Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A representation $\rho:G\to GL_k(\mathbb{F})$ is called fixed-point free if for every $1\neq g\in G$ and every $0\neq v\in \mathbb{F}^k$, $\rho(g)v\neq v$. Stated differently, it is a representation where $I-\rho(g)\in GL_k(\mathbb{F})\cup \{0\}$ for every $g\in G$. A group admiting such a representation is called a fixed-point free group.

I would like to find out which groups admit fixed-point free representations where also the sum of every 3 elements is invertible (or zero), i.e., $I_k+\rho(g_i)+\rho(g_j)\in GL_k(\mathbb{F})\cup\{0\}$ for every $g_i,g_j\in G$. Where can I find information about this problem?

Even more specifically, I'm interested in this property for representations of odd groups over fields with characteristic 2.

share|improve this question
    
you wrote "sum of every 3 elements", meaning "sum of every 2 elements and the identity"... –  Dima Pasechnik Mar 30 at 19:13
2  
I think it is equivalent because one can always divide by one of the elements since they are invertible. The product is invertible/0 iff it was already invertible/0, and the representation property gives this new sum. But sum of every 2 elements and the identity is actually what I'm interested in. –  A.B. Mar 30 at 19:18

1 Answer 1

up vote 8 down vote accepted

NEW EDIT: I MISREAD THE ORIGINAL QUESTION

If you intend to think about finite $G,$ then $|G|$ and the characteristic of $k$ must be relatively prime (for if the characteristic is $p,$ then an element of order $p$ would not act fixed point freely). Hence the representation is completely reducible. We might as well suppose that $k$ is algebraically closed. We might as well suppose that the representation is irreducible: any irreducible constituent is faithful by the fixed-point free condition.

Then all elements of order $3$ (if there are any) in $G$ are in $Z(G)$ for otherwise $I + x + x^{2}$ is not invertible for an element $x$ of order $3$ which is non-central. Then $G$ has a cyclic (possibly central) Sylow $3$-subgroup and a normal $3$-complement.

If the Sylow $3$-subgroup of $G$ is central, then it is a direct factor consisting of scalars. In that case, we might as well suppose that the group has order prime to $3$.

Given the remark at the end of the question, I assume now that the group $G$ has odd order.

Now $G$ is a Frobenius complement, and all Abelian subgroups of $G$ are cyclic. Hence $F(G)$ is cyclic and $G/F(G)$ is cyclic. Suppose that $G$ itself is not cyclic. Then the given representation of $G$ is induced from a representation $F(G)$ and is monomial.

I claim that $I + a+ b$ is either invertible or zero for $a,b \in F(G).$ For otherwise there
is $a \in F(G)$ of order $n,$ and there is a primitive $n$-th root of unit $\omega$ such that $1 + \omega + \omega^{j} = 0$ for some $j \neq 0,1.$ This forces $n = 3$ and $a$ to have order $3,$ (note that $b$ is a power of $a$ in this case),

The question reduces to considering monomial non-diagonal matrices $a,b$ of odd order such that $a+b$ has $-1$ as an eigenvalue. This splits into two case: $a^{-1}b$ diagonal and $a^{-1}b$ non-diagonal. I will return to this. An important case is when $a^{p}$ and $b^{p}$ are diagonal for some prime $p.$

Consider the eigenvalues of $I + c$ for $c$ a non-diagonal monomial matrix with $c^{p}$ diagonal of odd order , and $c$ of $p$-cycle shape. Then $(c+I)^{p} = \lambda I$ for some odd order root of unity $\lambda.$ The eigenvalues of $c$ are therefore of the form $\lambda \omega -1$ for some $p$-th root of unity $\omega.$ This can only be of the form $-\mu$ for some odd order root of unity $\mu$ if $p = 3.$

If $a^{-1}b$ is diagonal, again the only way that that $I + a^{-1}b$ can have $-\mu$ as an eigenvalue for some odd order root of unity $\mu$ is when that order is $3.$

We may conclude that there are no restrictions (apart from being a Frobenius complement) on the normal $3$-complement of $G.$ I omit the details but in general, when the group $G$ is a Frobenius complement of odd order, the extra condition of $I + a + b$ being invertible or zero imposes no further restriction. For $G$ necessarily has cyclic Sylow $3$-subgroups and a normal $3$-complement, and all elements of order $3$ are necessarily central in an odd Frobenius complement (eg by Hall-Higman Shult type results).

share|improve this answer
    
Why would 3 even divide $|G|$ ? And why must the normal 3-complement be Abelian? –  Dima Pasechnik Mar 30 at 20:20
    
I have to repeat @DimaPasechnik 's question - why does there exist an irreducible representation of dimension 3? I might take the quaternions for example (they do not interest me since this is an even group, but SmallGroup(275,1) on GAP also seems problematic). Another thing: notice that I allow for the sum to be 0, just not a non-zero singular matrix. –  A.B. Mar 31 at 5:13
    
@Dima Pasechnik: Sorry, I misread the question- for some reason, I thought that the question was about $3$-dimensional representations. –  Geoff Robinson Mar 31 at 8:55
    
So basically, if I work with odd fixed-point free groups then I'm safe? –  A.B. Mar 31 at 10:57
    
I believe that is the case –  Geoff Robinson Mar 31 at 10:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.