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https://mathoverflow.net/questions/42774 | 8 | Given a set $A\subset \mathbb{R}^n$ such that $A\cap (x+\mathbb{Z}^n)\ne \emptyset$ for any $x\in \mathbb{R}^n$ (that is, $p(A)=\mathbb{T}^n$ for the projection $p:\mathbb{R}^n\rightarrow \mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n$). Is it true that supremum of Eucledian distances between points of $A$ is not less then $\sqrt{n}$? (equality holds for the unit cube)
Or maybe even two points with $x=(x\_1,\dots,x\_n)$, $y=(y\_1,\dots,y\_n)$ with $|x\_i-y\_i|\geq 1-\epsilon$ for each coordinate?
It is not hard to check both claims for $n=2$, but already for $n=3$ I do not know.
| https://mathoverflow.net/users/4312 | minimal diameter of full preimage of torus | The second claim is false for $n=3$. Choose $\varepsilon$ small and $\delta\ll\varepsilon$. Let $A$ be the set of all points $(x,y,z)\in\mathbb R^3$ satisfying the following inequalities:
$$
\begin{cases}
-1.5+\varepsilon+\delta &\le x+y+z &\le 1.5+\varepsilon \\
-1.5+\delta &\le x+y-z &\le 1.5 \\
-1.5+\delta &\le x-y+z &\le 1.5 \\
-1.5+\delta &\le -x+y+z &\le 1.5 \\
\end{cases}
$$
The integer translates of this set cover the space, but its $\ell\_1$-diameter is no greater than $3-\delta$.
**Added.** The first claim is false too. In the above example, fix $\delta=\varepsilon/10$ and add the inequality
$$\max\{|x|,|y|,|z|\}\le 0.5+10\varepsilon$$
to the system.
| 4 | https://mathoverflow.net/users/4354 | 42792 | 27,224 |
https://mathoverflow.net/questions/42791 | 3 | I need an algorithm to solve a Quadratic Programming optimization problem where the unknowns are allowed to be negative.
I have an implementation of the Philip Wolfe simplex algorithm based on his article <http://pages.cs.wisc.edu/~brecht/cs838docs/wolfe-qp.pdf>, but it assumes x >= 0. In other places describing similar algorithms it always assumes non-negativity.
I have seen descriptions of the QP problems without such restriction, but no links to the algorithms there.
I tried to extend Wolfe's algorithm by allowing unrestricted values in the simplex tableau (the same way I would do it for the linear case), and it gives me some results, but I don't know whether it is correct or whether it will work reliably.
So basically I need to know if simplex algorithm can be enhanced to handle unrestricted unknowns, and if so, I want to see the mathematical proof that this enhanced algorithm works.
If the answer is no, then I would like some suggestions of how to approach this problem. Do I need to use a combinatoric approach where some variables are non-negative and others non-positive, and try all combinations? Or is there more efficient way?
| https://mathoverflow.net/users/10176 | Quadratic optimization without non-negativity restriction | You could replace each unrestricted variable x by (y-z) where y and z are each restricted to be positive.
| 1 | https://mathoverflow.net/users/9501 | 42793 | 27,225 |
https://mathoverflow.net/questions/42780 | 5 |
>
> **Question 1.** Given a topological space $X$ and two metrics $a$ and $b$ on it, compatible with the topology, what conditions should I impose on them so that box-counting (or other, for example Hausdorff) dimensions of $(X,a)$ and $(X,b)$ are equal?
>
>
> **Question 2.** Is there a notion of a dimension for topological (as opposed to metric) spaces which can assume non-integral values?
>
>
>
---
My motivation
-------------
Let $G$ be a finitely generated group and let $p$ be a prime number. Consider the compact topological space $X:=\prod\_G \mathbb Z/p$, infinite product of copies of the cyclic group of order $p$, indexed by elements of $G$. Let $T$ be an element of the integral group ring of $G$. Note that $T$ gives a map $X\to X$ in a natural way (in this context $T$ is sometimes called *cellular automaton*). Define $Y$ to be the subset of those points $x$ of $X$ such that $T(x)=0$.
I want to measure "how big" $Y$ is.
If we choose a generating set for $G$ then we get a metric $d$ on $G$, and we also get a metric on $X$: two sequences $x\_i$ and $y\_i$ of $X$ are $p^{-|B(1,k)|}$ apart if they agree on the ball $B(1,k)$ of radius $k$ around the neutral element $1$ of $G$, but they don't agree on any larger ball. It's straightforward to see that box counting dimension of $X$ is $1$.
Unfortunately the metric we just defined depends on the generators chosen for $G$, so I'd like to know if I have a chance to get any kind of "intrinsic size of $Y$" this way.
| https://mathoverflow.net/users/2631 | How indepenedent of a chosen metric is the box-counting dimension? Is there a non-integral dimension which is defined for topological spaces? | **Q2**. If you want that dimension for topological spaces to agree with Hausdorff dimension (for example) in case of metric space, then NO. For any $0 \le s \le \infty$, there is a metric on the Cantor set so that the Hausdorff dimension is $s$.
**Another topological result**. Let $X$ be a separable metrizable space. The infimum of the Hausdorff dimensions of all metric spaces homeomorphic to $X$ is the topological dimension of $X$. (Integer valued.)
**for the My Motivation comment**. See the notion "topological entropy".
| 5 | https://mathoverflow.net/users/454 | 42804 | 27,229 |
https://mathoverflow.net/questions/42809 | 19 | Let $k \ge 4$ be an even integer, and let $d$ be the dimension of the space $M\_k(\operatorname{SL}\_2(\mathbb{Z}))$ of modular forms of level 1 and weight $k$. Then the space of Hecke operators acting on $M\_k$ also has dimension $d$. Is it spanned by $T\_1, \dots, T\_d$?
Equivalently (more explicitly but also more messily): if $f \in M\_k(\operatorname{SL}\_2(\mathbb{Z}))$ satisfies $a\_i(f) = 0$ for $1 \le f \le d$, where $a\_i(f)$ are the $q$-expansion coefficients of $f$, with no assumption on $a\_0(f)$, then is it necessarily true that $f = 0$?
(Edit: See also this [follow-up question](https://mathoverflow.net/questions/42815/how-many-hecke-operators-span-the-hecke-algebra) which asks a related question for modular forms of higher level.)
| https://mathoverflow.net/users/2481 | How many Hecke operators span the level 1 Hecke algebra? | The answer is yes when $k$ is a multiple of $4$. There is a unique form
of weight $k$ of the form $f\_k=1+a\_dq^d+\cdots$. When $k$ is a multiple
of $4$ this is the theta series for a putative extremal even unimodular
lattice of rank $2k$. Theorem 20 in chapter 7 of Conway and Sloane's
*Sphere Packings, Lattices and Groups* asserts that $a\_d>0$. They give
several references for the proof, including a 1969 paper of Siegel.
| 12 | https://mathoverflow.net/users/4213 | 42811 | 27,232 |
https://mathoverflow.net/questions/42808 | 19 | This is partly inspired by answers to the question:
[Question about Hodge number](https://mathoverflow.net/questions/42709/question-about-hodge-number/42735#42735) .
Is there a family of compact complex manifolds, where the general fibres are
Kähler, but for which $E\_1$ degeneration of the Hodge to de Rham spectral sequence fails
at the special fibre? Or, even better, such that the special fibre has nonclosed
holomorphic forms?
I feel like I should know the answer, but somehow I don't. All
the examples I know where the spectral sequence doesn't degenerate are nilmanifolds\*,
so they aren't even homotopic to Kähler manifolds by standard rational homotopy theoretic obstructions (e.g. they aren't formal).
Also the famous Hironaka example [Ann. Math 1962] won't work either, because
the special fibre is an algebraic variety, so the spectral sequence will degenerate
(by an argument that can found in Deligne [Théorème de Lefschetz...]).
Obviously, I haven't thought about this deeply enough, but perhaps someone else has\*\*.
**Footnotes**
\*I was bit sloppy yesterday, since the examples I have in mind include
solvmanifolds. However, there are still topological obstructions to these being Kähler
due to Nori and myself.
\*\* From the answers, I gather that the work of Popovici suggests that
there may be no counterexample.
| https://mathoverflow.net/users/4144 | Deformations of Kähler manifolds where Hodge decomposition fails? | This is known, for projective (even Moishezon)
manifolds as shown by Dan Popovici in his
paper <http://arxiv.org/abs/1003.3605>
For general Kaehler manifold, this is conjectured.
Popovici has proved that a property of "strong Gauduchon"
is preserved in limits <http://arxiv.org/abs/1009.5408>
and (I think) there are no example of strong Gauduchon
manifold without Hodge decomposition.
| 12 | https://mathoverflow.net/users/3377 | 42823 | 27,238 |
https://mathoverflow.net/questions/42818 | 6 | At a high level, my question is the following: given a set of $k$ vectors in Euclidean space which are pairwise "almost orthogonal", can one find a set of $k$ orthogonal vectors which are pairwise close to the original ones? This could be seen as a stable version of Gram-Schmidt orthogonalization, in which, under the promise that the original set of vectors is not too far from being orthogonal, one has the guarantee that they do not need to be moved too much in order to become orthogonal.
More precisely, assume given vectors $v\_i$, $i=1,\ldots,k$ in a real (or complex) $k$-dimensional space, such that $\sum\_{i\neq j} \langle v\_i,v\_j\rangle \leq \epsilon k$ ($\epsilon$ should be understood as an arbitrarily small constant, or it could even be $o(k)$ if needed -- the weaker the assumption the better). Then, does there exist a set of orthogonal vectors $w\_i$ such that $\sum\_i \|w\_i - v\_i\|^2 \leq \epsilon' k$? (The norm is the usual Euclidean norm.) The interesting question is whether one can get an $\epsilon'$ which depends on $\epsilon$ only, not on $k$.
A related question (the one I am originally most interested in), is that of orthogonalizing $d$-dimensional projector matrices $P\_1,\ldots,P\_k$: assume $\frac{1}{d} \sum\_{i\neq j} \langle P\_i,P\_j \rangle \leq \epsilon$ (where now the inner product is the trace inner product), can you find orthogonal projectors $Q\_i$ such that $\frac{1}{d}\sum\_i \|P\_i-Q\_i\|\_F^2 \leq \epsilon'$? (Here the norm is the Frobenius norm, the sum of squares of the coefficients.) So far using various iterative procedures I can only get a bound where $\epsilon' = poly(\log k) \epsilon$, but I would like to know if the dependence on $k$ can be removed.
| https://mathoverflow.net/users/10183 | Stable orthogonalization procedure | Use a Procrustes rotation of the standard basis vectors onto your vectors. This gives the set of orthogonal vectors with the smallest sum of squares of distances to your vectors.
<http://en.wikipedia.org/wiki/Orthogonal_Procrustes_problem>
"The orthogonal Procrustes problem is a matrix approximation problem in linear algebra. In its classical form, one is given two matrices A and B and asked to find an orthogonal matrix R which most closely maps A to B."
In your case you want to find the orthogonal matrix R which most closely maps the standard basis to your matrix. Something like the columns of R should then be the set of orthogonal vectors which are nearest to your vectors, where 'nearest' is in the sense of sum of squares.
| 4 | https://mathoverflow.net/users/9501 | 42824 | 27,239 |
https://mathoverflow.net/questions/42837 | 7 | Suppose $T$ is an ergodic measure-preserving transformation on a measure space $(X,\Sigma,\mu)$, and $f\in L^1(\mu)$. Does the limit
$\lim\_{X\to\infty} \pi(X)^{-1}\sum\_{p\leq X} f(T^{p}x)$
exist almost everywhere? Is it constant almost everywhere? Here the sum runs over primes, and $\pi(X)$ is the prime counting function.
When $X=\mathbb{R}/\mathbb{Z}$ with Lebesgue measure, and $T:x\to x+\theta$ is an irrational rotation, the answers to these questions are "yes" and "yes", by Vinogradov.
| https://mathoverflow.net/users/1464 | A non-standard ergodic limit | Here is a partial answer: Mate Wierdl proved that the limit exists almost everywhere if $f \in L^r (\mu)$ for some $r>1$. See "Pointwise Ergodic Theorem along the Prime Numbers".
Also, there is a recent article by Trevor Wooley and Tamar Ziegler ("Multiple Recurrence and Convergence along the Primes") which proves $L^2$ convergence and multiple recurrence for more complicated ergodic averages of this kind.
| 3 | https://mathoverflow.net/users/7392 | 42842 | 27,251 |
https://mathoverflow.net/questions/42678 | 1 | ¿Could somebody tell me how can i write a zero order bessel function in an Hermite-Gauss basis?
Thanks
| https://mathoverflow.net/users/10155 | Bessel functions in an Hermite-Gauss basis | The Hermite-Gauss functions (suitably normalized) are orthonormal for an inner product defined by integration, just like $\{\sin(nx),\cos(mx)\}$ as $n,m$ run over positive integers. The coefficients of an expansion of a $J$-Bessel function in terms of Hermite-Gauss functions are computed by integrating the Bessel functions against the various Hermite-Gauss functions, in a way analogous to how a Fourier expansion is computed. You'll want to be looking at a table of integrals, or maybe *Mathematica*. (I'm not guaranteeing anything about the convergence of the series; I haven't thought about it that much.)
| 2 | https://mathoverflow.net/users/6756 | 42843 | 27,252 |
https://mathoverflow.net/questions/42847 | 12 | What kind of categories $C$ have the property that each slice category $C/c$ is a topos? Obviously, topoi have this property, but, the converse is not true. An example is the category $EtTop$ of topological spaces and only local homeomorphisms. $EtTop/X \cong Sh(X)$, but $EtTop$ is very far from being a topos! It doesn't have an initial or final object, nor pushouts...
Are there other of these "locally a topos" categories which are not topoi? Have they been studied before?
| https://mathoverflow.net/users/4528 | Locally a topos | The question of local toposes and similar categories was discussed a couple of years ago on the categories list by Peter Johnstone and others, if I recall correctly. I don't know anywhere that they appear in print, but I don't know the topos theory literature nearly well enough to be an authoratitive source on this.
---
On the local toposes themselves: $\newcommand{\Topos}{\mathbf{Topos}\_\mathit{slice}}$one other in-some-sense-trivial example, if I'm not mistaken, is the category $\Topos$, with objects all (small, fsvo small) elementary toposes, and with maps just the geometric morphisms that are (up to equivalence) induced by slicing, modulo natural isomorphism. (If we wanted to cover our tracks a little, we could say “the geometric morphisms whose inverse image functors are logical”; the equivalence of this is shown in Mac Lane and Moerdijk in their chapter on logical morphisms, iirc.)
But this is the *universal* example: any other local topos $\newcommand{\E}{\mathcal{E}}$has a unique-up-to-equivalence “local equivalence” $\E \to \Topos$, sending $A$ to $\E/A$, and any local equivalence into $\Topos$ must come from a local topos.[1]
But local equivalences into a fixed category $\newcommand{\C}{\mathcal{C}} \C$, in turn, correspond (up to equivalence-over-$\C$) to functors $\C^\mathrm{op} \to \mathbf{Sets}$. ($F : \mathcal{D} \to \C$ corresponds to the functor taking $C \in \C$ to the set of isomorphism classes of objects of $\mathcal{D}$ over $C$.) Of course, there's a size consideration: whatever size of $\mathbf{Sets}$ we use constrains the essential size of the fibers.
In terms of local toposes seen as categories in their own right, one point of interest is that you can interpret pretty much all the logic in them that you can in toposes (i.e. higher-order type theory; and geometric logic if you go Grothendieck-y)… except that of course you have to abandon the “empty context”, since you don't have a terminal object. (Everything else in the interpretation of logic is purely local.) In a local topos, there is no “global validity”: all truth is relative :-)
[1] The uniqueness issues here are a little subtle: if I'm not mistaken we need to either assume (large) choice, or use anafunctors instead of functors, or restrict the size of $\E$ enough that each slice will itself literally *be* an object of $\Topos$.
| 12 | https://mathoverflow.net/users/2273 | 42858 | 27,261 |
https://mathoverflow.net/questions/42851 | 2 | The [ETCS](http://ncatlab.org/nlab/show/ETCS) axioms give conditions on a category for it to be a category of sets. These axioms can be [written out in first order language](http://ncatlab.org/nlab/show/fully+formal+ETCS), resulting in a finite axiomatisation of the category of sets. Given a model of ZFC one can form the associated category of sets and this will satisfy the axioms. On the other hand, the axioms do not require a priori defined sets. (The existence of a model of the ETCS axioms then is probably on par with the assuption there exists a model of the ZFC axioms) The appropriate definition of map between ETCS categories is a [geometric morphism](http://ncatlab.org/nlab/show/geometric+morphism).
Is it possible to define geometric morphisms elementarily? In first order language?
Then one can define the category of ETCS categories and consider the relations between models. This is related to my [previous question](https://mathoverflow.net/questions/42710/how-do-we-compare-models-of-etcs).
| https://mathoverflow.net/users/4177 | Can we define geometric morphisms (between ETCS categories) elementarily? | Yes, it is possible. Precisely, we can write down a first-order theory for which a model is a pair of ETCS-models and a geometric morphism between them (am I right in thinking this is what you're asking for?).
To do this, on top of axiomatising “a pair of models of ETCS”, you add some extra function symbols for the adjunction. The conditions of functoriality, etc. are easily written algebraically; the adjunction can be expressed in various ways, of which the simplest to write down is probably the [triangle-inequalities form](http://ncatlab.org/nlab/show/triangle+identities). “Preserving finite limits”, when you write it out, is also just a scheme of first-order conditions; if you want to reduce it to a finite axiomatisation, note that it's enough to ask for preservation of finite products and equalisers (by the usual proof that all finite limits can be constructed from these).
This said, I disagree somewhat with an implicit premise of your question. You say: “Then one can define the category of ETCS categories…” But to do this, you don't need to show that geometric morphisms can be defined in first-order terms. To talk about “the category of ETCS categories”, you already need to be working in a meta-theory with some sort of notion of set or similar (eg types, etc.); and so don't need the definitions of the morphisms to be first-order.
The foundational advantage of a first-order axiomatisation of widgets is that you can then study a *single* widget without needing any meta-theory. But to study the collection of all widgets (as a category or whatever else), you still need a meta-theory.
| 6 | https://mathoverflow.net/users/2273 | 42860 | 27,263 |
https://mathoverflow.net/questions/42853 | 15 | Let $K$ be the imaginary quadratic field obtained by joining $\sqrt{-1}$ to the field of rational numbers $Q$. I would like to describe the extension $K^{ab}/Q^{ab}$, where for $F$ a number field, $F^{ab}$ denotes its maximal abelian extension (everything is taking place inside a big fixed field...).
More precisely I would like to know the Galois group and the ramification properties of such extension. Is this possible/easy? I suppose one should look at the kernel of the norm map between Idele class groups $N\_{K/Q}:I\_K\rightarrow I\_Q$. But at the moment it is not clear to me how to get the answer. Any hint or comment would be appreciated. Thanks.
EDIT: Probably the idele class group of a number field $F$ should be denoted by $J\_F$. Or by anything other than $I\_F$...
| https://mathoverflow.net/users/10189 | An explicit computation in class field theory | Given that you want to know the structure of the Galois group and ramification, I think that you are best off working with the kernel of the norm map between connected components of idele class groups, as you yourself suggest.
These groups are very explicit: for $K := \mathbb Q(i)$, one obtains $\hat{\mathcal O}\_K^{\times}/\{\pm 1,\pm i\}$,
and for $\mathbb Q$ one obtains $\hat{\mathbb Z}.$ (Here $\hat{}$ denotes the profinite
completion.) Apart from the diagonally embedded $\{\pm 1,\pm i\}$ quotient in the
group for $K$, both groups factor as a product over primes, and the norm map is given
component wise.
So the kernel of the norm map is equal to
$$\bigl(\prod\_p (\mathcal O\_K\otimes\_{\mathbb Z}\mathbb Z\_p)^{\times, \text{Norm } = 1}\bigr)/
\{\pm 1,\pm i\}.$$
This should be explicit enough to answer any particular question you have.
| 16 | https://mathoverflow.net/users/2874 | 42861 | 27,264 |
https://mathoverflow.net/questions/42778 | 0 | Suppose we are given a representation of a finite series of natural numbers:
$\sum\_{i=0}^N{c\_i x^i}$
The representation is essentially an expression that is a rational function of two polynomials.
Is it possible to add/subtract this series repeatedly to get a result that contains only part of the series?
>
> A simple formulation of the problem
>
>
>
We can forget that we have a series, for a moment, and consider it more like a tile. It contains a set of numbers on it that are in a particular order that cannot be changed. It's basically striaght; it just contains this set of numbers on a line.
Now we actually have many copies of this same tile, and some are oriented differently. The different orientation comes from the fact that the tile could be on a line into another dimension. For example, one tile may start at coordinates (0,0) and end at coordinates $(0,N)$. Another may start at (0,0) and end at $(N,0)$. All tiles are essentially the same ordered set of numbers that start at one point in $m$ dimensions and end at another point.
There is essentially one operation that we can perform. We can take the difference of numbers at a given point. For example, if the sequence is (1,2,3), we can take one tile that starts at (0,0) and proceeds to (0,2). Subtract another tile that starts at (0,0) and proceeds to (2,0). This would cancel out the ones in both tiles, since we subtract the one at (0,0) in the second tile from the one at (0,0) in the first tile.
I'm wondering if we can somehow add and subtract these tiles so that only a single number remains. It may be that more than one location contains this same number.
There are some rules, so an example and explanations will probably help.
>
> Could you give an example?
>
>
>
Here's an example. Consider the finite series $10 + 200x$.
We can repeatedly add and subtract (scalar multiples of) this series (and a similar series) in the $x,y$ "plane" to end up with $200x$, only part of the series.
Here's how:
(0) Start with $10+200x$.
(1) Subtract $10 + 200y$. This eliminates 10 and we're left with $200x-200y$.
(2) Add $(10+200x)\cdot y / x$. This eliminates the $-200y$ in the previous result and adds $10\cdot y / x$, resulting in $200x+10\cdot y / x$.
(3) Subtract$(10+200y)\cdot y / x$. This eliminates $10\cdot y / x$, subtracts $200\cdot y^2 / x$, and we're left with $200x - 200\cdot y^2 / x$
(4) Add $((10+200x)\cdot y / x)\cdot y / x$. This eliminates the $-200\cdot y^2 / x$ in the previous result and adds $(10\cdot y / x)\cdot y / x$, resulting in $((200x+10)\cdot y / x)\cdot y / x$.
...
We repeatedly add and subtract pieces of the series this way. It can be shown that the result of this infinite series is $200x$, which is only part of the original series.
I'd like to know if we can add and subtract larger series similarly (but possibly in more dimensions) to end up with only a single coefficient or piece of the series.
For example, If we consider the series $1 + 2x + 3x^2$, we may be able to eliminate $1$ and $3x^2$ from this series by a set of careful additions and subtractions.
I don't know which branches of mathematics this deals with, but I'm hoping an expert can provide me with some direction.
>
> Why are you doing this/ What's your motivation?
>
>
>
Results from this "puzzle" would help speed up an algorithm significantly, so I'm interested in publishing this result with whoever helps me. I know MO's position on algorithms, but I believe that this is more of a "tiling" or mathematics question than a question on an algorithm.
I'd like to know all branches of mathematics that deal with this question, and where I can go, or who I can go to, that will help me solve the general problem.
Some additional notes: The value of $x$ is not allowed to be modified. We can rewrite $x$ as another variable (I rewrote $x$ as $y$ above, for example). We are also allowed to "shift" the series (for example, multiply by $y / x$ as done above). Cancellation is, of course, allowed - and it may be done infinitely many times. These are the only allowable operations.
Also, I am particularly interested in the coefficient in the middle of the series.
One additional consideration: methods that use fewer variables are preferred.
| https://mathoverflow.net/users/3647 | Piece of a sequence | I'm not sure I fully understand the problem, but here goes. Start with $10+200x$. Multiply by $20x$ to get $200x+4000x^2$. Subtract $(10+200x)(400x^2)$ to get $200x-80000x^3$. Add $(10+200x)(8000x^3)$ to get $200x+1600000x^4$. Keep going. The result of this infinite procedure is, in some sense, $200x$, and you don't even need $y$ to do it.
All I'm doing is solving $(10+200x)Q(x)=200x$; you get $Q(x)=20x-400x^2+8000x^3+\dots$, which tells you what to add and subtract.
If you want to eliminate 1 and $3x^2$ from $1+2x+3x^2$, solve $2x=(1+2x+3x^2)Q(x)$ one way or another to get $Q(x)=2x-4x^2+2x^3+8x^4-22x^5+\dots$, and that tells you what you have to do.
If this doesn't solve your problem, then you may have to rethink your presentation.
EDIT: In view of OP's comment, maybe this is closer to what's wanted. Let $f(x)=a+bx+cx^2$ where $a,b,c$ are unknown. Here's how to pick out $bx$: $${\bigl(f(x)-f(y)\bigr)(x^2-z^2)-\bigl(f(x)-f(z)\bigr)(x^2-y^2)\over(x-y)(x^2-z^2)-(x-z)(x^2-y^2)}x=bx$$ I don't see how to do it with just $x$ and $y$, indeed, I think that if you start with a polynomial with $n$ terms you'll need $n$ variables. But I think that you can always pick out any term you like if you use that many variables, and I think you can work out how to do any particular case once you've worked out how the formula above does its job.
| 5 | https://mathoverflow.net/users/3684 | 42862 | 27,265 |
https://mathoverflow.net/questions/42820 | 2 | Is there a way to simplify the following expression:
$\lgroup{\int^A\_0 x(s)ds}\rgroup ^2$
I'm looking for an expression that can possibly get rid of the squared term, so that I can have just an integral of the first order.
| https://mathoverflow.net/users/10184 | Expressions for the Square of an Integral | I'm not sure about simplifying, but you can easily write your objective functional in Bolza form like this:
$$
\begin{align}
&\min\_{u(t) \in \Omega(t)} \, J = z(T)^2 + \int\_{0}^{T} s(t)u(t)dt \\
s.t. &\frac{dz(t)}{dt} = r(t)u(t),\quad z(0) = 0
\end{align}
$$
| 1 | https://mathoverflow.net/users/7851 | 42869 | 27,269 |
https://mathoverflow.net/questions/42875 | 3 | At first, if a group G is an infinite loop space (all are based), then `\pi_0(G)` must be Abelian. Therefore, if G is discret, then it must be Abelian. In fact, any Abelian group does be infinite loop space, by the EM space construction. But we have non-Abelian examples, the infinite groups U and O are infinite loop spaces, by Bott periodicity. (Does this contradict with the statement that the coefficient of a cohomology must be an Abelian group?) Are there any other examples?
| https://mathoverflow.net/users/7341 | Which Groups are Infinite Loop Spaces? | Edit: I just reread the question, and it says "*if* a group is an infinite loop space..." I realise the first paragraph of my answer is incorrect. The rest still stands.
---
Firstly, $\pi\_0(G)$ does not have to be abelian - it is $\pi\_1(G,e)$ which is abelian, as the Eckmann-Hilton argument shows.
The coefficients of cohomology do not *have* to be abelian groups, but I guess you are referring to the idea that the ordinary cohomology of a space (singular, or sheaf, say) only makes sense in all non-negative dimensions for abelian-group coefficients. One can define $H^1(X,G)$ ($X$ a space) for a nonabelian (topological) group, but for higher degree cohomology it is not straightforward, see [this MO answer](https://mathoverflow.net/questions/36466/do-we-have-non-abelian-sheaf-cohomology/36508#36508).
Now notice that one can use loop spectra as coefficients for cohomology, but one gets an extraordary cohomology theory: this is the case of $U$ and $O$, which represent spectra, and given $K$-theory and $KO$-theory respectively. (see the Wikipedia page on K-theory for example)
| 0 | https://mathoverflow.net/users/4177 | 42877 | 27,273 |
https://mathoverflow.net/questions/14803 | 4 | There are nonequivalent geometries, nonequivalent groups finite and infinite, nonequivalent logics ( fregean and nofregean <http://www.formalontology.it/suszkor.htm>), even nonequivalent logicians;-)
**Are there nonequivalent randomnesses?**
The main two theories we know dealing with randomness and probability is Kolmogorov Probability Theory ( by means of measure theory and Borel $\sigma$-algebras), and Bayesian a priori approach. Are they equivalent enough to say that they are the same in some deeper meaning?
---
**@ Johannes Hahn** - no, I am not asking about non isomorphic probability spaces as it would be trivial. rather I as about possible probability theories as different as different are geometries euclidean and noneuclidean.
The obvious generalization is to change linearity in the second axiom of probability ($P(A U B) = P(A) + P(B)$ when A,B are independent.
In fact I mention about it after reading probability overview by Terence Tao <http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/> He wrotes:
>
> probability theory is only “allowed”
> to study concepts and perform
> operations which are preserved with
> respect to extension of the underlying
> sample space.
>
>
>
Which in my opinion is something very deep ( but I am only a hobbyist;-). So probably if You have initial probability space, and You need to extent it to describe some additional phenomena, You have have done some kind of morphisms between structures of first space and another wider one. Is there a **unique**, canonical or any defined way of doing this? May we perform this kind of extension always in the same way or there are different ways of doing that? Gives it any predictable and interesting structure?
