For a lattice $\Lambda$ of rank $n$ in $\mathbb{Q}^{n\times n}$ whose non-zero elements are inversible matrices, let $$Z(s,\Lambda) = \sum_{a \in \Lambda^*} |\det(a)|^{-s}$$ I wonder if (and how to show) this Dirichlet series has a functional equation ?

The motivation is that $\zeta_K(s)$ the Dedekind zeta function of a number field is of the form $\zeta_K(s) =\frac{1}{|\mathcal{O}_K^\times|}\sum_{J \in C_K} N(J)^{s} Z(s,\Lambda_J)$ where $C_K$ is the ideal class group, $\mathcal{O}_K$ being isomorphic to such a lattice in $\mathbb{Q}^{n \times n}$.

Aurel commented that for most number fields $\sum_{a \in \mathcal{O}_K^*} |N(a)|^{-s}$ diverges (infinitely many units), so we have to look at $\sum_{a \in \mathcal{O}_K^*/\mathcal{O}_K^\times} |N(a)|^{-s}$, and on the lattice side, except when $\Lambda$ is (up to a constant) a subring of $\mathbb{Q}^{n\times n}$, it is not obvious how to translate it.

  • 1
    $\begingroup$ I assume $\Lambda^*$ means that you exclude all singular matrices. $\endgroup$ Jan 21, 2017 at 11:36
  • 1
    $\begingroup$ I have always thought that Dirichlet series satisfying functional equations must somehow be "motivic", in the very elementary sense that they should be connected to some arithmetico-geometric object having a certain notion of "duality" and that the functional equations reflects this duality. I would be very surprised to realise that an arbitrary $Z(s,\Lambda)$ as above verify such an equation. $\endgroup$ Jan 21, 2017 at 12:21
  • 4
    $\begingroup$ The first series you write does not converge in general since there could be infinitely many elements of $\Lambda$ having the same determinant. The equality you write about the Dedekind zeta function is incorrect for the same reason, unless there your field is $\mathbb{Q}$ or imaginary quadratic. $\endgroup$
    – Aurel
    Jan 21, 2017 at 14:40
  • 2
    $\begingroup$ @user1952009 Yes, take $\Lambda = \mathcal{O}_K$ where $K$ is a number field with infinite unit group (i.e. any number field that is not $\mathbb{Q}$ or imaginary quadratic). In such an $\mathcal{O}_K$ there are infinitely many elements of norm $1$, and therefore in the corresponding $\Lambda$ there are infinitely many elements of determinant $1$. $\endgroup$
    – Aurel
    Jan 21, 2017 at 15:48
  • 2
    $\begingroup$ @user1952009 watch out that unless $\Lambda$ is a subring, the set of elements with determinant $1$ might not be a subgroup. $\endgroup$
    – Aurel
    Jan 21, 2017 at 15:51


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy