User nn - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-19T01:02:58Zhttp://mathoverflow.net/feeds/user/24527http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/100981/ultrainfinitism-or-a-step-beyond-the-transfinite/101016#101016Answer by NN for ULTRAINFINITISM, or a step beyond the transfiniteNN2012-06-30T20:07:45Z2012-06-30T20:07:45Z<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#100971Answer by NN for Is there existing terminology for this technical condition on semilattices?NN2012-06-29T19:05:26Z2012-06-29T19:05:26Z<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#100966Answer by NN for Examples of conjectures that were widely believed to be true but later proved falseNN2012-06-29T17:48:24Z2012-06-29T17:48:24Z<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#100917Answer by NN for nontrivial isomorphisms of categoriesNN2012-06-29T09:09:17Z2012-06-29T09:09:17Z<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#100723Answer by NN for Name of a lattice-propertyNN2012-06-26T21:29:23Z2012-06-26T21:29:23Z<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#100720Answer by NN for Product-Decomposition of distributive latticesNN2012-06-26T21:04:51Z2012-06-26T21:04:51Z<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#100091Answer by NN for RS to RSK correspondenceNN2012-06-20T07:47:04Z2012-06-22T12:29:35Z<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#100345Answer 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 NN2012-06-22T11:06:59Z2012-06-22T12:12:21Z<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#100337Answer by NN for Conditions for non-triviality of Caratheodory measureNN2012-06-22T10:00:58Z2012-06-22T10:00:58Z<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#100336Answer by NN for Caratheodory and RieszNN2012-06-22T09:20:53Z2012-06-22T09:20:53Z<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#100058Answer by NN for Existence of an arbitrary Small positive continuous real Valued FunctionNN2012-06-19T23:08:23Z2012-06-20T23:15:43Z<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 < f < 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>$h<g$</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>$0<f<h$</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#100088Answer by NN for Are there infinite groups which have only a finite number of irreducible representations?NN2012-06-20T07:29:57Z2012-06-20T07:29:57Z<p>What happens for infinite simple groups?</p>
http://mathoverflow.net/questions/99831/total-ring-of-fractions-vs-localization/99982#99982Answer by NN for Total ring of fractions vs. LocalizationNN2012-06-19T09:30:36Z2012-06-19T19:51:11Z<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#99903Answer by NN for Examples of noncommutative analogs outside operator algebras?NN2012-06-18T14:36:37Z2012-06-18T14:36:37Z<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 (Ćukasiewicz 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#99872Answer by NN for Elementary proof wanted: every local principal ideal ring is a quotient of a PIDNN2012-06-18T05:43:12Z2012-06-18T05:43:12Z<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-spacesComment by NNNN2012-07-03T13:03:24Z2012-07-03T13:03:24ZAscoli 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-orderComment by NNNN2012-07-01T19:26:50Z2012-07-01T19:26:50ZInstead 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-aComment by NNNN2012-07-01T05:51:27Z2012-07-01T05:51:27ZTo check whether it is true, I would try to use the
Jordan normal form of a matrixhttp://mathoverflow.net/questions/100981/ultrainfinitism-or-a-step-beyond-the-transfinite/101016#101016Comment by NNNN2012-07-01T05:23:55Z2012-07-01T05:23:55ZThe 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#101016Comment by NNNN2012-06-30T20:58:55Z2012-06-30T20:58:55ZFor 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 "one level up", 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-approximationComment by NNNN2012-06-27T19:48:38Z2012-06-27T19:48:38ZI would start with the ultra-classic Hardy - Wright.
I vaguely remember some related discussion in the
final notes of the relevant chapters, with referenceshttp://mathoverflow.net/questions/100692/equicontinuity-of-continuous-families-of-maps-between-topological-vector-spacesComment by NNNN2012-06-27T17:28:06Z2012-06-27T17:28:06ZI 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#100723Comment by NNNN2012-06-27T14:13:57Z2012-06-27T14:13:57Z<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-semilatticesComment by NNNN2012-06-26T23:50:11Z2012-06-26T23:50:11ZFirst, 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 "finite breadth" 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#99387Comment by NNNN2012-06-26T23:14:27Z2012-06-26T23:14:27ZGleason 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.