Let $D$ be a central division (or maybe just simple) algebra over $\mathbb{Q}$. Let $\mathcal{O} \subset \mathcal{O}_m$ be an order inside a fixed maximal order and denote by $\mathcal{O}^1$ its group of units with norm $1$. If $[\mathcal{O}_m : \mathcal{O}] = N$, then can we say in general that $[\mathcal{O}^1_m : \mathcal{O}^1]$ is also of size $N$? More precisely, can we show that $N^{1-\epsilon} \leq [\mathcal{O}^1_m : \mathcal{O}^1] \leq N^{1+\epsilon}$ for (arbitrarily) small $\epsilon$?

This seems to be the case for $D = M_2(\mathbb{Q})$ (e.g. for Hecke congruence subgroups) and also more generally for quaternion algebras by the volume formulas (see e.g. Voight, relating the volume of the corresponding locally symetric space to the discriminant of the order). Are there more general volume formulas for all degrees? Is there a softer way to compute the "order of magnitude" of the index?

Edit: Using approximation theorems, it should be enough to show this locally. So we could assume to be working over the $p$-adic numbers. Are there additional classification results we could use now?


1 Answer 1


First, you need to avoid definite quaternion algebras over $\mathbb{Q}$: in this case, the unit groups are finite, so the index cannot grow with $N$.

With that out of the way, your algebra $D$ satisfies strong approximation (chapter 28 of my book, I'm sorry I only discuss the quaternion case, but see Remark 28.6.11; I also recommend the book "Units in skew fields" by Ernst Kleinert)! This means that the global index $[\mathcal{O}_m^1 : \mathcal{O}^1]$ is equal to the product of the local indices $[(\mathcal{O}_m)_p^1 : \mathcal{O}_p^1]$ over (finitely many) primes $p$, so it is enough to understand the local situation.

In the local case, an argument like the one in Lemma 26.6.7 applies. It starts out the same, and we are left to determine the ratio $[\mathcal{O}_m^1 : 1+p\mathcal{O}_m]/[\mathcal{O}^1 : 1+p\mathcal{O}]=\#(\mathcal{O}_m/p\mathcal{O}_m)^1/\#(\mathcal{O}/p\mathcal{O})^1$. Each term (numerator and denominator) is the calculation of the units in an $\mathbb{F}_p$-algebra, and already in the case of quaternions you can see that there are cases needed if we want an exact formula.

Actually, I think there's a more uniform thing to reduce to the semisimple case, using the Jacobson radical $J=\mathrm{rad} \mathcal{O}$ (20.4): we have a filtration $\mathcal{O} \supseteq J \supseteq J^2 \supseteq \dots \supseteq p\mathcal{O} \supseteq J^r$ for some $r \geq 1$ which gives $\mathcal{O}^\times \geq 1+J \geq 1+J^2 \geq \dots \geq 1+p\mathcal{O} \geq 1+J^r$ and $(1+J^i)/(1+J^{i+1}) \simeq J^i/J^{i+1}$ for $i \geq 1$. So you only need to count $(\mathcal{O}/J)^1$. Now $\mathcal{O}/J$ is semisimple (that's the job of the Jacobson radical), and we're over $\mathbb{F}_p$, so $\mathcal{O}/J$ is the product of matrix rings over finite fields, so the exact count can be given in terms of the number of factors, their degree, and the cardinality of the residue field.

Combining these should give you an explicit formula (with cases) or an estimate, like what you asked for above. (I didn't work out the details, hoping that you would be motivated to do so!) This kind of calculation shows up in the computation of zeta functions of orders in division algebras, but I couldn't quickly find a reference...

  • $\begingroup$ Thank you very much for the great answer! I am still trying to work out the details. One thing that I noticed is that, as far as I understand it, $1 + p \mathcal{O}$ need not lie in the norm 1 units. I computed that $[\mathcal{O}_m^\times : \mathcal{O}^\times] = [\mathcal{O}_m^1 : \mathcal{O}^1] \cdot [N(\mathcal{O}_m^\times) : N(\mathcal{O}^\times)]$. How can we also control the size of the quotient of norms? Have I made a mistake somewhere? $\endgroup$
    – Radu T
    Nov 1, 2021 at 13:11
  • $\begingroup$ Also, I have only found references for $J^r \subset p\mathcal{O}$ but you write $J^r = p\mathcal{O}$. Why can we assume that? $\endgroup$
    – Radu T
    Nov 1, 2021 at 16:22
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    $\begingroup$ You're quite right: the filtration is only on $\mathcal{O}^\times$. If you want norm 1 units, you need to consider the ratio $[\mathrm{nrd}(\mathcal{O}_m^\times):\mathrm{nrd}(\mathcal{O}^\times)]=[\mathbb{Z}_p^\times:\mathrm{nrd}(\mathcal{O}^\times)]$ (the reduced norm is surjective on the units of a maximal order, Theorem (33.4) of Reiner's "Maximal Orders"). This factor can be a bit delicate, but it is bounded: we have $\mathbb{Z}_p \subseteq \mathcal{O}$ so $\mathbb{Z}_p^{\times n} \leq \mathcal{O}^\times$ if the central simple algebra has degree $n$, so the index is bounded by $n$. $\endgroup$ Nov 2, 2021 at 14:44
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    $\begingroup$ To your second question about $J^r = p\mathcal{O}$: I think for quaternion orders you can ensure equality? But you shouldn't need this, there is still an isomorphism $(1+p\mathcal{O})/(1+J^r) \simeq p\mathcal{O}/J^r$ so everything past the first part of the filtration is computed by the index. $\endgroup$ Nov 2, 2021 at 14:59
  • $\begingroup$ Great! Thank you! I'll just add here to complete the answer to my question: using the argument above, the approximation in the question reduces to showing that $(\mathcal{O}/J)^1$ is of the same order of magnitude as $\mathcal{O}/J$, which is true by checking it on matrix algebras. $\endgroup$
    – Radu T
    Nov 2, 2021 at 15:28

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