User qfwfq - MathOverflow

Affinization and properization of algebraic varieties ?

2010-04-23T09:41:23Z

Given an algebraic variety $X$, I'm asking about the existence of a variety $X^{aff}$ and an "affinization" morphism

$f:X\rightarrow X^{aff}$

such that:

(a1) $f$ is injective when restricted to closed connected affine subvarieties of $X$,
(a2) Complete connected subvarieties of $X$ get shrinked by $f$ to points.

Condition (a2) may be strenghtened to

(a2') $f(x)=f(y)$ iff $x$ and $y$ are not separated by a regular function on $X$.

Then we may require that

(a3) whenever $f':X\rightarrow X'$ enjoys the above properties, then there is a closed embedding $j:X^{aff} \rightarrow X'$such that $f'=j\circ f$.

Q1: Does such an "affinization" morphism exist? Feel free to change the requirements (a1), (a2), (a3) in your aswer (i.e. answer a different question!), if it helps to get a better notion of what ought to be an "affinization" of $X$ (as my requirements may not be the best).

Q2: If it exists, is it unique?

I have something like a candidate for $f$, but I'm not sure the following makes sense. Consider the map

$X\rightarrow X^{aff}:=\operatorname{Spec} \mathcal{O}(X)=\operatorname{Spec}(H^{0}(X,\mathcal{O}_X))$

$x \mapsto \mathfrak{m}_x$

where $\mathfrak{m}_x$ is the ideal of functions in $\mathcal{O}(X)$ vanishing at $x$.

Q1': Is it even a morphism? Does this work as an "affinization" morphism?

An analogous question would involve a hypothetical "properization morphism"

$g:X\rightarrow X^{prop}$

such that:

(p1) $g$ is injective when restricted to complete connected subvarieties of $X$
(p2) closed connected affine subvarieties of $X$ get shrinked to points by $g$
(p3) an analogous "universal property" holds (if you like).

Condition (p2) maybe might be stregthened as:

(p2') closed quasi-affine varieties get shrinked to points.

Q3: Does such a "properization" morphism exist? Again, feel free to change my requirements in such a way that they meet a good heuristic definition of what a "properization morphism" should be, if it should exist at all.

Q4: In case it exists, what about uniqueness?

Then I could ask, in case of existence, if it would be the case that $f$ (resp. $g$) factors through a proper (resp. affine) morphism to $X^{aff}$ (resp. X^{prop}), but I feel that the above questions Qi are already sloppy enough! So, first I'm waiting for some answers or remarks that may point out some obvious things that I may have been missing.

Does Physics need non-analytic smooth functions?

2012-11-26T16:58:48Z

Observing the behaviour of a few physicists "in nature", I had the impression that among the mathematical tools they use a lot (along with possibly much more sofisticated maths, of course), there is certainly Taylor expansion. They have a quantity (function) that they need to approximate: they expand it in Taylor series, keep the order of approximation that is useful for their purposes, and discard the irrelevant terms.

Appearently, there is little preoccupation for mathematically justifying this procedure, even if the to-be-approximated quantity is not given by an explicit form which is clearly known to be analytic. As Physics clearly gets no problems from the above mathematical subtleties, this may just mean that the distinction between analytic and smooth functions is somehow irrelevant to the basic equations of physics, or rather to the approximations of their solutions that are empirically testable.

If non-analytic smooth functions are irrelevant to Physics, why is it so?

Are there equations of physical importance in which non-analytic smooth solutions actually are important and cannot be safely considered "as if they were analytic" for the approximation purposes?

Remark: analogous questions may arise about Fourier series expansions.

One possible way the practice goes might be:

Consider a (differential or otherwise) equation $P(f)=0$ usually with analytic coefficients.
Expand the coefficients in Taylor series around a point in the scale of physical interest.
Discard higher order terms obtaining an approximated equation with polynomial coefficients $\tilde{P}(f)=0$.
Make the ansatz that the solutions $f$ of interest must be analytic.
Find the coefficients of $f$ by hand or by other means.

This leaves open the question why the ansatz is mathematically justified, if the equation of interest was $P$ not $\tilde{P}$. Do analytic solutions of $\tilde{P}$ aptly approximate solutions of $P$? Edit: I understand now that these last two lines are not very well formulated. Perhaps, ignoring the $\tilde{P}$ thing, I should have just asked something like:

Given any $\epsilon>0$, does knowing the analytic solutions (i.e. knowing their coefficients, possibly up to an arbitrarily large but finite number of digits) of $P$ give all the information about all solutions of $P$ up to $\epsilon$-approximation? Are there physically well known classes of equations $P$ in which this may not happen (perhaps even up to taking very regular approximations of the coefficients/parameters of $P$ itself)?

What's the name of "twisted semidirect products"?

2013-03-30T20:07:09Z

Let $V$ be an $n$-dimensional real vector space, $\Lambda\subseteq V$ a lattice, and $K$ a subgroup of $Aut_{\mathbb{Z}}(\Lambda)\cong GL(n,\mathbb{Z})$. Let also $\sigma \in Z^1(K,V/\Lambda)$, $\sigma:K\to V/\Lambda$, be a cocycle, and $s:K\to V$ a lift of $\sigma$. Define

$K \ltimes^{\sigma} \Lambda:=\{ (k,\lambda + s(k))\in K\times V\; | \; k\in K,\; \lambda \in \Lambda \}\subseteq K \times V.$

Does the above construction have a name (maybe in more general form)? Is there a standard notation for it?

Remark: given a finite $K\subset GL(n,\mathbb{Z})$, it is used to construct the crystallographic group in the arithmetic crystal class $(K,\mathbb{Z}^n)$ corresponding to the cocycle $\sigma$.

Applications of n-dimensional crystallographic groups

2013-03-10T22:34:29Z

I would like to know what are the applications of the theory of $n$-dimensional crystallographic groups (aka space groups)

1) in mathematics

2) outside of mathematics,

besides the applications to $2$-dimensional and $3$-dimensional crystallography (or related fields like chemistry, or physics, of crystals).

One possible application of $4$-dimensional space groups is already reported in the wikipedia article I linked to (see "Magnetic groups and time reversal").

A little question on certain parallel-lines-preserving maps

2013-03-01T17:07:18Z

Let $\alpha:\mathbb{R}^n\to\mathbb{R}^n$, $n\geq 2$, be a $\mathbb{Q}$-linear bijection with the following properties:

1) $\alpha$ sends straight affine $\mathbb{R}$-lines to straight affine $\mathbb{R}$-lines

2) If $r$ and $s$ are parallel straight affine $\mathbb{R}$-lines, then so are $\alpha(r)$ and $\alpha(s)$.

Question: does $\alpha$ have to be $\mathbb{R}$-linear?

This (perhaps not exactly research level...) question came to me when I tried to answer the following (also elementary - Edit: as the comment by Misha points out, it was not elementary indeed! Just classical from nineteenth Century) question: does a set-theoretical bijection $\alpha:\mathbb{R}^n\to\mathbb{R}^n$ that sends lines to lines and preserves parallellism have to be $\mathbb{R}$-affine? When $n=1$ the answer is vacuously no, as there is only one line and every set-theoretical bijection would do. But when $n\geq 2$, assuming $\alpha(0)=0$, one can prove${}^*$ that $\alpha$ is actually $\mathbb{Q}$-linear, and then -if one assumes continuity- also that $\alpha$ is $\mathbb{R}$-linear. But what if we don't assume continuity?

${}^*$: First let's prove $\alpha$ is additive, that is $\alpha(u+v)=\alpha(u)+\alpha(v)$. If $u$ and $v$ are not multiples of each other, then the nondegenerate parallelogram corresponding to the points $0,u,v,u+v$ is sent to the nondegenerate parallelogram $0,\alpha(u),\alpha(v),\alpha(u+v)$, hence $\alpha(u+v)$ is the vector sum of $\alpha(u)$ and $\alpha(v)$. If $u,v$ are multiples of each other, write $u=u_1+u_2$ and $v=v_1+v_2$ generically enough that none of the following pairs $(w,w')$ of points is such that $w,w'$ are multiples of each other: $(u_1+v_1, u_2+v_2)$, $(u_1,v_1)$, $(u_2,v_2)$, $(u_1,u_2)$, $(v_1,v_2)$. Now,

$$\alpha(u+v)=\alpha(u_1+u_2+v_1+v_2)=\alpha(u_1+v_1)+\alpha(u_2+v_2)=$$

$$\alpha(u_1)+\alpha(v_1)+\alpha(u_2)+\alpha(v_2)=\alpha(u_1+u_2)+\alpha(v_1+v_2)=\alpha(u)+\alpha(v).$$

So $\alpha$ is additive. For every $u$, by additivity $\alpha(u)=n \alpha (\frac{1}{n} u)$, so for every $u$, $\alpha(\frac{1}{n} u)=\frac{1}{n} \alpha(u)$. The above implies $\mathbb{Q}$-linearity.

Undecidability and holomorphic functions (Reference request)

2013-02-20T14:34:52Z

The goal of this question is to recall a certain mathematical fact -not in my field- that I was once briefly told and that I have fogotten, and also to collect similar results.

The fact, I think, was about the undecidability (read: independence from ZFC or maybe ZF axioms) of a certain seemingly very "natural" sentence about the convergence of sequences of holomorphic functions in one variable.

So the question is:

What are some natural undecidable sentences about holomorphic functions? Where by "undecidable" I mean independent from the ZFC or the ZF axioms of set theory, and by "natural" I mean something that is not manifestly designed to be an independence result and possibly that arised quite autonomously from Logic.

Edit: I'm aware there are some independence results (I think by Kranz and Di Biase) related to the boundary behaviour of holomorphic functions. The "fact" I wanted to recall is not part of this theory, though independence examples related to this theory are well accepted in the answers.

Why is Set, and not Rel, so ubiquitous in mathematics?

2013-02-07T00:51:04Z

The concept of relation in the history of mathematics, either consciously or not, has always been important: think of order relations or equivalence relations.

Why was there the necessity of singling out a particular kind of relations, namely the functional ones? I guess (but I don't have data about this) historically the recognition that "operational" expressions like $x^3$ or $\sum_{i=0}^{\infty} \frac{x^n}{n!}$ could be formalized as functional relations led to devote more attention to functions understood in the modern set theoretical sense (i.e. as a special case of relations). That viewpoint permitted to consider things such as the Dirichlet function $\chi_{\mathbb{Q}}$ (which was previously not even considered to be a true "function"!) as fully legitimate objects, and to not dismiss them as pathological, with great theoretical advantage. The language and notation of functions was preferred even to deal with things that, technically, were relations: think of "multi-valued functions" in complex analysis such as $\sqrt x$ or $\log (x)$.

1) In which instances in modern mathematics are relations used as important generalizations of functions? One example that comes to mind is correspondences in the sense of algebraic geometry.

In modern Algebra the concept of homomorphism, a kind of function between algebraic structures, is central; we are used to see expressions like $f(x*y)=f(x)*f(y)$. But it would be equally possible to define a "homomorphic relation" $R$, for example on groups, by the requirement: $(xRz$ & $yRt)$ $\Rightarrow$ $(x*y)R(z*t)$, where $*$ is the group multiplication.

2) Has this kind of "homomorphic relations" been studied (on groups or other algebraic structures)? Why algebra is pervaded with homomorphisms but we never see "homomorphic relations"? Are there something more than just historical reasons?

Let Set be the usual category of sets, and Rel be the category of sets-with-relations-as-morphisms.

There is the faithful functor Set $\to$ Rel that simply keeps sets intact and sends a function to its graph. And there is also a faithful functor Rel $\to$ Set mapping $X\to 2^X$ and $R\subseteq X\times Y$ to $R_*:2^X\to 2^Y, A\mapsto R_*(A)=\{ y\in Y\; |\; \exists x \in A : (x,y)\in R \}$ .

Despite the trivial foundational fact that set theoretical functions are defined to be a special kind of relations, it seems that in category theory Set has priority on Rel. For example the Yoneda's lemma is stated for Set; and people talk of simplicial sets, not simplicial relations; and the category Rel is just retrieved as "the Kleisli category of the powerset endofunctor on Set" (I just learned this from wikipedia) and it doesn't seem to be so ubiquitous as Set (but this impression might just depend on my ignorance in category theory).

3) Are functions really more central/important than relations in category theory? If so, is it just for historical reasons or there are some more "intrinsic" reasons? E.g. is there an analogous of Yoneda's lemma for Rel?

Answer by Qfwfq for Ring with three binary operations

2013-02-05T19:00:47Z

An interesting non-example is the Eckmann-Hilton theorem, stating that if a set is endowed with two associative unital binary operations that "commute" (i.e., if I get it correctly, each multiplication operator $a\mapsto a*b$ is a homomorphism with respect to the other multiplication) then the two operations are the same.

This would exclude the existence of rings $(R, +,*,\circ)$ with two genuinely different "commuting" multiplications.

Noncontractible connected topological rings ?

2010-04-28T18:15:01Z

Are there any non-contractible connected topological rings?

Of course, such a thing cannot be a (topological) algebra over the reals.

(I have a vague memory of having a glance at an erticle by Lurie in which some (for me) rather esoteric theory of higher categorical structures gave rise to topological rings that would have some very nontrivial topology, but I know nothing about that field(s) and, well, I just don't remember... Maybe someone can provide less "esoteric" examples! :) )

Set-theoretical multiverse and foundations

2011-10-13T12:51:17Z

I just had a look to the article The set theoretical multiverse by (mo user) J.D.Hamkins. Not being a logician and not knowing forcing techniques, I couldn't fully appreciate the mathematical ideas, but I was fascinated by the possible philosophical perspective of being compelled (by mathematical practice of forcing in set theory) to consider a whole multiverse of sets as a natural "landscape" for set theory, without committing to any specific choice.

[There's a kind of abstract in the introduction of a n-category café blog post (which I haven't completely read yet) by David Corfield]

In the article it is also stated that it's possible to mimick the study of the "full multiverse" within ZFC. This is actually done in A natural model of the multiverse axioms: "we shall internalize the study of multiverses to set theory by treating them as mathematical objects within ZFC [...]"

Personally, I have more affinity for the formalistic viewpoint than for some version of platonism (multiversed or not). So, my first question (somehow dually to this one) is:

Is it conceivable that the "set theoretic multiverse principles" (which at the moment are, properly, ZFC sentences - see Hamkins and Gitman-Hamkins) could fit into a formal "multiverse theory" which is carried out in its own, i.e. not within a metatheory like ZFC, hence capturing the full-blown multiverse? Could such a theory be taken as the foundation (at least in some ZFC-flavoured sense) of mathematics?

(I will probably open one or more followup "philosophical" questions about multiverses, when I'll clarify to myself what to ask)

Where does the term "torsor" come from?

2012-12-29T16:50:06Z

Is there a heuristic reason why pricipal homogeneus spaces of a group (object) $G$ (in some categories) are called $G$-torsors? Does it have anything to do with the idea of "torsion", somehow? When and where did this piece of terminology originate?

Answer by Qfwfq for Math for a cake

2012-12-29T10:18:59Z

Euler's classical formula for convex polyhedra

$$v-e+f=2$$

where $v$ is the number of vertices, $e$ the number of edges and $f$ the number of faces of a convex triagulated polyhedron in $3$-space.

Answer by Qfwfq for Math for a cake

2012-12-29T10:38:11Z

My all-time favourite formula: Stokes theorem

$$\int_{M}\mathrm{d}\omega=\int_{\partial M}\omega$$

"Softness" vs "rigidity" in Geometry

2012-12-11T21:50:40Z

According to common wisdom, there are structures in Geometry that have a more "topological" flavor, others that are more "geometrical", and others that are halfway between. Usually, geometries${}^*$ that are closer to the topological end of the (possibly non totally ordered) spectrum are referred to as being "soft", whereas geometries closer to the geometric end are considered "rigid". For example, topological, differentiable and symplectic manifolds are considered soft, while Riemannian and complex analytic manifolds are considered rigid. And algebraic varieties are even more rigid geometric objects than analytic manifolds.

I've noted there's also a tendency in deformation theory to use the term "rigid" in the opposite way: rigid is something you cannot deform ${}^{**}$ , so e.g. compact differentiable manifolds are rigid in this sense (thanks to a well known theorem of Ehresmann), whereas usually algebraic varieties admit nontrivial deformations so they are usually not "rigid".

Also, near the ends of the spectrum (homotopy theory, geometric topology; finite geometries, arithmetic geometry) a more combinatorial/discrete/algebraic approach seems to prevail, while in the middle of the spectrum (metric spaces, Riemannian geometry, analytic and complex algebraic geometry) there seems to be a more "continuous" character.

How to make the above distinctions topology vs geometry and/or softness vs rigidity formally more rigorous?

What are other features of the softness vs rigidity phenomenon?

Let's start with some loose ideas about what I have the impression to be features typical of topology vs geometry.

Local invariants. Is the structure at a point distinguishable from the other points or from points of other spaces, or do they all locally look the same? In the case of topological, differentiable and symplectic manifolds there are no local invariants. This also happens for complex analytic manifolds, that we most often regard as instances of a rigid geometry, so it is certainly not a sufficient characterization. For Riemannian manifolds the curvature is a nontrivial local invariant (even a punctual one, in the sense that, even if its definition involves a neighbourhood, differences can be checked pointwise). In algebraic geometry we must be more precise as for the meaning of "locally": locally (w.r.t. some Grothendieck topology) or infinitesimally locally or formally locally? All smooth varieties over a field formally locally look like affine space, but look different locally in the Zariski topology. A condition often put on principal bundles is local isotriviality (i.e. local triviality in the étale topology); this subtlety doesn't appear for vector bundles (aka locally free sheaves).
Robustness vs deformability If you perturb the structure in some way, the resulting structure stays isomorphic. The perturbation can be a deformation in the sense of deformation theory, or just picking a close enough datum (e.g. a Riemannian metric close to the original one in the $\mathcal{C}^k$ topology). By Ehresmann, compact differentiable manifolds are invariant under deformations; but noncompact differentiable manifolds are not, indeed starting from dimension $4$ there may be fibrations with homeomorphic non diffeomorphic fibers, so in some sense it's a less purely topological feature. In analytic and algebraic geometry, projective spaces are invariant under deformation; and I have the impression that it also happens for combinatorially/algebraically defined varieties (e.g. toric).
Discreteness of moduli. This is somehow the global version of the previous point. A "deformation invariant" structure need not have trivial moduli, but while topological structures tend to have discrete moduli spaces (in whatever sense we intend "moduli space"), when moduli are nondiscrete it means there is something geometric going on. For example, if I remember correctly, line bundles on toric varieties have discrete moduli (the Jacobian is trivial), maybe because they depend only on the combinatorics of the orbits, which is -with some stretch of the meaning- a topological thing.
Homogeneity. In many topological categories the "generic" object tend to have a transitive group of isomorphisms (even $n$-transitive sometimes). This happens for topological, differentiable, and (differentiable or real analytic) symplectic manifolds. Of course transitivity of automorphisms also happens for homogeneous spaces in categories of rather "rigid" objects (Riemannian, analytic, algebraic homogeneous spaces), but they are very special objects, not a "random" representative of their category. Anyway they share with more topological categories the possibility of being described combinatorially/algebraically (Schubert cells of the Grassmannian, Lie algebras).
Obstructions. In some soft categories we have partitions of unity, which often allow us to patch local data together to obtain a globally defined thing from locally defined things; in more rigid categories this doesn't hold. Also, in rigid categories there are extension problems.

I would say, when two categories of geometric objects are to be compared in softness/rigidity, we are in the following situation. We have categories $\mathcal{C}$, $\mathcal{C}'$ concrete over some base category $\mathcal{S}$ (the latter can be sets, or topological spaces, or any fancy thing like presheaves of simplicial sets over a site). There is a "forgetful functor" $\mathbb{U} :\mathcal{C}\to\mathcal{C}'$ which commutes with the concretizations. We can say (mind that I'm only giving a vague suggestion and I have no idea of a more precise answer) that $\mathcal{C}'$ is obtained by "putting some geometric structure" on objects of $\mathcal{C}$ if one or more of the following holds:

$\mathbb{U}$ can make local invariants disappear (e.g. forgetting a Riemannian metric on a manifold).
$\mathbb{U}$ can turn a (nontrivially) deformable object into a deformation-rigid one.
If there is some notion of moduli spaces $\mathcal{M},\mathcal{M}'$ for objects of $\mathcal{C},\mathcal{C}'$, then the map $\mathcal{M}\to\mathcal{M}'$ induced by $\mathbb{U}$ has nondiscrete fibers.
$\mathbb{U}(\mathrm{Aut}(X))\subset\mathrm{Aut}(\mathbb{U}X)$.

${}^*$ (Unfortunately, the term "geometry" is, ambiguously, used both in the general sense of "pertaining to spaces of any kind" and in the more localized sense of "pertaining to structures that are metric or anyway more rigid than topological ones". E.g. the term "differential geometry" is sometimes used as a general header including differential topology, sometimes as a synonym of Riemannian geometry. I hope in the question the ambiguities will be cleared by the context)

${}^{**}$ (Rather than "rigidity", I think a more appropriate term would be "elasticity" or "flexibility", since after all one can fit a rigid variety as special fiber of a family, it's just that the deformed neighbours -i.e. the other fibers- turn out to be isomorphic to the variety you started with)

Answer by Qfwfq for Suggestions for good notation

2012-12-09T23:36:13Z

If $\mathcal{C}$ is a category and $X,Y\in\mathrm{obj}(\mathcal{C})$, I like the notation $\mathcal{C}(X,Y)$ to denote $\mathrm{Hom}_{\mathcal{C}}(X,Y)$.

So, $\mathcal{C}(X,X)=\mathrm{End}_{\mathcal{C}}(X)$.

What do you think of the notation $\mathcal{C}(X):=\mathrm{Aut}_{\mathcal{C}}(X)$ ?

This would be consistent with the notation (or similar notations) $\mathsf{DIFF}(S^1)$ (resp. $\mathsf{TOP}(S^1)$ ) for diffeomorphisms (resp. homeomorphisms) of the circle, i.e. the $\mathrm{Aut}$ in the category $\mathsf{DIFF}$ of smooth manifolds (resp. $\mathsf{TOP}$ of topological manifolds), sometimes used in topology (see e.g. here and here. And (see e.g. here) $\mathsf{TOP}(n)=\mathrm{Aut}_{\mathsf{TOP}}(\mathbb{R}^n)$.

What is the structure of the space of solutions of a non linear ODE?

2012-12-08T20:08:09Z

As is well known, the space of solutions of a linear ODE with, say, $\mathcal{C}^\infty$ coefficients on $\mathbb{R}$ is a finite dimensional affine space (a vector space, in the homogeneus case).

What is the structure of the space of solutions of a non linear ODE? In which cases does it have a "natural" structure of finite dimensional smooth manifold?

For the question to make sense, some conditions on the form of the equation must be put. Let's assume the ODE is of the form

$$F(t,u(t),u'(t),u''(t),\dots,u^{(n)}(t))=0$$

where $F:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}$ is a smooth function.

What is the algebraic geometry version of the spheres?

2012-12-07T00:09:05Z

In topology the spheres $S^n$ are the "simplest" closed manifolds, and they are like "Dirac's delta at $n$" for (reduced) cohomology groups. Furthermore they are boundaries of the simplest compact manifolds-with-boundary, i.e. the disks $D^{n+1}$, which are contractible. And $S^{n}$ is obtained by glueing two copies of $D^{n}$ along their boundary $S^{n-1}$. My question is:

Are there some objects of algebraic geometric nature that somehow reproduce the same pattern, or that are considerable as the equivalent of spheres from topology?

More generally, are there "homology spheres" for some homology theory like -say- Chow groups? What about an "algebraic Poincaré conjecture"?

If they do exist, I don't expect them to be standard varieties or schemes, otherwise they probably would have made their appearence "classically".

I imagine some experts in homotopy theory may find this question naive or trivial; well, I apologise if it is so...

How do fibers of the functor Algebraic Varieties $\to$ Complex Analytic Spaces look like?

2012-12-03T00:07:08Z

There's already a question (which got several interesting answers) asking about examples of the phenomenon of non (essential) injectivity of the functor $U:Alg\to AnEsp$, mapping each algebraic variety to its associated (reduced) complex analytic space.

I would like to complement that question by asking:

As far as it is currently known, do the fibers have any particular structure? Are there invariants classifying different algebraic structures on the same analytic space (like e.g. some cohomology space which bears some correspondence with algebraic structures, and is trivial iff the structure is unique)?

Answer by Qfwfq for A "holomorphic" Peano curve?

2012-11-29T17:40:13Z

Define $$\phi(z):=\frac{1}{2\pi i}\int_{S^1}(\zeta-z)^{-1}\cdot \varphi(\zeta)d\zeta$$

for $|z|<1$, where $\varphi: S^1\to\mathbb{C}$ is a Peano curve (i.e. its image has nonempty interior), and $\phi(z):=\varphi(z)$ for $z\in S^1$. [Edit: this construction doesn't work because $\phi$, as I defined it, may not be continuous up to the boundary - see the comments]

Answer by Qfwfq for Fundamental motivation for several complex variables

2012-11-27T17:35:12Z

So far, the motivations have been mostly of analytic flavor. Let me add an alien motivation.

Suppose you were an alien that only knows the complex number field $\mathbb{C}$ as a fundamental object (say, because it's Cauchy complete and algebraically closed) and considers $\mathbb{R}$ just as a minor object (much like we consider -say- the Eisenstein integers). And suppose you came up with the concept of increment and of derivative and then of differential: then, which Analysis would you study? And suppose you were able to imagine the notion of a manifold as a space with local coordinates in your favourite field: which geometry would you naturally study?

Of course, the answers are, respectively, "Complex Analysis" and "Holomorphic manifolds".

So, even if you're a Terrestrial, blinded by notions of linear order, isn't $\mathbb{C}$ a very fundamental object anyway? If so, then doing SCV and Complex Analytic Geometry is just doing the natural things with it.

Answer by Qfwfq for Sets = structured sets without structure

2012-11-22T01:09:45Z

I will call source objects in a category $C$ the objects you defined in the question, that is: objects $X$ such that, for every $Y$, $\mathrm{Hom}_{C}(X,Y)$ is nonempty.

1) The category of fields is concretizable and doesn't have source objects.

2) $HoTop_{*}$ (the homotopy category of pointed topological spaces) is not concretizable and every object is a source object.

Characterizing specific "concrete" mathematical objects by abstract general properties

2011-12-18T00:52:32Z

In this note by Tom Leinster the Banach space $\mathrm{L}^1[0,1]$ is recovered by "abstract nonsense" as the initial object of a certain category of (decorated) Banach spaces. So a function space, that would habitually be defined through the machinery of Lebesgue measure and integration, is uniquely described (up to isomorphism) in terms of abstract functional analysis and a bit of category theory.

I would be curious to see more results, ideally in diverse areas of mathematics, in the spirit of the above one, in which a familiar and important "concrete" mathematical object is recovered by a universal property (in the technical categorical sense) or -more generally- by a characterizing property that is abstract and general or doesn't delve into the "concrete" habitual definition of that object.

Community wiki, so put one item per answer please.

Answer by Qfwfq for Characterizing specific "concrete" mathematical objects by abstract general properties

2012-11-19T14:10:15Z

I happen to have just read Manes' theorem in the n-category café:

Theorem The algebras for the ultrafilter monad are the compact Hausdorff spaces.

The "ultrafilter monad" $X\mapsto \mathrm{U}(X)$ maps a set $X$ to the set of ultrafilters on it. The abstractness of the characterization of compact Hausdorff spaces lies in the fact that $\mathrm{U}$ is defined in purely set-theoretical (or, rather, category-theoretical) terms: it appears to be the "codensity monad" (don't ask me the meaning of this because I don't know!) of the inclusion $\mathrm{FinSet}\to\mathrm{Set}$.

The reals as continuous image of the irrationals

2012-11-11T23:16:54Z

In the wikipedia article about descriptive set theory I read that $\mathbb{R}$ (with its usual topology) is a Polish space, and that every Polish space

1) can be obtained as a continuous image of the Baire space $\mathcal{N}$

2) can be obtained as the image of a continuous bijection defined on a closed subset of the Baire space.

Then I learn from the wikipedia article on Baire space that it is actually homeomorphic to the irrational numbers with their usual subspace topology iherited from the real line.

So my questions:

Can we describe explicitely a surjective continuous map from the irrationals to $\mathbb{R}$?

Same for a continuous bijection from a closed subset of the irrationals to $\mathbb{R}$.

Do complete non-projective varieties arise "in nature"?

2012-11-05T01:44:54Z

I'm aware of the existence of complete (abstract) algebraic varieties that are not projective but, probably due to my ignorance, I have the impression that they arise only as very particular examples constructed just with the purpose of finding such an example. My question (perhaps a bit vague) is:

Are there exemples in the literature in which complete non-projective varieties appear without "being expected" from the beginning or without just being the goal of the construction or proof?

Answer by Qfwfq for The symmetry group of $\mathbb Z^d$

2012-10-13T21:24:34Z

As far as I read (See page 138 of R.W.Sharpe), the Erlangen program, strictly speaking, describes connected homogeneus manifolds $X$ as $G/H$ where $G$ is a Lie group considered as the "automorphism group" of a geometry on $X$ and $H$ is the stabilizer of a point.

First of all, the space $\mathbb{Z}^d$ is a non-connected zero dimensional manifold, so I don't know how much we can say it fits the Erlangen program. Anyways, the "full symmetry group" as a manifold (even as a Riemannian manifold, being zero dimensional) is then simply the full symmetric group (i.e. set theoretic permutations) $G=\mathfrak{S}(\mathbb{Z}^d)$, and $H$ the stabilizer of any point.

But you didn't say which structure on $\mathbb{Z}^d$ you want the symmetries to preserve...

If you want to preserve the distance induced by the Euclidean norm on $\mathbb{R}^d$, then you can take $G=O(d,\mathbb{Z})\ltimes\mathbb{Z}^d$ and $H=O(d,\mathbb{Z})$. [I see that S.Carnahan has posted the same suggestion right before me. Edit: and also Will Sawin]

Answer by Qfwfq for The sets in mathematical logic

2012-09-26T12:33:21Z

This is an answer to another question that unfortunately has been closed just before I could post it and that essentially revolves around the same, or very similar, points as the question of the above OP.

Even if, from the point of view of a logician (which I'm not), the question is elementary, I think it's worth trying to give an answer, so to clarify things (also to myself) and settle some doubts that many non-logicians like me often have about the foundations model theory and logic.

Forget for a moment about $ZFC$. When investigating any (let's say first order) formal theory (such as Peano Arithmetic $PA$, or Algebraically Closed Fields $ACF$) logicians tacitly assume to work within the framework of a metatheory $T$, which must be some kind of "set theory" (where the term is intended loosely, possibly including certain systems of second order arithmetic like $ACA_0$ or of category theory like $ETCS$ or of class theory like $NBG$, according to personal taste) otherwise they wouldn't be able to talk (in principle with rigour, i.e. formally) of all the classical syntactic concepts, such as: languages, signatures, sets of formulas, finite strings of symbols and concatenation thereof; but also of all the classical semantic concepts, like structures, interpretations, models, isomorphisms. Or at least they wouldn't be able to codify -in principle- all the above concepts in the metatheory. The metatheory $T$ is hence momentarily assumed to embody "ontological" concepts, i.e. to formalize the concept of the "real universe of sets" (as opposed to specific concepts of sets arising from theories that are under scrutiny within $T$).

As far as I understand, logicians (even if they usually don't like to explicitely spell out the metatheory they're tacitly using) are perfectly content in assuming the metatheory is the standard set theory $ZFC$ (after all, Logic is a part of Mathematics like any other).