**@sheldon-cooper** Bayesian approach to probability is sometimes seen as alternative ( not very well defined) to Kolmogorov axiomatic probability system, because it do not require given a priori probability space. For example in this approach we may say that probability that tomorrow will be could day ( say temp<-10) is defined, whilst in Kolmogorov approach You probably cannot define proper space ( because You cannot have equivalent population of Wednesdays which are tomorrow days, with different temperatures). Agree - when You have possibility to use properly defined probability space this two approach coincides. From Wikipedia: <http://en.wikipedia.org/wiki/Bayesian_probability>
>
> Bayesian probability interprets the
> concept of probability as "a measure
> of a state of knowledge",[1] in
> contrast to interpreting it as a
> frequency or a physical property of a
> system.
>
>
>
**@Qiaochu Yuan** - of course randomness here is colloquialism. Yes You have right: maybe I just should ask about different probability theories, but note that non-euclidean geometries in analogy are just geometries but in different spaces, with some special properties. So in fact they share the same meaning of geometric set, figure, space, even so complicated objects as coordinate systems, and angle. But they have different relations between them. So I ask about something similar: different kinds of randomness which are in scope of probability theory but describes different relations between for example different classes of ways of extensions of probability spaces. If the last procedure changes anything in resultants;-) Agree - maybe this is not very interesting question. Maybe it would be more interesting in scope of algorithmic information theory and its randomness concept?
| https://mathoverflow.net/users/3811 | Are there nonequivalent randomnesses? | A different answer from the ones so far: Quantum randomness is another kind of randomness that is a generalization of traditional randomness, i.e. classical or non-quantum probability. I think that it fits the question because you could likewise say that non-Euclidean geometry, interpreted as not-necessarily-Euclidean geometry, is a generalization of Euclidean geometry.
A classical probability space is usually defined as a $\sigma$-algebra $\Omega$ with a normalized measure. From the Bayesian viewpoint the measure could equally well be called a "state". Now, a $\sigma$-algebra is the algebra of Boolean random variables with a certain set of axioms. But you can just as well write down axioms for $L^\infty(\Omega)$, the algebra of bounded complex random variables. In favorable cases, it is a commutative von Neumann algebra. In quantum probability you instead allow a non-commutative von Neumann algebra $\mathcal{M}$. Also, in standard quantum probability you keep the usual completed tensor product $\mathcal{M} \otimes \mathcal{N}$ as the model of a joint system. (Free probability theory is still quantum probability, but with a certain free product instead of a tensor product.) You also still have states, conditional states, joint states, correlations, generalized stochastic maps, etc.
Some of the variant models mentioned so far lead to different theorems, but generally give the same answers in combinatorial probability, questions like the birthday paradox or modeling games of chance. Quantum probability leads to a significantly different picture of combinatorial probability, generalizing the old one, but also allowing new answers such as violation of Bell's inequalities, covariance matrices that are Hermitian rather than real symmetric, new complexity classes such as BQP, etc.
Other variant models mentioned so far no longer give any answers for combinatorial probability, for instance models of forcing. But, part of the interest in probability is that it models real life. Amazingly, so does quantum probability; that was the central discovery of quantum mechanics when it was defined in the 1920s and 1930s.
| 5 | https://mathoverflow.net/users/1450 | 42885 | 27,280 |
https://mathoverflow.net/questions/42888 | 5 | It is well known that in the category of all topological spaces, quotient maps aren't preserved by products (this follows from the simpler fact that $X\times (-):Top\to Top$ doesn't preserve quotients). The usual solution, if one is needed, is to change to the category of k-spaces and k-continuous maps. There are other categories where products and quotients 'get along' (e.g. $Set$, $Ab$).
Question 1: What is a large class of categories where quotient maps are preserved by products? Topoi? (Semi)abelian categories? Categories of algebras for a monad on a given category with this property?
Now one may be only interested in a certain class of quotient maps (like surjective submersions in $Diff$, the category of finite dimensional smooth manifolds). Say, regular epimorphisms, or maps admitting local sections (assuming we're in a site), or perhaps something like surjective topological submersions, where there are sections through every point in the domain. So in this case it is not a matter of putting restrictions on $X$ such that $X\times(-)$ preserves quotients, or changing the category, but narrowing the scope of the quotient maps one wants to preserve.
Question 2: Is there are large class of quotient maps (in $Top$, or in a general category - with finite products and enough colimits) that *are* preserved by products?
| https://mathoverflow.net/users/4177 | Categories with products that preserve quotients | Note: I have edited this answer further because I was being silly before (unnecessarily restrictive).
I take it you mean categories $E$ for which $- \times -: E \times E \to E$ preserves quotients. The word 'quotient' may be slightly ambiguous because sometimes people use it to mean'coequalizer', and sometimes just 'epi' (as in 'quotient object'), but I take it you mean 'coequalizer'.
A reasonably large class would be regular categories, which includes categories of algebras of monads on $Set$ and semi-abelian categories and toposes. Here quotients = regular epis are stable under pullback and in particular are closed under taking products on either side. Furthermore, in a regular category, every quotient is a *reflexive* coequalizer, meaning a coequalizer of a pair $f, g: X \to Y$ for which there exists $h: Y \to X$ with $f \circ h = g \circ h = 1\_Y$. In particular, the two projections $\pi\_1, \pi\_2: E \to Y$ of an equivalence relation $E$ on $X$, for example the kernel pair of a quotient, form a reflexive pair by the reflexivity property. So in a regular category, where quotients = coequalizers are necessarily coequalizers of their kernel pairs, quotients are quotients of reflexive pairs.
The reason reflexivity is relevant is a $3 \times 3$ lemma which says that in a (edit: commutative-in-parallel) diagram of $3 \times 3$ objects in which all rows and all columns are coequalizer diagrams of reflexive pairs, the diagonal is a coequalizer diagram. See the first page of Johnstone's Topos Theory. Then apply this lemma to the evident diagram whose rows are of the form
$$X\_i \times X\_{1}' \stackrel{\to}{\to} X\_i \times X\_{2}' \to X\_i \times X\_{3}'$$
and whose columns are of the form
$$X\_1 \times X\_{j}' \stackrel{\to}{\to} X\_2 \times X\_{j}' \to X\_3 \times X\_{j}'$$
In the category $Top$, it would therefore be natural to consider quotients by equivalence relations (or even just reflexive relations) which are preserved by taking products on each side. It's that latter condition which needs to be characterized (or at least discussed further), and I may come back to that later after I get the kids off to school. :-)
Edit: For a discussion of topological quotients which are stable under taking a product on either side, see the paper by Day and Kelly, On topological quotient maps preserved by pullback or product, Math. Proc. Cam. Phil. Soc. 67 (1970), 553-558. Or google Day-Kelly maps to find out more.
| 3 | https://mathoverflow.net/users/2926 | 42891 | 27,284 |
https://mathoverflow.net/questions/39823 | 13 | I am a graduate student trying to get involved in Ramsey theory. My question comes from:
Erdős on graphs: his legacy of unsolved problems
By Fan R. K. Chung, Paul Erdős, Ronald L. Graham
p.14 of this book is available as a google ebook.
They quote Erdos in a 1980/1981 paper
"Faudree, Shelp, Rousseau, and I needed recently a lemma stating:
(R(n+1,n)-R(n,n))/n --> infinity as n--> infinity. We could prove this without much difficulty."
My first question is how does one show this?? I have tried my hand at this using recursive formulas for bounds on ramsey numbers, but it seems the lower bounds of this type are particularly weak. Furthermore any proof of this is absent from the literature to my knowledge. Any insight will be very very much appreciated.
Also which is larger R(k,k-2)+1 or R(k-1,k-1)? Of course the second but can we prove it?
Also for m+n=r+s=v with m < r =< v/2 =< s < n can we prove R(r,s) >= R(m,n) ?
Bounty to the best response :)
| https://mathoverflow.net/users/9486 | Differences of near diagonal Ramsey numbers. |
>
> **Edit**: Erdős got three things wrong. First of all, it wasn't Faudree, Shelp, and Rousseau, it was Faudree, Shelp, and Burr. Second, it wasn't "recently", it was in the future (with respect to the quote you provide)! Third, they didn't prove that $(R(n+1,n)-R(n,n))/n \to \infty$, but only that $R(n+1,n)-R(n,n) \geq 2n-3$.
>
>
>
The relevant paper is [On the difference between consecutive Ramsey Numbers](http://www.renyi.hu/~p_erdos/1989-21.pdf), published in 1989. The proof is not long. On a (somewhat cursory) search I wasn't able to find any papers citing this one that address the same question, so it seems likely that this bound is still the best known.
---
**Old answer.**
For your first question: thanks to Miklos Simonovits and others, all of Erdős' papers are available from [this site](http://www.renyi.hu/~p_erdos/Erdos.html). I scanned through the papers by Erdős, Faudree, Rousseau and Schelp from up to 1982 but didn't see such a result. There are 12 papers with precisely these four coauthors, and another several that also have Burr as a coauthor, so it may take some time to find (especially if it isn't explicitly stated as a lemma but is embedded in a proof somewhere). But: if they published it, then you'll be able to find your result by scanning through the papers on that site.
| 7 | https://mathoverflow.net/users/3401 | 42893 | 27,285 |
https://mathoverflow.net/questions/42832 | 3 | Let $c$ be a $W^{1,2}$-curve into a (compact Riemannian) manifold $Q,$ defined on some open interval $I$. Let $t\_0\in I$ and $\xi\_0\in T\_{c(t\_0)}Q$ be arbitrary. I am looking for a citeable reference for the following statement: There is a unique $W^{1,2}$-vector field $\xi$ along $c$ satisfying
(i) $\nabla\_{\dot{c}}\xi=0$ a.e. on $I$
(ii) $\xi(t\_0)=\xi\_0.$
Notice that the statement above could be deduced from the standard Picard-Lindelöf theorem if $c$ was sufficiently regular (i.e. e.g. $c$ was $C^1$).
| https://mathoverflow.net/users/3509 | parallel transport along $W^{1,2}$-curves | Since $W^{1,2} \subset C^0$ and the zero-th order term of (i) depends linearly on $\dot{c} \in L^2$, the usual rewrite-as-integral-equation proof seems to work and is rather straightforward. I don't recall seeing this written down anywhere, but it's easy to verify and summarize.
| 3 | https://mathoverflow.net/users/613 | 42898 | 27,288 |
https://mathoverflow.net/questions/42900 | 19 | I have often read that the Riemann hypothesis is somewhat a statement like:
>
> The primes are as regularly distributed as we can hope for.
>
>
>
For example $\pi(x) = Li(x)+ O(x^{\sigma+\epsilon})$ for any $\epsilon>0$ as long as there are no zeros of $\zeta$ for any
$s \in \mathbb{C}$ with $\Re s > \sigma$. And of course $\sigma=\frac{1}{2}$ is the best we can hope for.
However there are instances/problems where the Riemann hypothesis does not give the "right conjectural"
answers about the distribution of primes.
Let me state one example I recently read about.
Cramer's conjecture <http://en.wikipedia.org/wiki/Cram%C3%A9r%27s_conjecture>
asserts that
$$ p\_{n+1} - p\_n = O(\log^2 p\_n ). $$
Here, the Riemann hypothesis only gives $p\_{n+1} - p\_n = O(\sqrt{p\_n} \log p\_n )$. So, in some sense RH does not give the best we can hope for.
I am now asking for for refinements that could achieve this. Could, for example, Cramer's conjecture be deduced from a refinement of $\pi(x) = Li(x)+ O(x^{\frac{1}{2}+\epsilon})$ (under RH), maybe making the O term more precise. And how would this be reflected in terms of the $\zeta$-function and its zeros.
I know there are generalizations to other $L$-functions and refinements like the pair correlation conjecture or predictions from random matrix theory(though I do not have any clue about this). But I do not know whether these
can help to resolve for example Cramer's conjecture. My question is somewhat unprecise (if anyone can write it up better, feel free to edit.)
: Is there a "Super Riemann Hypothesis" predicting stronger properties of the $\zeta$ or other $L$-functions that would settle most questions about the distribution of primes?
EDIT: Thank you for the answers so far. Since Charles pointed out that my question is really to imprecise as we could prove anything about the primes if we knew the zeros of $\zeta$ I am going to rephrase my question inspired by the interesting article of Heath-Brown mentioned by Idoneal:
What sort of properties do we have to know/assume about the zeros of $\zeta$ in order to deduce Cramer's conjecture?
| https://mathoverflow.net/users/3757 | Refinements of the Riemann hypothesis | Yes, your question is imprecise. If we knew exactly where the zeta zeroes were, we could answer any question about the primes that could be formulated by means of the explicit formulae. In crude terms, the primes are obtained from the zeroes by an integral transform. Specific questions can depend, for example, on rational dependencies between the imaginary parts of the zeroes. There is quite a large literature on conditional results on primes, and if you are asking whether there is a way of packaging it all up neatly I'd expect the more reasonable answer to be "no" rather than "yes". Several generations of analytic number theorists have contributed.
| 7 | https://mathoverflow.net/users/6153 | 42901 | 27,289 |
https://mathoverflow.net/questions/42836 | 18 | Does the braid group $B\_n, n\ge 3$, act properly by isometries on a CAT(0) cube complex?
**Update 1.** During a recent talk of Nigel Higson in Pennstate Dmitri Burago asked whether the braid groups are a-T-menable. I seem to remember that somebody proved that they do act properly by isometries on CAT(0) cube complexes. That would imply a-T-menability by Niblo and Roller or Cherix, Martin and Valette. Hence the question. Note: no co-compactness required.
**Update 2.** It looks like my question is still an open problem for $n\gt 3$. I think my confusion came from terminology. Dan Farley proved that all braided diagram groups (including the R. Thompson group $V$) act property by isometries on CAT(0)-cube complexes. But braided diagram groups (defined by Victor Guba and myself) are not related to braid groups, at least not explicitly because wires there intersect and do not form braids. One can define the notion of "really braided" diagram groups where wires form braids, but I do not think Farley's method will work. So I got confused by my own terminology. By the way, I do not see an obvious reason that $B\_n$ does not embed into $V$. $V$ is a big group with lots of complicated subgroups.
**Update 3.** As Bruce Hughes pointed out to me, even though the Haagerup property (a-T-menability) is unknown for $B\_n, n\ge 4$, all forms of Baum-Connes conjecture have been proved for it by Thomas Schick in [*Finite group extensions and the Baum-Connes conjecture*](https://arxiv.org/abs/math/0209165).
**Update 4** Concerning the question from Update 2. Collin Bleak and Olga Salazar-Diaz proved in [*Free products in R. Thompson's group V*](https://arxiv.org/abs/0911.0979) that $V$ does not contain subgroups isomorphic to ${\mathbb Z}^2\ast {\mathbb Z}$. Does $B\_n$ contain such subgroups?
| https://mathoverflow.net/users/nan | Braid groups acting on CAT(0)-complexes | As Sam Nead says, $B\_n$ contains $\mathbb{Z}^2 \* \mathbb{Z}$, and you can find an example the way he suggests.
If you'd like something much more explicit, you can simply take the first three standard generators $\sigma\_1$, $\sigma\_2$ and $\sigma\_3$, and then it follows from the solution (by Crisp and Paris) of a conjecture of Tits that the subgroup $\langle \sigma\_1^2, \sigma\_2^2, \sigma\_3^2 \rangle$ is isomorphic to $\mathbb{Z}^2 \* \mathbb{Z}$.
See [The solution to a conjecture of Tits on the subgroup generated by the squares of the generators of an Artin group](https://www.ams.org/mathscinet-getitem?mr=1839284) by Crisp and Paris.
Also, see Clay, Leininger and Mangahas' *[The geometry of right angled Artin subgroups of mapping class groups](https://arxiv.org/abs/1007.1129)* for more info and references about right angle Artin subgroups of mapping class groups.
| 9 | https://mathoverflow.net/users/1335 | 42905 | 27,291 |
https://mathoverflow.net/questions/42873 | 5 | Recently during a lecture, my professor mentioned that forcing over any poset which is countable, separative, and atomless, is essentially the same as forcing over the Cohen poset, that is to say results in adding a Cohen real.
My question is: Are there any other similar characterizations of "commonly used" forcing posets? Specifically, is there one for the Hechler poset?
The Hechler Poset/forcing notion $(H,\le)$ is given by setting $H=\omega^{\lt\omega} \times \omega^{\omega} $, and defining the relation $(t,v) \le (s,u)$ iff $ ( t \supset s \wedge (\forall n\in\omega) (u(n) \le v(n)) \wedge (\forall m \in dom(t \backslash s))(t(m) \gt u(m))$. When forcing with this poset, you end up adding an unbounded real to the ground model.
I understand that you cannot produce a model in which $\mathfrak{b}=\omega\_2$ using product forcing, and that you need iterated forcing to do so. Moreover, the iterated forcing construction I've seen that produces $\mathfrak{b}=\omega\_2$ in the forcing extension, used the finite support iteration of $\omega\_2$ many copies of the Hechler poset. Is this evidence for the lack of such a characterization?
(I apologize in advance if this is an ill-stated question, I will change it accordingly if it is.)
| https://mathoverflow.net/users/8843 | Strange question about Hechler | Here are a few additional examples of the kind you seek,
and in fact each of them directly generalizes the
characterization you mention of the forcing to add a Cohen
real. However, I know of no such characterization of Hechler forcing.
* The collapse forcing
$\text{Coll}(\omega,\theta)$ is, up to forcing equivalence, the unique forcing
notion of size $|\theta|$ necessarily collapsing
$\theta$ to $\omega$. (Note, this includes the case of
adding a Cohen real, since $\text{Add}(\omega,1)\equiv\text{Coll}(\omega,\omega)$.)
To see this, suppose that $\mathbb{Q}$ is a forcing notion
of size $|\theta|$ that necessarily collapses $\theta$ to
$\omega$. We may assume without loss of generality that
$\mathbb{Q}$ is separative, since the separative quotient
of $\mathbb{Q}$ is forcing equivalent to it and no larger
in size. Observe that below every condition in
$\mathbb{Q}$, there is an antichain of size $\theta$. Since
forcing with $\mathbb{Q}$ adds a function from $\omega$
onto $\theta$ and $\mathbb{Q}$ has size $\theta$, there is
a name $\dot g$ forced to be a function from $\omega$ onto
the generic filter $\dot G$. We build a refining sequence
of maximal antichains $A\_n\subset\mathbb{Q}$ as follows. Begin
with $A\_0=\{ 1\}$. If $A\_n$ is defined, then let
$A\_{n+1}$ be a maximal antichain of conditions such that
every condition in $A\_n$ splits into $\theta$ many elements
of $A\_{n+1}$, and such that every element of $A\_{n+1}$
decides the value $\dot g(\check n)$. The union
$\mathbb{R}=\bigcup\_n A\_n$ is clearly isomorphic as a
subposet of $\mathbb{Q}$ to the tree $\theta^{\lt\omega}$, and
so it is forcing equivalent to
$\text{Coll}(\omega,\theta)$. Furthermore, $\mathbb{R}$ is
dense in $\mathbb{Q}$. To see this, fix any condition
$q\in\mathbb{Q}$. Since $q$ forces that $q$ is in $\dot G$,
there is some $p\leq q$ and natural number $n$ such that
$p$ forces via $\mathbb{Q}$ that $\dot g(\check n)=\check
q$. Since $A\_{n+1}$ is a maximal antichain, there is some
condition $r\in A\_{n+1}$ that is compatible with $p$. Since
$r$ also decides the value of $\dot g(\check n)$ and is
compatible with $p$, it must be that $r$ forces $\dot
g(\check n)=\check q$ also. In particular, $r$ forces
$\check q\in\dot G$, and so by separativity it must be that
$r\leq q$. So $\mathbb{R}$ is dense in $\mathbb{Q}$, as
desired. Thus, $\mathbb{Q}$ is forcing equivalent to
$\mathbb{R}$, which is forcing equivalent to
$\text{Coll}(\omega,\theta)$, as desired.
* If $\kappa^{\lt\kappa}=\kappa$, then the forcing $\text{Add}(\kappa,1)$ to add a Cohen subset
to $\kappa$ with conditions of size less than $\kappa$ is,
up to forcing equivalence, the unique ${\lt}\kappa$-closed
necessarily nontrivial forcing notion of size $\kappa$.
For this, one similarly builds up a dense tree of conditions
inside the poset that is isomorphic to $\kappa^{\lt\kappa}$, which
is forcing equivalent to $\text{Add}(\kappa,1)$.
* More generally, if $\theta^{\lt\kappa}=\theta$, then the forcing $\text{Coll}(\kappa,\theta)$
is the unique ${\lt}\kappa$-closed forcing notion
necessarily collapsing $\theta$ to $\kappa$ and having size $\theta$. (The
straightforward generalization uses separativity, but the
separative quotient of a partial order may no longer be $\lt\kappa$-closed, so there is an
issue about it; but my
student Norman Perlmutter found a solution avoiding the issue.)
| 8 | https://mathoverflow.net/users/1946 | 42917 | 27,297 |
https://mathoverflow.net/questions/42915 | 5 | In any sort of type theory, there are a bunch of rules for constructing derivations of typing judgments such as $x:A,\; y:B(x) \;\vdash\; z:C(x,y)$. (I intend to include also judgments of the form $B:\mathrm{Type}$.) It's certainly possible to get to the same typing judgment using different derivations; for instance I could introduce an unnecessary variable with weakening, then substitute any term for that variable. But it feels as though such a derivation should be "$\beta$-equivalent" to a derivation which omits the unnecessary variable and substitution. So my question is:
*Is there a tractable (e.g. inductively generated) equivalence relation on derivations under which all derivations of the same typing judgment become equivalent?*
Although I want the answer to be yes, I suspect that it is no, because derivations are a lot like proofs, and I know that at least in intuitionistic logic, there can be multiple "essentially distinct" proofs of a given statement. If so, could it be true for some restricted class of type theories? Can one quantify its falsity?
---
**Edit:** Apparently there was a lot of room for misinterpretation of this question! To clarify: I was only talking about type theories in which inhabitation of types is witnessed by a specified term, as in the example typing judgment I gave above. (If you think of types as propositions and terms as proofs, then the question becomes "are all ways to derive a given proof-term equivalent?" But I don't generally tend to think of types only in that way.) Neel's answer seems to say: yes, as long as the type theory satisfies cut-elimination. Whether or not a given type can be inhabited by multiple distinct terms (e.g. one proposition can admit multiple distinct proofs) is a different question.
**Edit 2:** Maybe if I explain the origin of my question, it will make clearer what I was asking. When trying to define a semantics of DTT in a locally cartesian closed category, it seems to me that the natural approach is to define the interpretation of each type and term inductively using the corresponding categorical constructions, as in D4.4 of *Sketches of an Elephant*. Unfortunately, this doesn't quite work, because for instance substitition is "strict" in type theory, but pullback is only up to isomorphism in a category.
It seems that the standard way to deal with this is to first strictify the pullbacks in the lccc, e.g. by replacing its codomain fibration by a split one and maybe some additional data (people talk about "comprehension categories" and suchlike). However, looking at the direct construction that doesn't work, it seems to me that it can be interpreted meaningfully as an inductive construction acting not on types+terms, but rather on *derivations* of typing judgments. If that's right, it seems like one could then get a meaningful interpretation of type theory in an lccc by proving that any two derivations of the same typing judgment produce canonically *isomorphic* interpretations in the lccc. But that would require a notion of equivalence between derivations which is tractable (so that one can show that any equivalence of derivations gives an isomorphism in the lccc) and under which all derivations can be shown to be equivalent; hence my question.
| https://mathoverflow.net/users/49 | Can a typing judgment admit essentially different derivations? | This property is called "coherence", and no, it doesn't always hold.
Establishing this property holds for a given semantics of proofs is a proof obligation. An example of when it doesn't arises with coercive subtyping -- if the diagrams corresponding to possible coercions do not all commute, then the semantics is not coherent, in that the meaning of a term depends critically on its typing derivation. If all you have is the term and knowledge that it has a typing derivation, you can't tell what it means, necessarily.
For a proof-theoretic characterization, the thing to look for is a cut-elimination theorem. One way of reading the cut-elimination theorem is precisely that it says that eliminating intro/elim and elim/intro pairs (i.e., $\beta$ and $\eta$ equations) induces a genuine equivalence relation on proofs.
However, we can still have essentially different proofs. For example, $x : A \land A \vdash \pi\_1(x) : A$ and $x : A \land A \vdash \pi\_2(x) : A$ are two different proofs that $A$ and $A$ entails $A$. However, cut-elimination guarantees that the addition of proof terms does make the typing derivations unique -- any proof of this entailment must be equal to one or the other, since every proof is equivalent to a cut-free proof, and these are the only two cut-free proofs, and the proof terms let us distinguish the two possibilities.
EDIT: There must be a treatment of this subject in somebody's book (Barendregt?), but I don't know off-hand, since I learned it via osmosis. Probably the best reference I can point you at is Jan Schwinghammer's paper ["Coherence of Subsumption for Monadic Types"](http://www.ps.uni-saarland.de/Publications/documents/Schwinghammer_Coherence_09.pdf), which nicely illustrates the idea, and whose references also contain pointers to the best available literature.
Noam quite rightly points out that you might actually be interested in the case where all derivations of a judgement are equivalent. (This possibility honestly hadn't even occurred to me, since I'm too intuitionistically contaminated.) The keyword to search for in this case is "proof irrelevance", and I recommend Awodey and Bauer's paper ["Propositions as [Types]"](http://logcom.oxfordjournals.org/content/14/4/447.abstract).
EDIT 2: When cut-elimination holds, then the derivations of a calculus *with proof terms* become equivalent. If you don't have proof terms, then you can't guarantee equivalence of derivations.
For example, suppose we erased the proof terms from the derivation above, so that we simply had the judgment $A \land A \vdash A$. In this case, even if we know that this judgment holds, we still don't know whether the proof used the left or the right $A$, which are the two intrinsically different derivations of the same judgment. On the other hand, if we have proof terms, then we have a judgment $x : A \land A \vdash e : A$. In this case, the derivation does bebgcome unique, because now we can $\beta\eta$-normalize $e$ and just check whether it is equal to $\pi\_1(x)$ or $\pi\_2(x)$.
This is why lambda-calculus expressions are called "proof terms": these terms are evidence to establish which proof you meant.
To connect this back with proof irrelevance, a type is "proof irrelevant" when all the proof terms of that type are equivalent. That is, if $\Gamma \vdash e : A$ and $\Gamma \vdash e' : A$, then we know that the two derivations are equivalent, regardless of $e$ and $e'$. For example, the unit type $1$ is proof irrelevant, since every term in it is equivalent to the unit value $\left<\right>$.
| 8 | https://mathoverflow.net/users/1610 | 42921 | 27,300 |
https://mathoverflow.net/questions/42923 | 6 | Specifically, is it the case that (for $a,b\in\omega^\omega$) $a$ $\leq\_T$ $b$ implies $a$ $\leq\_c$ $b$?
I suspect it might be trivial, but not knowing much Recursion Theory, it's hard to see how it could.
Thank you in advance.
| https://mathoverflow.net/users/3462 | How does the Constructibility Degree of a real compare with its Turing Degree? | Yes, if $a\le\_T b$ then $a$ is first-order definable from $b$; in particular $a\in L(b)$ so $a\le\_c b$.
| 6 | https://mathoverflow.net/users/4600 | 42924 | 27,301 |
https://mathoverflow.net/questions/42553 | 13 | I'm finally at the end of Milnor's "On manifolds homeomorphic to the 7-sphere", and I stumbled upon something I cant figure out...
For those with the reference I'm talking about "lemma 5", it goes something like this, you have two $\mathbb{S}^3$ bundles over $\mathbb{S}^4$, we want to obtain the total space of this bundle, so you glue them via the transition function, one can think of this as having a pair of copies of $(\mathbb{R}^4 \setminus \{0\}) \times \mathbb{S}^3$ and gluing them by identifiying $(u,v) \mapsto (u',v')=(u / \|u\|^2, u^hvu^j/\|u\|)$ where $u$ and $v$ are quaternions, so far so good, now Milnor states that if $h+j =1$ then this manifold is a $7$-sphere, his reason is that the function $f(x) = \mathfrak{R}(v)/(1+\|u\|^2)^{1/2}$ is a morse function, this with the "first" coordinate chart, for the second he defines $u'' = u'(v')^{-1}$ and substitutes $(u',v')$ for $(u'',v')$ stating that the function $f$ is now given by $\mathfrak{R}(u'')/(1+\|u''\|^2)^{1/2}$.
He then says "It is easily verified that f has only two critical points (namely $(u,v) = \pm (0,1)$) and that these are nondegenerate".
That's where I get lost; I don't understand his change of coordinates $(u',v') \mapsto (u'',v')$, nor why he states the function is now the one stated... I tried developing the algebra but I can't get it to work out, I thought maybe he was using the involution $v \mapsto v^{-1}$ somehow but it doesn't add up either...
| https://mathoverflow.net/users/9187 | Morse Theory and Exotic Spheres | Milnor didn't explain the formula as much as maybe he should have, but the point is that the real part of a unit-length quaternion is invariant under both conjugation and inversion. Let $$r = ||u|| \qquad \hat{u} = u/r,$$
so that
$$v' = \hat{u}^h v \hat{u}^j \qquad u' = \hat{u}r \qquad ||u'|| = ||u''|| = 1/r.$$
Thus
$$v' \hat{u}^{-1} = \hat{u}^h v \hat{u}^{-h}$$
is conjugate to $v$. Thus
$$\mathfrak{R}(v) = \mathfrak{R}(v'\hat{u}^{-1}) = \mathfrak{R}(\hat{u} (v')^{-1}).$$
The first equality is conjugation, the second one is inversion. So then you get
$$\frac{\mathfrak{R}(v)}{\sqrt{1 + ||u||^2}} = \frac{\mathfrak{R}(v)}{\sqrt{1+r^2}} = \frac{\mathfrak{R}(\hat{u} (v')^{-1})}{\sqrt{1+r^2}} = \frac{\mathfrak{R}((\hat{u}/r) (v')^{-1})}{\sqrt{1+r^{-2}}} = \frac{\mathfrak{R}(u'')}{\sqrt{1+||u''||^2}}.$$
Note that, although $(u'',v')$ certainly is a valid parameterization of the second chart, it's enough to think of $u''$ as a convenient function rather than part of a coordinate frame.
