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https://mathoverflow.net/questions/12452 | 7 | [John's Theorem](http://en.wikipedia.org/wiki/John_ellipsoid) can be stated as "To every compact, convex body, there is a unique inscribed ellipsoid, whose volume is maximal among all inscribed ellipsoids." It goes on to classify this maximal ellipsoid. By using this theorem, one can prove that the ellipsoid of maximal volume which is contained in a square is a circle.
This strikes me as a problem which was probably studied well before Fritz John, and yet I have been unable to prove the statement about squares and circles in an elegant, but low-brow manner. Any thoughts?
| https://mathoverflow.net/users/2043 | Maximal Ellipsoid | Here's an attempt at a low-brow proof. Take a max area ellipse. Apply an affine transform to
make it a circle; then the problem becomes to show that a minimal area parallelogram containing a circle is a square. It is easy to see that both the height of the parallelogram
and its base are at least the diameter. Q.E.D.
| 12 | https://mathoverflow.net/users/2653 | 12459 | 8,435 |
https://mathoverflow.net/questions/12461 | 7 | Let $g$ and $g'$ be two $C^2$-smooth Riemannian metrics defined on neighborhoods $U$ and $U'$ of $0$ in $\mathbb R^2$, respectively. Suppose furthermore that the scalar curvature at the origin is $K$ under both metrics.
**My question:** Is there a coordinate transformation taking one metric to the other, such that they agree up to second derivatives at the origin? i.e., if $x : U \to U'$ is the transformation, we have
$g\_{ij}' = g\_{ab} ~x\_i^a x\_j^b$,
evaluating everything at $0$; there are similar equations for the first and second derivatives. Clearly this is false if the scalar curvatures aren't equal. I don't care what happens away from the origin.
In the excellent thread [When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$?](https://mathoverflow.net/questions/8023/when-is-a-riemannian-metric-equivalent-to-the-flat-metric-on-mathbb-rn), Greg Kuperberg says:
> If remember correctly, there is a more general result due to somebody, that any two Riemannian manifolds are locally isometric if and only if their curvature tensors are locally the "same".
If "local isometry" means that the metrics are equal on a neighborhood of the origin, then the metrics I have in mind are not locally isometric, since the only information I have is that their curvatures match at one point.
**Edit:** I'm pretty sure that Deane answered my question, but let me clarify. Let $g\_{ij}$ be some "reasonable" metric, e.g. a [bump surface metric](https://mathoverflow.net/questions/12341/jacobi-fields-on-a-bump-surface), and consider a point $p$ where the scalar curvature is $K$. Let $g\_{ij}'$ be an arbitrary metric on a neighborhood $U$ of the origin in $\mathbb R^2,$ with scalar curvature $K$ at $0$.
Then the question becomes: does there exist a coordinate change on the bump surface such that the equation $g\_{ij}'(0) = g\_{ab}(p) ~x\_i^a x\_j^b$ is satisfied, as well as the corresponding equations for the first and second derivatives? That is, there are $18$ pieces of pertinent information
$(\\*)~~~g\_{11}', g\_{12}', g\_{22}'; g\_{11,1}', g\_{12,1}', g\_{22,1}', g\_{11,2}', g\_{12,2}', g\_{22,2}'; g\_{11,11}', g\_{12,11}', g\_{22,11}', g\_{11,12}', g\_{12,12}', g\_{22,12}', g\_{11,22}', g\_{12,22}', g\_{22,22}'$.
I want to change coordinates on my nice surface such that the metric and its derivatives line up with $(\\*)$.
| https://mathoverflow.net/users/238 | Changing coordinates so that one Riemannian metric matches another, up to second derivatives | The answer is yes. Just use geodesic normal (also known as exponential) co-ordinates. If you have a book or two on Riemannian geometry, just look for that or a discussion of the exponential map.
[ADDITIONAL COMMENT]
For a 2-dimensional metric, it's a nice exercise to figure all of this out using Jacobi fields. In fact, in my opinion, the *best* way to work with and understand the exponential map (which is a natural parameterization of all radial geodesics emanating from a point) is via Jacobi fields and the Jacobi equation. For example, it leads to an easy proof of a standard result, namely that the coefficients of the Taylor series of the exponential map at the origin contain only the Riemann curvature tensor and its covariant derivatives. The answer to your question follows from this theorem.
| 11 | https://mathoverflow.net/users/613 | 12466 | 8,439 |
https://mathoverflow.net/questions/12442 | 1 | **Background**
Ray tracing is very common in computational geometry and the problem is then to find the point of intersection between the equation of a **line** and the equation of a **plane** in **3D**.
The parametric form of the line is given by
$\mathbf{p}\_\mathrm{line}=\mathbf{p}\_\mathrm{a} + \xi (\mathbf{p}\_\mathrm{b}-\mathbf{p}\_\mathrm{a})$
and the plane can be defined by
$\mathbf{p}\_\mathrm{plane} \cdot \mathbf{n}+\mathrm{d}=0$,
where $\mathbf{p}\_\mathrm{plane}$ is a point on the plane and $\mathbf{n}$ is the normal vector to the plane.
Combining these two equations $(\mathbf{p}\_\mathrm{line}=\mathbf{p}\_\mathrm{plane})$ gives a convenient expression for the desired point from
$\xi=\frac{-\mathrm{d}-\mathbf{p}\_\mathrm{a} \cdot \mathbf{n}}{(\mathbf{p}\_\mathrm{b}-\mathbf{p}\_\mathrm{a}) \cdot \mathbf{n}}$.
---
**Question**
I now consider the problem of finding the intsersection(s) between an ellipse and a plane in 3D. Is there an effective way to perform this without an iterative scheme?
| https://mathoverflow.net/users/3414 | Unusual ray tracing | The sets you mention (plane, ellipse) can be expressed as the zero sets of certain polynomial functions; you are asking about the set on which both of these polynomials vanish simultaneously -- this is a basic question in algebraic geometry. One common computational solution is to apply Buchberger's algorithm for computing a Gröbner basis for the corresponding ideal (the basis will give an explicit description of the zero set). An excellent resource on this subject -- even if you have no background in algebraic geometry -- is the book "[Ideals, Varieties, and Algorithms](http://rads.stackoverflow.com/amzn/click/0387946802)" by David Cox, John Little, and Donal O'Shea. Additionally, many computer algebra systems (such as [Mathematica](http://www.wolfram.com/products/mathematica/index.html)) provide implementations of these kinds of algorithms.
| 2 | https://mathoverflow.net/users/1557 | 12470 | 8,441 |
https://mathoverflow.net/questions/12473 | 0 | I'm in the middle of trying to prove something at the moment and am looking for a decomposition of the Lie algebra $\mathfrak{su}(3)$ into a tensor product of some algebra $A$, and another $B$ containing $\mathfrak{su}(2)$, or some such result. Does anyone know of anything?
| https://mathoverflow.net/users/2612 | Does there exist an $A$ and $\mathfrak{su}(2) \subset B$ such that $\mathfrak{su}(3) \simeq A \otimes B$ | $\mathfrak{su}(3)$ cannot be decomposed as a tensor product, since there are no nonabelian simple Lie algebras of dimension 4, 2 or 1 (thus, any 8-d Lie algebra which is a tensor product has a proper ideal, which $\mathfrak{su}(3)$ doesn't).
| 8 | https://mathoverflow.net/users/66 | 12476 | 8,445 |
https://mathoverflow.net/questions/12451 | 5 | There's a fair amount of literature comparing different models for the homotopy theory of homotopy theories, or the homotopy theory of $(\infty,1)$-categories. Julie Bergner has a survey of this literature at <http://www.math.ucr.edu/~jbergner/OneInfty.pdf>.
Does anyone know if similar comparisons have been made for different models of stable $(\infty,1)$-categories? In particular, have $A\_{\infty}$-categories (in characteristic 0) been compared to the stable infinity categories that are conceived of as infinity categories with extra properties? (For example, to stable quasicategories?)
| https://mathoverflow.net/users/3413 | references for models of stable infinity categories | There seems to be a little confusion in your question about what stable ($\infty$,1)-categories are, so I want to address that first.
The definition of a stable infinity category that I'm familiar with is Jacob Lurie's notion (definition 29 [DAGI:Stable Infty-Categories](http://www.math.harvard.edu/~lurie/papers/DAG-I.pdf)). The prototypical examples of stable infinity categories which this definition is designed to capture are the category of spectra (in the topological setting) and the category of chain complexes in an abelian category (in the algebraic setting). Rather this notion is designed to also capture the derived localization of the category of chain complexes, which is a robust version of the derived category.
The advantage of Jacob's succinct definition is that it is expressed categorically. This means that if you have equivalent notions of ($\infty$, 1)-category, then they will yield equivalent notions of stable infinity category. So in this sense Julie Bergner's is also comparing stable infinity categories.
The only other model of stable ($\infty$,1)-categories that I know which doesn't quite fit this idea but is close is that of stable model categories. These are to stable ($\infty$,1)-categories as ordinary model categories are to ordinary ($\infty$,1)-categories. That is they are roughly equivalent to a particularly nice class of stable ($\infty$,1)-categories.
Your question also asks about $A\_\infty$-categories as stable ($\infty$,1)-categories, and that is the part which is confusing me.
On the one hand, an infinity category can heuristically be defined as a category with topological spaces of morphisms and where composition is associative only up to higher coherence. All the various notions of $\infty$-category make this precise in one way or another. Since you mention characteristic zero, I take you are thinking of the algebraic/chain complex version of $A\_\infty$-category. This can be related to topological $A\_\infty$-categories via the Dold-Kan correspondence (assuming your complexes are connective).
One the other hand being enriched in chain complexes is like being enriched in topological abelian groups (or in HZ-spectra in the non-connective case). So maybe the concept you are after is that of spectral category, or category enriched in spectra? I know this has been studied, but I'm lean on references.
One last comment. Later in DAGI, Lurie shows that stable infinity categories are automatically enriched in the stable infinity category of spectra. This is totally analogous to the category of chain complexes in an abelian category being enriched in the category of chain complexes of abelian groups. However, just as being enriched in abelian groups is not enough for your category to be abelian, being enriched in spectra is not enough for your infinity category to be stable.
| 6 | https://mathoverflow.net/users/184 | 12480 | 8,448 |
https://mathoverflow.net/questions/12483 | 4 | Suppose I have two $n$-sided polygons A and B. Is there a non-trivial upper bound on the number of parameters (eg. area, perimeter, etc) of the two polygons, that need to be the same, for A and B to be identical?
(For example, if we consider a simpler case to the problem when say A is constrained to lie inside B, then if area(A) = area(B), then A and B are identical. Hence area is the only non-trivial parameter in this case)
| https://mathoverflow.net/users/3327 | Uniqueness of a polygon | I guess you are looking for a nice answer, but here is a stupid one.
On the other hand I'm sure that there is no "nice answer".
A polygon can is uniquely determined by length of sides $\ell\_i$ and angles $\alpha\_i$.
Thus we have to find a complete set of invariants for sequence $(\alpha\_1,\ell\_1,\alpha\_2,\ell\_2,\dots\alpha\_n,\ell\_n)$ which survive after even cyclic shifts and reversing order.
Then you prepare [symmetric polynomials](https://en.wikipedia.org/wiki/Ring_of_symmetric_functions) for your group.
Say take all monomials of degree at most one in each $\alpha\_i$ and $\ell\_i$ and take its mean value it along the group.
You obtain a big collection of polynomial expressions in $\alpha\_i$ and $\ell\_i$ which gives complete invariant (perimeter will be one of them).
| 5 | https://mathoverflow.net/users/1441 | 12487 | 8,450 |
https://mathoverflow.net/questions/12463 | 10 | What can be said about extensions à la $\mathbb{Q}\_p(\sqrt[n]{a})/\mathbb{Q}\_p$? Ramification behaviour, valuation ring, ...?
I find it hard to say anything general - for example, as a function of the $p$-adic valuation of $n$ and/or $a$. Of course some special cases are rather easy to handle, and I understand what happens when $v\_p(a) = 0$. This might be a hard question, or a question for which there is a standard reference - I didn't find one - or something rather easy, in which case I'm just missing something.
| https://mathoverflow.net/users/1107 | adding an n-th root to Q_p | If $n$ is prime to $p$, then ${\mathbb Q}\\_p(a^{1/n})$ is unramified if $n | v\_p(a)$,
and is tamely ramified otherwise.
To see this, we note that we may first of all divide $a$ by powers of $p^n$, and so assume
that $0 \leq v\_p(a) < n.$
If in fact $v\_p(a)=0$, i.e. $a$ is a unit, then the extension is unramified, and the ring
of integers is equal to ${\mathbb Z}\\_p[a^{1/n}]$ (by Hensel's lemma, since $x^n - a$
is then a separable equation mod $p$).
Otherwise, if $0 < v\_p(a) < n,$ we get a tamely ramified extension (essentially by the
definition of tamely ramified).
If $p | n$ then the situation is a little more complicated. For example, if $n = p$
and $0 < v\_p(a) < p,$ then the extension is wildy ramified.
If $a$ is a unit, then we may write $a = \zeta u,$ where $\zeta$ is a $(p-1)$st root of 1
and $u \equiv 1 \bmod p,$ and since $\zeta^p = \zeta,$ we see that
${\mathbb Q}\\_p(a^{1/p}) = {\mathbb Q}\\_p(u^{1/p}).$
Now (supposing that $p$ is odd, for simplicity) if $u \equiv 1 \bmod p^2,$ then $u$ is
in fact a $p$th power in ${\mathbb Q}\\_p,$ and so the extension is trivial. On
the other hand, if $u \equiv 1 \bmod p,$ but not mod $p^2$, then the extension is
wildy ramified of degree $p$, with ring of integers equal to ${\mathbb Z}\\_p[u^{1/p}].$
To see this last claim, note that if $X^p - u = 0,$ and we write $Y = X - 1$,
then $(Y + 1)^p - u = 0,$ i.e. $Y^p + pY^{p-1} + \cdots + p Y + (u-1) = 0,$
and so $Y$ satisfies an Eisenstein polynomial of degree $p$. This implies that the extension is wildly ramified of degree $p$, that $Y$ is a uniformizer in the extension,
and that the ring of integers is equal to ${\mathbb Z}\\_p[Y] = {\mathbb Z}\\_p[u^{1/p}].$
Added in response to Keith Conrad's comments below: As Keith points out, the extension
${\mathbb Q}\_p(a^{1/n})$ is not really well-defined unless ${\mathbb Q}\_p$ contains the
$n$th roots of $1$, or equivalently, if $n$ divides $p-1$ (or 2 if $p = 2$).
But note e.g. if $p$ does not divide $n$, then adding the $n$th roots of unity gives
an unramified extension of ${\mathbb Q}\_p(a^{1/n})$, and so the ramification behaviour
is independent of the choice of $n$th root, while in the case when $n = p$ also treated above,
adjoining the $p$th roots of unity is a tamely ramified extension of ${\mathbb Q}\_p$,
so the claims regarding wild ramification are independent of the choice of $p$th root.
| 11 | https://mathoverflow.net/users/2874 | 12488 | 8,451 |
https://mathoverflow.net/questions/12472 | 7 | Let X be a DM stack over a field k. We follow the definition in Laumon and Moret-Bailly's book, so that its diagonal is quasi-compact (and hence diagonal is of finite type). Then is the diagonal necessarily finite?
**Edit:** I meant to ask if the inertia $I\to X$ is finite. Recall that the inertia stack $I$ is defined to be the 2-fiber product of $X$ with $X$ over $X\times X,$ where the two maps $X\to X\times X$ are both the diagonal map. This question is equivalent (I think) to the following. Let $G\to S$ be an etale $S$-group scheme of finite type, where $S$ is a $k$-scheme of finite type. Then $G$ is finite over $S.$
| https://mathoverflow.net/users/370 | Is the inertia stack of a Deligne-Mumford stack always finite? | I'm not sure about the suggested equivalence in the last two sentences of your question, but at least the statement about etale group schemes has a negative answer.
That is, it is possible to have an etale group scheme $G \rightarrow S$,
with $G$ and $S$ both finite type over a field $k$, but $G$ not finite over $S$.
For example, let $H$ be the constant group scheme ${\mathbb Z}/2{\mathbb Z}$ over $S$,
let $s$ be some fixed closed point of $S$, and let $G := H \setminus 1\_s,$
where $1\_s$ is the non-zero element of the fibre $({\mathbb Z}/2{\mathbb Z})\_s$.
Then $G$ is open in $H$, hence etale over $S$. Assuming that $S$ is positive dimensional,
it is certainly not finite (we deleted one
point of one fibre), and it is a subgroup scheme of $H$. (If $T$ is an $S$-scheme,
then $G(T)$ is the subgroup of $H(T)$ consisting of points whose values at points of $T$ lying
over $s$ are trivial.)
| 6 | https://mathoverflow.net/users/2874 | 12490 | 8,452 |
https://mathoverflow.net/questions/12462 | 8 | how does limsup and liminf for a sequence of sets, apply to probability theory. any real world examples would be much appreciated
| https://mathoverflow.net/users/3421 | limsup and liminf for a sequence of sets | For a sequence of subsets $A\_n$ of a set $X$, the $\limsup A\_n$ $= \cap\_{N=1}^\infty ( \cup\_{n\ge N} A\_n )$ and $\liminf A\_n$ $= \cup\_{N=1}^\infty (\cap\_{n \ge N} A\_n)$.
If $ x \in \limsup A\_n$ then $x$ is in all of the $\cup\_{n\ge N} A\_n$, which means no matter how large you pick $N$ you will find an $A\_n$ with $n>N$ of which $x$ is a member. Thus members of $\limsup A\_n$ are those elements of $X$ that are members of infinitely many of the $A\_n$'s. If $A\_n$ are thought of as events (in the sense of probability) $\limsup A\_n$ will be another event. It corresponds exactly to the occurance of infinitely many of the $A\_n$'s. This is why $\limsup A\_n$ is sometimes written $x \in A\_n$ infinitely often.
Similarly, if $x\in \liminf A\_n$ then $x$ is in one of $\cap\_{n\ge N} A\_n$, which means $x \in A\_n$ for all $n > N$. Thus, for $x$ to be in the $\liminf$, it must be in all of the $A\_n$, with finitely many exceptions. This is how the phrase "ultimately all of them" comes up.
Both of these operations, similar to their counterparts in metric spaces, concern the tail of the sequence $\{A\_n\}$. I.e., neither changes if an initial portion of the sequence is truncated. As a previous response pointed out, often the sets $A\_n$ are defined to track the deviation of a sequence of random variables from a candidate limit by setting $A\_n = \{x: |Y\_n(x) -Y(x)| \ge \epsilon\}$. The members of $\limsup A\_n$ then represents those sequences that every now and then deviate $\epsilon$ away from $Y(x)$, which is solely determined by the tail of the sequence $Y\_n$.
Here is a conceptual game that can be partially understood using these concepts: We have a deck of cards, on the face of each card an integer is printed; thus the cards are $\{1,2,3...\}.$ At the nth round of this game, the first $n^2$ cards are taken, they are shuffled. You pick one of them. If your pick is 1, you win that round. Let $A\_n$ denote the event that you win the nth round. The complement $A\_n^c$ of $A\_n$ will represent that you lose the $n^{th}$ round. The event $\limsup A\_n$ represents those scenarios in which you win infinitely many rounds. The complement of this event is $\liminf A\_n^c$, and this represents those scenarios in which you ultimately lose all of the rounds. By the Borel Cantelli Lemma $P(\limsup A\_n)$ $=0$ or equivalently $P(\liminf A\_n^c)=1$. Thus, a player of this game will deterministically experience that there comes a time, after which he never wins.
| 23 | https://mathoverflow.net/users/3370 | 12497 | 8,459 |
https://mathoverflow.net/questions/12500 | 8 | I think it should be a standard procedure to construct such things, can anyone give a reference or give a hint? Can this be done over any base scheme?
| https://mathoverflow.net/users/1238 | (nontrivial) isotrivial family of elliptic curves | Hint: use quadratic twists.
**Edit**: So as not to drag things out, I hope it's okay if I just give you a standard example. Let
$E\_0: y^2 = x^3 + Ax + B$
be your favorite elliptic curve over $\mathbb{Q}$ (i.e., any will do). Consider the
elliptic curve
$E: t y^2 = x^3 + Ax + B$
over the rational function field $\mathbb{Q}(t) = \mathbb{Q}(\mathbb{P}^1)$. Spreading this out as a scheme over $\mathbb{P}^1\_{/\mathbb{Q}}$, we see that there are two singular fibers, at $t = 0$ and $t = \infty$. Discarding these we get an elliptic curve over $\mathbb{A}^1 \setminus \{0\}$ which is isotrivial -- the $j$-invariant over every fiber is $j(E\_0)$ -- but nontrivial: the isomorphism classes of the fibers are in bijection with $H^1(\mathbb{Q},\mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Q}^{\times}/\mathbb{Q}^{\times 2}$.
It's a good bet that you'll find this example somewhere in the chapter on elliptic surfaces in Silverman's *Advanced Topics in the Arithmetic of Elliptic Curves*.
Can it be done over any base scheme? Unless I misunderstand, of course not, e.g. not over the spectrum of a field.
| 3 | https://mathoverflow.net/users/1149 | 12501 | 8,460 |
https://mathoverflow.net/questions/12503 | 5 | Dear group theorists,
Let $n \geq 1$ and $I$ be an infinite set (you may assume $I$ to be countable). Is the abelian group $(\mathbb{Z}/n)^{(I)}$ (direct sum of copies $\mathbb{Z}/n$) a direct summand of $(\mathbb{Z}/n)^I$ (direct product of copies $\mathbb{Z}/n$)? This question is motivated by [that one](https://mathoverflow.net/questions/11767/infinite-tensor-products).
If $n$ is prime, this follows from Linear Algebra. Of course, this is not constructive at all. Thus it's also true when $n$ is squarefree (use the Chinese Remainder Theorem). What happens otherwise? The smallest example is $n=4$. I don't know how to start ...
| https://mathoverflow.net/users/2841 | (Z/n)^(I) is a direct summand of (Z/n)^I | Yes. The ring ${\mathbb Z}/n {\mathbb Z}$ is injective over itself, and over a Noetherian
ring, direct limits of injectives are again injective; thus ${\mathbb Z}/n{\mathbb Z}^{(I)}$
is injective over ${\mathbb Z}/n{\mathbb Z}$. Finally, any embedding of an injective splits,
as follows directly from the property of being injective. We can now apply this to
the embedding $({\mathbb Z}/n{\mathbb Z})^{(I)} \hookrightarrow ({\mathbb Z}/n{\mathbb Z})^I.$
| 12 | https://mathoverflow.net/users/2874 | 12504 | 8,461 |
https://mathoverflow.net/questions/12469 | 16 | I was inspired by the following algebraic topology orals question:
"Is $S^1$ the loop space of another space?"
This is easy to see if you recognize that $S^1$ is a $K(\mathbb{Z},1)$, and the loop space of any $K(G,n)$ is a $K(G,n-1)$.
I then also remembered that the loop space functor is a functor from pointed topological spaces and continuous maps to the category of H-spaces and continuous homomorphisms. H-spaces being topological spaces that satisfy the axioms of a group up to homotopy (see Spanier, Chapter 1, Section 5).
I have three questions:
1. Is there a useful criterion for when an H-space is actually a topological group?
2. Seeing that $S^1$,$S^3$, and $S^7$ are the only spheres that support group structures, it doesn't seem coincidental that $S^1$ is a loop space, because it is in fact an H-space. Since $CP^{\infty}$ is the loop space of $K(Z,3)$ it too is an H-space, but is it known if it is a topological group?
3. Even if not, is there a way (other than concatenation of loops) to "see" this structure on $CP^{\infty}$?
Thanks!
| https://mathoverflow.net/users/1622 | Group Structure on CP^infinty | Here's a few thoughts on your questions.
1. See algori's answer. (Incidentally, the "necessity is clear" step is because if $G$ is a topological group then it has a classifying space and then $G \simeq \Omega B G$, hence is homotopy equivalent to a loop space.)
2. For $CP^\infty$, here's a construction that makes it a topological group. Take the unitary group on a Hilbert space, $U(H)$. This is contractible by Kuiper's theorem ([MR0179792](http://www.ams.org/mathscinet-getitem?mr=0179792)). The centre of this group is the circle, $S^1$, acting by diagonal operators. As this is normal, the quotient $PU(H) = U(H)/S^1$ is a topological group. Since $U(H)$ is contractible, this is a $K(\mathbb{Z},2)$ and hence "is" $CP^\infty$.
3. The group structure on $CP^\infty$ can be viewed in a nice way using the fact that $CP^\infty$ represents $H^2(-,\mathbb{Z})$ and that $H^2(-,\mathbb{Z})$ classifies complex line bundles. Both of these views of $[-,CP^\infty]$ have obvious groups structures: for $H^2(-,\mathbb{Z})$ it is addition whilst for complex line bundles it is given by tensor product (the inverse operation is complex conjugation). (These two group operations are the same operation under the correspondence, by the way). As these are natural operations, they are represented by a group structure on the representing space, making $CP^\infty$ a group object in the $hTop$. That this is the "correct" group structure (aka, that coming from $CP^\infty \simeq K(\mathbb{Z},2) \simeq \Omega K(\mathbb{Z},3)$) is clearest from the characterisation of $CP^\infty$ as the representing space for $H^2(-,\mathbb{Z})$. The equivalence $CP^\infty \simeq \Omega K(\mathbb{Z},3)$ comes from the suspension isomorphism, $H^2(-,\mathbb{Z}) \cong H^3(\Sigma -, \mathbb{Z})$ which is additive and hence preserves the group structures on the representing spaces.
| 12 | https://mathoverflow.net/users/45 | 12517 | 8,471 |
https://mathoverflow.net/questions/12423 | 0 | Is there a univariate probability distribution $p\_{\lambda,\alpha}(\beta)$ over the reals, parameterized by $\lambda > 0$ and $1 >= \alpha >= 0$, such that $p\_{\lambda,\alpha} \propto \exp(-\lambda(\alpha \beta^{2} + (1 - \alpha)|\beta|)$? If so, is it a proper distribution (integrates to 1 over the real line)? Does the density function have a closed form expression? If not, does the density function have some other nice representation?
Background: One approach to avoiding overfitting in regression modeling with many predictors is to minimize the sum of prediction error plus a penalty based on the size of the coefficients. This is particularly used in machine learning problems with large numbers of features, as in computational linguistics and information retrieval. If an L2 penalty (i.e. a penalty proportional to the square of the coefficient) is used this is called ridge regression. It is equivalent to finding the Bayesian maximum a posteriori (MAP) fit with a prior that is the product of univariate gaussian distributions with mean 0. If an L1 penalty (proportion to absolute value of coefficient) is used this is called lasso regression, and is equivalent to finding the Bayesian MAP fit under a prior which is a product of univariate Laplace (double exponential) distributions with mean 0. This paper:
>
> Zou, Hui and Trevor Hastie. 2005. Regularization and Variable Selection via the Elastic Net. Journal of the Royal Statistical Society B. 67(Part 2):301–320.
>
>
>
introduced the "elastic net" which is a weighted combination of L1 and L2 penalties. My question is whether the elastic net corresponds to MAP estimation under a prior which is a product of univariate distributions, and if so what the nature of that distribution is.
| https://mathoverflow.net/users/3407 | univariate prior corresponding to weighted sum of L1 and L2 penalties? | I don't know if that particular density has a name, but it is a valid exponential family density. The normalisation constant does not have a closed form expression, but with a bit of work you can write as a function of the cumulative Gaussian density, by noting that, since the density is symmetrical around 0:
$\int{\exp(-\lambda(\alpha x^2 + (1-\alpha) |x|))}dx$=$2\int\_{-\infty}^0{\exp(Ax^2+Bx)}$
After some algebra you find, for $\alpha > 0$:
$\int{\exp(-\lambda(\alpha x^2 + (1-\alpha) |x|))}dx$=$2\Phi(0;(1-\alpha)/2\alpha,1/(2\alpha))\exp\(\lambda/(4\alpha(1-\alpha)^2)\)\sqrt{\pi/(\alpha\lambda)}$
That's finite for $\alpha,\lambda>0$ (of course for $\alpha=0$ you just recover the Laplace density so it works too).
As far as I can tell the elastic net is a valid MAP estimate.
| 1 | https://mathoverflow.net/users/3358 | 12523 | 8,476 |
https://mathoverflow.net/questions/12178 | 8 | Given 100 boxes. Each contains arbitrary number of red, blue and green balls, i.e., 100 non-negative integer triples $(r\_i,b\_i,g\_i)$.
Prove it's always possible to find 51 boxes so that the total number of balls of each color in these boxes is no less than the ones from the rest 49 boxes.
For n boxes, replace 51 with $\lfloor(n+3)/2\rfloor$, and prove that is the best lower bound.
This is a generalization of a high-school Olympiad question, which I was told to use pigeonhole principle. Could anyone shed light on how to apply it?
---
EDITED: Since this is not really related to the pigeon-hole principle, I have edited the title and the tag.
Besides the solution provided by domotorp, darij grinberg pointed to an elementary proof on [mathlinks](http://mathlinks.ro/Forum/viewtopic.php?t=32161/).
Also domotorp has found the [origin](http://turgor.ru/lktg/2005/4/index.htm) of the problem (which was sort of buried in the comments).
| https://mathoverflow.net/users/3350 | Balls in boxes (partition) | This proof uses a combinatorial equivalent of the Borsuk-Ulam theorem. I think that the proof is a little more complicated than the average proofs here, so please check my related [paper](http://www.cs.elte.hu/~dom/cikkek/necklace.pdf) if you have difficulties to understand.
Octahedral Tucker's lemma. If for any set-pair $A, B\subset [n], A\cap B=\emptyset, A\cup B\ne\emptyset$ we have a $\lambda(A,B)\in\pm[n-1]$ color, such that $\lambda(A,B)=-\lambda(B,A)$, then there are two set-pairs, $(A\_{1},B\_{1})$ and $(A\_{2},B\_{2})$, such that $(A\_{1},B\_{1})\subset (A\_{2},B\_{2})$ and $\lambda(A\_{1},B\_{1})=-\lambda(A\_{2},B\_{2})$.
We will use this lemma for n=100. If for the boxes in A, the sum of the red balls is more than half of the total number of red balls, then we set $\lambda(A,B)=+red$. If it is more than half in B, then we set $\lambda(A,B)=-red$. We do similarly for blue and green (if they are not set yet to red). We also set $\lambda(A,B)=\pm (|A|+|B|)$ if $|A|+|B|\le 96$ (if they are not set to anything else yet). This way the cardinality of the range of lambda is 99, just as in the lemma. It is easy to verify that it satisfies the conditions of the lemma, thus there must be a set-pair for which we did not set any value. But in that case either A or B must be bigger than 96/2=48, thus at least 49. We can put the remaining boxes to the other part and we are done.
Note that this proof easily generalizes to more colors.
| 3 | https://mathoverflow.net/users/955 | 12524 | 8,477 |
https://mathoverflow.net/questions/12502 | 1 | With my personal interest and hobby I started this ..
Given a sequence of numbers 1,2,3 .... N
where N is the highest among the sequence and length of the sequence as well ..
I tried my best to bring up a relationship where y=f(n) .. so that .. y (equal or not-equal to n) is an unique value for each value of n bounded within N,
for example:
for the sequence:
1, 2, 3, 4, 5
Corresponding sequence would be ..
3, 2, 5, 1, 4 (or Assume some other random sequence of same numbers) ..
where 3=f(1), 2=f(2), etc ..
The function f(n) must effectively work for all values of N .. ie we must be able to generate the random sequence of any value of N.
Initially I tried with `y = (n * 9) % N, y = (n * 7) % N and y = (n * 3) % N`
But they work only if the number is divisible by 10 and the max number if not divisible by 3, 7 and 9 ..
Is it possible to generalize the formula .. ? Please help me in deriving the same ..
| https://mathoverflow.net/users/2972 | Seeking for a formula or an expression to generate non-repeatative random number .. | When we hear random permutations, we bring in our intuition about permutations, and try to give a method which could generate a complicated permutation. Thus, I think we didn't pay enough attention to your examples like n\*3 mod N, which for most situations would not be an acceptable way of generating random numbers. The only problem is what to do if N is divisible by 3. As far as I can tell, divisibility by 10 is irrelevant, so I'm not sure why you mentioned it.
You say you don't want to write a program, just a simple formula in Excel. This is reasonable, and even something which makes sense mathematically: There are a few operations available in Excel formulas such as addition, exponentiation, factorial, conditional evaluation based on whether a statement is true or false (characteristic functions), etc. Can one create a formula with fixed complexity which takes in n and N, and which is a permutation of {1,...,N] for a fixed N? Trivially returning n works, but can one produce a permutation other than (+-n+k mod N)+1?
I suggest creating a formula which is equivalent to the following:
If N is not divisible by 71, return (71\*n mod N) + 1.
Otherwise N is divisible by 71. Permute the last digit base 71: return a + (3\*b mod 71) +1
where n-1 = a + b and a is divisible by 71 and $0 \le b \lt 93$, i.e.,
b = n-1 mod 71.
a = n-1 - (n-1 mod 71).
IF(MOD(N,71)!=0,n-(MOD(n-1,71)) + MOD(3\*(MOD(n-1,71),71),MOD(71\*n,N)+1).
(Debugging left to the reader.)
This would be lousy as a random permutation, but it may be acceptable for some purposes.
A better random permutation might be based on f(n), where f reverses the lowest binary digits of n if n is at most than the greatest power of 2 less than N, and does nothing if n is greater. Try f(N+1-f(n)). This can be done using the DEC2BIN and StrReverse functions, but you need a little Excel expertise to use those.
Once you have a few ways to generate random permutations, you can compose them, and even using unsatisfactory random permutations like adding floor(sqrt(N)) can improve the appearance of the resulting permutation.
| 2 | https://mathoverflow.net/users/2954 | 12543 | 8,489 |
https://mathoverflow.net/questions/12549 | 10 | Let $\pi:P \rightarrow M$ be a principal $G$-bundle, and let $A \in \mathfrak{g}$, where the Lie algebra of $G$ is indicated. The *fundamental field* $A$# used to define connections is given by
$A$#$(p) := \frac{d}{dt}(\exp(At)p)|\_{t=0}$.
$A$# is well defined since $e^{At}p$ can be regarded as a vector in $\pi^{-1}(\pi(p))$. Intuitively, I try to think of $A$# as the (vertical) direction of the displacement on the fiber generated by $A$.
By the defining properties of principal bundles (in particular, the free action of $G$), we have $\{A$# $: A \in \mathfrak{g}\} \simeq \mathfrak{g} \simeq \mathcal{V}(p)$, where $\mathcal{V}(p)$ is the vertical subspace at $p$. Something like
$A$#$(p) = \lim\_t t^{-1}(e^{At}p - p) = A \cdot p$
would be (to me) a nice way of thinking about $A$#$(p)$, except that this is (at best) a formal equality. More precisely, $L\_{\exp(At)}$ is the 1-parameter group of diffeomorphisms generated by $A$#, where $L\_g$ denotes left multiplication by $g$.
**My question**: is there a better way of thinking about a fundamental field than (either the definition itself) or the notional equation above?
| https://mathoverflow.net/users/1847 | What is a good way to think about a fundamental field on a principal G-bundle? | If you have a homogeneous space $X$ with structure group $G$ (in your case, the fiber passing through $p$) then left multiplication give you a nice action $L: X \times G \to X$. Then for a fixed $p \in X$, $L(p) : G \to X$. The differential of this guy is a map $L(p)\_\* : TG \to TX$ which takes an element of $T\_gG$ to an element of $T\_{g \cdot p} X$. At $g = 1$, this means you get a map $L(p)\_\* : \mathfrak{g} \to T\_pX$. This is actually a representation of $\mathfrak{g}$ on the space of vector fields on $X$, and will inherit nice properties depending on how nice the action of $G$ on $X$ is.
This map $L(p)\_\*$ is precisely the sharp map you describe: it takes an element $\xi$ of $\mathfrak{g}$ and gives a vector field on $X$ which coincides with infinitesimal multiplication by $\xi$. In your case, $G$ acts freely so $X$ looks like a copy of $G$.
If $G$ actually has a nice matrix representation then unrolling the definitions, you'll find that you really can write your proposed equation $A^\sharp(p) = A \cdot p$. For example, look at how the infinitesimal generators of $\mathfrak{so}(3)$ acts on points of the unit sphere by multiplication --- you'll easily see the vector fields whose flows are $\exp(\xi t)$.
| 10 | https://mathoverflow.net/users/2510 | 12552 | 8,493 |
https://mathoverflow.net/questions/12538 | 4 | EDIT: Proved it on my own. It easily follows from the Witt integrality theorem. Sorry for posting.
Let $P\in\mathbb{Z}\left[\Xi\right]$ be a polynomial (where $\Xi$ is a family of symbols that we use as indeterminates, for instance $\Xi=\left(X\_1,X\_2,X\_3,...\right)$). Let $n\in\mathbb{N}$.
Prove or disprove that $\displaystyle\frac{\delta}{\delta\xi}P\in n\mathbb{Z}\left[\Xi\right]$ for every $\xi\in\Xi$ if and only if there exist polynomials $P\_d\in\mathbb{Z}\left[\Xi\right]$ for all divisors $d$ of $n$ such that $\displaystyle P=\sum\_{d\mid n}dP\_d^{n/d}$.
A few remarks on this: The $\Longleftarrow$ direction is trivial. I can prove the $\Longrightarrow$ if $n$ is a prime power.
PS. No, this does not help in proving the Witt integrality theorem, even if it is true.
| https://mathoverflow.net/users/2530 | Do n-th Witt polynomials generate {P | P' is divisible by n} ? | Solved. The key is that the set of polynomials $P$ for which there exist polynomials $P\_d\in\mathbb{Z}\left[\Xi\right]$ for all divisors $d$ of $n$ such that $\displaystyle P=\sum\_{d\mid n}dP\_d^{n/d}$ is a subring of $\mathbb{Z}\left[\Xi\right]$ (by the Witt integrality theorem, which is 9.73 in [Hazewinkel's Witt vectors](https://arxiv.org/abs/0804.3888v1)).
EDIT: Wrote up [a proof](http://www.cip.ifi.lmu.de/~grinberg/algebra/witt5a.pdf).
| 5 | https://mathoverflow.net/users/2530 | 12555 | 8,495 |
https://mathoverflow.net/questions/12489 | -4 | Is the given expression, monotonically increasing or decreasing with increasing x?
$\frac{1}{x \log(x)} \sum\_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$
EDIT: This is the derivative of the prime counting function $\pi(x)$ w.r.t. x, ie.,
$\frac{d\pi(x)}{dx} \sim \frac{1}{x \log(x)} \sum\_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$
Also, you might check Bruce C. Berndt's "Ramanujan's NoteBooks Part IV" page 123, equation 10.16, which gives = sign instead of $\sim$
| https://mathoverflow.net/users/2865 | Is the given expression, monotonically increasing or decreasing with increasing x? | I'm not sure I should bother answering this question, because it seems like the original poster may not have asked the right question. However, it is a nice exercise in basic asymptotics.
---
For $x$ sufficiently large, the sum in question is decreasing.
First, note that this sum is equal to
$$\sum\_{k \geq 1} \frac{(\log x)^{k-1}}{x k! \zeta(k+1)}.$$
(See [here](http://mathworld.wolfram.com/GramSeries.html) for a very similar series; we are using the highly nontrivial identity $\sum \mu(i)/i=0$ to get rid of the "$k=0$" term.)
Substituting $x=e^u$, we want to know whether or not
$$e^{-u} \sum\_{k \geq 1} \frac{u^{k-1}}{k! \zeta(k+1)}$$
is increasing or decreasing in $u$. One can justify taking term by term derivatives, so we want to know whether
$$e^{-u} \left( \sum\_{k \geq 2} \frac{(k-1) u^{k-2}}{k! \zeta(k+1)} - \sum\_{k \geq 1} \frac{u^{k-1}}{k! \zeta(k+1)} \right)$$
is positive or negative.
Rearranging terms, we are interested in the sign of
$$e^{-u} \sum\_{\ell \geq 0} \frac{u^{\ell}}{\ell!} \left( \frac{ 1}{(\ell+2) \zeta(\ell+3)} - \frac{1}{(\ell+1) \zeta(\ell+2)} \right).$$
The quantity in parenthesis is $-1/(\ell+1)(\ell+2) + O(2^{- \ell})$. So we are interested in the sign of
$$e^{-u} \left( - \sum\_{\ell \geq 0} \frac{u^{\ell}}{(\ell+2)!} + \sum O(\frac{ u^{\ell} 2^{- \ell}}{\ell !}) \right) =$$
$$e^{-u} \left( - \frac{e^u -1-u}{u^2} + O(e^{u/2}) \right)=$$
$$-1/u^2 + O(e^{-u/2}).$$
This is negative for $u$ sufficiently large.
| 8 | https://mathoverflow.net/users/297 | 12557 | 8,497 |
https://mathoverflow.net/questions/12531 | 4 | Let $X$ be a Riemannian manifold and let $G$ be a (at most countable, if that matters) discrete group acting properly and by isometries on $X$. Let $\mathcal{O}$ be the sheaf of analytic functions on $X$. By analogy with what happens for finite groups acting on vector spaces, one is tempted to study a sheaf written $\mathcal{O} \rtimes G$. What is a good reference for an algebraist to learn about the various convergence conditions one might impose to define this sheaf, and their relationship with the quotient X/G?
[I'm fine with an answer that works in a different category---e.g. complex analytic spaces, but I want there to be some convergence conditions imposed at some point. The ideal reference is a short survey paper with precise definitions.]
| https://mathoverflow.net/users/nan | Non-commutative versions of X/G | Noncommutative versions of sheaves and holomorphic functions are not very well understood. Better understood are noncommutative versions of measurable, continuous, or smooth functions. I generally work with the continuous functions, i.e. $C^\* $-algebras, or various subalgebras that deserve to be called smooth. I'll describe things in the $C^\*$-framework.
What came to mind immediately for me is the notion of strong Morita equivalence, due to Rieffel. It works like this: suppose you have a locally compact group $G$ acting on a $C^\* $- algebra $A$ (think of $A$ as $C(X)$ here). You can form what is called the crossed product algebra, which is a $C^\*$-algebra containing $A$ and $G$, and where the action of $G$ on $A$ is implemented via conjugation by $G$; i.e. if $a \in A$ and $g \in G$, then $g a g^\* = \alpha\_g(a)$, where $\alpha$ is the action.
This can be done when $A$ is unital or not, and $G$ can be discrete or not. The resulting algebra, which I would denote $A \times\_\alpha G$, is unital if and only if $A$ is unital and $G$ is discrete.
Now suppose that $X$ is a compact Hausdorff space with an action of $G$. Then $G$ also acts on $A = C(X)$, and so we can make the crossed product algebra $C(X) \times\_\alpha G$. Here's the punchline: when the action of $G$ on $X$ is free and proper, so that the quotient $X/G$ is well-behaved, then the crossed product algebra is strongly Morita equivalent to the algebra $C(X/G)$ of functions on the quotient.
When the action is not free and proper, the quotient may be very bad (e.g. the integers acting on the circle by rotation by an irrational angle) and so the algebra $C(X/G)$ may be reduced to nothing more than scalars, and so be useless for obtaining any information about the quotient. In this case, one uses the crossed-product algebra as a sort of substitute for the algebra of functions on the quotient.
A reference for this is the paper "Applications of Strong Morita Equivalence to Transformation Group $C^\*$-algebras, by Rieffel, which is available on his [website](http://math.berkeley.edu/~rieffel/). Unfortunately it doesn't have the definitions of crossed products (which he calls transformation group algebras), but the [wikipedia page](http://en.wikipedia.org/wiki/Crossed_product) is ok, although phrased just for von Neumann algebras.
| 5 | https://mathoverflow.net/users/703 | 12560 | 8,499 |
https://mathoverflow.net/questions/12566 | 4 | I'm not sure if my question make sense, but it would also be interesting to know if it didn't, so I will ask anyway:
There exist a countable model of ZF. This means (if I understand it correctly) that we can find a model of ZF in ZF, such that the set of elements in the model is countable seen from the outer ZF.
**My question is:** Can every model of ZF be "seen from the outside" in a way that makes it countable?
It seems to me, that if we have a model A of ZF, the model will be a element in a countable model B of ZF. Now, if you look at B from "the outside" A is countable.
| https://mathoverflow.net/users/2097 | Is every model of ZF countable "seen from the outside"? | Technically, a model of ZF consists of a set with some relation, representing "being an element of". So the literal answer is no. An uncountable model is simply uncountable. Your argument that one can see every model as a model in a model which is countable doesn't work. Cardinality in a model depends on whether there exists certain bijections in the model. This has nothing to do with "outside cardinality".
However, the first-order statements true in some model of ZF are true in some countable model by the [Löwenheim-Skolem theorem](http://en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_theorem). Since we are usually not interested in differences between models that cannot be formulated in the first order theory of sets, we can say that every model of ZF has an, for practical purposes, equivalent countable model.
| 8 | https://mathoverflow.net/users/35357 | 12568 | 8,503 |
https://mathoverflow.net/questions/12079 | 6 | We can extend the binomial coefficient $\binom{n}{k}$ to $\mathbb{R}$ or $\mathbb{C}$ by defining $\binom{x}{y}=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}$. Do any the standard binomial coefficient identities have generalizations to this setting? Just as two simple examples, we have
$\sum\_{k=0}^n \binom{n}{k} = 2^n$ and $\sum\_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$
What are $\int\_0^x \binom{x}{y} dy$ and $\int\_0^x \binom{x}{y}^2 dy$, and are the answers analogous to the discrete case? Is there any combinatorial significance we can give to these integrals? Has this already been tried?
| https://mathoverflow.net/users/1916 | Gamma function versions of combinatorial identites? | Chapter 5.5 of *Concrete Mathematics* discusses generalizing binomial coefficient identities to the Gamma function. It doesn't discuss the two integrals you mention, though.
Doing a bit of thinking on my own, if $n$ is a positive integer then
$$\int\_{z=0}^n \binom{n}{z} dz = \int\_{z=0}^n \frac{n! dz}{\Gamma(1+z) \Gamma(n+1-z)}$$
$$\int\_{z=0}^{n} \frac{n! dz}{(n-z)(n-1-z) \cdots (1-z) \Gamma(1-z) \Gamma(1+z)}.$$
We have $\Gamma(1+z) \Gamma(1-z) = \pi z/\sin (\pi z)$, if I haven't made any dumb errors, so this is
$$\int\_{0}^n \frac{ n! \sin (\pi z) \ dz}{\pi z (n-z)(n-1-z) \cdots (1-z)}.$$
I suspect this integrand does not have an elementary anti-derivative, because it reminds me of $\int \sin t \ dt/t$. But there might be some special trick which would let you compute the integral between these specific bounds.
| 6 | https://mathoverflow.net/users/297 | 12573 | 8,506 |
https://mathoverflow.net/questions/12579 | 2 | I understand that every finite set is recursively enumerable, as I see that one could just encode each element of some finite set on a Turing Machines tape, and then have the machine check each member against any input to determine set membership....
...However, it isn't clear to me how there is an analog to this method in the domain of Diophantine representation, even though Matiyasevich's theorem assures us that one does exist. (That is, a set is recursively enumerable iff it is Diophantine. Every finite set is recursively enumerable, thus also Diophantine.)
In your answer, please give the explicit method by which any finite set can be written in a Diophantine representation.
| https://mathoverflow.net/users/3453 | Why is every finite set Diophantine? | Or, very simply stated, given the finite set $S = \{a\_1, \dots , a\_k\}$, consider the diophantine equation: $$(n-a\_1)\dots(n-a\_k)=0.$$
EDIT: Then we can write S as $\{ \ n \ | \ \exists x : (n-a\_1)\dots(n-a\_k)=0\ \}$. (Thanks David)
| 9 | https://mathoverflow.net/users/2693 | 12582 | 8,512 |
https://mathoverflow.net/questions/12527 | 1 | Let's assume we have a [simple linear regression](http://en.wikipedia.org/wiki/Simple_linear_regression) y(x) = a + bx. Is there a way to obtain a probability density function of y for any given x? Would the concept of confidence regions be useful here in any way?
I'm not a big expert in statistics and your help will be very much appreciated
Thank you
| https://mathoverflow.net/users/3432 | Distribution function of dependent variable, confidence regions | You need to make your answer more precise. First, the regression should be
y = a + bx + e
where e has some distribution, let's say described by a density g(e)
Now, do you mean:
1) Obtaining the density f(y|x) theoretically?
This one is easy. By the transformation of densities
f(y|x) = 1/(a+bx)\*g(y/(a+bx))
2) Estimating the density when you do not know the parameters of the regression model?
This one will depend on what you know and what you don't know, and what data you have at hand, and is a more complex issue, so you need to provide more information.
| 1 | https://mathoverflow.net/users/3458 | 12595 | 8,520 |
https://mathoverflow.net/questions/12530 | 16 | Let $f = a\_0 + a\_1 x + \ldots + a\_n x^n$ ($f \ne 0$), where $a\_i \in \{-1, 0, 1\}$. Let $p(f)$ be the largest number such that $f(x)$ is divisible by $y$ for any integer $x$ and for any $1 \leq y \leq p(f)$. Let $g(n)=max\_f\; p(f)$. Is it true that $g(n) = o(n)$? What is the best upper or lower bound on $g(n)$ can be derived?
For my application it would be great to prove that $g(n) = o(n)$ in order to obtain something non-trivial, or $g(n) = o(n^{2/5})$ in order to improve the best known result. Do you think it is real?
UPD It is an obvious consequence of Bertrand's postulate and Schwartz–Zippel lemma that $g(n) \leq 2n$. Using bruteforce I've got the following values:
$g(10) = 7$, $f = x^{10} + x^8 - x^4 - x^2$.
$g(15) = 10$, $f = x^{15} + x^{13} + x^{12} + x^{11} + x^{10} - x^7 - x^6 - x^5 - x^4 - x^3$.
$g(17) = 10$, $f = x^{16} + x^{15} + x^{14} + x^{13} + x^{12} + x^{11} - x^8 - x^7 - x^6 - x^5 - x^4 - x^3$.
| https://mathoverflow.net/users/3448 | Question about polynomials with coefficients in Z | I'll prove the upper bound $g(n) = O(n^{1/2+o(1)})$, which is essentially best possible if Kevin Costello's heuristics are correct.
Suppose that $q$ is a prime with $q > n^{1/2}+1$. Reducing $f(x)$ modulo $x^{q-1}-1$ in $\mathbb{Z}[x]$ amounts to reducing the *exponents* modulo $q-1$, so the result is a polynomial $h(x)$ of degree less than $q-1$ whose coefficients are at most $n/(q-1) + 1$ (which is less than $q$) in absolute value. On the other hand, if $q \le p(f)$, then $f(x) \bmod q$ is divisible by $x^q-x$ and hence also by $x^{q-1}-1$, so $h(x) \bmod q$ must be $0$; this is possible only if $h(x)=0$ in $\mathbb{Z}[x]$, which implies that $f(x)$ vanishes at the $\phi(q-1)$ primitive $(q-1)$-th roots of unity. Since $f(x)$ has at most $n$ complex zeros, we get $\sum\_{n^{1/2}+1 < q \le p(f)} \phi(q-1) \le n$ (the sum ranges over primes $q$ in the interval).
By the prime number theorem and Theorem 327 in Hardy and Wright (which states that $\phi(m)/m^{1-\delta} \to \infty$ for any $\delta>0$), this is a contradiction for sufficiently large $n$ if $p(f)>n^{1/2+\epsilon}$ for a fixed $\epsilon>0$.
**EDIT:** The more precise bound $\phi(m) \ge (e^{-\gamma} - o(1)) m/\log \log m$ given by Theorem 328 in Hardy and Wright leads to $g(n) \le (e^{\gamma}/2 + o(1)) n^{1/2} \log n \log \log n$.
| 20 | https://mathoverflow.net/users/2757 | 12596 | 8,521 |
https://mathoverflow.net/questions/12584 | 22 | In ordinary membership-based set theory, the **axiom schema of replacement** states that if $\phi$ is a first-order formula, and $A$ is a set such that for any $x\in A$ there exists a unique $y$ such that $\phi(x,y)$, then there exists a set $B$ such that $y\in B$ if and only if $\phi(x,y)$ for some $x\in A$. That is, $B$ is the "image" of $A$ under the "definable class function" $\phi$.
The related **axiom schema of collection** modifies this by not requiring $y$ to be unique, but only requiring $B$ to contain *some* $y$ for each $x$ rather than all of them. However, there are at least two different versions of this.
1. If for all $x\in A$ there exists a $y$ with $\phi(x,y)$, then there exists a set $B$ such that for all $x\in A$ there is a $y\in B$ with $\phi(x,y)$ (this is [Wikipedia's](http://en.wikipedia.org/wiki/Axiom_of_collection#Axiom_schema_of_collection) version; I'll call it "weak collection").
2. If for all $x\in A$ there exists a $y$ with $\phi(x,y)$, then there exists a set $B$ such that (1) for all $x\in A$ there is a $y\in B$ with $\phi(x,y)$, and (2) for all $y\in B$ there is an $x\in A$ with $\phi(x,y)$ (I'll call this "strong collection").
The third possibly relevant axiom is the **axiom schema of separation**, which states that for any $\phi$ and any set $A$ there exists a set $B\subseteq A$ such that $x\in B$ if and only if $x\in A$ and $\phi(x)$.
I know the following implications between these axioms:
* Strong collection implies weak collection, since it has the same hypotheses and a stronger conclusion.
* Strong collection implies replacement, since it has a weaker hypothesis and the same conclusion.
* Replacement implies separation (assuming excluded middle): apply replacement to the formula "($\phi(x)$ and $y=\lbrace x\rbrace$) or ($\neg\phi(x)$ and $y=\emptyset$)" and take the union of the resulting set.
* Together with AC and foundation, replacement implies weak collection: let $\psi(x,V)$ assert that $V=V\_\alpha$ is the smallest level of the von Neumann hierarchy such that there exists a $y\in V\_\alpha$ with $\phi(x,y)$, apply replacement to $\psi$ and take the union of all the resulting $V\_\alpha$.
* Weak collection and separation together imply strong collection: separation cuts out the subset of $B$ consisting of those $y$ such that $\phi(x,y)$ for some $x\in A$.
My question is: does weak collection imply replacement (and hence also separation and strong collection) without assuming separation to hold *a priori*? Feel free to assume all the *other* axioms of ZFC (including $\Delta\_0$-separation). I'm fairly sure the answer is "no," but several sources I've read seem to assume that it does. Can someone give a definitive answer, and ideally a reference?
| https://mathoverflow.net/users/49 | When does collection imply replacement? | The answer is no, if you allow me to adopt some weak-but-equivalent forms of the other axioms. And the reason is interesting:
* A shocking number of the axioms of set theory are true in the non-negative real line R+, with the usual order < being used to interpret set membership. (!)
Let's just check. For example, Extensionality holds, because if two real numbers have the same predecessors, then they are equal. The emptyset axiom holds, since there are no non-negative reals below 0. The Union axiom holds, since for any real x, the reals less than x are precisely the reals that are less than something less than x. (Thus, every set is its own union.) A weakened version of the Power set axioms holds, the Proper Power set, which asserts that for every x, there is a set p whose elements are the strict substs of x. This is because for any real number x, the reals below x are precisely the reals y (other than x), all of whose predecessors are less than x. (Thus, every real is its own proper power set.) An alternative weakening of power set would say: for every x, there is p such that y subset x implies y in p. This is true in the reals by using any p > x. A weakened pairing axiom states similarly: for every x,y, there is z with x ε z and y ε z. This is true in the reals by using any z above both x and y. The Foundation axiom is no problem, since 0 is in every nonempty set. Also, similarly AC holds in the form about families of disjoint nonempty sets, since this never occurs in this model. The Weak Collection Axiom holds since if every y < x has phi(x,y,w), then in fact any y in the same interval will work (since this structure has many automorphisms), and so we may collect witness with any B above x and w. Note that Separation fails, since, for example, {0} does not exist in this model. Also, Replacement fails for the same reason.
Similar interesting models can be built by considering the structure (ORD,<) built using only ordinals, or the class { Vα | α in ORD }. These also satisfy all of the weakened forms of the ZFC axioms without Separation, using Collection in place of Replacement.
Thus, part of the answer to your question is that it depends on what you mean by the "other axioms of ZFC".
Apart from this, however, let me say that the term Weak Collection is usually used to refer to the axioms that restrict the complexity of the formulas in the usual Collection scheme, rather than the axiom that you state. For example, in Kripke Platek set theory KP, a weak fragment of ZFC, one has collection only for Sigma\_1 formulas, and this is described as a Weak Collection axiom. (What you call Weak Collection is usually just called Collection.) And there is a correspondingly weakened version of Separation in KP.
But I am happy to adopt your terminology here. You stated that AC plus Replacement implies Weak Collection, but this is not quite right. You don't need any AC. Instead, as your argument shows, what you need is the cumulative Vα hierarchy, which is built on the Power set axiom, not AC. That is, If you have Replacement and Power set and enough else to build the Vα hierarchy, then you get Collection as you described, even if AC fails. For example, ZF can be axiomatized equivalently with either Collection or Replacement.
Your question is a bit unusual, since usually Separation is regarded as a more fundamental axiom than Replacement and Collection, and more in keeping with what we mean by set theory. After all, if one has a set A and a property phi(x), particularly when phi is very simple, it is one of the most basic set theoretic constructions to be able to form { x in A | phi(x) }, and any set theory violating this is not very set-like. We don't really want to consider models of set theory where many instances of Separation fail (for example, Separation for atomic formula is surely elemental).
Incidentally, there is another version of a weakening of Collection in the same vein as what you are considering. Namely, consider the scheme of assertions, whenever phi(x,y) is a property, that says for every set A, there is a set B such that for every a in A, if there is b with phi(a,b), then there is b in B such that phi(a,b).
---
OK, let me now give a positive answer, with what I think is a more sensible interpretation of your question. I take what I said above (and Dorais's comment) to show that we shouldn't consider set theories where the Separation Axiom fails utterly. Rather, what we want is some very weak set theory, such as the [Kripke Platek](http://en.wikipedia.org/wiki/Kripke%E2%80%93Platek_set_theory) axioms, and then ask the relationship between Weak Collection and Replacement over those axioms. And here, you get the postive result as desired.
