Commuting matrices in GL(n,Z)

2011-02-16T14:02:14Z

Suppose $M$ is a "hyperbolic" matrix in $GL(n,\mathbb Z)$, i.e., that its characteristic polynomial $p$ is irreducible over $\mathbb Z$ and has no roots of modulus 1.

Is there a closed description of the set of elements of $GL(n,\mathbb Z)$ which commute with $M$?

I have a vague recollection that it is somewhat similar to the Dirichlet theorem on the units of an algebraic field, but it is really vague so a reference would be appreciated.

The case I'm most interested in is when $p$ has only one root of modulus greater than 1. Can $M$ commute with another matrix $M'$ with the same property (and $M, M'$ not being powers of the same matrix in $GL(n,\mathbb Z)$)?

Answer by Pedro Martins Rodrigues for Commuting matrices in GL(n,Z)

2011-02-16T15:40:35Z

Suppose $M$ has a simple eigenvalue $\lambda$ with associated eigenvector $v$. Then $M'$ has also eigenvector $v$ associated to an eigenvalue $\rho$. This means that $\lambda$ and $\rho$ are units in the same ring $\mathbb{Z}[\lambda]$. The converse is clear, so a closed description of the $M' \in GL(n,\mathbb{Z})$ commuting with $M$ would be: The multiplicative group of units in the ring $\mathbb{Z}[\lambda]$. If this group is not cyclic then the answer to the second question is yes.

Answer by Igor Rivin for Commuting matrices in GL(n,Z)

2011-02-16T16:28:59Z

If you do a google search on "centralizer of semisimple element in linear group" you will learn more about this question than you ever wanted. (see, eg, Fleischmann/Janiszczak/Knorr).

Answer by anton for Commuting matrices in GL(n,Z)

2011-02-16T16:32:45Z

$M$ being hyperbolic means that it sits inside a torus, i.e., an algebraic group isomorphic to ${\mathbb G}_m^n$. Then its centralizer in the algebra of $n\times n$ matrices is semisimple, i.e., a product of smaller matrix algebras over skew fields. The $\mathbb Z$-valued points define an order in that algebra and the matrices in $GL_n({\mathbb Z})$ commuting with $M$ form the unit group of that order.

Proofs of these statements can be put together with the material in the books "Associative Algebras" by Pierce and "Maximal Orders" by Reiner.

Answer by David Speyer for Commuting matrices in GL(n,Z)

2011-02-16T17:37:21Z

This has very little to do with being hyperbolic; the key point is that the characteristic polynomial is irreducible. It is convenient to "close in" on $\mathbb{Z}$ be thinking about easier rings.

Let $M$ be a matrix in $GL(n, \mathbb{C})$ with distinct eigenvalues. Then I claim that the set of $\n \times n$ matrices which commute with $M$ is $\mathrm{Span}_{\mathbb{C}} (\mathrm{Id}, M, M^2, \cdots, M^{n-1})$.

Proof: This statement is clearly invariant under change of basis, so we may assume that $M$ is diagonal, with diagonal entries $\lambda_i$. Then $AM=MA$ if and only if $\lambda_i A_{ij} = \lambda_j A_{ij}$. Since the $\lamdba$'s are distinct, this forces $A$ to be diagonal. Since the Vandermonde matrix is invertible, the span of the first $n$ powers of $M$ is precisely the diagonal matrices.

Let $M$ be a matrix in $GL(n, \mathbb{Q})$ whose characteristic polynomial has distinct roots. Then the set of $n \times n$ rational matrices which commute with $M$ is $\mathrm{Span}_{\mathbb{Q}} (\mathrm{Id}, M, M^2, \cdots, M^{n-1})$.

Proof: Let $A$ be "the set of matrices which commute with $M$" and let $B$ be "the linear span of the first $n$ powers of $M$". Both of these are rational subvector spaces of $\mathrm{Mat}_{n \times n}(\mathbb{Q})$ . The previous section shows that $A \otimes \mathbb{C}$ and $B \otimes \mathbb{C}$ are the same subspace of $\mathrm{Mat}_{n \times n}(\mathbb{C})$ . A standard lemma is that, if $U$ and $V$ are both $K$-subspaces of a $K$ vector space $W$, and $U \otimes L = V \otimes L$ as subspaces of $W \otimes L$, then $U=W$.

Now, let $p$ be the characteristic polynomial, and assume furthermore that $p$ is irreducible. The $\mathbb{Q}$-span of the powers of $M$ is isomrophic, as a ring, to $\mathbb{Q}[t]/p(t)$. Since $p$ is irreducible, this is some number field, call it $K$. So the set of $\mathbb{Q}$-matrices which commute with $M$ is isomorphic to a number field. Since every element in a field, other than 0, is invertible, we get that the matrices in $GL(n, \mathbb{Q})$ which commute with $M$ are isomorphic to $K^*$ .

