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1h |
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A group G acting Properly Discontinuously and Cocompactly on a Proper geodesic space X A little context would help. Your formulation makes it sound like this is homework. |
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8h |
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fixedpoint or fixed point or fixed-point The underlying rule here is that you only need to hyphenate noun phrases that are being used adjectivally. Not doing so can lead to confusion - does the phrase 'fixed point theory' refer to a theory of fixed points or a fixed theory of points? |
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8h |
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Bound of prime pairs I think there's some agreement among the MO community that questions of the form 'is this recent preprint correct?' are off-topic. You need to dress your question up as something slightly more specific. (See quid's example, for instance.) I'm voting to close. |
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8h |
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Why don’t more mathematicians improve Wikipedia articles? Since it seems unlikely that this question has a single correct answer, I think it should be made community wiki. You can do this by checking a box on the 'edit' page. |
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May 15 |
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How to find quotients of infinite triangle groups or von Dyck groups? That's an improvement. Now I have a further question. You ask for conditions to ensure that 'the given quotients' are finite, but you haven't given us any quotients! In general, these groups have many finite quotients (because they are residually finite) and also many infinite quotients. Which quotients are you specifically interested in? |
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May 15 |
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How to find quotients of infinite triangle groups or von Dyck groups? Perhaps this needs emphasis. Subgroups and quotients are very different! If you want a good answer to your question, you should edit the title and the body to make this change. |
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May 15 |
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How to find quotients of infinite triangle groups or von Dyck groups? Ketan - it appears that Nick is exactly right. You are interested in finite quotients, not finite subgroups. |
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May 15 |
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How to find quotients of infinite triangle groups or von Dyck groups? Ketan - in the question I linked to above, it is explained that in the hyperbolic case these groups are always infinite. These groups are pretty famous in their own right. |
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May 15 |
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How to find quotients of infinite triangle groups or von Dyck groups? In particular, your group is finite precisely when $1/l+1/m+1/n>1$. |
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May 15 |
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How to find quotients of infinite triangle groups or von Dyck groups? Although your question is slightly more general, I think it was probably answered here: mathoverflow.net/questions/22459/x-y-xp-yp-xyp-1/… . |
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May 11 |
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The role of the Automatic Groups in the history of Geometric Group Theory (Oh, and we also need Agol's theorem.) |
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May 11 |
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The role of the Automatic Groups in the history of Geometric Group Theory Misha - a small correction. Fundamental groups of non-geometric 3-manifolds (graph manifolds, say) are known to be automatic but not known to be biautomatic. Therefore, this is not how you solve the conjugacy problem in this setting. The conjugacy problem was solved by Préaux. You can also deduce it from the fact that 3-manifold groups are conjugacy separable (Hamilton--W--Zalesskii), though unlike Préaux's work this uses some heavy machinery: Wise's deep work, and a very nice theorem of Minasyan. |
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May 3 |
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Extensions with trivial induced outer action (Where the action of $G$ on $C$ is trivial.) |
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May 3 |
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Extensions with trivial induced outer action Indeed, I think you answered your own question: there are such examples whenever $H^2(G,C)\neq 0$. |
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May 3 |
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Extensions with trivial induced outer action For your second question, consider for instance $1\to 2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 1$. |
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May 3 |
revised |
construct Seifert fibration on mapping torus of surface with monodromy a periodic mapping class Added arXiv tag. |
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May 2 |
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Extensions with trivial induced outer action If the outer action is trivial then the centre of $N$ is indeed central in $E$. |
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May 2 |
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Extensions with trivial induced outer action If $N$ is non-abelian, then it can't be central in $E$. On the other hand, $E=G\times N$ has the property that the outer action of $G$ is trivial. |
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May 1 |
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Distinguishing 3-manifolds by homologies of covers Ben - in this case, the homology never gets bigger than $\mathbb{Z}$, so probably the action is trivial for every finite quotient. |
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Apr 30 |
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Why isn’t $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group? I just want to point out that, although the map to $F_2\times F_2$ has been altered, it's still the case that the presentation you give can't embed into $F_2\times F_2$; by the Baumslag--Roseblade theorem, no subgroup is isomorphic to $\mathbb{Z}*\mathbb{Z}^2$. In general, it's a hard problem to compute a presentation for a subgroup from the generators. See Section 8 of arXiv:1003.5117v2 for a quick summary of what's known. There is an algorithm that works for finitely presentable subgroups of $F_2\times F_2$, as it happens. |
