User jean lecureux - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:45:00Z http://mathoverflow.net/feeds/user/915 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131416/definition-of-a-weakly-doubly-transitive-group-action/131418#131418 Answer by Jean Lecureux for definition of a weakly doubly transitive group action Jean Lecureux 2013-05-22T08:21:18Z 2013-05-22T08:21:18Z <p>It's defined in Définition 1.4.1. Weakly doubly transitive means for him that the action is transitive on pairs of vertices $(x,y)$ and $(x',y')$ such that $d(x,y)=d(x',y')$. Weakly $k$-transitive means the same with all the pairwise distances. </p> http://mathoverflow.net/questions/119224/irreducibility-of-coxeter-graphs-as-a-function-of-generating-sets/119316#119316 Answer by Jean Lecureux for Irreducibility of Coxeter Graphs as a Function of Generating Sets Jean Lecureux 2013-01-19T11:05:06Z 2013-01-19T11:05:06Z <p>For infinite Coxeter groups, there is a paper of Luis Paris, "Irreducible Coxeter groups", which proves that every infinite Coxeter group cannot be decomposed as a non-trivial direct product.In particular they must be irreducible.</p> http://mathoverflow.net/questions/90446/weyl-group-of-a-singular-torus Weyl group of a singular torus Jean Lecureux 2012-03-07T11:52:04Z 2012-03-08T13:26:20Z <p>Let $G$ be a semisimple algebraic group over an algebraically closed field, and let $T$ be a torus in $G$.</p> <p>If $T$ is a maximal torus, then $N_G(T)/Z_G(T)=N_G(T)/T$ is the Weyl group $W$ of $G$. If $T$ is not maximal, then what can be said aboug the "Weyl group" $W_T=N_G(T)/Z_G(T)$ ? Is there a relation between $W_T$ and $W$ ? I guess the answer is well-known, but I didn't find a relevant reference.</p> http://mathoverflow.net/questions/48740/geodesic-rays-and-horofunctions Geodesic rays and horofunctions Jean Lecureux 2010-12-09T10:20:54Z 2010-12-09T10:48:26Z <p>Let $(X,d)$ be a metric space.</p> <p>Let $(x_n)_{n\in\mathbb N}$ be a geodesic ray: $d(x_n,x_m)=\vert n-m\vert$. Is it true that, for all $y\in G$, the sequence $d(x_{n+1},y)-d(x_n,y)$ converges to 1 as $n$ goes to infinity ?</p> <p>I am particularly interested in the case of $\delta$-hyperbolic spaces. A positive answer to the above quetion would imply that any geodesic ray converges to a Busemann function (or horofunction).</p> <p>More generally, is anything known about Busemann functions on hyperbolic spaces ? In particular, how do the Busemann compactification relates to the visual boundary ? These two boundaries are the same for CAT(0) spaces, but need not be in general, as shown by the example $\mathbb Z\times \mathbb Z/2\mathbb Z$, with obvious generating set.</p> http://mathoverflow.net/questions/46685/distance-to-an-apartment-of-the-affine-building-of-gln/48364#48364 Answer by Jean Lecureux for Distance to an apartment of the affine building of GL(N) Jean Lecureux 2010-12-05T12:54:29Z 2010-12-05T12:54:29Z <p>I think that the answers to your three questions are negative. Here's an example for $n=3$.</p> <p>Choose first $x_A$ to be a vertex of $A$. In the link of $x_A$, it is possible to choose a chamber $d$ which is at distance $2$ from $2$ chambers in $A$, and at distance $3$ from the $4$ others. Choose $x$ in the alcove $d$ so that it projects on $x_A$. To fix the ideas, you can take the middle of the side opposite to $x_A$.</p> <p>Then the geodesic ray from $x$ to $x_A$ can be continuated in only four ways, namely via one of the four alcoves opposite $d$ in the link of $x_A$. This answers negatively to your second question. The retraction from the 4 points towards which these geodesics are going are easily calculated: you just have to retract the geodesic segment from $x$ to $x_A$ from somewhere further on your geodesic ray; the four points you get are the four points in the alcove opposite to the four alcoves which were opposite to $d$ (and, of course, on the middle of the side opposite to $x_A$).</p> <p>From this, it is possible also to answer negatively to the third question: there is some $x_U$ which is on in one of the two alcoves which are at distance $2$ from $d$. Then the distance from $x$ to this $x_U$ is the distance between the middle of two sides of a regular hexagon which are at distance $2$, and the distance from $x$ to $x_A$ is the distance to the center of this hexagon, so we do not have $d(x,x_A)=\frac 1 2 d(x,x_U)$. </p> <p>The calculation of the retractions centered at the two other points is not so easy. Let $\xi_1$ be one of them. The sector from $x_A$ to $\xi_1$ starts with some alcove $c$ which is at distance $2$ from $d$. Let $d'$ be the alcove adjacent to both $c$ and $d$, and let $H$ be the wall of $A$ which separates $c$ and $d'$. $H$ separates $A$ in two half-planes, say $\alpha$ and $-\alpha$, with $\xi_1$ in $\alpha$. Let $\beta$ be another half-plane bounded by $H$, such that $d'\in \beta$. Then $\beta\cup\alpha$ is another apartment $A_1$.