The question now in my mind is, how did Milnor think of this algebra? I do not know the answer. Maybe he started with a round 4-sphere with its quaternionic Hopf fibration, and the elementary Morse function that consists of one of the coordinates in $\mathbb{R}^8$. You immediately get that there are two critical points (the north and south pole) and that they lie on the same Hopf fiber, since opposite points on a sphere always do. Apparently this Morse function fits together in a similar way for all of these 3-sphere bundles over the 4-sphere.
| 10 | https://mathoverflow.net/users/1450 | 42930 | 27,302 |
https://mathoverflow.net/questions/19313 | 12 | Let $M$ and $N$ be "nice" model categories. I'm happy to have "nice" mean [combinatorial model category](http://ncatlab.org/nlab/show/combinatorial+model+category). Consider a Quillen pair
$$ L: M\rightleftarrows N: R.$$
I want the following result:
>
> There exists a set of maps $S$ in $M$, such that $L$ and $R$ descend to a Quillen pair
> $$ L: S^{-1}M \rightleftarrows N: R,$$
> with the property that a map $f:X\to Y$ between cofibrant objects in $M$ is a weak equivalence in $S^{-1}M$ if and only if $L(f)$ is a weak equivalence in $N$.
>
>
>
Here $S^{-1}M$ denotes the model category with the same underlying category $M$ obtained by localizing $M$ with respect to the set of maps $S$.
This result seems to encode a standard technique; in fact, the very first example of a localized model category (Bousfield's localization of spaces with respect to a homology theory) can be viewed (in retrospect) of a special case of this.
I think I could prove this result if I need to. But I would rather have a reference. I've looked in the usual places, but I can't seem to find anything exactly like it. I expected to find something like this in one of Dugger's papers on presentable model categories, but I don't find it there; he proves that under an additional condition, you can get a Quillen *equivalence* $S^{-1}M\rightleftarrows N$, but his construction of the set $S$ does not apply in my case.
| https://mathoverflow.net/users/437 | How to localize a model category with respect to a class of maps created by a left Quillen functor | I suspect that you already have one, but here is a proof. I will assume that $M$ and $N$ are combinatorial and that $M$ is left proper (otherwise, I don't think that the literature contains a general construction of the left Bousfield localizations of $M$ by any small set of maps). Everything needed for a quick proof is available in Appendix A of
J. Lurie, *Higher topos theory*, Annals of Mathematics Studies, vol. 170, Princeton University Press, 2009.
First, there exists a cofibrant resolution functor $Q$ in $M$ which is accessible: the one obtained by the small object argument (as accessible functors are closed under colimits, it is sufficient to know that $Hom\_M(X,-)$ is an accessible functor for any object $X$ in $M$, which is true, as $N$ is combinatorial). Let $W$ be the class of maps $f$ of $M$ such that $L(Q(f))$ is a weak equivalence in $N$. As $N$ is combinatorial, the class of weak equivalences of $N$ is accessible see Corollary A.2.6.9 in *loc. cit*. Therefore, by virtue of Corollary A.2.6.5 in *loc. cit*, the class $W$ is accessible. To Prove what you want, it is sufficient to check that $M$, $W$ and $C=${cofibrations of $M$} satisfy the conditions of Proposition A.2.6.8 in *loc. cit*. The only non trivial part is the fact that the class $C\cap W$ satisfies all the usual stability properties for a class of trivial cofibrations (namely: stability by pushout, transfinite composition). That is where we use the left properness. For instance, if $A\to B$ is in $W$ and if $A\to A'$ is a cofibration of $M$, we would like the map $A'\to B'=A'\amalg\_A B$ to be in $W$ as well. This is clear, by definition, if $A$, $A'$ and $B$ are cofibrant. For the general case, as $M$ is left proper, $B'$ is (weakly equivalent to) the homotopy pushout $A'\amalg^h\_A B$, and as left derived functors of left Quillen functors preserve homotopy pushouts, we may assume after all that $A$, $A'$ and $B$ are cofibrant (by considering the adequate cofibrant resolution to construct the homotopy pushout in a canonical way), and we are done. The case of transfinite composition is similar.
| 11 | https://mathoverflow.net/users/1017 | 42934 | 27,305 |
https://mathoverflow.net/questions/42928 | 6 | Let $G$ be a finitely generated group with the natural word length function ($|x|$ is the length of the shortest word in generators of $G$ representing $x$). We call a partial left invariant order $\le $ on $G$ *word order* if whenever $a\le b\le c$ we have $|b|\le C(|a|+|c|)$ for some constant $C$. Say, the standard order on $\mathbb Z$ is a word order.
**Question.** Is there a group $G$ with a left invariant linear order but without left invariant linear word order?
| https://mathoverflow.net/users/nan | orders and length functions on finitely generated groups | If $G$ is a finitely generated infinite group and $\leq$ is a linear word order, then for each $a, c \in G$ there are only finitely many elements $b \in G$ such that $a \leq b \leq c$. From this it follows that $(G, \leq)$ is order isomorphic to $\mathbb Z$. If $\leq$ is also left invariant, then this isomorphism must be a group isomorphism as well.
| 10 | https://mathoverflow.net/users/6460 | 42953 | 27,317 |
https://mathoverflow.net/questions/42966 | 17 | If $X$ a topological space one says that $X$ is *universally closed* if for every Hausdorff space $Y$ and every (continuous) map $f:X\rightarrow Y$, the image of $X$ is a closed subset of $Y$.
It is clear that every compact space is universally closed, but are there non compact universally closed spaces?
| https://mathoverflow.net/users/10217 | Topological spaces whose continuous image is always closed | If $Z$ is not compact, and $X=\{p\}\cup Z$ is the space whose nonempty open sets are of the form $\{p\}\cup V$ with $V$ open in $Z$, then $X$ is not compact, but every continuous function from $X$ to a Hausdorff space is constant.
| 15 | https://mathoverflow.net/users/1119 | 42973 | 27,330 |
https://mathoverflow.net/questions/41011 | 20 | What is the indefinite sum of the tangent function, that is, the function $T$ for which
$\Delta\_x T = T(x + 1) - T(x) = \tan(x)$
Of course, there are infinitely many answers, who all differ by a function of period 1. Ideally, I would like the solution to be of the form
$T(x) = $ nice\_function$(x)$ + possibly\_ugly\_periodic\_function$(x)$,
where nice is at least piece-wise continuous.
If any of the following sums can be found, the sum of tan can also be found:
* $\sum \sec x$
* $\sum \csc x$
* $\sum \cot x$
* $\sum \frac{1}{e^{ix} + 1}$
I have tried several methods without success, including using a newton series (which does not converge for non-integer $x$), and trying to guess possible functions.
I would also appreciate lines of attack if a solution is not known.
| https://mathoverflow.net/users/7540 | What is the indefinite sum of tan(x)? | I add more details for the solution in the distinguished answer due to Anixx. First, we need the **digamma** function
<http://en.wikipedia.org/wiki/Digamma_function>
which we will call $\Psi(x)$. Important properties (from that web page) are: $\Psi(x)$ is analytic in the complex plane except at the nonpositive integers where it has simple poles. $\Psi(x+1)-\Psi(x) = 1/x$. $\Psi(x) > 0$ for $x>2$. Asymptotics:
$$
\Psi(x) = \log x - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} + O(x^{-6})
\qquad\text{as } x \to \infty .
$$
So, define $T(z) ={}$
$$
-\sum\_{k = 1}^{\infty} \Biggl[\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) -
\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)\Biggr]
$$
For any fixed $z$, only finitely many preliminary terms involve $\Psi$ evaluated at a nonpositive argument, and the asymptotics of the remaining terms are computed (from the asymptotics given above) as
$$
\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) -
\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)
$$
$=z(1-z)/(k^2\pi^2) + o(k^{-2})$ as $k \to \infty$. So the series converges absolutely except when we are at a pole of one of the preliminary terms. Now, because of absolute convergence, we may subtract term-by-term and simplify to get
$$
T(z+1)-T(z) = \sum\_{k=1}^\infty\Biggl[\frac{8z}{(-\pi+2\pi k-2z)(-\pi+2\pi k+2z)}\Biggr] = \tan z .
$$
| 33 | https://mathoverflow.net/users/454 | 42975 | 27,332 |
https://mathoverflow.net/questions/42969 | 0 | Given an $m\times n$ 0-1 matrix A, I am interested in an efficient algorithm to locate all copies of a given $p\times q$ 0-1 submatrix B within it, where a permutation of rows and columns is allowed, i.e. find all collections of row indices $r\_1, r\_2,\ldots, r\_p$ and column indices $c\_1, c\_2,\ldots, c\_q$ (with $r\_i$'s and $c\_j$'s not necessarily in increasing order) so that A restricted to those rows and columns in that particular order yields B.
Any references will be useful.
Thanks.
| https://mathoverflow.net/users/10219 | Locating a submatrix within a matrix | Note that the problem of deciding whether there exist such row indices and column indices is already NP-complete. This is because the case where *B* is a square matrix entirely consisting of 1s is identical to the Balanced Complete Bipartite Subgraph problem, which is known to be NP-complete [Joh87].
[Joh87] David S. Johnson. The NP-completeness column: An ongoing guide. *Journal of Algorithms*, 8(3):438–448, Sept. 1987. <http://dx.doi.org/10.1016/0196-6774(87)90021-6>
| 3 | https://mathoverflow.net/users/7982 | 42993 | 27,344 |
https://mathoverflow.net/questions/41315 | 4 | Consider the Banach algebra $W^+=\ell^1(\mathbb{Z}^+)$, viewed upon as the analytic functions $f$ on the unit disc $\mathbb{D}$ such that $$\|f\|=\sum\_{k\ge0}|a\_k|<\infty$$ where
$$f(z)=\sum a\_kz^k$$
is the Taylor expansion of $f$. Clearly, $W^+\subset H^\infty(\mathbb{D})$. Now, it is well known that $H^\infty$ admits inner-outer factorization. Is there some similar factorization theorem for $W^+$?
| https://mathoverflow.net/users/8294 | Factorization in the Wiener algebra on the unit disc. | There is no factorization in Wiener algebra, it is easy to construct a counterexample.
Namely, if $B$ is a Blaschke factor with zeroes $z\_n$, $z\_n\to 1$ (of course $\sum(1-|z\_n|) <\infty$) and $g= (z-1)^2$ then
$
f= Bg$ has $C^1$ boundary values, and so is in the Wiener algebra.
On the other hand, $B$ is an inner part of $f$ (in $H^\infty$), and it is not in $W$, because it is not even continuous at $1$.
On the other hand, if $f(z)\ne 0$ for all $z:|z|=1$, then $f$ has only finitely many zeroes in the unit disc, so the factorization is trivial: the inner part is a finite Blaschke product.
| 2 | https://mathoverflow.net/users/10223 | 42996 | 27,346 |
https://mathoverflow.net/questions/42728 | 14 | If someone hands you a prime number $p$, and an algebraic number $x$ inside the Hasse-Weil bound, is there a normalized newform (say of weight two) so that $a\_p=x$, where $a\_p$ is the $p$th Fourier coefficient?
| https://mathoverflow.net/users/5730 | Fourier coefficient of a modular form | Some Remarks.
I parse the problem in the following way:
Start with a totally real algebraic integer $\alpha$ such that every conjugate of $\alpha$ has absolute value at most
$2 \sqrt{p}$. Then does there exist a normalized cuspidal Hecke eigenform $f$ of weight $2$ with $a\_p = \alpha$?
First, here is a heuristic reason why one should expect this to be false.
Suppose we ask a slightly stronger question, namely, that all the coefficients of $f$ are defined over the field $E = \mathbb{Q}(\alpha)$. Then, we are asking for the existence of an abelian variety of $\mathrm{GL}\_2$-type with endomorphisms by (some order in) the ring $\mathcal{O}\_E$. Such objects (ignoring issues of polarization) correspond to rational points on Shimura curves. But these curves, in general, will have large genus, and so there's no reason to expect that they have any rational points. It will probably be hard to prove anything this way, however.
A second heuristic is to ask the problem in different weights. For example, is there
a weight $12$ normalized cuspidal eigenform $f$ of level co-prime to $p$
with $a\_p = 0$? This sounds tricky. Maybe Serre even conjectured once that this never happened if $p$ was sufficiently large. Let's say he did. Are you going to contradict Serre?
Finally, let me show in a rather cheap way that the answer to the original question is
"not always". Suppose that $\alpha = 2 \sqrt{p}$, which satisfies the Weil bounds. Suppose that $a\_p = \alpha$, and let $\epsilon$ denote the nebentypus character of
$f$. Then the characteristic polynomial of Frobenius is
$$x^2 - 2 \sqrt{p} \cdot x + p \cdot \epsilon(p).$$
I claim that $\epsilon(p)$, which is a root of unity, is actually $1$. (This follows
easily from the fact that the roots of this polynomial are Weil numbers and the triangle inequality.)
In particular, the characteristic polynomial of Frobenius
is actually
$$x^2 - 2 \sqrt{p} \cdot x + p = (x - \sqrt{p})^2.$$
This doesn't happen!
Losely speaking, one knows that the action of crystalline Frobenius is semi-simple
on Abelian varieties, and yet the Eichler-Shimura relations implies
that $(\mathrm{Frob}\_p - \sqrt{p})^2 = 0$, which then implies that
$\mathrm{Frob}\_p = \sqrt{p}$ acts as a scalar, which contradicts how
one knows Frobenius to interact with the Hodge Filtration --- all
this is explained in (and is indeed the main point of) a paper
of Coleman and Edixhoven from the groovy 90's.
| 9 | https://mathoverflow.net/users/nan | 43016 | 27,355 |
https://mathoverflow.net/questions/43027 | 10 | Given a complete graph of n vertices (no three of which are no collinear) in the plane and straight edges, what is the maximal possible number of "incidental intersections" of edges, i.e., number of non-vertices at which two distinct edges intersect each other, not counting multiplicity?
This is a question that I pose to the students in my Mathematics for Elementary School Teachers as a way to understand mathematical conjecturing and proving -- and not always finding the solution. But it occurs to me that it might be handy to know whether the answer is actually known or not.
| https://mathoverflow.net/users/9102 | "incidental" intersections of a complete graph in the plane | Assuming straight line segment edges, any 4 vertices determine 6 edges with at most one non-vertex intersection, so you can't have more than $n$-choose-4. You will have $n$-choose-4 if no vertex is interior to any triangle of vertices, which is to say if all $n$ vertices lie on the boundary.
| 8 | https://mathoverflow.net/users/3684 | 43034 | 27,368 |
https://mathoverflow.net/questions/42976 | 6 | ### Motivation
I was re-reading parts of Grothendieck-Murre, and these questions came up naturally.
The situation in chapter 9 is that $S'=Spec(A)$ where $A$ is a complete local noetherian ring of dimension $2$ with an algebraically closed residue field, and $D$ is some divisor in $A$. Then they take a "desingularization" of $S'$, $T'$, such that $T'$ maps isomorphically to $S'$ away from $D$, and the divisor that goes to $D$ is normal crossings. Call the divisor that maps to $D$ also $D$. They first show that $\pi\_1^{(p)\,D}(T')$ is isomorphic to $\pi\_1^{(p)}$ of the completion of $T'$ with respect to $D$ where this $\pi\_1$ is the tame $\pi\_1$ with respect to $D$. I assume this goes through even if $D$ is not normal crossings (if I'm wrong about this, I would really like to know).
Then they prove that $\pi\_1^{(p)}(S'\setminus D)(=\pi\_1^{(p)}(T'\setminus D))$ is isomorphic to $\pi\_1^{(p),D}(T')$.
The question that arises is: why did they take this detour through this desingularization? Which step doesn't go through in the case that $D$ is not normal crossings? As I said, I think it's unlikely that it is the step that the tame fundamental groups are the same when completing, but I would like to know if I'm wrong.
### Question
Let $S'=Spec(A)$ where $A$ is a complete local noetherian ring of dimension $2$ with an algebraically closed residue field, and $D$ is some divisor in $A$. Could there exist a divisor, $D$ (necessarily not normal crossings), such that $\pi\_1^{(p)}(S'\setminus D)$ is not isomorphic to $\pi\_1^{(p),D}(S')$?
| https://mathoverflow.net/users/5756 | Relation between tame fundamental group w.r.t. to D, and the fundamental group of the complement of D | Your question as stated does not strictly speaking make sense. In Grothendieck-Murre sect.2.4 the tame fundamental group $\pi\_1^D(S,\xi)$ is only defined when $S$ is normal and $D$ is a DNC. Although in 2.2.2 they define tamely ramified covers in greater generality, they observe in Remark 2.2.3(4) that this is "certainly not the correct" definition when $D$ is not a DNC. However if you were to use their definition 2.2.2 (which requires in addition that $S'$ be normal) to define the tame $\pi\_1$, then the groups in your question would be isomorphic for any $D$. Indeed, you have an equivalence of the relevant categories of coverings, given by restriction to $S'\setminus D$ in one direction and by normalisation in the other.
The aim of Chapter 9 Grothendieck-Murre is to prove that $\pi\_1^{(p)}$ of a connected open subscheme of $S'$ (not necessarily normal) is topologically finitely presented. For this you really do need to pass to the desingularisation, so as to bring Kummer theory of tame covers to bear.
| 5 | https://mathoverflow.net/users/5480 | 43041 | 27,374 |
https://mathoverflow.net/questions/42959 | 4 | Let $X$ be a projective variety. Symmetric product of $X$ is the quotient of the product $X^n$ by the action of the symmetric group $\Sigma\_n$ permuting the factors.
When does it exist (as an algebraic variety)?
| https://mathoverflow.net/users/2234 | Symmetric products of projective varieties | To fix ideas, let $K$ be a field and $X/K$ be a seperated $K$-scheme of finite type. Let $G$ be a finite group operating on $X$ via $K$-morphisms. The operation is said to be admissible provided every orbit of $G$ is contained in an open affine subset of $X$.
If the operation is admissible, then
there is a pair $(Y, p)$ consisting of a seperated
$K$-scheme $Y$ of finite type and a finite, surjective morphism $p\in Hom(X, Y)^G$,
such that the
map
$$Hom(Y, Z)\to Hom(X, Z)^G,\ f\mapsto f\circ p$$
is bijective for all schemes $Z/K$. Then $(Y, p)$ is said to be the quotient of $X$ mod $G$. (Cf. SGA I, V.1 for such constructions.)
From now on assume that $X/K$ is projective. Then $X$ is a closed subscheme of $P\_n$ and every
finite subset of $X$ is contained in an open affine subset of $X$. We see: If a finite group $G$ acts on $X/K$ by $K$-morphisms, then the operation is automatically admissible.
Now take a projective $K$-scheme $V/K$. Then $V^n=V\times\cdots\times V$ is projective over $K$ (Segre-embedding) and hence the natural operation of the symmetric group
$S\_n$ on $V^n$ will be admissible. Consequently the quotient exists in that case.
Aside: It can happen that $V^n/S\_n$ is non-smooth, even if $V^n$ is smooth over $K$.
| 4 | https://mathoverflow.net/users/8680 | 43043 | 27,375 |
https://mathoverflow.net/questions/43042 | 5 | Given a nonprincipal ultrafilter $\mu$ on $\mathbb{N}$ and a sequence of groups $G\_i$, one can define its ultraproduct as:
$$ ^\*\prod\_{i\in \mathbb{N}}G\_i:=\{(x\_i)\_{i \in \mathbb{N}}| x\_i\in G\_i\}/\sim$$, where $(x\_i)\_i\sim (y\_i)\_i$, iff $x\_i=y\_i$ $\mu$-almost everywhere.
Suppose you are given also group homomorphisms $f\_{i+1}:G\_{i+1}\rightarrow G\_i$, then one could also consider something like an ultra- inverse limit:
$$\{[x\_i]\_i|f\_i(x\_i)=x\_{i-1} \quad \mu-\mbox{almost everywhere}\}$$
Has this been studied before? Is there a good source, where I could learn about this?
| https://mathoverflow.net/users/3969 | What is known about the ultra-inverse limit? | Either the even numbers or the odd numbers are $\mu$-large, so the condition degenerates to ultraproduct.
| 7 | https://mathoverflow.net/users/408 | 43046 | 27,377 |
https://mathoverflow.net/questions/43002 | 23 | Let $G$ be a topological group, and $\pi\_1(G,e)$ its fundamental group at the identity. If $G$ is the trivial group then $G \cong \pi\_1(G,e)$ as abstract groups. My question is:
If $G$ is a non-trivial topological group can $G \cong \pi\_1(G,e)$ as abstract groups?
About all I know now is that $G$ would have to be abelian.
| https://mathoverflow.net/users/5795 | Fundamental groups of topological groups. | Here is an example: a product of infinitely many $\mathbb{RP}^\infty$'s.
The crucial thing thing to see is that $\mathbb{RP}^\infty$ (or, easier to see, its universal cover $S^\infty$) has a group structure whose underlying group is a vector space of dimension $2^{\aleph\_0}$. This is not hard: the total space $S^\infty$ of the universal $\mathbb{Z}\_2$-bundle is obtained by applying a composite of functors to the group structure $\mathbb{Z}\_2$ in the category of sets:
$$\textbf{Set} \stackrel{K}{\to} \textbf{Cat} \stackrel{\text{nerve}}{\to} \textbf{Set}^{\Delta^{op}} \stackrel{R}{\to} \textbf{CGHaus}$$
($\textbf{CGHaus}$ here is the category of compactly generated Hausdorff spaces and continuous maps). Here $K$ is the right adjoint to the "underlying set of objects" functor; it takes a set to the category whose objects are the elements of the set and there is exactly one morphism between any two objects. The functor $R$ is of course geometric realization.
Each of these functors is product-preserving, and since the concept of group can be formulated in any category with finite products, a product-preserving functor will map a group object in the domain category to one in the codomain category. Even more: the concept of a $\mathbb{F}\_2$-vector space makes sense in any category with finite products since we merely need to add the equation $\forall\_x x^2 = 1$ to the axioms for groups, which can be expressed by a simple commutative diagram.
Thus $S^\infty$ is an internal vector space over $\mathbb{F}\_2$ in $\textbf{CGHaus}$. It can also be considered an internal vector space over $\mathbb{F}\_2$ in $\textbf{Top}$, the category of ordinary topological spaces, because a finite power $X^n$ in $\textbf{Top}$ of a CW-complex $X$ has the same topology as $X^n$ does in $\textbf{CGHaus}$ provided that $X$ has only countably many cells, which is certainly the case for $S^\infty$ (see [Hatcher's book](http://www.math.cornell.edu/~hatcher/AT/ATapp.pdf), Theorem A.6). Thus $S^\infty$ can be considered as an honest commutative topological group of exponent 2.
The underlying group of $S^\infty$ (in $\textbf{Set}$) is clearly a vector space of dimension $2^{\aleph\_0}$. We make take this vector space to be the countable product $\mathbb{Z}\_2^{\mathbb{N}}$. Modding out by $\mathbb{Z}\_2$ (modding out by a 1-dimensional subspace), the space $\mathbb{RP}^\infty$ is also, as an abstract group, isomorphic to this. And so is a countably infinite product $(\mathbb{RP}^\infty)^{\mathbb{N}}$ of copies of $\mathbb{RP}^\infty$.
Finally, the functor $\pi\_1$ is product-preserving, and so
$$\pi\_1((\mathbb{RP}^\infty)^{\mathbb{N}}) \cong \mathbb{Z}\_{2}^{\mathbb{N}}$$
and we are done.
| 43 | https://mathoverflow.net/users/2926 | 43047 | 27,378 |
https://mathoverflow.net/questions/43054 | 10 | Let $R$ be a dvr with residue field $k$ and quotient field $K$. Define $S=Spec(R)$. Let
$A/K$ be an abelian variety. To my knowledge the Neron model of $A$ is a group scheme
${\cal N}/S$ with generic fibre $A$, which represents the functor
$$Y\mapsto Mor\_K(Y\times\_S Spec(K), A)$$
on the category of smooth $S$-schemes. The morphism ${\cal N}\to S$ is smooth, in
particular it is flat and locally of finite type.
**Question 1.** Is it true that ${\cal N}\to S$ is of finite type?
I know that the special fibre ${\cal N}\times\_S Spec(k)$ is in general not connected and
that the component group of the special fibre is an important invariant.
But what about the scheme ${\cal N}$ itself?
**Question 2.** Is it true that ${\cal N}$ is connected?
I strongly assumed that the answer to question 2 is "yes". (My reason to believe this:
Assume ${\cal N}$ is not connected. Then ${\cal N}=U\cup V$ for nonempty disjoint
open subsets $U$ and $V$ of ${\cal N}$. Then $A=U\_K\cup V\_K$ and $U\_K$, $V\_K$ are
disjoint open subsets of $A$. Furthermore $U\to S$, $V\to S$ are flat morphisms, hence
$U\_K$ and $V\_K$ are non-empty. This is a contradiction, because $A$ is connected.)
However I saw in the book of Bosch-Lutkebohmert-Raynaud examples of non-connected Neron
models. And I saw in Deligne's articles on Hodge theory the expression "connected
Neron model of $A$" (as opposed to "Neron model of $A$"). Hence I am very puzzled...
(I have to admit that I did not yet go through the construction of a
representing object of the functor above. That is probably the reason why
I cannot help myself at the moment.)
| https://mathoverflow.net/users/8680 | Basic properties of Neron models | 1) Yes, that is part of Neron's theorem. (There also exist Neron models for semiabelian varieties, which are not of finite type in general.)
2) Yes, for the reason you give ($A$ is connected). When people say "connected component of the Neron model" they generally mean the open subgroup scheme which is in every fibre the connected component of the identity.
An alternative to BLR is the article "Neron models" by M. Artin in "Arithmetic Geometry" (ed. Cornell-Silverman). What you are asking is explained in the first section of that reference (look at 1.2 and 1.16).
| 13 | https://mathoverflow.net/users/5480 | 43056 | 27,383 |
https://mathoverflow.net/questions/42707 | 13 | I just started reading about Calabi-Yau manifolds and most of the sources I came across defined Calabi-Yau manifold in a different way. I can see that some of them are just same and I can derive one from other. But my question is the following :
"What is the most strict definition of Calabi-Yau Manifolds"
By that I mean the definition from which all the others follow.
| https://mathoverflow.net/users/9534 | Calabi - Yau Manifolds | There are several different "definitions" of Calabi-Yau manifolds, not all equivalent, and not all contained in one general definition. A good discussion of some of these inequivalent definitions can be found in Joyce's book:
[Compact Manifolds with Special Holonomy](http://books.google.ca/books?id=c3P-YUD8GZQC&dq=joyce+compact+manifolds&hl=en&ei=417ATLScGZSpnQeR5aWJCg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CDEQ6AEwAA)
The other answers you have gotten so far seem to be from the algebraic geometry side of things, and are fine in that context. From a Riemannian geometry point of view, the most natural definition of a Calabi-Yau manifold (whether compact or non-compact) is a $2n$-dimensional Riemannian manifold for which the holonomy of the Levi-Civita connection is exactly $SU(n)$. Allowing the holonomy to be a proper subgroup of $SU(n)$ is also common. In that case, hyperKahler (which is holonomy $Sp(n/2)$ in dimension $4n$) can also be considered as being Calabi-Yau, for example.
This Riemannian geometry definition is equivalent to the existence of the "Calabi-Yau package": a Riemannian metric $g$, an integrable complex structure $J$ (orthogonal with respect to $g$) together which induce the associated Kahler form $\omega$ by $\omega(X,Y) = g(JX, Y)$, and a *holomorphic volume form* $\Omega$, which is a holomorphic $(n,0)$-form on $M$. These tensors must satisfy:
(1) $\nabla \omega = 0$ (equivalent to $\nabla J = 0$, the Kahler condition) This is also equivalent to $d\omega = 0$ because we are assuming $J$ to be integrable.
(2) $\nabla \Omega = 0$
(3) $\frac{\omega^n}{n!} = c\_n \, \Omega \wedge \bar \Omega$ for some universal *constant* $c\_n$ depending only on the dimension which I can never remember.
These conditions imply, in particular that $g$ is Ricci-flat and $c\_1(M) = 0$. Also, if the holonomy is exactly $SU(n)$ rather than a proper subgroup, then it also follows that $h^{p,0} = h^{0,p} = 0$ for all $1 \leq p \leq n-1$.
| 10 | https://mathoverflow.net/users/6871 | 43062 | 27,385 |
https://mathoverflow.net/questions/43058 | 1 | I'm trying to better understand the manifold GL+(3,R)/S0(3) which is diffeomorphic to positive definite symmetric matrices. My motivation is to understand U in F = RU where F in GL+(3,R) = deformation gradient, R in S0(3), & U in GL+(3,R)/S0(3) = stretches & shears.
I think that:
(1) GL+(3,R)/SO(3) being diffeomorphic to SYMMETRIC positive definite matrices results from SO(3) not being SYMMETRIC.
(2) Any U in GL+(3,R)/SO(3) can be composed from the set of matrices consisting of all principal shears (identity matrices with 1 off-diag. non-zero entry), all principal stretches (diag. matrices with non-zero diag. entries), & the identity matrix.
(3) This set is not a group for reason given below.
An earlier discussion [Shear transformations](https://mathoverflow.net/questions/4103/shear-transformations) about shear transformations has been very helpful. It highlighted that:
(a) Any upper/lower triangular matrix with 1's on diag. is a composition of shears.
(b) Any matrix with det.= 1 including (a) but also S0(3) can be decomposed into upper/lower triangular matrices with 1's on diag..
(c) Given (b), shear transformations don't form a group as SO(3) not a shear but can be composed from shear transformations.
(d) Upper (or lower) triangular matrices with 1's on diag. (i.e. composed shears) are a group but obviously not containing ALL shears since lower (or upper) triangular shear matrices missing. And, I'm assuming, composing these two groups NOT a group as per (c).
So, I'm asking:
i) If my thoughts about GL+(3,R)/SO(3) in points (1) & (2) above are correct ?
ii) How, as in (b), can composing an upper & lower triangular matrix (both composed shears) result in a rotation in SO(3) ?
Sorry for such a long post !
| https://mathoverflow.net/users/9624 | Understanding manifold GL+(3,R)/SO(3) ? | $GL^+(3,R)/SO(3)$ is the space of 3 dimensional positive definite symmetric matrices because the [polar decomposition](http://en.wikipedia.org/wiki/Polar_decomposition) of $g \in GL^+(3,R)$ is $ g = o p $ , $o \in SO(3)$ and $p$ is positive definite symmetric .
The wikipedia page treats the complex case, but by repeating the computation for the real case you actually get a constructive proof why matrices $p$ are positive definite symmetric.
The answer to your second question (ii) is that almost any orthogonal matrix can be decomposed as a product of a lower triangular and upper triangular matrix (the LU decomposition which is the Bruhat decomposition of the big cell). As an exception you can take a permutation matrix which doesn't allow such a decomposition.
| 3 | https://mathoverflow.net/users/1059 | 43068 | 27,387 |
https://mathoverflow.net/questions/43065 | 1 | Let $M$ be a smooth manifold of 2n-dim, $v$ be a map from $M$ to the matrix of order $m\times m$.
We call $p\in M$ is the singularity, if $v(p)$ is non-invertible. Suppose $v$ is smooth and the singularity is submanifold. Let $C$ be a connected component of singularity, $U$ is the tubular neighborhood of $C$. How to computing the integration $$\int\_{\partial U}{\rm Tr}[(v^{-1}dv)^{2n-1}]$$
I guess its result is zero, may be that is wrong.
| https://mathoverflow.net/users/3896 | A question about to computing a integration | I think your guess is correct and the integral is zero, at least if you assume that $M$ is a closed oriented manifold and that U contains all singularities. Here is the proof, assuming closedness of M. Let $N = M - U$, a compact manifold with boundary. Let $G=Gl\_n (C)$ (in your problem, it does not matter whether you take real or complex matrices) and denote your map by $f: N \to G$. Consider the form $\omega= Tr((g^{-1}dg)^{2k-1})$ on $G$. Then your integrand is $f^{\ast} \omega$ and since $d\omega=0$ (see below), an application of Stokes gives you the answer you suspect. Why is $\omega$ closed? I am tempted to argue that on a compact Lie group, each bi-invariant differential form. Of course $Gl\_n (C)$ is not compact, but it is the complexification of the compact $U(n)$ and the unitary trick works.
For this special example, there is of course a direct computation (which you can do by yourself). It should be contained in Chern-Simons "Geometric invariant and characteristic forms).
| 3 | https://mathoverflow.net/users/9928 | 43071 | 27,388 |
https://mathoverflow.net/questions/43069 | 9 | Is there some place (on the internet or elsewhere) where I can find the number and preferably a list of all (isomorphism classes of) finite connected $T\_0$-spaces with, say, 5 points?
In know that a $T\_0$-topology on a finite set is equivalent to a partial ordering, and [wikipedia](http://en.wikipedia.org/wiki/Partial_order) tells me that there are, up to isomorphism, 63 partially ordered sets with precisely 5 elements. However, I am only interested in connected spaces, and I'd love to have a list (most preferably in terms of Hasse diagrams).
| https://mathoverflow.net/users/1291 | Is there a list of all connected T_0-spaces with 5 points? | There is a Java applet that displays all 5-element connected posets at
<http://www1.chapman.edu/~jipsen/gap/posets.html>.
| 9 | https://mathoverflow.net/users/2807 | 43074 | 27,391 |
https://mathoverflow.net/questions/42971 | 3 | In John Steel's paper "The derived model theorem",
<http://math.berkeley.edu/~steel/papers/dm.ps>
John Steel asserts that it is clear that $\mathrm{Hom}^{Y}\_{\kappa}$ is closed downward under continuous reducibility. Unfortunately that is not clear to me and I was wondering if anyone could help me understand it.
| https://mathoverflow.net/users/7966 | Question about John Steel's "The derived model theorem" | Rupert, I will explain the argument for $Y=\omega$ (this makes no difference, but you may find it easier to visualize) when the continuous function is particularly nice (in a way I will make precise. The general case requires a slight adaptation).
First, some basic background: The topology on $\omega^\omega$ is the product topology with the base set $\omega$ having the discrete topology. By definition of the product topology, it follows that $f:\omega^\omega\to\omega^\omega$ is continuous iff for any $x\in\omega^\omega$ and any $n$ there is an $s\_n=s\_n(x)\in\omega$ with $f(x)\upharpoonright n$ completely determined by $x\upharpoonright s\_n$. This means that if $x,y\in\omega^\omega$ and $x\upharpoonright s\_n=y\upharpoonright s\_n$, then $f(x)\upharpoonright n=f(y)\upharpoonright s\_n$.
The simplifying assumption I will make is that, for all $x$, $s\_n=n$. It follows that there exists a function $\hat f:\omega^{<\omega}\to\omega^{<\omega}$ such that:
1. ${}|\hat f(t)|=|t|$ for all $t\in\omega^{<\omega}$. Here, $|t|$ is the length of $t$.
2. Whenever $s$ is an initial segment of $t\in\omega^{<\omega}$, we have that $\hat f(s)$ is an initial segment of $\hat f(t)$.
3. For any $x\in\omega^\omega$, $f(x)=\bigcup\_n \hat f(x\upharpoonright n)$.
This is standard and any book in descriptive set theory and most books in set theory will mention it in passing at the very least. You may also want to take a look at the discussion of [meta-stability](http://terrytao.wordpress.com/2007/05/23/soft-analysis-hard-analysis-and-the-finite-convergence-principle/) in Tao's blog.
Ok. Suppose now that $A=S\_{\vec\mu}$ is a homogeneous subset of $\omega^\omega$, for some homogeneity system $\vec\mu=(\mu\_s\mid s\in\omega^{<\omega})$. [The full details of the definition are in the paper by Steel linked to in the body of the question.] The intuitive idea to keep in mind is that "membership in $A$ is continuously reduced to well-foundedness of towers."
This suggests that if $B$ continuously reduces to $A$, then "composing continuous maps" should give a continuous reduction of $B$ to well-foundedness of towers, i.e., verify that $B$ is homogeneous. Let's verify this precisely, in the case that $f$ is the reduction map.
We are thus assuming that $B\le\_W A$ via $f$, i.e., $B=f^{-1}(A)$.
>
> Let $\vec\rho=(\rho\_s\mid
> > s\in\omega^{<\omega})$ be given by
> $\rho\_s=\mu\_{\hat f(s)}$. This is also
> a homogeneity system. Then, blatantly
> from $A=S\_{\vec\mu}$, the definition of
> $\hat f$, and the fact that
> $B=f^{-1}(A)$, we have that
> $B=S\_{\vec\rho}$. This completes the
> proof.
>
>
>
[If this is not immediate: Note that $x\in S\_{\vec\rho}$ iff the ultrapower of $V$ via the tower $(\rho\_{x\upharpoonright n}\mid n<\omega)=(\mu\_{\hat f(x\upharpoonright n)}\mid n<\omega)=(\mu\_{f(x)\upharpoonright n}\mid n<\omega)$ is well-founded, i.e., iff $f(x)\in S\_{\vec\mu}=A$. This means that $S\_{\vec\rho}=f^{-1}(A)=B$.]
This needs to be slightly modified in the general case, since we stubbornly ask that each $\mu\_s$ concentrates on $Z^{|s|}$ rather than $Z^{n\_s}$ for some integer $n\_s$, and in general in our continuous functions, we do not have the 'fast convergence' condition that $x\upharpoonright n$ suffices to determine $f(x)\upharpoonright n$. But the case I explained should help you understand the general case.
| 5 | https://mathoverflow.net/users/6085 | 43075 | 27,392 |
https://mathoverflow.net/questions/41628 | 2 | In the Hopf algebra $SL\_q(N)$, it can be shown, using direct calculations, that $S(u^1\_i)u^j\_1 = q^{-1}u^j\_1S(u^1\_i)$. Can anyone see a more elegant way of establishing this?
Moreover, does anyone know of a similar relation in the more general case of $S(u^1\_r)u^i\_j$?
Edit (referneces): By $SL\_q(N)$ I mean the quantized coordinate algebra (not the quantized enveloping algebra). I am using the conventions of Klimyk and Schmudgen, Chpt 4 for the N=2 case, or Chpt 9 for the general case.
| https://mathoverflow.net/users/1095 | The relation $S(u^1_i)u^j_1 = q^{-1}u^j_1S(u^1_i)$ | This is a reasonably known result. That $S(u^1\_i)u^j\_1 = q^{-1}u^j\_1S(u^1\_i)$, was originally proven (to the best of my knowdledge) in FRT's '89 paper "Quantum Groups and Lie Algebras" - the paper is in Russian though. The only English write up of the proof that I known is in Theorem 1 of Vainermann and Podkolzin's '99 [paper](http://docs.google.com/viewer?a=v&q=cache%3AH1PD0VCip6MJ%3Aciteseerx.ist.psu.edu/viewdoc/download%253Fdoi%253D10.1.1.145.4800%2526rep%253Drep1%2526type%253Dpdf+quantum+Stieffel+manifold&hl=ga&pid=bl&srcid=ADGEESj18ZxTGm6y0t_mQ4GrrzMrIRnK_9B_Oynw9gDehj-Z6NEiQHuWDEhtgQxsfu8rEmlZnp1b-O6GGfNhD5mEJQK15EUlHs9WiQGsbIZJp61PjBfrTgJtrEnovntBZZBh_OmYcjXk&sig=AHIEtbRxdwbZXWeC01h1RfkqMawEnDG86w) on Quantum Stiefel Manifolds. It gives a general comm rel for $[S(u^i\_j),u^r\_s]$, for the general $N$ case, using just the $R$-matrix construction of the $SU\_q(N)$. I am sure there are other versions around somewhere though.
| 2 | https://mathoverflow.net/users/1867 | 43089 | 27,401 |
https://mathoverflow.net/questions/43070 | 3 | Is there any statistic that can tell how "hubby" is a graph?
By this I mean a number that is small when a graph has no hubs, that is, when all nodes are more or less equal degree-wise, and big when there are hubs, nodes that concentrate most of the connectivity.
I expect it to be zero (or minimal) for a fully connected graph and 1 (or maximal) for a "star shaped" graph with one node with degree N-1 and N-1 nodes with degree 1.
I was thinking to use some measure of the dispersion of the degree distribution (degree entropy, degree variance, ...). Or maybe to define a hub as a node that, when cut out, turns a connected graph in a non-connected graph and define its "hubbiness" as the number of connected components after I cut it from the graph (and define the hubbiness of the graph as the average hubbiness of the nodes).
But I'm not a specialist in this field, so I don't know if there is an already tried and established measure in the specific community, easy to estimate.
EDIT:
my graphs are usually small (in the worst case 100 graphs with 100 nodes each, but typically 20 graphs with 20 nodes each), so I'm slightly (not absolutely) more interested in theoretical soundness than efficiency.
| https://mathoverflow.net/users/757 | Hubbiness of a graph | Here are some concepts that might be helpful to you:
1. A vertex $v$ of a connected graph $G$ is an *articulation point* if the removal of $v$ from $G$ causes $G$ to be disconnected. My interpretation is that an articulation point corresponds to a "hub". This may or may not match your intuition. On the one hand, the center vertex in a star is an articulation point, but then so is any interior point in a path.
2. A connected graph $G$ is *biconnected* if it does not contain any articulation points (i.e.: the removal of any single vertex from $G$ does not disconnect $G$). By convention, the graph consisting of just two vertices and one edge is considered biconnected.
3. A *biconnected component* of a graph $G$ is a maximal, biconnected subgraph.
4. Finally, we can define the *block tree* $BT(G)$ of a connected graph $G$ as follows: The vertices of $BT(G)$ are the biconnected components of $G$ and an edge $(u,v)$ exists in $BT(G)$ iff the biconnected components corresponding to $u$ and $v$ share a vertex (in $G$). [This Wikipedia article](http://en.wikipedia.org/wiki/Block_tree) has a nice illustration of this concept and a description of an algorithm for constructing the block tree.
I would suggest that the degree of "hubbiness" of a graph $G$ is related to the size of $BT(G)$. Let $G$ be a graph with $n$ vertices. Then:
* If $G$ is biconnected, then $|BT(G)| = 1$
* If $G$ is a star, then $|BT(G)| = n$
* If $G$ is a path, then $|BT(G)| = n-2$
* In general, if $G$ is a tree, then$|BT(G)|$ equals the number of interior vertices
| 2 | https://mathoverflow.net/users/9840 | 43094 | 27,406 |
https://mathoverflow.net/questions/43083 | 11 | What is a good introduction in gradient flows in metric spaces?
I know the book *Gradient flows: in metric spaces and in the space of probability measures by Luigi Ambrosio, Nicola Gigli and Giuseppe Savaré*, but is too hard for an introduction (for me).
I'm looking for something with a similar content.
| https://mathoverflow.net/users/5295 | Textbooks or notes on gradient flows in metric spaces | Here are some links to the online lecture notes which are hopefully more accessible than the book you mentioned:
* [Lecture Notes on Gradient Flows and Optimal Transport](https://arxiv.org/abs/1009.3737) by S. Daneri
* [An Introduction to Gradient Flows in Metric Spaces](http://igk.math.uni-bielefeld.de/study-materials/sm-cle-09.pdf) and
[Introduction to Gradient Flows in Metric Spaces (II)](http://igk.math.uni-bielefeld.de/study-materials/notes-clement-part2.pdf) by P. Clément
| 11 | https://mathoverflow.net/users/2149 | 43097 | 27,407 |
https://mathoverflow.net/questions/43103 | 10 | if $x=d(n)$ is the number of divisors of $n$, what is the tightest lower-bound for $n$ only given $x$?
<http://en.wikipedia.org/wiki/Highly_composite_number>
| https://mathoverflow.net/users/10246 | What is the lower bound for highly composite numbers? | I will start off with the simplest type, $$ d(n) \leq \sqrt{3 n} $$ and $$ d(n) \leq 48 \left(\frac{n}{2520}\right)^{1/3} $$ and
$$ d(n) \leq 576 \left(\frac{n}{21621600}\right)^{1/4}. $$
The first one has equality only at $n = 12,$ second only at $n =2520,$ third only at $n= 21621600.$ Instead of continuing with fractional powers $1/k$ the better results switch to logarithms. Reference is a 1988 paper by J. L. Nicolas in a book called Ramanujan Revisited.
With equality at $n = 6983776800 = 2^5 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ and $d(n) = 2304,$
$$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1.5379398606751... \right)} = n^{ \left( \frac{1.0660186782977...}{\log \log n} \right) }. $$ Full details of the proof appear in J.-L. Nicolas et G. Robin. *Majorations explicites pour le nombre de diviseurs de n*, Canad. Math. Bull., 26, 1983, 485--492. The next two appear in the dissertation of Robin, are repeated in the 1988 Nicolas survey article indicated.
With equality at a number $n$ near $6.929 \cdot 10^{40},$
$$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)}. $$ Compare this one with Theorem 317 in Hardy and Wright, attributed to Wigert (1907),
$$ \limsup \frac{\log d(n) \log \log n}{\log n} = \log 2. $$
With equality at a number $n$ near $3.309 \cdot 10^{135},$
$$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1}{\log \log n} + \frac{4.762350121177...}{\left(\log \log n \right)^2} \right)} $$
Just to fill in one blank, the special integers $n$ here are "superior highly composite numbers" using Ramanujan's original recipe for prime factorization, which I like to write, with $ \delta > 0,$ as
$$ N\_\delta = \prod\_p \; p^{\left\lfloor \frac{1}{p^\delta - 1} \right\rfloor } $$
The first (largest) $\delta$ that assigns an exponent $k$ to a prime $p$ is
$$ \delta = \frac{\log \left(1 + \frac{1}{k} \right)}{\log p}. $$
See [Is there a formula that can predict the primes in the sequence of ratios of consecutive superior highly composite numbers? : $2, 3, 2, 5, 2, 3, 7,...$](https://mathoverflow.net/questions/185493/is-there-a-formula-that-can-predict-the-primes-in-the-sequence-of-ratios-of-cons/185529#185529) for some detail, with computations.
So $$ N\_{1/2} = 12, \; N\_{1/3} = 2520, \; N\_{1/4} = 21621600, $$
$$ N\_{0.23} = 6983776800, \; N\_{0.155} \approx 6.929 \cdot 10^{40}, \; N\_{0.1218} \approx 3.309 \cdot 10^{135}.$$
| 21 | https://mathoverflow.net/users/3324 | 43105 | 27,410 |
https://mathoverflow.net/questions/43015 | 12 | A (standard, real-valued) Brownian motion $W = \{W(t): t \geq 0\}$ is commonly defined by the following properties: 1) $W(0) = 0$ a.s., 2) the process has independent increments, 3) for all $s,t \geq 0$ with $s<t$, the increment $W(t) – W(s)$ is normally distributed with mean zero and variance $t-s$, and 4) almost surely, the function $t \mapsto W(t)$ is continuous.
As is well known, the above set of conditions can be reduced to 2), 3') for all $t \geq 0$, $W(t)$ has mean zero and variance $t$, and 4). [Note that, in 3'), $W(t)$ is not assumed to be normally distributed.] But what about omitting condition 2)? Can you find an example of a process $W$ satisfying conditions 1), 3), and 4), but not 2)? [Note that such $W$ must have (the Brownian motion) covariance $E[W(s)W(t)] = s$, $0 \leq s \leq t$; hence, it cannot be a Gaussian process, for otherwise it would be a Brownian motion.]
| https://mathoverflow.net/users/10227 | The conditions in the definition of Brownian motion | No, it is not true that a process *W* satisfying the properties (1), (3) and (4) has to be a Brownian motion. We can construct a counter-example as follows.
This construction is rather contrived, and I don't know if there's any simple examples.
Start with a standard Brownian motion *W*. The idea is to apply a small bump to its distribution while retaining the required properties. I will do this by first reducing it to the discrete-time case. So, choose a finite sequence of times 0 = *t*0 < *t*1 < ... < *t**n*. Then define a piecewise linear process *X* by *X**t**k* = *W**t**k* (*k* = 0,1,...,*n*) and such that *X* is linearly interpolated across each of the intervals [*t**k*-1,*t**k*] and constant over [*t**n*,∞).
Then, *Y* = *W* - *X* is a continuous process independent from *X*. In fact, *Y* is just a sequence of [Brownian bridges](http://en.wikipedia.org/wiki/Brownian_bridge) across the intervals [*t**k*-1,*t**k*] and is a standard Brownian motion on [*t**n*,∞). Also by linear interpolation, for any time *t* ≥ 0, *X**t* is a linear combination of at most two of the random variables *X**t*1,...,*X**t**n*. The increments of *W*,
$$
W\_t-W\_s = X\_t-X\_s + Y\_t-Y\_s,
$$
are then a linear combination of at most 4 of the random variables *X**t*1,...,*X**t**n* plus an independent term. So, choosing *n* ≥ 5, if it is possible to replace (*X**t*1,...,*X**t**n*) by any other ℝn-valued random variable without changing the joint-distribution of any 4 elements, then the distributions of the increments *W**t* - *W**s* will be left unchanged. So, properties (1), (3), (4) will still be satisfied but the new process for *W* will not be a standard Brownian motion. It is possible to change the distribution in this way:
>
> Let *X* = (*X*1,*X*2,...,*X**n*) be an ℝn-valued random variable with a continuous and strictly positive probability density *p**X*: ℝn → ℝ. Then, there exists a random variable *Y* = (*Y*1,*Y*2,...,*Y**n*) with a different distribution than *X* but for which the projection onto any *n* - 1 elements has the same distribution as for *X*.
>
>
>
That is, for any *k*1,*k*2,...,*k**n*-1 in {1,...,*n*}, (*Y**k*1,*Y**k*2,...,*Y**k**n*-1) has the same distribution as (*X**k*1,*X**k*2,...,*X**k**n*-1).
We can construct the probability density *p**Y* of *Y* by applying a bump to the probability distribution of *X*,
$$
p\_Y(x)=p\_X(x)+\epsilon f(x\_1)f(x\_2)\cdots f(x\_n).
$$
Here, ε is a fixed real number and *f*: ℝ → ℝ is a continuous function of compact support and zero integral, $\int\_{-\infty}^\infty f(x)\,dx=0$. Then, $\int\_{-\infty}^\infty p\_Y(x)\,dx\_k=\int\_{-\infty}^\infty p\_X(x)\,dx\_k$ for each *k*. So, the integral of *p**Y* over ℝ*n* is 1 and, by choosing ε small, *p**Y* will be positive. Then it is a valid probability density function. Finally, as the integral along the *k*th direction (any *k*) agrees for *p**X* and *p**Y*, the projection of *X* and *Y* onto ℝ*n*-1 along the *k*th direction give the same distribution.
| 11 | https://mathoverflow.net/users/1004 | 43111 | 27,414 |
https://mathoverflow.net/questions/43057 | 5 | Hi all,
I am looking to do some linguistic analysis of informal proofs. Therefore I am on a search for a collection of entry level proofs written in a clear, uninvolved style. I have one recommendation for Hardy and Wright's "An Introduction to the Theory of Numbers," and was wondering if there is something else you may add to this.
Many thanks in advance,
Nickolay
| https://mathoverflow.net/users/10234 | Looking for a collection of entry level proofs | You can try Aigner and Ziegler's book [Proofs from the book](http://rads.stackoverflow.com/amzn/click/3540636986)
| 2 | https://mathoverflow.net/users/5372 | 43115 | 27,418 |
https://mathoverflow.net/questions/43081 | 4 | The Kirchhoff's theorem is a classical result for counting the number of spanning trees in a graph.
However, what are the best known upper bounds on the number of spanning trees in a graph in terms of structural parameters (e.g., number of vertices, degrees, etc.) instead of algebraic quantities?
| https://mathoverflow.net/users/nan | Number of spanning trees: bounds from structural parameters | I had an email discussion with Russell Lyons a few years ago about maximizing the number of spanning trees among all graphs with a given number of vertices and edges. He had a simple argument for an upper bound of $(2e/v)^{v-1}$. There's an even simpler argument for an upper bound of $e\choose v-1$. Russell thought there was a good bound for regular graphs due to McKay.
As for lower bounds, if the graph is not connected, it has zero spanning trees, and even an $n$-vertex graph with just $n-1$ edges missing (compared to the complete graph) may not be connected. I suppose one could restrict to connected graphs and then ask for a minimum.
EDIT: Here are bibliographical details on two papers by McKay:
McKay, Brendan D., Spanning trees in regular graphs, European J. Combin. 4 (1983), no. 2, 149–160, MR 85d:05194.
McKay, Brendan D., Spanning trees in random regular graphs, Proceedings of the Third Caribbean Conference on Combinatorics and Computing (Bridgetown, 1981), pp. 139–143, Univ. West Indies, Cave Hill Campus, Barbados, 1981, MR 83g:05030.
| 5 | https://mathoverflow.net/users/3684 | 43116 | 27,419 |
https://mathoverflow.net/questions/43125 | 1 | Does anybody in here know how to get hold of this article:
"Tutte, W.T., A Theory of 3-connected graphs, Indag. Math. 23 (1961) 441-455"
or have it on paper?
| https://mathoverflow.net/users/1539 | A Theory of 3-connected graphs | It is in the volume "Selected Papers of W.T. Tutte" published by the Charles Babbage Research Centre about 20 years ago.
(Strangely the reference is slightly different, but the title and page numbers are identical.)
| 2 | https://mathoverflow.net/users/1492 | 43128 | 27,424 |
https://mathoverflow.net/questions/43124 | 19 | Let $A\in\mathcal M\_n$ be an $n\times n$ real [symmetric] matrix which depends smoothly on a [finite] set of parameters, $A=A(\xi\_1,\ldots,\xi\_k)$. We can view it as a smooth function $A:\mathbb R^k\to\mathcal M\_n$.
>
> 1. What conditions should the matrix $A$ satisfy so that its eigenvalues
> $\lambda\_i(\xi\_1,\ldots,\xi\_k)$,
> $i=1,\ldots,n$, depend smoothly on the
> parameters $\xi\_1,\ldots,\xi\_k$?
>
>
>
*e.g.* if the characteristic equation is $\lambda^3-\xi=0$, then the solution $\lambda\_1=\sqrt[3] \xi$ is not derivable at $\xi=0$.
>
> 2. What additional conditions should the matrix $A$ satisfy so that we can
> choose a set of eigenvectors
> $v\_i(\xi\_1,\ldots,\xi\_k)$,
> $i=1,\ldots,n$, which depend smoothly
> on the parameters
> $\xi\_1,\ldots,\xi\_k$?
>
>
>
**Update - important details**
* The domain is simply connected
* The rank of $A$ can change in the domain
* The multiplicities of the eigenvalues can change in the domain, they can cross
* The matrix $A$ is real symmetric
* $n$ and $k$ are finite
**Update 2**
* A relaxation of the conditions of the problem:
For fixed $p=(\xi\_{01},\ldots,\xi\_{0k})$, can we find an open neighborhood of $p$ in the domain and a set of conditions ensuring the smoothness of the eigenvalues and the eigenvectors?
| https://mathoverflow.net/users/10095 | Conditions for smooth dependence of the eigenvalues and eigenvectors of a matrix on a set of parameters | The fact that the entries of the matrix are real does seem to help. The state of the art is the following.
* The spectrum is continuous functions of $\xi$. However, it is not always possible to label the eigenvalues so that they individually are continuous functions.
* When the multiplicities $m\_1,\ldots,m\_r$ do not change as $\xi$ varies (no crossing of eigenvalues), then the eigenvalues are as smooth as the matrix. If the domain is simply connected, the eigenvalues may be labelled so as to be smooth functions.
* When the entries are analytic functions of a *single* variable ($k=1$) and the eigenvalues remain real, then the eigenvalues may be labelled so as to be analytic functions. However, in case of crossing, this nice labelling is not the obvious one (*i.e.* not $\lambda\_1\le\lambda\_1\le\cdots$). This become false for $k\ge 2$, as shown by the example
$$\left(\begin{array}{cc} \xi\_1 & \xi\_2 \\\\ \xi\_2 & -\xi\_1 \end{array}\right)\qquad\qquad (1).$$
* The situation is not that good concerning the eigenvectors. The following is called *Petrowski's example*,
$$\left(\begin{array}{ccc} 0 & \xi\_1 & \xi\_1 \\\\ 0 & 0 & 0 \\\\ \xi\_1 & 0 & \xi\_2 \end{array}\right).$$
The eigenvalues are real for every $\xi$, distinct when $ \xi\_1\ne0$. The matrix is diagonalisable for every $\xi$, but two eigenvectors have the same limit when $ \xi\_1\rightarrow0$.
If the domain is not simply connected, you may have additional difficulties with eigenvectors. Take example (1) above, with $\xi$ running over the unit circle $S^1$. When you follow continuously a unit eigenvector $V(\xi)$, it is flipped (*i.e.* multiplied by $-1$) after one loop
| 19 | https://mathoverflow.net/users/8799 | 43133 | 27,426 |
https://mathoverflow.net/questions/43095 | 8 | Fix positive integers $k, N$ and let $\omega$ be a Dirichlet character mod $N$.
Let $f\in S\_k(N,\phi)$ be a normalized newform (i.e. of weight $k$, level $N$ and character $\phi$) with fourier expansion $\sum\_{n\geq 1} a(n)q^n$. In her paper 'Newforms and Functional Equations', Winnie Li showed that if $q$ is a prime dividing $N$ and $\phi$ is not a character mod $N/q$, then $|a(q)|=q^{\frac{k-1}{2}}$. In particular, $a(q)\neq 0$. This is Theorem 3.(ii) of the paper. I should also note that special cases of this were proven earlier Atkin-Lehner, Hecke and Ogg.
I am interested in the Hilbert modular version of this result. More specifically, let $\mathfrak{N}$ be an integral ideal of a totally real number field $K$ of degree $d$ over $\mathbb{Q}$, $\Phi$ be a Hecke character induced by a numerical character mod $\mathfrak{N}$ and $k\in (\mathbb{Z}\_+)^d$.
Let $S\_k(\mathfrak{N},\Phi)$ be the space of Hilbert modular cusp forms of weight $k$, level $\mathfrak{N}$ and character $\Phi$, viewed adelically to circumvent class number issues (I make no assumptions regarding the class number of $K$). Shimura, in 'The special values of the zeta functions associated to Hilbert modular forms' defined "Fourier coefficients" $C(\mathfrak{m},\textbf{f})$ for cusp forms $\textbf{f}\in S\_k(\mathfrak{N},\Phi)$ (this on the bottom of page 649 of the paper).
Let $\textbf{f}\in S\_k(\mathfrak{N},\Phi)$ be a normalized newform. I am interested in knowing when $C(\mathfrak{q},\textbf{f})=0$, where $\mathfrak{q}\mid\mathfrak{N}$. In their paper 'Twists of Hilbert Modular Forms', Tom Shemanske and Lynne Walling showed that if the numerical character inducing $\Phi$ is not defined mod $\mathfrak{N}/ \mathfrak{q}$, then this coefficient is nonzero whenever $\mathfrak{q}$ has degree 1 over the rationals or $\mathfrak{q}$ exactly divides $\mathfrak{N}$ (this is part 2 of Theorem 3.3 in the paper). In a remark directly after the Theorem, Tom and Lynne say that the reason for the restrictions on the prime $\mathfrak{q}$ is that this is what is needed to make Ogg's proof in the classical case go through. The 'truth', however, is that the $\mathfrak{q}$-th coefficient should always be nonzero in this case, without any restrictions on the degree of the prime $\mathfrak{q}$. A referee told Tom (who is my advisor) that the representation theory implies this stronger result. Tom lost the referee's report some time ago however, and doesn't know of any reference for this representation theoretic result. Does anyone know of a paper in which this result is proven without any restrictions on the prime $\mathfrak{q}$?
$\textbf{Question}:$ Let $\textbf{f}\in S\_k(\mathfrak{N},\Phi)$ be a normalized newform, $\mathfrak{q}$ be a prime dividing $\mathfrak{N}$ and suppose that the numerical character inducing $\Phi$ is not defined mod $\mathfrak{N}\mathfrak{q}^{-1}$. Is it true that $C(\mathfrak{q},\textbf{f})\neq 0$?
| https://mathoverflow.net/users/nan | Hilbert Modular Newforms | If I've understood your question correctly, you're right that $C(q,f)\not=0$ always and there is a natural representation-theoretic proof of this result (before I start let me say that I don't know how to get these gothic $q$s and $N$s as in your question, so I am just using usual $q$s and $N$s, but they are ideals of $F$ just like yours). The one thing I am worried about is that I do not know what a "numerical character" is, probably because I think about Hilbert modular forms in a different way to you. For me, a Hilbert modular form is really just an automorphic representation of $GL(2,F)$ with certain properties, and the natural generalisation of the character of a classical modular form in this setting is the the following construction. Take the central character of this representation, which is a character of the ideles of $F$. This character decomposes as the product of a power of the norm character and a finite order character, and this finite order character is the natural generalisation of the character of the form. For me, the theorem is that if the conductor of $f$ is $N$, if $q^t$ is the exact power of $q$ dividing $N$, and if $q^t$ is also the exact power of $q$ dividing the conductor of the finite order character, then $C(q,f)\not=0$, where in this generality I am interpreting that as saying that the local $L$-factor attached to the automorphic representation at $q$ is $(1-c.Norm(q)^{-s})^{-1}$ with $c\not=0$.
So, as the referee states, this statement can be proved purely representation-theoretically. It is also a purely local assertion in fact. The automorphic representation is a tensor product of local representations and the local $L$-factor at $q$ can be computed from knowing $\pi\_q$, the factor at $q$. So we are done by the following purely local theorem, where now $K$ is the completion of the totally real field $F$ at the prime $q$, and $\pi$ is $\pi\_q$:
Thm) Say $K$ is a finite extension of $\mathbb{Q}\_p$, say $\pi$ is a smooth admissible irreducible representation of $GL(2,K)$, say $\pi$ is ramified, has conductor $q^t$ ($q$ a uniformiser of $K$), and say the central character of $\pi$ also has conductor $q^t$. Then $\pi$ is a ramified principal series representation associated to two character, one unramified and one ramified of conductor $q^t$.
The reason the result you want follows is that the $L$-function of $\pi$ is $(1-c.Norm(q)^{-s})^{-1}$ with $c$ equal to the value at a uniformiser of the unramified character.
These sorts of assertions (explicit computations of $L$-functions) can all be found in Jacquet-Langlands, a book which changed my life, but I am sure that there are references which are a gazillion times more readable nowadays.
So now all we have to do is to prove the theorem. Well there are probably purely representation-theoretic arguments, but I don't know them [edit: vytas does---see his answer], so I am going to use the following trick: hit everything with local Langlands. This translates the result we want into a question about 2-dimensional representations rather than infinite-dimensional ones, so we'll be in much better shape. To make this part of the argument work you need to have an explicit hold on what local Langlands says for $GL(2)$.
OK so apply local Langlands to $\pi$ and we get a Weil-Deligne representation $(\rho,N)$ of the Weil group of $K$. And we know that the conductor of this representation is $q^t$ and the conductor of its determinant is also $q^t$, and we want to prove that $\rho$ is reducible with one ramified and one unramified character on the diagonal, and that $N=0$. Then we're done.
OK so first I'll show $N=0$. This is because if $N\not=0$ then the definition of a Weil-Deligne representation forces $\rho$ to be $\chi+\chi|.|$ with $|.|$ the norm character. And we now compute conductors. If $\chi$ is unramified then the conductor of $(\rho,N)$ is $q$ but the determinant is unramified, so our hypotheses do not apply (this the situation for elliptic curves with multiplicative reduction, for example; curve has bad reduction but character is unramified at $q$). And if $\chi$ is ramified and has conductor $q^s$ with $s\geq1$ then $\rho$ has conductor $q^{2s}$ so again we can't be here because $s\not=2s$.
It remains to deal with the $N=0$ case. We have a representation $\rho$ with some conductor $q^t$ and its character also has conductor $q^t$---let me drop these $q$s and just talk about conductor $t$ out of laziness. Say first that $\rho$ is the sum of two characters $\sigma\_1$ and $\sigma\_2$ of conductors $t\_1$ and $t\_2$. Then the conductor of $\rho$ is $t\_1+t\_2$ and the conductor of its determinant is at most the max of $t\_1$ and $t\_2$, so if these are equal then one of the $t\_i$ had better be zero, and so the other one had better be non-zero, and this is the case that is really happening.
All that is left now is the case where $\rho$ is irreducible. [Edit: removed incomplete answer and replaced it with complete one]. To do this case one just looks at the definition of the conductor of a representation. It's a sum of the form $\sum\_i c\_i.\dim(V/V^{G\_i})$
where the $c\_i$ are rational and the $G\_i$ are running through a filtration on the inertia subgroup. The moment some $V^{G\_i}$ is zero then you're in trouble, because then the sum contributes 2 to the conductor of $\rho$ and at most 1 to the conductor of its determinant. So each $V^{G\_i}$ had better have dimension 1 or 2. In particular there are some inertial invariants. But these form a Galois-stable subspace, so the irreducible case cannot happen and we are finally done!
| 10 | https://mathoverflow.net/users/1384 | 43134 | 27,427 |
https://mathoverflow.net/questions/43131 | 7 | I saw a statement somewhere that for the Hirzebruch surfaces $F\_n:=\mathbb{P}\_{\mathbb{P}^1}(\mathcal{O}\oplus\mathcal{O}(n))$, $F\_n$ and $F\_m$ are symplectormorphic when $m$ and $n$ have the same parity.
My question is: Why is this true?
I can see that they are diffeomorphic by Freedman's Theorem: computing the intersection pairing on $\text{Pic}(F\_n)=H^2(F\_n, \mathbb{Z})$, which is an even form when n is even, and an odd form when n is odd.
But this result is deep and abstract, is there any easy way to construct a symplectomorphism? Certainly it is not given by polynomial maps, I'm wondering what it will be.
| https://mathoverflow.net/users/1657 | Why are the following varieties symplectomorphic? | In order to obtain an explicit description of the diffeomorphism, one can use the following argument.
Take $B=\mathbb{C}^{n-1}$, with coordinates $t\_1, \ldots, t\_{n-1}$, and consider the complex space
$\mathcal{X}$ obtained glueing $\mathbb{P}^1 \times \mathbb{C} \times B$ with $\mathbb{P}^1 \times \mathbb{C} \times B$ by the identification of
$(y\_0, y\_1, z, t\_1, \ldots, t\_{n-1})$ with $(y\_0', y\_1', z',t\_1, \ldots, t\_{n-1})$
if
$z'=z^{-1}, \quad y\_1'=y\_1z^{-n}, \quad y\_0'=y\_0+y\_1 \sum\_{i=1}^{n-1}t\_iz^{-i}$.
Let us denote by $\pi \colon \mathcal{X} \to B$ the family obtained in this way.
Let now $T\_k \subset B$ be the determinantal locus given by rank $M \leq k$, where $M$ is the matrix
\begin{bmatrix}
t\_1 & \ldots & t\_{k+1} \cr
t\_2 & \ldots & t\_{k+2} \cr
\cdot & \cdot & \cdot \cr
\cdot & \cdot & \cdot \cr
\cdot & \cdot & \cdot \cr
t\_{n-k-1} & \ldots & t\_{n-1}
\end{bmatrix}
Then, if $t \in T\_k - T\_{k-1}$ we have $X\_t :=\pi^{-1}(t) \cong \mathbb{F}\_{n-2k}$.
By using Ehresmann theorem, one concludes that
$F\_n$ is diffeomorphic to $F\_{n-2k}$.
Geometrically speaking, we are considering all the rank $2$ vector bundles $V$ which fit into the short exact sequence
$0 \to \mathcal{O}\_{\mathbb{P}^1} \to V\_n \to \mathcal{O}\_{\mathbb{P}^1}(n) \to 0$.
They are classified by $H^1(\mathbb{P}^1, \mathcal{O}(-n)) \cong \mathbb{C}^{n-1}$, and we consider the family of ruled surfaces $\mathbb{P}(V\_n)$, thus obtained, as a deformation of $\mathbb{F}\_n = \mathbb{P}(\mathcal{O} \oplus \mathcal{O}(n))$.
This argument shows that even Hirzebruch surfaces are diffeomorphic to $S^2 \times S^2$, whereas odd Hirzebruch surfaces are diffeomorphic to $\mathbb{CP}^2 \sharp \overline{\mathbb{CP}^2}$.
The uniqueness of the symplectic structrure in each case is a more difficult story, and was proven by Lalonde and McDuff [J-curves and the classification of rational and
ruled symplectic 4-manifolds, Contact and Symplectic Geometry (Cambridge,
1994)].
| 10 | https://mathoverflow.net/users/7460 | 43142 | 27,431 |
https://mathoverflow.net/questions/43139 | 6 | The title says it all, what's a good dense open of $\bar{M}\_g,n(X,\beta)$ which play the role of ${M}\_g$ in $\bar{M}\_g$?
My first (naive) guess is maps from a genus $g$ smooth curve to $X$ which represents the class $\beta$. But I'm a little bit concerned, is it dense for sure? Could it happen that in some cases one has singular curves only? Can a stable map from a singular curve always deform to a map from a smooth curve of genus $g$?
| https://mathoverflow.net/users/1657 | What's a good dense open of $\bar{M}_g,n(X,\beta)$? | If $X$ is convex and $g = 0$, then you can take smooth curves with distinct marked point, this will be dense. However, in general the locus of smooth curves is not dense. An easy example is that of degree 1 unpointed maps from a curve of genus $g > 0$ to $\mathbb P^1$; such a curve must consist of a copy of $\mathbb P^1$ mapping isomorphically and some vertical components, hence it can not be smooth.
In general I would guess that there is no analogue of $M\_g$. There are good philosophical reasons why it has to be so; the space $\overline{M}\_g,n(X,\beta)$ is much too big, in general, and badly singular, it can't be expected to behave nicely in any way. The “good” part of $\overline{M}\_g,n(X,\beta)$ is the virtual fundamental class; but that's just a class of cycles, it does not have open parts.
| 9 | https://mathoverflow.net/users/4790 | 43143 | 27,432 |
https://mathoverflow.net/questions/43147 | 29 | I have a basic question that others have definitely considered.
Often there are papers that originally appeared in a language that one might not understand (and I mean a natural language here). I would like to cite the original paper, because that is where the credit belongs. But on the other hand, doing so violates the golden-rule of *read that paper that you cite*! What should I do to overcome this dilemma?
So far, I have always cited the original, and if possible some other related work that has appeared in English---but sometimes, reviewers write back that I should not be citing papers written in a language different from English, which is what motivated me to ask this question.
Thanks for any useful advice.
| https://mathoverflow.net/users/8430 | Citing papers that are in a language that you do not read. | I think a common-sense approach is to cite the original paper (whatever the language) in order to give credit and attribution but only rely on arguments from papers you can understand in your proofs (so you don't violate the golden rule).
Regarding reviewers, the worst that can happen (I think) is that you use a crucial argument from a paper you can understand but that the reviewer cannot understand. In that case, I think the problem is the same whether the reviewer cannot understand it because he is unfamiliar with the math or because he is unfamiliar with the natural language. In both cases, you, as the author, should try to present relatively clear references, and that includes translations when appropriate I guess, but ultimately this is a failure of the reviewer. If I were reviewing a paper and found myself in this situation, I would politely ask the author if there is a translation available. If not, I would tell the editors I am not competent, but wouldn't blame the author.
It is a bad idea to upset reviewers, but banning reference in languages other than English (or any other language) even for attribution purpose is an outrageous suggestion that should not be complied with.
| 48 | https://mathoverflow.net/users/2284 | 43151 | 27,437 |
https://mathoverflow.net/questions/42922 | 3 | The following problem arises when we try to bound the expected offline optimal value of a simple online assignment problem with random values and unit weights, by its deterministic approximation.
The Problem
-----------
Consider a sequence $\{X\_i\}\_{i=1}^n$ of non-negative integrable i.i.d. random variables with absolutely continuous c.d.f. $F(x)$. Let $X\_{(i)}$ be the $i^{\rm th}$ order statistics, so that $X\_{(1)}$ is the minimum of the sequence and $X\_{(n)}$ is the maximum. Now, let $T\_k$ be the average of the top $k^{\rm th}$ order statistics, that is, $T\_k = \frac 1 k \sum\_{i=n-k+1}^n X\_{(i)}$. We would like to show that the expected value of the average of the top order statistics is upper bounded as follows:
>
> $$\mathbb{E} T\_{k} \le \mathbb{E} \left[ X \mid F(X) \ge 1 - k/n \right],$$
>
>
>
where the right hand size is the conditional expectation of X given that it is larger than the $(1-k/n)$-percentile. In order to keep things simple, we may assume that the c.d.f. $F(\cdot)$ is strictly increasing in its domain.
Moreover, we may we fix $\rho \in (0,1)$, and set $k=[\rho n]$; so that we are interested in the average of the top $\rho$ fraction of the sequence. If we scale both $n$ and $k$ to infinity while keeping the ratio $\rho$ fixed, it seems to be the case that the bound is asymptotically tight:
$$ \lim\_{n \rightarrow \infty} \mathbb{E} T\_{[\rho n]} = \mathbb{E} \left[ X | F(X) \ge 1 - \rho \right]. $$
Can you show if these results hold? I have done some numerical experiments with a couple of distributions (uniform, truncated normal, exponential) that confirm these results. Any help or pointer would be appreciated.
Thanks in advance.
More motivation
---------------
This is a stripped down version of a more complicated problem. Suppose that we have an incoming inventory of $n$ different items with unknown value arriving in an online fashion. We can only keep $k$ items of the $n$ total. The decision to keep an item has to be made at the moment of arrival, and once we decide to keep an item we need to stick to this decision.
The value of i-th item, denoted by $X\_i$, is unknown, and revealed when it arrives (before a decision needs to be made). However, we do have a prior for the values; they are drawn independently from an identical distribution $F(\cdot)$. The objective of the problem is designing an online policy that maximizes the expected total value of the assignment.
A useful benchmark, when comparing online policies, is the offline optimal solution. Given a realization of the values $X=\{X\_i\}\_{i=1}^n$, the optimal value of the offline problem, denoted by $P(X)$, is
$$\begin{align}
P(X) = \max\_y &\; \sum\_{i=1}^n y\_i X\_i \\
\text{s.t.}
& \sum\_{i=1}^n y\_i = k, \\
& y\_i \in \{0,1\}
\end{align}$$
In this simple case, the offline optimal solution is to keep the items with the k-th highest values. Equivalently, we have that $P(X) = \sum\_{i=n-k+1}^n X\_{(i)}$. We are interested in the expected optimal value of offline optimal solution, which is given by $\mathbb{E} P(X) = k \mathbb{E} T\_k$.
A simple online policy could instruct us to keep those items with value larger than the $(1-k/n)$-percentile. Such policy can be shown to attain an expected value of
$n \mathbb{E} \left[ X \mathbf{1} \{ F(X) \ge 1 - k/n\} \right]$ when $n$ is large.
The results that we want to prove would allows us to upper bound the expected optimal value of the offline problem by a bound that could be attained, asymptotically, by an online policy. This would confirm that our policy is good.
| https://mathoverflow.net/users/10203 | An Upper Bound for the Average of Top Order Statistics | Write
$$
E[T\_k]=E[E[T\_k|X\_{(n-k)}]]
$$
The distribution of $X\_{(n-k+1)},\dots,X\_{(n)}$ given $X\_{(n-k)}=x$ is the same as the conditional distribution of a monotone arrangement of $n-k$ independent r.v.'s with cdf $F$ given they all are not less than $x$. (This can be proved easily e.g. by using the quantile transform.)
Now the mean value of monotone arrangement of a sequence is equal to the mean value of the sequence itself. So by linearity of conditional expectation we can write
$$
E[E[T\_k|X\_{(n-k)}]]=E[E[X|X\ge X\_{(n-k)}]],
$$
where $X$ is an independent of $X\_{(n-k)}$ r.v. with cdf $F$.
Now it is again about the quantile transform. Let $U=F(X)$, $U\_{(n-k)}=F(X\_{(n-k)})$. We have (by the independence of $U$ and $U\_{(n-k)}$)
$$
E[X|X\ge X\_{(n-k)}]=E[F^{(-1)}(U)|U>U\_{(n-k)}] = \frac{1}{1-U\_{(n-k)}}\phi(U\_{(n-k)}),
$$
where $\phi(u) = \int\_u^1 F^{(-1)}(x)dx$. But we do know the density of $U\_{(n-k)}$, so the expectation of this expression is equal to
$$
\frac{n!}{(n-k-1)k!}\int\_0^1 u^{n-k-1}(1-u)^{k-1}\phi(u)du = \frac{n}{k}E[\phi(B)],
$$
where $B$ has $\mathrm{Beta}(n-k,k)$ distribution. But $\phi(x)$ is easily seen to be concave, so by Jensen's inequality $E[\phi(B)]\le \phi(E[B]) = \phi(1-k/n)$, as required.
| 3 | https://mathoverflow.net/users/8146 | 43159 | 27,443 |
https://mathoverflow.net/questions/43172 | 0 | I am looking at the automorphism group $G$ of a graph, represented as permutation matrices. The point in a proof I am trying to understand goes something like this:
"For any permutation matrix $P$ in $G$ there exists an orthogonal matrix $Q$ such that $Q^{-1}PQ=A$, where $A$ is a block-diagonal matrix representing a direct product of orthogonal groups. Hence there is an embedding of $G$ into this direct product."
Why is this? Is it always true that a morphism between two groups can be represented by a matrix $Q$ such that $Q^{-1}PQ=A$, where $P$ and $A$ are matrices representing elements of the domain and the image respectively? Or is this only the case for morphisms between two different representations of the same group?
If it is not true in general, why does it work in this specific case?
| https://mathoverflow.net/users/4078 | Morphisms between representations | Isn't this just a fancy way of talking about the cycle decomposition of a permutation? The language you use is a little imprecise. But, firstly, by change of basis P can be made into a block matrix form, with each block corresponding to a cycle. And secondly one needs only to think of a cyclic permutation in matrix form, say with 1's mostly just above the diagonal, to see it as an orthogonal matrix.
| 2 | https://mathoverflow.net/users/6153 | 43178 | 27,455 |
https://mathoverflow.net/questions/42731 | 2 | Let $V$ be a Hausdorff locally convex topological vector space over the field $\mathbb{K}$.
Let $B$ be a subset of $V$ such that
$\;$ for all functions $c : B\to \mathbb{K}$, if $\displaystyle\sum\_{b\in B} \; c(b)\cdot b = 0$, then $c$ is identically zero
and $f : B\times V \to \mathbb{K}$ be a function such that
$\;$ for all vectors $v$ in $V$, $\; \displaystyle\sum\_{b\in B} \; f(b,v)\cdot b = v$.
Let $b$ be a member of $B$, and $g : V \to \mathbb{K}$ be given by $g(v) := f(b,v)$. Does it follow that $g$ is continuous?
| https://mathoverflow.net/users/nan | Are coordinate functions on topological vector spaces always continuous? | Here is a non orthogonal example that is however natural and probably simpler than the other one I gave. Let $b\_1, b\_2,...$ be the character basis for $L\_2(-\pi,\pi)$. Let $b\_0$ be an $L\_1$ function whose Fourier series does not converge in the $L\_1$ norm. Then $b\_0,b\_1,b\_2,...$ is countably linearly independent because inner product with $b\_1,b\_2,...$ is continuous in the $L\_1$ norm, and $b\_0,b\_1,b\_2,...$ is thus an unconditional basis, in the $L\_1$ norm, for the linear span of $b\_0$ and $L\_2$. The coordinate functional for $b\_0$ is obviously discontinuous. For $n\ge 1$, the coordinate functional for $b\_n$ is discontinuous iff $\langle b\_0, b\_n \rangle \not= 0$. You can guarantee that this happens for every $n$ by perturbing the original $b\_0$ by an appropriate $L\_2$ function.
| 2 | https://mathoverflow.net/users/2554 | 43184 | 27,458 |
https://mathoverflow.net/questions/43180 | 3 | Hello,
I'm trying to bound an integral. I have a function $A(\nu) = | 1 + \exp(-I \nu) |$ (with $I$ being the imaginary unit) and I want to show that the condition (Paley-Wiener criterion for causality) applies
$$\int\_{-\infty}^{\infty} \frac{|\log(A(\omega))|}{1+\omega^2} \mathrm{d}\omega < \infty$$
(log is the natural logarithm) I used a transformation from $\omega$ to $\nu$: $\omega = \tan(\nu/2)$ and I converted the integral by substitution to
$$\int\_{-\pi}^{\pi} | \log(A(\nu)) |\mathrm{d}\nu < \infty$$
But I don't know how to show that this condition applies for the given function $A(\nu)$. I tried to simplify the problem by using $A(\nu) = | 1 + \exp(-I \nu) | = \sqrt{(1+\exp(-I\nu))(1+\exp(I\nu))}$ and thus simplifying the integral to
$$\frac{1}{2} \int\_{-\pi}^{\pi} | \log(1+\exp(-I\nu)) + \log(1+\exp(I\nu)) |\mathrm{d}\nu$$
But still I have trouble finding a bound. I also tried $A(\nu) = | 1 + \exp(-I \nu) | = \sqrt{2} \sqrt{\cos(\nu) + 1}$.
I though maybe the problem can be solved by providing an upper and lower bound function that converges. But because $A(\nu)$ has values in the range $[0,1]$ the logarithm assumes very large values (and there are actually points of singularity for $A(\nu)=0$).
Please help me solve this problem.
| https://mathoverflow.net/users/10256 | Integration problem: $\int_{-\pi}^{\pi} | \log( | 1 + \exp(- I \nu ) | ) | \mathrm{d}\nu < \infty$ | You want to show that
$$\int\_{-\pi}^\pi|\log|1+e^{-it}||dt$$
is finite. Now
$$|1+e^{-it}|=|e^{it/2}+e^{-it/2}|=2\cos(t/2)$$
so your integral is
$$\int\_{-\pi}^\pi|\log|2\cos(t/2)||dt
=2\int\_0^\pi|\log|2\cos(t/2)||dt.$$
Replacing $t$ by $\pi-2$ in the last integral gives
$$2\int\_0^\pi|\log|2\sin(t/2)||dt.$$
The integrand is nicely continuous away from $0$. Near $0$,
$\sin (t/2)=tf(t)$ where $f(t)\to1/2$ as $t\to0$. Then the
integrand is $|\log t+g(t)|$ where $g$ is continuous at $0$
and now finiteness follows since
$$\int\_0^1|\log t|dt$$
is finite (integration by parts).
| 3 | https://mathoverflow.net/users/4213 | 43185 | 27,459 |
https://mathoverflow.net/questions/43186 | 0 | Hi all,
what are the best strategies to find cutting edge papers and books on a field of mathematics?
..
Example:
2-3 years ago I had to analyze a time series. I found a paper and showed that to a mathematician who referred me to the REAL state-of-the art method how to do it.
Then I read the book ‘Analysis and Probability: Wavelets, Signals, Fractals’ by Palle E. T. Jorgensen, which is excellent and a good reference.
How can I find for example the latest advances in (applied) mathematical Logic (modal logic, high order logic, type theory, proof theory)?
How can I find the state of the art advances in time series analysis since mid 2007?
My current strategies are using google search and google scholar to find papers. Or to try to find a good mathematician in a field an ask in person for papers.
Thanks for any Tipps!
Steve
| https://mathoverflow.net/users/10258 | finding cutting edge papers and books | One method that is maybe more indirect is to regularly browse through the new articles posted on the arxiv in the section of your interests. I use an RSS reader for this, but you can also subscribe to an e-mail list I think.
Also, conferences for example are always a good way to keep up to date of what's going on in your field.
| 4 | https://mathoverflow.net/users/9545 | 43188 | 27,461 |
https://mathoverflow.net/questions/43175 | 5 | Suppose I have a family of Dirac operators over a compact base space B. From the paper of Atiyah and Singer about skew adjoint Fredholm operators we know that it has an index in $K^1(B)$.
Suppose furthermore I know "a lot" about these Dirac operators (like their spectrum, eigenspaces etc.) and $B$ is a simple space like e.g. a torus where I know everything about $K^1$ and the cohomology.
What methods are there to give an explicit description of the index in $K^1(B)$ or its image in the odd-dimensional cohomology?
Any suggestions or references welcome.
| https://mathoverflow.net/users/3816 | index of a family of Dirac operators in $K^1$ | Whether the following is useful might depend on your concrete example. Because you are mentioning $K^1 $ instead of $K^0$, I assume that your Dirac operator is ungraded (if it is graded, the index should be in $K^0$). The graded case is the ordinary Atiyah-Singer family index theorem. There is a version for the ungraded case, with index in $K^1$, which is expressed by the same formula, except that the symbol class is in $K^1 $ and its definition is slightly more complicated than in the $K^0$-case. This is explained in Atiyah-Patodi-Singer, "Spectral asymmetry and Riemann geometry III" (you do not have to read parts I and II, but only one section of part III, nevertheless, these are beautiful papers).
There is a cohomological formula that can be derived from the K-theory formula, much in the same way as in the ordinary $K^0$-case.
In other words, the family index in $K^1$ is as computable as the ordinary one. By the way: if you take a graded operator, you get - forget the grading - an ungraded operator. Its $K^1$-index is always zero.
If your base space is a circle and if you really know the spectrum of the operator at any point of $B$, there is a nice spectral-theoretic way for computing the index, by the so-called "spectral flow". The spectral flow of a map $f: S^1 \to Fred\_{sa}$ is the geometric intersection number of $f$ with the sub"manifold" of noninvertible operators in the space of selfadjoint Fredholms. Sounds complicated, but what you have to do is to count how often an eigenvalue crosses zero. A nice explanation of the spectral flow is in Booss-Woichechowski's book "Boundary value problems for Dirac-operators".
If $B$ is a $2$-torus, then $K^1 (B)$ should be $Z^2$, the factor being detected on the two circles, and this reduces the problem to the case $B=S^1$.
A last comment: if some miracle happens and the dimension of the kernel is constant, then your computation is VERY easy. The family index in $K^1 (B)$ will be zero in that case.
I wrote down a simple proof of this fact in <http://arxiv.org/PS_cache/arxiv/pdf/0902/0902.4719v3.pdf>, Theorem 4.2.1.
| 5 | https://mathoverflow.net/users/9928 | 43191 | 27,463 |
https://mathoverflow.net/questions/43024 | 8 | Hi,
Is the modular curve defined as the quotient of the upper half-plane by an arithmetic group $ \Gamma $ always a moduli space of elliptic curves with extra structure? I know this is true for $ \Gamma\_0(N), \Gamma\_1(N), \Gamma(N) $, but I'm interested in some of other groups, particularly those of the form $ \Gamma\_0(N|d) $ (notation from Conway and Norton). Are these curves moduli spaces, and if so for what classes of varieties?
Thanks!
| https://mathoverflow.net/users/4192 | Modular Curves as Moduli Spaces of Elliptic Curves | For the groups $\Gamma$ in Conway-Norton, there is always a moduli problem of $\Gamma$-structures, but since the groups always contain $\Gamma\_0(N)$ for some $N$, you won't be able to construct a universal family (because there is a $-1$ automorphism in the way). However, you will sometimes get a ``relatively representable'' problem in the sense of Katz-Mazur.
The upper half-plane quotients will be coarse spaces parametrizing objects of the following general form: You have a diagram of elliptic curves, with some isogenies of specified degrees between them, together with some data that tell you how much symmetry in the diagram you should remember. Since all of the groups normalize $\Gamma\_0(N)$ for some $N$, the diagrams will typically involve cyclic isogenies of degree $N$ in some way, and the symmetrization will involve a subgroup of the finite quotient $N\_{SL\_2(\mathbb{R})}(\Gamma\_0(N))/\Gamma\_0(N)$.
The standard example is $\Gamma\_0(p)^+$ for a prime $p$, which is generated by $\Gamma\_0(p)$ as an index two subgroup, together with the Fricke involution $\tau \mapsto \frac{-1}{p\tau}$. The $\Gamma\_0(p)$ quotient parametrizes diagrams $E \to E'$ of elliptic curves equipped with a degree $p$ isogeny between them. Taking the quotient of the moduli problem by the Fricke involution amounts to symmetrizing the diagram, so the $\Gamma\_0(p)^+$ quotient parametrizes tuples $( \{ E\_1, E\_2 \}, E\_1 \leftrightarrows E\_2)$ of unordered pairs of elliptic curves, with dual degree $p$ isogenies between them. Equivalently, you can ask for a set of diagrams $\{E\_1 \to E\_2, E\_2 \to E\_1 \}$ where the maps are dual isogenies.
A less well-known example is the 3C group, which is an index 3 subgroup of $\Gamma\_0(3|3)$, with Hauptmodul $\sqrt[3]{j(3\tau)} = q^{-1} + 248q^2 + 4124q^5 + \dots$. This group is labeled $\Gamma\_0(3|3)$ in the Conway-Norton paper, because $\Gamma\_0(3|3)$ is the eigengroup, namely the group that takes the Hauptmodul to constant multiples of itself. The 3C group contains $\Gamma\_0(9)$ as a normal subgroup, with quotient $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. You can view the upper half-plane quotient as a parameter space of quadruples of elliptic curves, with a rather complicated system of cyclic 9-isogenies and correspondences that get symmetrized (more on this in the last paragraph). A more succinct expression follows from using the matrix $\binom{30}{01}$ to conjugate $\Gamma\_0(3|3)$ to $\Gamma(1)$ and $\Gamma\_0(9)$ to $\pm \Gamma(3)$. Then you're basically looking at a moduli problem that parametrizes elliptic curves $E$ equipped with an unordered octuple of symplectic isomorphisms $E[3] \cong (\mathbb{Z}/3\mathbb{Z})^2$ that form a torsor under the characteristic 2-Sylow subgroup $Q\_8 \subset Sp\_2(\mathbb{F}\_3) \cong SL\_2(\mathbb{Z})/\Gamma(3)$.
In general, you can encode moduli problems attached to arithmetic groups using the fact that congruence groups like $\Gamma(N)$ and $\Gamma\_0(N)$ stabilize distinguished finite subcomplexes of the product of all $p$-adic Bruhat-Tits trees. Conway gives a explanation (that doesn't use the word "moduli") with pictures in his paper [Understanding groups like $\Gamma\_0(N)$](http://books.google.co.jp/books?id=-3D33eEkfXIC&lpg=PA443&ots=BloCCAeq9Q&pg=PA327#v=onepage). For example, when $N$ is a product of $k$ distinct primes, $\Gamma\_0(N)$ stabilizes a $k$-cube. Given a finite stable subcomplex, there is a standard way to make a moduli problem out of it by assigning elliptic curves to the vertices, isogenies to the edges, such that the induced transformations on the Tate module behave as you would expect from traversing the product of buildings. To symmetrize, just enumerate orbits of the transformations you want, and demand a torsor structure.
In the case of the 3C group in the above paragraph, $\Gamma\_0(9)$ pointwise stabilizes a subgraph of the 3-adic tree that is an X-shaped configuration spanned by 5 vertices. The edges coming out of the central vertex are in noncanonical bijection with points in $\mathbb{P}^1(\mathbb{F}\_3)$, and to symmetrize, you can make an unordered 4-tuple of diagrams of 5 elliptic curves, related by the action of the subgroup $V\_4 \subset PSL\_2(\mathbb{F}\_3)$ that preserves the cross-ratio.
| 11 | https://mathoverflow.net/users/121 | 43194 | 27,465 |
https://mathoverflow.net/questions/43138 | 10 | Let $A \otimes B$ be the algebraic tensor of two $C^{\ast}$ -algebras, and an element x in $A\otimes B$ is positive if $x=yy^{\ast}$. Then is it always possible to write x in the form $x=\sum a\_i\otimes b\_i$, where $a\_i$ and $b\_i$ are positive elements?
| https://mathoverflow.net/users/9858 | positive elements in tensor products | I think the answer is no. The matrix
$$
a=\begin{bmatrix}
1&0&0&1\\
0&0&0&0\\
0&0&0&0\\
1&0&0&1
\end{bmatrix}
$$
is positive in $M\_4(\mathbb{C})$. When we see this algebra as $M\_2(\mathbb{C})\otimes M\_2(\mathbb{C})$, it cannot be obtained as a sum of elementary tensors with positive entries. .
*(ok, several hours later, here is the argument)*
First, $a$ is positive because it is selfadjoint and $a=\left(\frac1{\sqrt2}a\right)^2$. Now, if we have a sum of elementary tensors in $M\_2(\mathbb{C})\otimes M\_2(\mathbb{C})$, it will look like
$$
\sum\_j\begin{bmatrix}
\alpha\_j&\overline{\gamma\_j}\\ \gamma\_j&\beta\_j\end{bmatrix}
\otimes
\begin{bmatrix}\alpha'\_j&\overline{\gamma\_j'}\\ \gamma\_j'&\beta\_j'\end{bmatrix}
=\begin{bmatrix}
\sum\_j\alpha\_j'\alpha\_j& \sum\_j \alpha\_j'\overline{\gamma\_j}& \sum\_j\overline{\gamma\_j'}\alpha\_j&\sum\_j\overline{\gamma\_j'}\gamma\_j\\
\sum\_j\alpha\_j'\gamma\_j& \sum\_j \alpha\_j'\beta\_j&\*&\*\\
\*&\*&\*&\*\\
\*&\*&\*&\*
\end{bmatrix}
$$
The assumption that each elementary tensor is made of the tensor of two positive matrices translates into $\alpha\_j\geq0$, $\beta\_j\geq0$, and $\alpha\_j\beta\_j\geq|\gamma\_j|^2$ for all $j$ (and the "prime'' version too). Now if the matrix on the right is going to be our $a$ above, then the 2,2 entry forces the following: for each $j$, the product $\alpha\_j'\beta\_j=0$. If $\alpha\_j'=0$, then $\gamma\_j'=0$; and if $\beta\_j=0$, then $\gamma\_j=0$. That is, for each $j$, $\overline{\gamma\_j'}\gamma\_j=0$, and this forces the 1,4 entry to be $0$; but it is not zero in $a$.
| 18 | https://mathoverflow.net/users/3698 | 43198 | 27,468 |
https://mathoverflow.net/questions/43211 | 5 | In Beauville's "Complex Algebraic Surfaces", given an elliptic surface $f : X \to C$ with a generic fiber $E$. Then either $\text{Alb}(X) \cong \text{Jac}(C)$ or there is an exact sequence of abelian varieties
$0 \to F \to \text{Alb}(X) \to \text{Jac}(C) \to 0$
with $F$ being isogenous to $E$.
Assume that $X$ is a properly elliptic that is in the second case. Does anyone know an example where $F$ is isomorphic to $E$ (and $X$ is not a product), and an example where $F$ is not isomorphic to $E$?
| https://mathoverflow.net/users/5197 | an exercise about elliptic surface in Beauville's book | Donu Arapura is right:
If you want $F$ isomorphic to $E$ you can do the following:
Consider the product $E \times P^1$. You find $F \simeq Alb(X) \simeq E$.
If you want $F$ and $E$ to be non isomorphic, one can do as follows.
Let $c$ be a 2-torsion point of $E$. Form the quotient $X$ of $E \times E$ by the involution defined by $(x, y) \mapsto (-x, y + c)$.
The first projection induced an elliptic fibration $X \to \mathbb P^1$. A generic fibre is $E$.
However, the albanese of $X$ is $E'$ = $E$/{0, $c$}.
The statement about the albanese follows by considering the elliptic fibration $X \to E'$ given by the second projection, and the statement of Beauville.
So let us continue this discussion with examples of Kodaira dimension one:
Take an elliptic curve $E$ and a curve $C$ with $g(C) \ge 2$.
Then $E \times C \to C$ is an example with $E$ and $F$ being isomorphic.
Now let $C$ be an hyperelliptic curve with involution $i \colon C \to C$.
Let $E$ be elliptic, and let $b \in E$ be a 2-torsion point.
We define $X$ to be the quotient of $E \times C$ by the involution given by $(x, y) \mapsto (x + b, i(y))$.
As before, the fibration $X \to \mathbb P^1$ gives an example for $E$ not isomorphic to $F$.
| 8 | https://mathoverflow.net/users/5273 | 43232 | 27,491 |
https://mathoverflow.net/questions/43104 | 2 | I need to bound the expectation of a nonnegative random variable that satisfies a Poisson-type tail bound:
$\mathbb{P}( X \geq t ) \leq \min( d \cdot (\frac{a}{t} )^{t}, \ 1)$ for $t > 0$
where $a > 0$ and $d \geq 3$. My guess for the mean:
$\mathbb{E} X \leq {\rm const} \cdot \max( a,\ \frac{\log d}{\log \log d} )$
The reference I checked (Ledoux & Talagrand, 1991) helpfully told me that this calculation is "standard". The argument apparently depends on integration by parts, but I can't figure out the trick.
| https://mathoverflow.net/users/10247 | Expectation of RVs with Poisson-type decay | It's a bit late now (so maybe I made a trivial mistake), but it seems to me that the statement is actually false. Take some large $d$ and set $a=\log d / \log \log d$. Let $t=Ka=K\log d / \log \log d$ for $K$ to fixed soon. Then the tail bound is
$$\mathbb{P}(X\ge t) \le d K^{-K\log d / \log \log d}=exp(\log d - \frac{K \log K \log d}{\log \log d})$$
so if we take $K$ not too large (say $K=\log \log \log d$) then the bound is more than 1 and hence the constant random variable $X=t$ satisfies the constraint but has expectation
$$Ka=K\frac{\log d}{\log \log d} >> \max(a, \frac{\log d}{\log \log d}) .$$
| 0 | https://mathoverflow.net/users/1061 | 43266 | 27,510 |
https://mathoverflow.net/questions/43209 | 7 | Dear all,
I have a probably rather simple question: Suppose we have a Matrix $ M\in SL\_2(\mathbb{Q}) $. Does the group $ M^{-1} SL\_2(\mathbb{Z}) M \cap SL\_2(\mathbb{Z})$ then always have finite index in $SL\_2(\mathbb{Z})$? Why? Why not?
I really was not able to solve this problem!
All the best
Karl
| https://mathoverflow.net/users/10264 | Has a conjugation of SL2(Z) finite index in SL2(Z)? (Modular group) | I, for one, am less than thrilled with snobbish kibitzing in the comments. Just answer the question already instead of dropping hints and passing judgment.
The answer is yes for $\text{SL}(n,\mathbb{Z})$. Let $d$ be the product of the denominators in the matrices $M$ and $M^{-1}$. Let $\Gamma\_d \subseteq \text{SL}(n,\mathbb{Z})$ be the subgroup of matrices of the form $I+dA$. This subgroup has finite index because it is the kernel of the congruence homomorphism
$$\text{SL}(n,\mathbb{Z}) \longrightarrow \text{SL}(n,\mathbb{Z}/d),$$
whose target is a finite group. On the other hand, $M\Gamma\_dM^{-1} \subseteq \text{SL}(n,\mathbb{Z})$ because $MIM^{-1} = I$ and $dMAM^{-1}$ is an integer matrix. Thus the intersection in question has finite index because it contains $\Gamma\_d$ as a subgroup.
The argument is quite general: You can replace $\text{SL}$ by other algebraic groups defined over $\mathbb{Z}$, and you can replace $\mathbb{Z}$ by any number field ring and $\mathbb{Q}$ by the corresponding number field.
| 22 | https://mathoverflow.net/users/1450 | 43272 | 27,513 |
https://mathoverflow.net/questions/43252 | 6 | In [A New Kind of Science: Open Problems and Projects](http://www.wolframscience.com/openproblems/NKSOpenProblems.pdf)(pg. 36).
>
> How can one extend recursive function definitions to continuous numbers? What is the continuous analog of the Ackermann function? The symbolic forms of the Ackermann function with a fixed first argument seem to have obvious interpretations for arbitrary real or complex values of the second argument. But is there a general way to extend these kinds of recursive definitions to continuous cases? Given a way to do this, how does it apply to recursive definitions like those on page 130? ... Stephen Wolfram
>
>
>
The following is an example of a flow of a map from MO [f(f(x))=exp(x)-1 and other functions “just in the middle” between linear and exponential.](https://mathoverflow.net/questions/4347/ffxexpx-1-and-other-functions-just-in-the-middle-between-linear-and-expo) .
Consider $g(x)=e^x-1$. Then
$g^n(x)= x+\frac{1}{2!}n x^2+\frac{1}{3!} \left(\frac{3 n^2}{2}-\frac{n}{2}\right)
x^3+\frac{1}{4!} \left(3 n^3-\frac{5
n^2}{2}+\frac{n}{2}\right) x^4 $
$+\frac{1}{5!}
\left(\frac{15 n^4}{2}-\frac{65 n^3}{6}+5
n^2-\frac{2 n}{3}\right) x^5 $
$ +\frac{1}{6!}
\left(\frac{45 n^5}{2}-\frac{385 n^4}{8}+\frac{445
n^3}{12}-\frac{91 n^2}{8}+\frac{11 n}{12}\right)
x^6 $
$ +\frac{1}{7!}\left(\frac{315 n^6}{4}-\frac{1827
n^5}{8}+\frac{6125 n^4}{24}-\frac{1043
n^3}{8}+\frac{637 n^2}{24}-\frac{3 n}{4}\right)
x^7 + \cdots$
Note that $g^0(x)=x, g^1(x)=e^x-1$ and that a symbolic mathematical program will also confirm that $g^m(g^n(x))=g^{m+n}(x) +O(x^8)$.
The [half-iterate](https://oeis.org/A052122) is also computed correctly,
$g^\frac{1}{2}(x)=x+\frac{x ^2}{4}+ \frac{x^3}{48} +\frac{x^5}{3840}-\frac{7 x^6}{92160} +\frac{x^7}{645120}$
See MO [What’s a natural candidate for an analytic function that interpolates the tower function?](https://mathoverflow.net/questions/20688/whats-a-natural-candidate-for-an-analytic-function-that-interpolates-the-tower-f/43003#43003) for more background.
**Questions**
1. What evidence is there for believing that maps do not have flows? Is there anything known that would prevent proofs to establish existence, uniqueness, and convergence? References would be nice but an explanation would be better.
2. Consider $f(f(x))=g(x)$ where $g: \mathbb{R} \rightarrow \mathbb{R}$. Can $f: \mathbb{R} \rightarrow \mathbb{C}$ be an appropriate solution or must $f: \mathbb{R} \rightarrow \mathbb{R}$?
3. Likewise, is there any reason beyond aesthetics for believing that maps have flows?
| https://mathoverflow.net/users/nan | Do maps have flows? | In general, there are obstructions for a map being the "time one map" of a flow (see [this question](https://mathoverflow.net/questions/31050/homomorphisms-from-r-to-diffeor-or-fractional-iterations/31056#31056)).
However, and I am not quite sure this is what you are looking for, there is a general procedure to construct flows out of maps, namely the suspention. For the map $g:M \to M$ you consider the flow in $M\times \mathbb{R}$ given by $\varphi\_t((x,s))= (x, s+t)$ and you quotient by $(x,s) \sim (g(x),s+1)$. This gives a flow, whose time one map preserves a set homeomophic to $M$ where the dynamics is $g$.
In the case of a map from $[0,\infty)$ to $[0,\infty)$ fixing $0$ you can work out flow to be
defined in $\mathbb{C}$ and such that the time one map is $g$, so this would give a partial answer to $2.$.
| 8 | https://mathoverflow.net/users/5753 | 43276 | 27,517 |
https://mathoverflow.net/questions/43269 | 2 | Can anyone give me the reference for this statement?:
Let $M$ be a closed oriented smooth 4-manifold. Any element $a\in H\_2(M)$ can be represented by a smoothly embedded, oriented surface.
I found this statement and the proof at Saveliev's book, Lectures on the topology of 3-manifold, but I think it is not a complete proof and I couldn't fill the gap. Let me know other reference.
| https://mathoverflow.net/users/6569 | Reference for the proof of this statement? | This result actually holds in all dimensions.
>
> Let $M^n$ be a closed smooth manifold $M^n$ of any dimension $n\geqslant 3$. Every element $\alpha \in H\_2(M,\mathbb Z)$ is represented by a smoothly embedded closed surface.
>
>
>
You can prove it as follows:
1. Take a cycle $a\_1\sigma\_1 + \ldots + a\_n\sigma\_n$ representing $\alpha$. The coefficients $a\_i$ are integers: by writing $a\sigma$ as $\pm(\sigma +\ldots +\sigma)$ you can suppose they are all $\pm 1$.
2. Since it is a cycle, restrictions on edges must cancel in pairs. You can glue correspondingly the triangles along these edges and get a map $f:S\to M^n$ from some (possibly disconnected) 2-dimensional complex $S$.
3. This complex $S$ is obtained from finitely many oriented triangles by gluing the edges in pairs (with orientation-reversing maps): it is necessarily a closed oriented surface (note: this is not true for higher-dimensional cycles where you only get onlu a kind of "pseudo-manifold" which might have singular codimension-2 stratum).
4. Up to homotopy you can take $f$ smooth. You can then put the map $f$ in general position. If the dimension $n$ of $M^n$ is $n\geqslant 5$ then $f$ is necessarily injective and you are done.
5. IF $n=4$ you may have isolated double points. These can be removed via some surgery which replaces locally the two intersecting transverse 2-discs with an annulus. The surgery *modifies* $S$ (genus increases by one) but not the class $\alpha$.
6. If $n=3$ you may have double and triple points. These can also be removed via some similar surgery.
For a reference, I suggest you the nice and readable book "*The wild world of 4-manifolds*" from A. Scorpan which treats the 4-dimensional case and also the general $n$-dimensional one (with further discussion and references inside concerning the general problem of realizing an integral class by a manifold).
| 7 | https://mathoverflow.net/users/6205 | 43278 | 27,519 |
https://mathoverflow.net/questions/43281 | 14 | I have some vague sense that certain types of categories are related to certain types of logic. I've been meaning to learn more about this, so I thought I'd ask about the simplest case, propositional logic. In particular, I'd be interested in a statement of the completeness theorem for propositional logic using these ideas (something like a representation theorem for a certain class of categories).
Here's what I think is a start: a propositional calculus should, at least, behave like a bicartesian closed category $C$, possibly satisfying some extra conditions. The product should be $\vee$, the coproduct should be $\wedge$, and the exponential object should be $\rightarrow$. I think the terminal object is $\top$ and the initial object is $\bot$, and the exponential $p \rightarrow \bot$ should then be $\neg p$. Maybe this means the same thing as "Heyting algebra."
We're specifically interested in the free bicartesian closed category $C\_P$ on a set $P$ of **primitive propositions**, and in functors from $C\_P$ to the truth-value category $\mathbf{2} = \{ \bot \rightarrow \top \}$. If I've set up the definitions correctly, then a subset $S$ of $C\_P$ **semantically entails** an object $t$ if and only if, for every functor $F : C\_P \to \mathbf{2}$ such that $F(s) = \top \forall s \in S$, it is also true that $F(t) = \top$.
If I've set up the definitions correctly, functors like $F$ are defined by what they send to $\top$. If we want to characterize such sets syntactically, we write down some set of **axioms** and some set of **inference rules** telling us when objects sent to $\top$ let us construct other objects sent to $\top$, and then the completeness theorem tells us that these axioms and inference rules characterize the functors $F$.
But I'm clearly missing something important: at some point we introduce the law of the excluded middle, and then our Heyting algebras should collapse to Boolean algebras. But I'm not sure exactly when that point is. Can someone help me out? I suspect this is an exercise in a book on topos theory somewhere.
| https://mathoverflow.net/users/290 | Propositional logic with categories | Qiaochu, let me see if this answers your question:
**Proposition:** Suppose $B$ is a cartesian closed category with finite coproducts such that the canonical double dual embedding
$$b \to (b \Rightarrow 0) \Rightarrow 0$$
is an isomorphism (excluded middle). Then $B$ is equivalent (as a category) to a poset, and hence to a Boolean algebra.
**Proof:** First, in a cartesian closed category with $0$, for any object $b$ there exists a morphism $i: b \to 0$ only if $i$ is an isomorphism. For given $i$, the composite
$$b \stackrel{\langle i, 1\_b \rangle}{\to} 0 \times b \stackrel{\pi\_2}{\to} b$$
is an isomorphism. But $\pi\_1: 0 \times b \to 0$ is an isomorphism since $- \times b$ preserves colimits by cartesian closure. Thus the composite of $i: b \to 0$ followed by the unique map $0 \to b$ is $1\_b$. Since $0$ is initial, the composite of $0 \to b$ followed by $i: b \to 0$ is $1\_0$. Hence $i$ is an isomorphism. In particular, there can be at most one morphism $b \to 0$ (again by initiality).
Next, for any two objects $b, c$ there can be at most one morphism $b \to c \Rightarrow 0$, since there are in bijective correspondence with morphisms $b \times c \to 0$, of which there is at most one. Finally, taking $c = b' \Rightarrow 0$ and assuming excluded middle, we have that for any two objects $b, b'$ there is at most one morphism $b \to b'$. $\Box$
The property of existence of $b \to 0$ forcing $b$ to be initial is called *strictness* of the initial object, and I think Mac Lane-Moerdijk prove that an initial object is strict in a cartesian closed category. I know this is in my notes on ETCS.
**Edit:** Qiaochu's instincts are correct. There is an old tradition in category theory, going back essentially to Lawvere's thesis, whereby one forms syntactic categories associated to theories and to various fragments of logic, and these are typically free categorical structures of various sorts. For example, the theory of groups is the free (initial) category with products containing a group object; the morphisms are well-formed terms in the equational language modulo the equational axioms imposed by the theory. A *model* of the theory will be a functor which preserves the categorical structure needed to express the logic (finite products in the case of finitary equational theories). A *completeness theorem* for a certain class of models says that whenever the semantic interpretations of two syntactic terms (or morphisms in the free categorical structure) are equal in each of the models of the class, then the syntactic terms are provably equivalent.
From the viewpoint of categorical logic, such completeness theorems can be seen as embedding or representation theorems. In spirit these go back to the Freyd-Mitchell embedding theorems; for example, each module category is a model of the abelian category axioms, and if a diagram commutes when interpreted in each such model, then it provably commutes on the basis of the abelian category axioms. Or, in categorical logic there is a notion of 'coherent theory' whose syntactic category is effectively its classifying topos, and there is a theorem due to Deligne that such a theory has 'enough points' (which are left exact left adjoints from the classifying topos to the category of sets). This can be interpreted either as saying that the classifying topos of the theory can be faithfully embedded (lex left-adjointly) into a product of copies of $Set$, or as saying that two morphisms are provably equivalent in the theory if their values under any set-theoretic model are equal.
Qiaochu's question has to do with bicartesian closed categories, seen as a natural categorification of the concept of Heyting algebra (the algebraic formalization of intuitionistic propositional logic, as Boolean algebras are the algebraic formalization of classical propositional logic). Here one can form appropriate syntactic categories as free bicartesian closed categories generated by a given set of sorts, or even by a given category of sorts. Now, there is still another tradition in categorical logic, due to Lambek and his school, whereby such free structures are constructed explicitly by taking suitable equivalence classes of formal Gentzen sequent deductions pertaining to the given fragment of logic, which here is intuitionistic propositional logic. The equivalence classes are dictated by naturality considerations and other compatibility conditions, reflecting the key conversions one sees in cut-elimination; see Girard's Proofs and Types. One can consult Manfred Szabo's Algebra of Proofs for a careful description of how this is done for quite a variety of logics, including, I believe, the case Qiaochu is interested in.
Off hand I am not sure what sort of completeness theorems there are for bicartesian closed categories (e.g., is there a faithful structure-preserving functor from a free bicartesian closed category to some power of $Set$? I have a sneaking suspicion finite sets are not enough). But perhaps the answer can be found in Freyd-Scedrov's *Categories, Allegories*, which contains a wealth or representation/embedding/completeness theorems [one of the great scientific themes of Freyd's career].
| 11 | https://mathoverflow.net/users/2926 | 43285 | 27,523 |
https://mathoverflow.net/questions/41292 | 0 | If you have non-Arch. local field F and E its finite extension, I am just wondering if anybody has any idea about the action of $\operatorname{Gal}(E/F)= \operatorname{Aut}\_F(E)$ on the lines in $k^2\_E$, Where $k\_E$ is a residue field of the extension?
Note:
* $\{ \text{Adjacent points} \} \simeq \mathbb{P}^1\_k$
* $\mathbb{P}\_{1}(k):=$ set of all dim 1 subspaces of $k\mathcal{P}^{1}$ a 2-dim k-vector space.
* $\mathbb{E}\mathcal{P^{1}}=$ lines in $\mathcal{k}\_{E}^{2}$.
| https://mathoverflow.net/users/9842 | Galois Action on the lines in the k^{2}_{E} | The action is fairly straightforward. Any line in $k\_E^2$ can be described by an equation of the form $ax+by=c$, where $a,b,c \in k\_E$, and $a$ and $b$ are not both zero. You can describe the action of $\operatorname{Gal}(E/F)$ on $k\_E$ by taking reduction modulo the maximal ideal. Alternatively, you can look at the action on the (prime-to-residue-characteristic) roots of unity in $E$ living over units in $k\_E$. You can then use the action on $k\_E$ to get the diagonal action on the coefficients $a,b,c$ in the equation $ax+by=c$.
If you want to restrict your view to lines through the origin, you get an action on pairs $(a,b)$, and the equivalence relation that sends proportional pairs to the same point in $\mathbb{P}^1$ is Galois-stable.
| 2 | https://mathoverflow.net/users/121 | 43290 | 27,526 |
https://mathoverflow.net/questions/43221 | 12 | Dennis Sullivan, "Infinitesimal computations in topology", Publ. IHES: At the end of section 8, he writes, among other things, roughly the following.
Let $\mathfrak{g}$ be a (finite-dimensional, real) Lie algebra and let $\Lambda \mathfrak{g}^{\ast}$ be the Chevalley-Eilenberg complex (i.e. the
exterior algebra,
with the differential dual to the Lie bracket). He considers the spatial realization $\langle \Lambda \mathfrak{g}^{\ast}\rangle $ with
respect to the simplicial d.g.a. of $C^{\infty}$-forms on the standard simplices. This is a simplicial set, the $p$-simplices are
the d.g.a.-homomorphisms $\Lambda \mathfrak{g}^{\ast} \to \mathcal{A}(\Delta^p)$ to the de Rham forms on the simplex.
Let $G$ be the simply-connected Lie group with Lie algebra $\mathfrak{g}$.
"Theorem" (8.1)' (the quotation marks are due to Sullivan)
says that the fundamental group of $\langle \Lambda \mathfrak{g}^{\ast} \rangle $ is isomorphic to $G$.
Now the set of $p$-simplices of $\langle \Lambda \mathfrak{g}^{\ast} \rangle $ has a topology, induced from the $C^{\infty}$-topology) on the space of $\mathfrak{g}$-valued forms on the simplices and so
$\pi\_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle )$ is a topological group. Sullivan also says that the above isomorphism is a homeomorphism.
It is not difficult to verify these assertions. Since it is rather clear how to describe the exponential map in this construction, we can also
recover the differentiable structure on $G$ from this method.
The upshot of this discussion is: once the existence of the simply-connected Lie group is known (this is Lie's third theorem, proven only decades after Lie by Cartan), it has the given abstract description.
It is well-known that Lie's third theorem is a pretty hard result (the standard proof goes via Ado's theorem).
It seems possible to reverse the logic of that argument and give a proof of Lie's 3rd theorem.
Given $\mathfrak{g}$, \emph{define} $G:=\pi\_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle)$
as a topological group. The exponential map $\mathfrak{g} \to G$ is given by the following formula: any $x \in \mathfrak{g}$ defines a constant
$1$-form on $\Delta^1$, hence a 1-simplex of $\pi\_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle)$. That was the easy part; here are the nontrivial
parts:
1. Show that $G$ is Hausdorff (probably difficult)
2. Put a smooth stucture on it, such that the exponential map is a local chart (probably the hardest part)
3. Once this is done, the simple connectivity of $G$ and the fact that $Lie (G)=\mathfrak{g}$ are probably both obvious.
After all these preliminaries, I can ask my question: has this approach been written down properly? I am aware that a generalization
of this argument has been used
by Getzler <http://arxiv.org/abs/math/0404003> and Henriques <http://arxiv.org/abs/math/0603563>, but in these papers, I do not find the details. It is of course also possible (maybe desirable) to banish all the fancy language from the discussion, leaving a definition of $G$ as the quotient of the space of $\mathfrak{g}$-valued 1-forms on the interval.
| https://mathoverflow.net/users/9928 | Lie's third theorem via differential graded algebras? | The details are here:
Marius Crainic, Rui Fernandes, *Integrability of Lie brackets*
<http://arxiv.org/abs/math/0105033>
(To connect this to your question, notice that a morphism $T X \to \mathfrak{g}$ of Lie algebroids, which is the language they use, is dually the same as a morphism $\Omega^\bullet(X) \leftarrow CE(\mathfrak{g})$ of dg-algebras. For more see [here](http://ncatlab.org/nlab/show/Lie+integration))
| 5 | https://mathoverflow.net/users/381 | 43291 | 27,527 |
https://mathoverflow.net/questions/43164 | 10 | Given an integer $n\ge 1$, what is the largest eigenvalue $\lambda\_n$ of the matrix $M\_n=(m\_{ij})\_{1\le i,j\le n}$ with the elements $m\_{ij}$ equal to $0$ or $1$ according to whether $ij>n$ or $ij\le n$?
It is not difficult to show that
$$ c\sqrt n \le \lambda\_n \le C\sqrt{n\log n} $$
for appropriate positive absolute constants $c$ and $C$, and numerical
computations seem to suggest that the truth may lie somewhere in between.
---
To make the problem a little bit "more visual", the first four matrices in question are as follows:
>
> $M\_1=\begin{pmatrix} 1 \end{pmatrix}$ $\quad$
> $M\_2=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ $\quad$
> $M\_3=\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}$ $\quad$
> $M\_4=\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}$
>
>
>
| https://mathoverflow.net/users/9924 | The largest eigenvalue of a "hyperbolic" matrix | I think, $C\sqrt{n}$ is an upper bound aswell. Take vector $x=(x\_1,\dots,x\_n)$ with $x\_j=j^{-1/2}$. Then $(Ax)\_i$ behaves like $C\sqrt{n/i}=C\sqrt{n}x\_i$. But we know that if $(Ax)\_i\leq C x\_i$ for vector $x$ with positive coordinates, then the largest eigenvalue of $A$ does not exceed $C$ (kind of Perron-Frobenius).
(This answer is very ugly displayed, I do not know why).
| 6 | https://mathoverflow.net/users/4312 | 43293 | 27,528 |
https://mathoverflow.net/questions/43019 | 3 | For a harmonic function $\Phi$ on a simply connected subset $\Gamma$ of $\mathbb{R}^3$, define a **guide curve** $\gamma: I \mapsto \Gamma$ of $\Phi$ as a simple regular $C^1$ curve such that
* all point in $\gamma(I)$ are critical points of $\Phi$, and
* for all points $p$ in $\gamma(I)$ there exists a neighborhood $V$ of $p$ so that all critical points of $\Phi$ within $V$ are also in $\gamma(I)$.
My question is whether there are any such guide curves which do not have an analytic parametrization?
For a concrete example, consider $\Phi(x,y,z)=x\ y\ z$ for which any part of a coordinate axis not including the origin is a guide curve.
| https://mathoverflow.net/users/6302 | Geometrical structure of critical points of harmonic functions | [This is an answer made of two comments and an example]
Since the laplacian is elliptic with real-analytic coefficients, a harmonic function $f$ is real-analytic in its domain of definition. Hence the set $C$ of critical points of $f$ is a real-analytic subset of $R^3$, and as such it admits a locally finite partition into real-analytic locally closed smooth submanifolds. Thus if $\dim C≤1$, it is locally a finite union of analytic open arcs and singular points (but the curves might not extend smoothly across those points).
A reference on real analytic functions (reedited in 2002) might be
S. Krantz, H. Parks, A primer of real analytic functions. Birkhäuser Verlag, 1992.
But maybe the "curve selection lemma" in Milnor's "Singular points on complex hypersurfaces" would be enough . edit: it concerns real *algebraic* subsets.
As an example of a curve of critical points not extending throuh a singular point, take the harmonic polynomial
$f(x,y,z)=y^3-3x^2y+y^3z-yz^3$, which has critical locus $y=0,z^3=-3x^2$. But of course you have the singular parametrization $x=3t^3, z=-3t^2$. I don't know if they exist in general.
Addendum : in fact the critical locus of a harmonic can polynomial can have an arbitrary (real) plane algebraic curve as a union of irreducible components. Let $P(x,y)$ be a real two variable polynomial,
and define $$f(x,y,z)=\Sigma\_k \frac{z^{2k+1}}{(2k+1)!}(-\Delta\_{x,y})^k P(x,y) \; .$$
It is easy to check that $f$ is harmonic, and $df$ vanishes on $z=0$, $P(x,y)=0$.
| 4 | https://mathoverflow.net/users/6451 | 43295 | 27,530 |
https://mathoverflow.net/questions/43249 | 35 | In the past couple years, I've read many words pertaining to "D-branes" without feeling I have fully comprehended them. In broad terms, I think I get what they're about: They're supposed to serve as habitats for the ends of open strings and can be conceived of as submanifolds (of the target manifold in a sigma model), possibly augmented with a vector bundle, or a sheaf of [somethings], or maybe some other kind of label/data. (Corrections welcome.)
In the hopes of tightening my grasp on the concept, here are some of the questions that have been nagging me during my reading.
1. What *specifically* is the definition of a D-brane, say in the context of a topological field theory? (Or what are the most promising provisional definitions?) What references are most accessible to a mathematical audience?
2. What picture should I have in my head when an author talks about "the moduli space of D-branes"?
3. What is the idea behind the "dynamics of D-branes" that researchers sometimes talk about? (Perhaps when I understand better how to think about these gadgets, it will be easier to conceive of how they should change over time.)
4. What goes into verifying (or at least asserting/conjecturing) that the elements of twisted K-theory classify "D-brane charges"?
(Question reposted from [here](https://math.stackexchange.com/questions/7435/what-are-d-branes-in-a-topological-field-theory).)
| https://mathoverflow.net/users/6005 | What are D-branes, really? | There is an *abstract algebraic* formulation of QFT: this says that an $n$-dimensional QFT is a consistent assignment of spaces of states and of maps between them to $n$-dimensional *cobordisms*.
If one allows cobordisms with boundary here, one speaks of *open-closed QFT*. A **D-brane** in this context is the type of data assigned by the QFT to these boundaries.
There is also a geometric aspect to this: many abstractly defined QFTs are imagined to be "sigma-models". They are supposed to be induced by a process called "quantization" from a functional (the "action") on a space of maps $\Sigma \to X$ from a cobordism $\Sigma$ into a smooth manifold $X$ equipped with extra geometric data (such as metric, connections, etc.)
Under this correspondence, one may ask which abstract algebraic properties of the QFT derive from which geometric aspects of these "background structures". One finds that the data that the QFT assigns to boundaries comes from geometric data on $X$ that *tends* to look like submanifolds with their own geometric data on them (but may be considerably more general than that!). If so, this geometric data on $X$ is called a **D-brane** of the sigma-model.
There are many instances of this that are understood at the rough level at which quantum field theory was understood in the 20th century. One special case that is by now under fairly complete mathematical control and which serves as a good guide to the general concept of D-branes is what is called "2d *rational* CFT" .
There is a complete mathematical classification of 2d rational CFTs in their abstract algebraic form: they are given by special symmetric Frobenius algebra objects *internal* to a modular tensor category of representation of a vertex operator algebra.
Under this classification theorem, the boundary data = D-branes in the algebraic formulation can be proven to be precisely *modules* over this Frobenius algebra object.
In special nice cases, one understands where these come from geometrically. The notable example is the Wess-Zumino-Witten model, where the target space is a group manifold. Here one finds that in the simplest case the geometric data corresponding to these D-branes are submanifolds given by conjugacy classes, and carrying twisted vector bundles. More generally, though, the D-branes are given by cocycles in the twisted differential K-theory of the group. So the identification "D-brane = submanifold" is too naive, in general. The correct identification is:
geometric D-brane = geometric data on the sigma-model target space that induces boundary data of the corresponding algebraically defined worldvolume QFT.
For more see
<http://ncatlab.org/nlab/show/brane>
| 21 | https://mathoverflow.net/users/381 | 43299 | 27,533 |
https://mathoverflow.net/questions/43007 | 2 | In maximum likelihood estimation, one typically needs to compute the log (natural log) of probability values. When a probability, say $p(x)$, becomes so close to zero, $log(p(x))$ returns -Inf. What is the usual trick to avoid these cases?
| https://mathoverflow.net/users/5223 | Numeric problem when evaluating log of a pdf | Work with the logs of probability directly, rather than trying to compute the probability and then compute the log. You can do arithmetic with the logs, as well; multiplication becomes addition, of course. Addition is somewhat more complicated, but it's not too hard to work out how to do it without taking the exponential back.
In response to JM's question, it's easy to come up with practical problems where the probability of something happening is indeed so low that it underflows a float or double.
| 5 | https://mathoverflow.net/users/5010 | 43301 | 27,535 |
https://mathoverflow.net/questions/43311 | 6 | I am struggling hard to understand the pushforwards and pullbacks of cosheaves. Are they also cosheaves? And what are quasicoherent cosheaves? Is there anything like coquasicoherent cosheaves? Please tell me a good refernce on theses topics, if there is some.
| https://mathoverflow.net/users/9492 | sheaves and cosheaves | **edit:** I was assuming you wanted an equalizer sheaf property, but this is not the definition of cosheaf, see comments - the following has nothing to do with cosheaves then!
If by "cosheaf" you mean a covariant functor from the opens of a space to sets/groups/etc., you could look at Moerdijk/MacLane's ["Sheaves in Geometry and Logic"](http://books.google.com/books?id=SGwwDerbEowC&printsec=frontcover&dq=%2522Sheaves+in+Geometry+and+Logic%2522&source=bl&ots=WxeQEqB_CR&sig=OTsIE4Vr1EdkGyDrLbHGDHYo5Do&hl=en&ei=VAnDTOGKJsaYOoKT1YAM&sa=X&oi=book_result&ct=result&resnum=2&ved=0CBoQ6AEwAQ#v=onepage&q&f=false) - there you can learn some general sheaf theory on sites, which includes the cosheaf case. In particular pushforward and pullback are transport along the the two functors comprising a "geometric morphism". The notion of quasicoherent (co)sheaf can maybe also be defined in this generality by saying that something should look locally like a pullback from the "base topos".
Sorry for the jargon, I didn't get from your question what exactly you are up to - just take a look at the book and see if it suits you.
| 2 | https://mathoverflow.net/users/733 | 43312 | 27,541 |
https://mathoverflow.net/questions/43298 | 15 | In the Wikipedia article on Hilbert's Nullstensatz,
<http://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz>
the following application of the Weak Nullstensatz is mentioned:
>
> Commuting matrices
>
>
> The fact that commuting matrices have a common eigenvector – and hence by induction stabilize a common flag and are simultaneously triangularizable – can be interpreted as a result of the weak Nullstellensatz, as follows: commuting matrices form a commutative algebra $K[A\_1,\ldots,A\_k]$ over $K[x\_1,\ldots,x\_k];$ the matrices satisfy various polynomials such as their minimal polynomials, which form a proper ideal (because they are not all zero, in which case the result is trivial); one might call this the '''characteristic ideal''', by analogy with the characteristic polynomial.
>
>
> One then defines an eigenvector for a commutative algebra as a vector $v$ such that for all $x \in A$ one has $x(v) = \lambda(x)\cdot v$ for a linear functional $\lambda\colon A \to K.$ This simply linearizes the definition of an eigenvalue, and is the correct definition for a common eigenvector, as if $v$ is a common eigenvector, meaning $A\_i(v)=\lambda\_i v,$ then the functional is defined as $\textstyle{\lambda(c\_0 I + c\_1 A\_1 + \cdots c\_k A\_k) := c\_0 + \sum c\_i \lambda\_i}$ (treating scalars as multiples of the identity matrix $A\_0 := I$, which has eigenvalue 1 for all vectors), and conversely an eigenvector for such a functional $\lambda$ is a common eigenvector. Geometrically, the eigenvalue corresponds to the point in affine $k$-space with coordinates $(\lambda\_1,\ldots,\lambda\_k)$ with respect to the basis given by $A\_i.$
>
>
> Then the existence of an eigenvalue $\lambda$ is equivalent to the ideal generated by (the relations satisfied by) $A\_i$ being non-empty, which exactly generalizes the usual proof of existence of an eigenvalue existing for a single matrix over an algebraically closed field by showing that the characteristic polynomial has a zero.
>
>
>
I am somewhat struggling to make sense of that. The Weak Nullstellensatz says that I find a functional $\lambda \colon K[A\_1, \dots, A\_k] \to K$, and $\lambda(A\_i)$ is an eigenvalue of $A\_i$. But how do I conclude that a common eigenvector exists?
| https://mathoverflow.net/users/5273 | Commuting Matrices and the Weak Nullstellensatz | My "solution" is a bit strange, but I hope it is correct.
We regard $V$ as a module over $A$. $X = {\rm Spec}\ A$ is zero-dimensional. Then (as a sheaf on $X$) $V$ decomposes into a direct sum of sheaves over the finitely many points of $X$. This corresponds to a decomposition $V$ into subspaces corresponding to different eigenvalues in the sense you described. We must have a point $p$ such that the piece $V\_p$ above it (which is the localization of $V$ at the maximal ideal $\mathfrak{m}$ of $A$ corresponding to $p$) is nonzero. Localizing at this ideal and restricting ourselves to the subspace $V\_p$, we may assume that $A$ is local. If there would be no common eigenvector in $V\_p$, then we would have $\mathfrak{m}V\_p = V\_p$, so by Nakayama's lemma $V\_p = 0$, a contradiction.
darij: Does it deserve to be "awfully sophisticated" by your standards?
| 6 | https://mathoverflow.net/users/3847 | 43315 | 27,542 |
https://mathoverflow.net/questions/43323 | 4 | I am talking about the principle that says that every set is the image of a [projective](http://planetmath.org/?op=getobj&from=objects&id=6437) set. For every set $x$ there is a surjection $f:y \twoheadrightarrow x$, such that for any set $u$ and function $j:y \to u$ and surjection $g:v \twoheadrightarrow u$ there is a function $h:y \to v$ such that $j=gh$.
There is a simple model of CZF in which this holds and the axiom of choice is violated. This also violates classical logic so is not a model of ZF (= CZF + classical logic). I am curious about the following comment of Andreas Blass on ZF:
>
> ...all the usual independence results involving weak axioms of choice remain true in
> the presence of the negation of EPSets. In particular, none of the axioms in the
> appendix of [Jech, the Axiom of Choice], except AC itself, implies EPSets.
> The following result ... is essentially all that is known about the strength of EPSets...
>
>
> Theorem 6.2 EPSets implies the axiom of dependent choice ...
>
>
> In particular, we do not know whether it implies AC ... we conjecture that it does not.
>
>
>
(Injectivity, Projectivity, and the Axiom of Choice, 1979)
[Consequence of the Axiom of Choice](http://consequences.emich.edu/CONSEQ.HTM) (1998) listed the question of whether it implies AC as still open, and in fact unless I missed something it listed no new non-trivial results.
So, in 2010, is there really still nothing to be said other than ZF+DC < ZF+EPSets $\le$ ZFC ?
| https://mathoverflow.net/users/6787 | Existence of enough projectives in the category of sets | I haven't heard any new information about "enough projective sets" in the classical ZF context. Your formulation, however, looks weaker. "Enough projective sets" should say that for every $x$ there is a surjection $f:y \to x$ such that every surjection from any $w$ to $y$ splits. That implies your formulation (by taking $w$ to be the pullback of $y$ and $z$ over $x$) but I don't see the converse.
The principle "enough projective sets" has been used in constructive mathematics by Peter Aczel (and probably others) under the name "presentation axiom".
| 2 | https://mathoverflow.net/users/6794 | 43330 | 27,551 |
https://mathoverflow.net/questions/43313 | 45 | Every now and then I attempt to understand better quantum mechanics and quantum field theory, but for a variety of possible reasons, I find it very difficult to read any kind of physicist account, even when the physicist is trying to be mathematically respectable. (I am not trying to be disrespectful or controversial here; take this as a confession of stupidity if it helps.) I am generally interested in finding online mathematical accounts which ideally would come close to being of "Bourbaki standard": definition-theorem-proof and written for mathematicians who prefer conceptual explanations, and ideally with tidy or economical notation (e.g., eschewing thickets of subscripts and superscripts).
More specifically, right now I would like a (mathematically trustworthy) online account of rigged Hilbert spaces, if one exists.
Maybe I am wrong, but the [Wikipedia account](https://en.wikipedia.org/wiki/Rigged_Hilbert_space) looks a little bit suspect to me: they describe a rigged Hilbert space as consisting of a pair of inclusions $i: S \to H$, $j: H \to S^\ast$ of topological vector space inclusions, where $S^\ast$ is the strong dual of $S$, $H$ is a (separable) Hilbert space, $i$ is dense, and $j$ is the conjugate linear isomorphism $H \simeq H^\ast$ followed by the adjoint $i^\ast: H^\ast \to S^\ast$. This seems a little vague to me; should $S$ be more specifically a nuclear space or something? My guess is that a typical application would be where $S$ is Schwartz space on $\mathbb{R}^4$, with its standard dense inclusion in $L^2(\mathbb{R}^4)$, so $S^\ast$ consists of tempered distributions.
I also hear talk of a nuclear spectral theorem (due to Gelfand and Vilenkin) used to help justify the rigged Hilbert space technology, but I don't see precise details easily available online.
| https://mathoverflow.net/users/2926 | Good references for Rigged Hilbert spaces? | Some time ago I was interested in rigged Hilbert space to get a better understanding of quantum physics. On that occasion I collected some references on this subject, see below. It's quite comprehensive. A good starting point for an overview could be the works of Madrid and Gadella. Note that there are different versions of "rigged Hilbert space" (in context of quantum physics) in literature.
J.-P. Antoine.
[Dirac formalism and symmetry problems in quantum mechanics. I.
General Dirac formalism.](https://doi.org/10.1063/1.1664761)
*Journal of Mathematical Physics*, 10(1):53–69, 1969. [Zbl 0172.56602](https://zbmath.org/?q=an:0172.56602)
N.Bogolubov, A.Logunov, and I.Todorov.
*Introduction to Axiomatic Quantum Field Theory*, chapters 1: Some
Basic Concepts of Functional Analysis, 4: The Space of States, pages 13–44,
112–128. Benjamin, Reading, Massachusetts, 1975. [Zbl 1114.81300](https://zbmath.org/?q=an:1114.81300)
R.de la Madrid.
*Quantum Mechanics in Rigged Hilbert Space Language*.
PhD thesis, Depertamento de Fisica Teorica Facultad de Ciencias.
Universidad de Valladolid, 2001.
Available at [`http://galaxy.cs.lamar.edu/~rafaelm/dissertation.html`](http://galaxy.cs.lamar.edu/%7Erafaelm/dissertation.html).
Also see: [The role of the rigged Hilbert space in quantum mechanics.](https://doi.org/10.1088/0143-0807/26/2/008) *European Journal of Physics*, 26(2):277–312, 2005. [arXiv:quant-ph/0502053](https://arxiv.org/abs/quant-ph/0502053). [Zbl 1079.81022](https://zbmath.org/?q=an:1079.81022)
M.Gadella and F.Gómez.
[A unified mathematical formalism for the Dirac formulation of quantum mechanics.](https://doi.org/10.1023/A:1016069311589)
*Foundations of Physics*, 32:815–869, 2002.
M.Gadella and F.Gómez.
[On the mathematical basis of the Dirac formulation of quantum mechanics.](https://doi.org/10.1023/B:IJTP.0000005956.11617.e9)
*International Journal of Theoretical Physics*, 42:2225–2254,
2003. [Zbl 1038.81020](https://zbmath.org/?q=an:1038.81020)
M.Gadella and F.Gómez.
[Dirac formulation of quantum mechanics: Recent and new results.](https://doi.org/10.1016/S0034-4877(07)80008-5)
*Reports on Mathematical Physics*, 59:127–143, 2007.
I.M. Gelfand and N.J. Vilenkin.
*Generalized Functions, vol. 4: Some Applications of Harmonic
Analysis*, volume 4, chapters 2–4, pages 26–133.
Academic Press, New York, 1964. [Zbl 0136.11201](https://zbmath.org/?q=an:0136.11201)
A.R. Marlow.
[Unified Dirac–von Neumann formulation of quantum mechanics. I. Mathematical theory.](https://doi.org/10.1063/1.1704352)
*Journal of Mathematical Physics*, 6:919–927, 1965.
E.Prugovecki.
[The bra and ket formalism in extended Hilbert space.](https://doi.org/10.1063/1.1666195)
*J. Math. Phys.*, 14:1410–1422, 1973. [Zbl 0277.47015](https://zbmath.org/?q=an:0277.47015)
J.E. Roberts.
[The Dirac bra and ket formalism.](https://doi.org/10.1063/1.1705001)
*Journal of Mathematical Physics*, 7(6):1097–1104, 1966.
J.E. Roberts.
[Rigged Hilbert spaces in quantum mechanics.](https://doi.org/10.1007/BF01645448)
*Commun. Math. Phys.*, 3:98–119, 1966. [Zbl 0144.23404](https://zbmath.org/?q=an:0144.23404)
D.Tjøstheim.
[A note on the unified Dirac–von Neumann formulation of quantum mechanics.](https://doi.org/10.1063/1.522629)
*Journal of Mathematical Physics*, 16(4):766–767, 1975.
**Edit.** I remember that there is also a discussion about Gelfand triples in physics in the [Funktionalanalysis](https://zbmath.org/?q=ai%3Agrossmann.siegfried%20la%3Agerman%20ti%3Afunktionalanalysis) books by Siegfried Großmann but I don't have a copy handy at the moment. Though it is in German it might be interesting for you, too.
| 31 | https://mathoverflow.net/users/7538 | 43332 | 27,552 |
https://mathoverflow.net/questions/43346 | 6 | Can any one help me in proving the following equality:
$$n^n= \sum\_{i=1}^n {n \choose i}\cdot i^{i-1}\cdot (n-i)^{n-i}$$
I tried some different ideas but none of them worked!
| https://mathoverflow.net/users/10286 | Combinatorial equation | Your equation can be written as an equation for exponential generating functions: $f(x) = g(x)(f(x)+1)$, where $$f(x) = \sum\_{n\ge1}n^nx^n/n!$$ and $$g(x) = \sum\_{n\ge1}n^{n-1}x^n/n!$$
We can see that for those $f(x)$ and $g(x)$, we have $f(x) = xg'(x)$. If we then solve the differential equation $$xg'(x) = \frac{g(x)}{1-g(x)}$$ with $g(0)=0$, we get that the solution satisfies $x=g(x)e^{-g(x)}$.
By the Lagrange inversion formula, the computational inverse of $xe^{-x}$ is exactly our $g(x)$.
I'm sure some permutation of the reasoning steps above gives a proof for your equation.
| 6 | https://mathoverflow.net/users/400 | 43349 | 27,559 |
https://mathoverflow.net/questions/43338 | 17 | Let X be a compact symplectic manifold with a form $\omega$. And $X \times X$ is equipped with the symplectic form $(\omega,-\omega)$. The diagonal $\Delta:X \mapsto X \times X$ is a Lagrangian submanifold. So, in this question, [Hochschild (co)homology of Fukaya categories and (quantum) (co)homology](https://mathoverflow.net/questions/11081/hochschild-cohomology-of-fukaya-categories-and-quantum-cohomology). Tim Perutz says "PSS is a canonical ring isomorphism from QH∗(X) to the Hamiltonian Floer cohomology of X, and the latter can be compared straightforwardly to the Lagrangian Floer cohomology of the diagonal." I have no doubt that this second assertion is straightforward, since I have consulted a couple of references and no one spells this out. But, I don't quite see it. I believe what I am missing is the relationship between holomorphic strips in $X\times X$ and holomorphic cylinders in X. (**Edit**: I would also like to understand the comparison of the product structures too)
**Edit:** here is a rough geometric idea which might have something to do with the truth. I want to assume that my Hamiltonian is time independent and that all orbits are actually fixed points. Given a map of a strip into $X\times X$ the two projections give us two strips into X. The idea is to glue the strips together to form a cylinder which is a map into X. Of course, this doesn't take into account issues of compactifications and so on... Anyways, if someone would be happy to spell it out I would appreciate it.
| https://mathoverflow.net/users/6986 | Comparison between Hamiltonian Floer cohomology and Lagrangian Floer cohomology of the diagonal | Let $f:X\to X$ be a (Hamiltonian) symplectomorphism. The claim is that the fixed point Floer homology of $f$ agrees with the Lagrangian intersection Floer homology of the graph of $f$ with the diagonal in $X\times X$. I think the argument is that if we choose almost complex structures for the two Floer theories in the following way then the two chain complexes are isomorphic. Or at least, the corresponding holomorphic curves agree. (I'm ignoring issues of transversality, which coefficient ring to use, etc.)
The Floer homology of $f$ is the homology of a chain complex which is generated by fixed points of $f$ and whose differential counts maps $u:{\mathbb R}\times {\mathbb R}\to X$ such that $u(s,t+1)=f(u(s,t))$ and $(\partial\_s + J\_t\partial\_t)u=0$ where $J\_t$ is a family of $\omega$-compatible almost complex structures on $X$ parametrized by $t\in{\mathbb R}$ such that $J\_{t+1}=f\_\*\circ J\_t\circ (f^{-1})\_\*$.
The Lagrangian Floer homology of the graph and the diagonal is the homology of a chain complex which is generated by intersection points and whose differential counts maps $v=(v\_-,v\_+):{\mathbb R}\times[0,1]\to X\times X$ such that $v\_-(s,0)=v\_+(s,0)$, $v\_+(s,1)=f(v\_-(s,1))$, and $(\partial\_s + \tilde{J}\_t\partial\_t)v=0$ where $\tilde{J}\_t$ is a family of $-\omega\oplus\omega$-compatible almost complex structures on $X\times X$ parametrized by $t\in[0,1]$.
To relate these, we first observe that there is an obvious bijection between the generators of the chain complexes. To get the holomorphic curves to match up, first choose the family $J\_t$, then define
$\tilde{J}\_t = (-J\_{\frac{1-t}{2}})\oplus J\_{\frac{1+t}{2}}$.
Now given a holomorphic cylinder $u$ as above, you can cut it into two halves to get a holomorphic strip $v$:
$v(s,t)=(u(\frac{s}{2},\frac{1-t}{2}),u(\frac{s}{2},\frac{1+t}{2})).$
Conversely, given a holomorphic strip $v$, one can glue together its two components $v\_-$ and $v\_+$ to get a holomorphic cylinder $u$ where $u(s,t)=v\_-(2s,1-2t)$ for $t\in[0,1/2]$ and $u(s,t)=v\_+(2s,2t-1)$ for $t\in[1/2,1]$. Because of the boundary conditions this is at least $C^1$ where we glue the pieces together, and so by elliptic regularity it is actually smooth.
| 21 | https://mathoverflow.net/users/6670 | 43357 | 27,565 |
https://mathoverflow.net/questions/43362 | 1 | Why does forcing seem to be so vacuously true?
It seems like you are just reversing the subset containment of the model of ZFC + CH to be the other way in the poset. So, why is this valid? Why are you allowed to just put the empty set at the top of the partial order\*?
\*I have just started learning set theory so I have only seen one example of forcing and it was the forcing of the set of reals to be equal to the second uncountable cardinal.
| https://mathoverflow.net/users/nan | The consistency of ZFC + CH gives the ability to travel to a universe which models ZFC + \neg CH? | The empty set $\emptyset$ here is the forcing condition that has the least amount of information about the generic object being constructed.
Since it has the least information, you might think it should be at the bottom of the partial order. There are two replies to this:
(1) Mathematically, we are free to define our partial order however we want, even if it confuses the reader.
(2) In this case, there is no intent to confuse the reader. Having the *least* information, the condition $\emptyset$ also leaves the *greatest* amount of possibilities still open. So it is really a 50-50 decision which way one should define it. (Others may know the history of this particular convention.)
A similar convention question is whether the Turing degree of the computable sets, $\mathbf 0$, is the smallest or the largest Turing degree. Conventionally one speaks of the degrees of *unsolvability* and so $\mathbf 0$ is the smallest, but if the convention were degrees of *solvability* it would be the greatest. Since none of the other degrees are very solvable at all, the current convention is perhaps the best in this case.
| 8 | https://mathoverflow.net/users/4600 | 43365 | 27,571 |
https://mathoverflow.net/questions/43366 | 6 | Kodaira embedding theorem says that a positive line bundle is ample, i.e. high tensor powers are holomorphically embeddable into complex projective space of high dimension.
However, ampleness is not stable under blow-ups. Usually a replacement is to consider big line bundles, which is stable under blow-ups.
Is there an embedding theorem for big line bundles? One cannot hope that an embedding of high tensor powers of a big line bundle to be everywhere injective, but can we have injectivity almost everywhere?
Here's the precise question. Let $X$ be a complex compact manifold. Let $L$ be an ample line bundle over $X$. Let $f:Y\to X$ be a blow up, or series of blow ups. Is there some condition for the pullback $f^\*L$ to be an embedding outside the exceptional locus of $f$?
| https://mathoverflow.net/users/nan | Embedding Theorem for big line bundles | If $X$ is normal, then the Iitaka fibration theorem implies that $L$ is big if and only if the rational map
$\phi\_m \colon X \dashrightarrow \mathbb{P}H^0(X, L^{\otimes m})$
is birational onto its image for some $m >0$, see [Lazarsfeld, Positivity in Algebraic Geometry I, p. 139].
I guess this is the "embedding theorem for big line bundles" you are looking for.
Regarding your last question, since $L$ is ample by assumption and $f$ is an isomorphism outside its exceptional locus $E$, some power of $f^\*L$ will surely give an embedding of $X \setminus E$.
| 7 | https://mathoverflow.net/users/7460 | 43368 | 27,573 |
https://mathoverflow.net/questions/43372 | 10 | This question is about p-adic representations of $\mathrm{Gal}(\overline{\mathbb{Q}}\_p / \mathbb{Q}\_p)$ and $(\varphi, \Gamma)$-modules. By theorems of Fontaine, Cherbonnier-Colmez and Kedlaya, the category of p-adic representations of $\mathrm{Gal}(\overline{\mathbb{Q}}\_p / \mathbb{Q}\_p)$ is equivalent to each of the following three categories:
* etale $(\varphi, \Gamma)$-modules over Fontaine's ring $\mathbb{B}\_{\mathbb{Q}\_p}$
* etale $(\varphi, \Gamma)$-modules over the subring $\mathbb{B}^{\dagger}\_{\mathbb{Q}\_p}$
* slope zero $(\varphi, \Gamma)$-modules over the Robba ring $\mathcal{R}$ (also known as $\mathbb{B}^{\dagger}\_{\mathrm{rig}, \mathbb{Q}\_p}$).
