5
$\begingroup$

I am interested in infinite order elements $A\in SL(3, {\mathbb Z})$ whose spectra are not contained in ${\mathbb R}$ (i.e. such $A$ has two distinct complex-conjugate eigenvalues which are not roots of unity); I will refer to them as NRS matrices.

Question. Is there a pair of commuting NRS matrices $A, B\in SL(3, {\mathbb Z})$ whose product is again an NRS matrix, such that $A, B$ generate a non-cyclic subgroup of $SL(3, {\mathbb Z})$?

As the last resort, one can look for such matrices by computer-search, but I would prefer to avoid doing this.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

There are no such pairs.

Let $\lambda_1$, $\lambda_2$, $\lambda_3$ be the eigenvalues of $A$, and $\mu_1,\mu_2,\mu_3$ be those of $B$ (with $\lambda_1,\mu_1\in\mathbb R$). Since the eigenvalues are distinct, $A$ and $B$ are diagonalizable; since they commute, they are simultaneously diagonalizable, i.e., an eigenbasis for $A$ is also that for $B$. We assume that in a common eigenbasis, $\lambda_i$ corresponds to $\mu_i$ (clearly, $\lambda_1$ corresponds to $\mu_1$ --- because only these eigenvectors may be chosen real).

Now let $e_2$ be an eigenvector for $A$ corresponding to $\lambda_2$. The linear system defining it has coefficients in $K=\mathbb Q[\lambda_2]$, so we may assume that the elements of $e_2$ are also in $K$. Writing down the condition that $e_2$ is an eigenvector of $B$ (with eigenvalue $\mu_2$) we get that $\mu_2\in K$. Thus $\mathbb Q[\mu_2]=K=\mathbb Q[\lambda_2]$. Moreover, $\lambda_2$ and $\mu_2$ belong to the units group of $K$ --- which, as is known, is cyclic.

Thus $\lambda_2^k=\mu_2^\ell$ for some $k$ and $\ell$, which yields also $\lambda_i^k=\mu_i^\ell$, and hence $A^k=B^\ell$. Thus the subgroup generated by $A$ and $B$ is cyclic.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thank you, Ilya! I was hoping for a positive answer, but alas... $\endgroup$
    – Misha
    Commented Sep 24, 2017 at 5:31
  • $\begingroup$ Perhaps a stupid question, but how do you go from $\lambda_2^k = \mu_2^l$ to $\lambda_1^k = \mu_1^l$ in the last line? Can we somehow assume wlog that $\lambda_1 = \mu_1 = 1?$ $\endgroup$
    – Vincent
    Commented Oct 13, 2017 at 7:31
  • $\begingroup$ @Vincent: We know that $\lambda_3=\overline{\lambda_2}$, so the relation holds for it. Next, $\lambda_1\lambda_2\lambda_3=1=\mu_1\mu_2\mu_3$, whence the relation for $\lambda_1$. (I could use the basic Galois theory technique, but I needed the fixed correspondence between the roots.) $\endgroup$
    – Ilya Bogdanov
    Commented Oct 13, 2017 at 7:55
  • $\begingroup$ @Vincent: IAs for your second question --- no, I do not know how to assume that. $\endgroup$
    – Ilya Bogdanov
    Commented Oct 13, 2017 at 7:56
  • $\begingroup$ @IlyaBogdanov Ah thank you! I missed the $\det(A) = \det(B) = 1$ condition. It all makes sense now. $\endgroup$
    – Vincent
    Commented Oct 13, 2017 at 8:41
0
$\begingroup$

The eigenvalues $\lambda_1,\lambda_2,\lambda_3$ of a matrix $M\in SL_3(\mathbb{Z})$ satisfy:

$\lambda_1\lambda_2\lambda_3=\det M=1$.

Let $f(x)\in\mathbb{Z}[x]$ be the characteristic polynomial of $M$, then $\lambda_i$ is a root of $f(x)$.

Let $f(x)=x^3+tx-1$ for some $t\in\mathbb{Z}$, then the discriminant $d_{f}=-4*t^3 - 27$. Now if you pick $t>0$, then $d_f<0$, then $f(x)$ has two complex root and one real root.

To find a choice for $M$, let $K$ be the number field defined by $K(\alpha)=\mathbb{Q}[x]/(f(x))$.

Then the ring of integer $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $3$. Now multiplication by $\alpha$ induces an Endomorphism $\phi(\alpha)\in End_{\mathbb{Z}}(\mathcal{O}_K)$. Since $\mathcal{O}_K$ is free $\mathbb{Z}$ module of rank $3$, we have $\phi(\alpha)\in GL_{3}(\mathbb{Z})$ since $\alpha$ is invertible in $\mathcal{O}_K$. Choose the sign of $\alpha$ such that $\det(\phi(\alpha))=1$, then we have $\phi(\alpha)\in SL_3(\mathbb{Z})$. Note that the characteristic polynomial of $\phi(\alpha)$ has $\alpha$ as a root and hence an irreducible polynomial. Then the conjugate of $\alpha$ also satisfies it. Now you can take $M=\phi(\alpha)$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ OK, you found one NRS matrix; how do you find three NRS matrices $A, B, C$ whose product is $1$ and which generate a subgroup isomorphic to ${\mathbb Z}^2$? $\endgroup$
    – Misha
    Commented Sep 23, 2017 at 16:48
  • $\begingroup$ You can use the fact that $\lambda_1\lambda_2\lambda_3=1$ to get $\phi(\lambda_1)\phi(\lambda_2)\phi(\lambda_3)$ is the identities map. Since you also have $\lambda_1\lambda_3\lambda_2=1$, you can switch order. Then the three matrices $\phi(\lambda_i)$ work. $\endgroup$
    – Dianbin Bao
    Commented Sep 23, 2017 at 16:54
  • $\begingroup$ $\phi(\lambda_1)\phi(\lambda_2)=\phi(\lambda_2)\phi(\lambda_1)$ since $\lambda_1\lambda_2=\lambda_2\lambda_1$. The three matrices commute. $\endgroup$
    – Dianbin Bao
    Commented Sep 23, 2017 at 17:08
  • 1
    $\begingroup$ Notice that $\mathbb Q[\lambda_1]\neq\mathbb Q[\lambda_2]$, so you cannot take their images under $\phi$ simultaneously --- or, if you define it exactly as you do, you will have $\phi(\lambda_1)=\phi(\lambda_2)$. $\endgroup$
    – Ilya Bogdanov
    Commented Sep 23, 2017 at 18:56
  • $\begingroup$ Dear Bogdanov, thanks for your comment, I see the mistake in my argument. $\endgroup$
    – Dianbin Bao
    Commented Sep 24, 2017 at 3:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .