Spectrum of a generic integral matrix.

2010-04-18T20:32:25Z

My collaborators and I are studying certain rigidity properties of hyperbolic toral automorphisms.

These are given by integral matrices A with determinant 1 and without eigenvalues on the unit circle.

We obtain a result under two additional assumptions

1) Characteristic polynomial of the matrix A is irreducible

2) Every circle contains no more than two eigenvalues of A (i.e. no more than two eigenvalues have the same absolute values)

We feel that the second assumption holds for a "generic" matrix. Is it true?

To be more precise, consider the set X of integral hyperbolic matrices which have determinant 1 and irreducible characteristic polynomial. What are the possible ways to speak of a generic matrix from X? Does assumption 2) hold for generic matrices?

Comments:

Assumption 1) doesn't bother us as it is a necessary assumption.
Probably it is easier to answer the question when X is the set off all integral matrices. In this case we need to know that hyperbolicity is generic, 2) is generic and how generic is irreducibility.

Answer by David Speyer for Spectrum of a generic integral matrix.

2010-04-19T22:53:56Z

Yes, a generic integer matrix has no more than two eigenvalues of the same norm. More precisely, I will show that matrices with more than two eigenvalues of the same norm lie on a algebraic hypersurface in $\mathrm{Mat}_{n \times n}(\mathbb{R})$. Hence, the number of such matrices with integer entries of size $\leq N$ is $O(N^{n^2-1})$.

Let $P$ be the vector space of monic, degree $n$ real polynomials. Since the map "characteristic polynomial", from $\mathrm{Mat}_{n \times n}(\mathbb{R})$ to $P$ is a surjective polynomial map, the preimage of any algebraic hypersurface is algebraic. Thus, it is enough to show that, in $P$, the polynomials with more than two roots of the same norm lie on a hypersurface. Here are two proofs, one conceptual and one constructive.

Conceptual: Map $\mathbb{R}^3 \times \mathbb{R}^{n-4} \to P$ by $$\phi: (a,b,r) \times (c_1, c_2, \ldots, c_{n-4}) \mapsto (t^2 + at +r)(t^2 + bt +r) (t^{n-4} + c_1 t^{n-5} + \cdots + c_{n-4}).$$

The polynomials of interest lie in the image of $\phi$. Since the domain of $\phi$ has dimension $n-1$, the Zariski closure of this image must have dimension $\leq n-1$, and thus must lie in a hyperplane.

Constructive: Let $r_1$, $r_2$, ..., $r_n$ be the roots of $f$. Let $$F := \prod_{i,j,k,l \ \mbox{distinct}} (r_i r_j - r_k r_l).$$ Note that $F$ is zero for any polynomial in $\mathbb{R}[t]$ with three roots of the same norm. Since $F$ is symmetric, it can be written as a polynomial in the coefficients of $f$. This gives a nontrivial polynomial condition which is obeyed by those $f$ which have roots of the sort which interest you.

Spectrum of a generic integral matrix. - MathOverflow

Spectrum of a generic integral matrix.

Answer by David Speyer for Spectrum of a generic integral matrix.