**Theorem.** If KP holds, then Weak Collection implies Replacement.
Proof. Assume KP plus (Weak) Collection. First, I claim that this is enough to prove a version of the Reflection Principle, since that proof amounts to taking successive upward Skolem hulls, which is what Collection allows. That is, I claim that for every set x and any formula phi, there is a transitive set Y such that x in Y and phi(w) is absolute between Y and the universe V. This will in effect turns any formula into a Delta0 formula using parameter Y.
Applying this, suppose we have a set A and every a in A has a unique b such that phi(a,b). By the Reflection Principle, let Y be a large set transitive containing A such that phi(a,b) and "exists b phi(a,b)" are absolute between Y and V. So Y has all the desired witnesses b for a in A. But also, now { b | exists a in A phi(a,b) } is a Delta0 definable subset of Y, since we can bound the quantifier again by Y. So the set exists. So Replacement holds. QED
I think we can get away with much less than KP. Perhaps one way to do the argument is to just prove Separation by induction on the complexity of formulas. One can collect witnesses by (weak) Collection, and this turns the formulas into lower complexity, using the new bound as a bound on the quantifiers.
| 34 | https://mathoverflow.net/users/1946 | 12597 | 8,522 |
https://mathoverflow.net/questions/12601 | 18 | This question originated from a conversation with Dmitry that took place here
[Is there a complex structure on the 6-sphere?](https://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere)
The Hirzebruch-Riemann-Roch formula expresses the Euler characteristic of a holomorphic vector bundle on a compact complex manifold $M$ in terms of the Chern classes of the bundle and of the manifold. On the other hand, for complex manifolds the $\bar\partial$ operator is a differential (i.e. it squares to zero) and hence, the complex of sheaves of smooth $(p,q)$ forms for a fixed $p$ equipped with the $\bar\partial$ differential is a resolution of $\Omega^p$; this is a complex of soft sheaves and so, by taking global sections we can compute the cohomology of $\Omega^p$. Moreover, since the de Rham complex resolves the constant sheaf, the alternating sum of the Euler characteristics of $\Omega^p$'s is the Euler characteristic of $M$.
For an arbitrary pseudo-complex manifold the only part of the above that makes sense is the "right hand side" of the Riemann-Roch formula (i.e. the one involving Chern classes) and the (topological) Euler characteristic of the manifold itself. So it seems natural to ask whether the relation between the two that is true in the complex case remains true in the almost complex case. In other words, is it true that for a compact almost complex manifold $M$ of dimension $2n$ we have $$\chi(M)=\sum\_{p=0}^n (-1)^p\sum\_{i=0}^{n\choose{p}}\mathrm{ch}\_{n-i}(\Omega^p)\frac{T\_i}{i!}$$ where $\chi$ is the topological Euler characteristic, $\Omega^p$ is the $p$-th complex exterior power of the cotangent bundle (i.e., the complex dual of the tangent bundle), $\mathrm{ch}$ is the Chern character and $T\_i$ is the $i$-th Todd polynomial of $M$?
In general, are there topological consequences of the existence of the Dolbeault resolution that would be difficult to prove (or, more ambitiously, would fail) for arbitrary pseudo-complex manifolds?
| https://mathoverflow.net/users/2349 | A topological consequence of Riemann-Roch in the almost complex case | I believe that the displayed equation is valid for almost complex manifolds. This is closely related to a computation I talked about [here](https://mathoverflow.net/questions/10630/why-do-todd-classes-appear-in-grothendieck-riemann-roch-formula).
Let $r\_1$, $r\_2$, ..., $r\_n$ be the chern roots of the tangent bundle. Then $\sum (-1)^p \mathrm{ch}(\Omega^p) = \prod (1-e^{-r\_i})$. Let $\mathrm{Td}$ denote the total Todd class, so $\mathrm{Td} := \sum T\_i/i! = \prod \frac{r\_i}{(1-e^{-r\_i})}$. The quantity you are interested in is
$$\int \mathrm{Td} \prod (1-e^{-r\_i}) = \int \prod r\_i.$$
In other words, the top chern class of the holomorphic tangent bundle.
So the question is "On an almost complex manifold, is it still true that the top chern class of the holomorphic tangent bundle is $\chi(M)$?" I believe the answer is yes. Take a generic smooth section $\sigma$ of the tangent bundle and integrate it to get a flow. I believe that the fixed points of that flow will precisely be, with multiplicity, the intersections of $\sigma$ with the zero section. So we are done by the Lefschetz fixed point theorem.
The reason I keep saying "I think" and "I believe" is that I don't spend much time working with nonintegrable complex structures, so I can easily believe that I missed some subtlety.
| 15 | https://mathoverflow.net/users/297 | 12604 | 8,526 |
https://mathoverflow.net/questions/12072 | 14 | Let $k$ be a field, and let $| \ |$ be a norm on $k$. The norm induces a metric. To construct the completion $\hat{k}$ as a normed field, the standard recipe is to take the quotient of the ring $\mathcal{C}(k)$ of all Cauchy sequences in $k$ -- viewed as a subring of $k^{\infty} = \prod\_{i=1}^{\infty} k$ -- by the maximal ideal $\mathfrak{m}\_0$ of all sequences converging to $0$.
This brings up the following [idle] question: what are the maximal ideals of $\mathcal{C}(k)$? The prime ideals?
My vague recollection had been that $\mathfrak{m}\_0$ was the unique maximal ideal of $\mathcal{C}(k)$, but this is evidently not the case: for every $n$, there is a maximal ideal $\mathfrak{m}\_n$ consisting of sequences whose $n$th coordinate is $0$, the residue field being $k$ again. It is easy to see though that $\mathfrak{m}\_0$ is the unique maximal ideal containing the ~~prime~~ ideal $\mathfrak{c} = \bigoplus\_{i=1}^{\infty} k$. (Edit: $\mathfrak{c}$ is prime iff the norm is trivial.)
Now this reminds me of filters. The prime ideals of the (zero-dimensional) ring $k^{\infty}$ correspond precisely to the ultrafilters on $\mathbb{Z}^+$. The principal ultrafilter of all sets containing $n$ pulls back to the maximal ideal $\mathfrak{m}\_n$. Since every nonprincipal ultrafilter contains the Frechet filter of cofinite sets, it follows that it pulls back to $\mathfrak{m}\_0$. But is it true that every maximal ideal of $\mathcal{C}(k)$ is pulled back from a prime (= maximal) ideal of $k^{\infty}$? If so, is this an instance of a general theorem?
---
Addendum:
Note that in the case that the norm is trivial -- so that the induced metric is the discrete metric -- a sequence converges iff it is eventually constant, so a sequence converges to $0$ iff it has only finitely many nonzero terms: $\mathfrak{c} = \mathfrak{m}\_0$. The converse also holds: for any nontrivial norm there exist nowhere zero sequences converging to $0$, e.g. $\{x^n\}$ for any $x \in k$ with $0 < |x| < 1$.
---
Once the original question is worked out, I am also curious about generalizations. What is the analogue for the ring of minimal Cauchy filters in an arbitrary topological ring?
| https://mathoverflow.net/users/1149 | What is the prime spectrum of a Cauchy series ring? | In this answer I will treat the case in which $|\text{ }|$ is not discrete.
I first claim that $\mathfrak m\_0$ is not the restriction of any proper ideal in
$k^{\infty}.$ Indeed, choose $x \in k$ such that $0 < |x| < 1$. Then $(x^i)$
is an element of $\mathfrak m\_0$ which is invertible in $k^{\infty}$ (with
inverse equal to $(x^{-i})$, and so $\mathfrak m\_0$ generates the unit ideal
of $k^{\infty}$.
This doesn't contradict anything; the maximal ideals of $k^{\infty}$ pull-back
to prime ideals in $\mathcal C(k)$ which are simply not maximal (as often happens
with maps of rings).
Furthermore, this pull-back is injective.
To see this, we first introduce some notation; namely, we
let $\mathfrak m\\_{\mathcal U}$ denote the prime ideal of $k^{\infty}$ corresponding to
the non-principal ultra-filter ${\mathcal U}$,and recall that $\mathfrak m\\_{\mathcal U}$ is defined as follows: an element $(x\_i)$ lies in $\mathfrak m\\_{\mathcal U}$ if and only
if $\{i \, | \, x\_i = 0\}$ lies in in $\mathcal U$.
Now suppose that $\mathcal U\_1$ and $\mathcal U\_2$ are two distinct non-principal ultra-filters. Let $A$ be a set lying in $\mathcal U\_1$, but not in $\mathcal U\_2$.
Then $A^c$, the complement of $A$, lies in $\mathcal U\_2$.
Choose $x \in k$ such $0 < | x | < 1,$ and let $x\_i = x^i$ if $i \in A$ and
$x\_i = 0$ if $i \not\in A$. Then $(x\_i)$ is an element of $\mathcal C(k)$,
in fact of $\mathfrak m\_0$, and it lies in $\mathfrak m\\_{\mathcal U\_2}$
but not in $\mathfrak m\\_{\mathcal U\_1}$.
Thus $\mathfrak m\\_{\mathcal U\_1}$ and $\mathfrak m\\_{\mathcal U\_2}$ have distinct pull-backs.
So the map
Spec $k^{\infty} \rightarrow $ Spec $\mathcal C(k)$
is injective
and dominant (since it comes from an injective map of rings), but is not surjective.
Choosing the valuation $|\text{ }|$ allows us to add to Spec $k^{\infty}$ (which is the
Stone-Cech compactification of $\mathbb Z\\_+$) an extra point dominating all the
other points at infinity (i.e. all the non-principal ultrafilters), because the valuation now gives
us a definitive way to compute limits (provided we begin with a Cauchy sequence).
| 5 | https://mathoverflow.net/users/2874 | 12607 | 8,528 |
https://mathoverflow.net/questions/12559 | -2 |
```
Cardinal Equivalence Theorem
```
For each boolean formula, |quantifications| = |assignments|.
The set of valid quantifications has some cardinality, call that |Q(B)|.
The set of satisfying assignments has some cardinality, call that |P(B)|.
Those two numbers are equal, |Q(B)| = |P(B)|, range from 0 through 2^n.
Question one:
Does anyone know the theorem by any other name?
++ Variable order
Changing the order of the variables of B changes the particulars
of each set, but their cardinalities are still the same.
If we knew more precisely what swapping two variables
does to the previously valid set of quantifications,
then perhaps some form of Zipper Theorem could Be.
However, my competency with quantifiers is less
than necessary or sufficient to even compose
any informally stated Zipper Theorem.
++ Question two:
Linear Corollary: Monotone QBFs are linearly decidable.
I only know this result as a followup to the Cardinal Equivalence.
Is there a well known name for the Linear Corollary as a theorem?
thank you, daniel.
pehoushek1 at gee mail dot com.
| https://mathoverflow.net/users/3446 | cardinal equivalence: for each boolean formula, |quantifications| = |assignments|. | I think Daniel might be asking about the following proposition:
>
> Let $f:\{0,1\}^n\to\{0,1\}$ be any function (i. e., an "n-ary boolean function"). The number of true formulas $$ Q\\_1 x\\_1 \ldots Q\\_n x\\_n : f(x\\_1,\ldots,x\\_n) = 1,$$ where each $Q\\_i$ is a quantifier $\forall$ or $\exists$, is equal to the number of $(x\\_1,\ldots,x\\_n)$ for which $f(x\\_1,\ldots,x\\_n) = 1$.
>
>
>
The proof is very easy (by induction on $n$). It's an amusing proposition, no doubt, but I don't know of any applications. It might make an interesting advanced exercise in a discrete mathematics course, though.
I've implicitly answered the question, but explicitly:
>
> Does anyone know the theorem by any other name?
>
>
>
I'm not aware of such.
>
> Have you or anyone you know ever heard of this equivalence?
>
>
>
I discovered it a few years ago, apparently about five years after you did. Nobody I tried to tell seemed interested by it, though.
>
> Do you prefer any other name, for casting into stone? (imo This theorem belongs in at least one major book...)
>
>
>
I prefer no name, actually. I don't think it's important enough to have the status of "theorem" (which is why I've been calling it a "proposition"), but I'm willing to be convinced otherwise.
| 11 | https://mathoverflow.net/users/302 | 12608 | 8,529 |
https://mathoverflow.net/questions/12606 | 19 | Today I was studying for a qualifying exam, and I came up with the following question;
>
> Is there a simple description in terms of the subspaces universal covers for the universal cover of a wedge product?
>
>
>
This question came about after calculating universal covers of the wedge of spheres ($\mathbb{S}^1 \vee\mathbb{S}^1$ and $\mathbb{S}^1 \vee\mathbb{S}^n$) and the wedge of projective space with spheres. In these cases, the universal cover looks like the cross product of the sheets of the universal covers of each space in the wedge.
For the case of wedging two spheres, we can use the fact that $\pi\_{n\geq2}\left(U\right)$ is isomorphic to $\pi\_{n\geq2}\left(X\right)$ for $U$ covering $X$.
I googled around a bit to try and find something, but nothing appeared.
Thanks in advance!
| https://mathoverflow.net/users/348 | Universal Covering Space of Wedge Products | If $X$ and $Y$ are two reasonable spaces with universal covers $\tilde{X}$ and $\tilde{Y}$, there is a nice picture of the universal cover $\widetilde{X \vee Y}$ which has the combinatorial pattern of an infinite tree. The tree is bipartite with vertices labeled by the symbols $X$ and $Y$. The edges from an $X$ vertex are bijective with the fundamental group $\pi\_1(X)$, and likewise for $Y$ vertices and $\pi\_1(Y)$. To make $\widetilde{X \vee Y}$, replace each $X$ vertex by $\tilde{X}$ and each $Y$ vertex by $\tilde{Y}$. The base point of $X$ lifts to $|\pi\_1(X)|$ points in $\tilde{X}$, and likewise for $Y$. In $\widetilde{X \vee Y}$, copies of $\tilde{X}$ are attached to copies of $\tilde{Y}$ at lifts of base points. For example, if $X = Y = \mathbb{R}P^2$, then the tree is an infinite chain and $\widetilde{X \vee Y}$ is an infinite chain of 2-spheres.
This tree picture nicely and dramatically generalizes to [Bass-Serre theory](http://en.wikipedia.org/wiki/Bass%E2%80%93Serre_theory).
| 23 | https://mathoverflow.net/users/1450 | 12611 | 8,532 |
https://mathoverflow.net/questions/12569 | 15 | Hi everyone,
I'm looking for a systematical introduction to (or treatment of) logarithmic structures on schemes. I am reading Kato's article ("Logarithmic structures of Fontaine-Illusie") at the moment, but I would like to have a more detailed source that goes through or gives an overview of the constructions of classical scheme theory that have analogs in the log-setup.
Are there any articles/books that in your opinion are required reading if I want to learn about log-geometry? What are beautiful examples of applications of this machinery?
| https://mathoverflow.net/users/259 | References for logarithmic geometry | I put up some old notes by Illusie [here](http://math.harvard.edu/~tdp/Illusie-Log.geometry.lecture.notes-single-page.pdf) for you; they're very detailed and treat log smoothness, the log de Rham complex, and other topics in their second exposé. They're my favourite first reference.
There is also Ogus's book, the latest draft of which is [here](http://math.berkeley.edu/~ogus/preprints/log_book/logbook.pdf).
| 13 | https://mathoverflow.net/users/307 | 12612 | 8,533 |
https://mathoverflow.net/questions/3030 | 1 | The famous Black-Scholes framework is usually derived using a hedging approach where a self-financing portfolio is constructed and the resulting stochastic differential equation is being solved under some conditions.
The self-financing portfolio is basically a dynamic trading strategy where according to the actual price development and the strike of the replicated option parts of the underlying are being bought or sold. For example when you want to replicate a call you would have to buy when the price rises and sell when it develops in the direction of the strike.
**My question:** I want to build a framework where you could match different dynamic trading strategies with derivatives. For example I would like to find the characteristics of a moving average approach in terms of a derivative, e.g. the P/L-diagram of this would in my opinion look like an covered call writing combination.
How would you do that?
First of all I guess in this example one could take the price process (e.g., a geometric Brownian motion) and e.g. convolute that with a rectangular function to model the moving average. After that one would have to find the resulting distribution density function...
But all of that is just speculation... (also quite literally ;-)
Has any one of you some ideas on how to do that - or are there even ready-to-use approaches (on the web, in articles, books, etc...)
| https://mathoverflow.net/users/1047 | Matching dynamic trading strategies with derivatives | At last I found two papers as a starting point that do exactly that:
* Asset allocation and derivatives by Martin B Haugh and Andrew W Lo
MIT Sloan School of Management and Operations Research Center, Cambridge, USA, 14 November 2000
<http://alo.mit.edu/wp-content/uploads/2015/08/AssetAllocationDerivatives2001.pdf> <http://dx.doi.org/10.1080/713665551>
* On the equivalence of the static and dynamic asset allocation problems by Robert V. Kohn and Oana M. Papazoglu, Courant Institute, New York, USA, 31 August 2004
<http://www.math.nyu.edu/faculty/kohn/papers/kohn-papazoglu-paper1.pdf>
<http://dx.doi.org/10.1080/14697680600580946>
| 2 | https://mathoverflow.net/users/1047 | 12615 | 8,535 |
https://mathoverflow.net/questions/12621 | 3 | In "Rings, Modules, and Algebras in Stable Homotopy Theory" Elmendorf, Kriz, Mandell and May introduce a notation of spectra indexed over an universe $\mathcal{U}$ as a collection of pointed topological spaces index by finite subspaces of the universe. Has anyone seen a definition using pointed simplicial sets instead? What would be a simplicial model for the one-point compactification $S^V$ of a real finite dimensional vector space? Just its singular complex?
| https://mathoverflow.net/users/2146 | Are there universe-indexed spectra over simplicial sets? | Yes to both interpretations of your question. It is not clear to me where you want to put pointed simplicial sets.
One interpretation of your question is that you want to replace pointed topological spaces with pointed simplicial giving the notion of a spectrum as a functor from supspaces of U to pointed sSet. This is a very common thing to do, and in some circles in homotopy theory is the standard definition of spectrum. Often "space" is interpreted as meaning simplicial set. This is because of the standard Quillen equivalence between Top and sSet.
The other possible interpretation of your question is that you are trying to replace vector spaces with simplicial sets. This is also (essentially) something which has been done. It gives a model of spectra known as *W-spaces*. This is one of the standard diagram category models of spectra. See the following paper for a comparison:
Model categories of diagram spectra, by M. A. Mandell, J. P. May, S. Schwede, and B. Shipley
| 8 | https://mathoverflow.net/users/184 | 12625 | 8,540 |
https://mathoverflow.net/questions/12629 | 3 | Is the category of groups with group-homomorphisms the same as the category of models of group theory with elementary maps?
If not so: why?
| https://mathoverflow.net/users/2672 | Category of groups = Category of models of group theory? | The Categories will be fundamentally different. The category of groups with group homomorphisms, (even with monomorphisms) enjoys a directedness property: any two groups can map monomorphically into their direct sum.
But in the category of models of group theory under elementary maps, the finite groups map elementarily only into isomorphic copies of themselves. The size of any particular finite group (or indeed any model in any theory) is first order expressible, and any elementary map will be an isomorphism.
| 3 | https://mathoverflow.net/users/1946 | 12635 | 8,542 |
https://mathoverflow.net/questions/12657 | 12 | This is an elementary question, but a little subtle so I hope it is suitable for MO.
Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$.
The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda\_iI$, and we have the Jordan canonical form:
$$ J = \begin{bmatrix} J\_1 \\\ & J\_2 \\\ & & \ddots \\\ & & & J\_n \end{bmatrix}$$
where each block $J\_i$ corresponds to the eigenvalue $\lambda\_i$ and is of the form
$$ J\_i = \begin{bmatrix} \lambda\_i & 1 \\\ & \lambda\_i & \ddots \\\ & & \ddots & 1 \\\ & & & \lambda\_i \end{bmatrix}$$
and each $J\_i$ has the property that $J\_i - \lambda\_i I$ is nilpotent, and in fact has kernel strictly smaller than $(J\_i - \lambda\_i I)^2$, which shows that none of these Jordan blocks fix any proper subspace of the subspace which they fix. Thus, Jordan canonical form gives the closest possible to a diagonal matrix. The elements in the superdiagonals of the Jordan blocks are the obstruction to diagonalization.
So far, so good. What I want to prove is the assertion that "Almost all square matrices over $\mathbb{C}$ is diagonalizable". The measure on the space of matrices is obvious, since it can be identified with $\mathbb{C}^{n^2}$. How to prove, perhaps using the above Jordan canonical form explanation, that almost all matrices are like this?
I am able to reason out the algebra part as above, but is finding difficulty in the analytic part. All I am able to manage is the following. The characteristic equation is of the form
$$(x - \lambda\_1)(x - \lambda\_2) \cdots (x - \lambda\_n)$$
and in the space generated by the $\lambda\_i$'s, the measure of the set in which it can happen that $\lambda\_i = \lambda\_j$ when $i \neq j$, is $0$: this set is a union of hyperplanes, each of measure $0$.
But here I have cheated, I used only the characteristic equation instead of using the full matrix. How do I prove it rigorously?
| https://mathoverflow.net/users/2938 | Proving "almost all matrices over C are diagonalizable". | The discriminant of the characteristic polynomial of a matrix depends polynomially on the coefficients of the matrix, and its vanishing detects precisely the existence of multiple eigenvalues. Therefore the set where the discriminant does not vanish is contained in the set of diagonalizable matrices.
Now the set where a non-zero polynomial vanishes is very, very thin (in many senses: it does not contain open sets, it has zero Lebesgue measure, etc) so in consequence the set of diagonalizable matrices is correspondingly thick.
| 36 | https://mathoverflow.net/users/1409 | 12659 | 8,559 |
https://mathoverflow.net/questions/786 | 9 | Hochschild homology gives invariants of (unital) $k$-algebras for $k$ a unital, commutative ring. If we let our algebra $A$ be the group ring $k[G]$ for $G$ a finite group, we get group homology. There are plenty of other connections to homological algebra. If we use cyclic homology, there are connections to geometry and topology involving the Chern character.
Von Neumann algebras are complex algebras, so we can take their Hochschild and cyclic homologies. When I have asked experts in the fields of von Neumann algebras and non-commutative geometry about what you get, I usually hear some approximation of the following: "There's also analysis in von Neumann algebras, so I wouldn't expect an algebraic invariant like Hochschild or cyclic homology to tell you anything useful."
Although this answer makes some sense, I find it very displeasing and cryptic. Why shouldn't it tell you something? Is there some way to make "it doesn't tell you anything" quantitative? Is there an example of a von Neumann algebra with nontrivial Hochschild or cyclic homology (different from that of the complex numbers)?
EDIT: After reading the responses so far, I should specify that I really want to know if there is a $II\_1$-factor with nontrivial Hochschild or cyclic (co)homology.
| https://mathoverflow.net/users/351 | Hochschild/cyclic homology of von Neumann algebras: useless? | The question is not well-posed. There are various versions of cyclic theory (for instance) which differ according to continuity conditions that are assumed. In Connes' original IHES papers he deals with both discrete (useful for arbitrary rings) and topological (useful in the $C^\infty $-setting.)
The basic problem is that cyclic theory is very very sensitive. Consider the following example (using Connes' topological cyclic theory.) Let $M$ denote a compact smooth manifold with a smooth foliation. Then there are three operator algebras that you can associate with the situatio:;
a) $C(M)$, the $C^\ast$-algebra of continuous complex functions on $M$.
b) $C^\infty (M)$, the smooth functions on $M$.
c) $C\_\tau ^\infty (M)$, the continuous functions on $M$ which are smooth in the leaf directions.
The cyclic cohomology of these three rings are all different typically (e.g. for the Kronecker flow on the torus).
a) leads to measures on $M$;
b) leads to de Rham cohomology on $M$;
c) leads to "tangential cohomology" on $M$ (cf. my book [Global analysis on foliated spaces](https://doi.org/10.1007/978-1-4613-9592-8) with Cal Moore).
| 10 | https://mathoverflow.net/users/3468 | 12664 | 8,562 |
https://mathoverflow.net/questions/12638 | 48 | Do you find it a good idea to take lecture notes (even detailed lecture notes) in mathematical lectures?
Related question: [Digital Pen for Math: Your Experiences?](https://mathoverflow.net/questions/12898/digital-pen-for-math-your-experiences)
| https://mathoverflow.net/users/1532 | Taking lecture notes in lectures | I usually bring a pad of paper with me to talks, but don't take notes. I do write down things I want to revisit later, and sometimes it turns into full blown note-taking. Taking notes reduces my ability to concentrate on the lecture, so if it's a really difficult lecture or if the material is totally new to me, taking notes is a sure way to not get much out of the lecture (unless I commit a large chunk of time to reviewing the notes afterwards).
If I *know* I'm going to review the material carefully (e.g. if I'm taking a class and I'm putting a lot of effort into it), then I live-TeX notes, which has a number of advantages:
* It's much faster to strike a key than it is to write a character by hand, so I TeX much better notes than I can take by hand. I find that I can even write down lots of things the speaker says but doesn't write.
* I can throw the notes up on my web page, which makes other people in the same class happy.
* I can grep (or otherwise search) my notes.
* If I have the time, I can edit the notes and end up with an awesome reference (e.g. I put a lot of effort into editing my Lie groups and Stacks notes). More often, I review my notes but not edit them very much, in which case they're still pretty nice to have later on.
If you're considering live-TeXing, try it! It's a lot easier than you might think. Also, look at [this page of advice](http://math.berkeley.edu/~anton/index.php?m1=me&m2=TeXadvice) I wrote up when I found myself repeating it to many people.
**Edit:** By the way, I [post all my source](http://math.berkeley.edu/~anton/index.php?m1=writings), and you're welcome to look at it if you want to see how I set up my headers or manage my files. If I have an svn repository for the notes, you can get them with the command `svn checkout svn://sheafy.net/courses/halg_sp2008` (to get my homological algebra notes from Peter Teichner's course for example). If there's no svn repository, you can download the tgz (tar+gzip) file and uncompress it.
| 65 | https://mathoverflow.net/users/1 | 12673 | 8,570 |
https://mathoverflow.net/questions/12652 | 26 | There are two ways to define smooth mapping spaces and I want to know how they compare.
Let's take the concrete special case of free loops spaces. I think this is the most studied example so will probably have the best chance of an answer. Roughly the free loop space is a space which is supposed to be the space of maps from the circle to a given space X. It is usually denoted LX. This is all fine and good for topological spaces. You get a nice mapping space LX by equipping the set of maps with the compactly generated compact open topology. It even satisfies the adjunction:
$ Map(Y, LX) = Map( Y \times S^1, X)$
However things get complicated when we want to work with manifolds. The first thing is that we want something that represents *smooth* maps from the circle into the manifold X.
There are basically two approaches to making such an object precise, and I want to know how they compare.
The first approach actually tries to construct an actual space of smooth maps. Here you start with the set of smooth maps LX, and with analytical muscle you give it the structure of an infinite dimensional Fréchet manifold. When X = G is a Lie group, this is an inifinite dimensional Lie group and is the thing whose (projective) representations make an appearence in conformal field theory.
The second approach is to study the loop space as a generalized smooth space. What is a generalized smooth space, you ask? Well there was a lot of discussion about it at the N-category Cafe, [here](http://golem.ph.utexas.edu/category/2008/01/comparative_smootheology.html) and [here](http://golem.ph.utexas.edu/category/2008/05/convenient_categories_of_smoot.html). Roughly LX is thought of as a kind of sheaf via the formula:
$LX(Y) = Maps(Y, LX) = Maps(Y \times S^1, X)$
In some models it must be a concrete sheaf (i.e. it has a underlying set of points and every map from it to anything else which is concrete is a particular set map. The technical details appear in the Baez-Hoffnung paper in the second link). Clearly this model has its own desirable properties.
>
> How does the manifold version of loop space compare to the sheaf theoretic version?
>
>
>
Presumably the Fréchet manifold model gives a sheaf (since we can just map into it). Does this agree with the sheaf defined by the adjunction formula? If they are not the same sheaf, they do seem to have the same points, right? And I think there is a comparison map from the manifold LX to the sheaf LX, which should be helpful. Can anyone explain their relationship? How similar/different are they?
| https://mathoverflow.net/users/184 | Loop Spaces as Generalized Smooth spaces or as Infinite dimensional Manifolds? | They are the same.
If you want the full gory details, read "A Convenient Setting of Global Analysis" by Kriegl and Michor (available as a free PDF via the AMS bookstore). For a gentler approach, read my online seminar notes "The Differential Topology of Loop Spaces". Michor has also written a fair bit on manifolds of mappings, and I've written the odd article on them as well.
But basically, the manifold structure on $LM$ is the "right" structure so that the exponential law:
$$
C^\infty(N \times S^1,M) \cong C^\infty(N, LM)
$$
holds.
I just (today, but before you posted this) started [smooth loop space](http://ncatlab.org/nlab/show/smooth+loop+space) with the intention of copying over some of my seminar notes and other oddments into the nLab. If there's anything you are particularly interested in, let me know and that'll give me some incentives to get on with the project!
