Timeline for Unbounded representations of groups
Current License: CC BY-SA 3.0
44 events
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| Mar 30, 2013 at 16:05 | comment | added | Misha | Yves: I was thinking about Lipschitz cocycles. The Lipschitz uniform Hilbert embedding I have in mind is a much more complicated one: Using a suitable infinite dimensional symmetric space and its logarithmic map. I do not have a detailed proof though since the argument also needs lower curvature bounds. | |
| Mar 30, 2013 at 15:34 | vote | accept | Kate Juschenko | ||
| Mar 30, 2013 at 10:42 | answer | added | YCor | timeline score: 10 | |
| Mar 30, 2013 at 9:57 | comment | added | Kate Juschenko | Yves, non-non-non, I don't claim that. | |
| Mar 30, 2013 at 9:30 | comment | added | YCor | @Misha, Kate: Taka noted one direction. You seem to claim the other direction is true. But I don't see any way, even for a proper cocycle $b$ over a (non-uniformly) bounded representation $\pi$, to deduce a coarse embedding. If you want the orbital map $g\mapsto b(g)$ to be a coarse embedding, you need a uniform bound on $\pi(g)b(s)$ when $s$ ranges over generators and $g$ ranges over $G$. | |
| Mar 30, 2013 at 1:21 | comment | added | Misha | I think the correct statement is that a fg group properly embeds in Hilbert space iff it admits a continuous proper affine action on such. Taka noted one direction in this statement. | |
| Mar 30, 2013 at 0:35 | comment | added | Kate Juschenko | Unless I am missing something, Taka's comment applies to both questions. | |
| Mar 30, 2013 at 0:08 | comment | added | YCor | Narutaka Ozawa answers question Q1 (for abstract representations) in the case of groups coarsely embeddable into Hilbert spaces, but it is not yet fully answered, or do I miss something? | |
| Mar 29, 2013 at 22:58 | history | edited | Kate Juschenko | CC BY-SA 3.0 |
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| Mar 29, 2013 at 22:35 | history | edited | Kate Juschenko | CC BY-SA 3.0 |
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| Mar 29, 2013 at 22:30 | comment | added | Kate Juschenko | Initially I was thinking exactly about the question that Mikael answered. But now I've got excited about the question which you guessed. I'll added it to the question. | |
| Mar 29, 2013 at 19:25 | comment | added | Mikael de la Salle | @Yves: yes, this was my interpretation of Alain's interpretation. It seems to be the only way to put together "unbounded operators" and "Domain is the whole space $H$". @Alain: I agree that although we still do not know the question, Taka's comment is a good answer. | |
| Mar 29, 2013 at 18:45 | comment | added | Alain Valette | @ Kate: I think that Taka's excellent comment comes very close to an answer... | |
| Mar 29, 2013 at 18:23 | comment | added | Alain Valette | @ Yves, Mikael: I feel psycho-analyzed, but you are right about how I interpreted Kate's question... (no unbounded operators in my interpretation). | |
| Mar 29, 2013 at 18:10 | comment | added | YCor | @Mikael: I can't guess the question to which Misha and Taka answered, (assuming it is uniquely defined)... also I'm not sure about the precise meaning of Alain's interpretation since continuity of linear maps is often implicit on affine actions on topological spaces. So in your interpretation of Alain's interpretation, we consider homomorphisms into $GL(H)\ltimes H$, where $GL(H)$ is the full group of invertible linear automorphisms of the abstract vector space $H$ and the subgroup $H$ in the semidirect product is the group of translations, and require it to define a proper action. | |
| Mar 29, 2013 at 17:50 | comment | added | Mikael de la Salle | @Yves. If one follows Alain, Kate calls unbounded operator a (not necessarily continuous) linear map on the vector space H. In this case the answer is yes: take an affine isometric action for which $\{g \cdot 0, g \neq 1\}$ are linearly independant vectors, extend this family to an algebraic basis of H, and conjugate your action by some suitably chosen linear invertible map that is diagonal in this basis). In my opinion the interesting question is the one to which Misha and Taka answered. | |
| Mar 29, 2013 at 17:32 | comment | added | user30035 | @Kate: sometimes when people edit their question, they leave the old question there and insert new words and write "EDIT" by the new words. The advantage of this would be that very early on I have a comment of the form "let $G$ be a group with huge cardinality" and you have now changed the question so $G$ is finitely-generated, so my comment now looks stupid. But this doesn't bother me -- you're in the hands of the experts now so it seems so I can stop thinking about this question :-) I still don't understand the quantifiers though! I don't know what is given -- the group, the Hilbert space... | |
