ho.history overview - What was Gödel's real achievement?

If I remember correctly, the notion of a model is already present in Hilbert-Bernays where finite structures are used for proving absolute consistency of some theories, and is probably older. Again, if I remember correctly, Frege, Russel, and Hilbert did have formal systems and the notion of a formal proof. Skolem's construction of a term model (which is now famous as Skolem's Paradox because the set of real numbers of the model turns out to be countable) is in his 1922 paper, where the Godel's completeness theorem is from 1929. In other words, it seems that Skolem did already have all the tools necessary for proving completeness in 1922. It seems that Hilbert had even stated the question of completeness for first-order theories before this date and Godel has learned about this problem in Carnap's logic course in 1928.

Hilbert's 10-th problem from his famous 23 problems asks for an algorithm to decide existence of solutions for Diophantine equations. I think there were attempts after this for understanding what is an algorithm. There were many definitions which came out before Turing's definition which were equivalent to his definition, although they were not philosophically satisfactory, at least Godel did not accept any of them as capturing the intuitive notion of computability before Turing's definition.

Godel's collected works can shed more light on these issues.

EDIT: Also

Solomon Feferman, "Gödel on finitism, constructivity and Hilbert’s program"
http://math.stanford.edu/~feferman/papers/bernays.pdf

Hilbert and Ackermann posed the fundamental problem of the completeness of the first-order predicate calculus in their logic text of 1928; Gödel settled that question in the affirmative in his dissertation a year later. [page 2]

Hilbert introduced first order logic and raised the question of completeness much earlier, in his lectures of 1917-18. According to Awodey and Carus (2001), Gödel learned of this completeness problem in his logic course with Carnap in 1928 (the one logic course that he ever took!). [page 2, footnote]

Martin Davis, "What did Gödel believe and when did he believe it", BSL, 2005

Godel has emphasized the important role that his philosophical views had
played in his discoveries. Thus, in a letter to Hao Wang of December 7,
1967, explaining why Skolem and others had not obtained the completeness
theorem for predicate calculus, ... [page 1]

mathematics education - Why is a topology made up of 'open' sets?

In this answer I will combine ideas of sigfpe's answer, sigfpe's blog, the book by Vickers, Kevin's questions and Neel's answers adding nothing really new until the last four paragraphs, in which I'll attempt to settle things about the open vs. closed ruler affair.

---

DISCLAIMER: I see that some of us are answering a question that is complementary of the original, since we are trying to motivate the structure of a topology, instead of adressing the question of which of the many equivalent ways to define a topology should be used, which is what the question literally asks for. In the topology course that I attended, it was given to us in the first class as an exercise to prove that a topology can be defined by its open sets, its neighbourhoods, its closure operator or its interior operator. We later saw that it can also be stated in terms of convergence of nets. Having made clear these equivalence of languages, its okay that anyone chooses for each exposition the language that seems more convinient without further discussion. However, I will mantain my non-answer since many readers have found the non-question interesting.

---

Imagine there's a set X of things that have certain properties. For each subset of S there is the property of belonging to S, and in fact each property is the property of belonging to an adequate S. Also, there are ways to prove that things have properties.

Let T be the family of properties with the following trait: whenever a thing has the property, you can prove it. Let's call this properties affirmative (following Vickers).

For example, if you are a merchant, your products may have many properties but you only want to advertise exactly those properties that you can show. Or if you are a physicist, you may want to talk about properties that you can make evident by experiment. Or if you predicate mathematical properties about abstract objects, you may want to talk about things that you can prove.

It is clear that if an arbitrary family of properties is affirmative, the property of having at least one of the properties (think about the disjunction of the properties, or the union of the sets that satisfy them) is affirmative: if a thing has at least one of the properties, you can prove that it has at least one of the properties by proving that property that it has.

It is also clear that if there is a finite family of affirmative properties, the property of having all of them is affirmative. If a thing has all the properties, you produce proofs for each, one after the other (assuming that a finite concatenation of proofs is a proof).

For example, if we sell batteries, the property P(x)="x is rechargeable" can be proved by putting x in a charger until it is recharged, but the property Q(x)="x is ever-lasting" can't be proved. It's easy to see that the negation of an affirmative property is not necessarily an affirmative property.

Let's say that the open sets are the sets whose characteristic property is affirmative. We see that the family T of open sets satisfies the axioms of a topology on X. Let's confuse each property with the set of things that satisfy it (and open with affirmative, union with disjunction, etc.).

Interior, neighbourhood and closure: If a property P is not affirmative, we can derive an affirmative property in a canonical way: let Q(x)="x certainly satisfies P". That is, a thing will have the property Q if it can be proved that it has the property P. It is clear that Q is affirmative and implies P. Also, Q is the union of the open sets contained in P. Then, it is the interior of P, which is the set of points for which Q is a neighbourhood. A neighbourhood of a point x is a set such that it can be proved that x belongs to it. The closure of P is the set of things that can't be proved not to satisfy P.

Axioms of separation: If T is not T0, there are x, y that can't be distinguished by proofs and if it is not T1, there are x, y such that x can't be distinguished from y (we can think that they are apparently identical batteries, but x is built in such a way that it will never overheat. So if it overheats, then it's y, but if it doesn't, you can't tell).

Base of a topology: Consider a family of experiments performable over a set X of objects. For each experiment E we know a set S of objects of X over which it yields a positive result (nothing is assumed about the outcome over objects that do not belong to S). If you consider the properties that can be proved by a finite sequence of experiments, the sets S are affirmative and the topology generated by them is the family of all the affirmative properties.

Compactness: I don't know how to interpret it, but I think that some people know, and it would be nice if they posted it. (Searchable spaces?)

Measurements: A measurement in a set X is an experiment that can be performed on each element of X returning a result from a finite set of possible ones. It may be a function or not (it is not a function if there is at least one element for which the result is variable). The experiment is rendered useful if we know for each possible result r a set T_r of elements for which the experiment certainly renders r and/or a set F_r for which it certainly doesn't, so let's add this information to the definition of measurement. An example is the measurement of a length with a ruler. If the length corresponds exactly with a mark on the ruler, the experimenter will see it and inform it. If the length fits almost exactly, the experimenter may think that it fits a mark or may see that it doesn't. If the length clearly doesn't fit any mark (because he can see that it lies between two marks, or because the length is out of range), he will inform it. It is sufficient to study measurements that have only a positive outcome and a negative outcome, a set T for which the outcome is certainly positive and a set F for which the outcome is certainly negative.

Imprecise measurements on a metric space: If X is a metric space, we say that a measurement in X is imprecise if there isn't a sequence x_n contained in F that converges to a point x contained in T. Suppose that there is a set of imprecise measurements available to be performed on the metric space. Suppose that, at least, for each x in X we have experiments that reveal its identity with arbitrary precision, that is, for each e>0 there is an experiment that, when applied to a point y, yields positive if y=x and doesn't yield positive if d(y,x)>e. Combining these experiments we are allowed to prove things. What are the affirmative sets generated by this method of proof? Let S be a subset of X. If x is in the (metric) interior of S, then there is a ball of some radius e>0 centered at x and contained in S. It is easy to find an experiment that proves that x belongs to S. If x is in S but not in the interior (i.e, it is in the boundary), we don't have a procedure to prove that x is in S, since it would involve precise measurement. Therefore, the affirmative sets are those that coincide with its metric interior. So, the imprecise measurements of arbitrary precision induce the metric topology.

Experimental sciences: In an experimental science, you make a model that consists of a set of things that could conceivably happen, and then make a theory that states that the things that actually happen are the ones that have certain properties. Not all statements of this kind are completely meaningful, but only the refutative ones, that is, those that can be proved wrong if they are wrong. A statements is refutative iff its negation is affirmative. By applying the closure operator to a non refutative statement we obtain a statement that retains the same meaning of the original, and doesn't make any unmeaningful claim.

An example from classical physics: Assume that the space-time W is the product of Euclidean space and an affine real line (time). It can be given the structure of a four-dimensional real normed space. Newton's first law of motion states that all the events of the trajectory of a free particle are collinear in space-time. To prove it false, we must find a free particle that incides in three non-collinear events. This is an open condition predicated over the space W^3 of 3-uples of events, since a small perturbation of a counterexample is also a counterexample. Assuming that imprecise measurements of arbitrary precision can be made, it is an affirmative property. I think that classical physicists, by assuming that these kind of measurements can be done, give exact laws like Newton's an affirmative set of situations in which the law is proved false. I also suspect (but this has more philosophical/physical than mathematical sense) that the mathematical properties of space-time (i.e. that it is a normed space over an Archimedean field) are deduced from the kind of experiments that can be done on it, so there could be a vicious circle in this explanation.

matrices - On a positivity of a matrix with trace entries.

Some basic observations lead me to ask the following quesiton

Let $A_1, cdots, A_m$ be $ntimes n$ complex matrices. For positive integer $kge 1$, show
$$left(begin{array}{cccc}Tr{(A_1^*A_1)^k}&Tr{(A_1^*A_2)^k}&cdots &Tr{(A_1^*A_m)^k}\Tr{(A_2^*A_1)^k}&Tr{(A_2^*A_2)^k}&cdots &Tr{(A_2^*A_m)^k}\cdots&cdots&cdots&cdots\Tr{(A_m^*A_1)^k}&Tr{(A_m^*A_2)^k}&cdots &Tr{(A_m^*A_m)^k}
end{array}right)$$
is positive semidefinite.

Remark

1). When $m=2$, it suffices to show $|Tr{(A_1^*A_2)^k}|^2le Tr{(A_1^*A_1)^k}cdot Tr{(A_2^*A_2)^k}$, which is a consequence of a unitarily invariant norm inequality appeared in p.81 of X.Zhan, Matrix inequalities, Springer, 2002.

2). It is easy to show $$left(begin{array}{cccc}(Tr{A_1^*A_1})^k&(Tr{A_1^*A_2})^k&cdots &(Tr{A_1^*A_m})^k\(Tr{A_2^*A_1})^k&(Tr{A_2^*A_2})^k&cdots &(Tr{A_2^*A_m})^k\cdots&cdots&cdots&cdots\(Tr{A_m^*A_1})^k&(Tr{A_m^*A_2})^k&cdots &(Tr{A_m^*A_m})^k
end{array}right)$$
is positive semidefinite, since it is $k$ Hadamard product of a Gram matrix.

ct.category theory - "Philosophical" meaning of the Yoneda Lemma

Here is an example on representable functors. Yoneda's lemma gives down-to-earth, morpshim oriented interpretation of representable functors, and vice versa.

I will explain this with an example.

In a category $mathscr{C}$, the product of $A$ and $B$ is the pair of object $Atimes B$ in $mathscr{C}$ and a fixed natural isomorphism
$$
sigma colon mathrm{Hom}(-,Atimes B)to mathrm{Hom}(-,A)times mathrm{Hom}(-,B).
$$

This definition of products only uses terminology of functors. By applying Yoneda's lemma, we arrive at a morphism oriented definiton of products. Yoneda's lemma says that there is a bijection
$$
Psi colon mathrm{Hom}left( mathrm{Hom}(-,Atimes B),mathrm{Hom}(-,A)times mathrm{Hom}(-,B)right) to mathrm{Hom}(Atimes B,A)times mathrm{Hom}(Atimes B,B).
$$
In particular, we apply this to $sigma$ and denote
$$
Psi(sigma)=sigma(Atimes B)(mathrm{id}_{Atimes B})=(pi^{A}colon Atimes Bto A,pi^{B}colon Atimes Bto B).
$$
Next, by applying the inverse of $Psi$, we compute
$$
sigma(X)=Psi^{-1}left( Psi(sigma)right)(X):mathrm{Hom}(X,Atimes B)to mathrm{Hom}(X,A)times mathrm{Hom}(X,B)
$$
$$
fcolon Xto Atimes Bmapsto (pi^{A}circ f,pi^{B}circ f).
$$
Since $sigma$ is a natural isomorphism, $sigma(X)$ is a bijection. This bijectivity is the usual definition of product based on morphisms (universality):

For any pair of morphisms $f^{A}colon Xto A$ and $f^{B}colon Xto B$, there exists a unique morphism $fcolon Xto Atimes B$ with $pi^{A}circ f=f^{A}$ and $pi^{B}circ f=f^{B}$.

I think the Philosophy behind Yoneda's lemma is that, it connects the world of functors (and natural transformations) $mathfrak{Set}^{mathscr{C}^{mathrm{op}}}$ and the world of morphisms $mathscr{C}$.

soft question - Major mathematical advances past age fifty

From A Mathematician’s Apology, G. H. Hardy, 1940:
"I had better say something here about this question of age, since it is particularly important for mathematicians. No mathematician should ever allow himself to forget that mathematics, more than any other art or science, is a young man's game. ... I do not know an instance of a major mathematical advance initiated by a man past fifty. If a man of mature age loses interest in and abandons mathematics, the loss is not likely to be very serious either for mathematics or for himself."

Have matters improved for the elderly mathematician? Please answer with major discoveries made by mathematicians past 50.

lo.logic - What do models where the CH is false look like?

Reading the comments, I just realized that part of your question asks about what sets of intermediate cardinality would look like. As Richard and John Goodrick pointed out, there are straightforward definitions of ordinals of intermediate cardinality. But I assume the question is more about whether there are any sets of real numbers with intermediate cardinality, and what such sets of reals would look like.

It's not too hard to use the ordinals to cook up some sets of real numbers. (For instance, there's probably a definable way to represent each countable ordinal by a real number, and then the set of representations will itself be of intermediate cardinality.) But as far as I know, no one has come up with any direct characterization of a set of real numbers that could have intermediate cardinality.

There are a lot of negative results though. The first is the Cantor-Bendixson theorem, which states that no closed set can have intermediate cardinality. (The theorem shows that every closed set is either countable or has a http://en.wikipedia.org/wiki/Perfect_set_property>perfect subset. Since perfect sets have cardinality of the continuum this means they can't be intermediate.)

In fact, some difficult work of Martin and others in the '70s showed (using just ZF) that Borel determinacy is true, which among other things entails that every Borel set is either countable or has a perfect subset. If we further assume projective determinacy (which set theorists tend to believe) then the same is true for projective sets.

Thus, any set of intermediate cardinality has to be pretty weird. It can't be closed, it can't even be Borel, and (if set theorists are right about projective determinacy) it can't even be projective. Thus, it must be pretty crazy, just as we know about non-measurable sets. (Though there's no guarantee that intermediate cardinality goes along with non-measurability - Martin's Axiom guarantees that in fact every set of intermediate cardinality is measurable and has measure 0.)

nt.number theory - Does (the ideal class of) the different of a number field have a canonical square root?

Let me point out that I would like to interpret "canonical" in a less strict sense. In order to explain what I mean, consider the following question:

Does there exist a canonical square root of $-1$ modulo primes $p = 4n+1$?

If canonical means that the result should be independent of any automorphism of the residue class group modulo $p$, then the answer is no: of $i$ is one answer, then the automorphism sending every residue class to its inverse will send $i$ to the other root $-i$.

Yet I would accept $i equiv (frac{p-1}2)! bmod p$ as an answer to the question.

Edit 2 (28.07.10) I'm still thinking about what canonical should mean in this context. The ideal class of the square root is defined up to classes of order $2$; doesn't this mean that a "canonical" choice of this ideal class should be an element in the group $Cl(K)/Cl(K)[2]$?

Edit. Trying to generalize frictionless jellyfish's example I ended up with the following results (which so far I have only partially proved).

Let $p$ and $q$ be two prime numbers with $p equiv q equiv 1 bmod 4$. There is a unique cyclic quartic extension $K/mathbb Q$ with conductor $pq$ and discriminant $p^3q^3$. Let $mathfrak p$ and $mathfrak q$ denote the prime ideals in $K$ above $p$ and $q$. Then $diff(K/mathbb Q) = {mathfrak p}^3 {mathfrak q}^3$. Moreover, ${mathfrak p}^2 {mathfrak q}^2 = (sqrt{pq},)$ is principal, so the ideal class of the different is either trivial or has order $2$.

Theorem If $(p/q) = -1$, then $ Cl_2(K) simeq [2]$ if $p equiv q equiv 5 bmod 8, $ and $Cl_2(K) simeq [4]$ otherwise. In both cases, the $2$-class field of $K$ is abelian over $mathbb Q$, hence is equal to its genus class field.

The ideal classes of each of the prime ideals above $mathfrak p$ and $mathfrak q$ generates the $2$-class group.

Taking $p=5$ and $q = 17$ gives an example of a "non-canonical" square root. I have not yet found a criterion that would tell me when the different is principal and when its class has order $2$. Both cases do occur.

The case where $p$ and $q$ are quadratic residues of each other is more involved, more interesting (and more conjectural):

Theorem Assume that $(p/q) = +1$, and that $p equiv q equiv 5 bmod 8$. Then
$Cl_2(K) simeq [2,2]$ if $(p/q)_4 (q/p)_4 = -1$, $Cl_2(K) simeq [4] $ if $(p/q)_4 = (q/p)_4 = -1$, and $Cl_2(K) simeq [2^n], n ge 3$ if $(p/q)_4 = (q/p)_4 = +1$.

If $(p/q)_4 = (q/p)_4 = -1$, then the ideal classes $[mathfrak p]$ and $[mathfrak q]$ are squares, but not fourth powers; in particular, the different is principal.

If $(p/q)_4 = (q/p)_4 = +1$, there are two cases:

1. $h_2(K) = h_2(k)= 2^n$; then $[mathfrak p]$ and $[mathfrak q]$ both have order $2$, and the different is principal.


2. $h_2(K) = 2h_2(k)= 2^n$; then either $mathfrak p$ or $mathfrak q$ is principal, whereas the other ideal generates a class of order $2$. In particular, the ideal class of the different has order $2$.

geometry - literature on geometrical viewpoint on calculus of variations for physics

There aren't a lot of up-to-date treatments of the calculus of variations for mathematicians nowadays,let alone physics students. To be fair,though,that's changing recently with the publications of the texts by van Brunt and Dracogna. Both those treatments are rather analytical and probably not what you're looking for.

The best and most complete introduction to the subject I've seen is in the last 4 chapters of S.P.Novikov and I.P Tiamanov's Modern Geometric Structures and Fields,available through the AMS.It walks the reader through not only modern formulations of classical and relativistic mechanics on both semi-Reimannian and symplectic manifolds with variational methods,the last chapter gives an in-depth derivation of the Yang-Mills equations using these methods.

I think you may find everything you're looking for in this book.

mirror symmetry - complexified kahler form

Since B-fields are being discussed in the other question, i'll try to answer your question on the complexified Kahler moduli. The set of Kahler classes of a Kahler manifold is an open cone in $H^{1,1}(X, C) cap H^2(X, R)$. There are results on the structure of this cone which roughly say that it is rational polyhedral away from some set (I think!). In the construction of the complexified Kahler cone you have to take a "framing" of this cone, in the end you get something isomorphic to $(Delta^{ast})^{h^{1,1}}$, where $Delta^*$ is the punctured disc in $C$. This matches the moduli of complex structures of the mirror, near a large complex structure limit point. These things are discussed in the book "Calabi-Yau manifolds and related geometries", by Gross, Huybrechts and Joyce.

big list - Examples of common false beliefs in mathematics

The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $sin z$ is a bounded function;
(iii) $sin z$ is defined and analytic everywhere on $mathbb{C}$;
(iv) $sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $mathbb{R}$ must be the whole of $mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

simplicial stuff - A question on a Davis complex of a Coxeter group

Let us have a look at p. 64 of M. Davis book "The Geometry and Topology of Coxeter Groups". The discussion preceeding Definition 5.1.3. shows that $mathcal{U}(G, X)/G$ is homeomorphic to $X$. Theorem 7.2.4. says that $mathcal{U}(W, K)$ is $W$-equivariantly homeomorphic to the Davis complex $Sigma$. So, $Sigma/W$ is homeomorphic to $K$. $K$ is the cone on the barycentric subdivision of the nerve $L$. $L$ can have topological type of any polyhedron. So $K$ can be a cone on any polyhedron (up to homeomorphism). But the action of $W$ on $Sigma$ is cocompact (p. 4, bottom). So $Sigma/W$ is compact, i.e., $K$ is compact. So a cone on any polyhedron is compact. What's wrong?

co.combinatorics - A number theoretic identity

Let $n$ be a positive integer such that $2n+1$ is prime.
The elements of the factor group $G = mathbb{F}^times_{2n+1}/{pm 1}$ can be represented by the integers $1,2,ldots,n$. For every $x in mathbb{F}^times_{2n+1}$, let $x' in {1,2,ldots,n}$ denote the representative of its image in $G$. For every $lambda in{2,ldots,n-1}$, let

$$ S_lambda = sum_{a=1}^n |a-(alambda)'|, $$

where $|cdot|$ denotes the usual absolute value. For example, when $n=6$,
and $lambda=3$,
$$ S_3 = |1-3| + |2-6| + |3-4| + |4-1| + |5-2| + |6-5| = 14. $$

Is it true that $S_lambda = n(n+1)/3$, for every $lambda geq 2$?

soft question - "In the sequel" - outdated mathematical jargon or precise technical term?

Possible Duplicate: http://math.stackexchange.com/questions/907/correct-usage-of-the-phrase-in-the-sequel-history-alternatives

As a non-native speaker of English, I have been perplexed by the phrase "in the sequel" as used in textbooks, lecture notes, and even research articles. None of my linguistically inclined native speaker friends have seen it outside of mathematical literature, and the relevant Oxford English Dictionary definition of "sequel" suggests mathematical usage is non-standard:

The ensuing narrative, discourse, etc.; the following or remaining part of a narrative, etc.; that which follows as a continuation; esp. a literary work that, although complete in itself, forms a continuation of a preceding one.

What I've inferred from context is that it means something along the lines of "for the rest of this book/paper/text", especially as it's usually used to introduce notation and/or convention.

Some examples:

We hope that the relation between linear transformations and matrices
is by now sufficiently clear that the reader will not object if in the sequel,
when we wish to give examples of linear transformations with various
properties, we content ourselves with writing down a matrix.

--Paul Halmos, Finite-Dimensional Vector Spaces, p. 86

[Here, and in the sequel, Card(S) denotes the number of elements in the
finite set S.]

--J.P. Serre, Local Fields, p. 64

In the sequel we shall denote by
∅ the empty set and by {pt} a set with
one element.

-- Pierre Schapira, Algebra and Topology course notes, p. 8

In the sequel, we will denote
by L(C ) the configuration space of any convergent CFG C.

--E. Goles et. al., Sandpile Models and Lattices: a comprehensive survey, Theoretical Computer Science, 2004, Vol 322, Issue 2, p. 398

So my actual question is two-fold.

1. What does the phrase actually mean?




2. When is its use warranted over the use of phrases such as "for the rest of the book/paper/text"?

arithmetic geometry - Questions about analogy between Spec Z and 3-manifolds

I'm not sure if the questions make sense:
Conc. primes as knots and Spec Z as 3-manifold - fits that to the Poincare conjecture? Topologists view 3-manifolds as Kirby-equivalence classes of framed links. How would that be with Spec Z? Then, topologists have things like virtual 3-manifolds, has that analogies in arithmetics?

Edit: New MFO report: "At the moment the topic of most active interaction between topologists and number theorists are quantum invariants of 3-manifolds and their asymptotics. This year’s meeting showed significant progress in the field."

Edit: "What is the analogy of quantum invariants in arithmetic topology?", "If a prime number is a knot, what is a crossing?" asks this old report.

An other such question:
Minhyong Kim stresses
the special complexity of number theory: "To our present day understanding, number fields display exactly the kind of order ‘at the edge of chaos’ that arithmeticians find so tantalizing, and which might have repulsed Grothendieck." Probably a feeling of such a special complexity makes one initially interested in NT. Knot theory is an other case inducing a similar impression. Could both cases be connected by the analogy above? How could a precise description of such special complexity look like and would it cover both cases? Taking that analogy, I'm inclined to answer Minhyong's question
with the contrast between low-dimensional (= messy) and high-dimensional (= harmonized) geometry. Then I wonder, if "harmonizing by increasing dimensions"-analogies in number theory or the Langlands program exist.

Minhyong hints in a mail to "the study of moduli spaces of bundles over rings of integers and over three manifolds as possible common ground between the two situations". A google search produces an old article by Rapoport "Analogien zwischen den Modulräumen von Vektorbündeln und von Flaggen" (Analogies between moduli spaces of vector bundles and flags) (p. 24 here, MR). There, Rapoport describes the cohomology of such analogous moduli spaces, inspired by a similarity of vector bundles on Riemann surfaces and filtered isocrystals from p-adic cohomologies, "beautifull areas of mathematics connected by entirely mysterious analogies". (book by R., Orlik, Dat) As interesting as that sounds, I wonder if google's hint relates to the initial theme. What do you think about it? (And has the mystery Rapoport describes now been elucidated?)

Edit:
Lectures by Atiyah discussing the above analogies and induced questions of "quantum Weil conj.s" etc.

This interesting essay by Gromov discusses the topic of "interestung structures" in a very general way. Acc. to him, "interesting structures" exist never in isolation, but only as "examples of structurally organized classes of structured objects", Z only because of e.g. algebraic integers as "surrounding" similar structures. That would fit to the guesses above, but not why numbers were perceived as esp. fascinating as early as greek antiquity, when the "surrounding structures" Gromov mentions were unknown. Perhaps Mochizuki has with his "inter-universal geometry" a kind of substitute in mind?

Edit: Hidekazu Furusho: "Lots of analogies between algebraic number theory and 3-dimensional topology are suggested in arithmetic topology, however, as far as we know, no direct relationship seems to be known. Our attempt of this and subsequent papers is to give a direct one particularly between Galois groups and knots."

cv.complex variables - Conformal maps of doubly connected regions to annuli.

By "see" I will assume you mean in a geometric sense. Then your question falls within a standard topic in geometric complex analysis. First some terminology: A doubly connected domain $R$ on the Riemann sphere is called a ring domain, and if you map it onto $r < |z| < s$ as a canonical domain, then $mathrm{mod}(R) = (2pi)^{-1}log(s/r)$ is called the conformal modulus or just modulus of the ring domain. By the way I have defined it, it is nearly trivially a conformal invariant, but it need not be defined this way. There is a geometric theory, the Ahlfors-Beurling theory of extremal length of curve families, within which the modulus of a ring domain can be defined directly and geometrically, without any preliminary conformal mapping onto some canonical domain. Extremal length can be proved to be a conformal invariant, and then one quickly sees that the two definitions coincide. There is an exposition of the theory of extremal length in Conformal Invariants by Ahlfors.

It would be unreasonable to expect to "see" the exact value of the modulus of a ring domain. The boundary of a ring domain can be extremely complicated geometrically, and every tiny wiggle impacts on the modulus. But extremal length yields inequalities for the modulus from geometric data. I will quote a single, rather striking, result of this kind:

If a ring domain $R$ contains no circle on the Riemann sphere separating its two boundary components, then $mathrm{mod}(R) leq 1/4$. The constant $1/4$ is sharp. The result is due to D. A. Herron, X. Y. Liu and D. Minda.

Small modulus means "thin" ring domain; if the modulus is large enough, the ring domain is so "fat" that it has to contain a separating circle.

ag.algebraic geometry - Twisted curves, admissible covers, and an algebraic analogue of a specific monodromy computation

This problem arose when trying to understand the stack of twisted stable maps into a stack (specifically BG), as introduced by Dan Abramovich, Angelo Vistoli and several co-authors (Olsson, Graber, Corti,...). However, my specific question can be formulated without mentioning such beasts and I will do so. If anyone wants background I can give some.

Suppose that we have the following set-up. One is given:

• A smooth projective curve C


• and a finite abelian group G acting on C,


• such that the quotient map $pi : C to C/G$ is an admissible cover.

Suppose moreover that $C/G cong mathbf{P}^1$. Over the complex numbers, one can make the following computation. For each ramification point of the covering, consider a small loop centered around that point, with orientation induced by the complex structure. Choose a lifting of that loop to a path on C. The path will start and end in the same fiber. Since the restriction of $pi$ away from the ramification locus is a G-torsor, the difference between start- and endpoint will give us a well-defined element of G. (This uses that G is abelian -- in general, one would only get an element up to conjugation, since there are several choices of liftings of the loop.) Now consider the product of all these elements over all ramification points. By considering the fundamental group of $mathbf{P}^1$ minus the ramification locus, it is clear that this product is the identity in G.

I would like to express this computation algebraically, i.e. without recourse to any monodromy or the classical fundamental group, working over an arbitrary base where the order of G is invertible. Unless I'm mistaken, one can still define an evaluation map associating an element of G to each ramification point, since the following (to me rather mysterious) construction should work. Consider the stack quotient $[C/G]$. This is a twisted curve in the sense of Abramovich et al, which basically means that the ramification points on C/G have been cut out and replaced by cyclotomic gerbes ("stacky points"). Now this twisted curve has an actual G-torsor over it, so we get a map $[C/G] to BG$. Restricting it to one of the stacky points, we obtain a cyclotomic gerbe over the base scheme with a G-torsor over it. But this is by definition an object of the rigidified inertia stack of BG, and the points of the (rigidified) inertia stack correspond to the elements of G.

Question 1: Is there a way of formulating this without using the language of twisted curves, i.e. associating an element of G to each ramification point only in terms of the admissible cover? There should be, since the moduli space of maps from twisted stable curves to BG is isomorphic to the moduli space of stable curves with an admissible cover which is a G-torsor away from the branch locus, but I don't see any sensible way of expressing such a construction algebraically.

Question 2: Can one show that in the algebraic setting, the product of the elements over all ramification points is the identity?

ag.algebraic geometry - Are schemes that "have enough locally frees" necessarily separated

Let me motivate my question a bit.

Thm. Let $X$ be a locally noetherian finite-dimensional regular scheme. If $X$ has enough locally frees, then the natural homomorphism $K^0(X)longrightarrow K_0(X)$ is an isomorphism.

A locally noetherian scheme has enough locally frees if every coherent sheaf is the quotient of a locally free coherent sheaf, $K^0(X)$ denotes the Grothendieck group of vector bundles on $X$ and $K_0(X)$ denotes the Grothendieck group of coherent sheaves on $X$.

The above theorem is shown as follows.

By the regularity (and finite-dimensionality!) of $X$, we can construct a finite resolution by a standard procedure. (Surject onto the kernel at each stage with a vector bundle.) Then the "Euler characteristic" associated to this resolution is inverse to the natural morphism.

Now, I was looking through the literature (Weibel's book basically) and I saw that this theorem appears with the additional condition of separability. (Edit: This is not necessary. The point is that noetherian schemes that have enough locally frees are semi-separated.)

Example. Take the projective plane $X$ with a double origin. Then $K^0(X) cong mathbf{Z}^3$ whereas $K_0(X) cong mathbf{Z}^4$.

Example. Take the affine plane $X$ with a double origin. Then $K^0(X) cong mathbf{Z}$, whereas $K_0(X) cong mathbf{Z}oplus mathbf{Z}$.

So I figured I must be missing something...Thus, I ask:

Q. Are locally noetherian schemes that have enough locally frees separated?

EDIT.

The answer to the above question is "No" as the example by Antoine Chambert-Loir shows.

From Philipp Gross's answer, we conclude that a noetherian scheme which has enough locally frees is semi-separated. This means that, for every pair of affine open subsets $U,Vsubset X$, it holds that $Ucap V$ is affine. Note that separated schemes are semi-separated and that Antoine's example is also semi-separated.

Taking a look at Totaro's article cited by Philipp Gross, we see that a regular noetherian scheme which is semi-separated has enough locally frees. (Do regular semi-separated and locally noetherian schemes have enough locally frees?)

This was (in a way) also remarked by Hailong Dao. He mentions the result of Kleiman and independently Illuzie. Recently, Brenner and Schroer observed that their proof works also with $X$ noetherian semi-separated locally $mathbf{Q}$-factorial. See page 4 of Totaro's paper. In short, separated is not really needed but semi-separated is.

Thus, we can conclude the following.

Suppose that $X$ is a regular and finite-dimensional scheme.

If $X$ has enough locally frees, then $K^0(X) longrightarrow K_0(X)$ is an isomorphism. For example, $X$ is noetherian and semi-separated.

Anyway, thanks to everybody for their answers. They helped me alot!

computer algebra - Software for rigorous optimization of real polynomials

I am looking for software that can find a global minimum of a polynomial function in R^n over a polyhedral domain (given by some linear inequalities say). The number of variables n is not more than a dozen. I know it can be done in theory (by Tarski's elimination of quantifiers in real closed fields), and I know that the time complexity is awful. However if there is a decent implementation that can handle a dozen variables with a clean interface, it would be great. I have tried builtin implementations in Mathematica and Maple, and they do not appear terminate on 4-5 variable instances.

If the software can produce some kind of concise "certificate" of its answer, it would be even better, but I am not sure how such a certificate should look like even in theory.

Edit: Convergence to the optimum is nice, but what I am really looking for is ability to answer questions of the form "Is minimum equal to 5?" where 5 is what I believe on a priori grounds to be the answer to optimization to be (in particular, it is a rational number). That also explains why I want a certificate/proof of the inequality, or a counterexample if it is false.

matroid theory - Has anyone implemented a recognition algorithm for totally unimodular matrices?

EDIT. Walter and Trümper have announced on arXiv their implementation, with source code available, of two methods for testing total unimodularity. Their paper describes the technical details of the implementation / algorithm, and also provides several experimental results.

---

I found the following link for an implementation in R, where they claim to have a function for testing whether a matrix is totally unimodular. I have not checked which particular algorithm they use.

Link: R package

na.numerical analysis - How well does a truncated fourier expansion of a stepfunction perform near the expansionpoint

Hey, i'am currently trying to make a stability analysis of a binary fluid at its phase border. As the governing equations in this double-diffusive problem are rather complicated i have to do a series expansion of my density profile. My problem is, that my PDEs contain a hevyside step function right at the border and i want to do a fourier expansion (i can then use symmetries of my problem and do some horrible algebra to actually get a "solution" which can be computed for up to 5 or 6 modes).

Now i know that fourier expansion performs quite horribly at a discontinous jump if only the first few modes are kept. Is it nervertheless possible to obtain a "margin of error"? Numerical precision is not of utmost concern, but the physics behind my problem should not all be thrown out by such an expansion. Do you know of any criteria or "better" expansion where its possible to make heavy use of symmetry/antisymmetry?

all the best,
jan

gr.group theory - Explicit cocycle for the central extension of the algebraic loop group G(C((t))).

Let G be a simple Lie group and let G(ℂ((t))) be its loop group.

The Lie algebra g[[t]][t^-1] has a well known central extension
(see e.g.
Wikipedia) given by the cocycle
c(f,g) = Res₀ < f dg >. Here, < > : g⊗g→ℂ denotes some invariant bilinear form on
g, and f dg is the (g⊗g)-valued differential given by multiplying f and dg.

Question: It there a similarly concrete cocycle for the central extension of G(ℂ((t))) by ℂ*?

To give you an idea of what I'm looking for, let me show
you a cocycle for central extension by S¹ of the smooth loop group $LG = mathit{Map} _ {C^infty} (S^1,G)$ of a compact Lie group $G$.

Pick a bounding disc D_γ : D² → G for each element γ ∈ LG. The cocycle is then given by

$$c(gamma,delta) = expbig(icdotbig(quadint langle D_gamma^*theta_L,D_delta^*theta_Rrangle
+int H^*etaquad big)big)$$

where $theta_L,theta_RinOmega(G,mathfrak{g})$ are the Maurer-Cartan 1-forms, $etainOmega^3(G)$ is the Cartan 3-form,
and $H:D^3to G$ in a homotopy between $D_gamma D_delta$ and $D _ {gammadelta}$.

---

References:
The cocycle for the smooth loop group can be found on page 19 of the paper
From Loop groups to 2-groups, by Baez, Crans, Schreiber, and Stevenson,
and also on page 8 of Mickelsson's paper From Gauge anomalies to Gerbes and Gerbal actions.

nt.number theory - P-adic local Langlands for non-unitary representations?

This is a natural question. For example, using Colmez's results, as completed by Paskunas (who shows that Colmez's p-adic local Langlands describes all topologically irreducible unitary admisisble Banach space representations of $GL_2(mathbb Q_p)$) one can start to prove purely representation-theoretic facts about unitary admissible Banach space reps.
of $GL_2(mathbb Q_p)$, using Colmez's description in terms of $(phi,Gamma)$-modules. Now while some of these might naturally be related to unitarity,
there are certainly results that now seem accessible in the unitary case, which I suspect
don't actually require unitarity in order to hold. However, if one is going to use Colmez's and Paskunas's results, one needs unitarity as a hypothesis.

One could imagine (and here I am talking at the vaguest level) working with some kind of Weil group representations rather than Galois representations in order to include the non-unitary representations. I think that Schneider and Teitelbaum may have pondered this at some point, but I don't know what came of it. And I don't know how reasonable it is to hope for such a correspondence. I am just making the most absolutely naive guess, which you've probably also made yourself!

(One thing that makes me nervous is that when one works with unitary reps., there is a natural way to go from locally analytic reps. to Banach ones, by passing to universal unitary completions, and this is sometimes sensibly behaved, e.g. in the case of locally analytic inductions attached to crystabelline reps., by Berger--Breuil. But if one starts
to imagine completions that are not unitary, then I could imagine that they are much more wild; but again, this is just speculation.)

Distribution of quadratic residues of a fixed number without using Dedekind zeta function

This answer summarizes the above discussion: Extend $p mapsto left( frac{n}{p} right)$ to a multiplicative function $chi$ on the positive integers. By quadratic reciprocity, $chi$ is periodic modulo $4n$, and it is multiplicative by construction, so it is a character. We know that $L(1, chi) neq 0$. Thus, $lim_{s to 1^{+}} |log L(s,chi)| < infty$.

We compute:
$$log L(s, chi) = - sum log left( 1-frac{chi(p)}{p^s} right) = sum frac{chi(p)}{p^s} + O(1).$$

So $sum left( frac{n}{p} right)/p^s$ is bounded as $s to 1^{+}$. A little more work shows that the limit as $s to 1^{+}$ exists.

If you want to prove results like that $sum left( frac{n}{p} right)/p$ converges, or that $|sum_{p < N} left( frac{n}{p} right)| = o(N)$, then you need Tauberian methods, as discussed in any book on analytic number theory.

nt.number theory - Zagier's one-sentence proof of Fermat's theorem.

It's been a while since I read Elsholtz's article, but after doing so I felt none the wiser. Below I have translated Heath-Brown's proof into the language of binary quadratic forms; Zagier's proof looks more interesting from this point of view (the connections to Gauss reduction are much closer), but when working out the details I got stuck in the middle.

One essential ingredient for the proofs by Heath-Brown and Zagier was pointed out already by Frick in 1918, who showed that if $p = a^2 + 4b^2$ is an odd prime number, then the indefinite binary quadratic form $Q = (-b,a,b)$ with discriminant $p$ is Gauss reduced and is contained in the principal cycle.

For proving that such a form exists without assuming that $p$ is a sum of two squares, we consider all forms $(A,B,C)$ with discriminant $p$ such that $A < 0$ and $C > 0$. From $p = B^2 - 4AC$ it then follows that the set
$$ S = {(A,B,C): B^2 - 4AC = p, A < 0, C > 0} $$
is finite. The obvious map
$$ mu: S to S, quad (A,B,C) to (-C,B,-A) $$
is an involution; if $S$ had odd cardinality, it would follow that $mu$ has a fixed point, say $(A,B,-A)$, from which we would get $p = B^2 + 4A^2$. Unfortunately, $S$ has even cardinality since the involution
$$ nu: S to S, quad (A,B,C) to (A,-B,C) $$
has no fixed points: this is because $B = 0$ implies $p = 4AC$, which is impossible for prime numbers $p$.

We now would like to find a subset $U subset S$ of $S$ with odd cardinality on which $mu$ is still defined. The most natural idea would be considering the forms with $B > 0$. For showing that this set of forms has odd cardinality, we have to define an involution $(A,B,C) to (A',B',C')$ on this subset that has exactly one fixed point. To find such an involution, we start with $(A,B,C) to (A,-B,C)$ and then apply reduction by changing the middle coefficient modulo $2A$ and then adjusting the last coefficient so that the discriminant is $p$. This gives
$$ (A,-B,C) to (A',B',C') = (A,-2A-B,A+B+C). $$
Now we are facing the problem that it is not clear at all that $B' = -2A-B > 0$, or that $C' = A+B+C > 0$. But if we set
$$ U = {(A,B,C) in S: A+B+C > 0 }, $$
then the map
$$ gamma: (A,B,C) to (A,-2A-B,A+B+C) $$
actually is an involution on $U$. Moreover, $(A,B,C)$ is a fixed point if and only if $-2A-B = B$ and $A+B+C = C$, which is equivalent to $A = -B$. Since $p = B^2 - 4AC = B^2 + 4BC = B(B+4C)$ is prime, we must have $|A| = |B| = 1$. Since $A < 0$, this implies that the fixed point is $(-1,1,frac{p-1}4)$; this form is equivalent to the principal form $(1,1,frac{p-1}4)$.

The involution $gamma$ on $U$ shows that $U$ has odd cardinality; the map
$$ (A,B,C) to (-C,-B,-A) $$
is an involution on $S$ sending $U$ to $S setminus U$, which impliesthat $|S| = 2 |U|$. The involution $nu$ on $S$ sends elements with $B > 0$ to elements with $B < 0$, hence
$$ T = {(A,B,C) in S: B > 0} $$
has the same number of elements as $U$, and in particular, it has odd cardinality. Finally, $mu$ is an involution on $T$, and now the Two-Squares Theorem follows.

References

1. H. Frick,
   Über den Zusammenhang der Perioden quadratischer Formen
   positiver Determinante mit der Zerlegung einer Zahl in die
   Summe zweier Quadrate, Diss. ETH Zürich, 1918

dg.differential geometry - Is there a complex structure on the 6-sphere?

Here is a shot in the dark (Disclosure: I really know nothing about this problem).

Let $G:=mathsf{SU}(2)$ act on $G^3$ by simultaneous conjugation; namely, $$gcdot(a,b,c)=(gag^{-1},gbg^{-1},gcg^{-1}).$$ Then the quotient space is homeomorphic to $S^6$ (see Bratholdt-Cooper).

The evaluation map shows that the character variety $mathfrak{X}:=mathrm{Hom}(pi_1(Sigma),G)/G$ is homeomorphic to $G^3/G,$ where $Sigma$ is an elliptic curve with two punctures.

Fixing generic conjugation classes around the punctures, by results of Mehta and Seshadri (Math. Ann. 248, 1980), gives the moduli space of fixed determinant rank 2 degree 0 parabolic vector bundles over $Sigma$ (where we now think of the punctures are marked points with parabolic structure). In particular, these subspaces are projective varieties.

Letting the boundary data vary over all possibilities gives a foliation of $mathfrak{X}cong G^3/Gcong S^6$. Therefore, we have a foliation of $S^6$ where generic leaves are projective varieties; in particular, complex.

Moreover, the leaves are symplectic given by Goldman's 2-form; making them Kähler (generically). The symplectic structures on the leaves globalize to a Poisson structure on all of $mathfrak{X}$.

Is it possible that the complex structures on the generic leaves also globalize?

Here are some issues:

1. As far as I know, the existence of complex structures on the leaves is generic. It is known to exist exactly when there is a correspondence to a moduli space of parabolic bundles. This happens for most, but perhaps not all, conjugation classes around the punctures (or marked points). So I would first want to show that all the leaves of this foliation do in fact admit a complex structure. Given how explicit this construction is, if it is true, it may be possible to establish it by brute force.


2. Assuming item 1., then one needs to show that the structures on the leaves globalize to a complex structure on all of $mathfrak{X}$. Given that in this setting, the foliation is given by the fibers of the map: $mathfrak{X}to [-2,2]times [-2,2]$ by $[rho]mapsto (mathrm{Tr}(rho(c_1)),mathrm{Tr}(rho(c_2)))$ with respect to a presentation $pi_1(Sigma)=langle a,b,c_1,c_2 | aba^{-1}b^{-1}c_1c_2=1rangle$, it seems conceivable that the structures on the leaves might be compatible.


3. Moreover, $mathfrak{X}$ is not a smooth manifold. It is singular despite being homeomorphic to $S^6$. So lastly, one would have to argue that everything in play (leaves, total space and complex structure) can by "smoothed out" in a compatible fashion. This to me seems like the hardest part, if 1. and 2. are even true.

Anyway, it is a shot in the dark, probably this is not possible...just the first thing I thought of when I read the question.

ag.algebraic geometry - Is the fixed locus of a group action always a scheme?

The question gives the "wrong" definition of Fix(T), hence the resulting confusion.

A more natural definition of the subfunctor X^G of "G-fixed points in X" is
(X^G)(T) = {x in X(T) | G_T-action on X_T fixes x}
               = {x in X(T) | G(T')-action on X(T') fixes x for all T-schemes T'}.
(Of course, can just as well restriction to affine T and T' for "practical" purposes.)

By way of analogy with more classical situations, if the base is a field k then a moment's reflection with the case of finite k shows that
{x in X(k) | G(k) fixes x}
is the "wrong" notion of (X^G)(k), whereas
{x in X(k) | G-action on X fixes x}
is a "better" notion, and is what the above definition of (X^G)(k) says.

From this point of view, if (for simplicity of notation) the base scheme is an affine Spec(k) for a commutative ring k then the "scheme of G-fixed points" exists whenever G is affine and X is separated provided that k[G] is k-free (or becomes so after faithfully flat extension on k). So this works when k is a field, or any k if G is a k-torus (or "of multiplicative type"). See Proposition A.8.10(1) in the book Pseudo-reductive groups.

gt.geometric topology - "Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known?

For 2) I think the following is an answer: Suppose $K$ is a compact Lie subgroup
of $mathrm{Diff}(S^n)$ of dimension $geq{n+1choose 2}$. Being compact it is
the group of isometries of some Riemannian metric of $S^n$ and we fix one such
metric. The stabiliser of a point therefore has dimension at most $nchoose 2$
(the dimension of the orthogonal group of $mathbb R^n$) and as an orbit has at
most dimension $n$, $K$ has at most dimension $n+{nchoose 2}={n+1choose 2}$
and hence we have equality. This means that the stabiliser contains
$mathrm{SO}_n$ and $K$ acts transitively. In particular the metric fixed by $K$
has constant curvature and then the curvature is positive and thus up to a
constant conformal factor is conjugate to the standard metric. This gives a
conjugation of $K$ into the isometry group of the standard sphere.

Addendum: Tom's example gives that the dimension of a non-compact group
acting (faithfully) on $S^n$ is unbounded. One question seems to remain namely
if the dimension of a semi-simple group action is bounded or not. By the above
we get a bound on the dimension of a maximal compact subgroup. In many cases
this seems to bound the dimension of the group itself. For instance, is the
dimension bounded if the center is finite? In that case the situation is
completely described by a Cartan decomposition of the Lie algebra and the
question is whether the $-1$-part (usually denoted $mathfrak p$) of it has
dimension bounded by the dimension of the $+1$-part (the Lie algebra of the
maximal compact). I myself don't know enough of the real Lie group theory to
decide that.

An example beyond that is the universal cover $G$ of $mathrm{SL}_2(mathbb R)$.
It is contractible and has no non-trivial compact connected subgroup. It also
acts on the universal cover of $S^1$ (compatibly with the projective action of
$mathrm{SL}_2(mathbb R)$ on $S^1$) and hence acts continuously on $[-1,1]$
(say) fixing the endpoints. However, the action at the endpoints is not flat so
this action does not extend to a smooth action on $mathbb R$ acting as the
identity outside if $[-1,1]$. If it could be modified to do so one could use
Tom's argument to get an action of any finite product of copies of $G$ (at least
on $S^1$). One could try to find three vector fields with support in $[-1,1]$
fulfilling the defining relations of $mathfrak{sl}_2(mathbb R)$ but it looks
tricky to me.

Addendum 1: To be more precise about the action on $[-1,1]$ we may use
$tanh$ as diffeomorphism from $mathbb R$ to $(-1,1)$. The rotation group of
$mathrm{SL}_2(mathbb R)$ lifts to the group of translations of $mathbb R$ and
they correspond under this diffeomorphism to $varphi_lambda(t)=(lambda t+1)/(t+lambda)$ of
$(-1,1)$ for $lambda>1$ or $lambda<-1$ (we miss the identity map which corresponds to
$lambda=pminfty$). Now, one possibility of getting an action of $G$ on $mathbb R$ which
is the identity outside of $(-1,1)$ is by trying to conjugate the one we have by
a diffeomorphism $gamma$ of $(-1,1)$. This requires every conjugate
diffeomorphism to be flat at $pm1$ where a diffeomorphism $psi$ of
$(-1,1)$ is flat at $1$ if $psi(x)=x+mathcal{O}(|x-1|^n)$ for all $n$ and $x$
close to $1$ (and similarly for $-1$). Now suppose that we have a $gamma$ such that
conjugating the given action of $G$ by it gives an action all of whose elements
are flat. In particular we have $gamma(varphi_lambda(gamma^{-1}(x)))=x+mathcal{O}(|x-1|^n)$
for all $n$. In particular, putting $n=2$ we get that
$lim_{tto1}gamma'(varphi_lambda(gamma^{-1}(t)))varphi_lambda'(gamma^{-1}(t))gamma'(gamma^{-1}(t))^{-1}=1$
and putting $s=gamma^{-1}(t)$, using that $lim_{tto1}gamma^{-1}(t)=1$ and that
$lim_{sto1}varphi_lambda'(s)=varphi_lambda'(1)=(lambda-1)/(lambda+1)$ this gives
$lim_{sto1}gamma'(varphi_lambda(s))/gamma'(s)=(lambda+1)/(lambda-1)$. This puts a very stringent
condition on $gamma$ and then also the same condition must be fulfilled for
$varphi_lambda$ replaced by any element of $G$ (as well as needing the flatness
condition for all orders $n$).

mp.mathematical physics - Motivating the Laplace transform definition

This answer is not exactly an answer to the original question, but this is for the benefit of MO user vonjd who wanted to know more details about the similarities between solving differential equations through Laplace transforms and solving recurrence relations using generating functions.

Since I was going to write it anyway, I figured I might as well post it here for anyone interested.

I will do an example of each, and this should be enough to show the similarities. In each case, we have a linear equation with constant coefficients; this is where both methods really shine, although they both can handle some variable coefficients more or less gracefully. Ultimately, the biggest challenge is to apply the inverse transform: always possible in the linear case, not so easy otherwise.

Differential Case

Take the function $y(t)=2e^{3t}-5e^{2t}$. It is a solution of the IVP:
begin{equation}
y''-5y'+6y=0; qquad y(0)=-3, y'(0)=-4.
end{equation}
If we apply the Laplace transform to the equation, letting $Y(s)$ denote the transform of $y(t)$,
we get
$$ s^2Y(s)-sy(0)-y'(0)-5[sY(s)-y(0)]+6Y(s)=0.$$
Substitute the values of $y(0)$ and $y'(0)$, and solve to obtain:
$$ Y(s)=frac{11-3 s}{s^2-5s+6};$$
and apply partial fractions to get:
$$ Y(s)= frac{2}{s-3}+frac{-5}{s-2}.$$
This is where you exclaim: "Wait a second! I recognize this, since it's well known that
$$mathcal{L}[e^{at}]= frac{1}{s-a}$$
for all $a$, then by linearity we recognize the function that I started from.

Recurrence case

Let $(a_n)$ be the sequence defined for all $ngeq 0$ by $a_n=2(3^n)-5(2^n)$.
It is a solution of the IVP:
begin{equation}
a_{n+2}-5a_{n+1}+6a_n=0 qquad a_0=-3, a_1=-4.
end{equation}
Define the generating function $A(x)$ to be:
$$ A(x)=sum_{n=0}^{infty} a_n; x^n. $$
Multiplying each line of the recurrence by $x^{n+2}$ gives:
$$ a_{n+2}; x^{n+2}-5a_{n+1}; x^{n+2}+6a_n; x^{n+2}=0 $$
You can sum those lines for all $ngeq 0$, do a small change of index in each sum, and factor out relevant powers of $x$ to get
$$ sum_{n=2}^{infty} a_n; x^n-5x sum_{n=1}^{infty} a_n; x^n+6x^2 sum_{n=0}^{infty} a_n; x^n=0.$$
Or in other terms:
$$ A(x)-a_1x-a_0-5x[A(x)-a_0]+6x^2A(x)=0.$$
Substituting $a_0$ and $a_1$ and solving for $A(x)$ then gives, with partial fractions:
$$ A(x)=frac{11 x-3}{6 x^2-5 x+1}=frac{2}{1-3 x}+frac{-5}{1-2 x}$$

Looks familiar? It should! If you substitute $x=1/s$, you will recover $sY(s)$ from the differential example.

For generating functions, the key fact we need here is the sum of geometric series:
$$ sum_{n=0}^{infty} (ax)^n=frac{1}{1-ax}.$$
Thus, by linearity again, we recognize the sequence we started from in the expression for $A(x)$.

Closing Remarks

In both theories, there is the notion of characteristic polynomial of a linear equation with constant coefficients. This polynomial ends up being the denominator of the Laplace transform, and the reversed polynomial $x^dp(1/x)$ is the denominator of the generating function. In both cases, multiple roots are very well managed by the theories and explain very naturally the appearances of otherwise "magical" solutions of the type $te^{lambda t}$ or $n(r^n)$.

The biggest mystery to me is the historical perspective: did one technique pre-date the other, and were the connections actively exploited, or did both techniques develop independently for a while before the similarities were noticed?

at.algebraic topology - Simplicial Model of Hopf Map?

Here is one thing to try. Start with the smallest simplicial model for S¹ (the 1-simplex modulo its boundary). Take the free group in each degree (but force the basepoint to be the identity). The resulting simplicial group FS¹ is a model for ΩS²; furthermore, being a simplicial group, it's a Kan complex. Thus, we know there must be some map f: S²->FS¹ which represents the generator of π₂ΩS²; the group of FS¹ in degree 2 is not too big, so it should not be hard to write this down explicitly (I haven't tried, though.)

Of course, you really want a map S³->X, where X models the 2-sphere. Since FS¹ is a simplicial group, let X=BF¹, it's classifying space. X is a model for the 2-sphere, and I expect that if you examine it closely, you will see the "suspension" of f corresponds to some explicit 3-simplex in X, which is your model.

I'm not sure this counts as a "combinatorial model", of course.

(I have a vague memory that Dan Kan did something like this in one of his papers in the 50s. Is that right?)

Introduction to modular property of affine alegebra and conformal vertex algebra

Yes -- see Cardy's paper Operator content of two-dimensional conformally invariant theories. In a nutshell, the torus partition function of a conformal theory can be computed using a transfer matrix in several ways. The equivalence of these different calculations is the modular invariance of the torus partition function.

ct.category theory - Is a groupoid determined by its Hopfish algebra?

This is a follow up to my question What is the precise relationship between groupoid language and noncommutative algebra language?. I will briefly review some definitions; for details, a good place to look is Christian Blohmann, Alan Weinstein. Group-like objects in Poisson geometry and algebra. 2007. arXiv:math/0701499v1. And actually, there are two versions of my question, one for groupoids and the other for categories. So that I can avoid all analysis, I will restrict my attention to finite things; if you know the answer in, say, topological spaces, or smooth manifolds, or..., then I'm also interested.

A category is a span of sets $C = { C_0 overset l leftarrow C_1 overset r rightarrow C_0}$ which is an algebra object in the category of $C_0,C_0$ spans. I.e. there are maps of spans $i: {C_0 = C_0 = C_0} to C$ and $m: C underset{C_0}times C to C$ making the usual diagrams commute. A category is a groupoid if additionally there is an involution ${^{-1}} : { C_0 overset l leftarrow C_1 overset r rightarrow C_0} to { C_0 overset r leftarrow C_1 overset l rightarrow C_0}$ satisfying some condition. A category $C$ is finite if both $C_0$ and $C_1$ are finite.

A finite-dimensional algebra $A$ (over a fixed field $mathbb K$) is sesqui if it is equipped with a bimodule ${_A Delta _{Aotimes A}}$ and an "associativity isomorphism" $$varphi: {_A Delta _{Aotimes A}} underset{Aotimes A}otimes bigl( {_A A _A} underset{mathbb K}otimes {_A Delta _{Aotimes A}}bigr) oversetsimto {_A Delta _{Aotimes A}} underset{Aotimes A}otimes bigl( {_A Delta _{Aotimes A}} underset{mathbb K}otimes {_A A _A} bigr) $$
of $A, A^{otimes 3}$ bimodules, which satisfies a pentagon. There should also be a "counit" bimodule $_A epsilon _{mathbb K}$, some triangle isomorphisms, and some more equations.
A sesquialgebra is hopfish if a hard-to-write-down condition is satisfied; see Xiang Tang, Alan Weinstein, Chenchang Zhu. Hopfish algebras. 2006. arXiv:math/0510421v2. Let ${_A {Delta^{rm flip}} _{Aotimes A}}$ denote the bimodule $Delta$ with the two right $A$-actions flipped. A sesquialgebra is symmetric if it comes equipped with a bimodule isomorphism $psi: {_A Delta _{Aotimes A}} oversetsimto {_A {Delta^{rm flip}} _{Aotimes A}}$ so that $varphi,psi$ satisfy two hexagons. A sesquialgebra is finite if $A,Delta, dots$ are finite-dimensional over $mathbb K$.

Let $C = { C_0 overset l leftarrow C_1 overset r rightarrow C_0}$ be a finite category. Then it gives rise to a finite symmetric sesquialgebra as follows. The algebra $A$ is given by the vector space $mathbb K C_1$ with the convolution product (given on the basis by $aotimes b mapsto ab$ if $(a,b)$ is a composable pair of morphisms, and $aotimes b mapsto 0$ otherwise). The bimodule $Delta$ is given as the vector space with basis all pairs $(a,b) in C_1 times C_1$ with $l(a) = l(b)$. I will let you work out the rest: the actions, the associator $varphi$ and symmetrizer $psi$, etc. If $C$ is actually a groupoid, then $mathbb K C_1$ is hopfish.
This construction extends to a 2-functor, and so sends equivalences of categories to Morita equivalences of sesquialgebras.

Question: It is well known that a groupoid $C$ cannot be recovered from the algebra $mathbb K C_1$; compare for example the group with two elements, thought of as a groupoid with one object, and the set with two elements, thought of as a groupoid with only identity morphisms. But the examples I know can be distinguished by remembering the hopfish structure.

• Can a finite category be recovered from its symmetric sesquialgebra?


• If not, can a finite groupoid be recovered from its symmetric hopfish structure?


• Can an equivalence class of finite categories be recovered from the Morita-equivalence class of finite symmetric sesquialgebras?


• If not, do we at least have the corresponding statement for groupoids/hopfish algebras?

In a footnote, Blohmann and Weinstein suggest that they do not know the answers to the above questions. But that was three years ago; perhaps there has been more recent work?

ct.category theory - Does linearization of categories reflect isomorphism?

In this answere I (try to) present the problem as a Algebraic Geometry one:

consider the category $mathscr{C} $ with two objects $X, Y$ and

$mathscr{C}(X, Y)$={$r_1, s'_1, r_2, s'_2$} ;
$mathscr{C}(Y, X)$={$s_1, r'_1, s_2, r'_2$} ; $mathscr{C}(X, X)$={$1_X, e_X$} ; $mathscr{C}(Y, Y)$={$1_Y, e_Y$} where $e_X, e_Y$ are idempotent, and any composition of a morphism by a a idempotent not alter the morphism, and $ 1_Y= r_1circ s_1= r_2circ s_2$, $ 1_X= r'_1circ s'_1= r'_2circ s'_2$, all other compositions give the (no identity) idempotent.
Suppose that $R$ is a commutative ring and in $Rmathscr{C} $ consider the morphims
$A:= a_1cdot r_1 + b'_1cdot s'_1 + a_2cdot r_2 + b'_2cdot s'_2: Xto Y$ and

$B:= b_1cdot s_1 + a'_1cdot r'_1 + b_2cdot s_2 + a'_2cdot r'_2: Yto X$.

Let $alpha :=a_1+b'_1+ a_2+ b'_2$, $beta :=b_1+a'_1+ b_2+ a'_2$,

Then we have $Bcirc A=1_X$ iff:

1) $ a'_1cdot b'_1+ a'_2cdot b'_2=1$ and

2) $beta cdot a_1+ (beta - a'_1)cdot b'_1+ beta cdot a_2+(beta -a'_2)b'_2=0$ i.e. $beta cdot alpha = a'_1cdot b'_1+ a'_2cdot b'_2 $

similarly we have $Acirc B=1_Y$ iff:

1') $ a_1cdot b_1+ a_2cdot b_2=1$ and

2') $alpha cdot beta = b_1cdot a_1+ b_2cdot a_2 $

all these equations are equivalent to the system of three equations:

$ a'_1cdot b'_1+ a'_2cdot b'_2=1, a_1cdot b_1+ a_2cdot b_2=1, alpha cdot beta = 1$

thinking these in $mathbb{C}[ a_1, b'_1, a_2, b'_2 , b_1, a'_1, b_2,a'_2] $ these represent three varieties on $mathbb{C}^8 $

If these varieties have an a intersections then $X, Y$ are isomorphic in $mathbb{C}mathscr{C} $ (but aren't isomorphic in $mathscr{C}) $.

differential equations - helmholtz zero in R^3

If $u$ is a solution to the equation $triangle u +k^2 u=0$ in a 3D domain $Omega$, then
for any $xinOmega$ and any $r>0$ such that ${yinmathbb R^3: |x-y|leq r }subsetOmega$, we have
$$u(x)=frac {p(r)}{4pi r^2}int_{|x-y|=r} u(y)dS_y,qquadqquadqquad(1)$$
where
$$p(r)=frac{rk}{sin rk}.$$

Formula (1) is an analogue of the mean value theorem for harmonic functions (in the case of spherical means).

Edit added: relation (1) is valid for all $r_1leq r$. If we multiply it by $4pi r^2/p(r)$ and integrate between $0$ and $r$ we will obtain that
$$u(x)=frac{k^3}{4pi(sin rk-rkcos rk)}int_{|x-y|leq r} u(y)dy.$$
The latter formula generalizes the property that the value of a harmonic function at $xinOmega$ is equal to function's average value over a ball with the center at $x$.

A short derivation of formula (1) can be found in chapter IV of Methods of Mathematical Physics (Vol. 2)
by Courant and Hilbert (or see Harald's comment below).

dg.differential geometry - differentiable structure and coordinates in R^2

There is only one differentiable structure permitted in R^2, meaning, I think, that all atlases in R^2 are diffeomorphic to the Cartesian atlas. But, doesn't the polar coordinate system represent an atlas that is not truly diffeomorphic to the Cartesian atlas, due to the coordinate singularity it has at its origin?

ap.analysis of pdes - Can an integral equation always be rewritten as a differential equation?

With Charles Matthews comments in perspective, these are some notes I made sometime ago on this topic. I dont have the books in front of me so I can't look up the details right now.

1) In Zabreyko's book Integral equations (902860393X), there is the method based on Green's functions in Appendix A.

2) Bellman in Perturbation techniques Sec 10 points out that the other way (ODE to integral equation) is actually better

Conversion of differential eqn to integral equation
is one of the powerful devices in
approximation theory. Its potency is
due to the fact that integration is a
smoothing op, while differentiation
accentuates small variations. If u(t)
and v(t) are close together, then
∫u(s)ds and ∫v(s)ds will be comparable
in value, but du/dt and dv/dt may be
arbitrarily far apart. Consequently,
when carrying out successive
approximations, we prefer integral
operators to differential operators.
On the other hand, in numerical
solutions, we prefer differential
operators to integral operators.

3) You can also look up Handbook of Integral Equations by Polyanin.
Sec 8.4.5, Sec 9.7 and sec 9.3.3 are three situations where the method reduces a specific integral equation to an ODE

co.combinatorics - Is the "diagonal" of a regular language always context-free?

It's unnecessary to assume that L is unambiguous: a regular language always is, because there exists a DFA that accepts it.

Following Richard's notation, it is easy to construct a DPDA for K, so it is a DCF language (a subset of the unambiguous CFLs). Looking at the construction that proves that the intersection of a CFL and a regular language is CF, we can see that the same property is also preserved for DCFLs, because no step in the construction would produce non-determinism if it isn't already.

So we can conclude that L'=K∩L is a DCFL, and in particular unambiguous.

sg.symplectic geometry - Hamiltonian displaceability of tori in symplectic balls

Here is my first try at a question, which is a really easy to state question
about displaceability:

Let $D$ be the unit disk in the complex plane $D = { |z| leq 1 }$ equipped
with its standard symplectic form and for $r in (0,1)$ let $S(r) subset D$ be the Lagrangian circle of radius $r$ centered at $0$, enclosing area $pi r^2$.

Question: for which $r_1,ldots, r_n$ is the Lagrangian torus
$S(r_1) times ldots times S(r_n)$ Hamiltonian displaceable in $D^n$?

Conjecture: for a given $r_1,ldots, r_n$, the Lagrangian is non-displaceable iff
$r_j^2 ge 1/2$ for all $j$.

Known: Using McDuff's probes, one can see that if some $r_j^2 < 1/2$, then the Lagrangian is displaceable. I think I can show that if each $r_j^2$ lies in the set ${ 1/2, 2/3,3/4, ldots }$ then the Lagrangian is non-displaceable, by embedding D^n in a product of weighted projective lines and showing the Floer homology of the Lagrangian is non-vanishing. But there must be a way of doing better.

Sub-question: (which came out of a discussion with Abouzaid): is there a Floer-theoretic
way of seeing the non-displaceability of $S(r)$ for $r^2 ge 1/2$? This is easy
to prove by elementary means.

lie algebras - Jordan decomposition in a classical group

Let $mathfrak{g} subset mathfrak{gl}_n$ be one of the classical real or complex semisimple Lie algebras. If $g in mathfrak{g}$, then $g$ has a Jordan decomposition $g = g_s + g_n$ with $g_s$ semisimple and $g_n$ nilpotent, and $[g_s,g_n]=0$.

The elements $g_s,g_n$, which a priori are just in $mathfrak{gl}_n$, are both in $mathfrak{g}$ again. There are various middle-brow general ways to see this (for one, use that $mathfrak{g}$ is algebraic), but for concrete choices of $mathfrak{g}$ it's basically elementary, as follows. One knows from the construction of the Jordan decomposition that $g_s,g_n$ are both polynomials in $g$ (different polynomials for different $g$, of course), and (EDIT) you can rig the construction so that these polynomials are odd. The Lie algebra $mathfrak{g}$ is the subspace of $mathfrak{gl}_n$ cut out by conditions like $mathrm{trace}(g)=0$, or $Jg = -g^{t} J$ for some matrix $J$, and so forth. The condition $mathrm{trace}(g)=0$ is always true for $g_n$, so it's true for $g_s$ if true for $g$. The condition $Jg=-g^t J$ is visibly true for odd $p(g)$ if true for $g$, so if true for $g$ then it's true for both $g_s$ and $g_n$. Thus $g_s$ and $g_n$ visibly satisfy whatever conditions $g$ is required to satisfy, and so are contained in $mathfrak{g}$.

(This might seem lowbrow but in fact I think this is basically the idea of the proof that Fulton-Harris give for general semisimple Lie algebras.)

Now suppose instead that $G$ is a real or complex linear Lie group with Lie algebra $mathfrak{g}$. This time the Jordan decomposition is $g = g_s g_u$ with $g_u$ unipotent, and indeed $g_s$ and $g_u$ are still in $G$. But if you try to make the same lowbrow argument as in the Lie algebra case, it appears to die horribly (a condition like $g^t = g^{-1}$ certainly need not be preserved by taking a polynomial in $g$). My question is, is there an elementary way to rescue it? (In particular, something other than just the general argument for algebraic groups.) Obviously you're fine for elements $g$ in the image of the exponential map, so the issue is passing to the whole group. A caveat is that I do $textit{not}$ want to assume that $G$ is connected.

dg.differential geometry - Can the geodesic flow be preserved by an inhomogeneous rescaling of a cross section?

Let $M$ be a compact Riemannian manifold with metric $g$ and associated Riemannian volume $nu$ and geodesic flow $G_t : UTM rightarrow UTM$, where the unit tangent bundle is indicated. Let $X_j subset UTM$ for $1 le j le n$ be open disjoint codimension one submanifolds transversal to $G_t$, i.e., local cross sections (a global cross section does not exist).

Is it possible to choose a metric $g'$
on $M$ with geodesic flow $G'_t = G_t$
and $nu'_1(X_j) equiv 1$?

NB. Here $nu'_1$ denotes the induced codimension one [relative] measure on $UTM$.

This question was prompted by a helpful comment to this one.

nt.number theory - Dirichlet series with integer coefficients as a UFD

First, I want to nail down a reference to the solution to the problem alluded to above: the Dirichlet ring of functions f: Z⁺ -> Z with pointwise addition and convolution product is a UFD. This was proved by L. Durst in his 1961 master's thesis at Rice University and independently by Cashwell and Everett. References:

Deckard, Don; Durst, L. K.
Unique factorization in power series rings and semigroups.
Pacific J. Math. 16 1966 239--242.

Cashwell, E. D.; Everett, C. J.
Formal power series.
Pacific J. Math. 13 1963 45--64.

Note that the latter is the second paper of Cashwell and Everett on the subject of factorization in Dirichlet rings. They also had a 1959 paper:

Cashwell, E. D.; Everett, C. J.
The ring of number-theoretic functions.
Pacific J. Math. 9 1959 975--985.

Ulam makes reference to the 1959 paper in his statement of the problem in the first (1960) edition of his book. In the 1964 paperback edition of his book, he announces the problem as solved by 1961 papers of Buchsbaum and Samuel.

I want to make the remark that since the original 1959 paper of Cashwell and Everett certainly makes the connection to formal power series rings, there is more to the solution of the problem than this. (Indeed, the 1961 result of Samuel, that if R is a regular UFD, then R[[t]] is again a UFD, seems to be the key.)

There is a 2001 arxiv preprint of Durst which claims a more elementary proof of the result:

http://arxiv.org/PS_cache/math/pdf/0105/0105219v1.pdf

As to whether the result has number-theoretic significance: not as far as I know. I should say that I first learned of the Cashwell-Everett theorem by reading an analytic number theory text -- Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, p. 26 -- but the author seems to mention it just for culture.

Finally, remember that this is a result about factorization in the ring of formal Dirichlet series, an inherently algebraic beast. So perhaps the following result is more relevant:

Bayart, Frédéric; Mouze, Augustin Factorialité de l'anneau des séries de Dirichlet analytiques. (French) [The ring of analytic Dirichlet series is factorial] C. R. Math. Acad. Sci. Paris 336 (2003), no. 3, 213--218

lie algebras - What is a Module over a Lie algebroid?

A $mathfrak{g}_A$-module $M$ is a $k$-module endowed with structures of $A$-module and $mathfrak{g}_A$-module satisfying the compatibility equations $(ax)m = a(xm)$ and $x(am) = x(a)m + a(xm)$ for any $ain A$, $xinmathfrak{g}_A$, and $min M$. Here $x(a)$ denotes the action of $mathfrak{g}_A$ in $A$, while the three other actions are denoted by $ax$, $am$, and $xm$.

A $mathfrak{g}_A$-module is the same that a module over the enveloping algebra $U_A(mathfrak{g}_A)$.