Timeline for Can the infinite von Dyck groups be subgroups of $SU(n)$?
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| Mar 16, 2012 at 21:47 | vote | accept | CommunityBot | ||
| Mar 16, 2012 at 17:54 | comment | added | Misha | @whistles: Answer to your simple question "the number of irreducible representations of a von Dyck group is infinite, is it not?" is "yes", furthermore, the cardinality is continuum. See item (b) of my updated answer. Finiteness holds for $SO(3)$ (and $SO(4)$) but fails in sufficiently higher dimensions. Here is a guess: Conjecture. For every hyperbolic von Dyck's group $\Gamma$ there exists a number $N$ so that for every $n\ge N$ the variety $Hom(\Gamma, U(n))$ contains (nonconstant) curves consisting of inequivalent irreducible representations. | |
| Mar 16, 2012 at 17:46 | comment | added | Misha | @whistles: I updated my answer to address some of your questions. Your question "only a finite number of irreducible representations of von Dyck groups that can be embedded in SO(3)" is poorly phrased. The correct statement is that there are only finitely many homomorphisms from von (any) Dyck group to $SO(3)$ (up to conjugation). Once you have a representation $\rho: \Gamma\to SO(3)$ and an irreducible representation $\pi: SO(3)\to U(n)$, then you also have compositions $\phi=\pi\circ \rho$. Some of them may fail to be irreducible. However, if $\rho(\Gamma)$ is dense, $\phi$ is irreducible. | |
| Mar 16, 2012 at 17:40 | history | edited | Misha | CC BY-SA 3.0 |
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| Mar 16, 2012 at 10:02 | comment | added | user22139 | that is, I will have matrices of dimension d that generate the von Dyck group for any $d$. Are you saying that these are reducible? A simple question: the number of irreducible representations of a von Dyck group is infinite, is it not? – whistles 0 secs ago | |
| Mar 16, 2012 at 9:57 | comment | added | user22139 | @Misha with respect to your second comment, there's something that I don't understand. For any given $q_1,q_2,q_3$ there's only a finite number of spherical triangles with angles of the form $A_i=\frac{k_i \pi}{q_i}$ so if I understand correctly, this implies that there are only a finite number of irreducible representations of von Dyck groups that can be embedded in $SO(3)$? But if I have one embedding into, say the defining representation of $SO(3)$, then there's a map that takes elements of a von Dyck group to matrices of any irrep of $SO(3)$, of which there's an infinite number... | |
| Mar 16, 2012 at 9:49 | comment | added | user22139 | I wrote a comment before, here is a better version of it. First of all thanks for taking interest in the question. Then, to your first comment, I hope that you understand that it is evident that (a) and (b) are different. Also, I consider (a) to be rather trivial once the original question has a positive answer. The question was (b) and what I want to know in general, without making any referencing to any embedding in a Lie group, is how to construct systematically ALL representations of von Dyck groups, or triangle groups for what matters just like it is known how to do it for Lie groups... | |
| Mar 15, 2012 at 14:26 | comment | added | Misha | @whistles: Three facts you need to know about representations of von Dyck groups $D(q_1,q_2,q_3)$ to $SO(3)$: They are parameterized by spherical triangles with angles $A_1,A_2,A_3$ of the form $A_i=\frac{k_i\pi}{q_i}$, where $k_i$ are positive integers. Necessary and sufficient conditions for existence of a spherical triangle with angles $A_i$ are: (i) angle sum $>\pi$ (Gauss-Bonnet) and (ii) triangle inequalities for the dual angles $A_i^*=\pi-A_i$ [see M.Berger, "Geometry"]. (3) Irreducible representations are equivalent to $A_i\ne 0, \pi$ and at most one right angle among $A_i$'s. | |
| Mar 15, 2012 at 14:07 | comment | added | Misha | @whistles: You have to be more precise with your questions, including the last one. Was it (a) that for every hyperbolic von Dyck group $D$ there exists $n$ so that both $D$ and $SO(3)$ admit an irreducible representation of dimension $n$? (Answer is "yes".) Or was it (b) that for every hyperbolic von Dyck group $D$ and for every $n$-dimensional irreducible representation of $SO(3)$ there exists an irreducible $n$-dimensional representation of $D$? (The answer is probably "no" already for $n=3$, but requires some computations to check.) Hope you understand that (a) and (b) are different. | |
| Mar 15, 2012 at 13:03 | comment | added | user22139 | Thanks Misha and Yves, the question referred in fact to all von Dyck groups - or at least a large quantity of them. So this answers that part. The other part of the question referred to representations. I don't mind representations being not faithful. The question comes from physics where faithful is not important. I do require them being irreducible. Do the von Dyck groups have irreducible representations of the same dimension than those of $SO(3)$? | |
| Mar 15, 2012 at 4:38 | comment | added | Misha | Thanks, Yves! My argument for $SL(n, {\mathbb Z})$ was more complicated and used superrigidity. Your argument is much better since it proves (finite-dimensional) non-unitarizability for all f.g. groups with distorted cyclic subgroups. | |
| Mar 15, 2012 at 2:01 | comment | added | YCor | About the question: the induction argument shows that if a group G has a finite index k subgroup H having a faithful representation in some O(n) then so does G. Induce a faithful rep of H to G. Get a representation; it's possibly not faithful but is faithful in restriction to G, but we can fix this by also considering a faithful rep of the finite group G/H. (In particular, G embeds into O(kn+k-1).) Note that in case G has no nontrivial finite normal subgroup intersecting H trivially then the induced representation is faithful for free. | |
| Mar 14, 2012 at 22:15 | comment | added | YCor | If $SL(n,Z)$, $n>2$ maps to $SO(m)$ then unipotents are mapped to distorted elements and therefore to finite order elements. Bounded generation of $SL(n,Z)$ (which is not trivial but much easier that the normal subgroup theorem or superrigidity) implies that the image is finite. | |
| Mar 14, 2012 at 22:06 | comment | added | Victor Protsak | A couple of thoughts, then: 1. GL(n) embeds diagonally into split O(n,n); 2. Superrigidity. | |
| Mar 14, 2012 at 21:49 | comment | added | Misha | Victor, I do not know, I just thought about it while typing my answer. | |
| Mar 14, 2012 at 21:30 | comment | added | Victor Protsak | Thank you for filling in some details, Misha! The construction with surface groups is really much easier (although one then needs to pull orthogonal representations of these groups out of a hat). What is the history of question in the last paragraph, is it a known problem? | |
| Mar 14, 2012 at 21:30 | history | edited | Misha | CC BY-SA 3.0 |
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| Mar 14, 2012 at 21:21 | history | edited | Misha | CC BY-SA 3.0 |
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| Mar 14, 2012 at 21:00 | history | answered | Misha | CC BY-SA 3.0 |