| Mar 29, 2013 at 16:26 | comment | added | YCor | So you need to assume that $c(h)$ belongs to the domain of definition of $\pi(g)$ for all $g,h$ in your group, right? | |
| Mar 29, 2013 at 15:37 | comment | added | Kate Juschenko | possibly by unbounded operators | |
| Mar 29, 2013 at 14:38 | comment | added | YCor | What do you mean by unbounded representation? possibly by unbounded operators, or by bounded operators, but with no uniform bound on their norms? | |
| Mar 29, 2013 at 14:00 | history | edited | user9072 | CC BY-SA 3.0 |
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| Mar 29, 2013 at 13:13 | history | edited | Kate Juschenko | CC BY-SA 3.0 |
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| Mar 29, 2013 at 13:01 | history | edited | Andreas Thom |
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| Mar 29, 2013 at 2:51 | comment | added | Misha | One approach would be to find a mteruc of nonpositive curvature on GL(H)/U(H). | |
| Mar 29, 2013 at 1:33 | comment | added | Narutaka OZAWA | If $G$ is finitely generated and coarsely embeddable into a Hilbert space, then the answer is yes. This follows from Bo\.zejko--Fendler's construction (Arch. Math. (Basel) 57 (1991), 290–298). Otherwise, as Misha, I bet on an expander counterexample. | |
| Mar 29, 2013 at 1:02 | comment | added | Joël | Alain, I am impressed by your guessing the right question from so few information. Kate, perhaps you should edit your question, adding enough details so that people who are not in the field but interested may have a chance to understand it. | |
| Mar 28, 2013 at 23:41 | comment | added | Misha | A natural class of counter-examples to try are groups containing expanders. It is quite likely that a proper affine continuous action would imply existence of a uniform embedding in a Hilbert space (possibly a different one), but I do not see how to prove this at the moment. | |
| Mar 28, 2013 at 22:51 | comment | added | Kate Juschenko | yes, Alain. Is this equivalent to something known? | |
| Mar 28, 2013 at 21:56 | comment | added | Alain Valette | My understanding of Kate's question is this: does any f.g. group admit a proper, affine action on a Hilbert space? Am I correct? | |
| Mar 28, 2013 at 21:51 | comment | added | Kate Juschenko | actually, the my remark about unbounded is bad, I want to include finite groups as well.. | |
| Mar 28, 2013 at 21:48 | history | edited | Kate Juschenko | CC BY-SA 3.0 |
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| Mar 28, 2013 at 21:44 | comment | added | Kate Juschenko | Yes, Alain, this is what I meant (both your remarks) | |
| Mar 28, 2013 at 21:43 | comment | added | Kate Juschenko | Domain is the whole space $H$ | |
| Mar 28, 2013 at 21:42 | history | edited | Kate Juschenko | CC BY-SA 3.0 |
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| Mar 28, 2013 at 21:40 | comment | added | Alain Valette | And my guess is that, here, a representation is just a homomorphism $G\rightarrow GL({\cal H})$. | |
| Mar 28, 2013 at 21:36 | comment | added | Alain Valette | I think that Kate meant UNbounded... | |
| Mar 28, 2013 at 21:26 | comment | added | Andreas Thom | wccanard, the question is whether there exists some representation with some proper cocycle. | |
| Mar 28, 2013 at 21:14 | comment | added | Andreas Thom | What do you mean precisely by an unbounded representation? One fixed dense domain for all group elements? | |
| Mar 28, 2013 at 21:09 | comment | added | Kate Juschenko | and cocycle is a map $c:G\rightarrow H$ such that $c(gh)=\pi(g)c(h)+c(g)$ for all $g,h$ in the group. | |
| Mar 28, 2013 at 21:08 | comment | added | user30035 | If it means what I think it means, then this question as it currently stands is trivial to answer in the negative. Let the Hilbert space be the complex numbers, and take any group whose cardinality is much larger than that of the complex numbers, acting trivially. Then a cocycle is a homomorphism and it the kernel must be huge for cardinality reasons. | |
| Mar 28, 2013 at 21:06 | comment | added | Kate Juschenko | $||c_g||<K$ is finite | |
| Mar 28, 2013 at 21:05 | comment | added | Kate Juschenko | $c$ is proper cocycle if the map $g\mapsto c_g$ is proper, I.e., for every constant K the number of elements $g$ in the group such that $||c_g||$ is finite. | |
| Mar 28, 2013 at 20:59 | comment | added | user30035 | What's a proper cocycle? Sorry to be so ignorant. | |
| Mar 28, 2013 at 20:42 | history | asked | Kate Juschenko | CC BY-SA 3.0 |