PS I notice you picked loops as an *example*. For the more general case of the exponential law, you should read the Kriegl and Michor book. Basically, it works fine where the object being exponentiated is a **compact** manifold. For non-compact manifolds you need to use a different structure which is a little weirder.
| 16 | https://mathoverflow.net/users/45 | 12685 | 8,578 |
https://mathoverflow.net/questions/12688 | 21 | I always had trouble remembering this. Is it true that a curve over a non-algebraically-closed field is normal implies that it's non-singular? How about a 1 dimensional scheme? How about dimension 2? I think I heard once that surfaces over a non-algebraically closed field is normal does imply that it's non-singular. Is it true for2 dimensional schemes? What is the reason that these theorems are true for small dimensions, but fail for higher dimensions?
| https://mathoverflow.net/users/3238 | Nonsingular/Normal Schemes | For curves over a field $k$, normal implies regular. (The point is that a normal Noetherian local ring
of dimension one is automatically regular, i.e. a DVR.) If $k$ is not
perfect, then it might not be smooth over $k$.
The reason is that in this case it is possible to have a regular local $k$-algebra of dimension one whose base-change to $\overline{k}$ is no longer regular. (On the other hand,
smoothness over $k$ is preserved by base-change, since it is a determinental condition
on Jacobians.)
Here is a (somewhat cheap) example: let $l$ be a non-separable extension of $k$ (which
exists since $k$ is not perfect), and let $X = \text{Spec} l[t],$ though of as a $k$-scheme.
This will be regular, but not smooth over $k$.
In dimension 2, even over an algebraically closed field, normal does not imply regular
(and so in particular, does not imply smooth). Normal is equivalent to having the singular locus be of codimension 2 or higher (so for a surface, just a bunch of points) (this is what Serre calls R\_1) together with the condition that if a rational function on some open subset has no poles in codimension one, it is in fact regular on that open set (this is Serre's condition S\_2).
For a surface in ${\mathbb P}^3$, which is necessarily cut out by a single equation, the
condition $S\_2$ is automatic (this is true of any local complete intersection in a smooth variety), so normal is equivalent to the singular locus being 0-dimensional.
For surfaces in higher dimensional projective space, $R\_1$ and $S\_2$ are independent
conditions; either can be satisfied without the other. And certainly both together (i.e. normality) are still weaker than smoothness.
From Serre's criterion (normal is equivalent to $R\_1$ and $S\_2$) you can see that normality
just involves conditions in codimension one or two. Thus for curves it says a lot,
for surfaces it says something, but it diverges further from smoothness the higher the dimension of the variety is.
Edit: As Hailong pointed out in a comment (now removed), I shouldn't say that S\_2 is a condition only in dimension 2;
one must check it all points. Never the less, at some sufficiently vague level, the spirit
of the preceding remark is true: $R\_1$ and $S\_2$ capture less and less information about the local structure of the variety, the higher the dimension of the variety.
| 40 | https://mathoverflow.net/users/2874 | 12690 | 8,582 |
https://mathoverflow.net/questions/11603 | 18 | Finite projective planes are fascinating objects from many perspectives. In addition to the geometric view, they can be viewed as combinatorial block designs.
From the geometric perspective, there are two very important structural properties for projective planes: the [Theorem of Desargues](http://en.wikipedia.org/wiki/Desargues%27_theorem), which holds exactly when the plane can be coordinatized by a division ring, and the [Theorem of Pappus](http://en.wikipedia.org/wiki/Pappus%27s_hexagon_theorem), which holds precisely when the plane can be coordinatized by a field. It is a famous theorem of Wedderburn that every finite division ring is a field, so the two properties are equivalent for finite projective planes.
Although they are both very combinatorial statements, I don't recall having seen anything similar to Desargues and Pappus for other classes of block designs.
>
> Are there interesting analogues or generalizations of the properties of Desargues and Pappus for other classes of block designs? Of particular interest would be analogues and generalizations that correspond (not necessarily exactly) to some form of coordinatization of designs.
>
>
>
**Comment.** This question has generated a fair amount of interest. I've been considering accepting the answer by John Conway and Charles Roque even though it is not quite satisfactory. (It only answers the first part of the question in a loose sense, and it does not address the second part.) So I decided to set up a small bounty to stimulate other answers.
| https://mathoverflow.net/users/2000 | Are there analogues of Desargues and Pappus for block designs? | I passed on your question to [John H. Conway](http://en.wikipedia.org/wiki/John_Horton_Conway). Here is his response: (**NB**. Everything following this line is from Conway and is written from his point of view. Of course, in the comments and elsewhere on the site, I am not Conway.)
I think it's wrong to focus on block designs in particular. This may not answer your question, but there are some interesting examples of theorems similar to Desargues's and Pappus's theorems. They aren't block designs, but they do have very nice symmetries.
I call these "presque partout propositions" (p.p.p. for short) from the French "almost all". This used to be used commonly instead of "almost everywhere" (so one would write "p.p." instead of "a.e."). The common theme of the propositions is that there is some underlying graph, where vertices represent some objects (say, lines or points) and the edges represent some relation (say, incidence). Then the theorems say that if you have all but one edge of a certain graph, then you have the last edge, too. Here are five such examples:
**Desargues' theorem**
*Graph*: [the Desargues graph](http://en.wikipedia.org/wiki/Desargues_graph) = [the bipartite double cover](http://en.wikipedia.org/wiki/Bipartite_double_cover) of [the Petersen graph](http://en.wikipedia.org/wiki/Petersen_graph)
*Vertices*: represent either points or lines
*Edges*: incidence
*Statement*: If you have ten points and ten lines that are incident in all of the ways that the Desargues graph indicates *except one incidence*, then you have the last incidence as well. This can be seen to be equivalent to the usual statement of Desargues's theorem.
**Pappus's theorem**
*Graph*: [the Pappus graph](http://en.wikipedia.org/wiki/Pappus_graph), a highly symmetric, bipartite, cubic graph on 18 vertices
*Vertices*: points or lines
*Edges*: incidence
*Statement*: Same as in Desargues's theorem.
**"Right-angled hexagons theorem"**
*Graph*: [the Petersen graph](http://en.wikipedia.org/wiki/Petersen_graph) itself
*Vertices*: lines in 3-space
*Edges*: the two lines intersect at right angles
*Statement*: Same as before, namely having all but one edge implies the existence of the latter. An equivalent version is the following: suppose you have a "right-angled hexagon" in 3-space, that is, six lines that cyclically meet at right angles. Suppose that they are otherwise in fairly generic position, e.g., opposite edges of the hexagon are skew lines. Then take these three pairs of opposite edges and draw their common perpendiculars (this is unique for skew lines). These three lines have a common perpendicular themselves.
**Roger Penrose's "conic cube" theorem**
*Graph*: [the cube graph](http://en.wikipedia.org/wiki/Hypercube_graph) Q3
*Vertices*: conics in the plane
*Edges*: two conics that are doubly tangent
*Statement*: Same as before. Note that this theorem is not published anywhere.
**Standard algebraic examples**
*Graph*: this unfortunately isn't quite best seen as a graph
*Statement*: Conics that go through 8 common points go through a 9th common point. Quadric surfaces through 7 points go through an 8th (or whatever the right number is).
Anyway, I don't know of any more examples.
Also, I don't know what more theorems one could really have about coordinatization. I mean, after you have a field, what more could you want other than, say, its characteristic? (Incidentally, the best reference I know for the coordinatization theorems is H. F. Baker's book "Principles of Geometry".)
In any case, enjoy!
| 10 | https://mathoverflow.net/users/1079 | 12720 | 8,606 |
https://mathoverflow.net/questions/12679 | 2 | What does a coequalizer in the category of primitive recursive functions look like? I know that in Set, a coequalizer is a *minimum* congruence but...what is it in particular in the category of primitive recursive functions (which is a subcategory of Set)?
| https://mathoverflow.net/users/3338 | Coequalizer in the category of primitive recursive functions | For those readers unfamiliar with the class of [primitive recursive functions](http://en.wikipedia.org/wiki/Primitive_recursive_function), let me say that you may simplify things somewhat by fruitfully thinking of them as poor cousins of the computable functions. They were introduced, before Turing, as a natural class of computable total functions. But the class of primitive recursive functions is too weak to carry out the unbounded search operation that is so essential to computability (and which leads to partial functions, when the search fails). Nothing in my answer to the question will depend on any difference between the primitive recursive functions and the class of computable functions.
My answer is that neither the category of primitive recursive functions nor of computable functions admits co-equalizers. The basic reason is that for some primitive recursive functions, the smallest congruence is simply not a computable relation.
If one thinks about how one might build the congruence, this is to be expected, since there is a hidden unbounded search operation involved, and seemingly no way to know in advance whether it will succeed. Thus, we don't expect it to be computable. More specifically, if f,g are maps from A to B, then the desired congruence on B is the smallest relation making f(a) ∼ g(a) for all a. One concrete way to think about it is that two points b and c in B become equivalent with respect to ∼, if there is a zig-zag pattern a1,a2,... such that b=f(a1), g(a1)=f(a2), g(a2)=f(a3), etc. ending at g(an)=c (or swapping f and g in this pattern). Thus, to say that b ∼ c, we seem to need to discover whether one can get from b to c by following such a zig-zag pattern through the graphs of f and g. But that is an existential search, and we seem to have no computable way to determine whether we will find one or not. (The congruence, however, will always be computably enumerable, if f and g are computable.)
Let me give now a more detailed proof. Let A be a primitive recursive subset of the plane ω2, such that the projection of A onto the first coordinate { a | exists b A(a,b) } is not primitive recursive. For example, let A be the set of pairs (a,b) such that a is a Turing machine program, and b is a code for a halting computation of program a on input 0. This is a primitive recursive set, since it is Delta0 definable, but the projection of A onto the first coordinate is precisely the halting problem, the set of halting programs, and this is definitely not primitive recursive, because it is not even computable.
Now, consider the function f(a,b)=(a,b+2), if ¬A(a,b), otherwise f(a,b)=(a,b+1) if A(a,b) holds. Let i(a,b)=(a,b) be the identity function.
Suppose towards contradiction that h:ω2 to N is a co-equalizer of f and i. According to my (crude) understandiing of what this means, which I just learned this afternoon at the Wikipedia page [here](http://en.wikipedia.org/wiki/Coequalizer), it means that h.f = h.i, and h has the universal property that wheneverr h'.f = h'.i, then there is u with u.h = h'. (I am using dot for composition.)
Consider the case that a is not in the projection of A, so that A(a,b) never holds. In this case, we will have f(a,b)=(a,b+2) for all b, and so h(a,b) = h(i(a,b)) = h(f(a,b)) = h(a,b+2). By the universal property, h will send all the even values h(a,2b) to one value, and h(a,2b+1) to another. In particular, in this case, h(a,0) is not equal to h(a,1).
In the other case, suppose that a is in the projection of A, so that A(a,x) for some smallest value x. Now, we will still have h(a,b) = h(a,b+2) for values of b below x. But now, once at x, we see that h(a,x) = h(i(a,x)) = h(f(a,x)) = h(a,x+1). So here, h is now agreeing at an odd and even value of the second coordinate. It follows that all the odd and even values below x must be sent to the same value by h, and so in this case, h(a,0) = h(a,1).
Thus, the function g(a) = 1 if h(a,0) = h(a,1) and g(a) = 0 if h(a,0) not= h(a,1) is precisely the characteristic function of the projection of A. But that projection was not computable! It follows that h cannot be computable either. Thus, there can be no co-equalizer of the functions f and i within the category of computable functions, and hence also no co-equalizer within the category of all primitive recursive functions. QED
Finally, let me apologize for my clumsy working with category theory here. I am a category theory newbie, coming from logic and set theory, with a little computability theory. But if I may say so, I find this question attractively on the boundary between logic and category theory. The audience for the question would seem to be two large groups of talented people, logicians and category theorists, who I believe might benefit from greater interaction.
| 8 | https://mathoverflow.net/users/1946 | 12721 | 8,607 |
https://mathoverflow.net/questions/12717 | 11 | In the familiar case of (smooth projective) curves over an algebraically closed fields, (closed) points correspond to DVR's.
What if we have a non-singular projective curve over a non-algebraically closed field? The closed points will certainly induce DVR's, but would all DVR's come from closed points? Is there a characterization of the DVR's that aren't induced by closed points?
And how about for a general projective variety that is regular in codimension 1 (both for algebraically closed and non-algebraically closed)? Point of codimension 1 induce DVR's. Do they induce all of them? What is the characterization of the ones they do induce?
How about complete integral schemes that are regular in codimension 1?
| https://mathoverflow.net/users/3238 | Points and DVR's |
>
> What if we have a non-singular projective curve over a non-algebraically closed field? The closed points will certainly induce DVR's, but would all DVR's come from closed points?
>
>
>
Yes: let $L/k$ be a function field in one variable, so it can be given as a finite separable extension of $K = k(t)$. Then the discrete valuations on $L$ which are trivial on $k$ are in canonical bijection with the closed points on the unique regular projective model $C\_{/l}$, where $l$ is the algebraic closure of $k$ in $L$ (since $l/k$ is finite, any valuation which is trivial on $k$ is also trivial on $l$).
By coincidence, this is almost exactly where I am in a course I am now teaching, although I won't insist on the geometric language: see Section 1.7, especially Exercise 1.22, of
[http://alpha.math.uga.edu/~pete/8410Chapter1.pdf](http://alpha.math.uga.edu/%7Epete/8410Chapter1.pdf)
>
> And how about for a general projective variety that is regular in codimension 1 (both for algebraically closed and non-algebraically closed)? Point of codimension 1 induce DVR's. Do they induce all of them? What is the characterization of the ones they do induce?
>
>
>
No, they do not induce all of them (even if the variety is smooth, which I will assume for simplicity). The problem here is that, unlike in dimension one, there is no unique nonsingular projective model, so e.g. there will be discrete valuations on $k(t\_1,t\_2)$ which correspond not to codimension one points on $\mathbb{P}^2$ but to points on (at least) some arbitrary blowup of $\mathbb{P}^2$. Because of this, I am pretty sure that there is no simple valuation-theoretic characterization of those valuations which correspond to closed points on a particular projective model of the function field.
By the way, I do not understand valuation theory on function fields in more than one variable very well, so I especially welcome further responses which elaborate on this issue.
| 11 | https://mathoverflow.net/users/1149 | 12723 | 8,609 |
https://mathoverflow.net/questions/12716 | 1 | Is it possible to express the mean curvature of a surface of revolution in terms of the first derivative of the polar tangential angle?
To be specific: Let $r=u(\theta)$ be a polar curve in the first quadrant of the xy-plane. The surface of revolution is generated by rotating this curve around the y axis. Let $\psi$ be the polar tangential angle, the angle between the tangent to the curve at a point and the ray from the origin to the point. Can the mean curvature be expressed as a first derivative of $\psi$?
Any suggestions or references appreciated.
| https://mathoverflow.net/users/3477 | mean curvature and polar tangential angle | Yes, it is true $H=F(u,\theta,\psi,\psi')$.
You ask for suggestion: *Calculate both principle curvatures,*
$k\_1=k\_1(u,\theta,\psi)$ *and* $k\_2=k\_2(u,\theta,\psi,\psi')$.
Here $k\_2$ is the curvature of original curve $r=u(\theta)$
and $k\_1$ is principle curvature in the normal direction.
| 2 | https://mathoverflow.net/users/1441 | 12727 | 8,613 |
https://mathoverflow.net/questions/12633 | 1 | Hi,
As part of one of my courses I need to simulate Gaussian, Student-T and Clayton copulas. The only way to do it that I am aware of uses the conditional copulas, so
$$C\_{1|2}(u, v) = P[X = F\_1^{-1}(u) | Y = F\_2^{-1}(v)]$$
What are the formulae for conditional copulas for Gaussian, Student-T and Clayton?
| https://mathoverflow.net/users/3160 | Conditional copulas | I understand that you want to generate and plot samples of Gaussian, t and Clayton copulas.
The method of sampling is different for the first two cases (elliptical copula) and the
third case (Archimedean copula).
Here is a brief description of the sampling. A reference for both cases is given:
In the first two cases, it is unnecessary (and not useful) to write a closed formula for
the copula density function. The algorithm for sampling the copula is as follows:
For the Gaussian case:
1. Generate two random vectors X1 and X2 from a multivariate Gaussian distribution having a given mean, correlation coefficient rho and standard deviation sigma.
2. Compute the marginal cumulative univariate Gaussian distribution functions
(U1 and U2) corresponding to each vector of samples (X1 and X2) given the appropriate mean and sigma.
3. Plot the samples of U2 as a function of U1
For the t-case, the first step is replaced by a multivariate t-distribution with given
correlation coefficient rho and given number of degrees of freedom v, and the second step
is replaced by cumulative univariate t-distribution with v degrees of freedom.
Here is a [Mathworks reference](http://www.mathworks.com/products/statistics/demos.html?file=/products/demos/shipping/stats/copulademo.html#15) showing examples of the upper two cases using Matlab functions from the statistics toolbox.
For the third case (Clayton), the algorithm is as follows:
1. Generate a random vector V from a Gamma distribution with the parameters (1/theta, 1).
2. Generate two uniform random independent vectors X1 and X2
3. The copula samples are computed as: U1 = psi(-ln(X1)/V) and U2 = psi(-ln(X2)/V), where psi
is the Clayton generator.
Here is the article by: [Alexander J. McNeil](http://www.ma.hw.ac.uk/~mcneil/ftp/NestedArchimedean.pdf), where this algorithm is described.
| 4 | https://mathoverflow.net/users/1059 | 12728 | 8,614 |
https://mathoverflow.net/questions/9474 | 31 | Reading [Serre's letter to Gray](http://books.google.de/books?id=4Vm4Lq1WSHgC&pg=RA1-PA722&lpg=RA1-PA722&dq=Jean-Pierre+Serre+%22Extensions+icosaedriques%22&source=bl&ots=CTXk-_vVqx&sig=kFq8FuyhYs7sGPOukgPe8kbQcBY&hl=de&ei=vXwvS8jxApjknAOiwY3PBA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CA0Q6AEwAA#v=onepage&q=&f=false)
, I wonder if now modern expositions of the themes in [Klein's book](http://digital.library.cornell.edu/cgi/t/text/text-idx?c=math;cc=math;view=toc;subview=short;idno=03070001)
exist. Do you know any?
| https://mathoverflow.net/users/451 | Do there exist modern expositions of Klein's Icosahedron? | "Geometry of the Quintic" is available for free at my website.
Jerry Shurman
| 37 | https://mathoverflow.net/users/3479 | 12729 | 8,615 |
https://mathoverflow.net/questions/12658 | 2 | Suppose you have a finite dimensional real Hilbert space $V$ and you form the tensor algebra
$$T(V) = \mathbb{R} \oplus V \oplus (V\otimes V) \oplus (V\otimes V \otimes V) \oplus \cdots$$
where the multiplication is given by the tensor product. Now imagine you have a finite set of vectors $S$ which you use to form an ideal $I(S)$. From this, we can get the quotient algebra $Q = T(V)/I(S)$ (see [this Wikipedia article](http://en.wikipedia.org/wiki/Quotient_algebra)).
I am interested in comparing elements of $Q$, ideally using an inner product, if it is possible to define one, or else using a distance, if it is possible to define a norm.
The approach I've taken so far is, given an element $x$ of $T(V)$ is to try to project it onto the orthogonal complement of $I(S)$ (call this $P(x)$) in the assumption that there will be a unique element of the equivalence class of $x$ that lives in this orthogonal complement, and that we can then use the inner product of $T(V)$. However I think this assumption is false. Consider the case where $S =$ {$a\otimes a - a$} for a one dimensional vector space $V$ with basis vector $a$. Then $I(S)$ is the vector space with basis $a - a\otimes a$, $a\otimes a - a\otimes a\otimes a$, $a\otimes a\otimes a - a\otimes a\otimes a\otimes a$, $\ldots$.
Intuitively, I would expect $Q$ in this case to be a two dimensional vector space with basis $1, [a]$. Is this correct? Projecting onto the orthogonal complement seems to give zero however.
Is there a standard way to define a norm or an inner product on a quotient algebra such as this? How can it be computed?
| https://mathoverflow.net/users/3467 | Norm on quotient algebra of a tensor algebra |
>
> Intuitively, I would expect Q in this case to be a two dimensional vector space with basis 1,[a]. Is this correct? Projecting onto the orthogonal complement seems to give zero however.
>
>
>
The first part is correct. This may be easiest to see by considering the isomorphism, in the present case where $V$ is one dimensional, of $T(V)$ with $\mathbb{R}[x]$. Then $Q$ is $\mathbb{R}[x]/(x^2-x)$, which is 2-dimensional by the division algorithm.
The orthogonal complement of $I(S)$ in this case is not zero, but it is one dimensional and your concern is valid. Thinking again in $\mathbb{R}[x]$ for convenience, the condition that $\sum\_k a\_k x^k$ is orthogonal to $x^{m+2} - x^{m+1}$ for all $m\geq0$ says that $a\_{m+2}=a\_{m+1}$ for all $m\geq0$; thus, $a\_k=0$ for $k\geq1$. Note that the conclusion is the same if we first complete to a Hilbert space by taking power series with square summable coefficients, so that unfortunately Andrew Stacey's desperate hope appears to be shattered. I have been assuming in this paragraph that $a$ (or $x$) has norm 1; it makes no difference for the conclusion in the non-completed case, but it simplifies the relation. (In other cases when $\dim(V)>1$ the Hilbert space quotient can be far too large for the vector space isomorphism to be possible, having dimension $2^{\aleph\_0}$, so completion is probably not the way to go.)
This shows that there is no hope in general of using the standard inner product to define a norm on $T(V)/I(S)$. However, the problem in the above case arose because the ideal was not homogeneous; this forces relations on coefficients of different degree for elements of the orthogonal complement, thereby forcing too many coefficients to be zero. For this construction to work I recommend considering the case where $S$ is a set of homogeneous elements of $T(V)$ (i.e., each element of $S$ lies in one of the tensor powers $V^{\otimes k}$). I don't know what else to tell you in the nonhomogeneous case.
### Gratuitous Add-on
The Hilbert space completion of the free algebra $T(V)$ is the full (or free) Fock space $F(V)$. The latter is not an algebra, but a Banach algebra completion of $T(V)$ can be obtained by first representing $T(V)$ on $F(V)$ by choosing an orthonormal basis $v\_1,\ldots,v\_n$ for $V$ and sending $v\_j\in T(V)$ to the "creation operator" $S\_j$ on $F(V)$ defined by $S\_j(w)=v\_j\otimes w$, and then closing the image in your favorite topology. These algebras were first studied by G. Popescu circa 1991 (but over $\mathbb{C}$ rather than $\mathbb{R}$ and with either norm or ultraweak closure), and a similar construction was earlier used by D. Evans to study the Cuntz algebra $\mathcal{O}\_n$ in the 1980 paper "On $\mathcal{O}\_n$".
| 3 | https://mathoverflow.net/users/1119 | 12736 | 8,620 |
https://mathoverflow.net/questions/11811 | 4 | ### Question
Say we have a map, C->D, of relative curves over a Dedekind scheme, S. What are some of the available methods for showing that this map has good reduction, or integral reduction, at some s∈S? By this I mean: what are some popular conditions that imply this? What are the tricks people usually use?
### Clarification
By a map having good reduction I mean that both Cs and Ds are regular integral curves. By integral reduction I mean that both Cs and Ds are integral curves.
You may assume whatever you want, this is part of the question. Assuming, for example, that C->D is generically Galois; or that D is smooth over S; is legitimate. This is pretty open-ended. Hence, community wiki.
| https://mathoverflow.net/users/2665 | Methods of showing a map has integral or good reduction | You might have a look [here](http://www.math.u-bordeaux1.fr/~liu/articles/tokyo.ps).
| 7 | https://mathoverflow.net/users/3485 | 12739 | 8,622 |
https://mathoverflow.net/questions/12676 | 9 | Is there a relative version of sheaf cohomology?
EDIT: I rather mean the cohomology of pairs.
| https://mathoverflow.net/users/3470 | Relative version of sheaf cohomology? | It turns out that my previous answer dealt with the wrong question. The answer to the
new question is also yes:
local cohomology $H\\_Z(X,\mathcal F)$ corresponds to cohomology of the pair $(X,X\setminus Z)$.
| 7 | https://mathoverflow.net/users/2874 | 12743 | 8,624 |
https://mathoverflow.net/questions/12745 | 5 | Is there a notion of exponentiation that subsumes the well known versions, and in particular the versions on
* tangent spaces (e.g., of Lie groups and Riemannian manifolds), in which the exponential map sends a vector to a point on a curve naturally defined in terms of the vector;
* unital Banach algebras?
(NB. I am not conversant with category theory beyond the words "morphism" and "functor". But a categorically flavored answer that takes my limited knowledge base into account would be preferable. An internet search led me to the notion of a "Cartesian closed category", which doesn't seem to be the sort of thing I have in mind.)
| https://mathoverflow.net/users/1847 | What is an exponential? | I think the exponential function for unital Banach algebras is a special case of the exponential functions for Lie groups modeled on topological vector spaces: the set of invertible elements in an unital Banach algebra naturally is a Banach Lie group, and the exponential function of this Lie group is the "classical" exponential function for the Banach algebra.
| 6 | https://mathoverflow.net/users/3108 | 12747 | 8,626 |
https://mathoverflow.net/questions/12742 | 7 | By Fermat's last theorem, the equation $u^3+v^3=w^3$ has no solutions
in positive integers $u,v,w$. Now consider the following variant : call $\rho(x)$
the distance between $x$ and the nearest integer, for any real number $x$
(thus $\rho(3)=0,\rho(3.2)=0.2,\rho(3.5)=0.5, \rho(3.9)=0.1$ etc).
An "approximate" version of the Fermat equation is to ask
for $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ to be arbitrarily small. A trivial
way to achieve this is to make one of the variables very small
compared to the other, say $v$ very small compared to $u$, so that
${( u^3+v^3 )}^{\frac{1}{3}}$ is very near to $u$.
It is therefore natural to ask if there is an absolute constant $C>0$
such that $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ can be made arbitrarily small
with $u,v$ positive and $u \leq C v, v \leq C u$ (so that neither of
$u$ or $v$ dominates).
Can a (reasonably) explicit $(u\_n,v\_n)$ sequence be found, such that
$\rho(u\_n^3+v\_n^3)$ tends to $0$ as $n$ goes to infinity and
$u\_n \leq C v\_n, v\_n \leq C u\_n$ ? "Closed-form" formula
would be the best, of course, but even a simple recursion would be nice.
| https://mathoverflow.net/users/2389 | Variant of Fermat's last theorem | The condition you want is very weak, and there are clearly many accidental solutions.
You can add severe restrictions and still find many solutions. For example, (as Steven Sivek pointed out) you can force $u\_n = v\_n$ and then by the theory of simple continued fractions, there are infinitely many $p\_n/q\_n$ so that
$|\frac{p\_n}{q\_n}-\sqrt[3]2|\lt q\_n^{-2}.$
Then $|p\_n - q\_n \sqrt[3]2| < 1/q\_n$ so $\rho(\sqrt[3]{(q\_n^3+q\_n^3)} )$ decreases to 0 rapidly.
This might not be viewed as explicit since the coefficients of the simple continued fraction for $2^{1/3}$ don't have a clear pattern, although you can define them recursively if you allow functions like $\lfloor1/\rho\rfloor$.
If you can find $x^3 + y^3 = z^3$ in numbers with known simple continued faction expansions, then you may be able to use this to construct closed form families of examples. For example,
$(5-\sqrt{6})^3 + 3\sqrt{6}^3 = (5+\sqrt{6})^3$.
Convergents of $\sqrt6$, $p\_n/q\_n$, satisfy $|\frac{p\_n}{q\_n} - \sqrt{6}| < 1/q\_n^2$.
Then $(5q\_n-p\_n)^3+(3p\_n)^3 = (5q\_n+p\_n)^3 + O(q\_n) = (5q\_n+p\_n)^3 + o((5q\_n+p\_n)^2)$,
so $\rho(\sqrt[3]{(5q\_n-p\_n)^3+(3p\_n)^3}) = o(1).$
For example, $\sqrt[3]{(5\times 881 - 2158)^3 + (3 \times 2158)^3} = 6552.99916...$
Since $\sqrt{6} = [2;2,4,2,4,2,4...]$ which is periodic, you can construct a closed form expression for the convergents $p\_n/q\_n$.
| 8 | https://mathoverflow.net/users/2954 | 12752 | 8,629 |
https://mathoverflow.net/questions/12758 | 18 | **Question.**
Let $C\_1,\dots,C\_k$ be conjugacy classes in the symmetric group $S\_n$. (More explicitly,
each $C\_i$ is given by a partition of $n$; $C\_i$ consists of permutations whose cycles
have the length prescribed by the partition.) Give a necessary and sufficient condition on
$C\_i$ that would ensure that there are permutations $\sigma\_i\in C\_i$ with
$$\prod\sigma\_i=1.$$
**Variant.**
Same question, but now $\sigma\_i$'s are required to be *irreducible* in the sense that
they have no common invariant proper subsets $S\subset\lbrace 1,\dots,n\rbrace$.
I am not certain how hard this question is, and I would appreciate any comments or observations. (I was unable to find references, but perhaps I wasn't looking for the right things.)
This question is inspired by [Jonah Sinick's question](https://mathoverflow.net/questions/12737/) via the simple
**Geometric interpretation.**
Consider the Riemann sphere with $k$-punctures $X=\mathbb{CP}^1-\lbrace x\_1,\dots,x\_k\rbrace$.
Its fundamental group $\pi\_1(X)$ is generated by loops $\gamma\_i$ ($i=1,\dots,k$) subject to the relation
$$\prod\gamma\_i=1.$$
Thus, homomorphisms $\pi\_1(X)\to S\_n$ describe degree $n$ covers of $X$, and the problem
can be stated as follows: Determine whether there exists a cover of $X$ with prescribed
ramification over each $x\_i$. The variant requires in addition the cover to be irreducible.
**Background.**
The Deligne-Simpson Problem refers to the following question:
Fix conjugacy classes $C\_1,\dots,C\_k\in\mathrm{GL}(n,\mathbb{C})$ (given explicitly by $k$ Jordan forms). What is the necessary and sufficient condition for existence of matrices
$A\_i\in C\_i$ with $$\prod A\_i=1$$
(variant: require that $A\_i$'s have no common proper invariant subspaces)?
There are quite a few papers on the subject; my favorite is [Simpson's paper](http://arxiv.org/abs/math/0610526), which has references to other relevant papers. The problem has a very non-trivial solution (even stating the answer is not easy): first there is a certain descent procedure (*Katz's middle convolution algorithm*) and then the answer is constructed directly (as far as I understand, there are two answers: Crawley-Boevey's argument with parabolic bundles, and Simpson's construction using non-abelian Hodge theory).
The same geometric interpretation shows that the usual Deligne-Simpson problem is
about finding local systems (variant: irreducible local systems) on $X$ with prescribed local monodromy.
So: any remarks on what happens if we go from $\mathrm{GL}(n)$ to $S\_n$?
| https://mathoverflow.net/users/2653 | Deligne-Simpson problem in the symmetric group | This question is a more than 100 years old problem and it is called in the litearture "Hurwitz exitence problem". This is an open problem. Though many partial cases are solved. For example you can check the following article
On the existence of branched coverings between surfaces with prescribed branch data, I
Ekaterina Pervova, Carlo Petronio
<http://arxiv.org/abs/math/0508434>
Of course, you can give an obvious formal answer, (in the first case that is not irreducible) that the cover exists if and only if the product of the elements in the group algebra of $S\_n$ corresponding to the permutation that you chose contains $1$ in their decomposition. But this is just a reformulation of the problem.
Here is a different example of a recent article on Hurwitz existence problem, it contains in partcular a lot of references on the research in this topic.
Solution of the Hurwitz problem for Laurent polynomials
<http://arxiv.org/abs/math/0611776>
Also notice that there is a whole branch of math nowadays where people try to compute the actual number of ramified covers, and not only to answer the quesiton wheather a cover exists or not. Here is a typical example
Gromov-Witten theory, Hurwitz theory, and completed cycles
Authors: Andrei Okounkov, Rahul Pandharipande
<http://arxiv.org/abs/math/0204305>
| 9 | https://mathoverflow.net/users/943 | 12759 | 8,633 |
https://mathoverflow.net/questions/12732 | 28 | There are a lot of theorems in basic homological algebra, such as the five lemma or the snake lemma, that seem like they'd be more easily proven by computer than by hand. This led me to consider the following question: **is the theory of categories decidable?**
More specifically, I was wondering whether or not statements about *abelian* categories can be determined true or false in finite time. Also, if they can be determined to be false, is it possible to explicitly describe a counterexample? If it is known to be decidable, is anything known about the complexity? (Other decidable theories often have multiply-exponential time complexities.) If it is known to be undecidable, say by embedding the halting problem, then can I change my assumptions a bit and make it decidable? (For example, maybe I shouldn't be looking at abelian categories after all.)
Thanks in advance.
**Edit**: It appears a clarification is needed. My goal was to consider the minimal theory that could *state* things like the five lemma, but not necessarily prove them. For example, I want to say:
>
> If in an abelian category, you have a bunch of maps $0\to A \to B \to C\to 0$ and $0\to A' \to B' \to C'\to 0$ which make up two short exact sequences and some more maps $a:A\to A'$, $b:B\to B'$, $c:C\to C'$ which commute with the previous maps, and $a$ and $c$ are isomorphisms, then $b$ is an isomorphism, too.
>
>
>
Sentences of this form would be inputs to a program, which decides if this statement is in fact true in ZFC (or your other favorite axiomatization of category theory). The point here is that I am restricting the *sentences* one can input into the program, but keeping ZFC or whatnot as my framework.
I hoped (perhaps naively) that if I restricted the class of sentences, it might be decidable whether or not these statements were true. For example, I imagined that every such theorem is either proven by diagram chasing, or it is possible to find a concrete example of maps among, say, R-modules that contradict the result.
| https://mathoverflow.net/users/1079 | Is the theory of categories decidable? | Thanks for clarifying your question. The formulation that
you and Dorais give seems perfectly reasonable. You have a
first order language for category theory, where you can
quantify over objects and morphisms, you can compose
morphisms appropriately and you can express that a given
object is the initial or terminal object of a given
morphism. In this language, one can describe various finite
diagrams, express whether or not they are commutative, and
so on. In particular, one can express that composition is
associative, etc. and describe what it means to be a
category in this way.
The question now becomes: is this theory decidable? In
other words, is there a computable procedure to determine,
given an assertion in this language, whether it holds in
all categories?
The answer is **No**.
One way to see this is to show even more: one cannot even
decide whether a given statement is true is true in all
categories having only one object. The reason is that group
theory is not a decidable theory. There is no computable
procedure to determine whether a given statement in the
first order language of group theory is true in all groups.
But the one-point categories naturally include all the
groups (and we can define in a single statement in the
category-theoretic language exactly what it takes for the
collection of morphisms on that object to be a group).
Thus, if we could decide category theory, then we could
decide the translations of the group theory questions into
category theory, and we would be able to decide group
theory, which we can't. Contradiction.
The fundamental obstacle to decidability here, as I
mentioned in my previous answer (see edit history), it the
ability to encode arithmetic. The notion of a [strongly
undecidable
structure](https://projecteuclid.org/euclid.lnl/1235423985)
is key for proving various theories are undecidable. A
strongly undecidable theory is a finitely axiomatizable
theory, such that any theory consistent with it is
undecidable. Robinson proved that there is a strongly
undecidable theory of arithmetic, known as Robinson's Q. A
*strongly undecidable structure* is a structure modeling a
strongly undecidable theory. These structures are amazing,
for any theory true in a strongly undecidable structure is
undecidable. For example, the standard model of arithmetic,
which satisfies Q, is strongly undecidable. If A is
strongly undecidable and interpreted in B, then it follows
that B is also strongly undecidable. Thus, we can prove
that graph theory is undecidable, that ring theory is
undecidable and that group theory is undecidable, merely by
finding a graph, a ring or a group in which the natural
numbers is interpreted. Tarski found a strongly undecidable
group, namely, the group G of permutations of the integers
Z. It is strongly undecidable because the natural numbers
can be interpreted in this group. Basically, the number n
is represented by translation-by-n. One can identify the
collection of translations, as exactly those that commute
with s = translation-by-1. Then, one can define addition as
composition (i.e. addition of exponents) and the divides
relation is definable by: i divides j iff anything that
commutes with si also commutes with
sj. And so on.
I claim similarly that there is a strongly undecidable
category. This is almost immediate, since every group can
be viewed as the morphisms of a one-object category, and
the group is interpreted as the morphisms of this category.
Thus, the category interprets the strongly undecidable
group, and so the category is also strongly undecidable. In
particular, any theory true in the category is also
undecidable. So category theory itself is undecidable.
| 37 | https://mathoverflow.net/users/1946 | 12760 | 8,634 |
https://mathoverflow.net/questions/12746 | 5 | Weak equivalences in the standard model structure on simplicial sets are allegedly closed under transfinite composition.
What's a reference for that?
| https://mathoverflow.net/users/381 | transfinite composition of weak equivalences in sSet | I don't have a complete reference (and like Tyler, I don't know exactly what result you want). But here are some observations:
* there is a functor $\mathrm{Ex}^\infty$, which replaces a simplicial set with a weakly equivalent fibrant replacement, and which commutes with filtered colimits. (See ch. 3 of Goerss-Jardine.)
* if you have a transfinite composition(s) in which all the simplicial sets are fibrant, it is straightforward to understand their behaviour with respect to weak equivalences, using the formula for simplicial homotopy groups, which gives the *right* homotopy groups for Kan complexes; (in particular, simplicial homotopy groups commute with filtered colimits in this setting.)
[Added later:] in particular, in any transfinite composition of weak equivalences between Kan complexes, the cocone componenent of the first object (i.e., the map from the first object to the colimit of the trasnfinite sequence) has to be a weak equivalence.
[Added:] These two facts taken together imply that a "transfinite composition" of weak equivalences is a weak equivalence.
| 7 | https://mathoverflow.net/users/437 | 12766 | 8,637 |
https://mathoverflow.net/questions/12769 | 3 | Given a partial order $R\_{\leq}$ over a set $D$, the set of upper bounds under $R$ of a subset $S$ of $D$ is commonly defined as $\{ y \in D | \ \forall x\in S, x R y \}$.
(The set of lower bounds of $S$ may be defined as the set of upper bounds of $S$ under the converse relation $R^{-1}$)
Is there a common name for the generalization of this notion where $R$ is not a partial order, and is possibly a heterogenous relation between domain $D$ and codomain $D'$ (hence the $y$ would be elements of the codomain)? This would be a subset of the image of $S$ under $R$ (and conversely, the dual notion would be a subset of the preimage).
| https://mathoverflow.net/users/3492 | Name for "lower/upper bounds" of arbitrary relations? | If your relation is at all order-like, then I would recommend just staying with the upper/lower bound terminology. And unless I misunderstand you, the example you describe is actually a (strict) partial order, no? If the relation only goes from disjoint sets D to D', then this is (vacuously) transitive, irreflexive and assymetric.
But in general, every binary relation is ultimately a directed graph. In this case, you would say that y is a *common direct successor* (or *common target*) of the elements of S if x R y for all x in X. And y is a *common direct predecessor* if y R x for all x in S.
If your relation is a tree, in one direction or the other, then you could use the *common parent* and *common child* terminology.
| 2 | https://mathoverflow.net/users/1946 | 12778 | 8,643 |
https://mathoverflow.net/questions/12767 | 9 | I have a little problem. I'm probably being just so careless..... Here k-varieties are all integral separated k-schemes of finite type over k, where k is a field.
Suppose $X, Y$ are $k$-varieties, and let $f :X \to Y$ be a morphism of $k$-varieties that is one to one and onto. Then, when can we say this $f$ is an isomorphism of $k$-varieties?
If this is too vague, let me add that the case I would like to see is when each fibre of $f$ (which is a singleton) is reduced. Under this assumption, would this give an isomorphism?
| https://mathoverflow.net/users/3168 | When two k-varieties with the same underlying topological spaces isomorphic? | The condition you are looking for is **seminormality**. A variety (or a reduced scheme) $Y$ is *seminormal* if any proper bijective morphism $f:X\to Y$, with $X$ reduced, *inducing isomorphisms on residue fields* $k(y)=k(x)$ for points $x\in X$, $y=f(x)\in Y$, is an isomorphism. A basic fact is that any variety has a unique seminormalization.
A related notion which differs only in positive characteristic is **weak normality** for which $k(y)\to k(x)$ is required to be purely inseparable and an isomorphism for each *generic point* $x\in X$.
One basic reference for this notion is the appendix to Chapter 1 of Koll'ar's "Rational curves on algebraic varieties", where you will find many standard facts and examples such as: normal implies seminormal; in dim 1 seminormal means analytically isomorphic to the $n$ axes in $A^n$; irreducible components of seminormal schemes need not be normal, etc. You will also find references to many papers where this notion was comprehensively investigated.
For clarity, let me add the standard fact: $f$ is proper and bijective $\iff$ it is *finite* and bijective (as opposed to quasifinite = finite fibers, which of course follows from bijective).
| 12 | https://mathoverflow.net/users/1784 | 12795 | 8,656 |
https://mathoverflow.net/questions/12763 | 2 | Given there is triangle: V in 3D space that transforms over time t -> t1 to V1, and a static point P is somewhere in 3d space, how can I determine if P ever collides with V, and if so at what value of t?
The transformation of the triangle from V -> V1 over time means the vertices Va,Vb,Vc each move linearly and independently in three dimensions to new vertex positions V1a,V1b,V1c. (Sorry about my notation). There are no constraints on the transformation of the vertices with respect to each other. The triangles at V and V1 representing the transformation may be any 2 triangles in 3d space.
| https://mathoverflow.net/users/3491 | find the collision of a particle with a swept triangle. | This type of problem is usually called "continuous collision detection". There is a substantial literature on this subject as you'll discover if you try a [google search](http://www.google.com/search?q=%22continuous+collision+detection%22) on those key words. It's rare that someone wants to do this for one triangle at a time. The literature has techniques for checking against large sets of triangles as well as generalisations to alternative shapes and more complex motion such as quadratic motion wrt time.
| 7 | https://mathoverflow.net/users/1233 | 12805 | 8,664 |
https://mathoverflow.net/questions/12804 | 24 | A cardinal $\lambda$ is weakly inaccessible, iff a. it is regular (i.e. a set of cardinality $\lambda$ can't be represented as a union of sets of cardinality $<\lambda$ indexed by a set of cardinality $<\lambda$) and b. for all cardinals $\mu<\lambda$ we have $\mu^+<\lambda$ where $\mu^+$ is the successor of $\mu$. Strongly inaccessible cardinals are defined in the same way, with $\mu^+$ replaced by $2^\mu$. Usually one also adds the condition that $\lambda$ should be uncountable.
As far as I understand, a "large cardinal" is a weakly inaccessible cardinal with some extra properties. In set theory one considers various "large cardinal axioms", which assert the existence of large cardinals of various kinds. Notice that these axioms are quite different from, say the Continuum Hypothesis. In particular, one can't deduce the consistency of ZFC + there exists at least one (uncountable) weakly inaccessible cardinal from the consistency of ZFC, see e.g. Kanamori, The Higher Infinite, p.19. I.e., assuming ZFC is consistent, these axioms can not be shown independent of ZFC.
The "reasonable" large cardinal axioms seem to be ordered according to their consistency strength, as explained e.g. here <http://en.wikipedia.org/wiki/Large_cardinal>. This is not a theorem, just an observation. A list of axioms according to their consistency strength can be found e.g. on p. 472 of Kanamori's book mentioned above. (Noticeably, it starts with "0=1", which is a very strong axiom indeed.)
Large cardinals appear to occur seldom in "everyday" mathematics. One such instance when they occur is when one tries to construct the foundations of category theory. One of the ways to do that (and the one that seems (to me) to be the most attractive) is to start with the set theory and to add Grothendieck's Universe axiom, which states that every set is an element of a Grothendieck universe.
(As an aside remark, let me mention another application of large cardinal axioms: incredibly, the fastest known solution of the word problem in braid groups originated from research on large cardinal axioms; the proof is independent of the existence of large cardinals, although the first version of the proof did use them. See Dehornoy, From large cardinals to braids via distributive algebra, Journal of knot theory and ramifications, 4, 1, 33-79.)
Translated into the language of cardinals, the Universe axiom says that for any cardinal there is a strictly larger strongly inaccessible cardinal. I have heard several times that this is pretty low on the above consistency strength list, but was never able to understand exactly how low. So I would like to ask: does the existence of a (single) large cardinal of some kind imply (or is equivalent to) the Universe axiom?
| https://mathoverflow.net/users/2349 | Large cardinal axioms and Grothendieck universes | A Grothendieck universe is known in set theory as the set Vκ for a (strongly) inaccessible cardinal κ. They are exactly the same thing. Thus, the existence of a Grothendieck universe is exactly equivalent to the existence of one inaccessible cardinal. These cardinals and the corresponding universes have been studied in set theory for over a century.
The Grothendieck Universe axiom (AU) is the assertion that every set is an element of a universe in this sense. Thus, it is equivalent to the assertion that the inaccessible cardinals are unbounded in the cardinals. In other words, that there is a proper class of inaccessible cardinals. This is the axiom you sought, which is exactly equivalent to AU. In this sense, the axiom AU is a statement in set theory, having nothing necessarily to do with category theory.
The large cardinal axioms are fruitfully measured in strength not just by direct implication, but also by their *consistency* strength. One large cardinal property LC1 is stronger than another LC2 in consistency strength if the consistency of ZFC with an LC1 large cardinal implies the consistency of ZFC with an LC2 large cardinal.
Measured in this way, the AU axiom has a stronger consistency strength than the existence of any finite or accessible number of inaccessible cardinals, and so one might think it rather strong. But actually, it is much weaker than the existence of a single Mahlo cardinal, the traditional next-step-up in the large cardinal hierarchy. The reason is that if κ is Mahlo, then κ is a limit of inaccessible cardinals, and so Vκ will satisfy ZFC plus the AU axiom. The difference between AU and Mahloness has to do with the *thickness* of the class of inaccessible cardinals. For example, strictly stronger than AU and weaker than a Mahlo cardinal is the assertion that the inaccessible cardinals form a stationary proper class, an assertion known as the Levy Scheme (which is provably equiconsistent with some other interesting axioms of set theory, such as the boldface Maximality Principle, which I have studied a lot). Even Mahlo cardinals are regarded as rather low in the large cardinal hierarchy, far below the weakly compact cardinals, Ramsey cardinals, measurable cardinals, strong cardinals and supercompact cardinals. In particular, if δ is any of these large cardinals, then δ is a limit of Mahlo cardinals, and certainly a limit of strongly inaccessible cardinals. So in particular, Vδ will be a model of the AU axiom.
Rather few of the large cardinal axioms imply AU directly, since most of them remain true if one were to cut off the universe at a given inaccessible cardinal, a process that kills AU. Nevertheless, implicit between levels of the large cardinal hierarchy are the axioms of the same form as AU, which assert an unbounded class of the given cardinal. For example, one might want to have unboundedly many Mahlo cardinals, or unboundedly many measurable cardinals, and so on. And the consistency strength of these axioms is still below the consistency strength of a single supercompact cardinal. The hierarchy is extremely fine and intensely studied. For example, the assertion that there are unboundedly many strong cardinals is equiconsistent with the impossibility to affect projective truth by forcing. The existence of a proper class of Woodin cardinals is particularly robust set-theoretically, and all of these axioms are far stronger than AU.
There are natural weakenings of AU that still allow for almost all if not all of what category theorists do with these universes. Namely, with the universes, it would seem to suffice for almost all category-theoretic purposes, if a given universe U were merely a model of ZFC, rather than Vκ for an inaccessible cardinal κ. The difference is that U is merely a model of the Power set axiom, rather than actually being closed under the true power sets (and similarly using Replacement in place of regularity). The weakening of AU I have in mind is the axiom that asserts that every set is an element of a transitive model of ZFC. This assertion is strictly weaker in consistency strength that even a single inaccessible cardinal. One can get much lower, if one weakens the concept of universe to just a fragment of ZFC. Then one could arrive at a version of AU that was actually provable in ZFC, but which could be used for most all of the applications in category theory to my knowledge. In this sense, ZFC itself is a kind of large cardinal axiom relative to the weaker fragments of ZFC.
| 31 | https://mathoverflow.net/users/1946 | 12809 | 8,668 |
https://mathoverflow.net/questions/12814 | 18 | I know this should be pretty simple, but right now the only way I can see how to prove it is to sit down and write out explicit formulae for the group law, and see that everything works out. What's the geometric or abstract-nonsense reason why the abelian group structure of elliptic curves behaves nicely under homomorphisms?
| https://mathoverflow.net/users/382 | Why does the group law commute with morphisms of elliptic curves? | This follows by a rigidity property of certain morphisms. It is important to note that elliptic curves are complete, that is, proper and integral schemes. Then we have the following "Rigidity Lemma" (see Mumford's Abelian Varieties, the beginning of chapter II (page 43 of the old edition), for example):
Let $X$ be a complete variety, $Y$ and $Z$ any varieties, and $f:X\times Y\rightarrow Z$ a morphism such that for some $y\_0\in Y$, $f(X\times\{y\_0\})$ is a single point $z\_0$ of $Z$. Then there is a morphism $g:Y\rightarrow Z$ such that if $p\_2:X\times Y\rightarrow Y$ is the projection, $f=g\circ p\_2$.
How does this help? Well, elliptic curves have a distinguished point (the origin), and a morphism of elliptic curves is a morphism of their underlying schemes taking one distinguished point to the other. We'd like to show that any such morphism is actually compatible with the group law. So consider a morphism of elliptic curves $f:E\_1\rightarrow E\_2$, and let $\Phi:E\_1\times E\_1\rightarrow E\_2$ be defined by $\Phi(x,y)=f(x+y)-f(x)-f(y)$. Then $\Phi(E\_1\times\{0\})=0$ so by the lemma I quoted, there is a morphism $g:E\_1\rightarrow E\_2$ such that $\Phi=g\circ p\_2$. And since $\Phi(\{0\}\times E\_1)=0$, as well, $g$ must be zero. But then $\Phi$ itself must be zero.
None of this relied on $E\_1$ and $E\_2$ being elliptic curves; it works exactly the same way for abelian varieties.
| 33 | https://mathoverflow.net/users/88 | 12818 | 8,673 |
https://mathoverflow.net/questions/12819 | 9 |
>
> A graph is Hamiltonian if and only if
> its closure is Hamiltonian.
>
>
>
I am looking for a simple (i.e. short) proof of the theorem, that I can use as part of an article on topological sorting.
I've not been able to find one in the literature I have access to. Any help would be appreciated.
| https://mathoverflow.net/users/3499 | Proof of Bondy and Chvátal Theorem | Let $G=G\_0, G\_1, G\_2$, etc. be a sequence of graphs where each $G\_i$ is formed by performing a single closure step to $G\_{i-1}$ — that is, add an edge $uv$ to $G\_i$ when $u$ and $v$ together have at least $n$ neighbors. If any graph in this sequence is Hamiltonian, let $k$ be the minimum $k$ such that $G\_k$ is Hamiltonian. Then I claim that $k=0$. For, otherwise let $uv$ be the edge added to form $G\_k$ from $G\_{k-1}$ and let $C$ be a Hamiltonian cycle in $G\_k$. There are $n-1$ other edges in $C$, and $n$ edges going out from $u$ and $v$ together, so by the pigeonhole principle there exists an edge $pq$ in $C$ (with the vertex labeling chosen so that $u$ is clockwise of $v$ and $p$ is clockwise of $q$) such that $G\_{k-1}$ contains edges $up$ and $vq$. But then $C + up + vq - pq - uv$ is a Hamiltonian cycle in $G\_{k-1}$ contradicting the assumption that $k>0$.
| 12 | https://mathoverflow.net/users/440 | 12821 | 8,675 |
https://mathoverflow.net/questions/12482 | 3 | I'm trying to understand the topology behind a certain group which fits into a truncated crossed complex, so I've been trying to understand Brown's construction of the classifying space of a crossed complex. I asked a similar question [here](https://mathoverflow.net/questions/2734/classifying-space-of-a-crossed-complex) and the answer cleared up some of my misunderstanding.
What I'm looking for now is an explicit construction of such a space. I've not found any papers that give explicit examples of such a construction, but hopefully there is one somewhere. I'm looking for something along the lines of the easy to follow of the construction of the classifying space of cyclic groups.
| https://mathoverflow.net/users/343 | Explicit classifying spaces for crossed complexes | It is not clear to me what you need / want. The classifying space of a cyclic group is constructed using a presentation and then killing off higher identities that may be around (there aren't any!). From that viewpoint the question you seem to ask is related to the combinatorial group theory of the group in question (or am I misderstanding the question.) There are examples that might help due to Loday in his paper on higher syzygies, but that may not quite fit the bill as he does not explicitly give the link with classifying spaces.
If you are happy with simplicial methods then you can build a simplicial T-complex from a crossed complex of groups by a modified Dold-Kan construction. The classifying space of that simplicial group (its Wbar) is something that has the same properties as Ronnie's classifying space. It is feasible if you know the crossed complex reasonably fully to construct this explicitly.
It depends what you need the construction for? How is your crossed complex arising precisely (and incidently what do you mean by the `topology', will a simplicial model do)?
| 2 | https://mathoverflow.net/users/3502 | 12850 | 8,695 |
https://mathoverflow.net/questions/12861 | 13 | Let G be a finite group and let F be an algebraically closed field. If the characteristic of F is 0, then the number of irreducible F-representations of G is given by the number of conjugacy classes of elements of G. A paper I'm reading says that if the characteristic of F is p>0, then the number of F-irreps of G is the same as the number of conjugacy classes of elements whose order is not divisible by p.
If G is abelian, it seems to me that this should say that the p-sylow subgroup of G acts trivially on every characteristic p irrep. This is because I can split G into G'x P (non-p and p-sylow subgroups), and then any irrep of G' extends to one of G by letting P act trivially. Since the formula mentioned above would say that they have the same number of irreps these must be all of them.
My question is: Is this true? If not, then where is my reasoning going off track?
| https://mathoverflow.net/users/493 | Representations in characteristic p | Yes, you are correct. The point is that a $p$-group acting in char. $p$ always has
a fixed point (and so acts trivially on an irrep.). So *every* irrep. of $G$
in char. $p$ factors through $G'$, as you anticipated.
The proof of the claim about $p$-groups is not hard. One approach (in general,
even when $P$ is not nec. abelian) is to prove it
by induction on the order of $P$, and so (using the fact that the descending central
series is non-trivial) reduce to the case when $P$ is cyclic of order $p$. In this
case, one can for example look at the group ring $k[P] = k[t]/(t^p - 1) = k[t-1]/(t-1)^p,$
and observe that it has a unique maximal ideal, and hence that $P$ has a unique
irrep., namely the trivial one.
| 19 | https://mathoverflow.net/users/2874 | 12862 | 8,703 |
https://mathoverflow.net/questions/12765 | 30 | I have a pretty good understanding of stacks, sheaves, descent, Grothendieck topologies, and I have a decent understanding of commutative algebra (I know enough about smooth, unramified, étale, and flat ring maps). However, I've never seriously studied Algebraic geomtry. Can anyone recommend a book that builds stacks directly on top of CRing in a (pseudo)functor of points approach? Typically, one builds up stacks segmentwise, first constructing Aff as the category of sheaves of sets on CRing with the canonical topology, which gives us CRing^op. Then, one constructs the Zariski topology on Aff, and from that constructs Sch, then one equips Sch with the étale topology and constructs algebraic stacks above that. (I assume that one gets Artin stacks if one replaces the étale topology there with the fppf topology?)
Does anyone know of a book/lecture notes/paper that takes this approach, where everything is just developed from scratch in the language of categories, stacks, and commutative algebra?
Edit: Some motivation: It seems like many of the techniques used to build the category of schemes in the first place are just less generalized versions of the constructions for algebraic stacks. So the idea is to develop all of algebraic geoemtry in "one fell swoop", so to speak.
Edit 2: As far as answers go, I'm not really interested in seeing value judgements about this approach. I know that it's at best a controversial approach, but I've seen all of the arguments against it before.
Edit 3: Part of the motivation for this question comes from a (possibly incorrect) footnote on [Wikipedia](http://en.wikipedia.org/wiki/Algebraic_space):
>
> One can always assume that U is an affine scheme. Doing so means that the theory of algebraic spaces is not dependent on the full theory of schemes, and can indeed be used as a (more general) replacement of that theory.
>
>
>
If this is true, then at least we can avoid most of the trouble Anton says we'll go through in his comment below. However, this being true seems to indicate that we should be able to do the same thing for algebraic stacks.
Edit 4: Since Felipe made his comment on this post, everyone has just been "voting up the comment". Since said comment was a question, I'll just post a response.
Mainly because I study category theory on my own time, and I've taken commutative algebra courses.
Now that that's over and done with, I've also added a bounty to this question.
| https://mathoverflow.net/users/1353 | Algebraic stacks from scratch | Another good place to look are the notes of [Master's course on stacks by Betrand Toen](https://ncatlab.org/nlab/show/Master+course+on+algebraic+stacks). I think they pretty much do exactly what you are looking for.
Here's the quick summary: You will want to read section 1 of Cours 2 where the term geometric context is defined. It's basically a category with a Grothendieck topology with a fixed class of morphisms that you call geometric. The main example are commutative rings with the etale topology or the smooth topology. This induces coverings in the presheaf category in the standard way.
Then skip straight to Cours 5. Although you said that you are comfortable with descent this section is definitely worth a close look. It introduces a homotopy theory on the category of groupoids and shows that there always is a weakly equivalent groupoid such that your functor becomes strict. It then reformulates descent via homotopy limits. The upshot is a nice category of stacks, Definition 4.4.
Then jump straight to Cours 8, Definition 1.4. and you've got algebraic stacks. The only point where you will need schemes or algebraic spaces is for representable morphisms, but judging from the remark after the definition you can get around that as well.
| 17 | https://mathoverflow.net/users/473 | 12866 | 8,705 |
https://mathoverflow.net/questions/12865 | 26 | As everybody knows, the ZFC axioms may serve as a foundation for (almost)
all of contemporary mathematics, and it is also well-known that several results
are "indecidable" in ZFC, which means that they cannot be proved or disproved within
ZFC.
It is therefore natural to look for "new axioms" to add to ZFC and make it a
stronger system. But by Godel's second incompleteness theorem, the consistency of
ZFC cannot be deduced from ZFC itself.
Therefore, we may add the axiom "ZFC is consistent" and obtain
a new system $ ZFC\_1 $ consisting of "ZFC+(ZFC is consistent)". We may
iterate this, and define $ ZFC\_2 $ as "$ZFC\_1$+($ZFC\_1$ is consistent)", etc,
and we may even define $ZFC\_{\omega}$, or $ZFC\_{\alpha}$ for any
ordinal $\alpha$.
This seems a little too easy, so my question to logicians is : is this construction
completely irrelevant to logic ans set theory questions ? If so why? Is it true that the results which are classically independent of ZFC are also
independent of $ZFC\_1$, $ZFC\_2$, $ZFC\_{\omega}$ etc ?
| https://mathoverflow.net/users/2389 | Using consistency to create new axioms in set theory | Such constructions are interesting! However, they are often done with PA instead of ZFC (see note). For an interesting discussion, I recommend Torkel Franzén's book *Inexhaustibility: a non-exhaustive treatment* (Lecture Notes in Logic 16, ASL, 2004). You can also read this excellent [blog article](http://xorshammer.com/2009/03/23/what-happens-when-you-iterate-godels-theorem/) by Mike O'Connor.
---
Note: The following is explained in Mike O'Connor's article, but I think I need to clarify why ZFC is not the ideal base theory to do this and why PA is a better candidate.
The idea is that Con(T) is usually understood as an arithmetical statement. More precisely, given a recursive presentation of the theory T the statement Con(T) is arithmetical formalization of "there is no proof of a contradiction from T" which is encoded using Gödel numbers for proofs and formulas. (This is the messy part of Gödel's Theorem.) This is why PA, or more generally any recursively axiomatizable extension of PA, provides a more natural environment for the analysis of such statements. For example, instead of ZFC, you may as well use the purely arithmetical part of ZFC.
There is also an even more fundamental problem with transfinite iterates. Given a recursive presentation of a theory T, the iterates T0 = T, T1 = T0 + Con(T0), T2 = T1 + Con(T1), etc. Can be continued into the transfinite, but only to a limited extent. It is easy to give a recursive presentation of Tω or Tω+ω+3 but there are only countably many ordinals for which this works. Indeed, these iterates are better defined in terms of [ordinal notations](http://en.wikipedia.org/wiki/Ordinal_notation) than in terms of proper ordinals. Ordinal notations can go pretty far up, but there are clear limitations.
These difficulties and their implications are discussed in great detail in Franzén's book. As Mike O'Connor explains, it is natural to go further and extend these to [subsystems of second-order arithmetic](http://en.wikipedia.org/wiki/Second-order_arithmetic), but there and in set theory the appropriate principles to study are reflection principles and large cardinal axioms which have better semantic interpretations and take advantage of their richer surroundings.
| 30 | https://mathoverflow.net/users/2000 | 12867 | 8,706 |
https://mathoverflow.net/questions/12847 | 15 | Suppose a compact Lie group $G$ acts on a manifold $M$ with only one orbit type $G/H$ ($H$ denotes the stabiliser group). Then the manifold $M$ becomes a fibre bundle over the quotient manifold $X:=M/G$ with typical fibre $G/H$ and structure group $G$.
On the one hand one could look at the cotangent bundle $T^\* X$ of the quotient (which carries a natural symplectic structure).
On the other hand consider the lifted action of $G$ on the cotangent bundle $T^\* M$ with moment map $\mu: T^\* M\to \mathfrak{g}^\* .$ The symplectic quotient $T^\* M//G:=\mu^{-1}(0)/G$ inherits the structure of a symplectic manifold. Here comes the question: Are $T^\* X$ and $T^\* M//G$ (canonically) symplectomorphic?
| https://mathoverflow.net/users/3509 | cotangent bundle symplectic reduction and fibre bundles | These two symplectic manifolds are canonically symplectomorphic.
Notice first, that the map $\mu$ vanishes on the sub-bundle of $T^\* M$ of 1-forms vanishing on the fibers of the fibration $M\to X$. Let us call this sub-bundle by $T\_h ^\* M$ (h- for horizontal).
To construct the symplectomorphism notice that there is an obvious projection $\pi: T\_h^\* M \to T^\* X$. The restriction of the symplectic form of $T^\* M$ to $ T\_h^\* M$ equals to the pullback of the symplectic form of $T^\* X$ under $\pi$. The projection $\pi$ commutes with the action of $G$ and $G$ preserves the symplectic form on $T^\* M$. Since the projection $\pi$ just produces the quotient of $T\_h^\*M$ by the action of $G$, now everything follows from definitions.
| 13 | https://mathoverflow.net/users/943 | 12868 | 8,707 |
https://mathoverflow.net/questions/12871 | 7 | Suppose $G$ is a finite group, and $l$ is a prime, with $l$ coprime to the order $|G|$. (Thus we have complete reducibility for $G$ representations.) Is there a straightforward condition on $l$ which ensures that every irreducible representation of $G$ is liftable to a characteristic zero representation? (For instance, does the fact that we assume $l$ coprime to $|G|$ suffice?)
| https://mathoverflow.net/users/3513 | When are all characteristic l representations liftable | Yes, $\operatorname{gcd}(l,|G|) = 1$ is sufficient. This is an easy consequence of Brauer's modular representation theory. See Serre's *Linear Representations of Finite Groups*, especially Chapter 18.
| 10 | https://mathoverflow.net/users/1149 | 12872 | 8,709 |
https://mathoverflow.net/questions/12889 | 6 | Let $h:M\to M$ be a homeomorphism of a compact manifold. Let $p:\tilde M\to M$ be a covering. 1) Is it always possible to lift $h$ to $H:\tilde M\to \tilde M$ so that everything fits into the commutative diagram?
2) Given such a diagram assume additionally that p is a self-covering. Is it true that $H$ is necessarily homotopic to $h$?
Thanks, Z.
1/I think I can see that the answer to the second question is "no". Any additional assumptions that would make it into a "yes"?
2/A reference to the proof of the statement in Ben's second paragraph is needed.
| https://mathoverflow.net/users/3375 | Lifting a homeomorphism, always possible? | No. Take any homeomorphism that doesn't preserve the subgroup of $\pi\_1$ that lift to closed paths in the covering. For example, take the 2:1 covering $S^1\to S^1$ take the product with the identity map on $S^1$. Let $h$ be the homeomorphism switching the factors.
In general, I believe a homeomorphism will lift if and only if the associated automorphism of $\pi\_1$ send the subgroup of the covering to a conjugate.
Another way of saying this is that the category of coverings is equivalent to the category of $\pi\_1$-sets, and a homeomorphism will lift if the corresponding twist of the $\pi\_1$-set preserves its isomorphism class.
| 9 | https://mathoverflow.net/users/66 | 12890 | 8,718 |
https://mathoverflow.net/questions/12892 | 5 | I'm posting my answer to [this question](https://mathoverflow.net/questions/5146/algebraic-geometry-versus-complex-geometry) as its own question:
Let $V$ be an irreducible projective variety over $\mathbb{C}$. Let $U$ be a Zariski open set in $V$. I'll use $V(\mathbb{C})$ and $U(\mathbb{C})$ to mean $V$ and $U$ equipped with their Euclidean topologies, respectively.
What is the easiest proof that $U(\mathbb{C})$ is connected?
Here's the proof I know: Suppose that $U(\mathbb{C})$ can be written as a disjoint union of two open sets $A$ and $B$. Since the complement of $U$ in $V$ is a variety of smaller dimension than $V$, a theorem of Remmert and Stein implies that the closures $\overline{A}$ and $\overline{B}$ of $A$ and $B$ in $V(\mathbb{C})$ are projective analytic sets. By Chow's theorem that projective analytic sets are algebraic, $\overline{A}$ and $\overline{B}$ are subvarieties of $V$. Since they're proper, $V$ is not irreducible, and we have a contradiction.
I guess I'm really asking for the most elementary argument, as I think the above argument is nice intuitively. A reference would be fine.
(To avoid going through the same discussion in the comments that happened at the other question, let me point out that I am aware that irreducible varieties are connected and that $U$ is itself a variety in the sense that it is locally affine. It is just not obvious to me that it is irreducible (without appealing to the above argument).)
| https://mathoverflow.net/users/1335 | Why can't subvarieties separate? | [This has been completely rewritten at the request of Autumn Kent.]
Let $X$ be an irreducible topological space and $U$ a non-empty open subset of $X$. Then $U$ is also irreducible -- see e.g. Proposition 141 on page 88 [here](http://alpha.math.uga.edu/%7Epete/integral.pdf). (Surely it's also in Hartshorne and lots of other places, but one of the advantages of typing up your own notes is to be able to easily point to a reference because you know exactly where it is.)
Thus the question reduces to the fact that if $X\_{/\mathbb{C}}$ is an irreducible complex variety, then $X(\mathbb{C})$ with its "Euclidean topology" is connected. For this, see e.g. Section VII.2 of Shafarevich's *Basic Algebraic Geometry II*. (Again, there are other places, but I think his discussion is especially good.) He gives two different proofs, one of which is a simple induction on the dimension.
| 10 | https://mathoverflow.net/users/1149 | 12896 | 8,721 |
https://mathoverflow.net/questions/12894 | 40 | I am writing a short paper in the area of combinatorics.
When the paper is complete, I would like to be able to submit it to arXiv.
The reasons that I would like to submit to arXiv are:
1. To obtain a date and time stamp from a central authority so that I can prove the work is mine.
2. To promote access to the paper under the Creative Commons Attribution-Noncommercial-ShareAlike license
Although I have created an account on arXiv, because I am not affiliated to any academic institution, it appears that it might be difficult to get the paper accepted by arXiv.
For example see,
arxiv.org/help/submit
>
> The following information is also required for submission:
>
>
> Institutional affiliation for the
> author(s) must be provided. Official
> report number(s) from the author(s)
> institution(s) must be provided.
>
>
>
Also the following blog article does not appear encouraging:
[Does The Arxiv Blacklist Authors ?](http://www.scientificblogging.com/quantum_diaries_survivor/does_arxiv_blacklist_authors_help_finding_out)
Does anyone on math*overflow* have any advice on how to submit to arXiv (or a similar database) without an academic affiliation?
Thanks in advance for any advice.
| https://mathoverflow.net/users/126024 | Submitting to arXiv when unaffiliated | I think you need to be endorsed first. See this link: <http://arxiv.org/help/endorsement>
| 17 | https://mathoverflow.net/users/2264 | 12897 | 8,722 |
https://mathoverflow.net/questions/12906 | 1 | The collection of all groups is a proper class, since every set gives rise to a group. But what about the collection of all isomorphism classes of groups? By which argument do I see, that it is a set or a proper class?
| https://mathoverflow.net/users/2672 | Is the collection of isomorphism classes of groups a proper class? | Since free groups are isomorphic if and only if they have a basis of the same cardinality (probably assuming some choice axiom here), the collection of all isomorphism classes of groups has at least the size of the collection of all isomorphism classes of sets, hence is not a set.
| 10 | https://mathoverflow.net/users/45 | 12909 | 8,729 |
https://mathoverflow.net/questions/12902 | 4 | This is a spin-off question from [How to select a journal?](https://mathoverflow.net/questions/7284/how-to-select-a-journal). Is there is any data available regarding processing time (acceptance time, time from submission to publication, or similar) specifically for combinatorics journals? Ideally, this data would be up-to-date.
| https://mathoverflow.net/users/2264 | Combinatorics journals processing time | One possibility is the "backlog" published annually in the AMS Notices. The latest version is [here (as PDF)](http://www.ams.org/notices/200910/rtx091001316p.pdf).
| 8 | https://mathoverflow.net/users/45 | 12917 | 8,734 |
https://mathoverflow.net/questions/12782 | 15 | **Situation**
Let $G$ be a finite group and provide $G\text{-mod} := {\mathbb Z}G\text{-mod}$ with the Frobenius structure of ${\mathbb Z}$-split short exact sequences. Denote by $\underline{G\text{-mod}}$ the associated stable category with loop functor $\Omega$.
For any Frobenius category $({\mathcal A},{\mathcal E})$ and a complete projective-injective resolution $P\_{\bullet}$ of some $X\in{\mathcal A}$, we have for any $Y\in{\mathcal A}$ a canonical isomorphism of abelian groups
$H^n(\text{Hom}\_{\mathcal A}(P\_{\bullet},Y))\cong [\Omega^n X,Y]$,
where $[-,-] := \text{Hom}\_{\underline{{\mathcal A}}}(-,-)$.
Applying this to $G\text{-mod}$ yields an isomorphism
$\widehat{H}^k(G;M)\cong [\Omega^k{\mathbb Z},M]$,
where $\widehat{H}^k(G;M)$ denotes the Tate-Cohomology of $G$ with values in $M$.
If I didn't mix things up, in this language Tate-Duality should mean that the canonical map
$[{\mathbb Z},\Omega^k{\mathbb Z}]\otimes\_{\mathbb Z}[\Omega^k{\mathbb Z},{\mathbb Z}]\to[{\mathbb Z},{\mathbb Z}]\cong{\mathbb Z}/|G|{\mathbb Z}$
is a duality.
**Question**
I'd like to know sources which introduce and treat Tate cohomology in the way described above, i.e. using the language of Frobenius categories and its associated stable categories. *In particular*, I would be interested in a proof of Tate Duality using this more abstract language instead of resolutions.
Does anybody know such sources?
**Remark**
It seems to be more difficult to work over the integers instead of some field, for in this case, the exact sequences in the Frobenius structure $G\text{-mod}$ are required to be ${\mathbb Z}$-split, which is not automatic. As a consequence, there may be projective/injective objects in $(G\text{-mod},{\mathcal E}^{G}\_{\{e\}})$ which are not projective/injective as ${\mathbb Z}G$-modules. Further, the long exact cohomology sequence exists only for ${\mathbb Z}$-split exact sequences of $G$-modules (not good, because Brown uses the exact sequence $0\to {\mathbb Z}\to{\mathbb Q}\to{\mathbb Q}/{\mathbb Z}\to 0$ in his proof of Tate duality); of course, one can choose particular complete resolutions of ${\mathbb Z}$ consisting of ${\mathbb Z}G$-projective modules, and such a resolution yields a long exact cohomology sequence for any short exact sequence of coefficient modules, but this seems somewhat unnatural and doesn't fit into the picture right now.
**Partial Results**
**(1)** For any subgroup $H\leq G$ there are restriction and corestriction morphisms
$[\Omega^k {\mathbb Z},-]^{\underline{G}}=\widehat{H}^\*(G;-)\leftrightarrows\widehat{H}^\*(H;-)=[\Omega^k{\mathbb Z},-]^{\underline{H}}$
defined as follows: for any $G$-module $M$, the abelian group $[{\mathbb Z},M]^{\underline{G}}$ is in canonical bijection with $M^G / |G| M^G$, and there are restriction and transfer maps
$\text{res}: M^G / |G| M^G\longrightarrow M^H / |H| M^H,\quad [m]\mapsto [m]$,
$\text{tr}: M^H / |H| M^H\longrightarrow M^G / |G| M^G\quad [m]\mapsto\left[\sum\limits\_{g\in G/H} g.m\right]$,
respectively. Now
$[\Omega^k{\mathbb Z},M]^{\underline{G}}\cong [{\mathbb Z},\Omega^{-k}M]^{\underline{G}}\stackrel{\text{res}}{\longrightarrow} [{\mathbb Z},\Omega^{-k}M]^{\underline{H}}\cong[\Omega^k{\mathbb Z},M]^{\underline{H}}$
$[\Omega^k{\mathbb Z},M]^{\underline{H}}\cong [{\mathbb Z},\Omega^{-k}M]^{\underline{H}}\stackrel{\text{tr}}{\longrightarrow} [{\mathbb Z},\Omega^{-k}M]^{\underline{G}}\cong[\Omega^k{\mathbb Z},M]^{\underline{G}}$
seems to be the natural thing to define restriction and transfer. (This is very similar to the usual method of giving a morphism of $\delta$-functors only in degree $0$ and extend it by dimension shifting, though a bit more elegant in my opinion)
Note that it was implicitly used that $\Omega^k$ commutes with the forgetful functor $G\text{-mod}\to H\text{-mod}$
**(2)** For any subgroup $H\leq H$, $g\in G$ and a $G$-module $M$ there is a map
$g\_\*:\ \widehat{H}^\*(H;-)\to\widehat{H}^\*(gHg^{-1};M)$
extending the canonical map
$M^H/|H|M^H\longrightarrow M^{gHg^{-1}}/|H|M^{gHg^{-1}},\quad [m]\mapsto [g.m]$.
(1) and (2) fit together in the usual way; there is a transfer formula and a lifting criterion for elements of Sylow-subgroups.
**(3)** The cup product on $\widehat{H}^\*(G;{\mathbb Z})$ is given simply by composition of maps:
$[\Omega^p{\mathbb Z},{\mathbb Z}]\otimes\_{\mathbb Z}[\Omega^q{\mathbb Z},{\mathbb Z}]\stackrel{\Omega^q\otimes\text{id}}{\longrightarrow}[\Omega^{p+q}{\mathbb Z},\Omega^q{\mathbb Z}]\otimes\_{\mathbb Z}[\Omega^q{\mathbb Z},{\mathbb Z}]\longrightarrow [\Omega^{p+q}{\mathbb Z},{\mathbb Z}]$
Does anybody see why this product is graded-commutative?
| https://mathoverflow.net/users/3108 | Tate Cohomology via stable categories | To address Hanno's question about checking that composition gives a graded-commutative ring structure on $End^{\*}(\mathbb{Z}) = \oplus\_i [\mathbb{Z}, \Omega^{-i} \mathbb{Z}]$ suppose first that
$a \stackrel{f}{\to} b \stackrel{g}{\to} c \stackrel{h}{\to} \Omega^{-1} a$
is a distinguished triangle in the stable category. Then we can produce from this two isomorphic triangles: one by rotation namely
$a \stackrel{-\Omega^{-1}f}{\to} b \stackrel{-\Omega^{-1}g}{\to} c \stackrel{-\Omega^{-1}h}{\to} \Omega^{-1} a$
and one by applying $\Omega^{-1}\mathbb{Z} \otimes\_\mathbb{Z}$
$\Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z} a \stackrel{1\_{\Omega^{-1}\mathbb{Z}}\otimes f}{\to} \Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z} b \stackrel{\Omega^{-1}\mathbb{Z}\otimes g}{\to}\Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z} c \stackrel{\Omega^{-1}\mathbb{Z}\otimes h}{\to} \Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z} \Omega^{-1}a$
The point of this is that the natural isomorphism commuting the loops functor across introduces a sign change, which is precisely the one you pick up by changing the order of composition since changing the composition order is equivalent to applying symmetry isomorphisms to the tensor product which is equivalent to commuting loops across.
To be completely explicit about this there are two functors naturally isomorphic to $\Omega^{-1}$ namely $\Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z}(-)$ and $\mathbb{Z}\otimes \Omega^{-1}(-)$ since $\otimes\_\mathbb{Z}$ is biexact there are natural transformations commuting $\Omega^{-1}$ with $\otimes\_\mathbb{Z}$ namely
$\Omega^{-1}(-)\otimes\_\mathbb{Z} (-) \stackrel{\sim}{\rightarrow} \Omega^{-1}((-)\otimes\_\mathbb{Z}(-)) \stackrel{\sim}{\leftarrow} (-)\otimes\_\mathbb{Z} \Omega^{-1}(-)$
Our example triangles above which can be obtained from one another by first moving the loops to the right and then applying the unit transformation show that there must be a sign attached to this map for this to give an isomorphism of these triangles. In particular for $\otimes\_\mathbb{Z}$ to be compatible with the triangulated structure the two natural isomorphisms moving $\Omega$ about must have different signs. These natural isomorphisms are the precise cause of the sign change.
| 4 | https://mathoverflow.net/users/310 | 12924 | 8,737 |
https://mathoverflow.net/questions/12920 | 24 | Maybe this is an elementary question, but I'm unable to find the appropriate reference for it. The *Stokes theorem* tells us that, for a $n+1$-dimensional manifold $M$ with boundary $\partial M$ and any differentiable $n$-form $\omega$ on $M$, we have $\int\_{\partial M} \omega = \int\_M d\omega $.
But Stokes theorem is also true, say, for a cone $M = \{(x,y,z) \in \mathbb{R}^3 \ \vert\ \ x^2 + y^2 = z^2, 0\leq z \leq 1 \}$, or a square in the plane, $M =\{(x,y) \in \mathbb{R}^2 \ \vert\ 0 \leq x, y\leq 1 \}$ which are not manifolds. So my questions are:
1. Are these cone and square examples of what I think are called "manifold with corners"?
2. If this is so, where can I find a reference for a version of Stokes' theorem for manifolds with corners?
3. If "manifold with corners" is not, which is the appropriate setting (and a reference) for a Stokes' theorem that includes those examples?
Any hints will be appreciated.
EDIT: Since thanking individually everyone would be too long, let me edit my question to acknowledge all of your answers. Thank you very much: I've found what I was looking for and more.
| https://mathoverflow.net/users/1246 | Stokes theorem for manifolds with corners? | The most general form of Stokes' theorem I know of is proved in the book
Partial Differential Equations 1. Foundations and Integral Representations by Friedrich Sauvigny.
The aim in the book is to provide a version of the divergence theorem which holds also in cases where the boundary has certain singularities (as you described: the singular boundary has to have zero capacity). As a precursor they also prove the Stokes' theorem (they credit the proof to E. Heinz!).
Note that this is much more general than manifolds with corners, it encompasses your cone as well!
| 22 | https://mathoverflow.net/users/3509 | 12927 | 8,740 |
https://mathoverflow.net/questions/12935 | 7 | It is well known that for any set A in R^d there exists a measurable set E such that E contains A and m\*(A)=m\*(E). Is it possible to go the other direction?
In other words, is it true that for any measurable set E (such that m(E)>0) there is a non-measurable subset A such that m\*(A)=m\*(E)?
| https://mathoverflow.net/users/3532 | Non Lebesgue measurable subsets with "large" outer measure | A set $E$ with positive Lebesgue measure can be decomposed as a union $E = A \cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^\*(A) = m^\*(B) = m(E)$.
An example for this construction is a [Bernstein set](http://en.wikipedia.org/wiki/Bernstein_set).
| 15 | https://mathoverflow.net/users/454 | 12938 | 8,750 |
https://mathoverflow.net/questions/12943 | 56 | For any set X, let SX be the symmetric group on
X, the group of permutations of X.
My question is: Can there be two nonempty sets X and Y with
different cardinalities, but for which SX is
isomorphic to SY?
Certainly there are no *finite* examples, since the symmetric
group on n elements has n! many elements, so the finite
symmetric groups are distinguished by their size.
But one cannot make such an easy argument in the infinite
case, since the size of SX is 2|X|,
and the exponential function in cardinal arithmetic is not
necessarily one-to-one.
Nevertheless, in some set-theoretic contexts, we can still
make the easy argument. For example, if the Generalized
Continuum Hypothesis holds, then the answer to the question
is No, for the same reason as in the finite case, since the
infinite symmetric groups will be characterized by their
size. More generally, if κ < λ implies
2κ < 2λ for all
cardinals, (in another words, if the exponential function
is one-to-one, a weakening of the GCH), then again
Sκ is not isomorphic to
Sλ since they have different
cardinalities. Thus, a negative answer to the question is
consistent with ZFC.
But it is known to be consistent with ZFC that
2κ = 2λ for some
cardinals κ < λ. In this case, we will
have two different cardinals κ < λ, whose
corresponding symmetric groups Sκ and
Sλ nevertheless have the *same* cardinality. But can we still
distinguish these groups as groups in some other (presumably more group-theoretic) manner?
The smallest instance of this phenomenon occurs under Martin's Axiom plus ¬CH, which implies 2ω =
2ω1. But also, if one just forces ¬CH by adding Cohen reals over a model of GCH, then again 2ω =
2ω1.
(I am primarily interested in what happens with AC. But if
there is a curious or weird counterexample involving
¬AC, that could also be interesting.)
| https://mathoverflow.net/users/1946 | Can the symmetric groups on sets of different cardinalities be isomorphic? | According to the Schreier–Ulam–Baer theorem, the nontrivial normal subgroups of $S(X)$ are *(i)* the subgroup $S\_\mathrm{fin}(X)$ of permutations of $X$ of finite support, *(ii)* the subgroup $A\_\mathrm{fin}(X)$ of $S\_\mathrm{fin}(X)$ of even permutations, and *(iii)* for each cardinal $\kappa$ the subgroups $S\_{<\beta}(X)$ and $S\_{\leq\beta}(X)$ of permutations which move strictly less than $\beta$ points and at most $\beta$ points, respectively.
Since, as you said, a cardinal is determined by the order type of the set of cardinals below it, looking at the lattice of normal subgroups of $S(X)$, then, lets you guess the cardinal of $X$.
| 53 | https://mathoverflow.net/users/1409 | 12946 | 8,756 |
https://mathoverflow.net/questions/6834 | 11 | I was wondering when the Kunneth formula holds for motivic cohomology:
$$
H^p(X,A(\alpha)) = \bigoplus\_{i+j=p;\beta+\gamma = \alpha} H^j(X,A(\beta)) \otimes H^i(X,A(\gamma))
$$
where $H^p(X,A(\alpha))$ is defined as you wish: by higher Chow groups, Hom groups in $DM(X)$, etc... The case I'm most interested in is $A= \mathbb{Q}$ and $M(X) \in DM(\mathbb{Q})\_{\mathbb{Q}}$ in the thick sub-triangagulated category generated by the $\mathbb{Q}(n)$, $n\in \mathbb{Z}$.
| https://mathoverflow.net/users/1985 | Kunneth formula for motivic cohomology | I now remember a nice argument, why there's no Kunneth formula for Chow groups of $X \times X$ unless $X$ has a Tate motive. Let $X$ be smooth projective of dimension $d$.
We start with a decomposition of a diagonal:
$$
[\Delta] = \sum\_{i,j} \alpha^i\_j \beta^{d-i}\_j \in \oplus\_i CH^i(X) \otimes CH^{d-i}(X)
$$
We can assume $\alpha^i\_j$ are linearly independent.
In this case we can show that $\alpha^i\_j$ form a basis of Chow groups and
$\beta^{d-i}\_j$ is the dual basis.
Indeed, as a correspondence $[\Delta]$ acts as identity on Chow groups,
so for any class c, $$c = [\Delta]c = \sum\_{i,j} \alpha^i\_j deg(\beta^{d-i}\_j \cup c),$$
and the claim follows if we substitute $c = \alpha^i\_j$.
Now $CH\_i(X) = Hom(\mathbb Z(i)[2i], M(X))$ and we can consider the set of $\alpha^i\_j$ as a morphism of motives $$\oplus\_{i,j}\mathbb Z(i)[2i] \to M(X).$$
A simple computation shows that it is an isomorphism with the inverse given by $\beta^i\_j$.
And of course, on the other hand, if $X$ has a Tate motive, then Kunneth formula for Chow groups follows (it doesn't answer the question, since I only consider smooth projective varieties).
| 11 | https://mathoverflow.net/users/2260 | 12948 | 8,757 |
https://mathoverflow.net/questions/12873 | 6 | Let $K$ be a field, $K^\times$ its multiplicative group and $I$ an infinite set. Is then $(K^\times)^{(I)} \subseteq (K^\times)^I$ a direct summand? If not, is it possible to characterize the fields for which this is true?
In any case, it's a pure subgroup. If $K$ is finite, the answer is [yes](https://mathoverflow.net/questions/12503/z-ni-is-a-direct-summand-of-z-ni). If $K$ has arbitrary roots, that is $K^\times$ is divisible, then it's also true. If $K^\times$ is the additive group of a vector space (i.e. it's elementary abelian for some prime or uniquely divisible), you can use linear algebra. If $K^\times$ is a finite direct sum of these types, then it also works; e.g. $\mathbb{R}^\times = \mathbb{Z}/2 \times \mathbb{R}^+$.
Now what about $K = \mathbb{Q}$. Here $K^\times = \mathbb{Z}/2 \oplus \mathbb{Z}^{(\mathbb{P})}$. If $I=\mathbb{N}$ and $\hom((\mathbb{Z}^{(I)})^I,\mathbb{Z})$ is countable, then it's false. But I don't know if this is true, the argument of Specker computing $\hom(\mathbb{Z}^\mathbb{N},\mathbb{Z})$ does not seem to take over. Another case would be that $K^\times$ is torsion, i.e. $K$ is an algebraic extension of $\mathbb{F}\_p$ for some prime $p$, e.g. $K = \mathrm{colim}\\_s \mathbb{F}\\_{p^{q^s}}$ for some prime $q$ and $K^\times = \mathrm{colim}\\_s \mathbb{Z}/(p^{q^s}-1)$. This is a subgroup of $\mathbb{Q}/\mathbb{Z}$, which does not have to be divisible.
I don't know an example of an abelian group $G$ such that $G^{(I)}$ is not a direct summand of $G^I$, but I'm pretty sure that there is one. But does this $G$ also arise as $K^\times$? (EDIT: I know that $G=\mathbb{Z}, I = \mathbb{N}$ does it, but $\mathbb{Z}$ is no $K^x$.) There are several characterizations1 when $G$ has the form $K^\times$ for some field $K$. Perhaps this is useful here. The whole question is motivated by the study of $K \otimes\_K \otimes\_K ...$ as defined [here](https://mathoverflow.net/questions/11767/infinite-tensor-products).
1R.M. Dicker, A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field, Proc. London Math. Soc., 18, 1968, p.114 - 124
| https://mathoverflow.net/users/2841 | Split powers of the multiplicative group of a field | Re: "I don't know an example of an abelian group $G$ such that $G^{(I)}$ is not a direct summand of $G^I$, but I'm pretty sure that there is one."
Let $G$ be the the integers, and $I$ a countable indexing set. If $G^{(I)}$ were a direct summand, let $P$ be a complement summand.
We arrive at a contradiction as follows: First, $P$ is isomorphic to $G^{I}/G^{(I)}$ which contains the element $(2,4,8,16,32,\ldots)$ modulo $G^{(I)}$, which (as we can peal off any of the initial terms) is a non-zero element which is infinitely divisible by 2. Second, $P$ is a subgroup of $G^{I}$, which has no infinitely divisible elements (other than zero).
---
I think this argument might be modified to show that the algebraic closure of a finite field will give you the counter-example you need (changing "divisibililty" to some sort of degree consideration), but I don't have a lot of time to think about it right now. I'll come back later if someone else doesn't answer your question fully.
---
Back now. Try the following. Let $K$ be the field obtained by adjoining to $\mathbb{Q}$ the $2^{p}$th root of each prime prime $p$ (in $\mathbb{Z}$). Let $G$ be the multiplicative group of $K$. Suppose by way of contradiction that $G^{(I)}$ is a direct summand of $G^{I}$, and let $C$ be a complement. As before, $C$ is isomorphic to $G^{I}/G^{(I)}$, and the element $(2,3,5,\ldots )$ modulo $G^{I}$ is infinitely divisible by $2$ (thinking of ``divisibility'' multiplicatively in this case--in other words, after chopping off the front, we can take square roots as many times as we want). However, I believe it is the case that there is no element of $G^{I}$ which is infinitely divisible in this sense. (I'll leave it to the experts to prove this, but I think some form of Kummer theory would suffice. But it may be difficult to prove it.) [One last edit: I think it may even be easier to look at $\mathbb{Q}(x\_1,x\_2,\ldots)$ and adjoin a $2^{n}$th root of $x\_{n}$, and modify the example accordingly.]
| 5 | https://mathoverflow.net/users/3199 | 12950 | 8,758 |
https://mathoverflow.net/questions/12957 | 4 | ### Question
Given an abelian variety $V$ and an integer $n$, is there a natural abelian category with a natural object $X$ and natural coefficients $F$ so that $V\simeq H^n (X,F)$?
### Motivation
Studying abelian varieties is awesome. Studying objects in long exact sequences is awesome. How do (somewhat forcefully) combine these two? I mean without taking cohomology of the variety like everyone else does...
### Possible answers
The abelian variety is a $G$-module, where $G=Gal(\bar{k}/k)$, $k$ the field over which the variety is defined. So, maybe there is an interesting $G$-module that answers the above? The cases of abelian varieties over number fields and finite fields are the most interesting, so $G$ is assumed to be interesting as well (i.e. not trivial).
Maybe it arises as the $n$-th cohomology of some interesting sheaf of some interesting related variety?
| https://mathoverflow.net/users/2024 | Can an abelian variety be represented as the cohomology of some other object? | Let $\mathcal{A}/S$ be an Abelian scheme. Then the dual Abelian scheme is given by $R^1\pi\_\*\mathcal{A}$, if I remember correctly. Also, $\mathcal{A}^\vee(V) = \mathrm{Ext}^1\_V(\mathcal{A},\mathbf{G}\_m)$.
| 2 | https://mathoverflow.net/users/nan | 12958 | 8,762 |
https://mathoverflow.net/questions/12956 | 6 | Is there a standard name for taking a homomorphism from the free object over an algebraic structure? Roughly speaking, this should amount to *evaluation* of any element of the free object under the homomorphism.
For example, take the free commutative monoid over the set of natural numbers; elements of this monoid are finite multisets of natural numbers. By taking commutative monoid homomorphisms we can express many common operations, such as cardinality (map every generator to 1 and take sum as the monoid operation), sum (take every generator to itself and map the monoid operation to sum of natural numbers), product (take product of natural numbers rather than sum) etc.
Since these operations are all abelian monoid operations, we can expect a sensible result which will not depend on our multiset being non-empty or listing its elements in a specific order.
Does this idea generalize to other algebraic structures? Any known references?
| https://mathoverflow.net/users/3539 | Terminology: Name for a homomorphism from the free object? | The nLab has proposed the name "adjunct", which seems OK to me. So, for example, you could say something like "let $F(S)$ be the free group on the set $S$, let $G$ be a group, let $f:S\to G$ be a set map, and let $g:F(S)\to G$ denote the adjunct of $f$." I think it would be even a bit better to say "left adjunct" instead of just "adjunct".
I think Mac Lane might have used the term "transpose".
| 3 | https://mathoverflow.net/users/1114 | 12964 | 8,766 |
https://mathoverflow.net/questions/12914 | 3 | How to define Kahler differential in an abelian category or more general category? Say exact category?
Is there any interesting example?
| https://mathoverflow.net/users/3156 | How to introduce Kahler differential in category? | It turns out that the whole complex of concepts
* [module](http://ncatlab.org/nlab/show/module)
* [derivation](http://ncatlab.org/nlab/show/derivation)
* [Kähler differentials](http://ncatlab.org/nlab/show/K%C3%A4hler+differential)
* [cotangent complex](http://ncatlab.org/nlab/show/cotangent+complex)
* [quasicoherent sheaf of modules](http://ncatlab.org/nlab/show/quasicoherent+sheaf)
* and all other aspects of [deformation theory](http://ncatlab.org/nlab/show/deformation+theory)
works in remarkable generality on pure category-theoretic grounds with respect to *any* category, and is nothing but different facets of one single general concept: that of the
* [tangent $(\infty,1)$-category](http://ncatlab.org/nlab/show/tangent+(infinity%2C1)-category).
This goes back to the old observation by Quillen, that the category of modules over a ring is equivalent to the category of abelian group objects in the given overcategory of rings. All other concepts follow from this: derivations are sections through the over-objects, and the assignment of Kähler differentials is the left-adjoint to the projection down from the overcategory.
The notion of "tangent $(\infty,1)$-category" takes this idea to its full generality: this is the over-$(\infty,1)$-category, fiberwise stabilized. See the above link for details.
This complete picture, based on Quillen's old idea, is fully developed and exposed in the first part of the very nice article
* Jacob Lurie, [Deformation Theory](http://ncatlab.org/nlab/show/Deformation+Theory).
So the answer to the question is: a notion of Kähler differentials exists with respect to *any* (oo,1)-category $C$! Here for given $C$, the resulting notion models universal modules for objects in $C^{op}$, regarded as function rings over the objects in $C$.
I can't quite tell what the abelian category is supposed to be that appears in the question. But notice that the plain vanilla version of the story is obtained by letting $C$ be the category of (simplicial) monoids in the abelian category $Ab$ of abelian groups.
So, indeed, for any abelian category whatsoever, it makes very good sense to regard the category of monoids inside it as a replacement for the category of rings, regard the category of abelian group objects in the slice-categories of that as the corresponding bifibration of modules, and take the corresponding Kähler differentials to be the corresponding universal modules with respect to derivations, just following the general nonsense linked to above.
| 7 | https://mathoverflow.net/users/381 | 12974 | 8,770 |
https://mathoverflow.net/questions/12966 | 3 | (I'm happy to work over an algebraically closed field....)
Let $\mathcal{C} \rightarrow Spec (R)$ be a (flat) family of proper, prestable curves where $R$ is a DVR. Suppose the generic fiber is smooth and the special fiber, $C\_0$, is reduced but may be reducible.
Given a finite map of curves $f\_0: D\_0 \rightarrow C\_0$ with $D\_0$ also prestable, can this be extended to some map on some family?
That is, is there a flat family of proper curves $\mathcal{D} \rightarrow Spec(R)$ and an $R$-morphism $f: \mathcal{D} \rightarrow \mathcal{C}$ which reduces to $f\_0$ on the special fiber?
Perhaps such an extension is possible only after a ramified cover of $Spec(R)$?
If so, can it be arranged that the generic fiber of $\mathcal{D}$ is smooth?
| https://mathoverflow.net/users/397 | Extending maps of curves | Stated as such, the question is really too vague.
For example, if $C\_0, D\_0$ are smooth of genus bigger than 1, and $f\_0$ is purely inseparable, then $f\_0$ can not lift to $R$ (even after extention) if $R$ has characteristic zero. Actually, as the generic fibers of $\mathcal C$ and $\mathcal D$ are equal to $g(C\_0)=g(D\_0)$, this will contradict Riemann-Hurwitz formula.
In [this survey paper](http://www.math.u-bordeaux1.fr/~liu/articles/tokyo.ps) you can find some sufficient conditions. For instance, the lifting property for finite generically étale morphisms of semi-stable curves are characterized.
| 6 | https://mathoverflow.net/users/3485 | 12975 | 8,771 |
https://mathoverflow.net/questions/12969 | 14 | Does anyone know of such a domain?
| https://mathoverflow.net/users/3186 | non-Dedekind Domain in which every ideal is generated by at most two elements | You may find Matlis' paper [The Two-Generator Problem for Ideals](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-17/issue-3/The-two-generator-problem-for-ideals/10.1307/mmj/1029000474.full) to be interesting, as its main theorem concerns the class of integral domains in which every ideal is generated by two elements.
It was proven by Cohen (in *Commutative Rings with Restricted Minimum Condition* ) that an integral domain with the property that there exists an integer $n$ such that every ideal can be generated with fewer than $n$ elements must be Noetherian and of Krull dimension 1.
Say that an integral domain R has property **FD** if every finitely generated torsion free R-module is direct sum of modules of rank 1. Moreover, say that R has property **FD locally** if RM has property FD for every maximal ideal M of R.
**Theorem** (simplified form) - Let R be an arbitrary integral domain. Then every ideal of R can be generated by two elements if and only if R is a noetherian ring that has property FD locally.
| 11 | https://mathoverflow.net/users/nan | 12978 | 8,773 |
https://mathoverflow.net/questions/12980 | 0 | Let $A$ be an algebra, $H$ a Hopf algebra, and
$$
\beta\_A: A \to A \otimes H, ~~~~~ a \mapsto a^{(1)} \otimes a^{(2)}
$$
a right $H$-coaction. This induces a right $H$-coaction on $A \otimes A$ defined by
$$
\beta\_{A \otimes A}: a \otimes b \mapsto a^{(1)} \otimes b^{(1)} \otimes a^{(2)}b^{(2)}.
$$
My question is: Does this restrict to a coaction on the universal calculus over $A$, namely to a $H$-coaction on the kernel of the multiplication map $m:A \otimes A \to A$? I feel this is a very simple question but I can't seem to find an answer.
If the construction does not work, does anyone know of a way to induce a coaction on the universal calculus over $A$ from $\beta\_{A}$?
| https://mathoverflow.net/users/1867 | Coaction on the Universal Calculus | If $A$ is an $H$-comodule algebra (that is, if the multiplication map $\mu$ is a map of comodules), then the answer is yes (trivially, because the category of $H$-comodules has kernels). If it isn't, then probably not, as then you have no compatibility between the algebra structure on $A$ and the comodule structure.
| 2 | https://mathoverflow.net/users/1409 | 12981 | 8,775 |
https://mathoverflow.net/questions/12962 | 2 | The blowdown of the zero section of the canonical bundle of the first del Pezzo surface $dP\_{1}$, the blowup of $CP^{2}$ at one point, is a Calabi-Yau cone. I was just wondering if this cone admitted a smoothing. Is the same thing true for the blowdown of the zero section of the canonical bundle of the second del Pezzo surface $dP\_{2}$, the blowup of $CP^{2}$ at two points. Thanks!
| https://mathoverflow.net/users/3566 | Existence of smoothing of Calabi-Yau cones over $dP_{1}$ and $dP_{2}$ | The answer to this question is contained in the article of Mark Gross, page 33
Deforming Calabi-Yau Threefolds
<http://arxiv.org/abs/alg-geom/9506022>
The first cone can not be smoothed the second one can be smoothed
(in the terminology of Gross, which is standard, $dP\_1$ is a del-Pezzo of degree 8, $dP\_2$ is the del Pezzo of degree 7).
| 3 | https://mathoverflow.net/users/943 | 12985 | 8,779 |
https://mathoverflow.net/questions/12973 | 109 | It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary operation of *symmetric difference* forms a group, and in ZFC there is a bijection between $S$ and the set of finite subsets of $S$, so the group structure can be taken to $S$. However, the existence of this bijection needs the axiom of choice.
So my question is
>
> Can it be shown in ZF that for any non-empty set $S$ there exists a binary operation $\ast$ on $S$ making $(S,\ast)$ into a group?
>
>
>
| https://mathoverflow.net/users/932 | Does every non-empty set admit a group structure (in ZF)? | In ZF, the following are equivalent:
(a) For every nonempty set there is a binary operation making it a group
(b) Axiom of choice
Non trivial direction [(a) $\to$ (b)]:
The trick is Hartogs' construction which gives for every set $X$ an ordinal $\aleph(X)$ such that there is no injection from $\aleph(X)$ into $X$. Assume for simplicity that $X$ has no ordinals. Let $\circ$ be a group operation on $X \cup \aleph(X)$. Now for any $x \in X$ there must be an $\alpha \in \aleph(X)$ such that $x \circ \alpha \in \aleph(X)$ since otherwise we get an injection of $\aleph(X)$ into $X$. Using $\circ$, therefore, one may inject $X$ into $(\aleph(X))^{2}$ by sending $x \in X$ to the $<$-least pair $(\alpha, \beta)$ in $(\aleph(X))^{2}$ such that $x \circ \alpha = \beta$. Here, $<$ is the lexicographic well-ordering on the product $(\aleph(X))^{2}$. This induces a well-ordering on $X$.
| 174 | https://mathoverflow.net/users/2689 | 12988 | 8,782 |
https://mathoverflow.net/questions/12949 | 19 | Let $\kappa$ be an infinite cardinal. Then there exists at least one [real-closed field](http://en.wikipedia.org/wiki/Real-closed_field) of cardinality $\kappa$ (e.g. [Lowenheim-Skolem](http://en.wikipedia.org/wiki/Lowenheim-Skolem); or, start with a function field over $\mathbb{Q}$ in $\kappa$ indeterminates, choose an ordering and a real-closure).
But I think there are many more, namely $2^{\kappa}$ pairwise nonisomorphic real-closed fields of cardinality $\kappa$. This is equal to the number of binary operations on a set of infinite cardinality $\kappa$, so is the largest conceivable number.
As for motivation -- what can I tell you, mathematical curiosity is a powerful thing. One application of this which I find interesting is that there would then be $2^{2^{\aleph\_0}}$ conjugacy classes of order $2$ subgroups of the automorphism group of the field $\mathbb{C}$.
**Addendum**: Bonus points (so to speak) if you can give a general model-theoretic criterion for a theory to have the largest possible number of models which yields this result as a special case.
| https://mathoverflow.net/users/1149 | Are there as many real-closed fields of a given cardinality as I think there are? | Hi Pete!
There's been a lot of study of this and similar problems. I believe that Shelah's theorem, from his 1971 paper "The number of non-isomorphic models of an unstable first-order theory" (Israel J. of Math) answers your question about real closed fields in the positive.
The best big result on such questions that I know of is in the 2000 Annals paper "The uncountable spectra of countable theories." by Hart, Hrushovski, Laskowski.
To answer the question on real closed fields specifically (and somewhat cautiously since I'm not a model-theorist):
The theory of real closed fields is a complete first order theory, with countable language. It is an unstable (an easy fact, I think, and [explained better on wikipedia](http://en.wikipedia.org/wiki/Stable_theory) than I could explain) theory as well. Hence Shelah's result applies, and the bound $2^\kappa$ is realized as you surmised.
Bonus points should go to Shelah (and perhaps also to Hart, Hrushovski, Laskowski, whose paper mentions the result of Shelah and proves other things) for proving that this bound is realized (for uncountable cardinals), **except** for theories $T$ which have **all** of the following properties:
1. $T$ has infinite models.
2. $T$ is superstable.
3. $T$ has prime models over pairs.
4. $T$ does not have the dimensional order property.
I have no clue what the fourth property means. But there are plenty of non-superstable theories to which Shelah's theorem applies, and hence which realize your bound (for uncountable cardinals).
For countable cardinality, I think there are still some open problems about how many non-isomorphic models there can be of a given theory, with cardinality $\aleph\_0$.
| 17 | https://mathoverflow.net/users/3545 | 12991 | 8,784 |
https://mathoverflow.net/questions/12992 | 11 | Let $F$ be a quasi-coherent sheaf on a scheme $X$. To check that $F$ vanishes it suffices to check that all the stalks of $F$ vanish. I would like to know whether it suffices to check that all the fibers of $F$ vanish.
(I think I am using standard terms: the tensor product over $O\_X$ with the local ring at $x$ is the stalk, and the tensor product over $O\_X$ with the residue field at $x$ is the fiber.)
It suffices to answer the question on an affine scheme. Let $R$ be a commutative ring. For each prime ideal $p$ of $R$ let $k(p)$ be the residue field of the local ring $R\_p$. Let $M$ be an $R$-module, and suppose that $\mathrm{Tor}\_i(k(p),M) = 0$ for all $i$ and all $p$. (Tor taken in the category of $R$-modules). Does it follow that $M = 0$?
If $M$ is finitely generated, the answer is yes. In that case $M$ vanishes even when $\mathrm{Tor}\_0(k(p),M) = 0$ for all maximal $p$, by Nakayama's lemma. My question is whether there is a good replacement for Nakayama's lemma when $M$ is not finitely generated.
| https://mathoverflow.net/users/1048 | When does a quasicoherent sheaf vanish? | If the scheme is locally noetherian, this is true and can be proved by noetherian induction. In fact, you can even replace $M$ with an object of bounded derived quasi-coherent category, if you are interested in such things.
The proof is relatively straightforward: For a complex of modules $M$ over the ring $R$, we may assume that any non-zero $f\in R$ acts by a quasi-isomorphism $f:M\to M$ (by the induction hypothesis), and then the cohomology of $M$ are defined over the field of fractions of $M$.
| 8 | https://mathoverflow.net/users/2653 | 12996 | 8,788 |
https://mathoverflow.net/questions/12968 | 5 | I am looking for a power series $\displaystyle f(z) = \sum\_{n=0}^{+\infty} a\_n z^n$ that converge uniformly on $\mathcal{D} = \Big\{ z \in \mathbb{C} \ / \ \vert z \vert \leq 1 \Big\}$ but not normally. Of course, a proof that this is impossible would be even better. It seems close to [this question](https://mathoverflow.net/questions/10831), but it's not quite the same.
It is well-known that normal convergence implies uniform convergence, and that the converse is false but I haven't found yet a counterexample in the form of a power series on $\mathcal{D}$.
In other words, I would like a complex sequence $(a\_n)\_{n \in \mathbb{N}}$ satisfying the three conditions :
* $\displaystyle\sum\_{n=0}^{+\infty} a\_n z^n$ converges for all $z$ such that $\vert z \vert \leq 1$
* $\displaystyle \sup\_{\vert z \vert \leq 1} \left\vert \sum\_{n=N}^{+\infty} a\_n z^n \right\vert \longrightarrow 0$ when $N\to+\infty$
* $\displaystyle\sum\_{n=0}^{+\infty} \vert a\_n \vert = +\infty$
| https://mathoverflow.net/users/3543 | Mode of convergence of a power series | An example, due to Fejér, appears in Hille's *Analytic function theory* on [page 122](http://books.google.com/books?id=4Qu41N_OT7gC&lpg=PA122&ots=RibiKvgZOJ&dq=%22uniformly%20but%20not%20absolutely%22&pg=PA122#v=onepage&q=&f=false) of the second edition, volume 1.
Erdős wrote a [paper](http://www.renyi.hu/~p_erdos/1952-09.pdf) showing the existence of classes of examples with many zero coefficients, and in that paper he states that Hardy was the first to find an example of what you're looking for. However, Erdős refers to [Landau's book](http://books.google.com/books?id=1TEPAAAAIAAJ&vq=hardy&dq=Darstellung+und+begr%C3%BCndung+einiger+neuerer+ergebnisse+der+funktionentheorie&source=gbs_navlinks_s) for evidence, and I don't read German, so someone more capable and willing may check that out.
**Edit**: I initially misnamed the author and book containing the example, as pointed out by Harald Hanche-Olsen.
**Update**: Coincidentally, while skimming through Sasane's [*Algebras of holomorphic functions and control theory*](http://books.google.com/books?id=1iSWPgAACAAJ&dq=amol+sasane+holomorphic&ei=0QB_S_6-FaGayATIxon7Cg&client=firefox-a&cd=1), I came upon another reference to Hardy's example. This time the reference was to Dienes's [*The Taylor series*](http://books.google.com/books?id=L4nQAAAAMAAJ&client=firefox-a&source=gbs_navlinks_s). I checked, and sure enough there is Hardy's example in the chapter "The Taylor series on its circle of convergence." I scanned it, along with a page containing a result used in the example. [Here it is](https://docs.google.com/fileview?id=0B_pEp00B111JZjBlM2JlOWUtNzk0YS00MDVkLWFhNmMtNTRhMmM0MjQ2NDNi&hl=en). (The example starts two thirds of the way down page 464, and the cited "105.IV" is the "IV" that appears on page 441.)
| 8 | https://mathoverflow.net/users/1119 | 13003 | 8,791 |
https://mathoverflow.net/questions/13001 | 2 | I'm trying to read Donoho's 2004 paper Compressed Sensing and am having trouble with a supposedly trivial statement (equation 1.2 on page 3).
He makes the sparsity assumption on $\theta \in \mathbb{R}^m$ that for some $0<p<2$ and $R>0$ we have $\|\theta\|\_p\leq R$. Then if $\theta\_N$ denotes $\theta$ with everything except the $N$ largest coefficients set to $0$ he claims that $\| \theta-\theta\_N \|\_2 \leq \zeta\_{2,p} \cdot \| \theta \|\_p \cdot (N+1)^{1/2-1/p}$ for $N=0,1,2,\ldots$ where $\zeta\_{2,p}$ depends only on $p$.
I've tried writing out the definitions of various things. I've noticed that the $N$th largest coefficient must satisfy $\mid\theta\_i\mid \leq RN^{-1/p}$ but I can't figure out how the result above follows.
I'm also having some difficulty thinking about $\ell^p$ spaces with $0<p<1$, in particular knowing what results from the $p>1$ theory apply. Does anyone know some good notes or a book that covers this?
| https://mathoverflow.net/users/3547 | l^p space inequality related to compressed sensing | Suppose that $\|\theta\|\_p=1$. Then, as you correctly observed, if we drop $N$ largest coefficients, we'll have $|\theta\_i|\le N^{-1/p}$ for all the remaining ones, whence $|\theta\_i|^2\le N^{-(2-p)/p}|\theta\_i|^p$. Now just add those inequalities up, use the fact that $\sum\_i|\theta\_i|^p\le\|\theta\|\_p^p=1$ and raise both parts to the power $1/2$.
| 5 | https://mathoverflow.net/users/1131 | 13004 | 8,792 |
https://mathoverflow.net/questions/12990 | 2 | so, you have, for any two members of the algebraic structure A and B and any nonnegative real values a, b:
two operations: \* and +, such that
a\*A + b\*A = (a+b)\*A is in the structure
A + B = B + A is in the structure
0\*A + B = B
but there is no guarantee that X s.t.
X + A = B
is in the structure.
---
As an example, the set of 2-dimensional Cartesian vectors that are in the first quadrant (i.e., x>=0 and y>=0) has the properties that I want. You can add them, scale them, but if you try to subtract them, you might leave the first quadrant.
Thanks very much!
| https://mathoverflow.net/users/634 | What is it called if a vector space doesn't have an additive inverse? | If the structure in question is a subset of a vector space, like it is in your example, I would call it a [convex cone](http://en.wikipedia.org/wiki/Convex_cone).
| 12 | https://mathoverflow.net/users/1096 | 13007 | 8,793 |
https://mathoverflow.net/questions/13002 | 2 | Consider the following pair of principal bundle descriptions of $\mathbb{CP}^2$:
$$
\mathbb{CP}^2 \simeq SU(3)/U(2) \simeq S^5/U(1).
$$
If I have a principal $U(2)$-bundle connection for $\mathbb{CP}^2$, will that correspond to a principal $U(1)$-bundle connection? (Where by correspond I suppose I mean give the same covariant derivative.)
I am also interested in how this generalises to the $n$-case
$$
\mathbb{CP}^{n} \simeq SU(n+1)/U(n) \simeq S^{2n+1}/U(1).
$$
| https://mathoverflow.net/users/1977 | Principal Bundle Connection Correspondence for two descriptions of the $\mathbb{CP}^2$ | The answer is no. You can see that for example from the connection one forms which are Lie algebra valued. In the first case they are u(2) valued and in the second case they are u(1) valued. However, in the case of CP2 (which also generalizes to CPn), the U(1) and SU(2) factors of the isotropy group U(2) commute, this means that given a U(2) connection on U(2)-->SU(3)-->CP2, you can project it to the U(1) factor to obtain a U(1) connection on U(1)-->S^5-->CP2, such that both connections will have the same horizontal subspaces.
| 1 | https://mathoverflow.net/users/1059 | 13009 | 8,794 |
https://mathoverflow.net/questions/13005 | 30 | The key step in Kontsevich's proof of deformation quantization of Poisson manifolds is the so-called formality theorem where 'a formal complex' means that it admits a certain condition. I wonder why it is called 'formal'. I only found the definition of Sullivan in Wikipedia: 'formal manifold is one whose real homotopy type is a formal consequence of its real cohomology ring'. But still I am confused because most of articles I found contain the same sentence only and I cannot understand the meaning of 'formal consequence'. Does anyone know the history of this concept?
| https://mathoverflow.net/users/3478 | What is 'formal' ? | I would guess that the terminology goes back to the work of Sullivan and Quillen on rational homotopy theory. You should probably also look at the paper of [Deligne-Griffiths-Morgan-Sullivan](https://doi.org/10.1007/BF01389853) on the real homotopy theory of Kähler manifolds. Actually, I think that at least some familiarity with the DGMS paper is an important prerequisite for understanding many of Kontsevich's papers.
I am not totally sure, but I believe that the definitions are as follows:
* A differential graded algebra $(A,d)$ is called formal if it is quasi-isomorphic (in general, if we work in the category of dg algebras and not, say, the category of A-infinity algebras, we need a "zig-zag" of quasi-isomorphisms) to $H^\ast(A,d)$ considered as a dg algebra with zero differential.
* A space X is called formal (over the rationals resp. the reals) if its cochain dg algebra $C^\ast(X)$ (with rational resp. real coefficients) with the standard differential is a formal dg algebra.
One of the things I'm not sure about is whether in the definition we should require $H^\ast(A,d)$ to be commutative; but for spaces this is not an issue since $H^\ast(X)$ is always (graded-)commutative.
The DGMS paper proves that if X is a compact Kähler manifold, then the de Rham dg algebra consisting of (real, $C^\infty$) differential forms on X with the standard de Rham differential is a formal dg algebra.
The phrase "the real (resp. rational) homotopy type of X is a formal consequence of the real (resp. rational) cohomology ring of X", which appears in e.g. the DGMS paper, simply means that the real (resp. rational) homotopy theory of X is determined by (and is probably explicitly and algorithmically computable from?) the cohomology ring of X. In other words, if X and Y are formal (over the rationals resp. the reals) and have isomorphic (rational resp. real) cohomology rings, then their respective (rational resp. real) homotopy theories are the same (and are explicitly computable, if we know the cohomology ring(s)?). For example, the ranks of their homotopy groups will be equal.
Actually I am not totally sure whether what I said in the last paragraph is true. I think it's true when X and Y are simply connected. I'm not sure about what happens more generally.
In the context of rational homotopy theory, I think the term "formal" is fine, for the reasons I've explained above. Perhaps in the more general context of dg algebras, the use of the term "formal" makes less sense. However, I think that it is still reasonable, for the following reasons. Let me use the more "modern" language of A-infinity algebras. In general, it is not true that a dg algebra $(A,d)$ is quasi-isomorphic to $H^\ast(A,d)$ considered as a dg algebra with zero differential. However, it is a "standard" fact (Kontsevich-Soibelman call this the "homological perturbation lemma" (for example, it's buried somewhere in [this paper](https://arxiv.org/abs/math/0011041)), and you can find it in the operads literature as the "transfer theorem") that you can put an A-infinity structure on $H^\ast(A,d)$ which makes $A$ and $H^\ast(A,d)$ quasi-isomorphic as A-infinity algebras. The A-infinity structure manifests itself as a series of $n$-ary products satisfying various compatibilities. Intuitively at least, these $n$-ary products should be thought of as being analogous to Massey products in topology. So $H^\ast(A,d)$ with this A-infinity structure does carry some "homotopy theoretic" information. In this language then, a dg algebra $(A,d)$ is formal if it is quasi-isomorphic, as an A-infinity algebra, to $H^\ast(A,d)$ with all higher products zero. In other words, all of the "Massey products" vanish\*, and thus the only remaining "homotopy theoretic" information is that coming from the ordinary ring structure on $H^\ast(A,d)$.
---
\*Don Stanley notes correctly that vanishing of Massey products is weaker than formality. However, I believe that triviality of the A-infinity structure is equivalent to formality. In the language of the DGMS paper, which does not use the A-infinity language, they say that formality is equivalent to the vanishing of Massey products "in a uniform way". I believe this uniform vanishing is the same as triviality of A-infinity structure. From the paper:
>
> ... a minimal model is a formal consequence of its cohomology ring if, and only if, all the higher order products vanish in a uniform way.
