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In the very beginning of the book "Introduction to Invariants and Moduli" Shigeru Mukai proves Molien's formula for the Hilbert series of the invariant ring of a finite group action on $\mathbb C^n$. For example, in the case of the standard action of Quaternions on $\mathbb C^2$ the Hilbert series is $\frac{1-t^{12}}{(1-t^4)^2(1-t^6)}$.

After this Mukai explains that this formula hints us that the ring of invariants can be generated by two elements of order $4$ and one element of order $6$, and indeed such elements can be found: $A=x^4+y^4$, $B=x^2y^2$, and $C=xy(x^4-y^4)$. Then one finds a relation $C^2=A^2B -4B^3$ and this gives a complete description of the ring of invariants.

Moreover the same procedure is shown to work in several other cases (e.g. binary icosahedral group).

My question is a follows: Is there some theorem that say that this heuristics works often? Namely, if we have an action of a finite group $G$ on $\mathbb C^n$, in order to describe the ring of invariants, we first look on the denominator of the Hilbert series (given by Molien's formula) and try to associate an invariant polynomial of degree $n$ to each factor $(1-t^n)$ (so that this gives us a full set of generators). Or at least, in practice, is this the first thing that one tries to do?

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Moeln = Molien? – darij grinberg Mar 12 '11 at 20:53
up vote 7 down vote accepted

As pointed out by Richard Stanley, there are some subtleties in finding the invariant generators.

The Molien series only tells you the Hilbert series of the ring of invariants up to reduced rational forms. So some information is lost, as illustrated in Richard's answer.

One can gain some information by looking at the actual series and a bit of work. For example, if your Molien series looks like:

$$1 + 2t^2 + 4t^4 +...$$

Then you know there are two invariant forms of degrees $2$. They generate three elements in degree $4$ ($f^2,fg,g^2$) so there must be another algebraically independent form in degree $4$, etc...

Once you know the degree of the generators, a decent way to guess them is to use the Reynolds operator (I assume the group is finite of order prime to characteristic of the field):

$$f \mapsto 1/|G|\sum_{g\in G} gf$$

Starting with any polynomial $f$, this always lands in the invariant subring, as you can easily check.

Once you know the first few generators, try to match the Hilbert series of the subalgebra generated by those with your Hilbert series. If they don't match, there must be another generator of degree equals to the first point the series mismatched, and on you go...

ADDED: I originally tried not to mention the word "Cohen-Macaulay", but it is particularly relevant here, so perhaps I should.

The Reynolds operator mentioned above give you a splitting $R^G \to R= \mathbb C[x_1,\cdots, x_n]$. Such splitting implies that $R^G$ is Cohen-Macaulay.

Why is that relevant? The point is that once you come up with $n$ invariant forms, say $f_1,\cdots f_n$ of degrees $d_1,\cdots d_n$, which you happen to know is a regular sequence on $R$ (hence also regular on $R^G$). Then as $R^G$ is Cohen-Macaulay, it is free over $R' = \mathbb C[f_1, \cdots, f_n]$. Assuming you can write

$$R^G = \oplus_1^m h_jR'$$

with each $h_j$ of degree $e_j$, then the Hilbert series of $R^G$ must be

$$\frac{\sum t^{e_j}}{\prod(1-t^{d_i})} $$

So by comparing with the Molien series you know the degree of each $h_j$. (See this survey by Richard Stanley!)

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The survey article can also be found at – Richard Stanley Mar 14 '11 at 0:51

The answer by Ottem shows that there is some way to write the generating function to correspond to a chain of syzygies. However, it is not true that every way works. For instance, let $G$ be the group of order eight generated by the two $3\times 3$ diagonal matrices with diagonals $(-1,-1,1)$ and $(1,1,i)$ (where $i^2=-1$). A "correct" way to write the generating function is $(1+t^2)/(1-t^2)^2(1-t^4)$. An "incorrect" way is $1/(1-t^2)^3$. Note that the series given in aglearner's question can also be written incorrectly by cancelling $1-x^6$ from the numerator and denominator.

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This follows from the Hilbert syzygy theorem. If $R=\mathbb{C}[x_0,\ldots,x_n]^G$ and $R^G$ is generated by elements $f_1,\ldots,f_r$ of degrees $a_1,\ldots,a_r$ respectively, $R^G$ has a free resolution as an $A=\mathbb{C}[f_1,\ldots,f_r]$-module, which implies that Hilbert series of $R^G$ is of the form $$ \frac{P(t)}{(1-t^{a_1})\cdots (1-t^{a_r})} $$where $P(t)$ is a polynomial. If moreover the $f_i$ have syzygies $g_1,\ldots,g_s$ of degrees $b_1,\ldots,b_s$ and there are no second syzygies, $P(t)$ will be of the form $1-\sum t^{b_i}$, so in your example is reasonable to expect that there is a syzygy in degree 12. In general, by the Syzygy theorem, $P$ will be a polynomial of the form $1-\sum t^{b_i}+\sum t^{c_i}-\ldots$ and so you can read off the degrees of the higher syzygies from the Hilbert series also.

EDIT: As pointed out by the answers above, the representation of the Hilbert series as a rational function is only unique up to cancellation of factors, so a priori you can not read off the degrees of the generators (in the above paragraph I assumed these were known). The Hilbert series does give you a useful hueristic though, since you can always use the Hilbert series to search for invariants $f_1,\ldots,f_r$ in small degrees and then compare the Hilbert series of $R^G$ with that of the subring spanned by the $f_i$ - if they are equal you have found the whole $R^G$, if not, search for generators in higher degrees.

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Thank you for the answer! Could you advise some place to read about this? (I understand why in the case when the ring is generated by homogeneous polynomials of degrees $a_i$ we have $(1-t^{a_i})$ in the denominator, but the rest is not clear for me – aglearner Mar 12 '11 at 22:37
Yes, try 'Using Algebraic Geometry' by Cox, Little and O'Shea or 'Invariant theory of Finite groups' by Smith. But there is nothing really mystic about the above, basically it's just writing up the resolution of $R^G$ and taking Euler characteristics. You could see the proof of the Hilbert-Serre theorem (in Atiyah-Macdonald) to see why $P$ has the form above. – J.C. Ottem Mar 12 '11 at 23:11

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