User sergei akbarov - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T13:34:42Z http://mathoverflow.net/feeds/user/18943 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126038/naive-question-about-the-representation-theory-of-algebraic-groups-and-hopf-algeb/126094#126094 Answer by Sergei Akbarov for Naive question about the representation theory of algebraic groups and hopf algebras Sergei Akbarov 2013-03-31T11:40:54Z 2013-06-03T06:38:31Z <p>There is a direct way to define group algebras for different classes of groups (locally compact groups, Lie groups, algebraic groups, etc.). </p> <p>For an (affine) algebraic group $G$ one should take the algebra ${\mathcal R}(G)$ of polynomials (i.e. regular functions) on $G$ and then consider the dual algebra ${\mathcal R}^\star(G)$ of linear functionals on ${\mathcal R}(G)$ with the natural topology (in this case the topology of poitwise convergence on ${\mathcal R}(G)$). Then ${\mathcal R}^\star(G)$ becomes a (topological) group algebra for $G$: each "polynomial representation" of $G$ corresponds bijectively to a continuous representation of ${\mathcal R}^\star(G)$. At the same time both ${\mathcal R}(G)$ and ${\mathcal R}^\star(G)$ are Hopf algebras in the monoidal category of stereotype spaces, see details in: S.S.Akbarov, Pontryagin duality in the theory of topological vector spaces and in topological algebra, <em>Journal of Mathematical Sciences</em>, 113(2): 179-349, 2003 (Theorem 10.12). </p> http://mathoverflow.net/questions/129688/a-sufficient-condition-for-a-probability-measure-to-have-compact-support/129690#129690 Answer by Sergei Akbarov for A sufficient condition for a probability measure to have compact support Sergei Akbarov 2013-05-05T02:27:37Z 2013-05-05T02:27:37Z <p>The Paley-Wiener theorem as it is presented in W.Rudin's "Functional analysis" (Theorem 7.23), will it be satisfactory for you? In your situation: $\mu$ has compact support if and only if $F_\mu$ can be extended as an entire function to $\mathbb C$, and for some $C>0$ and $r>0$ $$|F_\mu(z)|\le C\cdot e^{r\cdot |{\rm Im} z|},\qquad z\in {\mathbb C}.$$</p> http://mathoverflow.net/questions/107650/axiom-of-global-choice "Axiom of global choice" Sergei Akbarov 2012-09-20T05:04:15Z 2013-04-21T09:49:45Z <p>In some books on category theory (for example, in J.Adámek, H.Herrlich, E.Strecker "Abstract and concrete categories...") the authors use the idea of "big sets" ("conglomerates" or "collections") which can contain classes (as far as I understand, in the Goedel-Bernays sense) as elements, and they formulate the "generalized axiom of choice", where it is stated that the choice function exists (not only for families or classes of sets, but also) for families of classes (indexed by elements of those "big sets"). This approach allows to prove, in particular, the existence of a skeleton in each category, and some other useful things. </p> <p>This generalization of the axiom of choice is also mentioned In Wikipedia: <a href="http://en.wikipedia.org/wiki/Axiom_of_global_choice" rel="nofollow">http://en.wikipedia.org/wiki/Axiom_of_global_choice</a> (as the "strong form of the axiom of global choice").</p> <p>I wonder if there are any texts with the justification of this trick? The references I found (in particular, those mentioned in Wikipedia) give justification only for usual axiom of choice (for families of sets or for classes of sets, but not for "conglomerates of classes"). So actually I can't understand whether, for example, the existence of a skeleton, is true for all categories (in some interpretation of set theory) or for some special ones... Similarly the other corollaries of this "global axiom of choice" look doubtful for me. I would be grateful if anybody could clarify this.