Anyway, I presume much less is needed to "formalize Logic"; for example I think $ZF$ would be enough, or weaker theories such as $ACA_0$ would be ok, and I think (but some professional logician should say if my guess is correct) essentially nothing would be changed in Logic in switching from one suitable metatheory to another: that's why they usually don't bother themselves specifying the metatheory.

How does this formalization work, roughly? Let's consider for example the first order formal theory of groups (call it $GT$). It must be dealt with inside a metatheory (which is a "set theory") which we call $T$. So here we have sentences of $T$ talking about the "objects" $T$ is apt talking about (it could be sets, or sets of natural numbers...), and we have some way to codify the syntactic and the semantic concepts of Logic within $T$: for example some sentences of $T$ will talk about some particular sets that we (informally) interpret as being strings of symbols of $GT$, others as being elements of the language of $GT$, and so on. Of course there will also be sentences of $T$ talking about sets that are structures for the signature of $GT$ (that is, magmas equipped with an arbitrary element $1$ that need not be a unit, and maybe with an arbitrary unary function $x\mapsto (x)^{-1}$ which need not be the inverse), and sets that are models of $GT$ (that is, plain groups).

Now let's come back to $ZFC$. As in the case of $GT$, we must deal with the first order formal theory $ZFC$ within a metatheory ("set theory") $T$. Your question is: what about if $T$ is Zermelo Frenkel set theory with Choice?

For the syntactic aspect, nothing special happens.

You have sentences of $T$ talking about sets that we take to codify various syntactic notions of $ZFC$. Example: a certain sentence of $T$ will define a unique "set" (in the sense of "object of which $T$ is apt talking about" - in this case it means "set in the sense of Zermelo Frenkel set theory with Choice") which stands for the following finite concatenation of symbols in the language of $ZFC$:

$$\forall\forall\in\in\in\to\in x \in ((y(())\in\in = = \in, $$

others will be more meaningful, e.g. displaying an axiom, a theorem (or a conjecture) of $ZFC$.

For the semantic aspect, indeed, there are some subtleties.

In the case of $GT$ we consider models which are sets of $T$ equipped with some structure: $(G,\mu,\iota,1)$. Note that, in the language of $GT$, variables stand for (i.e. are interpreted as) elements of a group, and (the support of) a model is the set of all such elements. If we were asked to prove the existence of models of $GT$, we would have no problem: we would simply exhibit any group, for example $(\mathbb{Z}/ 2 \mathbb{Z},+,-,0)$ (or even the trivial group, for what it matters!).

In the case of $ZFC$, what would be a model? A model $(X,E)$, properly, would be a set $X$ of $T$ equipped with a relation $E \subseteq X \times X$ satisfying the axioms of $ZFC$ when we interpret (via $T$) the relation symbol "$\in$" of $ZFC$ as meaning $E$. Note that variables, in the language of $ZFC$, stand for sets, and the support $X$ of the model must be interpreted as the collection of all such elements, that is $X$ has to be interpreted as the class of all sets.

So far, nothing problematic: set theorists consider models (in the framework of Zermelo Frankel set theory with Choice) of $ZFC$ all the time. For example, if $\xi$ is an inaccessible cardinal, then $(V_{\xi},\in)$ will be a model of $ZFC$ (where $V_{\xi}$ is the slice of the cumulative hierarchy defined by $\xi$).

What about proving the existence of models of $ZFC$? This is the subtlety: when we take $T$ to be Zermelo Frenkel set theory with Choice, we cannot prove in $T$ that $ZFC$ has models, because otherwise we would contradict Goedel's second incompleteness theorem! For example, we don't know, on the grounds of $T$ alone, whether there exists an inaccessible cardinal. So, seen from the point of view of the metatheory $T$, the whole "ontologic" problem of existence of models of $ZFC$ is somehow "suspended". In order to be granted the existence of those models, we have to assume stronger axioms than the ones of $T$; for example, some large cardinal axioms. The existence of an inaccessible would be sufficient for having the mere existence of a model $(X,E)$ (in which, in this case, $E$ is the restriction of the "ambient" membership relation $\in$ itself) without further requirements.

There's another subtlety. In $T$, via some Russel like paradox, one can easily prove that there is no set $V$ such that every set belongs to $V$. This implies that, if for some reason we are handled a model $(X,E)$ of $ZFC$, then either the relation $E$ is not the restriction to $X$ of the "ambient" membership relation $\in$ of $T$ (those are the so called nonstandard models of set theory), or there are properties of some set $S$ defined in $ZFC$ that are not true "ontologically" about its interpretation in the model (i.e. the corresponding sentences of $T$ are false when applied to the interpretation $S^{\mathrm{int}}$ of $S$ in $(X,E)$). For example, if $(X,\in)$ is a countable (standard) model, then of course the notion of cardinality cannot be transferred literally from $ZFC$ to $(X,E)$, because $ZFC$ proves there are uncountable sets, yet the interpretation of any set in $(X,\in)$ will be "ontologically" countable just because $X$ is. To be more concrete, define $\mathbb{R}$ in $ZFC$ in one of the usual ways, then of course $ZFC$ proves « $\mathbb{R}$ is uncountable». In a countable standard model $(X,\in)$ there is a countable set $\mathbb{R}^{\mathrm{int}}$ which is the interpretation of $\mathbb{R}$. Being $X$ a model, the sentence «$\mathbb{R}^{\mathrm{int}}$ is $($ uncountable $)^{\mathrm{int}}$ » of $T$ is true, where «$($ uncountable $)^{\mathrm{int}}$» is the interpretation of the notion of uncountability in the model, but -this is the appearent paradox, which is called Skolem paradox- the sentence of $T$ «$\mathbb{R}^{\mathrm{int}}$ is uncountable» is false. The appearent confusion comes from mixing up the notion of uncountability in the theory and in the metatheory.

