gn.general topology - Is "compact implies sequentially compact" consistent with ZF?

Over at the nForum, we've been discussing sequential compactness. The discussion led me to realise that I naively assumed that nets were simply Big Sequences, and that I could make a reasonable guess at how nets would behave by thinking of them as such.

Not so. The crucial point, that I hadn't realised, was that subnets are not sub-nets in the way that subsequences are sub-sequences.

Where this came to light was in a discussion of the relationship between compactness and sequential compactness. Compactness can be expressed as:

Every net has a convergent subnet.

Sequential compactness as:

Every sequence has a convergent subsequence.

So, in my naivety, I assumed that compactness implied sequential compactness since I could take a sequence, think of it as a net, find a convergent subnet, and - ta-da - there's my convergent subsequence. The error, as Mike Shulman pointed out, is that not every subnet of a sequence is a subsequence.

And, indeed, there is a space that is compact but not sequentially compact. Writing $I = [0,1]$ then $I^I$ is compact but not sequentially compact. In particular, it is possible to find a sequence that has no convergent subsequence (the argument is a variant of Cantor's diagonal theorem) but that has plenty of cluster points and thus plenty of convergent subnets.

But the compactness of $I^I$ seems to require a Big Axiom (not quite the axiom of choice, or so I'm led to believe since $I$ is Hausdorff, but almost). I say "seems to" since I'm not an expert and there may be a way to prove that this specific space, $I^I$, is compact with only the basic axioms of ZF.

That's basically my question, except that I'm a topologist so I'm more interested in the implications for topological stuff than in the exact relationship between the Axiom of Choice and Tychanoff's theorem (and since I can just read the nLab page to learn that!). So, without further ado, here's the question:

Is "Compactness => Sequential Compactness" consistent with ZF?

This could be answered by a topologist since all it would require to show that this isn't so would be an example of a space that was compact but not sequentially compact and such that proving that didn't require any Big Axioms.

References:

1. nLab pages: sequential compactness (has more details on the above example), nets (contains the crucial definition of a subnet), Tychonoff's theorem (contains a discussion of the axiomatic strength of this theorem)


2. nForum discussion: sequential compactness

characteristic p - What is the family derived from the absolute Frobenius on the Hilbert scheme?

Let $f$ be a Hilbert polynomial, and $X := Hilb_h(P^d_{F_p})$ a Hilbert scheme defined over $F_p$. Then there is an absolute Frobenius map $F: X to X$. I'm even interested in the case $f equiv 1$, so $X = P^d$.

Which of the following is a sensical question?

1. What is the family over $X$ induced by pulling back the universal family along $F$? Is there a reasonable way to think about it, that makes it clear that it's not just again the universal family?




2. If that family is just the universal family again, then in what sense is the Hilbert scheme universal? (As it seems I've obtained the same family from two different maps, which I thought wasn't supposed to happen.)

ag.algebraic geometry - Birational invariants and fundamental groups

The paper of Keum and Zhang

"Fundamental groups of open K3 surfaces, Enriques surfaces and Fano 3-folds",
Journal of Pure and Applied Algebra vol. 170

provides some answers to Tom's question "What happens to the fundamental group of a singular variety when removing the singular points?"

It appears that the fundamental group of the smooth part can be quite complicate also in simple situations. For instance, one of the results in the paper is the following:

"THEOREM. Let X be a $K3$ surface with at worst Du Val singularities (then X is still simply connected), and let $X^0$ be its smooth part. The number $c$=#(Sing $X$) is bounded by $16$, and if $c=16$ then $pi_1(X^0)$ is infinite".

So, given for instance a quartic surface $X subset mathbb{P}^3$ with $16$ nodes (a Kummer surface) we have

$pi_1(X)={1}$, but $pi_1(X^0)$ is infinite!

This follows from the fact that $X^0$ has an étale $mathbb{Z}_2$-cover $Y^0 to X^0$, where $Y^0$ is an Abelian surface minus 16 points.

For smaller values of $c$, the group $pi_1(X^0)$ is finite, but not trivial in general.

In higher dimension, there is the following

"CONJECTURE. Let $V$ be a $mathbb{Q}$-Fano $n$-fold. Then the topological fundamental group $pi_1(V^0)$ of the smooth part $V^0$ of $V$ is finite".

(a normal variety $V$ with at worst log terminal singularities is $mathbb{Q}$-Fano if, by definition, the anti-canonical divisor $−K_V$ is $mathbb{Q}$-Cartier and ample).

mathematics education - Specializing early

Topic: this is a mathematics education question (but applies to other sciences too).

Assumptions: my first assumption is that most mathematical concepts used in research are not intrinsically more complicated to grasp than high-school and undergraduate maths, the main difference is the amount of prerequisites (and hence time and experience) involved. My second assumption is that some undergraduate topics currently taught compulsarily are a bit of a burden for someone focussed on a particular topic.

Now of course cognitive development is a constraint, but upon reaching the age of high-school, I would think that a fairly large proportion of the scientifically-enclined students could really understand things usually taught much later and indeed become active at research level within a few years, provided some shortcuts are introduced.

Early specialization: I'm wondering if a balanced curriculum already exists (or is planned) to provide such early specialization. What I'm looking for is this: a one-week panorama of maths (or physics, or biology) would be organized at the beginning, and then the students would decide which subtopic to study. For example someone interested by group theory (or quantum optics, or genetics) would thus start with basics at the age 15 or 16, and gradually learn more stuff and skills, but for a few years with a strong emphasis on things directly relevant for the chosen subtopic.

So for example the student specializing in group theory would only learn differential calculus and manifolds in passing in the context of Lie groups, and would skip most undergraduate real and functional analysis until it becomes relevant for his/her research topic, if at all. Of course other general courses would still be taught (history, sciences, programming, foreign languages...), but at least 50% of the student's week would be devoted to the research topic, ensuring satisfying progress.

Question: do you know of any active or planned educative curriculum (at a high-school or university, or maybe a specific home-schooling program) as outlined above? As an example of successful early specialization see e.g. the winners of the Siemens Foundation Prizes, but I haven't been able to learn much about their specific curriculum if any.

Note: Skipping grades in school to enter university earlier is not the point, I'm really interested in a subtopic-oriented curriculum.

What is a noncommutative fiber bundle?

One can consider torsors (principal bundles) within any sufficiently nice category and with respect to descent with respect to given Grothendieck topology and given fibered category over the ground site. See for example

• Tomasz Brzeziński, On synthetic interpretation of quantum principal bundles, AJSE D - Mathematics 35(1D): 13-27, 2010 arxiv:0912.0213.

where in the main motivational part the codomain fibration and regular epimorphism topology are implicitly used. For the fibered category of modules (viewed as quasicoherent sheaves) over noncommutative rings, the faithfully flat Hopf-Galois extensions are the answer provided we accept Hopf coactions as dual representations of group actions. The tensor product is not a monoidal product in the category of associative algebras so this is a bit of a problem. Next thing is that one needs to consider nonaffine objects, if one is in algebraic framework, what can allow for more general concept of noncommutative principal bundles over the covers by noncommutative localizations which are analogues of covers in a Grothendieck topology. This kind of noncommutative principal bundles were skecthed in my articles

• Z. Škoda, Localizations for construction of quantum coset spaces, math.QA/0301090, Banach Center Publ. 61, pp. 265--298, Warszawa 2003;




• Z. Škoda, Coherent states for Hopf algebras, Letters in Mathematical Physics 81, N.1, pp. 1-17, July 2007. (earlier arXiv version: math.QA/0303357),

based on general picture of actions in noncommutative algebraic geometry as explained in the newer survey

• Z. Škoda, Some equivariant constructions in noncommutative algebraic geometry, Georgian Mathematical Journal 16 (2009), No. 1, 183--202, arXiv:0811.4770.

See also the nlab:noncommutative principal bundle. I will hopefully release 2 more articles in this direction within next month or two.

Yes, for the spaces of sections of associated bundles with structure Hopf algebra one uses the cotensor product construction in the affine case; this is well known in the literature; the recipe can also be globalized by gluing along localizations. However this does not give the total spaces of associated bundles (in the category of noncommutative spaces) in satisfactory way in general, but only the spaces of sections.

Latex Template for a Popular Math Journal

Some journals provide class files and article templates for prospective authors, so possibly a good place to start would be to have a look at the submission guidelines pages for various journals. Geometry & Topology, for example, provides a public class file (which gives very similar results to the ones we use internally for producing the published versions) and an article template along with a page of LaTeX good practice advice and recommendations (more detailed documentation here). The template file is probably a good place to start, but will probably need a little alteration if you're using it for a different class file. (Alternatively, a web search for "LaTeX article template" should turn up something.)

Edit: Oh, I see. In that case, I guess you need to write a LaTeX class file which mimics the current journal style. And exactly how you go about that will depend on what the journal currently looks like, but I'd be inclined to start with a class file that gives almost the desired results (the basic article.cls in the absence of anything more suitable) and then start tweaking it until it looks right. Your original question seems to indicate that it looks a bit like the Notices of the AMS, but it doesn't look like they make their class files publicly available. Be warned also that some serious TeX hackery (HaXery?) will probably be involved.

The other thing to bear in mind is that if the production processes are currently built around a system that isn't LaTeX then you might have an uphill struggle persuading the editor and production team to switch, unless there are clear and obvious advantages to doing so. (I speak as someone who once wasted several weeks attempting, unsuccessfully, to convince a dyed-in-the-wool Fortran 77 programmer of the benefits of object-oriented programming.)

ac.commutative algebra - Do all Dedekind domains have the "Riemann-Roch property"?

Let $R$ be a Dedekind domain with fraction field $K$.

Say that a Dedekind domain $R$ has the Riemann-Roch property if: for every nonzero prime ideal $mathfrak{p}$ of $R$, there exists an element $f in (bigcap_{mathfrak{q} neq mathfrak{p}} R_{mathfrak{q}}) setminus R$, i.e., an element of $K$ which is integral at every prime ideal $mathfrak{q} neq mathfrak{p}$ and is not integral at $mathfrak{p}$.

Do all Dedekind domains have the Riemann-Roch property?

Motivation: for any subset $Sigma subset operatorname{MaxSpec}(R)$, put $R_{Sigma} := bigcap_{mathfrak{p} in Sigma} R_{mathfrak{p}}$. Then the maximal ideals of $R_{Sigma}$ correspond bijectively to the maximal ideals $mathfrak{p}$ of $R$ such that $mathfrak{p} R_{Sigma} subsetneq R_{Sigma}$. Thus $operatorname{MaxSpec}(R_{Sigma})$ may be viewed as containing $Sigma$. $R$ has the Riemann-Roch property iff for all $Sigma$,
$operatorname{MaxSpec}(R_{Sigma}) = Sigma$. Equivalently, the mapping $Sigma mapsto R_{Sigma}$ is an injection.

Remarks: $R$ has the Riemann-Roch property if its class group is torsion: then for every $mathfrak{p} in operatorname{MaxSpec}(R)$ there exists $n in mathbb{Z}^+$ and $x in R$ such that $mathfrak{p}^n = (x)$, so take $f = frac{1}{x}$. Also the coordinate ring $k[C]$ of a nonsingular, integral affine curve $C$ over a field $k$ has the Riemann-Roch property...by the Riemann-Roch theorem. Unfortunately this already exhausts the most familiar examples of Dedekind domains!

ca.analysis and odes - How helpful is non-standard analysis?

The other answers are excellent, but let me add a few
points.

First, with a historical perspective, all the early
fundamental theorems of calculus were first proved via
methods using infinitesimals, rather than by methods using
epsilon-delta arguments, since those methods did not appear
until the nineteenth century. Calculus proceeded for
centuries on the infinitesimal foundation, and the early
arguments---whatever their level of rigor---are closer to
their modern analogues in nonstandard analysis than to
their modern analogues in epsilon-delta methods. In this
sense, one could reasonably answer your question by
pointing to any of these early fundamental theorems.

To be sure, the epsilon-delta methods arose in part because
mathematicians became unsure of the foundational validity
of infinitesimals. But since nonstandard analysis exactly
provides the missing legitimacy, the original motivation
for adopting epsilon-delta arguments appears to fall away.

Second, while it is true that almost any application of
nonstandard analysis in analysis can be carried out using
standard methods, the converse is also true. That is,
epsilon-delta arguments can often also be translated into
nonstandard analysis. Furthermore, someone raised with
nonstandard analysis in their mathematical childhood would
likely prefer things this way. In this sense, the
preference between the two methods may be a cultural matter
of upbringing.

For example, H. Jerome Keisler wrote an introductory calculus
textbook called Elementary Calculus: an infinitesimal
approach, and
this text was used for many years as the main calculus
textbook at the University of Wisconsin, Madison. I
encourage you to take a look at this interesting text,
which looks at first like an ordinary calculus textbook,
except that in the inside cover, next to the various
formulas for derivatives and integrals, there are also
listed the various rules for manipulating infinitesimals,
which fill the text. Kiesler writes:

This is a calculus textbook at the college Freshman
level based on Abraham Robinson's infinitesimals, which
date from 1960. Robinson's modern infinitesimal approach
puts the intuitive ideas of the founders of the calculus
on a mathematically sound footing, and is easier for
beginners to understand than the more common approach via
limits.

Finally, third, some may take your question to presume that
a central purpose of nonstandard analysis is to provide
applications in analysis. But this is not correct. The
concept of nonstandard models of arithmetic, of analysis
and of set theory arose in mathematical logic and has grown
into an entire field, with hundreds of articles and many
books, with its own problems and questions and methods,
quite divorced from any application of the methods in other
parts of mathematics. For example, the subject of Models
of
Arithmetic
is focused on understanding the nonstandard models of the
first order Peano Axioms, and it makes little sense to
analyze these models using only standard methods.

To mention just a few fascinating classical theorems: every
countable nonstandard model of arithmetic is isomorphic to
a proper initial segment of itself (H. Friedman). Under the
Continuum Hypothesis, every Scott set (a family of sets of
natural numbers closed under Boolean operations, Turing
reducibility and satisfying Konig's lemma) is the
collection of definable sets of natural numbers of some
nonstandard model of arithmetic (D. Scott and others).
There is no nonstandard model of arithmetic for which
either addition or multiplication is computable (S.
Tennenbaum). Nonstandard models of arithmetic were also
used to prove several fascinating independence results over
PA, such as the results on
Goodstein sequences,
as well as the
Paris-Harrington theorem on the
independence over PA of a strong Ramsey theorem. Another
interesting
result
shows that various forms of the pigeon hole principle are
not equivalent over weak base theories; for example, the
weak pigeon-hole principle that there is no bijection of n
to 2n is not provable over the base theory from the weaker
principle that there is no bijection of n with
n². These proofs all make fundamental use of
nonstandard methods, which it would seem difficult or
impossible to omit or to translate to standard methods.

nt.number theory - Computing zeta(k), for k odd, using Fourier coefficients

I'm not really sure what topics exactly this falls under, so I apologize if I've misclassified this question.

There is a neat way of computing $sum_{n=1}^{infty}frac{1}{n^{2}}$ using Fourier analysis: Compute the Fourier series of $t^2$ extended $2pi$-periodically, which turns out to be

$$frac{pi^{2}}{3}+4sum_{n=1}^{infty}frac{1}{n^{2}}$$

By Fejer's theorem (I think), the Fourier series around $pi$ converges, so we get an equation that can be solved for the $zeta(2)$.

I think a similar approach can be taken for $zeta(2k)$ by taking $t^{2k}$ extended $2pi$-periodically, but all my attempts to do something like this for odd integers fail.

On the other hand, since $1/n^k$ for $k$ odd is in $ell^2$, there should be an $L^2$ function that has the sequence as its Fourier coefficients. Can one be explicitly constructed? What if we allow the entries in the sequence to alternate, or let finitely many of them deviate from $1/n^k$?

Basically, I want to find an $L^2$, $2pi$-periodic function whose Fourier coefficients would give a relatively straightforward computation of $zeta(k)$ when $k$ is odd.

analytic continuations

To answer your specific questions, a) a maximal (by inclusion) domain does not have to exist. Consider $sqrt{1-z}$ in the unit disk. That is actually the reason why Riemann introduced Riemann surfaces.

But there are maximal domains in various other senses. Suppose you have an analytic germ at infinity. Then there is a unique (!) set $K$ in the plane of
of minimal logarithmic capacity, such that our germ has a (single-valued) analytic continuation to $Cbackslash K$. This is due to H. Stahl.

b) The answer depends on what you exactly mean by a criterion. There are necessary and sufficient conditions for existence of an analytic continuation
(multiple valued) to a given region. Not surprisingly they are difficult to check. See, for example
MR1711032
Atzmon, A., Eremenko, A., Sodin, M.
Spectral inclusion and analytic continuation.
Bull. London Math. Soc. 31 (1999), no. 6, 722–728.

There is a necessary and sufficient condition for a function defined in the unit disk to have an analytic continuation to a given point on the circumference (Euler).

And there are many separate necessary (but not sufficient) and sufficient (but not necessary) conditions in terms of coefficients in various special cases. This is a very vast subject. A good survey of these conditions can be found in Bieberbach, Analytische Fortsetzung, Springer, 1955, and
V. Bernstein, Lecons sur les progres recents de la théorie des séries de Dirichlet, Paris: Gauthier- Villars. XIV, 320 S. (1933).

ag.algebraic geometry - Algebraic data and purity associated to codimension greater than 2

Consider the following statement: Let $X$ be a smooth and geometrically integral variety over a field $k$ and let $U$ be any open subset of $X$ whose complement is of codimension greater or equal to $2$. Then the following statements hold

$k[X] cong k[U]$ (clear)

$Pic(X) cong Pic(U)$ (restriction of codimension $1$ irreducibles)

$Br(X) cong Br(U)$ (Grothendieck purity for the Brauer group, see 'Le groupe de Brauer III')

$pi_1^{et}(X) cong pi_1^{et}(U)$ (SGA1 Corollary X.3.3)

The middle two statements are essentially referring to cohomology groups with $mathbb{G}_m$ coefficients.

The question is (albeit rather vague) whether there is some reason why one would expect a lot of the algebraic data associated to a scheme to be in some sense encoded in lower dimension. This is in line with various purity statements such as that in etale cohomology with coefficients in locally constant sheaves of $mathbb{Z}/nmathbb{Z}$ modules as can be found in Milne's book. Do many other statements of this type exist?

ag.algebraic geometry - equivalence of Grothendieck-style versus Cech-style sheaf cohomology

Let $X$ be a topological space, and $T$ its category of open sets with the usual Grothendieck topology. Let $T'$ be any sieve of $T$ (a subcategory of $T$ such that if $U$ is in $T'$ then any subset of $U$ is also in $T'$). For example, $T'$ might be the collection of open subsets subordinate to the open subsets in a cover $mathcal{U}$. Any sheaf on $T$ induces a functor on $T'$ which can be viewed as a sheaf on $T'$ if $T'$ is given the minimal topology (the only covers are the identity maps). This determines a morphism of topoi $f : T rightarrow T'$, hence a spectral sequence

$H^p(T', R^q f_ast F) Rightarrow H^{p+q}(T, F)$ .

(One could surely also convince oneself that such a spectral sequence exists without any reference to topoi.)

The Cech cohomology of $F$ with respect to some covering family $mathcal{U}$ is

$H^p(mathcal{U}, F) = H^p(T', f_ast F)$

where $T' = T'(U)$ is the sieve associated to the cover $mathcal{U}$. The Cech cohomology is then the filtered colimit

$check{H}^p(T, F) = varinjlim_{(T',f)} H^p(T', f_ast F)$

taken over the projections $f : T rightarrow T'$ associated as above to covering families $mathcal{U}$.

One evidently has edge homomorphisms

$check{H}^p(T, F) rightarrow H^p(T, F)$

from the spectral sequence, and the question is when these induce an isomorphism. If we could somehow eliminate the $R^p f_ast F$, $p > 0$, by passing to a "small enough" cover we would have equality. This condition already holds in many cases; the following condition is more general (but I haven't checked carefully that it actually works!):

For every cover $mathcal{U}$ of $X$, every $U_1, ldots, U_n in mathcal{U}$, and every class in $alpha in H^p(U_1 mathop{times}_X cdots mathop{times}_X U_n, F)$, $p > 0$, there exists a refinement $mathcal{U}'$ of $mathcal{U}$ such that the restriction of $alpha$ under the map

$H^p(U_1 mathop{times}_X cdots mathop{times}_X U_n, F) rightarrow H^p(U'_1 mathop{times}_X cdots mathop{times}_X U'_n, F)$

is zero.

To make sense of this, one must use some convention for the covers $mathcal{U}$ and $mathcal{U}'$ to ensure there is a map as above. For example, one could work only with covers indexed by the points of $X$ (a cover is then a collection of neighborhoods of each point of $X$).

A more refined version of the above condition would say that Cech cohomology equals cohomology in degrees at most $q$ if the above condition holds for $p leq q$. Since it always holds for $p = 0,1$ this implies that

$check{H}^1(T, F) = H^1(T, F)$

in general.

Edit in response to David's comment:

The Cech complex always computes cohomology correctly in a presheaf category (i.e., when the topology is "chaotic": an object has no covers by anything except itself). Trying to compute cohomology in an arbitrary site using the Cech complex is (heuristically) something like trying to approximate the site by a presheaf category.