</p> <p><strong>UPDATE 21.09.2012</strong></p> <p>From the comments I see that there is a risk of misunderstanding, so I want to explain that by <strong>justification</strong> I mean an accurate (rigorous) definition of the new tool together with the analysis of whether it is compatible with the other tools of the theory. </p> <p>As an illustration, in the case of the usual axiom of choice (I mean its "weak form", in terms of Wikipedia), there are many textbooks (I can recommend E.Mendelson "Introduction to mathematical logic" or J.Kelly "General topology", the appendix), where the fundamental objects of the theory (in this case, the classes) are accurately introduced (here, axiomatically) and the necessary constructions (like functions) are rigorously defined in the theory. This makes possible to give rigorous formulation to the axiom of choice (again, to its "weak form") inside the theory, and moreover, this presentation of a new axiom is followed by a thorough investigation of whether it contradicts to the previous axioms of the theory. Only after receiving the answer that no contradictions can appear (in fact, a more strong thing is true: the new axiom is independent from the others, that was the result by P.Cohen) mathematicians can use this new axiom without worrying that something is wrong here. </p> <p>So my question is whether there is something similar for the "strong form of the axiom of choice"? Is it possible that nothing lies behind these words? On the contrary, if there is a justification, where can I read about it?</p> <p><strong>UPDATE 21.04.2013</strong></p> <p>Dear colleagues, from what I learn on this subject in the textbooks which I found, in Wikipedia and here in MO, I deduce that what people call "axiom of global choice" is just the usual axiom of choice (as it is presented in Kelly's book) applied to some special classes of sets arising in consideration of what is called the Grothendieck Universe. It's a puzzle for me</p> <p>1) why people call this special case "a stronger form of the axiom of choice", and </p> <p>2) why they don't want to give references, where this construction is accurately introduced.</p> <p>With the aim to accelerate the clarification of this question, I now nominated for deletion the article in Wikipedia devoted to his topic: <a href="http://en.wikipedia.org/wiki/Wikipedia:Articles_for_deletion/Axiom_of_global_choice" rel="nofollow">http://en.wikipedia.org/wiki/Wikipedia:Articles_for_deletion/Axiom_of_global_choice</a>. As I wrote there, I don't exclude that the partisans of the idea will rewrite the article in Wikipedia for endowing "global choice" with some mathematical sense, but you should agree that in its present form this article and the other mentionings of "global choice" available for external observers, look indecently vague. I invite all comers to share their opinion here or at the Wikipedia page.</p> http://mathoverflow.net/questions/111193/monoidal-structure-on-a-category-with-products-and-with-terminal-object Monoidal structure on a category with products and with terminal object Sergei Akbarov 2012-11-01T19:13:01Z 2012-11-02T09:24:09Z <p>Let $K$ be a category with products $(X,Y)\mapsto X\sqcap Y$ and with a terminal object $T$. It seems obvious to me that $\sqcap$ and $T$ define a structure of a monoidal category on $K$, but I can't find a reference. When I try to prove this myself I come to amazingly bulky constructions. Is there a text where this is accurately proved, or at least formulated? </p> http://mathoverflow.net/questions/96661/the-richardson-theorem-and-the-base-identities-problem the Richardson theorem and the base identities problem Sergei Akbarov 2012-05-11T12:18:05Z 2012-05-12T21:45:19Z <p>In the fields related to school mathematics there is some acitivity on proving (or disproving) deducibility/decidability for some classes of school identities. In particular, </p> <p>1) In logic they considered not long ago the <a href="http://exsolver.narod.ru/Artical/Mathemat/probtotog.html" rel="nofollow">base identities problem</a> (this term is the translation from Russian, I am not sure that it is correct). The problem was the following. Let $N$ be the set of positive integers, and $\mathcal K$ a class of all functions from $N^k$ into $N$ ($k$ runs over $N$) which can be represented as compositions of usual algebraic operations $x+y$, $x\cdot y$ and $x^y$. Let us call <em>a base of identities</em> in $\mathcal K$ a set $B$ of identities for functions in ${\mathcal K}$, such that any identity for functions in $\mathcal K$ can be deduced from $B$. The question was, does there exist a finite base of identities for ${\mathcal K}$? This question