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:


1. $| M^j v|$ is bounded
2. 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:


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


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



Growth of an integer vector under the action of a matrix in $GL_n(\mathbb{Z})$
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.




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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.


