User nn - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:02:58Z http://mathoverflow.net/feeds/user/24527 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100981/ultrainfinitism-or-a-step-beyond-the-transfinite/101016#101016 Answer by NN for ULTRAINFINITISM, or a step beyond the transfinite NN 2012-06-30T20:07:45Z 2012-06-30T20:07:45Z <p>You might like the peculiar set theory NFU (Quine New Foundation, but with atoms) extended with axioms which are quite natural in the NFU context and that turn out to be equiconsistent with ZFC plus suitable large cardinal axioms. You can follow the wikipedia page about NF to reach the references. In this world, the strongly cantorian sets make a natural model for ZFC plus large cardinals, and there is a universal set (yes, set, not a proper class).</p> <p>However, I sometimes dream of something even stronger. Extend ZF (without choice) with something like Reinhardt cardinals, the ones that are incompatible with choice by Kuhnen's proof (even recent attempts were unable to show incompatibility with ZF, even if incompatibility seems not far away). Then this should correspond, in the NFU world, to something where the atoms this time are not much more than sets (something that must happen in NFU with choice, by Specker's refutation of choice in NF), so that suitable models of ZF with suitable Reinhardt cardinals should correspond to suitable models of NFU with so many sets (in comparation to atoms) that a model of NF (without atoms!) should be possible (consistency of NF without atoms relative to some standard set theory is an open problem).</p> <p>This would be a world where extremely large sets exist, so large that choice functions in the largest collections cannot exists (in italian I would say "assioma dell'imbarazzo della scelta", I have no idea of a proper english translation). A world where Specker's refutation of AC in NF corresponds to Kunen's refutation of AC in ZF plus Reinhardt cardinals (despite the fact that the sequences of cardinals which the two proofs use go in opposite directions). A world that actual set theorists do not consider as "real" (they like choice too universally to restrict it to only to an initial segment of the universe; to model failures of AC they prefer inner models rather than extensions), a world whose consistency is infact unknown. But you asked for people with strong faith in the strong infinity ... [incidentally, bishop Berkley would have been happy with Soloway - Shelah theorem about Lebesgue measurability of every set of reals: he probably would have said that an Analyst can chose to live in a choiceless world, if he like so, but can do this reasonably iff he has faith in the inaccessible infinity]</p> http://mathoverflow.net/questions/83111/is-there-existing-terminology-for-this-technical-condition-on-semilattices/100971#100971 Answer by NN for Is there existing terminology for this technical condition on semilattices? NN 2012-06-29T19:05:26Z 2012-06-29T19:05:26Z <p>in lattice theoretic terms, the condition is "finite breadth" (more on this in the comment above and with search engine requests for "breadth lattice")</p> http://mathoverflow.net/questions/95865/examples-of-conjectures-that-were-widely-believed-to-be-true-but-later-proved-fal/100966#100966 Answer by NN for Examples of conjectures that were widely believed to be true but later proved false NN 2012-06-29T17:48:24Z 2012-06-29T17:48:24Z <p>Two examples from lattice theory: is every lattice with unique complements distributive? [no] is every distributive algebraic lattice isomorphic to the lattice of congruences of a lattice? [no] See <a href="http://www.ams.org/notices/200706/tx070600696p.pdf" rel="nofollow">http://www.ams.org/notices/200706/tx070600696p.pdf</a></p> http://mathoverflow.net/questions/13995/nontrivial-isomorphisms-of-categories/100917#100917 Answer by NN for nontrivial isomorphisms of categories NN 2012-06-29T09:09:17Z 2012-06-29T09:09:17Z <p>This is really a comment, not an answer. But since it is a not-so-short comment to many answers together, it had to become an answer.</p> <p>It has been observed that (first-order) definitional equivalences give categorical isomorphisms, at least for categories of first order structures with isomorphisms as their only morphisms. In my opinion the fact that two equivalent definitions of a mathematical structure give the same isomorphisms but possibly different morphisms (which maps between [complete] lattices should one consider: isotone? meet-semilattice morphisms? join-semilattice morphisms? lattice morphisms? [complete join, or meet, or both, morphisms?]) is a big virtue: it means that two definitions give really different points of view on the same kind of structure (in a way, they formalize a kind of non-triviality of the equivalence). This also happens for second-order structures (complete lattices, uniform and topological spaces); the definitional equivalences are expressed in the natural language of Bourbaki's "scale of sets" (or natural model of type theory) above the base sets of the (multisorted) structure (detractors of Bourbaki and/or lowers of category theory would instead speak of the topos "somewhat freely" generated by the (sorts for the) base stets; when the equivalence of definitions is completely constructive one can really take a free topos, but depending of the principles of classical logic which are needed to prove the equivalence of definitions, one considers the topos freely generated in more restricted classes). </p> <p>So in summary: syntactically defined equivalences induce isomorphisms between categories of structures. As Hodges notes (for example in his book "model theory"), pratically everything which in mathematics can "really" be considered a "construction" is formalizable as a interpretation or at least a "word-construction" (and moreover it is the syntactical form itself which shows what kind of morphisms more general than isomorphisms are "preserved" by the construction. I understand that few lovers of category theory would approve such a extreme syntactical view, but note that even the "categories, allegories" book by Freyd and Scedrov insists on the "Galois correspondence" between syntactical and semantical aspects; I simply happen to prefer the syntactical side). From this point of view, Hodges'remarks about (cases slightly more general than) adjunctions among quasivarieties (and universal Horn classes) induced by forgetful functors are related to the already given remark about monadic adjunctions.</p> <p>Besides, the book "abstract and concrete categories" by Adameck, Herrlich, Strecker conains many examples of "concrete isomorphisms"; some of them shoulb be interesting (and all of them, if I remember correctly, can be seen as syntactically defined as above). </p> <p>Incidentally, the three authors say that non reasonable concept of "concrete equivalence" can be given; I disagree since cases exist where two categories can be concretely reflected on full subcateories of objects "in normal form", and the subcategories are concretely isomorphic [for example, take affine geometry of dimension at least three: form affine spaces algebraically defined by points, group of translations, sfield of scalars one "normalizes" to the particular case where translations are a subgroup of the group of permutations of the points and scalars are a subring of the ring of endomorphisms of the group of translations. For affine spaces geometrically defined in Hilbert's Grundlagen style, the general case can be reflected onto the "normal" case with the same set of points where lines and planes are sets of points and incidence is the set-theoretic one]</p> <p>It has already been observed that, in presence of choice, "isomorphic categories" means "equivalent categories where corresponding isomorphism classes of objects have the same cardinality". Freyd and Scedrov observe that, even in absence of choice, the "correct" notion of equivalence is: to have isomorphic inflations. This means that all usual examples of equivalence of categories induce examples of isomorphisms (without the trick with arbitrary choices to consider skeletons, but instead using canonical "inflations" of the isomorphism classes)</p> http://mathoverflow.net/questions/88516/name-of-a-lattice-property/100723#100723 Answer by NN for Name of a lattice-property NN 2012-06-26T21:29:23Z 2012-06-26T21:29:23Z <p>Faigle and Herrmann call them point-lattices. They are useful in the modular, algebraic case (Faigle embedding theorem) and also more generally in the (strongly) semi-modular algebraic case (generalized matroid lattices).</p> http://mathoverflow.net/questions/97844/product-decomposition-of-distributive-lattices/100720#100720 Answer by NN for Product-Decomposition of distributive lattices NN 2012-06-26T21:04:51Z 2012-06-26T21:04:51Z <p>The following can surely be found in Birkoff, lattice theory. Almost surely also in Gratzer.</p> <p>The decompositions of a poset in finite direct products are the same as the "partitions of unity" in a certain Boolean algebra. When the poset has universal bounds (a mimimum element 0 and a maximum element 1) the Boolean algebra is the Boolean algebra of central elements of the poset. When the poset is a distibutive lattice with 0 and 1, the central elements are exactly the complemented elements of the lattice, and so the center is the largest Boolean subalgebra of the lattice. The poset is a direct product of n directly indecomposable components iff the center is a Boolean algebra with N atoms. So your claims are correct, with the N atoms of the center being the N elements of the distributive lattice which are complemented but are not <em>disjoint</em> union of two smaller nonzero elements. However, also in the finite case, an element can be directly irreducible without being join irreducible; consider the following Hasse diagram:</p> <p>1 / \ a b \ / c | 0</p> <p>i.e. the length 3 distributive lattice with two co-atoms a,b and one atom c. In the dual of the above lattice, join irreducible elements are indecomposable. In a finite distributive lattice, coincidence of join irreducible elements with directly indecomposable elements, plus the dual condition, happens iff the lattice is a direct product of chains. </p> <p>So I do not see real advantages in the use of join irreducible elements in comparation with indecomposable ones to describe direct decompositions. (But note that I never have been interested in combinatorics, so that you might see things differently for your specific application) </p> <p>As you note, to express the results in terms of the poset of join-irreducible elements (instead of the center of the lattice) you can use the Bikhoff transform. If you want to use Birkhoff transform (categorical dual equivalence between finite posets and finite distributive lattices, so that disjoint unions of posets [of join-irreducible elements] correspond to direct products for the lattices), the infinite case is the following (and in particular note that it does not apply to all distributive lattices, only to special ones): </p> <p>posets are dually equivalent to algebraic and dually algebraic distributive lattices and also to Alexandroff discrete topological spaces: the poset is the poset of points with the specialization order; the lattice is the lattice of open sets. Also, the lattice is the lattice of order ideals in the poset, and the poset is the poset of (completely) join-irreducible elements of the lattice. Central elements of the lattice correspond to clopen sets of the topological space.</p> http://mathoverflow.net/questions/100085/rs-to-rsk-correspondence/100091#100091 Answer by NN for RS to RSK correspondence NN 2012-06-20T07:47:04Z 2012-06-22T12:29:35Z <p>Warning: I do not know nothing about combinatorial problems, so what I say now might be completely wrong. However: Rota in his talk at the Birkhoff memorial conference (The many lives of lattice theory, easily available online) has, in the section about semiprimary lattices, something which seemes strongly related.</p> <p>Edit: citing from the article: each of the two chains is associated with a standard Young tableau, hence we obtain the statement and proof of the Schensted algorithm, which precisely associates a pair of standard Young tableaux to every permutation.</p> http://mathoverflow.net/questions/100342/are-there-subsets-l-in-rn-such-that-it-is-easy-to-find-closest-point-in-l-to-a/100345#100345 Answer by NN for Are there subsets L in R^n such that it is "easy to find" closest point in L to a given P in R^n ? Vague question motivated by error-correcting codes NN 2012-06-22T11:06:59Z 2012-06-22T12:12:21Z <p>From your requirements I cannot understand whether this qualifies: point that realizes the distance between a closed convex set and a given point (distance for a complete uniformly convex norm). Perhaps you should be more specific in your question.</p> <p>Edit: I do not understand the meaning of "easy" for you. But I also know pratically nothing about coding theory. If you clarify "easy" for your context, perhaps someone else can give you more useful answers.</p> http://mathoverflow.net/questions/95814/conditions-for-non-triviality-of-caratheodory-measure/100337#100337 Answer by NN for Conditions for non-triviality of Caratheodory measure NN 2012-06-22T10:00:58Z 2012-06-22T10:00:58Z <p>The trivial necessary and sufficient condition is that the initially given external measure is the external measure of some sigma-measure. In other terms, a closed object for the Galois correspondence between positive sigma-measures (on sigma-algebras) and positive external measures (defined on all subsets). Is this trivial condition completely useless? Well, you learn that "regular" external measures are the essentially same as "comp0lete" measures. You can also obtain that when you start from a finitely additive measure on a semiring, or a lattice, of sets, then all the initially given measurable sets (and even the Peano - Jordan measurable ones) are still measurable for the sigma-measure that you obtain with the two-step Caratheodory process. Some books on measure theory also consider the case of a initially given finitely subadditive measure (on a ring of sets).</p> <p>There is another process, again with a Galois correspondence, to "bi-complete" a measure: not only you want that a subset of a zero-measure set again has measure zero, but also that if a set has infinite measure then it has a subset with finite nonzero measure (or, equivalently, the measure is the sup of the measure of the integrable subsets).</p> <p>I think that standard measure theory text should have all this (Fremlin should be freely available online). A more sophisticated question is the distinction between the above measures (complete and locally finite) and the measures that are direct sum of finite measures. A sigma-finite and complete measure is a direct sum of finite complete measures and a direct sum of finite measures is complete and locally finite, but neither of the implications can be reversed (the counterexamples for the first are very easy, for the second they are not). In standard books, you can check also the relation of the above conditions with the following possible properties of a complete (positive sigma-additive) measure: (a) the Boolean algebra of measurable sets modulo measure zero sets is complete; (a') an analogue for the vector lattice of measurable functions modulo almost everywhere zero functions; (b) the natural duality between integrable and bounded measurable functions gives the dual of the Banach space of integrable functions (modulo "almost everywhere". The dual of bounded measurable functions is never reduced to the integrable ones [i.e. absolutely continuous sigma-measures], except for trivial [finite dimensional] cases; the dual of the other usual spaces is the "expected one" for all complete measures); (c) there is a "lifting" for the bounded measurable functions modulo "almost everywhere" to the bounded measurable functions. [Hint: (c) is subtly different from the preceding properties, check in the standard books]</p> <p>I added this last part about "locally finite" measures becouse it is related to the other question of yours, but note that it is not exactly the same as the Bourbaki distinction (some measure theory books write about the distinction between the tau extension and the sigma-extension for a Daniell integral). I hope that with all these keywords you can check in your measure theory books.</p> http://mathoverflow.net/questions/100327/caratheodory-and-riesz/100336#100336 Answer by NN for Caratheodory and Riesz NN 2012-06-22T09:20:53Z 2012-06-22T09:20:53Z <p>I might misunderstand your question, but it seems that you are asking about the distinction between the measure and "essential measure" in Bourbaki measure theory ("presque partut" vs. "localment presque partut"). In locally compact paracompact spaces there is not such a "byzantine" distinction (where the term "byzantine" is used by A.Weil is the comments in his collected papers). However, the collections of measurable sets is the same for the two measures; essentially, the "local" version is a "smaller" measure with some sets with infinite (total variation) measure for the initial measure which became of (total variation) measure zero for the local version.</p> http://mathoverflow.net/questions/100032/existence-of-an-arbitrary-small-positive-continuous-real-valued-function/100058#100058 Answer by NN for Existence of an arbitrary Small positive continuous real Valued Function NN 2012-06-19T23:08:23Z 2012-06-20T23:15:43Z <p>So you are asking: which topological spaces, besides the discrete ones, are such that for every strictly positive real function $g$ there is a strictly positive continuous real function $f$ with $ 0 &lt; f &lt; g $? Only some hints.</p> <p>Others have already noted that there cannot be nontrivial convergent sequences.</p> <p>You can note that if there are no nontrivial convergent sequences, then replace $g$ with the largest <code>$h&lt;g$</code> which takes only values of the form $1/n$, and then note that this $h$, even if not continuous, at least does not give the above noted problem (forcing a possible continuous $f$ to have value $0$). Taking the closures of the sets where $h>1/n$ one can then use Urhyson's lemma / Tietze extension theorem (when the space is normal) and so obtain a continuous $f$ with <code>$0&lt;f&lt;h$</code>.</p> <p>Is there a non-discrete normal space with no nontrivial converging sequence? You can easily find the answer in any standard book on general topology which treats Stone compactification.</p> <p>Added: please replace "no nontrivial convergent sequences" with the stronger "every countable subset is closed". And examples are even more exotic, but Gillman and Jerison, rings of continuous functions, should have them (and perhaps also Engelking).</p> <p>Concerning TeX commands: I do not use them on purpose, but if it this the rule to always use them here, then I will in future only give answers that do not require mathematical notation, so that we all will be able to read in the way we like.</p> http://mathoverflow.net/questions/100087/are-there-infinite-groups-which-have-only-a-finite-number-of-irreducible-represen/100088#100088 Answer by NN for Are there infinite groups which have only a finite number of irreducible representations? NN 2012-06-20T07:29:57Z 2012-06-20T07:29:57Z <p>What happens for infinite simple groups?