</p> <p>The retraction on $A_1$ centered at $\xi_1$ sends $d$ to a chamber which is at distance $2$ from $c$ and adjacent to $d'$, so it is the alcove in $\beta$ which is adjacent to $d'$. The retraction of $d'$ on $A$ is then the retraction of this alcove is the alcove which is at distance 2 from $c$ and in $-\alpha$. Since this alcove is already opposite to an alcove which is opposite to $d'$, we get the same point as some other $x_U$.</p> <p>Of course, the retraction centered at the last point is treated in a similar way. So, in conclusion, there are only 4 different retractions of $x$. The two "double" points form a segment whose middle is $x_A$. The two other are symmetric with respect to $H$, but are not in the same sector. So the barycenter of our 6 points is some point of $H$ which is not $x_A$.</p> http://mathoverflow.net/questions/18212/spectral-properties-of-cayley-graphs/20376#20376 Answer by Jean Lecureux for Spectral properties of Cayley graphs Jean Lecureux 2010-04-05T09:43:39Z 2010-04-05T09:43:39Z <p><a href="http://www.springerlink.com/content/bvl3421x255653k8/" rel="nofollow">This paper</a>, by A. Valette, is a survey devoted to this question, although he's more interested in infinite groups. In the infinite case, the "adjacency matrix" is a bounded operator on $\ell^2(\Gamma)$, and its spectrum makes sense. Of course, it depends on the generating set.</p> <p>One of the first results he mentions is a theorem of Kesten : it is possible to recover the fact that $G$ is amenable, or free, by looking at this spectrum.</p> http://mathoverflow.net/questions/15948/inverse-limit-in-metric-geometry/18265#18265 Answer by Jean Lecureux for Inverse limit in metric geometry Jean Lecureux 2010-03-15T13:30:20Z 2010-03-15T13:30:20Z <p><a href="http://dx.doi.org/10.4171/CMH/168" rel="nofollow">This paper</a> by P.-E. Caprace, uses a "refined boundary" of a CAT(0) space. This boundary is constructed in the following way : given a point $\xi$ in the boundary at infinity of your space $X$, you construct a point $X_\xi$, which is the inverse limit of the horoballs centered at $\xi$. Here the maps $\phi_{m,n}$ are the CAT(0) projections. Then the space $X_\xi$ is itself CAT(0), and you can iterate the construction. Under reasonable hypotheses, the construction stops after a finite number of steps, and the refined boundary is the union of all the spaces you get.</p> <p>(In the case of symmetric spaces, this construction has been already considered by Karpelevic in 1965, but with different definitions, and I don't think he saw it as inverse limits).</p> http://mathoverflow.net/questions/4547/definitions-of-hecke-algebras/4915#4915 Answer by Jean Lecureux for Definitions of Hecke algebras Jean Lecureux 2009-11-10T20:03:02Z 2009-11-10T20:03:02Z <p>A few more words to explain how (Iwahori)-Hecke algebras arise as spaces of functions. The key point in identifying the Hecke algebra with an algebra of functions on a group is the presence of a Tits system, or BN-pair. You can find the proof for example in Bourbaki, Lie Groups and Algebras, Chap.IV.1, Exercises 21 to 23 if I remember correctly. (The exercises are doable). You can also find it in the book of Davis on Coxeter groups, and probably in many other places.</p> <p>Let me say that a group G admits a Bruhat decomposition if there is a group B such that the double cosets in A BN-pair is, roughly, a way to express the Bruhat decomposition, together with the behavior of double cosets.</p> <p>The Bruhat decomposition tells you that $B\backslash G/B$ is in bijection with the Coxeter group $W$ (eg. a finite group for an algebraic group over any field, or an affine group for an algebraic group over a local field), for some suitable subgroup $B$.</p> <p>Now, consider the algebra $L(G,B)$ of functions $G$ to $\mathbb C$ which are bi-B-invariant. If $W$ is infinite, you may want to restrict yourself to "compactly" supported functions, in the case when $B$ is compact (eg for a group over a local field). More generally, there is a "bornology" on $G$ which makes B bounded, and $L(G,B)$ is the space of functions bi-$B$-invariant with bounded support.</p> <p>Then a basis of $L(G,B)$ (as a vector space) consists in characteristic functions of double cosets, so it is obviously parametrized by elements of $w$. Then you have to calculate the convolution product of two such functions. Then the convolution product can be expressed as the cardinality of some intersection of double cosets, and this is the Bourbaki exercises. If the double coset $BsB$ has cardinality $q$ for every generator $s$ of $W$, then you will get the relations you expect.</p> http://mathoverflow.net/questions/2229/iwasawa-decomposition/2793#2793 Answer by Jean Lecureux for Iwasawa Decomposition Jean Lecureux 2009-10-27T08:31:16Z 2009-10-27T08:31:16Z <p>The original reference for this is the paper of Bruhat-Tits (available on [NUMDAM]), see Prop. 4.4.3. Another reference, probably easier to read, is the book of Macdonald, "Spherical functions on a group of p-adic type", Theorem 2.6.11.</p> <p>There is a nice proof of this fact using the geometry of buildings, which goes as follows. You can think only about trees (eg for G=SL(2)) to get the main ideas.