User andrea ferretti - MathOverflow

The egg and the chicken

2010-04-29T19:52:27Z

After posting this question (in particular after Carl's and Peter's answers) I have realized that the answer seems to depend on a basic problem in foundations.

Most mathematicians accept as given the ZFC (or at least ZF) axioms for sets. These are treated at an intuitive level. Using set theory they define languages, axioms, theories, models and all the logic toolbox. Then they define (formalized) set theory again, using this language.

The second point of view is typical of logicians. They realize that in order to talk of logic they don't need the full power of set theory, so they take logic as God-given instead. Then set theory is formalized in this framework.

I always thought that the points of view were interchangeable, as far as one was interested in the mathematical consequences. But comparing Carl's and Peter's answers it seems that actual (but still foundational) mathematic may depend on the point of view accepted. I'd like to understand this better.

Are there any mathematical consequences of choosing one of the two points of view?

Consequences of the Riemann hypothesis

2010-03-05T20:11:50Z

I assume a number of results have been proven conditionally on the Riemann hypothesis, of course in number theory and maybe in other fields. What are the most relevant you know?

It would also be nice to include consequences of the generalized Riemann hypothesis (but specify which one is assumed).

Quick proof of the fact that the ring of integers of Q(\zeta_n) is Z[\zeta_n]?

2010-03-06T15:48:52Z

I cannot find a good reference for the proof that the ring of integers in a cyclotomic field $\mathbb{Q}(\zeta_n)$ is $\mathbb{Z}[\zeta_n]$. The proof I usually find does an induction on the number of prime factors of $n$, coupled with a lengthy and somewhat computational proof in the case where $n$ is the power of a prime.

Do you know a quicker and possibly more conceptual approach?

Can we prove set theory is consistent?

2010-04-26T18:54:01Z

Disclaimer

Of course not, I'm aware of Gödel's second incompleteness theorem. Still there is something which does not persuade me, maybe it's just that I've taken my logic class too long ago. On the other hand, it may turn out I'm just confused. :-)

Background

I will be talking about models of set theory; these are sets on their own, so a confusion can arise, since the symbol $\in$, viewed as "set belonging" in the usual sense, may have a different meaning from the symbol $\in$ of the theory. So, to avoid confusion, I will speak about levels.

On the first level is the set theory mathematicians use all day. This has axioms, but is not a theory in the usual sense of logic. Indeed, to speak about logic we already need sets (to define alphabets and so on). In this naif set theory we develop logic, in particular the notions of theory and model. We call this theory Set1.

On the second level is the formalized set theory; this is a theory in the sense of logic. We just copy the axioms of the naif set theory and take the (formal) theory which has these strings of symbols as axioms. We call this theory Set2.

Now Gödel's result tells us that if Set2 is consistent, it cannot prove its own consistence. Well, here we need to be a bit more precise. The claim as stated is obvious, since Set2 cannot prove anything about the sets in the first level. It does not even know that they exist.

So we repeat the process that carried from Set1 to Set2: we define in Set2 the usual notions of logic (alphabets, theories, models...) and use these to define another theory Set3.

A correct statement of Gödel's result is, I think, that

if Set2 is consistent, then it cannot prove the consistence of Set3.

The problem

Ok, so we have a clear statement which seems to be completely provable in Set1, and indeed it is. This doesn't tell us, however that

if Set1 is consistent, then it cannot prove the consistence of Set2.

So I'm left with the doubt that what one can do "from the outside" may be a bit more than what one can do in the formalized theory. Compare this with Gödel's first incompleteness theorem, where one has a statement which is true for natural numbers (and we can prove it from the outside) but which is not provable in PA.

So the question is:

is there any reason to believe that Set1 cannot prove the consistence of Set2? Or I'm just confused and what I said does not make sense?

Of course one could just argue that Set1, not being formalized, is not amenable to mathematical investigation; the best model we have is Set2, so we should trust that we can always "shift our theorems one level". But this argument does not convince me: indeed Gödel's first incompleteness theorem shows that we have situations where the theorem in the formalized theory are strictly less then what we can see from the outside.

Final comment

In a certain sense, it is far from intuitive that set theory should have a model. Because models are required to be sets, and sets are so small...

Of course I know about universes, and how one can use them to "embed" the theory of classes inside set theory, so sets may be bigger than I think. But then again, existence of universes is not provable from the usual axioms of set theory.

Riemann surface disconnected at infinity

2010-02-12T15:37:25Z

This question may be trivial, I did not think hard about it.

A friend of mine is looking for an irreducible (reduced) analytic subspace $C \subset \mathbb{C}^2$ with the following property. Let $f \colon C \rightarrow \mathbb{C}$ be the projection on the first factor. He wants that

1) All singular points of $C$ and all ramification points for $f$ lie in a limited set, so removing that set we obtain a topological covering from some open set of $C$ to $\mathbb{C}$ with a ball removed.

2) That covering should be trivial (even better if it is finitely-sheeted).

So the curve $C$ is connected, but only if one passes near the origin. Sufficiently far from that ther should be no way to jump between sheets. Is it possible to find such a $C$?

Ubiquity of the push-pull formula

2010-03-19T22:09:08Z

The push-pull formula appears in several different incarnations. There are, at least, the following:

1) If $f \colon X \to Y$ is a continous map, then for sheaves $\mathcal{F}$ on $X$ and $\mathcal{G}$ on $Y$ we have $f_{*} (\mathcal{F} \otimes f^{*} \mathcal{G}) \cong f_{*} (\mathcal{F}) \otimes \mathcal{G}$ .

A similar formula holds for the derived functors and for $f^{!}$.

2) If $f \colon X \to Y$ is a proper map of schemes, with $Y$ smooth, both $f^{*}$ and $f_{*}$ are defined on the Chow groups, and $f_{*}(\alpha \cdot f^{*} \beta) = f_{*} \alpha \cdot \beta$ for classes $\alpha \in CH^{*}(X)$ and $\beta \in CH^{*}(X)$ .

Of course a similar results holds in cohomology if $f$ is a proper map of smooth manifolds, using Gysyn map for push-forward.

3) If $H < G$ are finite groups, we have two functors $\mathop{Ind}_{H}^{G}$ and $\mathop{Res}_{H}^{G}$ , which can be seen as pull-back and push-forward maps between the representations rings $R(G)$ and $R(H)$. Again we have $\mathop{Ind}(U \otimes \mathop{Res} V) \cong \mathop{Ind} U \otimes V$ .

Edit: one more example appears in the book linked in Peter's answer. It is a bit complicated to state, but basically (if I understand well)

4) for a compactly generated topological group $G$ and for $G$-spaces $A$ and $B$ one considers the category $G \mathcal{K}_A$ of $G$-spaces over $A$ with equivariant maps (up to homotopy). Then for a $G$-map $f \colon A \to B$ one has functors $f^{*} \colon G\mathcal{K}_B \to G\mathcal{K}_A$ and $f_{!} \colon G\mathcal{K}_A \to G\mathcal{K}_B$ satisfying $f_{!}(f^{*}Y \wedge_A X) \cong Y \wedge_B f_{!} X$ .

There are probably several other variations which now I fail to recall. I should mention that in some situations 2) can be obtained by 1), but not always, as far as I know.

Is there a unifying principle (even informal) which explains why in these diverse settings we should always have the same formula?

Books about history of recent mathematics

2010-05-05T23:57:19Z

I draw on this question to ask something that has always been a pet peeve of mine. It is very easy to find books about the history of mathematics, much less so if one wants books about the recent (say > 1850) one.

Of course I know that this is difficult because not so many people would understand what's going on; to learn about the history of a subject, one should better know the subject beforehand. On the other hand, my feeling is that more or less all mathematics I know has been developed after 1850, and the growth, like in many other sciences, has been exponential. So the amount of mathematics which appears in history book seems negligible to me.

Can you point me to any good resources about the recent history of maths?

Why is addition of observables in quantum mechanics commutative?

2010-02-06T11:35:30Z

I am no expert in the field. I hope the question is suitable for MO.

Background/Motivation

I once followed a quantum mechanics course aimed at mathematicians. Instead of the usual motivations coming from experiment at the turn of the 19th century, the following argument (more or less) was given to show that the QM formalism is in some sense unavoidable.

When one does physics, he is interested in measuring some quantity on a given state of the universe. The quantity (say the speed of a particle) is defined experimentally by the tool used to do the measure. We define such an instrument, with a given measure unit, an observable. So for every state and every observable we get a real number.

We can now define a sum and a product of observables. These are obtained by performing the two measures and then adding or multiplying their values. Similarly we can define scalar multiplication. These operations are then associative, but there is no reason why they should be commutative, since performing the first measure can (and indeed does) change the state of the universe. For some reason I cannot understand, anyway, addition is assumed commutative. I also see no reason why multiplication should distribute over addition. We can now also consider observables with complex values, by linearity.

At this point observables form an $\mathbb{R}$-algebra. We intoduce a norm it as follows. The norm of an observable is the sup of the absolute values of the quantities which can be measured. Every instrument will have a limited scale, so this is a real number. By definition this is a norm. Moreover it satisfies $\|A B \| \leq \|A\| \| B \|$. We can now formally take the completion of our algebra and obtain a Banach algebra.

Finally we define an involution * on our algebra by complex conjugation of observables. This yields a Banach * -algebra, and the third assumption which is mysterious to me is that the $C^*$ identity holds.

Finally we can use the Gelfand-Naimark theorem to represent the given algebra as an algebra of operators on a Hilbert space. If this turns out to be separable, it is isomorphic to $L^2(\mathbb{R}^3)$ and we recover the classical Schrodinger formalism.

The problems

In this approach I see three deductions which seems arbitrary: addition is commutative, multiplication is distributive and the $C^*$ identity holds. Is there any kind of hand-waving which can jusify these? In particular

Why is addition of observables commutative, while multiplication is not?

Elliptic curves with Mordell-Weil group Z/2Z over Q

2011-05-13T09:17:31Z

This question is not very precise; I hope it is suitable for the site.

I have come to a situation where I have to study rational points on an elliptic curve defined over $\mathbb{Q}$. I don't know much about the curve, let alone its equation. I already have one rational point, which sits on a bounded real connected component. What I want to avoid is that this is the only rational point (other than the marked point).

I am not sure what to use about my curve that will help me get there, so I turn the questio the other way round:

What is known about elliptic curves $E$ over $\mathbb{Q}$ such that $E(\mathbb{Q}) \cong \mathbb{Z}/2 \mathbb{Z}$?

Can skeleta simplify category theory?

2010-01-13T17:51:15Z

I am not by any means an expert in category theory. Anyway whenever I have studied a concept in category theory I have always had the feeling that most of the subtleties introduced are artificial.

For a few examples:

-one does not usually consider isomorphic, but rather equivalent categories

-universal objects are unique only up to a canonical isomorphism

-the category of categories is really a 2-category, so some natural constructions do not yield functors into categories, but only pseudofunctors

-cleavages of fibered categories do not always split

....

My question is: can skeleta be used to simplify all this stuff? It looks like building everything using skeleta from the beginning would remove a lot of indeterminacies in these constructions. On the other hand it may be the case that this subtleties are really intrinsic, and so using skeleta, which are not canonically determined, would only move the difficulties around.

Homologically trivial submanifolds

2010-06-10T14:43:44Z

Unuseful prequel

Let $M$ be a (compact, oriented, differentiable) manifold. Before knowing anything about homology theory a naif but clever mathematician may want to measure the holes in $M$ by the following procedure. One lets $B_k(M)$ be the abelian group generated by embedded submanifolds of $M$ of dimension $k$ with border. There is an obvious border operator $\delta \colon B_{k}(M) \to B_{k-1}(M)$ and trivially $\delta^2 = 0$ since the border of a manifold with border is a closed manifold. So one may define $H^K_{naif}(M)$ by taking the quotient of cycles modulo boundaries.

Ok, this does not really work, for at least two reasons. The first there is no way to make this functorial without considering more degenerate objects (singular simplices or currents being two possible choices). The second is that even if one adjusts the definition to get, for instance, the bordism groups of $M$, these will be highly nontrivial for differential geometric reasons, rather than for the topology of $M$ (for instance they will be nontrivial for a point).

Anyway, one finally chooses a working (co)homology theory, for instance singular, and then can use the fundamental class of submanifolds to link this theory to the naif one. Two questions naturally arise:

Are fundamental classes of manifolds enough to generate the cohomology? This is nicely answered here.
Are submanifolds with border enough to generate homology relations? This is the purpose of the present question.

The actual question

Let $M$ be a (compact, oriented, differentiable) manifold and let $N \subset M$ be a (closed) submanifold. The problem is to test whether $N$ is the boundary of submanifold with boundary of $M$. There are two obvious obstructions:

$N$ should be bordant as an abstract variety
The class $[N] \in H_{*}(M, \mathbb{Z})$ should be $0$.

Assume 1 and 2 hold. Are there any conditions on $M$, $N$ or the embedding $N \to M$ which guarantee that $N$ is the boundary of submanifold with boundary of $M$?

There are a few classical cases which come to mind:

If $N$ and $M$ are spheres, 1 and 2 are vacuous and the thesis is true by the (generalized) Jordan curve theorem.
If $N = S^1$ and $M = \mathbb{R}^3$, the thesis is still true by the existence of Seifert surfaces.
If $\dim N = 1$ and $\dim M = 2$ the result seems true to me by the classification of surfaces as quotients of polygons and the usual Jordan curve theorem (but I did not check the details).

On the other hand I'm sure there are plenty of negative example even in $3$-dimensional topology. Is there anything nontrivial which can be said about this question, whether on the positive or negative side?

Answer by Andrea Ferretti for Why nilpotent elements must be allowed in modern algebraic geometry?

2011-02-12T23:58:05Z

As others have mentioned, nilpotent elements show up (at least) in the structure rings of varieties counted with multiplicities. Why should we want to have such objects? I can think of at least three reasons:

The concept of family is easier to deal with. For instance, in the context of schemes, it is easy to speak of a family of conics degenerating to a double line. If we replace the double line with the same line counted once, the family behaves more badly (it is not flat)
As rings have fibered coproducts (tensor products), schemes have fibered products. This is a general construction with good categorical properties, and it generalizes a variety of contexts (fibers, intersections, pullbacks of vector bundles...). If you want to stick with varieties, this construction will not be available, as the tensor products of reduced rings can be non-reduced
A particular non-reduced scheme $k[x]/(x^2)$ is very useful in deformation theory. In deformation theory you want to study a given map up to the first order (or maybe higher orders, so rings like $k[x]/(x^n)$ appear). The existence of non-reduced schemes allows you indentify such an object (a first order approximation to some map) with an actual map from $k[x]/(x^2)$. This is quite handy and simplifies many arguments.

One more reason to be happy in keeping schemes the way they are is the existence of the Quot scheme. This is a general construction due to Grothendieck which allows you to have schemes which parametrizes a manifold of objects: subschemes, morphisms and so on. Moreover, most other moduli space in use in algebraic geometry are constructed starting from a Quot scheme, typically as a GIT quotient.

There is no corresponding general existence theorem in the context of varieties, so moduli theorists would have a pretty hard time abandoning schemes. Of course often it happens that the relevant Quot schemes are actually varieties, but we do not know how to construct them directly. It is easier to produce something (a scheme) and then show that it is nice (a variety), than producing a nice object in one step.

Answer by Andrea Ferretti for Collecting proofs that finite multiplicative subgroups of fields are cyclic.

2011-02-08T09:45:54Z

Let $n = |G|$ and let $m$ be the l.c.m. of the orders of the cyclic factors of $G$. Then $x^m = 1$ for all $x \in G$; since we are in a field this equation has at most $m$ roots, which shows that $m \geq n$. It follows that $m = n$ and $G$ is cyclic.

Of course here one uses the classification of finite abelian groups as product of cyclic groups, which you may want to avoid.

Answer by Andrea Ferretti for Identifying the orientation bundle uniquely

2011-02-04T13:01:06Z

If I interpret your question correctly, you are asking which class in $H^1(S; \mathbb{Z}/2 \mathbb{Z})$ corresponds to the orientation covering. This is the first Stiefel-Whitney class of $TS$, and there are many constructions for it.

Reference book for commutative algebra

2010-02-25T17:08:57Z

I'm looking for a good book in commutative algebra, so I ask here for some advice. My ideal book should be:

-More comprehensive than Atiyah-MacDonald

-More readable than Matsumura (maybe better organized?)

-Less thick than Eisenbud, and more to the point

To put this in context, I'm an algebraic geometer, so I know enough commutative algebra, but I didn't study it systematically (apart from a first course on A-M which I followed as an undergraduate); rather I learned the things I needed from time to time. So I would like to give me an occasion to get a better grasp on the subject.

EDIT: I will be more specific about the level. As I said I already had a course on Atiyah-MacDonald, and I know that material well, so I'm not interested in books of a comparable level. But I'm not completely familiar with Cohen-Macaulay rings and the relationship between regular sequences and the Koszul complex for example. And I know very little of Gorenstein rings and duality. So I'm looking for something a little bit more sophisticated than what has been already proposed. Yes, I know Eisenbud does these things but it's easy to get lost in that book. Something more to the point would be nice.

Stacks and sheaves

2010-02-03T01:48:47Z

I'm a bit confused by the double role which sheaves play in the theory of stacks.

On the one hand, sheaves on a site are the obvious generalization of a sheaf on a topological space. On the other hand a sheaf on a site is (or better its associated category fibered in sets is) a very particular stack itself, so a generalization of a space. This is not completely confusing: more or less it amounts (I believe) to identifying a space X with the sheaf of continuos functions with values in X.

But now my question is the following. An equivalent condition for a fibered category to be a prestack is that for any two objects (over the same base object), the associated functor of arrows should be a sheaf. In particular this is true for a stack, so for any stack and any two objects in it we have a sheaf, and so a stack (over a comma category).

What is the meaning of this geometrically?

For instance take the stack $\mathcal{M}_{g,n}$.

Giving two objects in the stack (over the same base object) means giving two families $X$ and $Y$ of stable pointed curves over the same scheme $S$, and the associated functor of arrows maps every other scheme $f \colon T \rightarrow S$ to the set of morphism between $f^* X$ and $f^* Y$. How should I think of the associated stack as a space?

To avoid misunderstandings I give the defition of the functor of arrows. Let $\mathcal{F}$ be a fibered category over $\mathcal{C}$. Take $U \in \mathcal{C}$ and $\xi, \eta \in \mathcal{F}(U)$. Then there is a functor $F \colon \mathcal{C}/U \rightarrow Set$ defined as follows. For a map $f \colon T \rightarrow U$ we put $F(f) = Hom(f^* \xi, f^* \eta)$. The action on arrows requires some diagrams to be described, but it's really the only possible one.

A form of cohomology and base change

2010-11-19T14:30:02Z

Let $f \colon X \to Y$ be a proper morphism of (Noetherian) schemes, $\mathcal{F} \in \mathop{Coh}(X)$. Let $i_Z \colon Z \hookrightarrow Y$ be a closed subscheme and take the inverse image $W := X \times_Y Z \overset{i_W}{\hookrightarrow} X$.

If we assume that $\dim X_y \leq k - 1$ for all $y \in Y \setminus Z$, then the sheaf $R^k f_{*}(\mathcal{F})$ is concentrated on $Z$ by flat base change for the open inclusion $Y \setminus Z \to Y$.

Is it true in this case that $$R^k f_{*}(\mathcal{F}) \cong i_{Z*} R^k (f|_W)_{*}(\mathcal{F}|_{W})$$ where $\mathcal{F}|_{W} = i_W^{*}(\mathcal{F})$?

I was not able to derive this from the standard results about base change, nor to find any counterexamples.

Special case: What I actually need is a rather special case of the former. Namely in the case I am insterested in:

all objects are varieties over an algebraically closed field
$X$, $Z$ and $W$ are smooth
$\mathcal{F}$ is a line bundle
$k = 1$
the map $X \setminus W \to Y \setminus Z$ is an isomorphism.

It interesting to know something about the general case, though.

Answer by Andrea Ferretti for Most memorable titles

2010-10-31T14:30:31Z

One that comes immediately to mind is Can one hear the shape of a drum?

Minimal model which is necessarily singular

2010-04-11T15:23:10Z

I was told during a summer school on the MMP a nice example (which I have also mentioned here on MO) that I'm not able to figure out anymore.

The example (due, I think, to Miles Reid) is a smooth compact threefold $X$ such that the number of sections of $\mathcal{O}_X(m K_X)$ grows like $m^3/4$ (if I recall correctly). The nice thing about this example is the following.

Assume $X$ is birational to a smooth variety $Y$ such that $K_Y$ is nef. Then sections of $K_Y$ grow in the same fashion, in particular $K_Y$ is big, so by Kawamata-Viehweg vanishing we have $h^0 (Y, \mathcal{O}_Y(m K_Y)) = \chi(Y, \mathcal{O}_Y(m K_Y)) \sim m^3/4$, and by Riemann-Roch we find $K_Y^3 = 3/2$, which is not possible, since that number must be integer.

So, if one wants to have a minimal model for $X$, one has to allow singular varieties into the picture.

Can anyone tell me how the variety $X$ is obtained (or another example in a similar flavour)?

EDIT: As I wrote in the comments to VA's answer, I'm looking for an elementary example, where $\mathcal{O}_X(m K_X)$ can be computed and compared to the Riemann-Roch expansion, in order to have a completely intersection-theoretic argument. In particular I'd like to avoid using the concepts of canonical and terminal singularities, since I view this example mainly as a motivation to introduce exactly those concepts. It would also be nice if one could directly find a smooth $X$, rather then using Hironaka to resolve a singular variety with fractional $K_Y^3$ (which one secretly knows is terminal).

Answer by Andrea Ferretti for Why do we care about the Hilbert scheme of points?

2010-09-27T16:04:01Z

One of the reasons is the following. If $X$ is a smooth curve the symmetric product $X^{(n)}$ is smooth. If $X$ is a smooth surface, $X^{(n)}$ is singular, but it is a theorem of Fogarty that $X^{[n]}$ is smooth. So it is a (rather natural) resolution of singularities of $X^{(n)}$.

If moreover $X$ is a symplectic surface (i.e. either a $K3$ or an abelian surface), $X^{[n]}$ has a symplectic form. As Beauville showed, in the $K3$ case it is even an irreducible symplectic variety; in the abelian case a certain subvariety $K_{n-1}(X) \subset X^{[n]}$ is irreducible symplectic.

There are really few known examples of irreducible symplectic varieties in higher dimension: up to deformation, I have listed them all, except for two sporadic examples constructed by O'Grady! This shows that the Hilbert schemes $X^{[n]}$ are indeed very relevant to people who study holomorphic symplectic geometry.

Restriction of Ext sheaves

2010-09-06T14:42:53Z

Let $f \colon X \to Y$ be a map of schemes, $\mathcal{F}, \mathcal{G}$ two coherent sheaves on $Y$. I'm interested in conditions which guarantee an isomorphism $$f^{*} \mathcal{E}xt^i(\mathcal{F}, \mathcal{G}) \cong \mathcal{E}xt^{i}(f^{*} \mathcal{F}, f^{*} \mathcal{G}).$$

I know this is true for $f$ flat by [EGA III.12.3.4]. Moreover it seems to me that this is true for $f$ arbitrary when $\mathcal{F}$ is locally free and $i = 0$, essentially by the argument in [EGA I.0.6.7.6]. I am also able to prove this in some other cases by ad hoc methods, but I could not find a general statement.

What are conditions which guarantee the existence of such an isomorphism? I am also interested in having some counterexamples at hand.

Interdependence between A^1 homotopy theory and algebraic cobordism

2010-08-25T12:02:16Z

I would like to learn something about $\mathbb{A}^1$-homotopy theory. I know about standard references on the subject, but before dwelling into studying them I have a doubt which some expert could clear.

Having a look at Levine's survey it seems that algebraic cobordism has been very succesful at solving open problems in $\mathbb{A}^1$-homotopy theory.

How much the two topics are independent of each other? Is it possible to understand the full implications of $\mathbb{A}^1$-homotopy theory without knowing about algebraic cobordism? Has the latter theory in some sense superseded the former?

Answer by Andrea Ferretti for Which is the correct generalization of Euler sequence to the projectivization of a vector bundle?

2010-08-25T12:07:34Z

This sequence is indeed exact. Once you check that the maps are globally defined, exactness can be checked fiberwise (remember that we are dealing with a sequence of vector bundles, not an arbitrary sequence of sheaves), and in this case it follows by the usual Euler sequence for the projective space.

Answer by Andrea Ferretti for Do all graphs of C1 functions have Hausdorff dimension 1?

2010-08-20T22:13:39Z

Let $f \colon I \to \mathbb{R}$. Since $f$ is $C^1$, the graph $\Gamma_f$ is locally bilipschitz to $I$, via the projection. It follows that Hausdorff dimension is the same as that of $I$ (being defined in terms of the metric space structure only), so it is $1$.

Disclaimer: I haven't seen these topics for quite a while, so I may have said something stupid.

Fundamental group of Lie groups

2010-08-17T13:18:32Z

Let $T$ be a torus $V/\Gamma$, $\gamma$ a loop on $T$ based at the origin. Then it is easy to see that $$2 \gamma = \gamma \ast \gamma \in \pi_1(T).$$

Here $2 \gamma$ is obtained by rescaling $\gamma$ using the group law, while $\ast$ denotes the operation in the fundamental group. The way I can check this is rather direct: one lifts the loop (up to based homotopy) to a segment in $V$ and uses the identification of $\pi_1(T)$ with the lattice $\Gamma$.

