# units in distinct division algebras over number fields---are they definitely not isomorphic as abstract groups?

This is really an irrelevant question in the sense that the answer isn't remotely "logically crucial for the Langlands programme" or whatever---it's just something that occurred to me when writing some lectures on the subject.

Let $F$ be a number field and let $D$ and $D'$ be division algebras over $F$ with centre $F$. Let me put myself in the following simplified situation: assume $D$ and $D'$ both have dimension $p^2$ over $F$ (so are forms of $M_p(F)$) with $p$ an odd prime, and let me also assume that $D$ and $D'$ are both ramified at precisely the same finite places of $F$ (and at no infinite places of $F$) but that $D$ and $D'$ are not isomorphic.

### The question: Are $D^\times$ and $(D')^\times$ sometimes non-isomorphic as groups?

Let me put this slightly strange question into context. Experts will already know what is coming: $D^\times$ and $(D')^\times$ are group-theoretically similar in certain ways; this is what I'm about to explain. This group-theoretic similarity would be trivially explained if the groups were isomorphic! But I think the whole point should be that the groups aren't isomorphic, although I don't know an instance where I can prove this.

OK so here are the details. Given $D$ and $D'$ as above, there is a link between automorphic forms on $D^\times$ and $(D')^\times$. When proving this link using the trace formula, one has to check that two complicated formulae for traces, one associated to $D$ and one to $D'$, coincide. The formulae are of the form "sum over conjugacy classes of certain orbital integrals". The strategy I understand to check these sums coincide is first to use general theory of central simple algebras to explicitly list the conjugacy classes in the groups $D^\times$ and $(D')^\times$, and then to write down an explicit natural bijection between them, and then to check that things add up in the sense that each term on the $D$ side is equal to the corresponding term on the $D'$ side, and then to deduce that certain traces are equal, and then you follow your nose to the answer (modulo a technicality that has to be dealt with using a Plancherel measure argument but which isn't relevant here).

Crucial in this strategy is the identification of the conjugacy classes in $D^\times$ with the conjugacy classes in $(D')^\times$. The way this works in this situation is that a conjugacy class in $D^\times$ is either an element of $F^\times$ (the central classes) or contains some $e\not\in F$; in this case $E:=F(e)$ is a field extension of $F$ of degree $p$ which splits $D$, and the fields that split $D$ of this form are precisely the degree $p$ extensions of $F$ which have one prime above $v$ for all $v\in S$ our bad set. The miracle that has occurred here is that this argument (when fleshed out) shows that the conjugacy classes in $D^\times$ are parametrised by a set that depends only on $F$, $p$ and $S$, and in particular this set is the same for $D$ and $D'$, which have the same degree and which ramify at the same primes.

I will present this argument in detail in class today, and my instinct is to stress that a miracle has occurred, because $D^\times$ and $(D')^\times$ are non-isomorphic groups whose conjugacy classes are completely naturally in bijection with one another. Of course if $D$ and $D'$ are isomorphic then then $D^\times$ and $(D')^\times$ will be isomorphic. Moreover, if $D'=D^{opp}$ then again $D^\times$ and $(D')^\times$ will be isomorphic as groups. In both cases it is hardly surprising that I have managed to canonically biject the conjugacy classes of $D^\times$ and $(D')^\times$! But in general I am still convinced that a miracle has occurred. However to really convince the audience of this I want to assert confidently that the groups $D^\times$ and $(D')^\times$ really are not always isomorphic. Is this definitely true??

{\bf EDIT}: Here's an even stronger question, which is even less likely to be true, and given that I'm hoping/guessing that such things aren't true, it seems worth formulating (the stronger it is, the easier it should be to falsify).

