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Let $G$ be a finite group, and let $k$ be a field whose characteristic divides $\left|G\right|$. Let $\rho:G\to \mathrm{End} V$ be a (finite-dimensional) representation of $G$ over $k$. Prove or disprove that

$\sum\limits_{g\in G} \frac{1}{\det\left(\mathrm{id}-T\rho\left(g\right)\right)} = 0$ as an equality between power series in $k\left[T\right]$.


Let $G$ be a finite group, and $k$ be any field. Let $\rho:G\to \mathrm{End} V$ be a (finite-dimensional) representation of $G$ over $k$. For every $d\geq 0$, let $n_d$ denote the dimension of the space $\left(\mathrm{Sym}^d V^{\ast}\right)^G$ of the $G$-invariant symmetric $d$-ary forms on $V$. Molien's formula states that

$\left|G\right|\cdot \sum\limits_{n=0}^{\infty} n_dT^d = \sum\limits_{g\in G} \frac{1}{\det\left(\mathrm{id}-T\rho\left(g\right)\right)}$ as an equality between power series in $k\left[T\right]$

if $\mathrm{char} k$ does not divide $\left|G\right|$. I am wondering whether this formula still holds if $\mathrm{char} k\mid \left|G\right|$. The standard proof, using the Maschke projection, does not make much sense in this case...

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For those unfamiliar with the term 'Molien', I found en.wikipedia.org/wiki/Molien_series to be helpful. –  Stopple Feb 23 '11 at 16:08
Good point - post edited. (Somehow I forget that not everybody knows invariant theory.) –  darij grinberg Feb 23 '11 at 16:19

3 Answers 3

up vote 6 down vote accepted

A simple-minded argument:

Pick any element $g\in G$ of order $p$, the characteristic of $k$, and let $C(g)$ be the centralizer of $g$.

Now $G\setminus C(g)$ is the disjoint union of orbits under the action of the inner automorphism $\iota_g:h\in G\mapsto ghg^{-1}\in G$, and those orbits are of size $p$. The sum of the terms in your sum corresponding to the elements of $g$ in one of those orbits is then zero, for those terms are all equal.

If $h\in C(g)$, then the terms in your sum corresponding to the elements $h$, $gh$, $g^2h$, $\dots$, $g^{p-1}h$ are all equal---because $g$ and $h$ commute, you can take them simutaneously to Jordan canonical form and $g$ has only $1$ as an eigenvalue---so that their sum is also zero.

We have thus partitioned $G$ into, on one hand, the orbits of $\iota_g$ in $G\setminus C(g)$, and, on the other, the cosets of $\langle g\rangle$ in $C(g)$, and checked that the sum of the terms in each part of this partition is zero. Therefore your sum is zero.

NB: Notice that the specific form of the terms in your sum does not really matter, as long as it only depend on the eigenvalues of the $g\in G$. Thus, for example, exactly the same reasoning shows that the "other" Molien formula $$\sum_{g\in G}\det(I-t\rho(g))$$ also vanishes.

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Wow, this is beautiful! –  darij grinberg Feb 23 '11 at 16:23

By a well known lemma in group theory, we can write any $g\in G$ uniquely as $h_1h_2$ where $h_1$ has order a power of $p$, $h_2$ has order prime to $p$, and $h_1h_2=h_2h_1$. Thus we may write $G$ as a disjoint union

$G=\cup_g P_g\cdot g$

where $g$ ranges over the elements of order prime to $p$ and $P_g$ denotes the set of elements of $p$-power order which commute with $g$. Fix a char. p representation $\rho$. Note that every element of $\rho(P_g\cdot g)$ has the same characteristic polynomial as $\rho(g)$. Also, $|P_g|$ is either divisible by $p$ or equal to one (copy the proof of Cauchy's Theorem). In the former case, we clearly have

$\sum\limits_{\tilde{g}\in P_g\cdot g} \det\left(\mathrm{id}-T\rho\left(\tilde{g}\right)\right)^{-1} = 0$.

The remaining $g$'s all have centralizer of order prime to $p$, so each of their conjugacy classes has order divisible by $p$. It follows from this observation that the sum over the remaining elements must vanish as well.

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Thanks. I didn't even read your post to the end; after the "well known lemma" it was immediately clear to me. I am going to accept your solution as soon as you add a reference to this lemma. –  darij grinberg Feb 23 '11 at 12:55
@darij: The reference is "folklore". Write $ord(g)=ab$ with relative prime $a,b$ and choose $s,t$ such that $sa+bt=1$. Then $g=g^1=(g^a)^s \cdot (g^b)^t$. Obviously the two factors commute and $ord((g^a)^s) | b$ and $ord((g^b)^t) | a$. In fact the orders equal those terms because $ab=ord(g)=ord((g^a)^s)ord((g^b)^t) | ba$. One the other hand: If $g=xy$ with $ord(x)=b, ord(y)=a$ and $xy=yx$ then $g^{sa} = (x^{sa})(y^{sa})=x^{sa}=x^{sa \mod b}=x$ and $g^{tb}=(x^{tb})(y^{tb})=y^{tb}=y^{tb \mod a}=y$. –  Johannes Hahn Feb 23 '11 at 15:16
Ah, thanks! I was stupid. I confused the "well known lemma" with a lemma I wanted to use in my proof and couldn't show. They only see what they want to see. –  darij grinberg Feb 23 '11 at 15:58
Oh. Wait. In $G=\bigcup_g P_g\cdot g$, why is the union disjoint? –  darij grinberg Feb 23 '11 at 16:08
The union is disjoint by uniqueness of the decomposition $g=h_1h_2$. Anyway, Mariano's argument is cleaner. –  Kevin Ventullo Feb 23 '11 at 18:25

I am not specialist of the subject but seems that the equality does not work in the special case:

$k=\mathbb{F}_2,$ $G=\{0,1\}$ the additive group of the same field $\mathbb{F}_2,$ the vector space still the same field $V=k,$ and the representation $$ \rho(0)=id,\;\;\rho(1)=0 $$

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This is not a representation: $\rho(1)\cdot\rho(1)\neq \rho(1+1)$. –  Kevin Ventullo Feb 23 '11 at 6:10
@Kevin: You are right. I misunderstood the definition... –  Luis H Gallardo Feb 23 '11 at 8:11
@Kevin, more importantly, $\rho(1)$ is not an automorphism :) –  Mariano Suárez-Alvarez Feb 23 '11 at 16:25

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