appeared when A.Wilkie gave a counterexample for the <a href="http://en.wikipedia.org/wiki/Tarski%27s_high_school_algebra_problem" rel="nofollow">Tarski high school algebra problem</a> (where a list of identities was suggested by Tarski, and the question was whether this list is a base). In 1980-es R.Gurevich proved that there is no finite base of identities, so the problem of base identities is solved in negative. At the same time, as far as I understand, R.Gurevich proved that instead of <em>finite</em> base of identities, there exists a <em>recursive</em> base of identities, and as far as I understand this is an example of what logicians call <a href="http://en.wikipedia.org/wiki/Decidability_%28logic%29" rel="nofollow">decidability</a>. </p> <p>2) In computer algebra there is the so-called <a href="http://www.math.upenn.edu/~wilf/AeqB.html" rel="nofollow">Richardson theorem</a>, which states that if $\mathcal R$ is a class of expressions generated by</p> <p>-- the rational numbers and the two real numbers $\pi$ and $ln 2$,</p> <p>-- the variable $x$,</p> <p>-- the operations of addition, multiplication, and composition, and</p> <p>-- the sine, exponential, and absolute value functions,</p> <p>then for $F\in {\mathcal R}$ the predicate $F=0$ is <em>recursively undecidable</em>.</p> <p>My question is whether these two fields are related to each other? Is decidability for Richardson the same as <a href="http://en.wikipedia.org/wiki/Decidability_%28logic%29" rel="nofollow">decidability for logicians</a>? If yes, then which exactly logical system does Richardson mean?</p> <p>I am not a specialist here, I am interested in this because I write a textbook on mathematical analysis (I am sorry, this happens sometimes with mathematicians), and when describing elementary functions I faced a problem analogous to the base identities problem above, but the difference is that the list of operations (and elementary functions) is wider (for example, both $x-y$ and $x^y$ are included), and as a corollary the arising functions are defined not everywhere on $R$ (one can look at the details at page 197 in the <a href="http://arxiv.org/abs/1010.0824" rel="nofollow">draft</a> of the first volume of my textbook -- unfortunately, it is in Russian). </p> <p>This is strange, but I can't find anyone who could explain me this. I asked this question in <a href="https://groups.google.com/forum/#!topic/sci.math.research/dt3GA2H4S_w" rel="nofollow">sci.math.research</a> some time ago, but the problem of overcoming the Kevin Buzzard resistance turned out to be undecidable for me there. So I would be much obliged to MO if my question will hang here for some time so that, perhaps, some specialitsts in logic could clarify me something.</p> http://mathoverflow.net/questions/37610/demonstrating-that-rigour-is-important/96004#96004 Answer by Sergei Akbarov for Demonstrating that rigour is important Sergei Akbarov 2012-05-04T18:04:45Z 2012-05-04T18:53:56Z <p>I think, the Cosmonut mentioning of Stokes' theorem (as a response to Gowers' specification of the answer) must be generalized to a wider statement which can't be bypassed here: in fact, <strong>exactly the impossibility to make Calculus rigorous can be considered as a cause of why Analysis appeared</strong>. </p> <p>In modern language the problem is the following. One could expect that the operations over what is called "elementary functions" in Calculus can be axiomatized independently from the axioms of real numbers, so that one gets a closed purely algebraic system, where the equalities of elementary functions are derived from the list of "axiomatic identities" between $x^y$, $\sin x$, etc., and the operations like derivative and integral are conceived in purely algebraic way - the formulation of the problem can be found at page 197 in my textbook in arxiv: <a href="http://arxiv.org/abs/1010.0824" rel="nofollow">http://arxiv.org/abs/1010.0824</a> (unfortunately, in Russian).</p> <p>But as it turns out, this is impossible, at least for the complete list of elementary functions: even the equality of elementary functions can't be defined axiomatically. And, what is amazing, this is not a classical result, it is quite new. However, I can't give a reference, what I write here is what Sergei Soloviev from Toulouse told me not long ago.</p> http://mathoverflow.net/questions/93225/topology-of-the-normal-spectrum-of-a-commutative-von-neumann-algebra Topology of the "normal spectrum" of a commutative von Neumann algebra Sergei Akbarov 2012-04-05T15:41:45Z 2012-04-05T18:42:56Z <p>Kadison and Ringrose define normal states $\omega$ of a von Neumann algebra $A$ as such that $\omega(H_\alpha)\to \omega(H)$ for each monotone increasing net of operators $H_\alpha$ with least upper bound $H$ (definition 7.1.11) </p> <p>Let $A$ be a commutative von Neumann algebra and $NS(A)$ be its set of normal characters, and let us endow $NS(A)$ by some natural topology, for example, by the the weak topology generated by elements of $A$. Did anybody try to describe the topological properties of $NS(A)$? As far as I understand, usually this space is not compact, but from the construction of the von Neumann envelope it follows that such spaces are "natural covers" for all Hausdorff compact spaces. So I wonder how this picture can be explained from the topological point of view.</p> http://mathoverflow.net/questions/84054/a-variant-of-the-stone-weierstrass-theorem A variant of the Stone-Weierstrass theorem? Sergei Akbarov 2011-12-21T23:07:12Z 2012-03-21T22:04:59Z <p>I would like to ask specialists in C*-algebras if the following variant of the Stone-Weierstrass theorem is true.</p> <p>Suppose $A$ is a C*-algebra and $C$ is its center. Since $C$ is a commutative C*-algebra, there exists a compact space $T$ such that $C$ is isomorphic to the algebra $C(T)$ of continuous functions on $T$. Does this mean that there exists a C*-algebra $B$ such that </p> <p>1) $A$ is isomorphic to a closed subalgebra in the algebra $C(T,B)$ of continuous mappings $f:T\to B$ with the pointwise algebraic operations (and the topology of uniform convergence on $T$), and</p> <p>2) this isomorphism turns $C$ into the algebra of scalar mappings, i.e. the mappings of the form $f(x)=\lambda(x)\cdot 1_B$, where $1_B$ is the identity in $B$, and $\lambda(x)$ $\in$ $\mathbb{C}$ for all $x\in T$.</p> <p>EDIT 21-03-12: All the C*-algebras here are supposed to be unital, excuse me for not mentioning this from the very beginning!</p> http://mathoverflow.net/questions/81645/weakened-notion-of-extremal-epimorphism Weakened notion of extremal epimorphism? Sergei Akbarov 2011-11-22T19:47:35Z 2011-11-23T19:30:54Z <p>An epimorphism $f$ is said to be <em>extremal</em>, if for any decomposition $f=i\circ p$ with $i$ a monomorphism, the morphism $i$ is automatically an isomorphism. (This is from the textbook by F.Borceux.) </p> <p>Let us say that $f$ is <em>weakly extremal</em>, if for any decomposition $f=i\circ p$ with $i$ a monomorphism <em>and $p$ an epimorphism</em>, the morphism $i$ is automatically an isomorphism.</p> <p>Are these definitions equivalent?</p> http://mathoverflow.net/questions/80146/is-the-mendeleev-table-explained-in-quantum-mechanics Is the Mendeleev table explained in quantum mechanics? Sergei Akbarov 2011-11-05T19:36:52Z 2011-11-17T23:16:52Z <p>Does anybody know if there exists a mathematical explanation of Mendeleev table in quantum mechanics? In some textbooks (for example in "F.A.Berezin, M.A.Shubin. The Schrödinger Equation") the authors present quantum mechanics as an axiomatic system, so one could expect that there is a deduction from the axioms to the main results of the discipline. I wonder if there is a mathematical proof of the Mendeleev table?</p> <p>P.S. I hope the following will not be offensive for physicists: by a mathematical proof I mean a chain of logical implications from axioms of the theory to its theorem. :) This is the standard approach everywhere in mathematics. For instance, in Griffiths' book I do not see axioms at all, therefore I can't treat the reasonings at pages 186-193 as a proof of Mendeleev table. By the way, that is why I did not want to ask this question at a physical forum: I do not think that people there will even understand my question. However, after Bill Cook's suggestion I made an experiment - and you can look at the results here: <a href="http://theoreticalphysics.stackexchange.com/questions/473/is-the-mendeleev-table-explained-in-quantum-mechanics" rel="nofollow">http://theoreticalphysics.stackexchange.com/questions/473/is-the-mendeleev-table-explained-in-quantum-mechanics</a> </p> <p>So I ask my colleagues-mathematicians to be tolerant.