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Apr 29 |
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Why isn’t $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group? Brian - yes, I think that's right. You should get the orientable surface of the relevant genus, with one point identified. |
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Apr 29 |
accepted | Why isn’t $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group? |
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Apr 28 |
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Why isn’t $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group? Perhaps you're right, Misha. But the same considerations apply when you're checking if an arbitrary compact 2-complex is a surface: every edge should be contained in exactly two faces, and every vertex link should be a cycle. |
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Apr 28 |
answered | Why isn’t $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group? |
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Apr 28 |
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Why isn’t $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group? (added to clarify)... in some free-cross-free subgroup. (Of course, the factors may be cyclic or trivial, or the subgroup may be diagonal.) |
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Apr 28 |
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Why isn’t $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group? Brian, your claim that $F_2\times F_2$ contains a surface group is incorrect. A theorem of Baumslag--Roseblade says every finitely presented subgroup of of $F_2\times F_2$ is of finite index. |
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Apr 26 |
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Distinguishing 3-manifolds by homologies of covers (See, for instance, Corollary 1.4 of L. Funar, 'Torus bundles not distinguished by TQFT invariants'.) |
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Apr 26 |
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Distinguishing 3-manifolds by homologies of covers Neil - there exist pairs of Sol manifolds with isomorphic profinite completions. I suspect they will answer your question in the integral case. |
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Apr 26 |
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Distinguishing 3-manifolds by homologies of covers It's natural to include in your data the action of $\pi_1M_i$ on $H_{i,k}$, which factors through a finite quotient. If you do, then a positive answer to either question would imply that $\pi_1M_1$ and $\pi_1M_2$ have isomorphic profinite completions. For 3-manifolds with non-solvable fundamental group, this is Question 9.28 in our survey article on 3-manifold groups (arXiv:1205.0202v3). Without this data, it's not so clear to me, but it seems likely that the two questions are equivalent (ie you can reconstruct the action on homology). |
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Apr 25 |
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Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators? Tiny quibble: the easiest example of a non-elementary hyperbolic subgroup of a RAAG is $F_2\subseteq F_2$. I agree that yours is the easiest interesting example. |
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Apr 25 |
answered | Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators? |
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Apr 22 |
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Compelling evidence that two basepoints are better than one Ronnie, this last comment begs the question. The reason Higgins' paper 'has been ignored by the specialists' is that it isn't particularly useful: the groupoid normal form isn't importantly different from the usual group normal form. (Also, the groupoid normal form is already implicit in Serre's work.) |
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Apr 20 |
accepted | Are virtual cubulated groups cubulated? |
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Apr 19 |
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Are virtual cubulated groups cubulated? n the final paragraph, I point out that the main purpose of the Hruska--Wise paper is to develop these ideas in the relatively hyperbolic setting. That said, I think the paper was around in one form or another a lot earlier than 2012. Indeed, it's cited (as in preparation) in Bergeron--Wise. |
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Apr 19 |
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Are virtual cubulated groups cubulated? Ian - I agree with both your comments. As I say, the $H_i$ that we use are precisely the hyperplane stabilizers in $N$. And I didn't mean to suggest that you need to use the Hruska--Wise result, only that it's a convenient statement to apply. I suppose, strictly speaking, you could do it without starting with the induced action on $X^{|G:N|}$, but I find that a helpful way of thinking about it. |
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Apr 19 |
accepted | Thompson’s group T |
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Apr 19 |
answered | Are virtual cubulated groups cubulated? |
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Apr 19 |
answered | Thompson’s group T |
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Apr 7 |
awarded | ● Nice Answer |
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Apr 3 |
awarded | ● Nice Answer |
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Apr 3 |
awarded | ● Nice Answer |
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Apr 2 |
awarded | ● Enlightened |
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Apr 2 |
awarded | ● Nice Answer |
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Apr 2 |
awarded | ● Nice Answer |
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Mar 30 |
revised |
Finite vertex-transitive graphs that look like infinite vertex-transitive graphs Rolled back after mistaken rewrite. |
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Mar 30 |
revised |
Finite vertex-transitive graphs that look like infinite vertex-transitive graphs Completely rewritten. |
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Mar 30 |
accepted | Finite vertex-transitive graphs that look like infinite vertex-transitive graphs |
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Mar 30 |
revised |
Finite vertex-transitive graphs that look like infinite vertex-transitive graphs Removed one implication. |
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Mar 29 |
revised |
Finite vertex-transitive graphs that look like infinite vertex-transitive graphs Added missing hypothesis. |
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Mar 29 |
answered | Finite vertex-transitive graphs that look like infinite vertex-transitive graphs |