</p> <p>Consider the affine building X associated to G(K). The buildings at infinity of X (which is also the space of flags) can be seen as equivalence classes of sectors in X. Then U(K) is the union of fixators of sectors in such a class \xi. The group T(K) is the group of translation in some apartment A, and B:=T(K)U(K) is the stabiliser of the equivalence class of sector. G(O) is the stabiliser of some vertex o.</p> <p>The main point is that the building is the union of all apartments containing a sector pointing towards \xi. It follows that, for every x in X, there is an element u of U(K) such that u.x is in A.</p> <p>Let g in G. Applying this to the element x=g.o, we see that the vertex ug.o is in A. By transitivity of the action of T(K) on vertices of A, there is an element t in T such that tug.o=o. Thus tug is in G(O), which gives the decomposition of g.</p> http://mathoverflow.net/questions/2283/iwasawa-and-cartan-decompositions/2442#2442 Answer by Jean Lecureux for Iwasawa and Cartan Decompositions. Jean Lecureux 2009-10-25T08:59:41Z 2009-10-26T18:31:27Z <p>I think the key point is the proposition 4.4.2, where "good" subgroups are caracterised geometrically as stabilisers of special subgroups (ie, stabilisers of a point o such that the Weyl group W is the semidirect product of its translations and of the stabiliser of o in W).</p> <p>Then the group G is the product of B (the stabiliser of a class of sector, a minimal parabolic group for an algebraic group) and the group K. Moreover, the group B itself is the product of B^0 (which is the union of pointwise fixators of sectors) and the group of translations (acting on some apartment containing o and a sector in this class). </p> <p>The Cartan decomposition is, as usual, the decomposition of an element g in kvk', where k and k' are element of K and v is an element which sends o to a vertice of the sector starting at o in the class defined by B.</p> <p>The proposition 4.4.4 is meant to explain the relation between the two decompositions (ie, when you know the translation part in the Iwasawa decomposition, can you deduce it in the Cartan decomposition ?)</p> <p>If you know how to attach a building to a reductive group, then the book "Buildings" by Abramenko and Brown is a good reference (see chap. 11), much easier to read. They treat every building, but construct only the affine building associated to SL(n). Another reference is the small book of Macdonald, "Spherical functions on a group of p-adic type" (chapter II, Theorem 2.6.11)</p> http://mathoverflow.net/questions/2640/longest-element-of-an-affine-weyl-group/2658#2658 Answer by Jean Lecureux for Longest Element of an Affine Weyl Group Jean Lecureux 2009-10-26T18:29:58Z 2009-10-26T18:29:58Z <p>If you consider just a Weyl group, then I guess there is nothing replacing the longest element (of course, every element also has a reduced decomposition, though).</p> <p>However, as pointed by Greg Muller, there are some situations when there is a good analogue, namely in the theory of twin buildings. For example, if you consider a semisimple algebraic group G defined over a field K, then G(K[t,t^(-1)]) acts on two affine buildings, which are the buildings associated to G(K((t)) ) and G( K((t^(-1))) ). Then there is something called "codistance" between chambers of these two buildings. Two chambers at codistance 1 are called "opposite". This opposition relation shares the same kind of property, and should really be thought as an analogue of, the opposition relation in finite buildings and Coxeter complexes.</p> http://mathoverflow.net/questions/46685/distance-to-an-apartment-of-the-affine-building-of-gln Comment by Jean Lecureux Jean Lecureux 2010-11-22T14:26:44Z 2010-11-22T14:26:44Z The sector $C$ in your second question is not well-defined. I guess you should consider the sets of all fixed points of $u$ in $A$ ; if you don't, I think you can't always choose $c$ so that $c_0=x_A$. http://mathoverflow.net/questions/27609/an-example-of-a-non-amenable-exact-group-without-free-subgroups Comment by Jean Lecureux Jean Lecureux 2010-09-15T14:56:21Z 2010-09-15T14:56:21Z @Agol: Exactness of $Gamma$ is equivalent to the topological amenability of some action on sa compact space. The definition is the following : the action of $Gamma$ on $X$ is exact if there is a map $\mu_n$ from $X$ to the set of probability measures on $\Gamma$, such that $\Vert \mu_n(sx)-s\mu_n(x)\Vert$ converges to 0 uniformly in $x$, as $n$ goes to infinity. See <a href="http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_74.pdf" rel="nofollow">icm2006.org/proceedings/Vol_II/contents/&hellip;</a> for references. http://mathoverflow.net/questions/2229/iwasawa-decomposition/2793#2793 Comment by Jean Lecureux Jean Lecureux 2009-10-27T19:38:59Z 2009-10-27T19:38:59Z Sure. The new book of Abramenko and Brown, &quot;Buildings&quot;, is quite good. There is also a book by Paul Garett, available online at <a href="http://www.math.umn.edu/~garrett/m/buildings/" rel="nofollow">math.umn.edu/~garrett/m/buildings</a>