Closed forms and trajectories of vector fields

2012-09-21T07:17:01Z

This question is inspired by this recent one and this one; I hope it's not too elementary.

Let $M$ be a (closed) smooth manifold and $X$ a vector field on $M$. Fix any Riemannian metric $g$ on $M$ and let $X^{\flat}=g(X,\cdot)$ be the metric-dual differential form of $X$.

If $dX^{\flat}=0$, what can be said about the trajectories of $X$?

Convenient definition of "category of Riemannian manifolds"?

2012-09-01T12:59:29Z

Has a notion of "category of Riemannian manifolds" been defined and used in the literature?

For which reasons is it or would it (not) be a useful notion?

I think the objects should be all (perhaps complete) Riemannian manifolds, and two objects should certainly be isomorphic if they are isometric as Riemannian manifolds. Which "should" be the morphisms of such a category?

I think some possibilities are:

1) isometries

2) local isometries

3) finite compositions of local isometries and Riemannian submersions

4) conformal maps

5) any of the above localized at local isometries

I tagged it "soft question" because I don't have in mind any specific application of this notion.

Does the algebra of bounded variation functions have a "noncommutative geometric" meaning and generalization?

2012-08-30T13:09:13Z

According to Gelfand-Naimark theory, $C^*$ -algebras of continuous functions $\mathcal{C}^0(X,\mathbb{C})$ on a compact Hausdorff topological space completely capture its topology. Furthermore, every commutative $C^*$ -algebra arises in this way. In noncommutative geometry (NCG), noncommutative $C^*$ -algebras generalize the notion of compact (locally compact, if the algebra is possibly non unital) Hausdorff topological space.

Given a measure space $(X,\mu)$, the algebra $L^\infty (X,\mu)$ of essentially bounded functions on $X$ is a von Neumann algebra which completely describes the "measure theory" on $(X,\mu)$. In NCG, noncommutative von Neumann algebras are considered, which somehow generalize measure theory to the NC setting.

I learn from this wikipedia entry that a certain "chain rule" holds for the space $\mathrm{BV}(\Omega)$ of bounded variation functions on an open subset $\Omega\subseteq\mathbb{R}^n$, making it an algebra, and even a Banach algebra.

I would like to know:

1) Which geometric aspect of $\Omega$ -if any- is completely described by $\mathrm{BV}(\Omega)$ ?

2) Which is -if there is any- the "right" NCG generalization of the $\mathrm{BV}(\Omega)$ algebra?

Comment by Qfwfq

2013-05-08T19:10:51Z

The question would be more interesting if we knew what motivates it...

Comment by Qfwfq

2013-05-07T23:44:57Z

Which sets ??

Comment by Qfwfq

2013-05-05T19:09:14Z

@xuhan: yes, I think it's $\mathrm{Aff}(1,\mathbb{F}_8)$

Comment by Qfwfq

2013-04-25T11:08:59Z

@Zsbán: I don't agree, as this is intended to be a mathematical question (a question in mathematical physics, if you want). +1 for the xkcd quote! ;)

Comment by Qfwfq

2013-04-04T14:56:08Z

If I remember correctly, in the book "Infinite Grassmannians and moduli spaces of G-bundles" by S.Kumar there's a chapter or at least a paragraph on Ind-varieties.

Comment by Qfwfq

2013-03-11T13:04:04Z

Thanks Gerry M. for correcting the typos - it was late here and I typed too hastily.

Comment by Qfwfq

2013-03-06T00:02:12Z

As S.H. says algebraic numbers are countable, and the reason is pretty obvious: the set of polynomials with rational coefficients is clearly contable and each one has finitely many roots.

Comment by Qfwfq

2013-03-01T19:42:25Z

Thanks Misha, I was not aware of that '800 classical result.

Comment by Qfwfq

2013-02-28T17:29:32Z

(I took the liberty of editing the title because, adding the " " )

Comment by Qfwfq

2013-02-24T15:54:37Z

@AdamEpstein: amazon.com/… (I'm quite sure F. Di Biase has obtained some results related to my question; actually, I'm not sure if also the work with Krantz had some "undecidability" aspects)

Comment by Qfwfq

2013-02-24T15:49:23Z

@AdamEpstein: amazon.it/Fatou-Type-Theorems-Functions-ebook/dp/…

Comment by Qfwfq

2013-02-22T21:21:19Z

In the last sentence: is it the germ ($\in$ the stalk $\mathcal{G}_x$) or the "value" ($\in$ the fiber $\mathcal{G}\otimes_{\mathcal{O_Y}} \kappa (x)$)?

Comment by Qfwfq

2013-02-17T00:20:00Z

I think you meant " vector bundle associated with [...]", not " $G$-bundle associated with [...]".

Comment by Qfwfq

2013-02-17T00:14:49Z

@Martin: a morphism is a map $f: E\to F$ over $B$ such that for every $p\in B$, for every $\alpha$ of a covering that trivializes both bundles, the map $\phi_{\alpha}^F \circ f|_{E_p} \circ (\phi_{\alpha}^E)^{-1} : \mathbb{R}^k \to \mathbb{R}^k$, with the obvious notations, is linear. Are there problems with this definition?

Comment by Qfwfq

2013-02-15T16:25:05Z

Why do you apologize for asking a big list question?