>
>
>
and also
>
> [Choosing a quasi-isomorphism from a minimal dg algebra to its cohomology] is a way of saying that *one may make uniform choices so that the forms representing all Massey products and higher order Massey products are exact*. This is stronger than requiring each individual Massey product or higher order Massey product to vanish. The latter means that, given one such product, choices may be made to make the form representing it exact, and there may be no way to do this uniformly.
>
>
>
(Sorry for the proliferation of parentheses, and sorry for my lack of certainty on all of this, I have not thought about this in a while. People should definitely correct me if I'm wrong on any of this.)
| 28 | https://mathoverflow.net/users/83 | 13019 | 8,800 |
https://mathoverflow.net/questions/13017 | 16 | Inspired by [a recent question](https://mathoverflow.net/questions/12873/split-powers-of-the-multiplicative-group-of-a-field) on the multiplicative group of fields. Necessary conditions include that there are at most $n$ solutions to $x^n = 1$ in such a group and that any finite subgroup is cyclic. Is this sufficient? (Edit: Well, no, it's not, since the only such groups which are finite are the cyclic groups of order one less than a prime power. Hmm.)
| https://mathoverflow.net/users/290 | Which commutative groups are the group of units of some field? | The following paper claims an answer to this question:
>
> Dicker, R. M.
> A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field.
> Proc. London Math. Soc. (3) 18 1968 114--124.
>
>
>
You can find it here:
[http://alpha.math.uga.edu/~pete/Dicker1966.pdf](http://alpha.math.uga.edu/%7Epete/Dicker1966.pdf)
One might hope for a more aesthetically appealing characterization. I don't know if such a thing has ever been given.
| 18 | https://mathoverflow.net/users/1149 | 13024 | 8,803 |
https://mathoverflow.net/questions/13027 | 17 | Freyd's [Abelian Categories](http://www.tac.mta.ca/tac/reprints/articles/3/tr3.pdf) is the only textbook I know where the primacy of arrows over objects is taken seriously already in the axioms: there is no talk of objects at all. Only later one sees, that an *object* is what arrows with the same *domain* (another derived-only concept) have in common. This helps thinking abstractly from the very beginning.
>
> Question #1: Why isn't Freyd's axiomatization standard?
>
>
>
I guess it's mainly a matter of convenience, but this seems to be rarely mentioned explicitly.
---
The same goes for graph theory: the usual and beginner's way of thinking is that arcs (seen as pairs of vertices) are 'ontologically' secondary to vertices, just as arrows (seen as functions between sets) would be secondary to objects (seen as sets). But in the most general setting for graph theory - multidigraphs or quivers - it becomes obvious that it is the other way around: arcs can be the only primary objects and vertices can be thought of equivalence classes of arcs according to the two equivalence relations "has the same source (resp. target) as".
>
> Question #2: Is there an introductory
> textbook to graph theory that
> emphasizes that general graph theory is
> "nothing but a theory of two arbitrary
> equivalence relations".
>
>
>
[Addendum:] The *structure* of a graph comes in by identifying some equivalence classes of the first equivalence relation with some of the second.
| https://mathoverflow.net/users/2672 | Primacy of arcs/arrows over vertices/objects | Qu 1: At about the time of Freyd's book there were two approaches to defining categories. One came from algebraic topology and homological algebra, thus from Eilenberg and MacLane and used the objects and arrows definition, the other was motivated by differential geometry and used the arrows only formalism. This second one is perhaps better suited to those areas where the arrows are what is seen first. For instance, when introducing the concept of the fundamental groupoid of a space then you can think of the idea as a set with a partially defined composition satisfying certain rules and that has distinct advantages for that setting. Ehresmann wrote his book on categories from this viewpoint, but some of the basic ideas do end up being less clear from that viewpoint, others are perhaps clearer. That approach is also linked to more algebraic ideas such as inverse semigroups.
The first 'objects plus arrows' approach is thought to be more accessible to researchers with a `standard' mathematical background e.g. from algebra or algebraic topology. There the objects are usually thought of as being what is being studied and the morphisms are a tool for that study.
The recent use of categorification, quasicategories, internal categories, enriched categories and other similar ideas, tends to show that both viewpoints are best kept in balance, meeting in that higher categorical area.
Put simply I thing the answer to your question is: historically the comparison of objects using arrows was the more pressing application to start with. It came to dominate. More researchers came to use it.
Now the question arises `which to use with a given audience? and that is a hard one to answer.
| 15 | https://mathoverflow.net/users/3502 | 13033 | 8,808 |
https://mathoverflow.net/questions/12970 | 12 | Suppose we are given a triangulable topological space $X$. If $X$ has the fixed point property (FPP), then obviously for every triangulation $K$ of $X$ and every simplicial map $f:K\to K$ a simplex $\sigma\in K$ exists such that $f(\sigma)=\sigma$. This will be called the fixed simplex property (FSP).
One can give examples of simplicial complexes (included below) with FSP whose geometric realizations do not have FPP. Moreover, triangulations of spaces without FPP exist whose iterated barycentric subdivisions all have FSP.
Is it true that if $X$ is a triangulable space without FPP, then a triangulation of $X$ exists that does not have FSP? A related question: given a triangulation $K$ of $X$, can we find a (non-barycentric) subdivision of $K$ that does not have FSP?
---
**Example** (taken from J.A. Barmak's [thesis](http://mate.dm.uba.ar/~jbarmak/tesisfinal2.pdf) (p. 101), who in turn cites K.Baclawski and A.Björner "Fixed points in partially ordered sets" (Adv. Math. 1979)):
Consider the regular CW-complex (square + 4 triangles) $C$ which is the border of a pyramid with square base. $C$ is homeomorphic to $S^2$, so it doesn't have FPP. Let $K$ be the following subdivision of $C$: divide each of the 3 triangular sides into 3 triangles by adding a vertex in the middle and divide the bottom square into 4 triangles the same way.
Name the top vertex of the pyramid $x$. Let $f:K\to K$ be simplicial. If $f$ is onto (on the vertices), then, by finiteness of $K$, $f$ is an automorphism. Since $x$ is the uniqe vertex that belongs to exactly eight 1-simplices, it is a fixed point. If $f$ is not onto, then $|f|$ is nullhomotopic and thus the Lefschetz number $\lambda(f)=1$. $|f|$ has a fixed point, so $f$ fixes a simplex.
The same argument (just count the 1-simplices that $x$ and other vertices belong to) applies to arbitrary barycentric subdivisions of $K$.
However, if in $C$ we subdivide only the square base into 4 triangles by adding a vertex in the middle, then the obtained simplicial complex doesn't have FSP.
| https://mathoverflow.net/users/2578 | Does a triangulation without fixed simplex property always exist? | EDITED. The arugment related to Mostov rigidity is completed according to a nice suggestion of Tom Church
The answer to the first question is no. There exsit manifolds of dimension 3 such that every simlicial map of the manfiold to itself (for any simplicial decomposition) has a fixed point (and hence a fixed simplex). At the same time every 3-mafiold adimits a smooth self-map without fixed points.
Namely, take $M^3$ with vanishing first and second homology ($H\_1(M^3,R)=H\_2(M^3,R)=0$) and that is a hyperbolic 3-manifold. Moreover take such $M^3$ that does not have isometries. Existence of such manifolds is a standard result of 3-dimensional hyperbolic geometry. Let us prove that every such manifold gives us an example.
Proof. All compact 3-manifolds have zero Euler characteristics, so on $M^3$ there is a non-vanishing vector field $v$. Take the flow $F\_t$ generated by $v$ in small time $t$. This will give us a family of diffeo $F\_t$ of $M^3$ that don't have fixed points for small $t$. So $M^3$ in not FPP.
Now, let us show that $M^3$ has FSP. Take any simplicial decomposition of $M^3$. First we state a simple lemma (without a a proof)
Lemma. Consider a simplicial decomposition of a compact orienable manifold. Suppose we have a simpicial map from it to itself, that send simplexes of highest dimensions to simplexes of higher dimentions (i.e. don't collapse them) and don't indentify them. Then this is an automorphism of finite order.
Corollary. Every non-identical simplicial map $\phi$ from $M^3$ to itself either collapses a simplex of dimension 3 or identifies two such simplexes. In paricular the generator of $H^3(M^3,\mathbb Z)$ is sent to zero by this map.
This corollary together with Lefshetz fixed point theorem implies immediately that $\phi$
has a fixed point, and so it proves FSP for $M^3$ (we use that $H\_1(M^3)=H\_2(M^3)=0$).
Proof of corollary. If $\phi$ does not collapse 3 simplexes of $M^3$ or identify them, then it is a homeomorphism of $M^3$ of finite order (Lemma above). From Mostov rigidity it follows that this automorphism is homotopic to the identity. In order to show that it IS in fact the identity we need to use a more involved statement suggseted by Tom Church below. Namely, a paritial case of Proposition 1.1 in <http://www.math.uchicago.edu/~farb/papers/hidden.pdf>
says that for a hyperbolic 3-manifold the group of isometries of any Riemanninan metric on it is isomorphic to a subgroup of the group of hyperbolic isometries. By our choice the group of hyperbolic isometries of $M^3$ is trivial. It is clear that $\phi$ preserves a Riemannian metric on $M^3$. So by Prop 1.1 it is the identity.
From this it immediately follows that $\phi$ sends $H^3(M^3, Z)$ to zero (since the volume is contacted). End of proof.
| 11 | https://mathoverflow.net/users/943 | 13034 | 8,809 |
https://mathoverflow.net/questions/13020 | 5 | Let $\mathcal{M}$ be a real, compact, orientable manifold and let $X$ be a vector field on $\mathcal{M}$. Consider the functional
$$E(f) = \int\_{\mathcal{M}} X\_p(f)^2 dV$$
where $X\_p(f)$ is the directional (Lie) derivative of $f$ along $X$ at the point $p$ and $dV$ is a volume form on $\mathcal{M}$ -- this functional essentially measures the total amount of change in $f$ along $X$ over all of $\mathcal{M}$ in the $L^2$ sense. Then $\delta E(f)$ is a differential operator whose eigenspectrum
$$\delta E(f) = \lambda f$$
(for $\lambda \in \mathbb{R}$) yields the critical points of $E$ over the set of functions with unit norm. Is there an established name for this operator (or functional) and/or its corresponding eigenspectrum?
The prototype for this operator is Dirichlet's energy
$$\int\_{\mathcal{M}} ||\nabla f||^2 dV$$
which has as its (unit-norm) critical points the Laplacian eigenspectrum
$$\nabla^2 f = \lambda f,$$
the main difference being that Dirichlet's energy measures the total *gradient*, i.e., the change in all directions, rather than just the change along a particular direction at each point.
| https://mathoverflow.net/users/1557 | Is there a name for this differential operator and/or its corresponding spectrum? | There is an simple explicit formula for your operator in terms of known operators. To see this, note that $\delta E\_f(g)$ (the differential of $E$ at the function $f$ in the direction $g$) is equal to
$$
2 \int X(f) X(g) dV = 2\int \left[L\_X(g X(f) dV )-g\left(X(X(f)) dV+X(f)L\_X(dV)\right)\right]
$$
where $L\_X$ is the Lie deriviative with respect to $X$. Now, $\int L\_X(\alpha) = 0$ for any top degree form $\alpha$, so we get
$$
\delta E\_f(g) = - 2 \int\_X g \left[X(X(f)) + X(f) div(X)\right]dV
$$
where the divergence of $X$ is defined by $div(X)= L\_X(dV)/dV$. So, using the $L^2$-inner product on functions, we can interpret the 1-form $\delta E$ as the differential operator
$$
D :f \mapsto - X(X(f)) - X(f) div(X).
$$
(This is the same way that the differential of Dirichlet energy is seen as the Laplacian.)
Note that the leading order part of D is just $X^2$ and so, in particular, $D$ is not elliptic. You'd expect this, of course, because $E$ only sees the change of $f$ in the $X$-direction.
| 5 | https://mathoverflow.net/users/380 | 13036 | 8,810 |
https://mathoverflow.net/questions/13008 | 20 | What is the expected value of the determinant over the uniform distribution of all possible 1-0 NxN matrices? What does this expected value tend to as the matrix size N approaches infinity?
| https://mathoverflow.net/users/3551 | Expected determinant of a random NxN matrix | As everyone above has pointed out, the expected value is $0$.
I expect that the original poster might have wanted to know about how big the determinant is. A good way to approach this is to compute $\sqrt{E((\det A)^2)}$, so there will be no cancellation.
Now, $(\det A)^2$ is the sum over all pairs $v$ and $w$ of permutations in $S\_n$ of
$$(-1)^{\ell(v) + \ell(w)} (1/2)^{2n-\# \{ i : v(i) = w(i) \}}$$
Group together pairs $(v,w)$ according to $u := w^{-1} v$. We want to compute
$$(n!) \sum\_{u \in S\_n} (-1)^{\ell(u)} (1/2)^{2n-\# (\mbox{Fixed points of }u)}$$
This is $(n!)^2/2^{2n}$ times the coefficient of $x^n$ in
$$e^{2x-x^2/2+x^3/3 - x^4/4 + \cdots} = e^x (1+x).$$
So $\sqrt{E((\det A)^2)}$ is
$$\sqrt{(n!)^2/2^{2n} \left(1/n! + 1/(n-1)! \right)} = \sqrt{(n+1)!}/ 2^n$$
| 33 | https://mathoverflow.net/users/297 | 13040 | 8,813 |
https://mathoverflow.net/questions/13038 | 9 | This question was prompted by my interpretation of [a question](http://letterstonature.wordpress.com/2010/01/26/genus-analogues/) by cosmologist Berian James.
**Background**
Some cosmologists have suggested using the cosmological dark matter density, which defines a function $f:M\to \mathbb{R}$ with $M$ the spatial universe, in order to probe the topology of $M$. (**edit**: Berian comments below that this may not be what this is about! Listen to him... not to me!) The original reference seems to be [this paper by Gott et al.](http://go2.wordpress.com/?id=725X1342&site=letterstonature.wordpress.com&url=http%3A%2F%2Fadsabs.harvard.edu%2Fcgi-bin%2Fbib_query%3F1986ApJ...306..341G). Although the paper does not mention it explicitly, it seems that the natural mathematical framework for this proble is Morse theory.
Berian is interested not just in functions, but also e.g., vector fields or more generally sections of bundles on $M$. Hence the following
**Question**
Can one extend Morse theory beyond functions to sections of bundles? or perhaps to differentiable maps $f: M \to X$?
Pointers to the literature would be most welcome.
Cheers.
| https://mathoverflow.net/users/394 | Is there a Morse theory for sections of bundles or more generally for maps? | You are asking a very classical question. It seems to me, that general answer is No. But there were some interesting attempts (could not find referencies..). In some cases one can prove that there is no "Morse theory" for a class of maps, since they satisfy h-principle and one can eliminate (almost all) singularities.
I formulate a result (due to Gromov) of a Morse-type estimates of a singular points of smooth mappings to the plane. Consider a generic smooth map of a closed manifold $M$ to the plane. Consider the image of its singular points. It is a curve having cusps and double self-intersections as singular points only. Denote by $C$ a number of its cusps, by $X$ a number of double self-intersections and by $Comp$ a number of components. Then the sum $B(M)$ of Betti numbers of $M$ is restricted by $C+2X+2Comp$.
| 11 | https://mathoverflow.net/users/2823 | 13041 | 8,814 |
https://mathoverflow.net/questions/13021 | 8 | The following graph theoretic notion appeared in an [economics paper entitled: "Prize competition under limited comparability,](http://www.homepages.ucl.ac.uk/~uctprsp/framing2.pdf) by Michele Piccione and Ran Spiegler which studies models of economics were the firms are rational but the consumers are not.
A graph is **weighted-regular** if vertices (nodes) can be assigned positive weights, such that each node has the same total weighted-expected-number of edges (links). Every regular graph is, of course, weighted regular (give all vertices weight 1). A star, tree with center vertex connected to k leaves is weighted-regular: Give the center weight 1 and every leaf weight 1/k.
(In the paper, every vertex is also contained in a loop. Therefore its weight also counts when we compute its weighted-expected-number of edges. The star is still weighted-regular: give weight 1 to the center and weight 0 to the leafs.)
Is there a characterization of weighted-regular graphs? Is this notion known? Does it have known applications or connections?
| https://mathoverflow.net/users/1532 | Weighted Regular Graphs | The examples you gave can be extended. If a graph has a transitive group then it is weighted regular. Also if the graph such that one point is connected to all other points and if that point is removed the graph remaining has a transitive group then it is weighted regular.
Also there exists graphs that are not weighted regular Look at the graph on 5 points and look at the following graph 1 is connected to all other points for $n$ equal to 2 through 4 $n$ is connected to $n+1$. Then the sum of the weights on 3 and 4 is equal to the sum of the weights on all points of the graph plus the sum of weight one. The sum of the weights on 2 and 5 is the sum the weights on 3 and 4 plus twice the weight of 1. Combining these two results forces the weight on 2 and 5 to be zero and hence not positive so not all graphs are weighted regular.
Also for any graph if a point $x$ connects only to a point $y$ then if $z$ connects to point $y$ it cannot connect to any other point t if it does since the weights of $z$ and $x$ are the same the value assigned to $y$ must equal the sum of the values assigned to $y + t$ and hence $t$ is forced to be zero. This means that the only tree which can be weighted regular is a star and the only forest one with each component a star and if any graph has a component with a vertex connected to only one point it must be a star.
The join of any two weighted regular graphs is weighted regular.
Let the graphs be $G$ and $H$ with $m$ and $n$ points respectively.
Let there weights assigned to each graph so that each node in each graph has the same
weight. Let the weight of each node in $G$ be $w\_1$ and the weight of each node
in $H$ be $w\_2$. Let the sum of the weights of $G$ be $w\_3$ and the sum of the
weights of $H$ be $w\_4$. multiply the weights of $G$ by $a$ and $H$ by $b$. Then
the weight of any node of $G$ in the join is $aw\_1 + bw\_4$ and of any node of H in
the join is $aw\_2 + bw\_3$
so for the join to be a weighted regular graph $aw\_1 + bw\_4 = bw\_2 + aw\_3$ or $a(w\_1-w\_3)= b (w\_2-w\_4)$ or $a/b =(w\_1-w\_3)/(w\_2-w\_4)$ we
can find such $a$ and $b$ iff $w\_1$ is not equal to $w\_3$ and $w\_2$ is not equal
to $w\_4$.But the sum of the weights on all points is not equal to the sum of the weights
on all points on graph so that condition is satisfied.
Now using the comment by Douglas Zare that the convex hull of the rows of the adjacency matrix contains a multiple of the all-1s vector we can find whether any graph is weighted regular in polynomial time. We can convert it to a linear programming problem and there algorithms such as the ellipsoid method that solve linear programming problems in a polynomial amount of time.
In this paper: "On quantum perfect state transfer in weighted join graph" available at the URL: <http://arxiv.org/abs/0909.0431> the concept of a weighted regular graph is applied to perfect state transfer on quantum networks.
| 3 | https://mathoverflow.net/users/1098 | 13065 | 8,831 |
https://mathoverflow.net/questions/13016 | 3 | This is prop 2.6b on p.28 of Silverman's the Arithmetic of Elliptic curves.
It says that let $\phi: C\_1 \rightarrow C\_2$ be a non-constant map of projective non-singular irreducible curve. (probably over an algebraically closed field, but I am not too sure) Then for all but finitely many $Q \in C\_2$, #$\phi^{-1} (Q) = deg\_s (\phi)$, where RHS is the separability degree of the function fields.
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> I don't understand Silverman's proof.
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The proof just says that it is Hartshorne II.6.8, and I don't understand how it is related to this proposition at all. Hartshorne II.6.8 roughly states that if $f: X \rightarrow Y$ is a morphism where $X$ is a complete nonsingular curve over an algebraically closed field $k$, and $Y$ is any curve over $k$, then either $f(X) = pt$ or $Y$, and in the latter case, $f$ is finite morphism and $[K(X):K(Y)] < \infty$.
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> Can anyone show a proof of the proposition?
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I failed to show that the set of all such $Q$ is open myself, can anyone shed some light on this? Thanks!
| https://mathoverflow.net/users/nan | Generic fiber of morphism between non-singular curves | Here is a complete proof: as remarked in the answer by Norondion, we can reduce to
the case when $C\\_1 \rightarrow C\\_2$ is generically separable, i.e. $k(C\\_1)$ is separable
over $k(C\\_2)$. Let $A \subset k(C\\_1)$ be a finite type $k$-algebra consisting of the regular
functions on some non-empty affine open subset $U$ of $C\\_2$ (it doesn't matter which one
you choose), so that $k(C\\_2)$ is the fraction field of $A$.
By the primitive element theorem, we may write $k(C\\_1) = k(C\\_2)[\alpha]$, where
$\alpha$ satisfies some polynomial $f(\alpha) = \alpha^n + a\_{n-1}\alpha^{n-1} + \cdots
+ a\_1 \alpha + a\_0 = 0,$ for some $a\_i$ in $K(C\\_2)$.
Now the $a\_i$ can be written as fractions involving elements of $A$, i.e. each
$a\_i = b\_i/c\_i$ for some $b\_i,c\_i \in A$ (with $c\_i$ non-zero). We may replace
$A$ by $A[c\\_0^{-1},\ldots,c\\_{n-1}^{-1}]$ (this corresponds to puncturing $U$ at
the zeroes of the $c\_i$), and thus assume that in fact the $a\_i$ lie in $A$.
The ring $A[\alpha]$ is now integral over $A$, and of course has fraction field equal
to $k(C\_2)[\alpha] = k(C\_1)$. It need not be that $A[\alpha]$ is integrally closed,
though. We are going to shrink $U$ further so we can be sure of this.
By separability of $k(C\_1)$ over $k(C\_2)$, we know that the discriminant $\Delta$
of $f$ is non-zero, and so replacing $A$ by $A[\Delta^{-1}]$ (i.e. shrinking $U$
some more) we may assume that $\Delta$ is invertible in $A$ as well.
It's now not hard to prove that $A[\alpha]$ is integrally closed over $A$. Thus
$\text{Spec }A[\alpha]$ is the preimage of $U$ in $C\_1$ (in a map of smooth curves,
taking preimages of an affine open precisely corresponds to taking the integral closure
of the corresponding ring).
In other words, restricted to $U \subset C\_2$, the map has the form
$\text{Spec }A[\alpha] \rightarrow \text{Spec }A,$ or, what is the same,
$\text{Spec }A[x]/(f(x)) \rightarrow \text{Spec A}$.
Now if you fix a closed point $\mathfrak m \in \text{Spec }A,$ the fibre over this point
is equal to $\text{Spec }(A/\mathfrak m)[x]/(\overline{f}(x)) = k[x]/(\overline{f}(x)),$
where here $\overline{f}$ denotes the reduction of $f$ mod $\mathfrak m$.
(Here is where we use that $k$ is algebraically closed, to deduce that $A/\mathfrak m = k,$
and not some finite extension of $k$.)
Now we arranged for $\Delta$ to be in $A^{\times}$, and so $\bar{\Delta}$ (the reduction
of $\Delta$ mod $\mathfrak m$, or equivalently, the discriminant of $\bar{f}$)
is non-zero, and so $k[x]/(\bar{f}(x))$ is just a product of copies of $k$,
as many as equal to the degree of $f$, which equals the degree of $k(C\_1)$ over
$k(C\_2)$. Thus $\text{Spec }k[x]/(\bar{f}(x))$ is a union of that many points,
which is what we wanted to show.
| 5 | https://mathoverflow.net/users/2874 | 13067 | 8,832 |
https://mathoverflow.net/questions/13058 | 7 | I have seen the use of the word "proper Deligne-Mumford stack". Now, it is clear to me what it means for a morphism f of stacks to be proper: as usual it should be representable, and every morphism between schemes obtained from f by base change should be proper.
Now, first I guess that "proper" here actually means "complete". A scheme over a field is complete when the structural morphism to a point is proper. But it does not make sense for a stack to ask that the morphism to the point is proper. Indeed it would be in particular representable, and since a point is a scheme this would imply that the stack itself is a scheme.
Another possibility is that the sentence means "a stack with a proper atlas", so that one cannot speak of proper stacks, but only of proper Deligne-Mumford stacks.
So I am asking here what the standard terminology is.
| https://mathoverflow.net/users/828 | What is a proper stack? | As requested, an answer on terminology
My favorite reference on basics for DM stacks is [Edidin's paper](http://arxiv.org/abs/math/9805101), which I find much easier to read than Laumon & Moret-Bailly (who of course deal with Artin stacks).
Short summary. Suppose $P$ is a property of morphisms $f:X\to Y$ in the category of schemes:
* If $P$ is local on both $X$ and $Y$ (`local' in appropriate topology, e.g., etale for DM stacks), it makes sense for morphisms of stacks (pass to compatible presentation of both stacks).
* If $P$ is local on $Y$ only, it is easy to define for representable morphisms $F:{\mathcal X}\to{\mathcal Y}$ by changing base to a presentation of ${\mathcal Y}$.
* If $P$ is local on $Y$, but you want to make sense of it for all morphisms, you have to make a special definition --- there is no general approach that works for all properties.
This is what happens with definitions of separated/proper morphism of stacks.
So proper morphism of stacks need not be representable.
| 9 | https://mathoverflow.net/users/2653 | 13076 | 8,836 |
https://mathoverflow.net/questions/13032 | 13 | In the symplectic topology view on Gromov-Witten-Invariants some authors use what they call a Kuranishi structure on the moduli of stable maps. These were introduced by Fukaya and Ono and are also used in their big book on Fukaya categories.They are also used a lot in recent papers by Joyce.
The key feature of the Kuranishi structure is that there locally exists a homeomorphism or a diffeomorphism, depending on whether you follow Fukaya or Joyce, to a zero set of a section of the "obstruction bundle".
On the algebraic geometry side there also is something involving the term obstruction, namely the perfect obstruction theory on the moduli space of maps. This is a morphism [E-1 -> E0] -> LX, where L is the cotangent complex.
Here's my question: To what does the obstruction bundle on the symplectic topology side correspond on the algebraic side?
There are three candidates I can think of:
* E-1
* the kernel of E-1 -> E0
* the thing the kernel of E-1 -> E0 maps surjectively to, mostly called the T2
The definition of the obstruction bundle on the symplectic side is a finite dimensional supspace of the cokernel of the linearized operator of the pseudoholomorphic curve equation, p.978 of Fukaya-Ono:"Arnold Conjecture and Gromov Witten Invariant." As a die hard algebraic geometer, that's just to hard to digest...
| https://mathoverflow.net/users/473 | Obstruction bundle for spaces with Kuranishi structure | Here's a view of the symplectic side of the bridge.
The Kuranishi model (see Donaldson-Kronheimer, *The geometry of four-manifolds*, ch. 4) goes like this. You're interested in a (moduli) space $M$ cut out as $\psi^{-1}(0)$, for some smooth but nonlinear map of Banach spaces, $\psi \colon (E,0) \to (F,0)$ such that $\delta:=D\_0\psi$ is a Fredholm operator (finite dim kernel and cokernel). That means that $\delta$ is "almost" an isomorphism, and Kuranishi's principle is that one can construct a non-linear map $\kappa \colon \ker(\delta) \to \mathrm{coker}(\delta)$, such that $\kappa(0)=0$ and $D\_0 \kappa =0$, and (locally near $0$) a homeomorphism $M \to \kappa^{-1}(0)$. This gives a finite-dimensional model of $M$. "Kuranishi structures" are a formalism in which one can say that $M$ is everywhere-locally given as the zeros of maps like $\kappa$.
In the case of genus 0 GW theory, $M$ is the moduli space of (say) parametrized pseudo-holomorphic maps from $S^2$ to an almost complex manifold $X$; $\psi$ is a non-linear Cauchy-Riemann operator, and, for a pseudo-holomorphic map $u\in M$, $D\_u \psi$ is a linearized C-R operator - the $(0,1)$-part of a covariant derivative acting on sections of $u^\ast TX$. Its kernel can be identified with the holomorphic sections $H^0(S^2,u^\ast TX)$ of the holomorphic structure on the vector bundle $u^\ast TX$ defined by the C-R operator. Its cokernel is isomorphic to $H^1(S^2,u^\ast TX)$. If you're lucky, you have a Zariski-smooth moduli space $M$ whose Zariski tangent space at $u$ is $\ker D\_u\psi$. In this case, one could take $\kappa=0$, and then the spaces $\mathrm{coker} (D\_u\psi)$ form a vector bundle $Obs \to M$, which is what symplectic geometers usually call the obstruction bundle. One can now try to divide everything by $Aut(S^2)$, and quite possibly get into orbi-mathematics.
In the integrable case, still with non-singular but excess-dimensional $M$, I could write $Obs$ as $R^1\pi\_\* \Phi^\*T\_X$, where $T\_X$ is the holomorphic tangent sheaf, $\Phi$ the evaluation map $\mathbb{P}^1\times M \to X$, and $\pi$ the projection to $M$. At this point, if what I've said is accurate, you and other algebraic geometers out there are better placed than me to answer your question. Is it your $(E\_{-1})^\vee$?
Of course, you didn't really want to assume $M$ smooth. A place where these deformation theories are compared in generality is Siebert's 1998 paper [Algebraic and symplectic Gromov-Witten invariants coincide](http://arxiv.org/abs/math/9804108).
| 13 | https://mathoverflow.net/users/2356 | 13078 | 8,838 |
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