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It is well known that on every $d$-dimensional torus there exists linear Anosov automorphisms.

My question is the following:

Given $k< d$ 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 $1$?

This can be phrased in terms of matrices with integer coeficients (please add the corresponding relevant tags) as:

Given $k< d$ 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 $\mathbb{Q}$?.

Some relevant related information can be found in this paper ( where some results of W. Duke, Z. Rudnick, P. Sarnak as well as of Nevo and Sarnak are refered to.

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up vote 2 down vote accepted

Given $k < d$, 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].

Then you can simply take the companion matrix of such a polynomial.

Or do you need an explicit construction?

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This seems to answer my question. However, sorry for my ignorance, but is it obvious that one can make such polynomial with integer entries and such that it yields a matrix with integer coeficients and determinant equal to 1? – rpotrie Mar 18 '12 at 17:45
1. I suggest you read about the Diriclet unit theorem. A good starting point is the article on Wikipedia, I guess: 2. Once you have a polynomial $p$ with the leading coefficient 1 and the constant term $\pm1$, you may construct the companion matrix from this polynomial ( whose determinant is also $\pm1$ and whose characteristic polynomial is exactly $p$. Hope this helps. – Nikita Sidorov Mar 18 '12 at 18:17
Thanks. I guess that I don't have problem with 2. But for 1. it seems that I need to understand a bit more why should the existence of units gives the desired integer polynomials with leading coeficient 1 and constant term $\pm 1$. I will accept this answer since it is my ignorance which does not allow me to fully understand the answer yet. – rpotrie Mar 18 '12 at 18:30
If you really need an explicit example for your research, I could ask around. I think I know a guy who might know this. – Nikita Sidorov Mar 18 '12 at 19:08
No, thanks. The question is for a talk I must give in a seminar about what is known on Anosov diffeomorphisms and this question naturally came up. With the references you give I already understand much more, and toghether with the Pisot number commented below I already can answer the first question (about existence of Anosov with any splitting, maybe not irreducible). – rpotrie Mar 18 '12 at 19:48

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 $\frac{n}{6}\le j \le\frac{n}{2}$.

If instead $k=d-1$, take any Pisot number. Edit (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' multinacci.

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Thanks, this is useful. I guess it should not be deleted, some key words are of importance, at least for me (in order to look into references is important to have those key words). – rpotrie Mar 18 '12 at 17:42
Denis, for $k = d -1$ one can take the companion matrix for $x^d-x^{d-1}-\dots-x-1$. This is a Pisot number (called the multinacci number), hence $k=d-1$. – Nikita Sidorov Mar 18 '12 at 18:24

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