# Yves Cornulier

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 Name Yves Cornulier Member for 2 years Seen 40 secs ago Website Location Age
 1d comment Characterization of amenable actionsWhat is $G$? a discrete group? countable? what does "Borel" mean in an abstract measure space? 1d revised Group action on the real lineadded 242 characters in body 1d comment Group action on the real line@Harry: it's unclear from your question if you want an action such that some or every point $p$ has a trivial stabilizer. This dramatically changes the answer! 1d accepted Volume growth of covers and growth of deck-transformation groups 1d answered Group action on the real line 1d comment Group action on the real lineI don't think that $\{g(p)| g\in G\}$ is always dense or discrete. May19 answered Volume growth of covers and growth of deck-transformation groups May19 comment Volume growth of covers and growth of deck-transformation groupswhat do you mean by equal? is $\exp(2n)=\exp(3n)$? May10 comment Actions of Thompson group F. IIAn exotic action of $F$ is the following: let $F_4$ be the set of elements in $F$ with all slopes integral powers of $4$ and $X=F/F_4$. May10 comment Actions of Thompson group F. IIactually I'm not sure I know any bounded degree graph with a subgraph isomorphic to a 3-regular tree but with no coarsely embedded 3-regular tree. May7 comment An ultrafilter is a set of subsets containing exactly one element of each finite partition: reference request@Pete: write $X=X_1\sqcup X_2\sqcup X_3$ where $X_1=X$ and $X_2=X_3=\emptyset$. For a unique $i$ we have $X_i\in\mathcal{U}$, so necessarily $i=1$ (otherwise uniqueness would fail). So $\emptyset\notin\mathcal{U}$. May6 comment Which topological spaces are coset spaces of locally compact groups?@Ramiro: indeed, probably (3) should be replaced by: $X$ is a disjoint union of clopen subsets with the Suslin condition. May6 comment Which topological spaces are coset spaces of locally compact groups?These conditions are not sufficient, the Sierpinski carpet, etc, are not homogeneous spaces of LC-groups (although compact, metrizable and homogeneous). The only 1-dimensional compact connected spaces occuring as homogeneous spaces of LC-groups are solenoids (projective limits of circles). Except the circle, these spaces are not path-connected. Also, all connected compact manifolds are homogeneous, but in dimension $\ge 2$ most of them are not homogeneous under a locally compact group (it is not hard to check that this implies homogeneous under a connected Lie group) May6 comment Actions of Thompson group F@Kate: a homomorphism from $F$ to any group is either injective of factors through the abelianization. In other words, any action (on any set) of $F$ not factoring through $\mathbf{Z}^2$ is faithful. Still equivalently, any nontrivial normal subgroup of $F$ contains $[F,F]$. May6 comment Actions of Thompson group F@Kate: OK to call this the standard action... but this is not a transitive action on a discrete set. When you say that for the standard action there is a binary tree on the schreier graph, which action do refer to? May5 comment Actions of Thompson group F@Kate: this is vague. May5 comment Actions of Thompson group FNote that you haven't yet said what is the "standard" action. If you look at actions on orbits on the open interval, the stabilizers are pairwise distinct and therefore, since $Aut(F)$ is countable, you also have continuum many actions that are pairwise non-conjugate by automorphisms of $F$. It mayby happens, however, than none of these actions is fine for you. May5 comment Actions of Thompson group FCould you clarify if you are concerned with transitive actions on discrete sets (as the fact that you consider the Schreier graph suggests)? the action on the interval is not one, and you didn't say what is the standard action (the action on the dyadics in the open interval?) May4 comment Unbounded metrics on groups@Wlodzimierz: how do you prove that the 0-1 metric is unbounded? :) May4 comment Unbounded metrics on groupsThis is actually originally due to Bergman (Bull London Math Soc) arxiv.org/abs/math/0401304, Rosendal considers the version for a group endowed with a topology. May4 comment Actions of Thompson group FThe question seems to concern transitive actions on discrete sets. Actually, I don't know if $F$ admits a Schreier graph of subexponential growth (or even with no bilipschitz binary tree) besides the ones factoring through the action of an abelian quotient of $F$. Apr27 comment Torsion version of HNN extensions. This is the quotient of the usual HNN extension $\langle x,H\mid\dots\rangle$ by the normal subgroup generated by $t^n$. This is probably the best way to view it. Unlike in the HNN extension, it is most likely not true that the natural homomorphism $G\to G_\{\phi,n}$ is always injective. Apr27 answered Rank gradients and HNN-extensions Apr24 comment Ordered groups - examplesSide remark: for $n\le -1$, $BS(1,n)$ is left-orderable but not bi-orderable. Apr24 comment Ordered groups - examples@Alain: a countable group admits a *left*-invariant ordering iff it embeds in $Homeo^+(\mathbf{R})$. This proves that the (positive!) affine group of $\mathbf{R}$ is left-orderable. But it is indeed bi-orderable by a general fact on group extensions, here using the fact that the action of the multiplicative group of positive reals on the additive group of reals preserves an ordering. Apr24 comment Ordered groups - examplesIt is safe to write bi-orderable group instead of ordered group. Because orderable groups denote, according to authors, either left-orderable or bi-orderable groups and is thus blatantly ambiguous. Also left/bi-ordered group is more suitable to mean a group endowed with a (left or bi...)