute the series:

n=0ai,ntn=(#G)1gGχi(g)α(λ1(g)t)α1(λm(g)t)αm=(#G)1gGχi(g)(1λ1(g)t)1(1λm(g)t)1=(#G)1gGχi(g)det(1tg|V*)1.

Taking χi to be the trivial character yields Molien's formula.

Example

Consider the symmetric group S3 acting on R3 by permuting the coordinates. We add up the sum by group elements, as follows. Starting with the identity, we have

det(1t0001t0001t)=(1t)3.

There is a three-element conjugacy class of S3, consisting of swaps of two coordinates. This gives three terms of the form

det(1t0t10001t)=(1t)(1t2).

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

det(1t001tt01)=(1t3).

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

M(t)=16(1(1t)3+3(1t)(1t2)+21t3)=1(1t)(1t2)(1t3).

On the other hand, we can expand the geometric series and multiply out to get

M(t)=(1+t+t2+t3+)(1+t2+t4+)(1+t3+t6+)=1+t+2t2+3t3+4t4+5t5+7t6+8t7+10t8+12t9+

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

σ1=x+y+z
σ2=xy+xz+yz
σ3=xyz

we can see for example that in degree 5 there is a basis consisting of σ3σ2, σ3σ12, σ22σ1, σ13σ2, and σ15.

(In fact, if you multiply the series out by hand, you can see that the tk term comes from combinations of t, t2, and t3 exactly corresponding to combinations of σ1, σ2, and σ3, 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

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Further reading

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