OK so same set-up: $D$ and $D'$ division algebras of dimension $p^2$, $p$ prime, and ramified at the same set of finite primes $S$, but $D$ and $D'$ not isomorphic. Let $G$ be the form of $PGL_p$ associated to $D^\times/Z(D^\times)$ and let $G'$ be the form corresponding to $D'$. If $\mathbf{A}^S$ denotes the adeles of $F$ away from $S$ then choosing isomorphisms $D_v=D'_v$ for all $v\not\in S$, sending $O_v$ to $O'_v$ for two chosen maximal orders $O$ and $O'$ in $D$ and $D'$ when $v$ is finite, gives us an induced isomorphism $G(\mathbf{A}^S)=G'(\mathbf{A}^S)$. Call this group $X$. Now $\Gamma:=D^\times/F^\times$ and $\Gamma':=(D')^\times/F^\times$ are two discrete subgroups of $X$ and using the trace formula one can prove that not only do the conjugacy classes of $D^\times$ and $(D')^\times$ match up, but (if my understanding is correct) the covolumes of $\Gamma$ and $\Gamma'$ are the same. This would be explained by the highly unlikely statement that $\Gamma$ and $\Gamma'$ were actually conjugate within $X$! This is perhaps even stronger than just being abstractly isomorphic, in this setting? One might also ask to give an explicit example where this cannot possibly be the case?

I mention this stronger statement because people might find it easier to use high-powered methods to refute it.

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"as abstract groups" inserted :-) I suspect that more generally if ram(D) contains ram(D') then there's an injection from the conj classes of D^* to the conj classes of (D')^* which in some sense is even more miraculous for me. – Kevin Buzzard Jun 16 '10 at 19:01
Not to downplay the miracle, but in the context of the Langlands lifting, we are really looking at $D$ and $D'$ together with all their completions $D_v$ and $D'_v,$ so we might as well include in the question compatible isomorphisms between their units. – Victor Protsak Jun 16 '10 at 22:32
One can certainly get some instances of nonisomorphism just by considering elements of finite order. For instance, if $B$ is a quaternion algebra over $\mathbb{Q}$, then $B^{\times}$ admits an element of order $4$ (resp. $6$) iff it admits $\mathbb{Q}(\sqrt{-1})$ (resp. $\mathbb{Q}(\sqrt{-3})$) as a splitting field. Using this, you can easily show that there are at least four distinct isomorphism classes of unit groups of quaternion algebras over $\mathbb{Q}$... – Pete L. Clark Jun 16 '10 at 22:50
Kevin, let $F/\mathbf{Q}$ be a quad. ext'n split at $p$, $D$ the deg-9 central div. alg. over $F$ ram. only at primes $P$ and $P'$ over $p$, with resp. local inv'ts 1/3 and 2/3. Let $g$ be nontriv. aut. of $F$, and $D'$ the twist $g^{\ast}(D)$ (deg.-9 central div. alg. over $F$ ram. only at $P$ and $P'$, with resp. local inv'ts 2/3 and 1/3). $D$ and $D'$ not isom., so their alg. unit groups not $F$-isom. But $D \simeq D'$ via $x \mapsto 1 \otimes x$ is ring isom. (not over $F$), so defines isom. of unit gps. Upshot: better to ask if $D'$ an ${\rm{Aut}}(F,S)$-twist of $D$ up to opposite! – BCnrd Jun 17 '10 at 2:23
Brian's example is a general feature: since the local algebra with invariant $i$ is the opposite of the local algebra with invariant $-i$, if the global algebra $D$ has invariants $i_v$ at the various places $v$, then $D^{op}$ has invariants $-i_v$ at the various places, so is not isomorphic (outside the quaternion algebra setting). But $x \mapsto x^{-1}$ gives an isomorphism between $D^{\times}$ and $(D^{op})^{\times}$. (This seems to have also come up in some form in some of the comments below.) – Emerton Jun 17 '10 at 3:59

