Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi everyone

Let $p$ be a prime number. I am interested to classify $\{ A\in {\rm GL}_{p-1}(\mathbb{Z}): {\rm ord}(A)=p \}$ up to conjugacy. One reason to consider this problem is its relation to class number of $\mathbb{Q}(\zeta_p)$, more precisely one can show $|\{ [A]\in {\rm GL}_{p-1}(\mathbb{Z}): {\rm ord}(A)=p \}|=h(\mathbb{Q}(\zeta_p))$ where $[A]$ is denoted for conjugacy class.

It seems $\{ [A]\in {\rm GL}_{p-1}(\mathbb{Z}): {\rm ord}(A)=p \}$ is more easier to study than class number of $\mathbb{Q}(\zeta_p)$. At least its seems is easier to show is finite without invoking Geometry of number which shows $h(\mathbb{Q}(\zeta_p))$ is finite. For example Borel and Harish-Chandra theorem ("Arithmetic subgroups of algebraic groups") shows that this set should be finite, (they proved much more general theorem which implies this, I have never read that paper so I don't know their proof for the general theorem. So they might reduce it to class number or some adelic version of class number).

Ok, now is time to ask a question. Is there any paper considering this problem which give us some finiteness theorem (beside Borel and Harish-Chandra paper)?

share|improve this question
Who is this Borel-Hirsh-chandra of whom you speak? –  Igor Rivin May 13 '11 at 22:08
@Igor Rivin: I typed it wrongly. I meant Armand Borel and Harish-chandra. –  M.B May 13 '11 at 22:14
@M.B.: It was Harish-Chandra with a capital "C". Right? There were also papers by Grunewald and Segal and by Sarkisyan, proving the Hasse local-global principle. –  Mark Sapir May 13 '11 at 22:55
I agree that the end result is arguably cleaner but I think that if you trace the prehistory of the Borel/Harish-Chandra result then it has its origin in one of the early proofs of finiteness of class number. –  Torsten Ekedahl May 14 '11 at 6:11
You might know all this already, but just in case: the eigenvalues of any matrix of order $p$ must be $p$th roots of unity. In addition one of them must be not equal to 1, or else the matrix doesn't have actually order $p$. If a matrix with integer entries has any primitive $p$th root of unity as an eigenvalue, then it has all primitive $p$th roots of unity as eigenvalues. Since the matrices we're discussing here have size $p-1$, that implies that each primitive $p$th root of unity is an eigenvalue of multiplicity 1. All such matrices are conjugate over the complex numbers, at least. –  Greg Martin May 15 '11 at 22:23
show 4 more comments

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.