Growth of an integer vector under the action of a matrix in $GL_n(\mathbb{Z})$ I have some questions regarding the dynamics of elements of $GL_n(\mathbb{Z})$ acting on $\mathbb{Z}^n$. In particular, given an invertible integer matrix $M \in GL_n(\mathbb{Z})$, and given an integer column vector $v \in \mathbb{Z}^n$ which is not equal to the zero vector, I want to know about the asymptotics of the sequence of vector norms $| M^j v|$ for positive integers $j$, with particular emphasis on boundedness and polynomial growth.
One can deduce some things about this using the Jordan decomposition for the action of $M$ on $\mathbb{C}^n$. For example, here is a nice, precise characterization of boundedness. The following are equivalent:


*

*$| M^j v|$ is bounded

*The vector $v$, regarded as a complex vector, is contained in the direct sum of the eigenspaces corresponding to eigenvalues of $M$ that are roots of unity. 


What one uses to deduce $1 \implies 2$ is a discreteness argument: $\mathbb{Z}^n$ has only finitely many elements with a given bound, and hence boundedness of $|M^j v|$ implies the existence of $j$ such that $M^j v = v$.
I would like a similarly precise characterization of polynomial growth. The best I know at the moment is that the following are equivalent:


*$| M^j v|$ is bounded above by a polynomial function of $j$, for positive integers $j$.

*$v$ is contained in the direct sum of generalized eigenspaces corresponding to eigenvalues $\lambda \in \mathbb{C}$ such that $\lambda \le 1$, i.e. to those $\lambda$ that lie on or inside the unit circle of $\mathbb{C}$.


But statement 4 is somewhat loose, in that it allows some seeming possibilities that turn out to be impossible using a discreteness argument. For example, it is impossible that $v$ be contained in the direct sum of the generalized eigenspaces corresponding to eigenvalues that lie inside the unit circle, for in that case one can show that the sequence $M^j v$ converges to the zero vector, an impossibility for $M \in GL_n(\mathbb{Z})$ and $v \in \mathbb{Z}^n$.
What stronger statements are there which characterize polynomial growth of $|M^j v|$? For example, given $M \in GL_n(\mathbb{Z})$ and $v \in \mathbb{Z}^n$, are Statements 3 and 4 above equivalent to the following nice, strong statement?


*$v$ is contained in the direct sum of generalized eigenspaces corresponding to eigenvalues of $M$ that are roots of unity.

 A: The linear algebra needed to answer this question is spelled out in my paper Dynamical Properties of Quasihyperbolic Toral Automorphisms (Ergodic Th. & Dynam. Sys. 2 (1982), 49-68), in particular Section 2.
The observation that a monic polynomial with integer coefficients all of whose roots lie on the unit circle must be cyclotomic goes back to Kronecker: Zwei Sätze über Gleichungen mit ganzzahligen Coefficienten, J. reine angew. Math. 53 (1857), 173-175. David Boyd proved a generalization of this for polynomials of several variables, which says that a polynomial whose logarithmic Mahler measure vanishes (for one variable this is equivalent to having all roots on the unity circle) must be a product of generalized cyclotomic polynomials (Kronecker's Theorem and Lehmer's Problem for Polynomials in Several Variables, J. Number Theory 13 (1981), 116-121.
There are some much deeper results along the following lines. Let $E$ be the direct sum of the generalized eigenspaces of $M$ corresponding to eigenvalues on or inside the unit circle, and assume there are no eigenvalues that are roots of unity. Katznelson showed that there is a constant $C>0$ such that if $v$ is a nonzero integer vector then its Euclidean distance to $E$ is greater than $C\|v\|^{-\dim E}$ (Ergodic Automophisms of $T^n$ are Bernoulli, Israel J. Math. 10 (1971), 186-195). This is the key diophantine component in his proof that all ergodic toral automorphisms are measurably isomorphic to Bernoulli shifts. Subsequently this was extended by me and others to show that all ergodic group automorphisms are Bernoulli.
A: Showing that 4 implies 5 reduces to showing:

Let $A\in\mathrm{GL}_n(\mathbf{Z})$. Let $V_-,V_{\mathrm{ru}}\subset\mathbf{C}^n$ be the sum of characteristic subspaces of $A$ relative to eigenvalues of modulus $\le 1$ (resp., to eigenvalues that are roots of unity). Then $V_-\cap\mathbf{Q}^n=V_{\mathrm{ru}}\cap\mathbf{Q}^n$.

Separating roots of unity and other eigenvalues, this reduces to showing:

Let $A\in\mathrm{GL}_m(\mathbf{Z})$, assume that $A$ has no root of unity as eigenvalue. Then $V_-\cap\mathbf{Q}^m=\{0\}$.

Suppose we have a counterexample. Passing to either a smaller invariant subspace or to a quotient allows to assume that the action of $A$ on $\mathbf{Q}^m$ is irreducible. Then $V_-\cap\mathbf{Q}^m$ is clearly a invariant subspace and is nonzero. So it is all of $\mathbf{Q}^m$. Hence $A$ has all its eigenvalues of modulus $\le 1$. Since the product of its eigenvalues, given by the determinant, is $\pm 1$, all its eigenvalues have modulus 1 (and are not roots of unity). This cannot hold, by standard "geometry of numbers" argument à la Minkowski.

Here's the simple argument for the last point:

Let $A\in\mathrm{GL}_m(\mathbf{Z})$ act irreducibly on $\mathbf{Q}^m$, with only complex eigenvalues of modulus 1. Then $A$ has finite order.

Proof:The irreducibility assumption implies that the minimal polynomial of $A$ is irreducible, hence $A$ is $\mathbf{C}$-diagonalizable. Since $A$ has only eigenvalues of modulus 1, it follows that the subgroup $\langle A\rangle$ is bounded in $M_m(\mathbf{C})$. Since it lies in $M_m(\mathbf{Z})$, it is therefore finite.
