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In an informal talk I heard a statement:

"Any cyclic subgroup in a linear group is at most exponentially distorted"

with a vague reference to a work of Lubotzky with coauthors.

The works of Lubotzky that may be relevant to this quiestion and that I was able to find reference for:

Lubotzky, Alexander; Mozes, Shahar; Raghunathan, M. S. Cyclic subgroups of exponential growth and metrics on discrete groups. C. R. Acad. Sci. Paris Sér. I Math. 317 (1993), no. 8, 735–740. http://www.ams.org/mathscinet-getitem?mr=1244421

Lubotzky, Alexander; Mozes, Shahar; Raghunathan, M. S. The word and Riemannian metrics on lattices of semisimple groups. Inst. Hautes Études Sci. Publ. Math. No. 91 (2000), 5–53 (2001). http://www.ams.org/mathscinet-getitem?mr=1828742

seem to deal with lattices in Lie groups, not with arbitrary linear groups. So the question is:

Is the statement above true in such generality? If not, then for which classes of linear groups cyclic subgroups are at most exponentially distorted?

EDIT: As far as I understood, the distortion of a cyclic subgroup $H$ inside $G\subset GL(n,F)$ is meant here to be with respect to the respective word metrics on $H$ and $G$, not the underlying metric on $GL(n,F)$.

EDIT2: Sorry for the possible confusion. I guess, the distortion here is meant in the following sense:

Let $H$ be a subgroup of $G$ and fix some generating sets for $H$ and $G$. Consider a ball $B(n,1)$ of radius $n$ in the Cayley graph of $G$ centered at $1$, i.e. the set of all elements of $G$ having word length $\le n$ with respect to the generating set of $G$. Define a function $$ f(n):=\max\{|h|_H \mid h\in H\cap B(n,1)\}, $$ where $|h|_H$ is the length of the element $h$ with respect to the generating set of $H$.

Then $f(n)$ measures the distortion of subgroup $H$ inside $G$, and its order of growth is independent on the choice of the generating sets.

For example, in the group $G=\langle a,b\mid aba^{-1}=b^2\rangle$, subgroup $H=\langle b\rangle$ has exponential distortion function, as the element $h=a^mba^{-m}=b^{2^m}$ lies in the ball $B(2m+1,1)$ in $G$, but the word length of $g$ in $H$ is $2^m$.

But if we consider a group $G=\langle a,b,c\mid aba^{-1}=b^2, bcb^{-1}=c^2\rangle$ then a similar reasoning shows that $H=\langle c \rangle$ has double exponential distortion in $G$: $f(n)=2^{2^n}$.

The statement I am inquiring about stipulates that cyclic subgroups $H$ of linear groups $G\subset GL(n,F)$ cannot have distortion functions that grow faster than exponential functions.

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Do you mean cyclically distorted with respect to an invariant metric on $GL_n(\mathbb{R})$? (If that is the case you would at least need to assume that it is discrete.) I think the main point of Lubotzky-Mozes-Raghunathan is to prove that higher-rank lattices are not distorted; a proof of "at most exponentially distorted" would likely be much simpler. For example, if $\gamma$ generates a discrete subgroup in a real Lie group then you can put it inside a $SL_2({\mathbb R})$ and I think this yields the answer in this case. –  Jean Raimbault Sep 27 '13 at 7:39
    
@JeanRaimbault: As far as I understood, the distortion here is meant to be with respect to the word metrics on a cyclic subgroup $C$ and the linear group $H$ containing $C$. –  mathreader Sep 27 '13 at 13:40
    
It would be very helpful if you could add a precise definition of distortion. –  Nick Gill Sep 27 '13 at 13:59
    
@NickGill: I added more info, sorry for the confusion. –  mathreader Sep 27 '13 at 15:15
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1 Answer 1

up vote 9 down vote accepted

Yes it's true. First recall the definitions:

1) if $G$ is a group (discrete), a finitely generated subgroup $H$ is at most exponentially distorted (resp. undistorted) if for some/every finite generating subset $S$ of $H$ and any finite subset $T$ of $G$ such that $H$ is contained in $\langle S\rangle$, there exists a function with exponential growth (resp. with linear growth) $f$ such that $|g|_S\le f(|g|_T)$ for all $g\in H$. [If $G$ is finitely generated it's enough to check with a given $T$.] Note it is trivially true if $H$ is finite.

2) A linear group is a subgroup of $\mathrm{GL}_d(K)$ for some $d$ and field $K$.

Now to answer your question:

Proposition: in a linear group, every cyclic subgroup is at most exponentially distorted.

Lemma: let $\mathbf{K}$ be a nondiscrete locally compact field, and $g\in\mathrm{GL}_d(\mathbf{K})$. Endow the latter with the word length $\ell$ with respect to some/any compact generating subset. Then exactly one of the following holds:

a) $(\ell(g^n))_{n\in\mathbf{Z}}$ is bounded. This holds iff all eigenvalues of $g$ (over a finite extension) have modulus one, and in case $\mathbf{K}$ is Archimedean you require in addition that $g$ is semisimple.

b) $(\ell(g^n))$ grows logarithmically. This holds iff $\mathbf{K}$ is Archimedean, all eigenvalues of $g$ have modulus 1, and $g$ is not semisimple.

c) $(\ell(g^n))$ grows linearly. This holds iff $g$ has an eigenvalue (over some finite extension) of norm $\neq 1$.

I leave the proof of the lemma as an instructive exercise.

Now let $G\subset\mathrm{GL}_d(K)$ be a linear group and $\langle g\rangle$ be a cyclic subgroup and let us show the result. Clearly, we can assume that $\langle g\rangle$ is infinite and $G$ is finitely generated; actually we can then assume that $K$ is a finitely generated field.

If some eigenvalue of $g$ is not a root of unity, we can embed $K$ in a nondiscrete locally compact field so that this eigenvalue has norm $\neq 1$ (this is a theorem of Tits, used in the proof of Tits' alternative). It follows that $\langle g\rangle$ is undistorted.

If otherwise every eigenvalue of $g$ is a root of unity, after passing to a power we can suppose that $g$ is unipotent. Then since $g$ has infinite order, we deduce that the characteristic of $K$ is 0 and $g$ is not semisimple. There exists an field embedding of $K$ into $\mathbf{C}$, so by the lemma the growth of $g^n$ in $\mathrm{GL}_d(\mathbf{C})$ is logarithmic. So $g$ is at most exponentially distorted.

Note that the proof also shows that in a linear group in positive characteristic, every cyclic subgroup is undistorted.

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PS: All this is very basic and comes much before the Lubotzky-Mozes-Raghunathan theorems, in which a nontrivial result is to show that under suitable hypotheses, elements that are exponentially distorted in the ambient group actually remain exponentially distorted in the lattice. –  YCor Sep 27 '13 at 15:31
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