## Spectrum of a generic integral matrix.

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?

• 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.
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Unless all eigenvalues are collinear, there must be a circle containing 3 of them. Or does 2) mean that no more than two eigenvalues have the same modulus? – Gjergji Zaimi Apr 18 2010 at 22:46
Yes circle centered at origin, you are right, 2) just means that no more than two eigenvalues have the same absolute value. – Andrey Gogolev Apr 18 2010 at 23:07

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.

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Although this does not really affect the answer, there is one more component: one of the roots is real and the other two are complex conjugate. – damiano Apr 20 2010 at 7:22
Good point. So one needs a second component, parameterized by $(t-r)(t^2+at+r^2)(t^{n-3}+\ldots)$ or, from the second perspective, one needs to consider $\prod (r_i r_j - r_k^2)$. – David Speyer Apr 20 2010 at 11:51
Thank you very much! This is really nice, especially the "conceptual proof". I understand that the estimate $O(N^{n^2-1}$ should follow from the fact that "bad" matrices lie on an algebraic hypersurface. This is because all the "folding" occurs in a compact core, outside of which the hypersurface is sufficiently "straight". But is it really so obvious? – Andrey Gogolev Apr 20 2010 at 17:08