Timeline for units in distinct division algebras over number fields---are they definitely not isomorphic as abstract groups?
Current License: CC BY-SA 2.5
19 events
| when toggle format | what | by | license | comment | |
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| Feb 28, 2017 at 13:25 | answer | added | Venkataramana | timeline score: 6 | |
| Jun 17, 2010 at 20:03 | answer | added | Ben Wieland | timeline score: 0 | |
| Jun 17, 2010 at 14:21 | comment | added | moonface | BCnrd mentions superrigidity, and it seems likely it follows from that -- I don't see why the isomorphism needs to preserve arithmeticity. Such an isomorphism gives a representation $f: \Gamma \rightarrow (D')_v^{\times}$, where $\Gamma$ is the $T$-arithmetic subgroup of $D$ and $v$ a place. I think if one chooses $T$ large enough the key assumptions of superrigidity are satisfied (rank bigger than 2, big enough image); it then asserts that $f$ is ``of algebraic origin,'' and this should be enough. There are many little details to check of course. | |
| Jun 17, 2010 at 9:28 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
Edit: added an even stronger probably-false statement.
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| Jun 17, 2010 at 9:17 | comment | added | Kevin Buzzard | ...and in my mind this could theoretically be explained by the existence of an injection of groups $D^\times\to(D')^\times$ (it would have to be injective because any non-trivial element of the kernel would give a non-tri conj class which is killed), but I bet it almost never is explained in this way. Again though, because of the ramification conditions, I bet you can't use torsion to kill this (any torsion in $D^\times$ gives an explicit quadratic field embedding into $D$ but this same quadratic field will embed into $D'$). | |
| Jun 17, 2010 at 9:14 | comment | added | Kevin Buzzard | @Pete: yeah, I've rigged it so that the degree $p$ fields that embed into $D$ are precisely the degree $p$ fields that embed into $D'$, so I don't think one can find some torsion trick. If you want to stick to quaternion algebras here's another question: if $D$ and $D'$ are quaternion algebras and the ramified primes of $D'$ are strictly contained within the ramified primes of $D$, then I think the same arguments give a completely canonical injection from the set of conj classes of $D^\times$ into the set of conj classes of $(D')^\times$.... | |
| Jun 17, 2010 at 7:14 | comment | added | Emerton | Oops, after Brian's last comments, I realized that the opposite issue is discussed in the question. Sorry for the redundancy. | |
| Jun 17, 2010 at 6:19 | comment | added | Pete L. Clark | @Brian: Thanks for the comment. I didn't read the whole question, only "the question". Kevin's question is indeed more subtle than I had realized. | |
| Jun 17, 2010 at 5:45 | comment | added | BCnrd | Last comment for the evening on this: for the question of the natural salvage that I suggested (${\rm{Aut}}(F,S)$-twist or opposite of such a twist), Gopal pointed out to me that super-rigidity might help to overcome the anisotropicity of the norm-1 unit groups (i.e., of the derived groups of the unit groups). This comes down to whether a group isomorphism $D^{\times} \simeq {D'}^{\times}$ preserves arithmeticity of some subgroups, up to ${\rm{Aut}}(F,S)$-twist or opposite stuff. Gopal is tentatively optimistic and will think about it; I'll let you know if I hear back from him on it. | |
| Jun 17, 2010 at 5:35 | comment | added | BCnrd | Upon seeing Matt's comment I realize I should have given a better example (i.e., I missed that 1/3 and 2/3 are additive inverses in $\mathbf{Q}/\mathbf{Z}$: 3 is too small, as I meant to do something beyond the "opposite" issue which Kevin noted in his initial question). OK, use two rational primes $p$ and $q$ split in $F$ and take $D$ to be the degree-25 central division algebra with local invariants 1/5 and 3/5 at the respective primes $P$ and $P'$ over $p$, and 2/5 and 4/5 at the respective primes $Q$ and $Q'$ over $q$. Then $D' \ne D^{\rm{opp}}, D$ but ${D'}^{\times} \simeq D^{\times}$. | |
| Jun 17, 2010 at 3:59 | comment | added | Emerton | 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.) | |
| Jun 17, 2010 at 2:23 | comment | added | BCnrd | 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! | |
| Jun 16, 2010 at 23:45 | comment | added | BCnrd | @Pete: in Kevin's setup, the two division algebras are assumed to have the same bad primes, so in the quaternion algebra case isomorphism is forced. The interesting case is higher rank. | |
| Jun 16, 2010 at 22:50 | comment | added | Pete L. Clark | 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 $\mathbb{Q}$... | |
| Jun 16, 2010 at 22:32 | comment | added | Victor Protsak | 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. | |
| Jun 16, 2010 at 19:01 | comment | added | Kevin Buzzard | "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. | |
| Jun 16, 2010 at 19:00 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
edited title for BCnrd.
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| Jun 16, 2010 at 12:54 | answer | added | Victor Protsak | timeline score: 4 | |
| Jun 16, 2010 at 10:03 | history | asked | Kevin Buzzard | CC BY-SA 2.5 |