Timeline for Is the fundamental group of any compact hyperbolic 3-manifold embeddable into a p-adic group?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2017 at 21:55 | comment | added | Ian Agol | Example 5.2.8 (2) of MacLachlan-Reid (which is (10,1) surgery on the $5_2$ knot) might be a candidate for having no faithful integral $PSL_2$ representation. They prove this for the discrete faithful representation and Galois conjugates, and it appears from their discussion that there may not be any other representations. | |
| Apr 23, 2017 at 20:39 | comment | added | Ian Agol | An interesting refinement of your question is whether a hyperbolic 3-manifold has a faithful representation into $PSL_2(\mathbb{C})$ with all integral traces. This is true for arithmetic and non-Haken hyperbolic 3-manifolds (actually, ones that contain no geometrically finite embedded surface). But there may be Haken 3-manifolds such that every representation is either non-integral or not faithful. | |
| Apr 23, 2017 at 16:56 | vote | accept | Pablo | ||
| Apr 23, 2017 at 16:36 | answer | added | Ian Agol | timeline score: 11 | |
| Apr 23, 2017 at 14:15 | answer | added | YCor | timeline score: 11 | |
| Apr 23, 2017 at 13:26 | comment | added | YCor | The naive definition $PSL_2(R)=SL_2(R)/\{\pm 1\}$ is not satisfactory because it would not satisfy $PSL_2(\mathbf{Z}/6\mathbf{Z})=PSL_2(\mathbf{Z}/2\mathbf{Z})\times PSL_2(\mathbf{Z}/3\mathbf{Z})$. One option (not nice in the schematic point of view) is to quotient by the center, which is the set of scalar matrices $\mathrm{diag}(t,t)$ with $t^2=1$; probably it's fine here. | |
| Apr 23, 2017 at 13:25 | comment | added | YCor | There's a confusion in the conjecture in Luo's version and in the AFW version. In the first, the target is $PSL_2(R)$, but the latter is not defined. It leaves a little ambiguity (how do you define $PSL_2$ over a ring?). In the AFW survey, $PSL_2$ is replaced with $SL_2$, which leaves no ambiguity but it might have trivial counterexamples then. | |
| Apr 23, 2017 at 13:06 | comment | added | YCor | A group that is residually embeddable in $PSL_2$ of finite commutative rings will be embeddable in $PSL_2$ of a finitely generated (pass to the product and then restrict to suitable entries). So it maps with nilpotent kernel into a finite product of $PSL_2$ over fields. | |
| Apr 23, 2017 at 13:03 | comment | added | Pablo | @YCor - Exactly! | |
| Apr 23, 2017 at 13:02 | comment | added | YCor | OK the conjecture in the paper is for arbitrary compact 3-manifolds. In the hyperbolic case (which was part of your assumptions) it follows by the above argument, and more generally if $\pi_1(M)$ has a faithful rep in $PSL_2$ of a field. | |
| Apr 23, 2017 at 13:02 | comment | added | Pablo | @YCor is there somewhere a hyperbolicity assumption in your proof? Maybe when you assume that the group is linear? | |
| Apr 23, 2017 at 12:58 | comment | added | YCor | Proof. First let $G'$ be the inverse image of $G$ in $SL_2$. Let $R$ be the ring generated by entries of $G'$. Let $g\in G$ with $g\neq 1$ and $g'$ a lift in $G'$ of $g$. Let $b$ be a nonzero entry of $g'-1$ and $c$ a nonzero entry of $g'+1$. Since every f.g. domain is residually a finite field (this is part of the usual proof that f.g. linear groups are residually finite), there exists a finite quotient field $K$ of $R$ in which $bc\neq 0$. So the image of $g'$ in $SL_2(K)$ is neither 1 nor $-1$. So $G'\to PSL_2(K)$ factors through $G$ and maps the image of $g$ to a nontrivial element. | |
| Apr 23, 2017 at 12:56 | comment | added | Pablo | @YCor Why such a field $K$ exists? By the way, this conjecture is also mentioned in the survey of Aschenbrenner, Friedl, and Wilton. | |
| Apr 23, 2017 at 12:53 | comment | added | YCor | I'm saying that Conjecture 1 in Luo's paper sounds to me as a simple standard exercise. Namely, for every finitely generated subgroup $G$ of $PSL_2(\mathbf{C})$ and every $g\in G-\{1\}$, there exists a finite field $K$ and a homomorphism $G\to PSL_2(K)$ mapping $g$ to a nontrivial element. | |
| Apr 23, 2017 at 12:51 | comment | added | Pablo | @YCor I am sorry but I do not understand your comment. Do you suggest that the implication I mention in the question is wrong? | |
| Apr 23, 2017 at 12:49 | comment | added | YCor | Conjecture 1 in Luo's paper you mention says that $\pi_1(M)$ is residually embeddable into $PSL_2$ of finite commutative rings. But this follows from $\pi_1(M)$ being embeddable into $PSL_2$ of a field, not necessarily in the rings of integers of a $p$-adic field. So it's true (with old methods) unless I have a misunderstanding. | |
| Apr 23, 2017 at 12:44 | history | asked | Pablo | CC BY-SA 3.0 |