Is there a more conceptual way to prove this identity that will extend to more general (real or complex) Lie groups, or maybe to linear algebraic groups? Or is this fact false in more generality?

Answer by Andrea Ferretti for Origins of names of algebraic structures

2010-08-12T01:29:30Z

I'm not sure that I'm historically accurate, but that is how I always thought about algebraic nomenclature.

1) Group actually comes from group of substitutions. I guess that Galois could have introduced any other word, like "set" of substitutions or "flock" of transformations. Set theory was not yet established, so I guess a collection of functions could be called 'group', 'set' and so on according to the taste.

2) For field, I guess it comes from the meaning of field as "sphere", "subject", "area". It makes sense that such a word could come in talking about "solving an equation in the real field" rather than "solving an equation in the complex field". Then the concept of an abstract field could have followed.

3) Ring comes from "Zahlring", ring of numbers. This, as far as I know, is a terminology due to Dedekind. He was actually working with number rings, of the form $\mathbb{Z}[\alpha]$, where $\alpha$ is integral over $\mathbb{Z}$. So for some $n$, $\alpha^n$ can be expressed in terms of lower powers of $\alpha$; in some sense the components of the basis of $\mathbb{Z}[\alpha]$ over $\mathbb{Z}$ cycle, although this is accurate only when $\alpha$ is a root of unity. Hence the name ring of numbers.

4) Ideal is easy. When Dedekind realized that in a ring like $\mathbb{Z}[\sqrt{-5}]$ unique factorization does not hold, he searched for a substitute. He then realized could restore unique factorization allowing something more general than elements, the ideals. These are now called this way since he thought of them as "ideal elements" of the ring. useful to restore unique factorization. It is a fortunate coincidence that indeed for the rings he was working with (which are now called Dedekind rings), unique factorization for ideals actually holds.

5) Idéle has the same origin, being the contraction of the French "idéal élement", although the wording is inverted with respect to French use.

Answer by Andrea Ferretti for The definition of homotopy in algebraic topology

2010-08-11T17:28:34Z

The answer to (1) is no: the exponential law $C(X \times Y, Z) \cong C(X, C(Y, Z))$ is only valid for some nice topological spaces; for instance I think that Hausdorff and compactly generated is ok.

One of the reasons why the category of simplicial sets is nicer than that of topological spaces is exactly the fact that the exponential law is always valid.

So what happens for spaces which are not nice? Well, for these spaces the path space $C(I, X)$ can behave bad, so the usual definition which does not involve $C(I, X)$ works better.

Answer by Andrea Ferretti for Examples of great mathematical writing

2010-08-11T12:23:58Z

I an algebraic geometer, so the book I'm going to propose is about as far from my subject as it can be. Still I think that Steele's book on stochastic calculus is one of the best written mathematical books I know. It really makes you enjoy probability, starting from the simplest examples of random walks and building a lot of theory, like martingales, Brownian motion and Ito's integral. I almost wanted to change my subject when I was reading it! :-)

Answer by Andrea Ferretti for Continuous + holomorphic on a dense open => holomorphic?

2010-08-06T13:27:26Z

If the image of the curve has measure 0, it seems true. Indeed it is enough to prove that both the real and imaginary part are harmonic. This will be true if they satisfy the mean value identity on small balls.

The mean value identity is clear outside of $C$; on points of $C$ it follows by continuity and the fact that the integral of $f$ on $C$ with respect to the $2$-dimensional Lebesgue measure is $0$.

EDIT: I'm sorry, according to Mohan Ramachandran comment below, this answer is wrong.

Answer by Andrea Ferretti for Is there a Whitney Embedding Theorem for non-smooth manifolds?

2010-08-05T17:28:23Z

I don't know if you can get the same bound, but you have the embedding in some big $\mathbb{R}^N$. The proof is the same as in the smooth case, even simpler, Let me show how it works assuming $M$ is compact, say of dimension $n$.