This is close to the famous paper of Borel and Tits, Homomorphismes abstraits'' de groupes algébriques simples, Ann. of Math. (2) 97 (1973), 499 – 571 MR0316587 (47 #5134). Unfortunately, one cannot appply their strongest result (abstract homomorphism $\implies$ algebraic homomorphism composed with a field embedding, valid for general infinite fields) because $D$ is anisotropic. Nonetheless, in 9.13 (i) they prove that if $G$ is a semisimple algebraic group over a local field $k\ ({char } k=0)$ and $G'$ is a reductive algebraic group over a local field $k'$ and $G'$ has no nontrivial complex factor, then every abstract homomorphism $G(k) \to G'(k')$ is continuous. Consequently, if $D^*_v \simeq D'^*_v$ as abstract groups then $SL_1(D_v) \simeq SL_1(D'_v)$ as topological groups and I thought1 that this implied that $D_v \simeq D'_v$. If that holds for all $v$ then $D \simeq D'.$2

Footnotes
1 Wrongly, see Kevin's and BCnrd's examples in the comments.
2True but irrelevant. For Kevin's question, it would be sufficient to find a single pair of $D$ and $D'$ with abstractly non-isomorphic unit groups, and the conclusion is too strong in view of the previous comment.

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Victor, perhaps it would be good to clarify that the abstract hom. may also "imply" a field hom. mixed up with the algebraic hom. (e.g., the abstract hom. could just be a field aut. applied to rational points of a group defined over a subfield, so no nontrivial alg. aut. in sight). Also, it is unclear (to me) how to promote an isomorphism $D^{\times} \simeq {D'}^{\times}$ of rational points over the number field to one of rational points over local fields, so I don't see how Kevin could access your suggestion. (I had considered a related idea using the BT paper, but gave up due to this.) – BCnrd Jun 16 '10 at 13:07
I agree. I started optimistically with "I think that this follows from BT", then replaced it with "the corresponding local fact follows from BT", and finally with conservative "this is close to... unfortunately...". Does $D^{\times}_v \simeq D'^{\times}_v$ as topological groups imply $D_v\simeq D'_v$, at least? – Victor Protsak Jun 16 '10 at 13:29
The Lie algebra of $D_v^{\times}$ as an $F_v$-analytic group is the same as for the analogous algebraic group: direct product of 1-dimensional central part and 3-dimensional semisimple part that must be $\mathfrak{sl}_ 2$ since we're in characteristic 0. The same goes with $D'_v$. Hence, $D_v^{\times}$ and ${D'}_v^{\times}$ have $F_v$-isomorphic Lie algebras, so they're $F_v$-analytically isomorphic near the identity. So it seems a fine line to infer anything about the algebraic group from the analytic group (equivalently from the topological group, when $F_v = \mathbb{Q}_p$!). – Boyarsky Jun 16 '10 at 17:03
If D'=D^{opp} then I think D_v^* and (D'_v)^* are isomorphic as topological groups but D_v and D_v' may not be isomorphic as F_v-algebras. So I think that your "I think" might be too optimistic---possibly this is the only thing that can go wrong though... – Kevin Buzzard Jun 16 '10 at 19:04
Kevin, $g \mapsto g^{-1}$ is an algebraic group isomorphism! (Doesn't extend to an algebra isomorphism, so no contradiction.) I was mistaken about the 3-dimensional Lie algebras: I had in mind Jacobson-Morozov, but can't use it unless there's a nonzero nilpotent in the Lie algebra, and in the anisotropic case there isn't any such element (since 14.26 of Borel's book + Galois descent with intersections in char. 0 would produce a split smooth connected unipotent subgroup inside, which forces rank $> 0$ in any connected reductive group). Good news in the local case in the next comment! – Boyarsky Jun 16 '10 at 19:41

Second try. Victor points out that my original application of super-rigidity was faulty. This edit is to remove the false claim and leave only the lemma I believe. While I'm at it, I have added some stuff I don't understand.

