Eigenvalues for toral Anosov automorphisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:03:38Z http://mathoverflow.net/feeds/question/91544 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91544/eigenvalues-for-toral-anosov-automorphisms Eigenvalues for toral Anosov automorphisms rpotrie 2012-03-18T15:58:56Z 2012-03-19T15:42:37Z <p>It is well known that on every $d$-dimensional torus there exists linear Anosov automorphisms. </p> <p>My question is the following:</p> <p>Given $k&lt; d$ does there exists a linear Anosov automorphism of $\mathbb{T}^d$ with exactly $k$ eigenvalues smaller than $1$? If true (which I expect), does there exists an \emph{irreducible} linear Anosov automorphism of $\mathbb{T}^d$ with exactly $k$ eigenvalues smaller than $1$? </p> <p>This can be phrased in terms of matrices with integer coeficients (please add the corresponding relevant tags) as:</p> <p>Given $k&lt; d$ does there exists a matrix in $SL(d,\mathbb{Z})$ such that all eigenvalues have modulus different from $1$ and $k$ of them are of modulus smaller than $1$? What about if the characteristic polynomial is irreducible over $\mathbb{Q}$?. </p> <p>Some relevant related information can be found in this paper (http://arxiv.org/pdf/1009.2994v2.pdf) where some results of W. Duke, Z. Rudnick, P. Sarnak as well as of Nevo and Sarnak are refered to. </p> http://mathoverflow.net/questions/91544/eigenvalues-for-toral-anosov-automorphisms/91552#91552 Answer by Denis Serre for Eigenvalues for toral Anosov automorphisms Denis Serre 2012-03-18T17:12:28Z 2012-03-19T15:42:37Z <p>This is only a partial answer, which I shall delete if I find a better one. Every pair $(k,d)$ of the form $$d=\frac12\phi(n),\qquad k={\rm card}(\frac{n}{6}\le j \le\frac{n}{2},j\wedge n=1)$$ is OK: take the cyclotomic polynomial $\Phi_n$ and form the irreducible polynomial $P_n\in{\mathbb Z}[X]$ defined by $$\Phi_n(t)=t^{\frac{n}{2}}P_n\left(t+\frac1t\right).$$ The roots of $P_n$ are the numbers $2\cos\frac{2j\pi}{n}$ with $j\wedge n=1$, smaller than $1$ if and only if $\frac{n}{6}\le j \le\frac{n}{2}$.</p> <p>If instead $k=d-1$, take any Pisot number. <strong>Edit</strong> (after Nikita's comment below): One may take the companion matrix of $X^d-X^{d-1}-\cdots-X-1$. Its only root of modulus greter than $1$ is a Pisot number, also called a multinacci number. If $d=2$, this is just the golden ratio, at the basis of the Fibonacci sequence, hence the `word' <em>multinacci</em>.</p> http://mathoverflow.net/questions/91544/eigenvalues-for-toral-anosov-automorphisms/91554#91554 Answer by Nikita Sidorov for Eigenvalues for toral Anosov automorphisms Nikita Sidorov 2012-03-18T17:25:06Z 2012-03-18T17:25:06Z <p>Given $k &lt; d$, one can always construct a monic polynomial irreducible over $\mathbb Q$ with exactly $k$ roots less than 1 in modulus and $d-k$ roots greater than 1 in modulus. This follows from the general construction of algebraic units, namely, each group of units of an algebraic field contains a unit with a given $k$ -- see, e.g., [Borevich and Shafarevich]. </p> <p>Then you can simply take the companion matrix of such a polynomial. </p> <p>Or do you need an explicit construction?</p>