</p> http://mathoverflow.net/questions/99831/total-ring-of-fractions-vs-localization/99982#99982 Answer by NN for Total ring of fractions vs. Localization NN 2012-06-19T09:30:36Z 2012-06-19T19:51:11Z <p>[I might have misunderstood your equivalent question]. Some hints: a commutative reduced ring is the same as a subring of a direct product of fields (i.e. a ring of field-valued functions, but the co-domain can change with the point); you also wants that never zero functions have an inverse. So typically one considers continuous, of differentiable (or whatever) real valued functions. Then you want to invert also a function f that is sometimes zero; this "kills" the zero set Z of f (the localization is a ring of functions defined outside Z). But then you want that every g which is nonzero outside Z has a inverse, i.e. inverting one particular f should invert every g. Taking the sequences which have limit (i.e. the continuous functions on the Alexandroff compactification of the naturals), suppose that you want to invert f(n)=1/n ; does this invert all positive and converging to zero sequences? Inverting f gives functions that are of polynomial growth when n tends to infinity, so what happens for a g with exponential decrease?</p> <p>Added: as it was kindly noted, it happens that the ring of continuous functions (on a generic completely regular space) is not a classical ring of quotients; its ring of quotients is obtained considering functions which are definite and continuous on a dense co-zero set (dense open set in the normal case), two such functions being identified when they coincide on a dense co-zero set. But then this ring of quotients is von Neumann regular (and it is naturally identified with the ring of continuous functions on a certain almost P-space associated to the original space); being von Neumann regular there is no hope to obtain the kind of example wanted (as it was implicitly remarked by the requester). However, the idea that "regular" functions can be modified on a ideal of "small" sets (in this case, the sets whose closure has empty interior) still works, and in another answer it was well used (with polynomial functions as regular functions and finite sets as ideal of small sets. One can use other choices of regular functions and/or small sets: meager sets, measure zero sets, ... however, for algebraic geometry, polynomials and finite sets are the most natural choices)</p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/99903#99903 Answer by NN for Examples of noncommutative analogs outside operator algebras? NN 2012-06-18T14:36:37Z 2012-06-18T14:36:37Z <p>Some non-commutative analogs in lattice theory. </p> <p>von Neumann's coordinatization theorem is the non-commutative analogue of Stone's definitional equivalence between Boolean algebras (complemented distributive lattices) and Boolean rings (associative rings with 1 where all elements are idempotent). </p> <p>The Baer - Inaba - Jonnsson - Monk coordinatization theorem gives the non-commutative analogue of direct products of finite chains (Ɓukasiewicz propositional logics); for this one uses not the usual formulation of the theorem (which coordinatizes primary lattices with modules over a artinian ring where one-sided ideals are two-sided and form a chain), but a reformulation that gives a true equivalence between the lattice (of submodules of the module) and the ring (of endomorphisms of the module); this way one has a true analogue of the (dual) equivalence between commutative C^*-algebras and compact Hausdorff spaces (or better, their lattice of open sets). </p> <p>One has a common generalization of the two cases above: equivalence between lattices of subobjects and rings of endomorphisms for finitely presented modules (of geometric dimension at least 3) over a "auxiliary" ring which is WQF (weakly quasi Frobenius, a.k.a. IF, injectives are flat). The categories of finitely presented modules over such auxiliary rings are exactly (up to equivalence) the abelian categories with an object which is injective, projective and finitely generates and finitely cogenerates every object.</p> <p>The ultimate generalization is G.Hutchinson's coordinatization theorem, a correspondence between arbitrary abelian categories and modular lattices with 0 where each element can be doubled and intervals are projective (in lattice theory, sense i.e. the classical projective geometry meaning) to initial intervals. When one looks at this theorem together with the Freyd - Mitchell embedding theorem, one has that three languages are fully adequate and equivalent ways to do linear algebra: (1) the usual language of sums and products (modules over associative rings); (2) the language of category theory (abelian categories); (3) the old fashioned language of synthetic geometry of incidence (joins and meets in suitable modular lattices).</p> <p>In these equivalences, the lattices are the (pointless, noncommutative) spaces, and the rings are the rings of coordinates or functions over the space (I am not considering the distinction between equivalences and dual equivalences because I am more interested in structures, with their unique concept of isomorphism attached, rather than more general morphisms, which depend upon the particular way to define a structure. But the complementarity between the structural and categorical views, where neither subsumes the other, is another long theme).</p> <p>Since all these ideas have their origin in von Neumann works about continuous geometries and rings of operators, I now explain the relation with operator algebras.</p> <p>First note that the usual definition of pointless topological space as complete Heyting algebra is not a true generalization of the "topological space" concept: they are a true generalization of sober spaces, but to generalize topological spaces one must consider pairs: a complete boolean algebra (which in the atomic case is the same thing as a set) with a complete Heyting subalgebra (the lattice of open sets). To obtain the non-commutative analogue, complete boolean algebras are generalized to meet-continuous geometries, and algebras of measurable sets are replaced with suitable structures (projection ortholattices of von Neumann algebras) which are embedded in the meet-continuous geometries in the same way as a right nonsingular ring is embedded in its maximal ring of right fractions (a regular right self-injective ring).</p> <p>A meet continuous geometry is a complete lattice which is modular, complemented and meet distubutes over increasing joins (not arbitray joins, like Heyting algebras). These structures were introduced by von Neumann and Halperin in 1939; they are a common generalization of (possibly reducible) continuous geometries (the subcase where join distrubutes over decreasing meets) and (possibly reducible and infinite dimensional) projective geometries (the atomic subcase). For them one has a dimension and decomposition theory much like the one for continuous geometries and rings of operators (the theory of S.Maeda, in its last version of 1961, is sufficient; one does not need the 2003 theory by Wehrung and Goodearl). One can define the components of various types, in particular I_1 (the boolean component, i.e. classical logic), the I_2 component (the 2-distibutive component i.e. subdirect product of projective lines; physically these are "spin factors" and quantum-logically it is the non-classical component which nonetheless has non-contextual hidden variables), the I_3 nonarguesian component (subdirect product of projective nonarguesian planes i.e. irreducible projective geometries that cannot be embedded in larger irreducible projective geometries; quantum-logically this means that interacion with other components is only possible classically, without superposition). Once these bad low dimensional components are disregarded, von Neumann coordinatization theorem gives a equivalence between the meet-continuos geometries and the right self-injective von Neumann regular rings. So meet-continuous geometries are pointeless quantum (i.e. non-commutative) sets in the same way as complete Boolean algebras are (commutative) pointless sets. Regular rings are the coordinate rings of these quantum sets in the same way as (commutative, regular) rings of step functions (with values in a field) are the ring equivalent of a boolean algebra (classical propositional logic); the important new fact is that in the "truly non-commutative case" the regular ring is uniquely and canonically determined by the lattice (in the distributive case, on the contrary, it is not: one can use step functions with values in any field, and one can change the field with the point; commutative [resp. strongly] regular rings are the subrings of direct products of [skew] fields which are stable for the generalized inverse operation).</p> <p>The above "propositional logics" are without the negation operator; on the other hand, the projection ortholattices of von Neumann algebras are complete orthomodular lattices with sufficiently many completely additive probability measures and sufficiently many internal simmetries (von Neumann said that the strict logic of orthocomplementation and the probability logic of the states uniquely determine each other by means of his symmetry axioms in his characterization of finite factors as continuos geometries with a transition probability. One should also note how much more physically meaningful are von Neumann axioms when compared with the "modern" ones based on Soler's theorem, but this is another large topic). Using Gleason's theorem (and