Let $H$ and $G$ be algebraic groups of units of division algebras over $\mathbb Q$ (because super-rigidity is of limited use over bigger fields). If $f\colon H(\mathbb Q)\to G(\mathbb Q)$ is an isomorphism of abstract groups, then super-rigidity implies that for all $S$, $H(\mathbb Z[S^{-1}])$ and $G(\mathbb Z[S^{-1}])$ contain finite index subgroups identified by $f$. One can't ask for more, since the $S$-integral points of $H$ and $G$ are only well-defined up to finite index. These groups are insensitive to inclusion of the ramified places in $S$.

I don't know how much this buys us.

Perhaps one can use the congruence subgroup property to reconstruct the local points of the group? I believe that the CSP is conjectured but open for these groups (specifically, that there should be a finite congruence kernel, perhaps for large $S$). Worse, I'm not sure what it says, it particular whether it allows reconstruction of the compact factors. (If this does work, then the compact factors would be the profinite completion of the rational points and one would not need the first paragraph. Maybe this should have been a separate answer.)

If we can reconstruct the compact local factors from the ($S$-)integral points, and thus from the rational points, we can reconstruct the local invariants, up to a sign (the local version of this question, addressed by Victor and Boyarsky). This would not be enough to show that the two discrete groups of rational points are isomorphic as algebraic groups, but it would be enough to show that some examples of groups of rational points are not isomorphic. A division algebra with local invariants $\frac15$ and $\frac45$ has non-isomorphic factors to one with $\frac25$ and $\frac35$. But a division algebra with invariants $(\frac 15,\frac 25,\frac 35,\frac 45)$ is not isomorphic or opposed to one with invariants $(\frac 45,\frac 25,\frac 35,\frac 25)$, even though locally it is always isomorphic or opposed. Thus, such a method could not distinguish their unit groups.

Proof of the claim in the first paragraph that isomorphism of rational points must preserve integral points: Super-rigidity says that maps of lattices extend to maps of their ambient locally compact groups. It requires that the $S$-rank of the source is least $2$, where the $S$-rank is the sum over places in $S$ of the local ranks. Under Kevin's assumptions, we already have rank $2$ from the infinite place; in general we would need to use large $S$ to draw conclusions about small $S$. $H(\mathbb Q)$ is not finitely generated, but it is filtered by groups of $S$-integers, which are finitely generated. Thus an isomorphism $f\colon H(\mathbb Q)\cong G(\mathbb Q)$ yields a homomorphism $H(\mathbb Z)\to G(\mathbb Z[S^{-1}])$, for some $S$. By super-rigidity, $H(\mathbb Z)$ contains a finite index subgroup on which this extends to a (continuous) homomorphism $H(\mathbb R)\to G(\mathbb R\times \mathbb Q_p\times\ldots)$, which has to land in $G(\mathbb R)$. And similarly back, so we get an isomorphism $H(\mathbb R)\cong G(\mathbb R)$ which restricts to be $f$ on a finite index subgroup of $H(\mathbb Z)$. That is, $f$ preserves the integral points. Similarly for the $S$-integral points.

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I've had a hard time following: can you, please, state your claim precisely? Also, all rigidity theorems have some hypotheses (rank at least 2, Zariski dense, no relatively compact factors), how do you deal with that for anisotropic groups? – Victor Protsak Jun 17 '10 at 20:26
Oops...no compact factors is a problem. Super-rigidity doesn't tell us anything about compact factors. The compact factors are the ones with ramification, the ones we really want to know about, so I retract most of it. But I still claim that if $H(\mathbb Q)\cong G(\mathbb Q)$, then for all $S$, $H(\mathbb Z[S^{-1}])$ and $G(\mathbb Z[S^{-1}])$ contain finite index subgroups which are identified by the given map. I don't know if that buys us anything. Rank is at least 2 because we're split at $\mathbb R$, but in general, we could take $S$ large. – Ben Wieland Jun 18 '10 at 0:15