Here is how Cech cohomology computes cohomology of presheaves. Consider any category $T'$. If $F$ is a presheaf of groups on $T'$ then the sheaf cohomology groups of $F$ are the derived functors of the inverse limit for diagrams of shape $T'$. They are also computed as

$Ext(mathbf{Z}, F)$

where $mathbf{Z}$ is the constant sheaf associated to the integers. Remarkably, in a presheaf category, $mathbf{Z}$ has a canonical projective resolution associated to any cover of the final presheaf. A cover of the final presheaf is a collection of objects $U$ of $T'$ such that every object of $T'$ has a map to at least one object of $U$. The $i$-th term of this complex is the direct sum, over all choices of $i$ elements $U_1, ..., U_i$ of $U$, of the groups $mathbf{Z}_{U_1 times cdots times U_i}$. (You can check this is projective by noting it is the extension by $0$ of $mathbf{Z}$ from the slice category $T' / U_1 times cdots times U_i$ and extension by $0$ preserves projectives (since it has an exact right adjoint) and $mathbf{Z}$ is projective on the slice category since all higher cohomology of all sheaves vanishes (since it has a final object). It's also easy to check by a direct calculation.)

Denote this complex by $K$. Since this is a projective resolution of $mathbf{Z}$, $mathrm{Hom}(K, F)$ computes the cohomology of $F$. But it is also easy to see that this is just the Cech complex of $F$.

set theory - Determinacy interchanging the roles of both players

Let me refer to Jech's "Set Theory" Chap. 33 Determinacy:

"With each subset A of $omega^omega$ we associate the following game $G_A$, played by two players I and II. First I chooses a natural number $a_0$, then II chooses a natural number $b_0$, then I chooses a1, then II chooses b1, and so on. The game ends after $omega$ steps; if the resulting sequence $<a_0, b_0, a_1, b_1, ...>$ is in A, then I wins, otherwise II wins.

A strategy (for I or II) is a rule that tells the player what move to make depending on the previous moves of both players. A strategy is a winning strategy if the player who follows it always wins. The game $G_A$ is determined if one of the players has a winning strategy.

The Axiom of Determinacy (AD) states that for every subset A of $omega^omega$, the game $G_A$ is determined."

Now, there is some apparent lack of symmetry in the definition of the $G_A$ game: the player who plays first (I) attempts for a sequence in A.

What happens if we interchange the roles of both players? I. e. if we let the player who plays first attempt for a sequence not in A? Let us call this game $G'_A$

Is it the case that for every subset A, A is determined wrt $G_A$ iff A is determined wrt $G'_A$?

ap.analysis of pdes - What are the interesting cases of the generalized Korteweg-de Vries equation?

José is correct in his comment. Just to elaborate: in the linear case, one can easily study the equation using Fourier methods. Let $tilde{u}$ denote the space-time Fourier transform and $hat{u}$ denote the spatial Fourier transform, the equation with $p=1$ can be written as
$$ (tau - xi^3 + xi)tilde{u} = 0$$
or
$$ partial_that{u} = i(xi^3 - xi)hat{u} $$

The first formulation tells you that the space-time Fourier transform of a solution is a measure supported on the curve $tau = xi^3 - xi$ in frequency space. That the frequency support has curvature implies that the solution should decay in time in physical space, justifying José's comment. (Look up Fourier restriction theorems or Strichartz estimates in the literature for more details.)

We can also see the temporal decay directly from the second formulation using oscillatory integral techniques. The solution can be written as
$$ hat{u}(t,xi) = e^{i(xi^3 - xi)t}hat{u}_0(xi) $$
It is then a standard exercise to show that, given initial data of sufficient decay in frequency space (say, the frequency has compact support), taking the inverse Fourier transform of the above solution gives you something with decay in $L^infty$.

If a solution decays in $L^infty$, it cannot be a soliton.

We can also see the lack of solitons by posing the traveling wave ansatz $u(t,x) = f(omega t + x)$. A simple computation shows that the function $f$ must solve
$$ -(omega + 1)f' = f''' implies f = exp [ i sqrt{omega + 1}(omega t + x) ] $$
so that traveling waves cannot be spatially localized. In fact, the traveling wave ansatz gives us a dispersion relation for this equation: that waves of spatial frequency $beta$ travels with velocity $omega = beta^2 - 1$. The fact that different frequency components of the wave tend to travel at different velocities illustrates why, starting with a pulsed wave packet, the solution will become wider and wider while its height gets smaller and smaller.

In short, I think the reason why this equation is not heavily studied in the literature is because that as a linear PDE in (1+1) dimensions, it is not really all that interesting to look at.

st.statistics - Expected distance between two points with missing coordinates

You can rephrase your question as follows: first we subtract the known vector from both and then take care of the known coordinates. So assuming the coordinates of the two points are $(alpha,beta)$ and $(gamma,X)$ where $alpha,gamma in mathbb{R}^m$ are known, and $beta in mathbb{R}^n$ is known, but $X$ represent the unknown coordinates constrainted to lie inside the cube $[-1,1]^n$, the integral to evaluate becomes
$$frac{1}{2^n}int_{[-1,1]^n} sqrt{ |alpha-gamma|^2 + |beta - X|^2 } dX$$
In the lower dimensional case this can be integrated. But an analytical expression in higher dimensions is elusive. For the case $alpha = gamma$ and $beta = 0$, some bounds were obtained in an old paper of Anderssen et al. http://dx.doi.org/10.1137/0130003 For more general probability distributions there is a recent paper with some bounds by Burgstaller and Pillichshammer. http://journals.cambridge.org/action/displayAbstract?aid=6622208

Of course, one can get a fairly trivial bound by Cauchy-Schwartz
$$ int_{[-1,1]^n} f(X) dX leq 2^{n/2} left( int_{[-1,1]^n} f(X)^2 dX right)^{1/2} $$
and that
$$ int_{[-1,1]^n} R^2 + |beta - X|^2 dX = 2^n (R^2 + beta^2) + int_{[-1,1]^n} X^2 dX$$
the last term is simply evaluated as $n 2^n / 3$, so putting it all together we have the upper bound for the expected value by
$$ frac{1}{2^n}int_{[-1,1]^n} sqrt{ |alpha-gamma|^2 + |beta - X|^2 } dX leq sqrt{ |alpha -gamma|^2 + beta^2 + frac{n}{3}}$$
which is slight improvement over the utterly trivial upper/lower bound of $sqrt{|alpha-gamma|^2 + beta^2 pm n}$ if you just maximize/minimize each coordinates.

nt.number theory - Number of integral solutions to multi-variable polynomials

This question follows the article discussed here

---

Problem

Suppose we're trying to bound the number of integral solutions to a system of multi-variable polynomials,
say

$$ sum_{i=1}^n x_i^t = sum_{i=1}^n y_i^t, $$
where each $x_i,y_i in mathbb N$ and for each $t < c$ for some constant $c$.

If we do not put any constrains on the solution,
there are infinitely many possible solutions even when $n=C=1$.
So if we put some constrains on {$x_i,y_i$} like $x_i,y_i in$ {$0,1,ldots,n$},
then how many possible solutions can we get?
Naively there are $O(n^n)$ choices, but it seems highly unlikely that there are many solutions to the system of equations. Is there any exist bound on the number of solutions,
say $O(n^k)$ for fixed $k$ or even better bounds? Are there some well-known approaches to bound the number of solutions of an equation?

Motivation

This question arose when I'm trying to come up with some reasonable constrains with the equation in Prouhet-Tarry-Escott Problem.
It seems like if we restrict the maximum value of variables, there aren't many solutions to the equation. I tried to add more constrains to get rid of the already few solutions,
but it seems that there is no direct way making the solution set empty, that is, no possible solutions under such constrains.

So I turn to find some existing bounds for the equation, but sadly nothing occurred.
Can it be still hard to find such results, or there are some theorems like the Fundamental Theorem of Algebra, concerning the number of solutions to a multi-variable equation? Any information is useful. Thank you all!

Edited

According to Felipe Voloch (Thanks!), the general approach to the question is the Hardy-Littlewood method, which considers the number of solutions to an equal-power Diophantine equation. But it seems that the method gives a lower bound on the number of solutions (is this correct?), rather than an upper bound. Or there are some ways to give upper bounds by the same method?

One more question: How about further restricting the solutions to be prime numbers?
Does this make any difference?

homotopy theory - Do the signs in Puppe sequences matter?

There is a specific sort of situation I know about where that sign matters. Suppose you have $f:X rightarrow Y$ and $g:Z rightarrow W$ cofibrations (if the maps are not cofibrations, all the same things work - you just replace the quotient spaces by mapping cones). You extend both maps to their Dold-Puppe sequences, so you get the sequences

$X rightarrow Y rightarrow Y/X rightarrow Sigma X rightarrow Sigma Y ldots$

and

$Z rightarrow W rightarrow W/Z rightarrow Sigma Z rightarrow Sigma W ldots$

Now suppose you have maps $a: X rightarrow W$ and $b: Y rightarrow W/Z$ making the obvious square commute up to homotopy. You can then extend these to make a commutative ladder from the first Dold-Puppe sequence to the second. (Notice that the sequences are deliberately offset from each other by one spot.)

Using the usual parameters and the obvious choices of homotopies you will get a square involving $Y/X, Sigma X, Sigma Z, Sigma W$. This square will commute if it includes the map $-Sigma g: Sigma Z rightarrow Sigma W$, but not generally with the map $Sigma g$. (To check all this, I recommend doing the Dold-Puppe sequences with mapping cones rather than quotient spaces but keeping the homotopy equivalences with the quotient spaces in mind, which is the only way I know to calculate what the right maps should be.)

At this point, if you were feeling stubborn, you could replace the map in your ladder $Sigma X rightarrow Sigma Z$ with $-1$ times that map, and that would allow you to have used $Sigma g$ in the square I mention in the above paragraph, but that creates other issues; if you choose not to simply use suspensions of your original maps to go from one Dold-Puppe sequence to the other then you run into problem when you are mapping between Dold-Puppe sequences without the shift of this example.

I hope this helps unravel Greg's answer (which is correct - you need the sign to get good mapping properties).

Of course one sees exactly the same phenomenon in the category of chain complexes of abelian groups (where homotopy is chain homotopy) and other such categories. I agree with Theo and Mark that one thinks about the suspension as "odd" (in the sense of parity not the sense of peculiar).

The published paper that Mark refers to that has an error of exactly this sort (which is unfortunately fundamental to the paper) is by Lin Jinkun in Topology v. 29, no. 4, pp. 389-407. I read this paper in preprint form in 1988 and missed this error, but discovered it in 1992 when reading another paper by the same author with the same error. In the Topology paper the error is made in diagram 4.4 on the right hand square (proof of Lemma 4.3).

mg.metric geometry - Volumes of Sets of Constant Width in High Dimensions

Background

The n dimensional Euclidean ball of radius 1/2 has width 1 in every direction. Namely, when you consider a pair of parallel tangent hyperplanes in any direction the distance between them is 1.

There are other sets of constant width 1. A famous one is the Reuleaux triangle in the plane. The isoperimetric inequality implies that among all sets of constant width 1 the ball has largest volume. Lets denote the Volume of the n-ball of radius 1/2 by $V_n$.

The question

Is there some $epsilon >0$ so that for every $n>1$ there exist a set $K_n$ of constant width 1 in dimension n whose volume satisfies $VOL(K_n) le (1-epsilon)^n V_n$.

This question was asked by Oded Schramm who also asked it for spherical sets of constant width r.

A proposed construction

Here is a proposed construction (also by Schramm). It will be interesting to examine what is the asymptotic behavior of the volume. (And also what is the volume in small dimensions 3,4,...)

Start with $K_1=[-1/2,1/2]$. Given $K_n$ consider it embedded in the hyperplane of all points in $R^{n+1}$ whose (n+1)th coordinate is zero.

Let $K^+_{n+1}$ be the set of all points $x$, with nonnegative (n+1)th coordinate, such that the ball of radius 1 with center at $x$ contains $K_n$.

Let $K^-_{n+1}$ be the set of all points $x$, with nonpositive (n+1)th coordinate, such that $x$ belongs to the intersection of all balls of radius 1 containing $K_n$.

Let $K_{n+1}$ be the union of these two sets $K^-_{n+1}$ and $K^+_{n+1}$.

Motivation

Sets of constant width (other than the ball) are not lucky enough to serve as norms of Banach spaces and to attract the powerful Banach space theorist to study their asymptotic properties for large dimensions. But they are very exciting and this looks like a very basic question.

References and additional motivation

In the paper: "On the volume of sets having constant width" Isr. J Math 63(1988) 178-182 Oded Schramm gives a lower bound on volumes of sets of constant width. Schramm wrote that a good way to present the volume of a set $K subset R^n$ is to specify the radius of the ball having the same volume as $K$, called it the effective radius of the set $K$ and denote it by $er K$. Next he defined $r_n$ as the minimal effective radius of all sets having constant width two in $R^n$. Schramm proved that $r_n ge sqrt {3+2/(n+1)}-1$. He asked if the limit of $r_n$ exists and if $r_n$ is a monotone decreasing sequence.

Our question is essentially wheather $r_n$ tends to 1 as $n$ tends to infinity.

In the paper: O. Schramm, Illuminating sets of constant width. Mathematika 35 (1988), 180--189. Schramm proved a similar lower bound for the spherical case and deduced the best known upper bound for Borsuk's problem on covering sets with sets of smaller diameter.

ho.history overview - Characterizing triangles unembeddedly

The mathedu mailing list has a recent longish thread at

http://www.nabble.com/Why-do-we-do-proofs--to25809591.html

which discussed among other things whether we should teach triangles as labeled or unlabeled to high school students (this is a vast oversimplification of the thread). I have long been concerned with how we think (informally and formally) about mathematical objects, so naturally I started to consider how we think about triangles.

Consider circles. Most informal and formal descriptions involve an embedding into R^2, but they can be characterized as manifolds (even as Riemannian manifolds) of dimension 1 with specific properties, independent of any embedding. This sort of thing has turned out to be a major way to think about all sorts of spaces. So can we describe triangles in a similar way?

Unfortunately, manifolds are far removed from my usual mathematical work (category theory). What I think I understand is that there can be piecewise linear manifolds, even Riemannian ones. So perhaps we can say a triangle is a piecewise linear manifold of dimension 1 with certain properties. Now, I want to define a triangle so that it comes complete with information about the lengths of its sides and what the three angles are. Riemannian manifolds have a way to specify length and angles, and I can believe you can make the sides have specific lengths. But the angles? It seems to me that the tangent spaces (like those on a circle) result in all angles being 0 or pi, except at the corners where they don't exist. But I may not understand the situation correctly.

So my question is: Is there a known methodology that allows triangles to be characterized independent of embeddings in such a way that incorporates information about side lengths and angles?

nt.number theory - Why Is 163/ln(163) a Near-Integer?

What make us confident about some mystery in these observations?

1st note:
"An example discovered by Srinivasa Ramanujan around 1913 is $exp(pisqrt{163})$,
which is an integer to one part in $10^{30}$, and has second continued fraction term
$1,333,462,407,511$. (This particular example can be understood from the fact
that as $d$ increases $exp(pisqrt{d})$ becomes extremely close to
$j((1 + sqrt{-d})/2)$, which turns out to be an integer whenever there
is unique factorization of numbers of the form $a + b sqrt{-d}$ --- and $d=163$
is the largest of the 9 cases for which this is so.) Other less spectacular examples
include $e^{pi}-pi$ and $163/log(163)$."

2nd note:
"Any computation involving 163 gives an answer that is close to an integer:
$$
163pi = 512.07960dots, quad
163e = 443.07993dots, quad
163gamma = 94.08615dotstext{"}
$$
and
$$
text{"}67/log(67)=15.9345774031dots, quad
43/log(43)=11.432521184dots
$$
...nah, with class number 1 it's not connected.
It's just the same 163. $ddotsmile$"

A synthetic example of my own:
$$
root3of{163}-frac{49,163}{9,000}
=0.0000000157258dots
$$
(note the double appearance of 163).

So, let's feel that the prime 163 is a supernatural number. $ddotsmile$

EDIT. Another interpretation the original question is related to
the observation of Kevin O'Bryant who computed the first successive maxima
of the sequence $|n/log(n)|$ where $| cdot |$ denotes the distance
to the nearest integer. The existence of infinitely many terms
is guaranteed by the following

Problem.
For any $epsilon>0$, there exists an $n$ such that $|n/log(n)|<epsilon$.

See solution by Kevin Ventullo to this question. I hope that this fact demystifies the original problem in full.

nt.number theory - Are Q-curves now known to be modular?

I really should know the answer to this, but I don't, so I'll ask here.

A Q-curve is an elliptic curve E over Q-bar which is isogenous to all its Galois conjugates. A Q-curve is modular if it's isogenous (over Q-bar) to some factor of the Jacobian of X_1(N) for some N>=1 (here X_1(N) is the compact modular curve over Q-bar).

Has current machinery proved the well-known conjecture that all Q-curves are modular yet?

Remark: I know there are many partial results. What I'm trying to establish is whether things like Khare-Wintenberger plus best-known modularity lifting theorems are strong enough to give the full conjecture yet, or whether we're still waiting.

lo.logic - Order types of positive reals

Yes, one can have any countable ordering. Indeed any countable totally
ordered set can be embedded in $mathbb{Q}$. Write your ordered set as
$ lbrace a_1,a_2,ldots rbrace $
and define the embedding recursively: once you have placed $a_1,ldots,a_{n-1}$
there will always be an interval to slot $a_n$ into.

mathematical finance - Transformation of the Black-Scholes PDE into the diffusion equation - shift of coordinate system

I'm afraid the Planetmath page put my browser into an infinite reload loop, so I can't help you with the formalism there.

I would recommend instead looking at the change of variables in the Wikipedia article. The last time I checked it, it seemed to work.

Edit: Okay, I have a formulation that works. I'll write s for sigma, so the equation is initially:

V_t + (1/2)s²S²V_SS = rV - rSV_S.

Since S follows a lognormal random walk (in particular the stochastic diff eq governing S involves a logarithmic derivative), it is natural to change to x = log S, or S = e^x, so the log price x follows normal Brownian motion. This yields the equation:

V_t + (1/2)s²(V_xx - V_x) = r(V - V_x).

Black-Scholes is a final-value problem, i.e., we know the value of the option at time T, and diffusion works backwards. It is therefore natural to negate the time variable (and multiply by a suitable scalar to make things neater). tau = (1/2)s²(T-t). Then we get:

(1/2)s²(V_xx - V_x - V_tau) = r(V - V_x).

Finally we rescale the value function to remove exponential growth effects. u = e^{ax + b(tau)}V for undetermined coefficients a and b. We can substitute, multiply the equation by 2e^ax+b(tau)/s², and we get:

u_xx + (something)u_x + (something else)u = u_tau.

(something) is a degree one polynomial in a and is independent of b. (something else) is a degree one polynomial in b, so we can choose a and b to kill those terms. This yields the diffusion equation.

Hope that helps.

nt.number theory - Factorizing polynomials of several variables (in a different perespective)

I am looking for factorization of polynomials of several variables in the way outlined below.

Consider a second degree polynomial of two variables over the complex numbers.

"P(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F" (see the edit below)

Experimenting with some polynomials of this sort showed me that factorization is possible in the following way.

     "P(x,y) = (ax + by + c)(dx + ey + f)"  (see the edit below) ,

the coefficients being over the complex numbers.

So, given an nth degree polynomial in n variables without a constant term, is it always possible to factorize it into n linear factors each having n variables in the above way? (This rings bells about the fundamental theorem of algebra).

Please suggest a reading material or journal, if any.

[EDIT: I am sorry, I erred. I have edited my question. The edit is that the polynomial has no constant term:
P(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey
and in the expected factorization, the last linear factor does not have a constant term, too:
P(x,y) = (ax + by + c)(dx + ey + f) ]

I kept the original question as it is for documentation purposes.

gr.group theory - Does Burnside's Lemma / Counting Formula have a Cousin?

Burnside's Lemma / Counting Formula says that the number of orbits of an action is equal to the average number of fixed points of the acting permutations. In my case, I'm particularly interested in the sizes of the orbits themselves for a particular action.

Is there a result as general as Burnside's Lemma but that deals with the sizes of orbits? Or, more specifically, does the following setup seem familiar to anyone?

Additional Motivation: The motivation for my question is something like the following.

Suppose we have a set of $m$ sharply transitive permutations $Pi = {pi_1, ldots, pi_m}$ on a set $X$. Here I mean that for $x,y in X$, there is a unique $pi in Pi$ such that $pi(x) = y$.

The permutations cooperate "nicely" in the following sense. Let $A$ be an $m times m$ matrix with off-diagonal entries drawn from ${1, ldots, m}$ such that for $i neq j$ whenever $pi_i (x) = pi_j (y)$, it follows that $pi_j (x) = pi_{A(i,j)} (y)$.

Let $G$ be the permutation group generated by $Pi$. In the general case I'm working on, I really only know $A$, and I'm shooting for a statement of the form: "If $A$ has a certain property, then $|G(x)|$ is divisible by $m$."

The origins of the matrix $A$ while of course essential to proving anything like this statement remain sufficiently messy that I'd rather not get into it.

pr.probability - Random Walks and Lyapunov exponents

Random dynamical systems by Ludwig Arnold contains a thorough discussion of various multiplicative ergodic theorems (including the Furstenberg-Kesten result), but not the central limit theorems. As far as I remember, the case of stationary sequences of linear stochastic iterations is also included there.

Edit. Concerning central limit theorems for products of random matrices, a quick search yields this reference.

ag.algebraic geometry - Rational map defined over K leads to algebra question

Hello,

Concrete algebraic question : Let $K$ be a perfect field, $bar{K}$ a fixed algebraic closure and let $f in bar{K}[x_1,ldots,x_n]$. I was wondering when there exists another polynomial (non-zero) $g in bar{K}[x_1,ldots,x_n]$ such that $fg in K[x_1,ldots,x_n]$ ?

Another formulation would be, when $f cdot bar{K}[x_1,ldots,x_n] cap K[x_1,ldots,x_n]$ is stricly larger than {0} ?

My motivation : Let $V subseteq mathbb{P}^n(bar{K}) $ be a variety defined over $K$, i.e., its ideal $I(V)$ can be generated (over $bar{K}[x_1,ldots,x_n]$) by polynomials in $K[x_0,ldots,x_n]$.
Let $phi = [f_0, ldots ,f_n] : V_1 to V_2$ be a rational map between projective varieties defined over $K$. The arithmetic of elliptic curves, by Silverman, say $phi$ is defined over $K$ when there exists some $lambda in bar{K}^{ast}$ such that the $lambda f_0, ldots, lambda f_n in K(V_1) = frac left( K[x_0,ldots,x_n] / I(V) right)$, where $I(V)$ is generated by polynomials in $K[x_0,ldots x_n]$.

Hence if $phi = [i, i] : mathbb{P}^1(mathbb{C}) to mathbb{P}^1(mathbb{C})$ and $psi : [X-i,X-i] : mathbb{P}^1(mathbb{C}) to mathbb{P}^1(mathbb{C})$, I hope I'm not wrong if I say they are not the same rational maps (the first being defined over $mathbb{R}$ but not the second one). But they both are the morphism $[1,1]$.

co.combinatorics - Is 8 the largest cube in fibonacci sequence?

This is only a partial answer, but: one characterization of Fibonacci numbers is that an integer $a$ is Fibonacci if and only if $5a^2 pm 4$ is a perfect square.

Because of this: finding Fibonacci cubes is equivalent to solving the Diophantine equation(s) $5a^6 pm 4 = b^2$. There are standard techniques for doing this: for example, in the case of $5a^6 + 4 = b^2$, the equation can be rewritten and factored as $5a^6 = (b-2)(b+2)$, and using the fact that any common factor of $b-2$ and $b+2$ must divide 4, you can get a lot of information out of this situation by looking at the prime factorizations of both sides.

In the case of $5a^6 + 4 = b^2$, you have to pass to the Gaussian integers (numbers of the form $a+bi$ with $a$ and $b$ both integers) to obtain a factorization: $5a^6 = (b+2i)(b-2i)$. Again, you can apply the same approach (Gaussian integers also have unique factorization): it will be a little bit trickier, but I've seen problems like this solved by the same method.

Anyone want to try to fill in this sketch of a proof?

Edit: Having thought about it a bit more, I'm less optimistic. The problem is dealing with the factors of 5: the methods I describe are excellent for dealing with questions like $a^6 pm 4 = b^2$ or even $a^3 pm 4 = b^2$, but those arguments break down when you take the 5 into account. Also, any method that uses either of the factorizations I mentioned will also have to say something about solutions to the equation $5a^2 pm 4 = b^2$: but we know that this equation has many solutions (since a can be any Fibonacci number). There are other potential ways we could rearrange this equation to factor it, but analyzing them won't be easy.

However, rephrasing this question as a Diophantine equation is still helpful. There's a general and rather hard theorem (Siegel's Theorem) that any Diophantine equation of genus $ge1$ has only finitely many solutions. What "genus" is is a bit technical, but most equations of degree $ge3$ have this property, and in particular Siegel's equation applies to equations of the form $y^2 = P(x)$ where $P(x)$ is a polynomial in $x$ of degree $ge3$ (with no repeated roots), which includes our two equations.

So we can deduce from this that there are only finitely many Fibonacci numbers that are cubes. Unfortunately, Siegel's result isn't very helpful for the original problem because it doesn't give you any way of telling when you've found all of them.

Additionally, we know that if we have an solution to $5a^6 pm 4 = b^2$, setting $c = a^2$ gives us a solution to the equation $5c^3 pm 4 = b^2$. This is the equation of an elliptic curve, and there is a fair amount known about how to find the integer points on an elliptic curves (and more generally, rational points on elliptic curves), although the computations involved are usually not fun to do by hand. Anyone know how hard it would be to get a computer to solve this problem?

co.combinatorics - Given n k-element subsets of n, is there a small subset A of n which intersects them all?

I believe, reading the abstract, that the paper "Transversal numbers of uniform hypergraphs", Graphs and Combinatorics 6, no. 1, 1990 by Noga Alon answers your question in the affirmative, for some definition of ``your question''. Namely, the worst case is that $A$ has to have size about $2log k/k$ times $n$, and this multiplier tends to zero as $k$ tends to infinity.

Here's a free copy of the paper.

http://www.cs.tau.ac.il/~nogaa/PDFS/Publications/Transversal%20numbers%20of%20uniform%20hypergraphs.pdf

I'm certainly no expert on these matters and my advice would be to look at this and related literature on transversals of hypergraphs. Your collection $C$ of sets is the same thing as a $k$-uniform hypergraph, and the property that you want from $A$ is equivalent to it being a transversal.

Reading Alon's paper a little more I see that what you want is the easier direction of his argument (which gives a tight dependence on $k$). The basic idea is to choose your transversal randomly by picking elements of ${1,dots,n}$ with an appropriate probability $p$. That way, with high probability, you'll hit most of the sets from your collection $C$, and then you just add in one extra element of $A$ for each un-hit set from $C$.

Reading a little further still, I see that the upper bound is probabilistic as well: that is, to make a collection $C$ which is ``bad'', the best plan is to choose sets in $C$ at random from amongst all $k$-element subsets of ${1,dots,n}$.

There's probably literature on your ``almost transveral'' question, but I'll leave someone else to find it. My guess is that random does best in both directions there too.

nt.number theory - Relation between indefinite quadratic forms and continued fractions

Recently, I got obsessed with working out this story, to the detriment of my mathematical work. Here is a quick crib sheet for relating properties of continued fractions to properties of real quadratic fields. Warning: I haven't checked this against a standard reference, so there may be errors.

Preliminary notation: Continued fractions

It is convenient to convert a continued fraction to a sequence of binary symbols; I've been using red and blue dots, which I'll represent here as $r$ and $b$. For example, $sqrt{13}=[3,1,1,1,1,6,1,1,1,1,6,cdots]$; I'll write this as
$$r r r b r b r b b b b b b r b r b r r r r r r cdots$$

This makes the continued fraction purely periodic instead of having that peculiar $3$ at the beginning; the period starts again in the middle of that last block of six $r$'s.
Note also that the period of the fraction now appears to be twice what it was; when we get to the old period, we have switched colors. Finally, a sequence which starts with $b$'s should be thought of as a continued fraction for a number in $(0,1)$; the sequence starts with zero copies of $r$.

Preliminary notation: Real quadratic fields

Let $K$ be a quadratic field and let $Lambda$ be a rank $2$ sublattice of $K$, with $mathbb{Q} Lambda =K$. Let $mathrm{End}(Lambda)$ be the ring of $theta in K$ such that $theta Lambda subseteq Lambda$. This is an order in $mathcal{O}_K$.

Let $K$ be a real quadratic field, with fixed embedding $K to mathbb{R}$, and write $z mapsto overline{z}$ for the Galois symetry of $K$. We'll say that two lattices $Lambda_1$ and $Lambda_2$ with CM are "strictly equivalent" if there is an element $alpha in K$ such that $Lambda_1 = alpha Lambda_2$ with $alpha$ and $overline{alpha}$ both positive.

Summary of results:

$bullet$ Take any periodic sequence $a_i$ of $r$'s and $b$'s and turn it into a continued fraction. Let $z$ be the value of that continued fraction. Then $z$ is a quadratic irrational, with $z>0$ and $overline{z}<0$. We have $a_0=r$ if $z>1$ and $a_0=b$ if $z<1$. Extending the periodicity to negative indices, $a_{-1}=r$ if $overline{z} < -1$ and $a_{-1} =b$ if $overline{z} > -1$.

$bullet$ Switching the colors changes $z$ to $1/z$. Reversing the sequence changes $z$ to $-overline{z}$.

Strict ideal classes

$bullet$ Let $K = mathbb{Q}(z)$ and let $Lambda = langle 1, z rangle$. Shifting the periodic sequence does not change the strict equivalence class of $Lambda$.

$bullet$ The above is a bijection between periodic sequences of $r$'s and $b$'s, up to shift, and strict equivalence classes of lattices in real quadratic fields.

$bullet$ Switching the colors corresponds to multiplying by our lattice by an element of
negative norm. So producing a lattice which is equivalent, but not strictly equivalent.

$bullet$ Reversing the sequence sends $Lambda$ to $overline{Lambda}$. If $Lambda$ is a lattice in $K$, and $R = mathrm{End}(Lambda)$, then $Lambda overline{Lambda}$ is a strictly principal fractional ideal for $R$. (This is a special property of quadratic fields, which I know of no generalization of in higher degree number fields.) So, with the understanding that we treat a fractional ideal as a fractional ideal for its full endomorphism ring, reversing the sequence sends $Lambda$ to $Lambda^{-1}$.

Units

$bullet$ Let $R$ be the endomorphism ring of $Lambda$. Let $p/q$ be the convergent obtained by truncating the continued fraction just before the first repetition of the block which contains $a_0$. Let $u=p-qz$. Then $u$ is a unit of $R$ with norm $1$, and is the fundamental generator of the group of such units.

For example, $langle 1, sqrt{13} rangle$ has endomorphism ring $mathbb{Z}[sqrt{13}]$. We truncate the above sequence to
$$r r r b r b r b b b b b b r b r b$$
or
$$[3,1,1,1,1,6,1,1,1,1] = frac{649}{180}$$
and $649-180 sqrt{13}$ is the fundamental positive unit of $mathbb{Z}[sqrt{13}]$.

$bullet$ The color reversal of our sequence is a shift of itself if and only if $R$ has units of norm $-1$. We can recover them by the same recipe, truncating before the color reversed copy of $a_0$.

For example,
$$[3,1,1,1,1] = frac{18}{5}$$
and $18-5 sqrt{13}$ is the fundamental unit of norm $-1$ in $mathbb{Z}[sqrt{13}]$.

Note, by the way, that we have not yet seen the fundamental unit of $mathbb{Q}(sqrt{13})$, which is $(3-sqrt{13})/2$. That's because $langle 1, sqrt{13} rangle$ doesn't have CM by this unit.

We can obtain the same unit from two quite different looking continued fractions. For example
$$sqrt{10} = r r r b b b b b b r r r cdots quad sqrt{10}/2 = r b r b b r b r cdots$$
where I have given a full period for each fraction. Truncating to before the middle block gives
$$r r r = frac{3}{1} quad r b r = frac{3}{2}.$$
Both of these give the unit $3-sqrt{10} = 3-2 frac{sqrt{10}}{2}$.

Fixing the endomorphism ring; working with triples $(a,b,c)$

$bullet$ Let $R = mathbb{Z}[sqrt{D}]$, for $D>0$ and not square. The continued fractions which give rise to rings containing $R$ correspond to ordered triples $(a,b,c)$ of integers with $D=b^2+ac$ and $a$, $c>0$, by the recipe $(a,b,c) mapsto (b+sqrt{D})/a$.

The corresponding ring is exactly $mathbb{Z}[sqrt{D}]$ if and only if $(a,2b,c)$ have no common factor.

$bullet$ Let $R = mathbb{Z}[(1+sqrt{D})/2]$ with $D equiv 1 mod 4$, positive and not square. The continued fractions which give rise to rings containing $R$ correspond to ordered triples $(a,b,c)$ with $D=b^2+ac$, $a$ and $c>0$, and the additional condition that $b$ is odd and $a$ and $c$ are even.

Example: If we want to get the ring $mathbb{Z}[(1+sqrt{13})/2]$, we need to pick $z$ so that $langle 1, z rangle$ has CM by this ring. An obvious choice is $z=(1+sqrt{13})/2$, with $(a,b,c) = (2,1,6)$. The continued fraction is
$$r r b b b r r r b b b r r r b b b $$
or $[2, 3,3,3,3,ldots]$ in conventional notation.

$bullet$ The corresponding ring is exactly $mathbb{Z}[(1+sqrt{D})/2]$ if and only if $(a,b,c)$ have no common factor.

$bullet$ There are only finitely many $(a,b,c)$ for any $R$.

$bullet$ Adding an $r$ at the beginning of the sequence changes $(a,b,c)$ to $(a,a+b,c-a-2b)$. Adding a $b$ at the beginning changes $(a,b,c)$ to $(a-c-2b,b+c,c)$. Reversing the sequence changes $(a,b,c)$ to $(a,-b,c)$; color switching the sequence sends $(a,b,c)$ to $(c,-b,a)$.

Continued fractions with special symmetry

$bullet$ A continued fraction is a shift of its color switch if and only if $R$ contains a unit with norm $-1$; we have already described how to find this unit.

$bullet$ A continued fraction is a shift of its reversal if and only if it is a $2$-torsion class in the strict ideal class group.

Consider continued fractions which equal their reversal, so the periodic sequence starts at the middle of an even block of $r$'s or $b$'s, like the sequence for $sqrt{13}$ above. These correspond to $z = sqrt{D}/a$ for some divisor $a$ of $D$.

Consider continued fractions which are off from one by a shift of their reversal, so the periodic sequence starts in the middle of an odd block of $r$'s or $b$'s. These correspond to $z = (b+sqrt{D})/(2b)$. If $D$ is odd, then we can take $b$ to be any divisor of $D$. If $D$ is $2 mod 4$, then there are no solutions to $b^2+2bc=D$. If $D$ is $0 mod 4$, then we can take $b$ of the form $2 b'$, where $b'$ is a divisor of $D/4$.

$bullet$ Let $(-)$ denote the strict ideal class of principal $Lambda$ ideals generated by elements of negative norm. A continued fraction is a shift of its color switched reversal if and only if $Lambda^2 = (-)$ in the strict ideal class group.

A continued fraction actually equals its color switched reversal if and only if $a=c$. In other words, such continued fractions for $R = mathbb{Z}[sqrt{D}]$ are in bijection with solutions to $a^2+b^2=D$ with $a$ and $b>0$, and $GCD(a,2b)=1$. Such continued fractions for $R=mathbb{Z}[(1+sqrt{D})/2]$ are in bijection with solutions to $a^2+b^2=D$ with $a$, $b>0$, such that $a$ even and $b$ odd.

Example: We have $34=3^2+5^2$. So take $z=(3+sqrt{34})/5$. Take $Lambda$ to be the lattice $langle 1, (3+sqrt{34})/5 rangle$, which is strictly equivalent to the ideal $I=langle 5, 3+sqrt{34} rangle$ in $mathbb{Z}[sqrt{34}]$. This is a non-principal prime ideal dividing $5$. We have $I^2 = langle 3+sqrt{34} rangle$, which is principal, but not strictly principal.

The corresponding continued fraction is
$$r b r r r b b b r b r b r r r b b b r b r b cdots$$
or $[1,1,3,3,1,1,1,1,3,3,1,1,1,1,cdots]$ in conventional notation. This sequence is its own color switched reversal, reflecting that $I^2=(-)$. However, it is not a shift of its own reversal, reflecting that $I^2$ is not strictly principal, and it is not a shift of its color switch, reflecting that $mathbb{Z}[sqrt{34}]$ does not have a unit of norm $-1$.

nt.number theory - Prime numbers and strings of symbols

I think there's an easier way (well, easier if you assume Dirichlet's Theorem - I think it's simpler than the strong form of Betrand's Postulate). Interpret the string as an integer M in base N; if it happens to be relatively prime to N, you're done - use Dirichlet to find a prime congruent to M modulo N^k where k is the length of the string. David's method places the given string as the most-significant digits of the prime, whereas this approach places it as the least-significant ones.

If M happens to have a common divisor with the base N, you can just pad it with a "1" at the end (e.g. replace 31415 with 314151 in base 10). The padding replaces M by NM+1 which is obviously relatively prime to N.

ag.algebraic geometry - étale fundamental group of projective space

It is a birational invariant (for smooth proper connected schemes over a field, ultimately due to Zariski-Nagata purity of the branch locus), and its formation is compatible with products (for proper connected schemes over an algebraically closed field), so we can replace projective $n$-space with the $n$-fold product of copies of the projective line to conclude. Likewise, due to limit arguments and invariance of the etale site with respect to finite radiciel surjections (such as a finite purely inseparable extension of a ground field), it suffices to take the ground field to be separably closed rather than algebraically closed. This is all in SGA1.

nt.number theory - Number theoretic sequences and Hecke eigenvalues

What are some number theoretic sequences that you know of that occur as (or are closely related to) the sequence of Fourier coefficients of some sort of automorphic function/form or the sequence of Hecke eigenvalues attached to a Hecke eigenform?

I know many such sequences, but am always looking for more.

Some examples

(1) The sequence a(n) deriving from the traces a(p) of the Frobenius elements in a Galois representation (Langlands reciprocity conjecture)

(2) Number of representations of a natural number as a sum of k squares (theta functions)

(3) The sum of powers of divisor functions (Eisenstein series)

(4) The central critical values of L-functions attached to all quadratic twists of a Hecke eigenform (Kohnen, Waldspurger)

(5) Intersection numbers of certain subvarieties of Hilbert modular surfaces (Hirzebruch-Zagier)

I'll end with a question that is ill-posed but nevertheless very interesting (at least to me personally): why do so many familiar and yet diverse sequences appear in this fashion? Note that many of them have a history of study that precedes the recognition that they are essentially coefficients of automorphic functions/forms.

gr.group theory - Counting the Groups of Order n Weighted by 1/|Aut(G)|

Computational evidence suggests $g(n)$ varies wildly with $n$. When $n$ is a power of a prime, or has lots of small factors, $g(n)$ can be very large (I would guess $g(2^k)$ is superexponential in $k$), and $a(n)$ contributes negligibly. In particular, for the prime power case, three-step nilpotent groups seem to dominate, and a good theoretical reason for this is that they asymptotically dominate in number of isomorphism types by a huge factor.

If $n$ has only a few factors, then $g(n)$ is a little bigger than $1/n$. In this case, $a(n)$ contributes nontrivially.

For those of you who are interested in groups of order $2^n$ (and who isn't?), I've computed a decomposition of $g(2^k), k leq 7$ by nilpotence class, so the class one columns on the left indicate $a(2^k)$, the abelian contribution.

g(2) = a(2) = 1.

nilpotence class |   1  
isom. types      |   1
weighted count   |   1

$g(4) = a(4) = 2/3 sim 0.67$.

nilpotence class |   1  
isom. types      |   2
weighted count   |  2/3

$g(8) = 23/42 sim 0.55$

nilpotence class |     1      |     2    
isom. types      |  3 (60%)   |  2 (40%) 
weighted count   | 8/21 (70%) |  1/6 (30%) 

$g(16) = 1247/2520 sim 0.49$

nilpotence class |       1      |     2     |   3
isom. types      |    5 (36%)   |  6 (43%)  |   3 (21%)
weighted count   | 64/315 (41%) | 1/6 (34%) | 1/8 (25%) 

$g(32) = 149297/312480 sim 0.48$

nilpotence class |        1        |       2      |      3     |   4
isom. types      |     7 (14%)     |   26 (51%)   |  15 (29%)  |  3 (6%)
weighted count   | 1024/9765 (22%) | 37/240 (32%) | 3/16 (39%) | 1/32 (7%) 

$g(64) = 48611383/78744960 sim 0.62$

$begin{array}{rccccc} text{nilpotence class} & 1 & 2 & 3 & 4 & 5 \
text{isomorphism types} & 11, (4%) & 117, (44%) & 114 , (43%) & 22, (8%) & 3, (1%) \
text{weighted count} & frac{32768}{615195} , (9%) & frac{3}{20} , (24%) & frac{5}{16} , (51%) & frac{3}{32} , (15%)& frac{1}{128} , (1%) end{array}$

$g(128) = 8999449693/8000487936 sim 1.12$

$begin{array}{rccccc} text{nilpotence class} & 1 & 2 & 3 & 4 & 5 & 6 \
text{isomorphism types} & 15 , (1%) & 947 , (41%) & 1137 , (49%) & 197 , (8%) & 29 , (1%) & 3 , (0%) \
text{weighted count} & frac{2097152}{78129765} , (2%) & frac{275929}{1451520} , (17%) & frac{2295}{3584} , (57%) & frac{7}{32} , (19%) & frac{3}{64} , (4%) & frac{1}{512}, (0%) end{array}$

The groups of order 64 took about 15 minutes on my ancient computer running GAP, and the groups of order 128 took about 40 hours. I'll leave the question of groups of order 256 to someone else, since a modern desktop computer should be able to work it out in a little under a year.

You might be curious about the three isomorphism types of nilpotence class $k-1$ for $k geq 4$. They are the dihedral, quasidihedral, and quaternion groups.

nt.number theory - Maximal extension almost everywhere unramified and totally split at one place

Nope.

I'm lacking a reference in front of me at the moment (see NSW's Cohomology of Number Fields, or Gras's Class Field Theory -- I'll update with a precise reference later), but there are remarkably clean formulas for the generator and relation ranks for the Galois group of the maximal $ell$-extension of $mathbb{Q}$ unramified outside $S$ and completely split at $T$, for finite sets of primes $S$ and $T$. Throwing out some silly cases, these depend only on $|S|$ and $|T|$ (and, in your problem, maybe even just $|S|-|T|$). In your case, where $|T|=1$, it's just a matter of making $S$ big enough (again, a reference will say how big, but right now, I think $|S|=4$ does the trick.)