Cover $M$ by finitely many charts $U_1, \dots, U_k$ homeomorphic to $\mathbb{R}^n$. For every $i$ consider the map $f_i \colon M \to S^n$ which collapse the complement of $U_i$ to a point. Of course you can see $f_i$ as a map to $\mathbb{R}^{n+1}$. Then $f = (f_1, \dots, f_k)$ is the desired embedding.

Comment by Andrea Ferretti

2012-01-26T22:22:58Z

Wow, an impressive answer. I must admit that it requires me some time to actually follow it all in detail, but it seems a very clarifying point of view. Of course, this leads to the question: why all this different objects satisfy some kind of base change formula? :-)

Comment by Andrea Ferretti

2012-01-26T22:14:45Z

Well, the point of my question is in big part metamathematical. That is, the explanation I received was meant as a motivation to study quantum mechanics via C*-algebras, and a posteriori, via the Gelfand-Naimark-Segal theorem, via operator theory. I admit that the explanation does not convince me completely, but I think it has some points. The reason why I asked here was trying to fill the gaps. Instead you assume that observable are represented by operators, which is meant to be the conclusion

Comment by Andrea Ferretti

2011-11-03T18:54:24Z

Those functions have all derivatives 0 in a point, not on a whole interval

Comment by Andrea Ferretti

2011-06-08T10:31:58Z

I aked basically the same question some time ago mathoverflow.net/questions/18799/… Some answers gave interesting links.

Comment by Andrea Ferretti

2011-05-13T11:59:20Z

The fact is, I don't know much more than stated in the question. I can probably extract more information, if I know what to look for. So, in a sense, any theorem starting like "Let $E/\mathbb{Q}$ be a an elliptic curve with $E(\mathbb{Q}) = \mathbb{Z}/(2)$; then..." would help me.

Comment by Andrea Ferretti

2011-05-13T10:35:54Z

Yes, I know my question is vague. That is because I do not know much about this point. I was hoping that having Mordell-Weil group of order 2 would imply some restrictions, and then work backwards.

Comment by Andrea Ferretti

2011-04-12T16:47:20Z

How did this change history?!??!?

Comment by Andrea Ferretti

2011-04-08T11:39:29Z

Please, share these videos also on MathOnline, in the video section! :-) mathonline.andreaferretti.it

Comment by Andrea Ferretti

2011-02-27T17:29:44Z

No, it is not. The problem is to verify the mean value identity on small balls centered on points of $C$, which is not as trivial as I envisaged. A simple continuity argument does not work, as I found by trying to add more details.

Comment by Andrea Ferretti

2011-02-26T18:23:08Z

Uhm... I understand the point. It would be very interesting to see a proper family which is analitically, but not algebraically, locally trivial.

Comment by Andrea Ferretti

2011-02-26T16:47:41Z

So you are claiming that every isotrivial family of compact complex manifolds is actually locally trivial? I have to admit I do not have a proof of non-local-triviality for any isotrivial family (over $\mathbb{C}$) I can think of (and foor good reasons, apparently!) but then I wonder why the terminology... I wish I was always just told "here is a fiber bundle with fiber $F$".

Comment by Andrea Ferretti

2011-02-26T15:19:16Z

I am confused about the Grauert-Fischer theorem, and I don't have a reference at hand (not to mention my german is not in very good shape). Certainly there exist proper maps $\pi \colon X \to S$, with both $X$ and $S$ compact, which are isotrivial on an open subset $U \subset S$, but not locally trivial. It seems to me that the restriction $\pi^{-1}(U) \to U$ is still proper, but does not satisfy the conclusion of the G-F theorem. :-?

Comment by Andrea Ferretti

2011-02-22T17:00:20Z

I think you should rewrite your question and explain your notation. I do not understand what $\omega$, $g_j$ and $h_j$ are. Furthermore I do not understand what your last comment about deriving a function has to do with the issue at hand.

Comment by Andrea Ferretti

2011-02-22T16:40:57Z

Indeed it is often the case that there more than one complex structure on a complex manifold. In any case, once you fix a complex structure, the almost complex structure associated to it is multiplication by i. How is this definition not global?

Comment by Andrea Ferretti

2011-02-08T12:41:52Z

@Pete: Yes, I'm aware that classification of finite abelian groups may be a cannonball here. On the other hand, in the introductory algebra course in Italy one usually encounters groups before more complicated structures like fields (at least, it used to be like that) so one may have the result at hand anyway. In any case the question was about collecting proofs of this results, so I thought it may be worth to add this one. :-)