User vahid shirbisheh - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:28:47Z http://mathoverflow.net/feeds/user/23860 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128906/torsion-version-of-hnn-extensions Torsion version of HNN extensions. Vahid Shirbisheh 2013-04-27T08:32:17Z 2013-04-30T02:54:25Z <p>I am thinking of a version of HNN extensions as follows: </p> <p>Assume $H,K$ are subgroups of a group $G$ and $\phi:H\to K$ is an isomorphism. We define $G_{\phi,n}$ to be the group generated by $G$ and $x\notin G$ satisfying the conditions $x h x^{-1}=\phi(h)$ and $x^n=1$. </p> <p>I was wondering if such a construction has appeared in the literature before? What are the main properties (and the name) of these groups? </p> <p>I appreciate any references.</p> http://mathoverflow.net/questions/125470/what-information-about-a-locally-compact-group-g-is-encoded-in-c-r-astg-wh What information about a locally compact group $G$ is encoded in $C_r^\ast(G)$ which is not in $L^1(G)$? Vahid Shirbisheh 2013-03-24T17:03:44Z 2013-03-24T17:03:44Z <p>Let $G$ be a locally compact group and let $C_r^\ast(G)$ denote its reduced group $C^\ast$-algebra. Many features of a $G$ can be realized from $L^1(G)$ or $C_r^\ast(G)$. For example, $G$ is discrete iff $L^1(G)$ (resp. $C_r^\ast(G)$) is unital, or $G$ is abelian iff $L^1(G)$ (resp. $C_r^\ast(G)$) is commutative.</p> <p>I was wondering what kind of information about the group $G$ is encoded in $C_r^\ast(G)$ which is not in $L^1(G)$?</p> <p>Is it correct if we say $C_r^\ast(G)$ is just a $C^\ast$-completion of $L^1(G)$ and in fact we lose some information about $G$ by considering $C_r^\ast(G)$ instead of $L^1(G)$?</p> http://mathoverflow.net/questions/119456/orthonormal-basis-for-l2g-h Orthonormal basis for $L^2(G/H)$. Vahid Shirbisheh 2013-01-21T10:59:05Z 2013-03-13T14:26:05Z <p>Let $G$ be a locally compact group and $H$ be a closed subgroup of $G$. Is there any way to define a reasonable orthonormal basis for $L^2(G/H)$? By "reasonable" I mean elements of the orthonormal basis should be continuous and whose supports are compact and of bounded measure (i.e. there is a large number $M$ such that $\mu(supp f)\leq M$ for all $f$ in the basis).</p> <p>At least, do you know any reference to learn about concrete orthonormal bases on $L^2(G)$, where $G$ is a locally compact group?</p> http://mathoverflow.net/questions/118422/induced-representations-for-profinite-groups Induced representations for profinite groups Vahid Shirbisheh 2013-01-09T06:44:58Z 2013-01-15T20:18:40Z <p>I was wondering if there is any application for induced representations of profinite groups, for example in Galois theory or number theory? Is it a good research idea? Do you know any paper discussing this problem or related problem or at least unitary representations of profinite groups?</p> http://mathoverflow.net/questions/118931/stabilization-in-banach-algebras Stabilization in Banach algebras Vahid Shirbisheh 2013-01-14T23:41:28Z 2013-01-15T12:36:22Z <p>In $C^\ast$-algebras we use $K(H)$, the algebra of compact operators on a separable Hilbert space, for stabilization of a $C^\ast$-algebra, i.e. $S(A):=A\otimes K(H)$. Is there any similar stabilization functor in Banach algebras? What is the substitute of $K(H)$?</p> http://mathoverflow.net/questions/118639/why-is-mathcalm-textsl-2-mathbbz-spanned-by-e-4-and-e-6/118651#118651 Answer by Vahid Shirbisheh for Why is $\mathcal{M}(\text{SL}_2(\mathbb{Z}))$ spanned by $E_4$ and $E_6$? Vahid Shirbisheh 2013-01-11T17:27:56Z 2013-01-11T17:27:56Z <p>See Proposition 1.3.4 of Bump's book "Automorphic forms and representations". </p> http://mathoverflow.net/questions/118226/seeing-topological-geom-properties-of-the-space-via-corresponding-c-algebra/118228#118228 Answer by Vahid Shirbisheh for Seeing topological (geom.) properties of the space via corresponding C^*-algebra Vahid Shirbisheh 2013-01-06T22:01:25Z 2013-01-06T22:01:25Z <p>The answer to your first (and second) question is negative, because commutative C*-algebras only reflect global features of the underlying space.