as always in absence of the bad low-dimensional components) one obtains an equivalence between von Neumann's "rings of operators" (i.e. <em>real</em> von Neumann algebras, or their self-adjoint part, real JBW-algebras) and their projection ortholattices (the normal measures on the ortholattice give the predual of the ring of operators). One can see these logics inside a meet-continuos geometry by equipping the geometry with a linear orthogonality relation which has for each element a maximum orthogonal element (pseudo-orthocomplementation in part analogous to "external" in a stonean topological space, in part anti-analogous as it happens with Lowere closure when compared to Kuratowski closure). The regular ring of the lattice is the ring generated by all complementary pairs in the lattice (which are the idempotents of the ring: kernel and image) with the relations corresponding to the partial operation e+f-ef which is defined on idempotents whenever fe=0 (at the lattice level this partial operation is implemented with disjoint join of the images and co-disjoint meet of the kernels); in the case associated to a "ring of operators", this regular ring is the ring of maximal right quotients of the von Neumann algebra, and conversely the algebra is recovered from the lattice with orthogonality by taking the subring generated by orthogonal projections (idempotents whose kernel and image are orthogonal); by a theorem of Berberian the algebra is ring generated by its self-adjoint idempotents, and there is clearly at most one involution on the algebra which fixes such generators.</p> <p>Hence, in summary, in absence of bad low dimensional components (whose exclusion is physically meaningful, see their meanings above) one has equivalences between the following concepts:</p> <p>(0) right self-injective regular rings with a suitable additional structure (to associate a orthogonal projection onto the closure of the image to any element) (1) real von Neumann algebras (2) real JBW-algebras (the Jordan algebra of self-adjoint operators i.e. observables) (3) the effect algebra (of operators with spectrum in [0,1]) i.e. unsharp quantum logic (4) the projection ortholattice (the sharp quantum logic) (5) the pointeless quantum set (meet-continuous geometry) with a suitable orthogonality (6) the convex compact set of normal states</p> <p>Any of the above is a adequate starting point for quantum foundation since all the other points of view can be canonically recovered. </p> <p>All this is restricted to the level of non-commutative measure spaces; a locally compact topological space is something more precise (like a C^* algebra when compared to a von Neumann algebra), and a (Riemannian) metric space is something still more precise (and in particular a differentiable structure, which can be seen as an equivalence class of riemannian structures: note that a isometry for the geodesic metric between complete Riemannian manifolds is automatically differentiable, so the differentiable structure must be definable from the geodesic metric, and infact Busemann and Menger had such a explicit definition of the tangent spaces from the global metric. The topological structure is then another equivalence class of metrics, for another weaker equivalence).</p> <p>Given the equivalence (0)--(6) above, one can note that all the concepts which are used in Connes definition of spectral triples, and analogues structures, can be seen equivalently from each of the above points of view. In this way, all of the above points of view are a possible starting point for non-commutative geometry.</p> <p>[Sorry for the too long post and for my bad pseudo-english language]</p> http://mathoverflow.net/questions/25663/elementary-proof-wanted-every-local-principal-ideal-ring-is-a-quotient-of-a-pid/99872#99872 Answer by NN for Elementary proof wanted: every local principal ideal ring is a quotient of a PID NN 2012-06-18T05:43:12Z 2012-06-18T05:43:12Z <p>Theorem 5.2 in <a href="http://www.emis.de/journals/BAG/vol.46/no.1/b46h1her.pdf" rel="nofollow">http://www.emis.de/journals/BAG/vol.46/no.1/b46h1her.pdf</a> gives an answer (take the projective limit. The paper has a related one with corrections, but not for the part that is related to your question). This is for a non-commutative case, and the theorem has a non-commutative extension: a PIR is a finite direct product of prime and artinian indecomposable cases, which are matrix rings over CPU rings (Faith, Algebra II should contain all the needed references)</p> http://mathoverflow.net/questions/100692/equicontinuity-of-continuous-families-of-maps-between-topological-vector-spaces Comment by NN NN 2012-07-03T13:03:24Z 2012-07-03T13:03:24Z Ascoli theorem in Engelking, general topology, where domain is a k-space. Note that X and Y can be supposed to be complete (linear functions extend