-invariant total order. For instance, there is a Chabauty space of bi-ordered groups on $k$ generators, which surjects onto the Chabauty space of bi-orderable groups on $k$ generators, which is a closed subset of the space of marked groups on $k$ generators. Apr13 accepted Extensions of Groups Apr13 answered Extensions of Groups Apr13 comment Extensions of GroupsBy Mostow, $G$ has a maximal compact subgroup $K$ and $KG_0=G$. In particular, if you assume in addition that $G_0$ is simply connected, then $G=G_0\rtimes K$ so your extension is split (just assuming $G_0$ nilpotent). In general, since $G$ is nilpotent and $K$ is compact, the action of $K$ on the Lie algebra of $G$ is unipotent and hence trivial, so $[K,G_0]=1$. Note that there are easy non-split extensions, e.g. with $G_0$ the circle and $G/G_0$ noncyclic group of order 4. Apr12 comment Extensions of Groups... But to understand the general case (with your assumptions: $G$ virtually connected nilpotent Lie group), you can probably reduce to the case when $G_0$ is a torus and in general, I expect that if $W$ is the maximal (compact) torus in $G_0$ then the nilpotent extensions of $G_0$ by a finite nilpotent group $F$ should be classified by the same object as the central extensions $W$-by-$F$. Apr10 comment Definition of a uniformly bounded dual of a groupOK, well, I'd rather had written the question as "What is the topology considered by Cowling in...". It seems to me that the natural approach to such a question is just to ask him, if he hasn't written the definition in any of his papers. Apr9 comment Definition of a uniformly bounded dual of a group@Piotr: I'm confused about the meaning of your question. Could you specify a little more what you would like? can you at least define the nets converging to the trivial representation? (Besides, the unitary dual usually means the sets of classes of irreducible reps, although I agree the Fell topology is natural to define on all reps, leaving aside that they don't form a set.) Apr9 comment Syndetically separated topological groups@Sebastian. In 2) I assumed $G$ connected. So the answer is yes for (3) if you assume $G$ connected or more generally with finitely many components. But false in general (since arbitrary discrete groups are particular instances of Lie groups). Apr9 comment Syndetically separated topological groups@Sebastian 2) It's complicated. Sufficient conditions: $G$ abelian, $G$ semisimple. Necessary condition: $G$ unimodular. For a simply connected nilpotent Lie group (that's always unimodular), a necessary and sufficient condition is that the Lie algebra be $\mathbf{Q}$-defined, i.e. isomorphic to $\mathfrak{g}\otimes_\mathbf{Q}\mathbf{R}$ for some Lie algebra $\mathfrak{g}$ over $\mathbf{Q}$. Also, $\text{SL}_2(\mathbf{R})\ltimes\mathbf{R}^2$ has no cocompact lattices. Reference: Raghtnathan's book for most of this. Apr9 comment Syndetically separated topological groups@Sebastian 1) $\Gamma$ is finitely generated, as is any cocompact lattice in a connected Lie group $G$. This holds because it acts properly cocompactly on a connected manifold (namely $G$, with the action by left translations). On the other hand, Malcev's theorem is that any finitely generated linear group is residually finite. Combining, it follows that any cocompact lattice of a connected Lie group is residually finite. Apr8 answered Syndetically separated topological groups Apr7 comment Definition of a uniformly bounded dual of a groupNote that for unitary representations it is usually easier to describe neighborhoods of the trivial representation than of an arbitrary unitary rep (there are subtleties with convex combinations...). Here you might want to specify how you expect the operator norm to converge. If a sequence $\pi_i$ converges to the identity, and $K$ is a compact neighbourhood of 1 in the group, do you want $\sup_K\|\pi_i(g)\|$ to tend to 1? or just to be bounded? Or none? Apr4 comment Examples of Amenable Groups other than Z_nfree groups are not inverse limits of finite groups neither (at least not in the category of groups). They are residually finite, anyway (that is, in the category of marked groups, the same as inverse limits of finite groups). Apr2 awarded ● Yearling Mar30 revised Unbounded representations of groupsI mentioned uncountable groups Mar30 accepted Unbounded representations of groups Mar30 answered Unbounded representations of groups Mar21 revised Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ?added 8 characters in body Mar19 awarded ● Nice Answer Mar19 revised Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ?1 typo Mar19 answered Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ? Mar8 answered Hyperbolic groups with infinitely generated commutator subgroups Feb24 revised abelianization of adelic points of an algebraic groupadded 39 characters in body Feb14 awarded ● Nice Answer Feb9 answered isometric embeddings of Cayley graphs in “nice” spaces