<br> For the third question: we have the Riesz representation theorem which says: For locally compact and Hausdorff topological space $X$, there is and isometric isomorphism between the dual of $C_0(X)$ and the space of radon measures on $X$. See Theorem 7.17 in Folland's book "Real analysis". </p> http://mathoverflow.net/questions/117156/morita-equivalence-for-algebras/117161#117161 Answer by Vahid Shirbisheh for Morita equivalence for *-algebras Vahid Shirbisheh 2012-12-24T18:35:16Z 2012-12-24T18:35:16Z <p>I suggest you look at the book "Morita Equivalence and Continuous-Trace C*-algebras" by I. Raeburn and D. P. Williams. They define much of the construction of a Morita equivalence for <em>-subalgebras of C</em>-algebras without assuming the completeness. You can also look at S. Echterhoff's notes arXiv:1006.4975, where the notion of linking algebra is used to construct Morita equivalence. I have a feeling you can use linking algebras for your purpose as well. Good Luck! </p> http://mathoverflow.net/questions/117037/extending-length-functions-regarding-certain-group-extensions Extending length functions regarding certain group extensions. Vahid Shirbisheh 2012-12-22T17:38:47Z 2012-12-22T17:38:47Z <p>Consider the group extension $1\rightarrow H\rightarrow G\rightarrow \mathbb{Z}^n\rightarrow 1$ ($H$ is discrete). Assume we have a proper length function $L$ on $H$ (proper means the kernel of $L$ is finite). How can we extend $L$ to a proper length function $L'$ on $G$?</p> <p>I am also interested in any comment (or introducing references) helping me to understand the geometry or algebraic structure of $G$ in terms of $H$ and $n$. </p> http://mathoverflow.net/questions/116734/publishing-a-bad-paper/116737#116737 Answer by Vahid Shirbisheh for Publishing a bad paper? Vahid Shirbisheh 2012-12-18T20:41:46Z 2012-12-18T20:51:40Z <p>Besides the effect of bad papers in job search, I consider some kind of scientific dignity for myself and I try to do the right thing in the similar situations. So, I suggest you do the same. </p> <p>For example I have experienced at least two similar situations: </p> <ol> <li><p>When I was a graduate student and even at the present time, I knew if I share my papers with some of the old professors by putting their names as the coauthor they are going to help me in my career and hopefully do not cause official difficulties for me. But I was not able to convince myself to do this, first because it is a lie, secondly because doing this kind of things makes me hate myself and that's the worst thing can happen for some one. </p></li> <li><p>One of my colleagues suggested once we share our papers with each other so that by writing one paper, it will be counted two publications for us. I know this suggestion sounds juvenile and stupid, but it really happened. Because of scientific dignity, again, I rejected this suggestion. </p></li> </ol> <p>I think you should care about long term and about how you will feel about your today action in the future. </p> http://mathoverflow.net/questions/110017/injective-morphisms-of-cd-the-c-algebra-of-continuous-functions-on-the-clos/110021#110021 Answer by Vahid Shirbisheh for injective *morphisms of C(D), the C^* algebra of continuous functions on the closed unit disk D Vahid Shirbisheh 2012-10-18T15:34:40Z 2012-10-18T15:34:40Z <p>Assume $D$ lies in $\mathbb{R}^2$ and define $f:D\rightarrow D$ by $f(x):=2x$ if $\|x\|\leq 1/2$ and $f(x):=\frac{x}{\|x\|}$ if $\|x\|\geq 1/2$. Then, $f$ is onto and continuous, but it is not injective. It is easy to see that $f^\ast :C(D)\rightarrow C(D)$ is an injective $^\ast$-homomorphism, but it is not onto.</p> http://mathoverflow.net/questions/105147/measure-theory-treatment-geared-toward-the-riesz-representation-theorem/105188#105188 Answer by Vahid Shirbisheh for Measure theory treatment geared toward the Riesz representation theorem Vahid Shirbisheh 2012-08-21T19:39:06Z 2012-08-21T19:39:06Z <p>Have you looked at G.B. Folland's book "Real analysis, Modern techniques and their applications"? Chapter 7 of this books covers this topic. I have taught it once and I totally recommend it. </p> http://mathoverflow.net/questions/101746/commutator-subgroup-group-theory/101835#101835 Answer by Vahid Shirbisheh for Commutator Subgroup - Group Theory Vahid Shirbisheh 2012-07-10T09:50:01Z 2012-07-10T09:50:01Z <p>There is a very general theorem (see Proposition 4 in Chapter 1 of J-P. Serre, Trees. Springer-Verlag Berlin Heidelberg (1980).) which says: Let $A$ and $B$ be two groups. The kernel of the natural quotient map $A\ast B\rightarrow A\times B$ is a free group generated by all commutators of the form $[a,b]$ where $a\in A - {1}$ and $b\in B - {1}$. Your question is the special case that $A=B=\mathbb{Z}$. In short, it coincides exactly with Andreas' answer. </p> http://mathoverflow.net/questions/98821/how-often-do-people-read-the-work-that-they-cite/98841#98841 Answer by Vahid Shirbisheh for How often do people read the work that they cite? Vahid Shirbisheh 2012-06-05T04:30:45Z 2012-06-05T04:30:45Z <p>Mostly, I read those part of the papers that I need to cite in my works. Sometimes though, I read them almost completely before starting the paper (to get some ideas and learn the techniques). There are also occasions that I read only a few pages of the paper before citing it. </p> http://mathoverflow.net/questions/98118/when-a-name-is-used-for-two-different-notions When a name is used for two different notions! Vahid Shirbisheh 2012-05-27T14:38:35Z 2012-05-28T02:33:41Z <p>What would you do if you see a notion which you used to work with has two names and the same name you use for this notion is used for another notion too. For example, after the work of J.B. Bost, A. Connes, (Hecke algebras, type III factors and phase transition with spontaneous symmetry breaking in number theory. Selecta Math. (New Series) Vol.1, no.3, 411-456, (1995)), in operator algebras community people call a subgroup $H$ of a group $G$ almost normal if every double coset of $H$, like $HgH$ is the union of finitely left cosets. It is equivalent to say that $H$ is almost normal if it is commensurable with its conjugates. Therefore in group theory people call such a subgroup conjugate commensurable. In stead, in group theory people call a subgroup $H$ of a group $G$ almost normal if its normalizer is of finite index in $G$. Now if I use the word conjugate commensurable for the above notion it is confusing in operator algebras community and if I use almost normal it can be confused with group theorists' notion of almost normal subgroups. What is the best way to deal with these sort of problems.</p> http://mathoverflow.net/questions/98082/etiquette-question-how-to-acknowledge-bugs-bunny/98084#98084 Answer by Vahid Shirbisheh for Etiquette question: how to acknowledge Bugs Bunny? Vahid Shirbisheh 2012-05-27T03:23:06Z 2012-05-27T03:23:06Z <p>I think it depends how important is the help you have received. If it is a crucial step in your work and without it you were not able to complete the paper, you may want to suggest the helper to be your co-author. But if it is not that important you can send him/her an email and thank him/her and let him/her know that you will acknowledge his/her help at the end of the paper. </p> http://mathoverflow.net/questions/97971/haar-measure-for-profinite-groups-reference-needed Haar measure for profinite groups (reference needed) Vahid Shirbisheh 2012-05-25T18:23:36Z 2012-05-25T18:40:36Z <p>I was wondering if anybody knows a good reference book or exposition for Haar measures over profinite groups (with some concrete examples and computations)?</p> http://mathoverflow.net/questions/97909/commutativity-of-the-fundamental-group-of-any-lie-group/97912#97912 Answer by Vahid Shirbisheh for Commutativity of the fundamental group of any Lie Group Vahid Shirbisheh 2012-05-25T04:54:59Z 2012-05-25T13:54:04Z <p>It is actually true for all topological groups. Topological groups possess a structure which makes them H-spaces and fundamental group of every H-space is abelian. The formulation and the proof is given in Algebraic Topology, Homotopy and Homology, by Switzer Pages 14-16.</p> http://mathoverflow.net/questions/97681/non-elementary-examples-of-nearly-normal-subgroups Non-elementary examples of nearly normal subgroups Vahid Shirbisheh 2012-05-22T16:56:45Z 2012-05-22T17:02:55Z <p>$H$ is called a nearly normal subgroup of a group $G$ if it is of finite index in its normal closure, $H^G:=\cup_{g\in G} gHg^{-1}$, in $G$. Clearly, every normal subgroup or subgroup of finite index of $G$ is nearly normal. I am not interested in finite subgroups either. I am looking for infinite nearly normal subgroups of infinite index in well-known groups. By a well-known group, I mean a group that some of its basic properties like amenability, growth or its geometric nature (the spaces it acts) have been studied in the literature. </p> http://mathoverflow.net/questions/97581/property-rd-for-mathbbq/97622#97622 Answer by Vahid Shirbisheh for Property (RD) for $\mathbb{Q}$ Vahid Shirbisheh 2012-05-22T03:40:38Z 2012-05-22T03:40:38Z <p>Thanks to 'Yves Cornulier's answer to my other question about the growth of $\mathbb{Q}$, we now know (1) there is a length function on the additive group of $\mathbb{Q}$ which makes $\mathbb{Q}$ of polynomial growth. (2) there is no length function on $\mathbb{Q}^\times$ making it of polynomial growth. </p> <p>We can modify a theorem by Jolissaint which says: if $G$ is an amenable (finitely generated) group, then $G$ has (RD) if and only if $G$ is of polynomial growth. To generalize this theorem to infinitely generated groups one only needs to show that if $G$ has (RD) w.r.t. some length function $L$ then ${ g\in G; L(g)\leq r}$ is finite for all $r\geq 0$. This is easily done by introducing to sequence of functions in $\mathbb{C}G$ (I will give details in the next few days). </p> <p>Now, since $\mathbb{Q}$ and $\mathbb{Q}^\times$ are both amenable, $\mathbb{Q}^\times$ does note have (RD) and $\mathbb{Q}$ has (RD). </p> http://mathoverflow.net/questions/97581/property-rd-for-mathbbq Property (RD) for $\mathbb{Q}$ Vahid Shirbisheh 2012-05-21T17:51:55Z 2012-05-22T03:40:38Z <p>Do the additive group or the multiplicative group of $\mathbb{Q}$ have property (RD) (Rapid Decay)?</p> http://mathoverflow.net/questions/97576/growth-of-infinitely-generated-groups growth of infinitely generated groups Vahid Shirbisheh 2012-05-21T17:23:18Z 2012-05-22T03:14:44Z <p>Is there any length function on additive group of $\mathbb{Q}$ such that $\mathbb{Q}$ is of polynomial growth WRT this length function? What about the multiplicative group of $\mathbb{Q}$ instead?</p> http://mathoverflow.net/questions/97573/non-commuative-geometry-and-its-prerequisites/97577#97577 Answer by Vahid Shirbisheh for Non-commuative geometry and its prerequisites Vahid Shirbisheh 2012-05-21T17:29:14Z 2012-05-21T17:29:14Z <p>$C^\ast$-algebra would be a good place to start. Afterwards, you can learn $K$-theory of $C^\ast$-algebras. </p> http://mathoverflow.net/questions/126364/books-on-advanced-galois-theory Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-04-03T09:19:11Z 2013-04-03T09:19:11Z I think inverse Galois problems and Galois embedding problems could be interesting subjects to continue. Just google these phrases to find reading materials that suits you. You can download my book on Galois embedding problem at researchgate: (<a href="https://www.researchgate.net/publication/200151467_Galois_embedding_problems_with_abelian_kernels_of_exponent_p?ev=prf_pub" rel="nofollow">researchgate.net/publication/&hellip;</a>). http://mathoverflow.net/questions/126106/are-there-any-nontrivial-ring-homomorphisms-m-n1r-rightarrow-m-nr Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-31T18:17:12Z 2013-03-31T18:17:12Z When $R=\mathbb{C}$ (and probably most fields), it is easy to see that every homomorphism from $M_n(R)$ into another algebra is either an injection or the zero homomorphism, because $M_n(R)$ (in this case) is a simple algebra. For an easy proof, see Proposition 4.3.23 of arXiv:1211:3404. Certainly, by modifying the proof, you can relax some of the conditions on $R$. http://mathoverflow.net/questions/125945/quasinilpotent-elements-of-group-c-star-algebras/125951#125951 Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-30T13:12:25Z 2013-03-30T13:12:25Z For a proof for the above statement see, Page 110 (Proposition II.6.4.14) of Blackadar's book &quot;Operator algebras&quot;. http://mathoverflow.net/questions/125470/what-information-about-a-locally-compact-group-g-is-encoded-in-c-r-astg-wh Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-24T18:05:43Z 2013-03-24T18:05:43Z @Martin: That's exactly my point. I'd like to know what makes mathematicians to consider $C_r^\ast(G)$ instead of $L^1(G)$. When $G$ is abelian, $C_r^\ast(G)$ appears in the Pontryagin duality, $C_r^\ast(G)$ is a $C^\ast$-algebra and therefore easier to work with. What else can be said about the benefits of $C_r^\ast(G)$. For example, I am tempted to say the computation of $K$-theory of $C_r^\ast(G)$ is easier, but I am not sure about it. http://mathoverflow.net/questions/124811/can-we-generalize-the-result-of-urysohns-lemma-to-countable-collection-of-pairwi Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-17T19:29:42Z 2013-03-17T19:29:42Z No, take $A_n=\{1/n\}\subset \mathbb{R}$. http://mathoverflow.net/questions/124804/how-many-mathematicians-are-working-in-how-many-fields-how-long-per-task Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-17T18:54:27Z 2013-03-17T18:54:27Z Do you have any reason to assume there is a set of distinctive mathematical fields? http://mathoverflow.net/questions/124773/shall-i-send-this-paper-to-the-arxiv Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-17T17:58:26Z 2013-03-17T17:58:26Z I'd also like to add to Martin's comment that there are only freaky actions. http://mathoverflow.net/questions/119456/orthonormal-basis-for-l2g-h/124420#124420 Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-14T07:33:54Z 2013-03-14T07:33:54Z Thanks for references. The original problem is to show the Hecke algebra $\mathcal{H}(G,H)$ has a left regular representation on $L^2(H\backslash G)$. When $H\backslash G$ is discrete, there is a proof based on an orthonormal basis which I wanted to generalize to locally compact case. However, when $H$ is cocompact, one can use standard techniques like the Fubini theorem to show that the convolution product defines a representation. http://mathoverflow.net/questions/119456/orthonormal-basis-for-l2g-h/124420#124420 Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-14T06:21:08Z 2013-03-14T06:21:08Z Your suggestion looks promising (in special cases), but my original problem has an easier solution when $H\backslash G$ is compact. Anyway, I would like to learn more about the technique you suggested. Could you give some references? http://mathoverflow.net/questions/119456/orthonormal-basis-for-l2g-h/124411#124411 Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-03-14T06:20:30Z 2013-03-14T06:20:30Z I am interested in groups or homogeneous spaces, because I am looking for a suitable way to define a convolution like product and define a regular representation for certain algebras on $L^2(H\backslash G)$. I have been thinking about a similar construction as you described in your answer. But it does seem have several problems: 1. There is no general recipe to partition a general group as the union of cubes in $\mathbb{R}^n$. The Gram-Schmidt orthogonalisation process does not give us an explicit orthonormal basis to work with, but it proves the existence of an orthonormal basis. http://mathoverflow.net/questions/119910/how-similar-different-are-dense-subgroups-of-a-compact-group Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-01-26T09:29:45Z 2013-01-26T09:29:45Z A very simple example of two completely different dense subgroups of $\mathbb{T}$, the circle in $\mathbb{C}$ can be considered by letting $H_1$ be the subgroup of all roots of unity and $H_2$ be the cyclic infinite subgroup generated by $e^{2\pi i \lambda}$ where $\lambda$ is an irrational number. http://mathoverflow.net/questions/118931/stabilization-in-banach-algebras Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-01-25T20:58:14Z 2013-01-25T20:58:14Z @Yemon: You are welcome. Following Vincent Lafforgue's works, Walter Paravicini has studied Morita equivalence of Banach algebras too. http://mathoverflow.net/questions/119456/orthonormal-basis-for-l2g-h Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-01-21T11:49:39Z 2013-01-21T11:49:39Z Thanks Alain for sharing your insight. http://mathoverflow.net/questions/119402/why-all-irreducible-representations-of-compact-groups-are-finite-dimensional-e Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-01-20T16:02:12Z 2013-01-20T16:02:12Z I suggest the book &quot;Principles of Harmonic Analysis&quot; by Deitmar and Echterhoff. Chapter 7 of this books answers your first question, see Thm. 7.2.4. http://mathoverflow.net/questions/118931/stabilization-in-banach-algebras Comment by Vahid Shirbisheh Vahid Shirbisheh 2013-01-17T05:58:32Z 2013-01-17T05:58:32Z @Yemon: I list some instances that clarify the importance of stabilization in $C^*$-algebras: 1. Both K-theory and KK-theory are stable functors meaning $K(A)\simeq K(A\otimes K(H))$. 2. Two separable $C^*$-algebras $A$ and $B$ are Morita equivalent if and only if they are stably isomorphic, i.e. $A\otimes K(H)\simeq B\otimes K(H)$. 3. Tensoring by $K(L^2(G))$ also appears in some theorems too, for example see Takai-Takesaki duality. So, it is nice to have a similar notion in Banach algebras, for instance proving item 2 for Banach algebras would be a good start.