</p> <p>P.P.S. After closing this topic and reopening it again I received a lot of suggestions to reformulate my question, since in its original form it might seem too vague for mathematicians. So I suppose it will be useful to add here, that by the Mendeleev table I mean (not just a picture, as one can think, but) a system of propositions about the structure of atoms. For example, as I wrote here in comments, the Mendeleev table claims that the first electronic orbit (shell) can have only 2 electrons, the second - 8, the third - again 8, the fourth - 18, and so on. Another regularity is the structure of subshells, etc. So my question is whether it is proved by now that these regularities (perhaps not all but some of them) are corollaries of the system of axioms like those from Berezin-Shubin book. Of course, this assumes that the notions like atoms, shells, etc. must be properly defined, otherwise the corresponding statements could not be formulated. I consider this as a part of my question -- if experts will explain that the reasonable definitions are not found by now, this automatically will mean that the answer is 'no'.</p> <p>The following reformulation of my question was suggested by Scott Carnahan at <a href="http://meta.mathoverflow.net/discussion/1202/should-a-mathematician-be-a-robot/#Item_0" rel="nofollow">http://meta.mathoverflow.net/discussion/1202/should-a-mathematician-be-a-robot/#Item_0</a> : "Do we have the mathematical means to give a sufficiently precise description of the chemical properties of elements from quantum-mechanical first principles, such that the Mendeleev table becomes a natural organizational scheme?" </p> <p>I hope, this makes the question more clear.</p> http://mathoverflow.net/questions/80007/topology-on-the-space-of-schwartz-distributions/80168#80168 Answer by Sergei Akbarov for Topology on the space of Schwartz Distributions Sergei Akbarov 2011-11-05T23:39:49Z 2011-11-06T01:26:52Z <p>Concerning this: "I am likewise interseted in the operator algebra of operators on S* that restrict to operators on S. In particular, I would like to abstractly characterize this space"</p> <p>I am not sure that I understand you correctly, but if you want to characterize the operators which are extensions from S to S*, then the following can be the answer. First, let us use some notations: for any x,y$\in$S we set $(x,y)=\int x(t)y(t)dt$, for any operator (everywhere operator will be linear and continuous) $A:S\to S$ we define a transposed operator $A^T:S\to S$ by formula $(Ax,y)=(x,A^Ty)$ (it does not always exist), the dual operator A*:S*$\to$ S* by formula A*f(x)=f(Ax) (it always exists) and for any operator B:S*$\to$S* its dual B*:S$\to$S by formula B*x(f)=Bf(x) (it also always exists). Let us also consider an operator F:S$\to$S*, Fx(y)=(x,y). </p> <p>Then </p> <p>1) B:S*$\to$S* is an extension of $A:S\to S$ iff $B\circ F=F\circ A$;</p> <p>2) an operator A:S$\to$S can be extended to some operator B:S*$\to$S* iff there exists a transposed operator $A^T:S\to S$; in this case B=A$^T$*;</p> <p>3) an operator B:S*$\to$S* is an extension of some operator A:S$\to$S iff for B* there exists a transposed operator B*$^T:S\to S$; in this case A=B*$^T$.</p> <p>Or were you asking about something else?</p> http://mathoverflow.net/questions/80007/topology-on-the-space-of-schwartz-distributions/80015#80015 Answer by Sergei Akbarov for Topology on the space of Schwartz Distributions Sergei Akbarov 2011-11-04T07:31:50Z 2011-11-05T08:17:35Z <p>Perhaps, the following will be interesting for you: the strong topology on S*, obviously, coincides with the topology of uniform convergence on totally bounded sets, and this means that S*, being endowed with this topology, is what is called stereotype space. :) The definition is as follows. For a locally convex space X let us denote by X* the dual space (of functionals) endowed with the topology of uniform convergence on totally bounded sets. Then X is said to be {\it stereotype}, if