Edit to add in in a precise reference (though the above book references certainly contain the results as well): Christian Maire's "Finitude de tours et p-tours T-ramifiees moderees, S-decomposees".

ag.algebraic geometry - Why the rank of a locally free sheaves is well defined?

Let $mathcal{F}$ be a locally free sheaf on $X$. For any $x$ in $X$ there exists $x in U subset_{open} X $ such that

$mathcal{F}|_U cong mathcal{O}_X|_U^{(I)}$ $ (star)$.

In particular, for each $y$ in this particular $U$, one has $mathcal{F}_y cong mathcal{O}_{X,y}^{(I)}$ (which is given by the isomorphism above!!!).

Suppose now $X$ is connected and $mathcal{F}$ is locally free (we need this). Fix an indexing set $I$ (and I think I need to take this $I$ to be one of the indexing sets from $(star)$ above). The properties of $mathcal{F}$ show that the set

$S_I = left(x in X : mathcal{F}_x cong mathcal{O}_{X,x}^{(I)}right)$

is both closed and open in $X$. We know that there exists

$x$ in $X$ with $mathcal{F}_x cong mathcal{O}_{X,x}^{(I)}$,

we have $S_I = X$.

In particular, $text{rank}_{mathcal{O}_{X,x}}(mathcal{F}_x)$ is constant as $x$ varies in $X$.

ag.algebraic geometry - Theta Functions and Cousins

I'm having trouble understanding this, because under your definition, T(-z) requires T to be defined on the lower half-plane. The series doesn't converge there.

Also, N(v) should be half of the squared norm. Equivalently, N(v) should be the value of the quadratic form that defines the inner product on the lattice.

I can think of one interpretation, where the theta function is actually a Jacobi theta function, which is a Jacobi form, i.e., a section of a line bundle on the universal elliptic curve. This can be viewed as a function T(t,z) on HxC, with some invariance properties under translation by lattice elements in C and SL(2,Z) transformations in H. Then you can negate the z variable.

Eichler and Zagier have a book on Jacobi forms, called The theory of Jacobi forms.

gr.group theory - Do decidable properties of finitely presented groups depend only on the profinitization?

OK, I think I have an example of two groups with the same profinitization and a computable property which distinguishes them. The point is that very fine detail about the commutator subgroups can't be seen in the profinitization.

Let $q$ be prime and let $K$ be the $q$-th cyclotomic field.
Choose $q$ such that the class group of $K$ is not trivial. Let $I$ be a trivial ideal of $mathcal{O}_K$ and $J$ a nontrivial ideal. Our groups $G$ and $H$ will be $(mathbb{Z}/q) ltimes I$ and $(mathbb{Z}/q) ltimes J$.

For any group $B$, let $B' = [B,B]$ and $B'' = [B', B']$. Note that $B/B'$ acts on $B'/B''$ by conjugation. Our computable criterion is the following:

$B/B' cong mathbb{Z}/q times mathbb{Z}/q =: A$, the action of the group ring $mathbb{Z}[A]$ on $B'/B''$ factors through a map $mathbb{Z}[A] to mathcal{O}_K$ and, as such, $B'/B''$ is a free $mathcal{O}_K$ module.

We leave it as an exercise that $G$ satisfies this condition and $H$ does not.

I believe this condition should be computable. We can go from a finite presentation of $B$ to one of $B'$. (UPDATE I have revised this argument.) Abelianizations are computable, so we can check whether $B/B'$ has the right format. If it does, then $B'$ has finite index in $B$. I think we can use this to get a finite presentation of $B'$: Let $Delta$ be a two-dimensional $CW$-complex with one vertex, an edge for each generator of $B$ and a two cell for each relation. Let $Delta'$ be the cover of $B$ corresponding to $B'$. Since $B$ has finite index in $B'$, $Delta'$ will have finitely many cells, and we get a finite presentation of $B'$.

We can the compute the abelianization of $B'$ and, I think, the action of the abelianization of $B$ on that of $B'$ should be computable. Note that there are only $q^2$ maps from $mathbb{Z}[A]$ to $mathcal{O}_K$, so we can just check them each in turn. The class of a finite generated module for a Dedekind domain should be computable by standard number theory methods, although I admit I couldn't describe them.

The fact that these two groups have the same profinitization is relatively well known. Let $hat{I}$ and $hat{J}$ denote the profinite completions of $I$ and $J$. The profinite completions of $G$ and $H$ are $mathbb{Z}/n ltimes hat{I}$ and $mathbb{Z}/n ltimes hat{J}$.

We can identify $hat{I}$ and $hat{J}$ with submodules of $mathbb{A}^0_K$, the integral adeles of $K$. Since $I$ and $J$ are locally principal, these are principal ideals in the ring $mathbb{A}^0_K$. They are thus equivalent as $mathbb{A}^0_K$ modules, and thus as $mathcal{O}_K$ modules.

at.algebraic topology - Mayer-Vietoris homotopy groups sequence of a pull-back of a fibration

A low level Mayer-Vietoris sequence for a pull-back of a fibration of groupoids is in
(R. BROWN, P.R. HEATH and H. KAMPS), ``Groupoids and the
Mayer-Vietoris sequence'', {em J. Pure Appl. Alg.} 30 (1983)

and you will also find a version for coverings of groupoids in
`Topology and Groupoids', R. Brown (available on amazon.com).

I've set as an exercise in my new coauthored book `Nonabelian algebraic topology' (see my web pages) to get a Mayer-Vietoris sequence for a pullback of a fibration of crossed complexes.

computational complexity - Does IP = PSPACE work over other rings?

Background: It is possible (see e.g., this) to define a Turing machine over an arbitrary ring. It reduces to the classical notion when the ring is $mathbb{Z}_2$; the key difference is that elementary algebraic computations are allowed to be performed in one step (and one has infinite precision). It is possible to define analogs of classical complexity classes (e.g. $mathrm{P}_R, mathrm{NP}_R, mathrm{BPP}_R$, etc. with respect to a ring $R$). Sometimes the ring may be required to be ordered, and the machine is allowed to test for ordering (it is always allowed to test for equality).

It should, therefore, be possible to define a class $mathrm{IP}_R$ of problems that can be solved by a polynomial-time interactive proof system over $R$ with error probability $leq frac{1}{3}$, a class $mathrm{PSPACE}_R$ of problems that can be solved in polynomial space by a (deterministic) Turing machine. (Probably this has already been done, but I just haven't been able to find it.) When $R = mathbb{Z}_2$ (i.e. the classical case), it is a known result that $mathrm{IP}=mathrm{PSPACE}$.

Question: Does the same work over other rings $R$?

Perhaps there might be a problem, since the only proof I've seen (e.g. in Sipser's Introduction to the Theory of Computation) uses the $mathrm{PSPACE}$-completeness of $TQBF$, and I don't know whether that would work over an arbitrary ring.

at.algebraic topology - Topological description of Manifold with boundary

For simplicity I want to stick to the compact, orientable case. Further i want to assume, that $N$ has the structure of a CW-complex.

It might still be interesting to consider the case, where the dimension of $M$ and the dimension of $N$ are different; for example $D^2simeq pt$ and $[0;1]times S^1simeq S^1$.

To find the dimension of $M$ one has to consider the structure of the (co-)jhomology and the $cap$-product. The dimension of $M$ is the dimension of the highest non-vanishing homology $H_*(N)$. Futhermore $H_*(N)$ and $H^*(N)$ have to satisfy Poincare duality, so that the highest non-vanishing homology has to be $mathbb{Z}$ and if one picks a generator, the $cap$-product with this generator has to give isomorphisms.
If this is satisfied, $N$ is called a "Poincare complex". (One can consider some simple examples of manifolds with boundaries like a disk bundle over a manifold or $S^1times S^1setminus D^2$).

The question whether a given Poincare complex is homotopy equivalent to a manifold (the other way round) is one classical problem in surgery theory.

There is a involved obstruction process coming, which is very roughly described in the wikipedia. More details can be found for example in the books mentioned there.

There should also be twisted versions of Poincare duality int the nonorientable case, which should also give a complete answer.

I do not know, whether the additional assumption, that the Poincare complex $N$ is indeed a manifold with boundary has any consequences for the surgery obstructions.

nt.number theory - The large sieve for primes

Let $Lambda(n)$ be the von Mangoldt function, i.e., $Lambda(n) = log p$ for $n$ a prime power $p^k$ and $Lambda(n) = 0$ for all $n$ that not prime powers. Let

$$S(alpha) = sum_{n leq N} Lambda(n) e(alpha n).$$

Now, using, say, Lemma 7.15 in Iwaniec-Kowalski (or the same result in Montgomery), we get

$$sum_{q leq q_0} sum_{a pmod{q}: gcd(a,q)=1} lvert S(a/q)rvert^2
leq frac{(N + Q^2) N log N}{sum_{substack{qleq Q text{ squarefree} \ gcd(q,P(q_0))=1}} phi(q)^{-1}},$$

where $Q$ is arbitrary and $P(z):=prod_{p leq z} p$.

In practice, we would choose $Q$ slightly smaller than $sqrt{N}$, and obtain

$$sum_{q leq q_0} sum_{substack{a pmod{q} \ gcd(a,q)=1}} lvert S(a/q) rvert^2 leq (1+epsilon) 2 e^gamma N^2 log q_0,$$

where gamma is Euler's constant $0.577cdots$ and $epsilon$ is very small.

Now, the 2 in the bound $leq (1+epsilon) 2 e^gamma N^2$ is due to the parity problem,
and thus should be next to impossible to remove (except for very small $q_0$).
However, the factor of $e^gamma$ clearly has no right to exist. The true asymptotic should be simply $N^2 log q_0$.

Can we remove that nasty $e^gamma$? That is, can you prove a bound of type

$$sum_{q leq q_0} sum_{substack{a pmod{q} \ gcd(a,q)=1}} lvert S(a/q)rvert^2 leq (1+epsilon) 2 N^2 log q_0 ?$$

Harald

soft question - Most helpful math resources on the web

Sloane's OEIS has already been mentioned.

A similarly useful site is ISC, Simon Plouffe's Inverse Symbolic Calculator.

Here you enter the decimal expansion of a number to as many places as you know, and the search engine makes suggestions of symbolic expressions that the expansion might be derived from. The answer might involve pi, e, sin, cosh, sqrt, ln, and so on.

Sometimes, it becomes difficult to calculate symbolically. Therefore, you can proceed numerically instead, and hope to recover the exact symbolic solution at the end, using ISC: sometimes proving that an answer is correct can be easier than calculating, or discovering, it in the first place.

It can also be useful for discovering simplifications of nested radicals, for example.

fourier analysis - Estimation of DFT

The answer is no, if you mean an uniform bound in $j$. Here is the example:

Fix $j$ and define
$$
a_r = begin{cases} frac{1}{N}, & Re(exp(-2pi i j r/ K)) geq 0;\
0, & otherwise.end{cases}
$$
It is than easy to estimate that the number of $a_r = frac{1}{N}$ is comparable to $K$. Even more is true, one has that the number of $a_r = frac{1}{N}$ such that $Re(exp(-2pi i j r/ K)) geq sigma$ is comparable to $K$ for any $sigma > 0$.

