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Does there are exist simple proof for the following statement? Let $\rho,V$ be an irreducible representation of group $G$ of dimention $n$. Assume that there are exist $g \in G$ such that $\rho(g)$ just flips two coordinates. (that is $\rho(g)e_1=e_2,\ \rho(g)e_2=e_1,\ \rho(g)e_i=e_i$) Then $|G|\geq 2^n$, where $n$ is a dimention of the representation.

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What are the precise assumptions here? Apparently $G$ should be finite and nonabelian, with $n \geq 2$. Is there any restriction on the underlying field? Also, in the second sentence it's assumed that $g$ has the special property relative to some basis of $V$ called $e_1, \dots, e_n$. –  Jim Humphreys Mar 8 '11 at 12:29
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up vote 11 down vote accepted

[I assume characteristic 0, and then I can extend the scalars to an algebraically closed field, which I can take to be $\mathbb{C}$ as $G$ is finite and $V$ is finite dimensional. It may appear that I then need the starting $V$ to be absolutely irreducible for the argument below, but actually since the eigenvector $e_1 - e_2$ is rationally defined, just irreducible is enough.]

$\rho(g)$ has a single $-1$ eigenvalue, and the rest are $1$, i.e. it is a (complex) symmmetry in the sense of Serre's book Complex Semisimple Lie Algebras. Let $v$ be a $-1$ eigenvector. Since $V$ is irreducible, the orbit of $v$ spans $V$. So the orbit forms a (complex) root system, where the symmetry corresponding to the root $\rho(h)(v)$ is $\rho(hgh^{-1})$ (note we've dropped the integrality hypothesis from Serre's definition of root system). Now complex root systems in this sense correspond to complexifications of real root systems. So the image under $\rho$ of the subgroup of $G$ generated by these symmetries is isomorphic to a (possibly reducible) finite real reflection group of rank $n$. Since reducible Coxeter groups are direct products of their parts, we need only check that irreducible finite Coxeter groups have order at least $2^r$, where $r$ is the rank, which is true by the classification, or by using the normal form for words in Coxeter groups to show that the $2^r$ words formed from omitting any subset of simple reflections from $s_1 s_2 s_3 \dots s_r$ give distinct elements.

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What's the normal form you are alluding to? –  fherzig Mar 8 '11 at 22:50
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Well, not exactly a normal form. But two words representing the same element both lead to some common word upon making a sequence of replacements of the forms $s_a s_a \mapsto 1$ and $s_a s_b s_a s_b s_a \dots \mapsto s_b s_a s_b s_a s_b \dots$. But since the words I used don't contain more than one of a given simple reflection, you can never make the first type of replacement (so in particular, they are minimum length representatives), and the second type of replacement merely changes the ordering. Hence they represent distinct elements of the Coxeter group. –  ndkrempel Mar 9 '11 at 0:24
    
Thanks! I now remembered the 1969 paper of Tits solving the word problem for Coxeter groups (MR0254129). You are using his result mentioned in the MR review about when two words are equal, right? –  fherzig Mar 9 '11 at 15:46
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Yes, that's right. –  ndkrempel Mar 9 '11 at 16:15
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