X** is naturally isomorphic (as a locally convex space) to X. </p> <p>By the way, this kind of duality allows to consider linear continuous operators X*$\to$X* as linear continuous operators X$\to$X. :)</p> http://mathoverflow.net/questions/79708/is-the-category-of-rings-co-well-powered Is the category of rings co-well-powered? Sergei Akbarov 2011-11-01T12:54:12Z 2011-11-03T21:13:44Z <p>Dear colleagues,</p> <p>Can anybody explain me if a category of (associative) rings is co-well-powered (this is the MacLane definition, in Russian literature this is called "locally small from the right side")? I mean, it is well-powered, of course, since for any ring A one can easily find a skeleton in the category Mono(A) (of all monomorphisms with values in A) and this will be a set (the set of all subrings in A). But is it true, that it is co-well-powered, i.e. for any ring A there exists a skeleton in the category Epi(A) (of all epimorphisms from A into other rings) and is this skeleton again a set?</p> <p>Thank you in advance, Sergei Akbarov</p> http://mathoverflow.net/questions/80146/is-the-mendeleev-table-explained-in-quantum-mechanics/80155#80155 Comment by Sergei Akbarov Sergei Akbarov 2013-06-20T11:38:45Z 2013-06-20T11:38:45Z Ben, you should explain yourself, this sounds strong: &quot;The answer to 1 is yes&quot;. And this also: &quot;physical theories aren't axiomatic systems&quot;. What about classical mechanics? Or probability theory? http://mathoverflow.net/questions/133030/defining-a-topology-in-the-power-set Comment by Sergei Akbarov Sergei Akbarov 2013-06-07T06:35:26Z 2013-06-07T06:35:26Z A remark to the construction of Dan Ramras: it becomes much more interesting if you endow the two point space, let us denote it by $2=\{0,1\}$, with the connected topology, where $\{0\}$ is closed and $\{1\}$ is open. Then the pre-image of $\{0\}$ is closed in $T$, and the pre-image of $\{1\}$ is open. And the set $2^T$ of maps $f:T\to 2$ is in one-to-one correspondense with the set of all closed (/open) subsets in $T$, and you can endow $2^T$ with different interesting topologies. So actually, I think you should understand first, whether you need all subsets in $T$ or, say, just closed ones. http://mathoverflow.net/questions/131603/homomorphisms-preserving-constant-functions Comment by Sergei Akbarov Sergei Akbarov 2013-05-23T15:01:38Z 2013-05-23T15:01:38Z You should explain your notations (and the notion of homomorphism). http://mathoverflow.net/questions/127889/is-rigour-just-a-ritual-that-most-mathematicians-wish-to-get-rid-of-if-they-could/129799#129799 Comment by Sergei Akbarov Sergei Akbarov 2013-05-06T06:06:40Z 2013-05-06T06:06:40Z Fedja, too venomously in the last paragraph. In my opinion. http://mathoverflow.net/questions/129688/a-sufficient-condition-for-a-probability-measure-to-have-compact-support/129690#129690 Comment by Sergei Akbarov Sergei Akbarov 2013-05-05T10:58:24Z 2013-05-05T10:58:24Z No, actually, I don't know... http://mathoverflow.net/questions/129092/the-topology-of-power-series-ring Comment by Sergei Akbarov Sergei Akbarov 2013-04-29T16:50:19Z 2013-04-29T16:50:19Z A naive question: what is DVR? http://mathoverflow.net/questions/107650/axiom-of-global-choice/127472#127472 Comment by Sergei Akbarov Sergei Akbarov 2013-04-26T05:52:29Z 2013-04-26T05:52:29Z Joel, I understand that the choice for classes is stronger than the choice for sets, there is no need to persuade me. But (logically) this doesn't mean that everybody must use different names for these two axioms. There is some per cent of people (and I belonged to them) who (after reading Kelley) believe that there is just one axiom of choice, the axiom for classes of sets, and the other its formulations (for sets of sets, for countable sets of sets, etc.) are just special cases. For those people it was impossible to understand what was written in that Wikipedia article before 21/04/2013. http://mathoverflow.net/questions/107650/axiom-of-global-choice/127472#127472 Comment by Sergei Akbarov Sergei Akbarov 2013-04-24T19:41:52Z 2013-04-24T19:41:52Z &quot;I would regard it as a mistake to refer to both axioms as the &quot;axiom of choice&quot;&quot; - however, Kelley and many others do this in