Let $M$ be a matrix in $GL(n, \mathbb{Z})$ whose characteristic polynomial is irreducible. Let $K$ be the field of $\mathbb{Q}$-matrices which commute with $M$, as discussed above. The set of the matrices whose entries are in $\mathbb{Z}$ forms a lattice $\mathcal{O}$, of rank $n$, in $K$, which is also a subring. Such a subring of a number field is called an order; I don't think there is much to say about this order which is not true of general orders.

Finally, you want to understand those matrices of $\mathcal{O}$ which are in $GL(n, \mathbb{Z})$, meaning that their inverses are also in $\mathcal{O}$. This is the unit group of $\mathcal{O}$. And, indeed, Dirichlet's unit theorem applies to orders: if $K$ has $r$ real places and $s$ complex places, then $\mathcal{O}^*$ is a finite group times $\mathbb{Z}^{r+s-1}$.

Finally, you want to know whether or not you can have $M$ hyperbolic, $N$ hyperbolic commuting with $M$, but $N$ not a power of $M$. The answer is YES. I'll first give a theoretical proof, and then sketch an actual computation. Let the eigenvalues of $M$ be $\lambda_1$, ..., $\lambda_n$, with $|\lambda_1|>1$. Let $N=f(\lambda_i)$, with $f$ a polynomial with rational coefficients. Note that the $\lambda$'s are the Galois orbit of $\lambda_1$. For any Galois automorphism $\sigma$, we have $f(\sigma(\lambda_1)) = \sigma(f(\lambda_1))$. So the eigenvalues of $N$ are the Gaois orbit of $f(\lambda_1)$. In short, your question is equivalent to the following:

Let $\mathcal{O}$ be an order in a number field. Suppose that $\lambda$ and $\mu$ are units such that $|\lambda|$ and $|\mu|>1$, but their Galois conjugates are less than $1$. Can this hapen without $\mu$ not a power of $\lambda$?

It certainly can. I'll give the conceptual proof, then sketch a computation. If you recall the standard proof of Diricihlet's unit theorem, it goes as follows: Map $\mathcal{O}^*$ to $\mathbb{R}^{r+s}$ by $u \mapsto (\log |u|, \log |\sigma_2(u)|, \cdots, \log |\sigma_{s+r}(u)| )$, where the inputs to the logs are the Galois orbit of $u$. Clearly, the image lands in the hyperplane where the coordinates sum to $0$. One proves that the image is a discrete lattice, of rank $r+s-1$, in this hyperplane.

In particular, we are interested in units where $\log |u|>0$ but where all the other coordinates are negative. This is the intersection of our discrete lattice with a full dimensional cone; once $r+s-1>1$, this will be larger than the one dimensional sublattice of the powers of any single unit.

If you want an explicit example, lets take $K=\mathbb{Q}[\cos (2 \pi/7)]$. The Galois action permutes $(\cos (2 \pi/7), \cos (4 \pi/7), \cos (6 \pi/7))$ cyclically. (Note that $\cos (4 \pi/7) = 2 \cos^2 (2 \pi/7) -1$, so it is in the same field, and similarly for $\cos (8 \pi /7 ) = \cos (6 \pi /7 )$.

Set $u=(1-\cos(2 pi/7))/(1-\cos (4 \pi/7))$, $v=(1-\cos(4 pi/7))/(1-\cos (6 \pi/7))$ and $w=(1-\cos(4 pi/7))/(1-\cos (8 \pi/7))$. So the Galois group permutes $(u,v,w)$ cyclically and $u*v*w=1$. You can check that $u$, $v$ and $w$ obey the equation $$t^3 - 6 t^2 + 5t -1=0$$ so $u$, $v$ and $w$ are units. I think they generate the unit group; in any case, the have finite index in it. Also note that $|u|$ and $|v| < 1$, while $|w|>1$. So any matrix with eigenvalues $u$, $v$ and $w$ is hyperbolic.

Numerical experimentation reveals that $|u w^4|>1$, while $|v u^4|$ and $|w v^4|<1$.

So, write down explicit matrices for the action of $w$ and $u w^4$ on the ring of integers of $K$. Then these will be two hyperbolic matrices which commute, but where neither is a power of the other.

Commuting matrices in GL(n,Z) - MathOverflow

Commuting matrices in GL(n,Z)

Answer by Pedro Martins Rodrigues for Commuting matrices in GL(n,Z)

Answer by Igor Rivin for Commuting matrices in GL(n,Z)

Answer by anton for Commuting matrices in GL(n,Z)

Answer by David Speyer for Commuting matrices in GL(n,Z)