to the completion). However, I forgot that the k-modification of the weak topology is something well known to be useful only for Banach spaces (Eberlein-Smullyan). In any case, the obtained equicontinuity is for a different topology on X (I realize now that you can change the topology on the space of functions, not in X). Sorry (also for the delay; I had lost the question until a posted answer re-put today the question in the first page). http://mathoverflow.net/questions/101088/lexicographical-order Comment by NN NN 2012-07-01T19:26:50Z 2012-07-01T19:26:50Z Instead of a direct answer, a universal pseudo-algorithm to extend definitions for posets to preorders: apply the definition to the associated posets (identify elements that are related in the preorder and its dual) then apply the map from the set construction (here the cartesian product of the two sets where the preorders are defined) to the poset construction (here the lexicografical ordering). Note: a preorder is essentially the same thing as a surjection from a set to a poset. http://mathoverflow.net/questions/101023/ad-an-is-a-polynomial-in-ad-a Comment by NN NN 2012-07-01T05:51:27Z 2012-07-01T05:51:27Z To check whether it is true, I would try to use the Jordan normal form of a matrix http://mathoverflow.net/questions/100981/ultrainfinitism-or-a-step-beyond-the-transfinite/101016#101016 Comment by NN NN 2012-07-01T05:23:55Z 2012-07-01T05:23:55Z The problem is not forcing vs. inner models. The problem is thinking that the universe should satisfy AC, or even ZF (separation implies that Russell class is big, but it's not so in NFU). Perhaps the best way to produce natural models of the many set theories is abandoning the idea that ZFC axioms are really universal. At the moment I like type theory with typical ambiguity, plus universes (with the axiom that Tarski explicitly noted to be compatible with type theory). In summary: to realize the project of the requester, abandon either AC or even ZF for the global theory of the extension. http://mathoverflow.net/questions/100981/ultrainfinitism-or-a-step-beyond-the-transfinite/101016#101016 Comment by NN NN 2012-06-30T20:58:55Z 2012-06-30T20:58:55Z For the very little I know, even now it might be that NF (without atoms) is inconsistent, or that on the opposite it is as weak as Zermelo (without replacement). My way to look at NF(U) is however by means of 3 level type theory with type-level pairs, plus typical ambiguity. Example: the structure of real numbers is naturally obtained &quot;one level up&quot;, then a structure of elements must exist by typical ambiguity. Note: category theorists do not know <i>why</i> sets - classes - conglomerates are sufficient, but in NF one knows why. http://mathoverflow.net/questions/100799/liouvilles-theorem-in-diophantine-approximation Comment by NN NN 2012-06-27T19:48:38Z 2012-06-27T19:48:38Z I would start with the ultra-classic Hardy - Wright. I vaguely remember some related discussion in the final notes of the relevant chapters, with references http://mathoverflow.net/questions/100692/equicontinuity-of-continuous-families-of-maps-between-topological-vector-spaces Comment by NN NN 2012-06-27T17:28:06Z 2012-06-27T17:28:06Z I have not checked, what follows might be completely wrong. H is compact in a space of continuous functions, subspace of the continuous functions on X (with the weak topology) with the compact-open topology. Use Ascoli - Arzela. http://mathoverflow.net/questions/88516/name-of-a-lattice-property/100723#100723 Comment by NN NN 2012-06-27T14:13:57Z 2012-06-27T14:13:57Z <a href="http://www.springerlink.com/content/h38122511085g364/" rel="nofollow">springerlink.com/content/h38122511085g364</a> http://mathoverflow.net/questions/83111/is-there-existing-terminology-for-this-technical-condition-on-semilattices Comment by NN NN 2012-06-26T23:50:11Z 2012-06-26T23:50:11Z First, note that finitely generated join-semi-lattices with 0 are finite lattices. Then check the definition of lattice of breadth N, and conclude: you are exactly considering semi-lattices of breadth at most N. So call them semi-lattices of finite breadth. PS: nice, Hyers - Ulam stability for multiplicative homomorpisms on semigroups. Curiously (or perhaps not so curiously), for additive functions on groups there is an analogue &quot;finite breadth&quot; sufficient condition for stability (every element of the derived subgroup is product of at most N commutators) http://mathoverflow.net/questions/64492/continuity-of-borel-measurable-gleason-frame-functions/99387#99387 Comment by NN NN 2012-06-26T23:14:27Z 2012-06-26T23:14:27Z Gleason theorem (for von Neumann algebras, and even JBW algebras, without type I_2 part) has been extended to signed / complex measures (and even suitable vector valued measures). S.Maeda, Bunce and Maitland wright wrote surveys around 20 years ago.