their texbooks. On reading that article in Wikipedia (the version of 7 months ago) one could think that a great theory was invented, an extension of G&#246;del-Bernays, something indeed powerful, which allows to operate with &quot;conglomerates of classes&quot;! Who could expect that the explanation of that peal will be so banal - just terminology: we'll just call this special version of choice a &quot;global choice&quot;, and everybody is free to think what he wants about &quot;conglomerates&quot;! http://mathoverflow.net/questions/107650/axiom-of-global-choice/127472#127472 Comment by Sergei Akbarov Sergei Akbarov 2013-04-24T18:58:44Z 2013-04-24T18:58:44Z And, Joel, I don't agree with you that this is completely standard material. I was also studying logic at the university in Moscow, and I never met the term &quot;global choice&quot;. What Fraenkel, Bar-Hillel and Levy call &quot;global choice&quot; our professors (and the authors of the books we were reading) called just &quot;choice&quot;. To say nothing about &quot;conglomerates of classes&quot;. http://mathoverflow.net/questions/107650/axiom-of-global-choice/127472#127472 Comment by Sergei Akbarov Sergei Akbarov 2013-04-24T18:50:05Z 2013-04-24T18:50:05Z The more so that even specialists can't explain some details in those hints. For example, I would be very grateful to you, if you could add also some references where the term which attracted me from the very beginning is explained -- the &quot;conglomerates of classes&quot;. It was mentioned in the Wikipedia article 7 months ago when I created this thread, and for me, an external observer, the impression was that it is exactly &quot;global choice&quot; which describes the properties of those &quot;conglomerates&quot;. But up to now nobody explained the meaning of this term. http://mathoverflow.net/questions/107650/axiom-of-global-choice/127472#127472 Comment by Sergei Akbarov Sergei Akbarov 2013-04-24T18:32:59Z 2013-04-24T18:32:59Z Joel, when making impression of this topic I followed the references of the article in Wikipedia devoted to this &quot;Axiom of global choice&quot;. That was before 21/04/2013 when I initiated that investigation on whether this term indeed has sense. There were only two textbooks there, namely, By Jech and by Kelley. None of them contains this term &quot;global choice&quot;. My idea was that when people explain something mathematical it is their duty to provide the reader the necessary references, it's not the reader who must catch their hints. http://mathoverflow.net/questions/107650/axiom-of-global-choice Comment by Sergei Akbarov Sergei Akbarov 2013-04-21T17:05:45Z 2013-04-21T17:05:45Z doesn't matter! http://mathoverflow.net/questions/107650/axiom-of-global-choice Comment by Sergei Akbarov Sergei Akbarov 2013-04-21T16:59:47Z 2013-04-21T16:59:47Z Asaf, you exaggerate my ignorance. The problem was that instead of writing simply and clearly that &quot;usual choice&quot; is for sets of sets, while &quot;global choice&quot; is for classes of sets, they pretended that there is a &quot;more powerful choice&quot; for &quot;collections of classes&quot; or for &quot;conglomerates of classes&quot; – that was the main puzzle, and finally it turned out to be a hoax. The aim of my initiative in Wikipedia now was to force them to throw away any rudiments of these sermons. And they did it at last. – http://mathoverflow.net/questions/107650/axiom-of-global-choice Comment by Sergei Akbarov Sergei Akbarov 2013-04-21T12:26:15Z 2013-04-21T12:26:15Z @Asaf: In Wikipedia <a href="http://en.wikipedia.org/wiki/Wikipedia:Articles_for_deletion/Axiom_of_global_choice#Axiom_of_global_choice" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a> they have just explained me that &quot;axiom of global choice&quot; is exactly what Kelley calls &quot;axiom of choice&quot;. It is a great surprise for me that people used to divide this general result into classes of special statements depending on different &quot;Universes&quot;. http://mathoverflow.net/questions/107650/axiom-of-global-choice/128226#128226 Comment by Sergei Akbarov Sergei Akbarov 2013-04-21T10:35:07Z 2013-04-21T10:35:07Z Appendix of SGA? What is it? In which translation is this? English?