This implies that
$$
Re(hat{a}_j) geq c N^{1 - alpha}
$$
for some different $c$.

nt.number theory - Examples of asymptotic formulas with optimal error term

Let $omega(n)$ denote the number of distinct prime factors of $n$ and $Phi(cdot)$ denote the Normal distribution with mean $0$ and variance $1$. Then, uniformly in $t$, the number of integers $n leq x$ with $omega(n) leq loglog n + t sqrt{loglog n}$ is $$Phi(t) + Oleft(frac{1}{sqrt{loglog x}}right)$$ as $x rightarrow infty$. The error term is sharp.

This particular error term was conjectured to hold by Leveque in the 40's. His conjecture was settled a few years later by Erdos and Renyi.

Before Erdos and Renyi's paper the best error term was $O(logloglog x / sqrt{loglog x})$ and was due (if I recall correctly) to Kubilius. Kubilius's method was in its origin probabilistic and relied in an essential way on truncating the "random variable" $omega(n)$. This introduced an additional factor of $logloglog x$ to the error term, as inevitably, truncation leads to loss of information.

In contrast, Renyi's and Erdos's method was purely analytic: the idea was to estimate $sum_{n leq x} exp(text{i}t omega(n))$ uniformly in $t$ in a certain range, and extract the desired conclusion from the behavior of this sum. To this end, they apply Berry-Esseen's theorem, but in principle one could use an more hands-on approach: smooth the indicator function of $omega(n) leq loglog n + t sqrt{loglog n}$ and express it in terms of a variant of Perron's formula, after summing the resulting expression over $n leq x$ one could proceed with the saddle-point method.

However when purely analytic methods are not directly accessible, Kubilius's method is the canonical method. For example suppose that you want to investigate the distribution of $omega(n)$ over a peculiar subset of $[1,x]$ on which sieve methods -- but not "heavy" analytic methods -- are applicable, then Kubilius's method is still your best bet.

(As far as terminology is concerned it's "Kubilius model's" rather than "Kubilius's method"; more details about "Kubilius's model" can be found in Volume 1 of Elliott's "Probabilistic Number Theory").

schemes - Is there an example of a formally smooth morphism which is not smooth?

Here's an elementary example. For any field $k$, consider the ring $k[t^q|qinmathbb Q_{>0}]$, which I'll abbreviate $k[t^q]$. I claim that the natural quotient $k[t^q]to k$ given by sending $t^q$ to $0$ is formally smooth but not flat, and therefore not smooth.

First let's show it's formally smooth. Let $A$ be a ring with square-zero ideal $Isubseteq A$, and suppose we have maps $f:k[t^q]to A$ and $g:kto A/I$ making the following square commute (I drew it backwards because you're probably thinking of Spec of everything)

$$
begin{array}{ccc}
A/I & xleftarrow g & k \
uparrow & & uparrow\
A & xleftarrow f & k[t^q]
end{array}
$$

We'd like to show that there's a map $kto A$ filling the diagram in. For any $qin mathbb Q_{>0}$, note that $f(t^q)in I$ by commutativity of the square, so $f(t^{2q})in I^2=0$. But every $q$ is of the form $2q'$ for some $q'$, so we've shown that $f(t^q)=0$ for all $qin mathbb Q_{>0}$. So $f$ factors through $k$, as desired.

Now let's show that $k$ is not flat over $k[t^q]$. Consider the exact sequence
$$0to (t)to k[t^q]to k[t^q]/(t)to 0.$$
When you tensor with $k$, you get
$$0to kto kto kto 0,$$
which is obviously not exact. So $k$ is not flat over $k[t^q]$.

Non-commutative algebraic geometry

In fact, I revised some of problems in Chater II Scheme theory in Hartshorne using Kontsevich-Rosenberg's machine. I have to notice, what you should deal with is probably module category over noncommutative ring. In noncommutative algebraic geometry, this is just category of quasi coherent sheaves on noncommutative affine schemes. But I did not restrict myself to noncommutative ring case. I try to do it in general noncommutative scheme, say a Grothendieck category or an abelian category. Noticed that, one can take grothendieck category as category of quasi coherent sheaves on quasi compact and quasi separated "would be scheme". So one should consider category of grothendieck category as category of "space" and morphism between spaces as iso class of inverse image functor. Rosenberg developed algebraic geometry in this 2-category. He introduced various spectrum for various destination. I should mention, spectrums for abelian category in his sense coincides with prime spectrum of a commutative ring when you take module category over commutative ring. In fact, one can define Zariski topology on this 2-categories using a family of conservative(faithful)exact localization functors(Serre subcategory in dual language). Then one can introduced the associated topology on the spectrum of abelian category. Then one can continue to introduce the "fiber" at each point of spectrum as stack of local category.(as fibered category) This is called geometric realization of an abelian category or Grothendieck category. Then one can take category of quasi coherent sheaves on this fibered category. At last, we get reconstruction theorem for noncommtative scheme. If we take the original category as quasi coherent sheaves of quasi compact(or not in general)quasi separated commutative scheme. Then we get reconstruction theorem for commutative scheme which means commutative algebraic geometry can be fully embedded into noncommutative algebraic geometry.

Beacuse of this "Justify" theorem, we can develop various notions correspondent to commutative algebraic geometry. One can define noncommutative affine scheme (can be seen as category with projective cogenerator, then by Gabriel cheating theorem, equivalent to a module category). One can also define affine morphism, open/closed immersion/coimmersion(for the motivation of representation theory) picard group, vector bundles

One can also define differential operators in abelian category, monoidal category(for motivation of representation theory of quantum group and math phy), in particular, Noncommutative D-modules on noncommutative space, in particular quantum D-module on quantized flag variety. (I think this is related to the problem mentioned by siegels).

As is well known to all, Beilinson Bernstein's framework aiming to representation theory lives in triangulated category. Actually, there is indeed whole abelian picture developed mainly by Rosenberg and Lunts-Rosenberg-Tanisaki later.

In fact,for most(I think it should be all)problems in Hartshorne (facts in commutative algebraic geometry)has correspondence version in noncommutative algebraic geometry(in paticular, what you mentioned, noncommutative ring)

There is indeed noncommutative flat descent theory in Konstevich-Rosenberg's work. I think the more accurate name should be categorical flat descent theory(Beck's theorem)

One more comment: What I mentioned above is ONE framework they developed.(Mainly for representation theory). There is ANOTHER framework introduced by them base on presheave view point(proposed by Gabriel-Grothendieck). They develop algebraic geometry in this view point which is NOT equivalent to the CATEGORICAL GEOMETRY I mentioned above in general. They coincides in affine case and then go to completely different direction.The mainly motivation for this view point is from Konstevich,he wanted to consider noncommutative grassmannian which might be helpful in understanding M-theory in Physics. Along this direction, they define noncommutative algebraic space, stack(DM and Artin) and so on.

last comment: One guy mentioned above, the category of rings did not have good property as commutative ones.But I guess this is not a very big deal. Rosenberg define so called right exact category(say category of rings, category of affine schemes,category of vector bundles). He developed whole homological algebra in this settings and Universal algebraic k theory, algebraic cycles. chow group and so on

I am sorry I have to go back to work instead of typing here. There are various noncommutatve algebraic geometry. If you are dealing with projective scheme, you might be interested in work of Artin's School on NC projective geometry

Several other comments: we have the notion of locally noetherian category whose objects is generated by noetherian object. For example, category of quasi coherent sheaves on noetherian commutative scheme is a locally noetherian category.We can play game in this setting. Then we can get whole commutative algebraic geometry of noetherian scheme

derived category - Freyd-Mitchell for triangulated categories?

Is there a nice analog of the Freyd-Mitchell theorem for triangulated categories (potentially with some requirements)? Freyd-Mitchell is the theorem which says that any small abelian category is a fully faithful, exact embedding into the module category of some ring.

Therefore, I'd like a theorem like this:
Any small triangulated category is a fully faithful, triangulated subcategory of the unbounded derived category of modules on some ring.

My guess is that this fails to be true, for similar reasons to a triangulated category not always being the derived category of its core. Is there a simple example of this? -and- Is there a set of properties that do imply the above theorem?

cv.complex variables - Interpreting the Famous Five equation

The answers so far give interpretations of the exponential as a limit of discrete approximations. An alternative interpretation is that any continuous map that takes addition to multiplication on the complex line and takes reals to reals has a purely imaginary kernel isomorphic to the integers. The constant $e$ arises from a normalization for which unit speed paths on the imaginary axis are taken to unit speed paths on the unit circle, and $pi$ then shows up as a path length. One way to emphasize the additive-multiplicative relationship is to expand the formula as: $e^{pi i-0i} = -1/1$.

Here is a more formal treatment: Both $(mathbb{C}^times, times)$ and $(mathbb{C}, +)$ are one dimensional analytic groups, and the latter is simply connected, so there is a universal covering homomorphism $exp: mathbb{C} to mathbb{C}^times$ from the additive group to the the multiplicative group. The homomorphism is unique once we choose a normalization, e.g., by demanding that it is analytic and satisfies the differential equation $partial_z exp = exp$. The differential equation can be related to the homomorphism, after choosing coordinates, by considering the respective formal group laws, or just reasoning heuristically with infinitesimals.

Claim 1: The function $exp$ takes purely imaginary numbers to elements of unit norm.

Proof: The function $exp$ takes additive inverses to reciprocals (because it is a homomorphism), and complex conjugates to complex conjugates (because it is defined over the reals). By composing, we find that reflection in the imaginary axis is taken to unit circle inversion, and fixed points are taken to fixed points.

Remarks: Note that the only part of the normalization we used here was the fact that $partial_z exp$ is a real multiple of $exp$. The "defined over the reals" bit may be unsatisfying to some, but the conjugation behavior can be verified directly by expanding as a power series that converges everywhere, and checking that the coefficients are real. One can also prove the claim by more direct methods, such as applying the above differential equation to grind out the identity $frac{partial}{partial y} | exp(iy) |^2 = 0$.

Claim 2: $exp(pi i) = -1$.

Proof: By combining the previous claim concerning unit norms with the differential equation $partial_z exp = exp$, we conclude that $exp$ takes any unit speed path on the imaginary axis to a unit speed path on the unit circle. We have $exp(0) = 1$ by the homomorphism assumption, and the length of a minimal path from $1$ to $-1$ on the unit circle is $pi$.

Remarks: Depending on how $z mapsto e^z$ is defined, one may still have to check that it agrees with $exp$, but this isn't a big deal. I tried to avoid choosing square roots of minus one as much as possible, but the statement of the identity makes it a bit difficult to maintain such discipline.