It's well known that slope 0 $(\varphi, \Gamma)$-modules over the Robba ring can sometimes be written as extensions of other Robba-ring $(\varphi, \Gamma)$-modules which are not themselves of slope 0. (Indeed there is the whole rich theory of trianguline representations, whose Robba-ring $(\varphi, \Gamma)$-modules are built up entirely from rank 1 pieces.)
**My question**: does this happen for either of the other two categories of $(\varphi, \Gamma)$-modules? Can one have a short exact sequence of $(\varphi, \Gamma)$-modules over $\mathbb{B}\_{\mathbb{Q}\_p}$ or $\mathbb{B}^{\dagger}\_{\mathbb{Q}\_p}$ where the middle term is etale but the two end terms are not?
| https://mathoverflow.net/users/2481 | Can an etale (phi, Gamma) module be an extension of non-etale ones? | In the first two cases, the slopes of $\varphi$-modules are given by the "standard" Dieudonné-Manin decomposition. In particular, subobjects of étale objects are étale.
For more info, see (for example) chapter 4.5 of Kedlaya's "Slope Filtrations Revisited".
| 9 | https://mathoverflow.net/users/5743 | 43373 | 27,577 |
https://mathoverflow.net/questions/43380 | 3 | It is well known that the upper bound on the number of quadratic residues mod n is approximately n/2 and it reaches this bound for n prime.
Is there any similar lower bound on the number of quadratic residues mod n?
Some numerical experiments indicate that it would be somewhere at n^0.65 for highly composite n with many small factors, but can you point me at any more rigorous treatment of the subject?
| https://mathoverflow.net/users/10292 | Numbers with few quadratic residues | Edit: First version of this answer had a silly mistake (forgot to multiply by $n$ the first estimate). Argument is still the same but the final result changes.
If you take $n$ to be the product of the first $r$ odd primes, then the number of quadratic residues modulo $n$ is bounded below by ${(1/2)}^rn$, by the Chinese remainder thm. On the other hand, $n$ is about $(1/2)e^{r\log r}$ by the prime number theorem, so you get a lower bound of $n^{1-c/\log \log n}$ for the number of quadratic residues. So you have a lower bound of the form $n^{1-\epsilon}$ for any $\epsilon>0$ if $n$ is large. I think the numbers I constructed will be minimal so you'll get your lower bound.
What did the drowning analytic number theorist say?
| 5 | https://mathoverflow.net/users/2290 | 43384 | 27,582 |
https://mathoverflow.net/questions/43381 | 10 | More precisely, what real numbers $r$ have the following property: for any $\epsilon > 0$ there exist infinitely many pairs $(p, q)$ of integers such that
$$\left| \frac{p}{q} - r \right| < \frac{\epsilon}{q^2}.$$
I think that this is impossible if $r$ is a quadratic irrational. On the other hand, it's certainly possible for any number with [irrationality measure](http://mathworld.wolfram.com/IrrationalityMeasure.html) strictly greater than $2$.
What I really want to know is if the real numbers which don't have the property above have measure zero. If that's true, it would answer [the last part of this math.SE question](https://math.stackexchange.com/questions/7634/you-are-standing-at-the-origin-of-an-infinite-forest-holding-an-infinite-bb-gu/).
| https://mathoverflow.net/users/290 | What numbers can be approximated "pretty well" by rationals? | This is a well-studied question in diophantine approximation. You can look up Markov or Lagrange spectrum for a "description" of the numbers for which you cannot take $\epsilon$ arbitrarily small. For the answer to your last question, look up Khinchin's theorem (the answer is no, they have full measure).
| 13 | https://mathoverflow.net/users/2290 | 43385 | 27,583 |
https://mathoverflow.net/questions/16794 | 5 | In this joint paper that I should be working on, we make significant use of a certain generalization of a triangulated disk. Many of our important examples are triangulated disks, but we are also interested in certain simplicial complexes that are singular disks, or even more generally singular disks tiled by polygons. It is easy to describe the disks of interest to us as simplicial complexes which are contractible, compact subsets of the plane. The embedding in the plane only matters at all up to isotopy, and it also does not matter all that much; the most important condition is that the complex is contractible and at most two triangles meet at every edge. For instance, you can have two triangulated disks that meet at a vertex, trees, "barbells", etc.
One could more generally look at those topological spaces in the plane that are an intersection of nested, closed disks, or maybe those that are locally connected. For instance, the Mandelbrot set is one. We do not need them for what we are doing if they are not simplicial complexes, but this is an interesting class of topological spaces that should have a good name.
The following names have been proposed:
1. Van Kampen diagram - standard but ugly; very close to what we use although our edges are not labeled by group elements.
2. Contractible plane continuum - descriptive but clumsy.
3. Diskoid - a good name, by analogy with a dendroid which is the 1-dimensional case, but it doesn't seem to be standard.
4. Cactus - also a good name, by analogy with trees in graph theory, but it seems non-standard.
5. Singular disk - livable but not specific enough.
The reason that I want a short name is that the object X itself is used to make a moduli space or an algebraic variety. The moduli space is easy to describe: Assign fixed lengths to the edges of X and look at its rigid embeddings into a metric space. So we wouldn't want to say "contractible plane continuum variety". Also, for no particular reason I've been thinking of the triangulation as an extra decoration and instead name the underlying topological space.
I can certainly think of livable names, but the general idea appears in several places in mathematics and I would prefer a good name. Could someone with a good sense of the literature argue for a particular term, not necessarily one of the ones listed above? Or any useful opinion would be welcome.
| https://mathoverflow.net/users/1450 | Better term for a (simplicial) contractible plane continuum | In the end, we (Joel Kamnitzer, his student Bruce Fontaine, and I) agreed on the term "diskoid". In the abstract, the word "cactus" seemed too clever by half. When I actually wrote it into the paper, it was awkward and it didn't emphasize the main property of interest, that our shapes are meant as a mild generalization of disks. Moreover, there have been papers that used the term "cactus" in similar but slightly different ways, and the mathematical etymology of this term is already somewhat confused. Actually, I don't have a good explanation of why the term that we picked seems good (or least bad), but somehow that is the feeling.
| 1 | https://mathoverflow.net/users/1450 | 43387 | 27,585 |
https://mathoverflow.net/questions/43051 | 3 | I am trying to compare some homomorphism groups over different base rings, so given a commutative local ring $(A,\mathfrak{m})$ and a finite dimensional Azumaya algebra $R$ over $A$.
If $M$ and $N$ are two left $R$-modules which are finitely generated and torsion free over $A$, is there an $A$-isomorphism $R \otimes\_A Hom\_R(M,N) \rightarrow Hom\_A(M,N)$?
So for example if $M=N=A^n$ and $R=M\_n(A)$, then we have $Hom\_R(M,N)\cong A$ and $Hom\_A(M,N)\cong M\_n(A)$ so we have an isomorphism in this case.
What about "non trivial" examples? Or do we need to have put stronger conditions on $M$ and $N$?
$\textbf{New idea}$: I think the Hom-Tensor adjunction won't help here. So next try:
Put $M=R$, this gives $Hom\_R(M,N)=Hom\_R(R,N)\cong N$ as $A$-modules. So we have $R\otimes\_A Hom\_R(M,N) \cong R\otimes\_A N$. Now because $R$ is Azumaya we have $R\cong Hom\_A(R,A)$. So we really have $R\otimes\_A Hom\_R(M,N) \cong Hom\_A(M,N)$.
So we have an isomorphism for $M=R^k$. Now what about if $M$ is a projective $R$-module? Then there is some $R$-module $P$ such that $M\oplus P\cong R^k$. Can we somehow conclude that we have an isomorphism for M in this case?
| https://mathoverflow.net/users/3233 | Comparing homomorphisms over different base rings | I think this is true. There is a homomorphism $R \otimes\_A Hom\_R(M,N) \rightarrow Hom\_A(M,N)$, given from the obvious $R$-module structure of $Hom\_A(M,N)$. To check that this is an isomorphism we can make a faithfully flat extension and split $R$; so we may assume that $R$ is a matrix algebra $M\_n(A)$.
Now we use Morita equivalence: tensoring with the $(R-A)$-bimodule $A^n$ gives an equivalence of categories between $A$-modules and $R$-modules. If $M = A^n \otimes\_A M'$ and $N = A^n \otimes\_A N'$, we have
$$
Hom\_A(M,N) = Hom\_A(A^n \otimes\_A M', A^n \otimes\_A N') = M\_n(A) \otimes\_A Hom\_A(M',N') = R \otimes\_A Hom\_R(M,N)
$$
and it should be easy to see that this is the desired isomorphism.
Does this work?
| 3 | https://mathoverflow.net/users/4790 | 43390 | 27,587 |
https://mathoverflow.net/questions/43397 | 16 | Finding primes in signals is seen as a sign of some kind of intelligence - see e.g. the role of primes in the search for extraterrestrial life (see e.g. [here](http://www.math.dartmouth.edu/~carlp/PDF/extraterrestrial.pdf)).
This is because there are relatively few examples of numbers that appear in nature because they are prime.
One example of the use of prime numbers in nature is as an evolutionary strategy used by cicadas of the genus Magicicada (see e.g. [here](http://en.wikipedia.org/wiki/Prime_number#Prime_numbers_in_nature) or here: [1])
**My question:**
Do you know of any other instances where prime numbers occur in nature? Could you please also give a source/link - and perhaps some background. Thank you.
[1] Goles, E., Schulz, O. and M. Markus (2001). "Prime number selection of cycles in a predator-prey model", Complexity 6(4): 33-38
**Edit**: Obviously many people misunderstood me. I didn't mean the occurrence of prime numbers just by coincidence - but *because they are prime*. The cicadas example - although being controversial - at least hints at some kind of evolutionary strategy.
| https://mathoverflow.net/users/1047 | Examples of prime numbers in nature | Somewhat longish to be a comment, so here goes:
How about examples like:
* Polygons in nature? Unfortunately, the most famous one of them is a hexagon ;-)
But I am certain there are several chemical compounds, physical structures such as crystals, and so on that exhibit polygonal structures with prime number of edges / facets --- perhaps because the "primeness" there leads to a physically / chemically more stable configuration.
Is this an acceptable kind of primeness? Perhaps not. It seems to me that the question you pose is almost at a "meta" level!
| 2 | https://mathoverflow.net/users/8430 | 43401 | 27,595 |
https://mathoverflow.net/questions/43400 | 10 | This should be really well known but I don't seem to find a statement about it nor a question in MO answering this.
Consider a Compact Hausdorff topological space $X$. The [cohomological dimension](http://en.wikipedia.org/wiki/Cohomological_dimension) of $X$ is the first natural number where the cohomology vanishes. The [topological dimension](http://en.wikipedia.org/wiki/Inductive_dimension) is as defined in wikipedia (see also Hurewicz Wallman's book).
For CW complexes, it seems easy to prove that finiteness of topological dimension implies finiteness of the other. However, for general spaces I don´t figure out how to prove (or construct a counterexample) to
Q) Does finite topological dimension of $X$ implies finite cohomological dimension?
Maybe there are well known inequalities also (for example, it seems that for spaces homotopy equivalent to CW complexes one should have that the topological dimension is bigger than or equal to the cohomological one). Information about this kind of inequalities would be welcome.
| https://mathoverflow.net/users/5753 | Topological dimension versus cohomological dimension | Well, I think it depends on which dimension you mean and which cohomology. The best fit I think is covering dimension and Čech cohomology. The Čech cohomological dimension is indeed bounded (more or less by definition) by the covering dimension.
**Addendum**: Just to comment on BCnrd's comment. The usual definition of cohomological dimension is the vanishing of all sheaf cohomology above that dimension. I however got the impression that the OP interpreted it in the sense of all cohomology with constant coefficients vanishing. In any case the Čech cohomology of sheaves also vanish above the covering dimension so for that it doesn't matter what one means, unless sheaf cohomology is meant. However, a Hausdorff space (I am asuming that that is part of the OP's assumption) that either has finite covering dimension or is compact is paracompact and hence it doesn't matter after all.
| 11 | https://mathoverflow.net/users/4008 | 43405 | 27,598 |
https://mathoverflow.net/questions/43305 | 18 | Let say that an infinite subsets $A$ of $\mathbb{N}$ is "nice w.r.to ergodic limits", if it can replace $\mathbb{N}$ in the individual ergodic theorem, that is, if it is such that the following statement is true:
>
> For any probability space
> $(X,\Sigma,\mu),$ for any
> measure-preserving transformation $T$
> on $X,$ for any $f\in L^1(X,\mu)$ the
> ergodic means along $A,$
>
>
> $$M(f,T,A,t,x):=\big|\{j\in A\, : j\leq t\}\big|^{\,-1}\sum\_{j\in A,\, j\leq t}f(T^{j}x)$$
>
>
> converge for a.e. $x\in X$ to the conditional expectation w.r.to the $T$ invariant $\sigma$-algebra, $\mathbb{E}(f|\Sigma\_T),$ as $t \to +\infty.$
>
>
>
So $\mathbb{N}$ itself is nice in this sense, by Birkhoff's theorem; if $A$ is nice and $m$ is a positive integer, the set of translates $A+m$ is nice (the set of convergence with $T$ along $A+m$ coinciding with the $T^{\, m}$ pre-image of the set of convergence with $T$ along $A$). Also, a disjoint union of two nice sets is nice.
Is there any other structure on the family of these sets? What about e.g. the union of two of them? (at a glance it seems to me that something more can be said for the analogous cases of other ergodic theorems, e.g. for the $L^p$ convergence.
Looking at this very related [question](https://mathoverflow.net/questions/42837/a-non-standard-ergodic-limit), and its answer, make me think that the situation may be non-trivial and interesting enough, so that it should have been studied.
| https://mathoverflow.net/users/6101 | Ergodic limits along subsets of $\mathbb{N}.$ | These are called good universal sets.
Bourgain (1987) proved that sequences of the form $p(n)$, $n \in {\bf N}$, $p$ a non constant polynomial, are good.
He also proved (1988) that the set of primes is a good universal set for $L^p$ functions, $p> {(1+\sqrt{3})\over 2}$. This was later improved to $p>1$ by Wierdl, see its article for a short introduction to the problem <http://www.springerlink.com/content/e027w4211n7784h1/fulltext.pdf> . There is now an extensive litterature on the problem, following Bourgain's articles.
In another direction, note that a transformation T is mixing iff the ergodic theorem for T holds for all subsequences (see e.g. the book of Krengel, "ergodic theorems").
| 10 | https://mathoverflow.net/users/6129 | 43414 | 27,604 |
https://mathoverflow.net/questions/43336 | 12 | Suppose that $X$ is a Cohen-Macaulay normal scheme/variety and $\pi : Y \to X$ is a proper birational map with $Y$ normal.
**Question:** Is $Y$ also Cohen-Macaulay? Are there common conditions which imply it is?
If $Y$ is not normal I know of several ways to show that the answer to the first question is no.
There are obvious spectral sequences but I don't see how to deduce what I want from them, perhaps I'm being dumb (or maybe there is an obvious example).
| https://mathoverflow.net/users/3521 | Blowups of Cohen-Macaulay varieties | An example was given in Section 3 of this [paper by Cutkosky](https://doi.org/10.1007/BF01233425 "Invent Math 102, 157–177 (1990). zbMATH review at https://zbmath.org/?q=an:0718.14025"): "A new characterization of rational surface singularities." (The scheme $Z$ in the last page, which is a blow up of some $m$-primary ideal of a regular local ring of dimension $3$, is normal but not Cohen–Macaulay.)
The algebraic side of this example has been studied quite a bit, so perhaps more explicit examples are known. I am not an expert here, but you can check out a paper by [Huckaba–Huneke](https://doi.org/10.1515/crll.1999.049 "Normal ideals in regular rings. J. Reine Angew. Math. 510, 63–82 (1999). zbMATH review at https://zbmath.org/?q=an:0923.13005") ([Wayback Machine](https://web.archive.org/web/20070117005728/http://www.math.fsu.edu/%7Ehuckaba/normalideals-jour-vers.pdf)), or papers by Vasconcelos (he has a [book](https://doi.org/10.1017/CBO9780511574726 "London Mathematical Society Lecture Note Series. 195. Cambridge: Cambridge University Press. 329 p. (1994). zbMATH review at https://zbmath.org/?q=an:0813.13008") called "Arithmetic of Blowup Algebras" which discussed, among other things, Serre's condition $(S\_n)$ on Rees algebras), and the references there.
| 11 | https://mathoverflow.net/users/2083 | 43415 | 27,605 |
https://mathoverflow.net/questions/43408 | 12 | In customary formulations of the Singular Value Decomposition or SVD that I have seen,
(e.g., Wikipedia or Gil Strang's textbooks) it is always stated in terms of writing an
$m \times n$ matrix $M$ (say of rank $r$) as a product $U \Lambda V$, where $U$ and $V$ are
orthogonal $m \times m$ and $n \times n$ matrices and $\Lambda$ is a diagonal $m \times n$ matrix
with non-negative "singular values" on the diagonal, $s\_1 \ge s\_2 \ge \ldots \ge s\_r >0$ and the
rest zero. Looking back over the many times I have taught linear algebra to both undergraduates
and graduate students, I realized that I have not once covered the SVD, and even though I consider
myself pretty knowledgeable about linear algebra, I have never felt comfortable with the statement
of SVD or felt that I understood it in a more than formal way. (I might add that many theoretical
linear algebra texts do not mention the SVD and many of my more theoretically minded colleagues do
not even recognize the term.) But a few days ago, a social scientist friend of mine asked me about
a problem he was interested in; one that involved the SVD in an essential way. After thinking about
it for a while, I realized that SVD can be reformulated as a statement about linear transformations
that, to me at least, seems a lot more conceptual and geometric:
If $T$ is a linear map, say of rank $r$, between finite dimensional inner-product spaces $V$ and
$W$, then there are orthonormal bases $v\_1, \ldots, v\_m$ for $V$ and $w\_1, \ldots, w\_n$ for $W$
and $r$ positive numbers $s\_1 \ge s\_2 \ge \ldots \ge s\_r$, such that $T v\_i$ equals $s\_i w\_i$ if
$i \le r$ and equals zero if $i > r$.
I certainly realize that this is a pretty obvious reformulation of SVD, once you see it (and
those poor misguided souls who prefer matrices to linear transformations may even see it as
a step backwards :-), but my question is whether there is some standard source for this
reformulation that I can reference.
| https://mathoverflow.net/users/7311 | Is this formulation of the Singular Value Decomposition standard? | I just looked in Wikipedia (<http://en.wikipedia.org/wiki/Singular_value_decomposition>). There is a very thorough discussion, including a section "Geometric meaning" in which your interpretation is clearly explained. Well, there $K^n$, where $K = \mathbb R$ or $K = \mathbb C$ is used, instead of arbitrary inner product spaces, but I suppose you'll agree that the difference is not essential.
| 6 | https://mathoverflow.net/users/4790 | 43416 | 27,606 |
https://mathoverflow.net/questions/43411 | 5 | Is there any subset of $R^n$ homotopically equivalent to the wedge product of countable many circles.
In particular, is the union of circles in $R^2$ with center (0,n) and radius n for n=1,2 ... n
(sort of inverse Hawaiian earring) homotopically equivalent to that wedge product.
Note that the answer is no if we ask about topological equivalence: the wedge product is closed but not locally compact.
| https://mathoverflow.net/users/10299 | homotopical immersion of the wedge product of countable many circles in $R^n$ | You can certainly embed the real line, with perpendicular circles at the integer points, into $\mathbb{R}^3$, and this is homotopy equivalent to the wedge you want.
Let $W$ be the expanding `inverse Hawaiian earring'.
Then the obvious map $\bigvee S^1\to W$
is a homotopy equivalence. To define the inverse,
let $Z$ be the intersection of $W$ with
a small closed rectangle around the origin;
collapse $Z$ to a point, then map the resulting quotient back to
$\bigvee S^1$ by the inverse of the obvious map.
We have to check that $W\to \bigvee S^1$ is continuous.
If $U\subseteq \bigvee S^1$ is open
and does not contain the basepoint $\star$,
then the preimage is obviously open.
If $\star\in U$, then the preimage is the
preimage of $U- \star$ (which is open) together with
a neighborhood of $Z$, which is also open.
Since $Z$ is contractible and its
inclusion into $W$ is a cofibration,
the composite $W \to \bigvee S^1\to W$
is homotopic to $\mathrm{id}\_{W}$.
In essentially the same way, the composite
$\bigvee S^1\to W\to \bigvee S^1$
is homotopic to the identity.
| 5 | https://mathoverflow.net/users/3634 | 43432 | 27,616 |
https://mathoverflow.net/questions/43428 | 1 | Suppose we have two sets of discrete events, $A$ and $B$. Then I think it is true that:
$$2\sum\_{i \in A, j \in B}\Pr[i\ \textrm{AND}\ j] \leq \sum\_{i \in A}\Pr[i]+ \sum\_{j \in B}\Pr[j] +\sum\_{i, j \in A}\Pr[i\ \textrm{AND}\ j] + \sum\_{i, j \in B}\Pr[i\ \textrm{AND}\ j]$$
My intuition for why this should be true is just by analogy to the simple Cauchy-Schwarz like inequality for $a, b \in \mathbb{R}:$
$$2ab \leq a + b + a(a-1) + b(b-1)$$
To see why this identity is true, note:
$$a + b + a(a-1) + b(b-1) = a^2 + b^2 = 2ab + (a-b)^2 \geq 2ab$$
Is my inequality true, and is there a simple proof?
| https://mathoverflow.net/users/10303 | Probability space analogue of Cauchy-Schwarz inequality | yes, if $\sum\_{i,j\in A}$ means double summation (each pair $i,j$ is taken twice)
then denote $f\_i$ and $g\_i$ characteristic functions of your events from $A$ and $B$ respectively, LHS equals $2 \sum \int f\_i g\_j=\int 2(\sum f\_i)(\sum g\_j)$, RHS equals $\sum \int f\_i^2+\sum \int g\_j^2+2\sum \int f\_i f\_j+2\sum \int g\_i g\_j=\int (\sum f\_i)^2+(\sum g\_j)^2$, and RHS-LHS equals $\int (\sum f\_i-\sum g\_j)^2\geq 0$.
| 3 | https://mathoverflow.net/users/4312 | 43437 | 27,618 |
https://mathoverflow.net/questions/43436 | 0 | Hi Mathoverflow
I hope you bear with me that my linear algebra knowledge is a little rusty, but I have a question that might potentially very easy to answer. Nevertheless it's been bugging me for a good 3 hours now.
I'm trying to find the LU decomposition of a matrix in order to find the determinant. This would be a straightforward process if it weren't for the fact that my matrix library (Meshcach) only gives me back one matrix when use their LU factor method (together with a pivot). So when I have the matrix A:
```
Matrix: 3 by 3
2 4 1
4 -10 2
1 2 4
```
and I call LUfactor, I get the pivot p:
```
Permutation: size: 3
0->0 1->1 2->2
```
and the matrix LU:
```
Matrix: 3 by 3
2 4 1
2 -18 0
0.5 -0 3.5
```
My question is: How do I find the upper matrix from LU? It seems that the library stores the upper and lower matrix as one single matrix, but I can't figure out how to separate them.
The true upper and lower matrices should be:
```
Lower matrix
1 0 0
0.5 1 0
0.25 0.5 1
Upper matrix
4 -10 2
0 -9 0
0 0 3.5
```
But I'm not sure how the output of meschach would be used to obtain them. I'm aware that this might partly be an algorithm question and could be argued to belong in stackoverflow instead, but I really think there is a higher chance of running in to somebody math-savy who can figure out what's going on, than there is of running in to a programmer who dealt with exactly this problem before.
I hope somebody can help
| https://mathoverflow.net/users/8672 | Finding the determinant of a matrix with LU composition | The factorization you give at the end ("the true upper and lower matrices") is incorrect. To see why, just check the (1,1) element in your original matrix. Multiplying your $L$ by your $U$ gives 4 for that element, but your original matrix has a 2 there.
Meshcach's factorization is correct. The right $L$ and $U$ matrices are
```
L = 1 0 0
2 1 0
0.5 0 1
```
```
U = 2 4 1
0 -18 0
0 0 3.5
```
You should be able to read that off of Meshcach's output pretty easily.
| 2 | https://mathoverflow.net/users/9716 | 43440 | 27,620 |
https://mathoverflow.net/questions/36810 | 3 | Let $X$ be a subcomplex of a CW-complex $Y$. Is $(Y/X)^{\wedge k}$ homotopy equivalent to $Y^{\wedge k}/X^{\wedge k}$, where $\wedge k$ is the $k$-fold smash product? I know it is not true for products but am having a hard time visualizing for smash products.
| https://mathoverflow.net/users/8658 | Do Smash Products and Quotients Commute? | The easiest way I know to say what is going on is to resort to looking at
"products" of pairs:
$$
(X, A) \times (Y, B) = ( X\times Y , A\times Y \cup X\times B).
$$
The point of this notation is that the functor $(X, A) \mapsto (X/A, \*)$
carries $(X, A) \times (Y, B)$ to $X/A \wedge Y/B$. We can iterate this procedure,
and I'll write $T^n(Y,X)$ for the subspace of $Y^n$
satisfying
$$
(Y, X)^n = ( Y^n, T^n(Y, X)).
$$
Thus $(Y/X)^{\wedge n} = Y^n /T^n(Y,X)$.
You can easily check that
$$
T^n( Y, X) = \lbrace (y\_1, \ldots, y\_n) \mid y\_i \in X\ \mbox{for at least one $i$}\rbrace.
$$
On the other hand $Y^{\wedge n}/X^{\wedge n}$ is the quotient
of $Y^n$ by the subspace
$$
T^n(Y,\*) \cup X^n,
$$
which is different (unless $X = \*$).
| 4 | https://mathoverflow.net/users/3634 | 43442 | 27,621 |
https://mathoverflow.net/questions/43446 | 4 | I am looking for concrete examples of a complete discrete valuation ring $R$ of characteristic 0, residue characteristic $p$ and ramification index $e$. By residue characteristic, I mean the characteristic of the field obtained by the quotient of $R$ with its unique maximal ideal $M$ and by ramification index I mean the largest positive integer $e$ such that $M^e \supseteq pR$.
Without the restriction on the ramification, a simple example is the $p$-adic integers $\mathbb{Z}\_p$. However, when we try to fix the ramification index, this becomes more challenging. For example, with $e = 2$ we can take $R = \mathbb{Z}\_p[\sqrt{p}]$. The maximal ideal of this ring is $M = \sqrt{p}R$ which has ramification index 2.
My question: is there a simple construction for such a ring with arbitrary $p$ and $e$? If not, can an infinite family of such rings be constructed that have a known ramification index $e > 2$?
| https://mathoverflow.net/users/10204 | Examples of DVRs of residue char p and ramification e | The family of rings $\mathbb{Z}\_p[p^{\frac{1}{e}}]$ does what you want.
Because of your question, I gather you do not yet know the correspondence between totally ramified extensions and Eisenstein polynomials. For this see e.g. Serre's *Local Fields*, Lang's *Algebraic Number Theory*, or Section 4.3 of
[http://alpha.math.uga.edu/~pete/8410Chapter4.pdf](http://alpha.math.uga.edu/%7Epete/8410Chapter4.pdf)
| 6 | https://mathoverflow.net/users/1149 | 43448 | 27,623 |
https://mathoverflow.net/questions/43447 | 5 | Suppose $(C,\otimes)$ is a symmetric monoidal finitely-cocomplete category such that $\otimes$ preserves colimits. Given two morphisms $a:A\_1\to A\_2$ and $b:B\_1\to B\_2$, define $a\Box b$ to be the induced "lower right corner map" $A\_1\otimes B\_2\coprod\_{A\_1\otimes B\_1} A\_2\otimes B\_1\to A\_2\otimes B\_2$.
Is $\Box$ associative? Does it give a monoidal product on $Arr(C)$?
I'm having trouble figuring out what the diagram for this should look like if it's true (although a quick "arithmetic check" shows that it should be).
If it's not associative for some stupid reason that I've overlooked in the general case presented here, is it true in the case that $C$ is a presheaf category and $\otimes$ is the cartesian product?
| https://mathoverflow.net/users/1353 | Is the box product of morphisms associative? | Yes, that gives a monoidal product (but don't forget the monoidal unit! it's $0 \to I$ where $I$ is the monoidal unit of $\mathcal{C}$).
To simplify the description of the associativity, imagine that $\mathcal{C}$ is strict monoidal to begin with. Draw the entire cube of possible paths that lead from $A\_1 \otimes B\_1 \otimes C\_1$ to $A\_2 \otimes B\_2 \otimes C\_2$, except leave off the three edges that terminate at the corner $A\_2 \otimes B\_2 \otimes C\_2$. The colimit of this diagram is what you want to consider. By taking pushouts one at a time (and taking advantage of preservation of colimits by tensoring on either side), you should be able to see that one way of deriving the colimit is by pushing out $(AB) \otimes C\_1 \to A\_2 \otimes B\_2 \otimes C\_1$ along $(AB) \otimes C\_1 \to (AB) \otimes C\_2$, and another way is by pushing out $A\_1 \otimes (BC) \to A\_2 \otimes (BC)$ along $A\_1 \otimes (BC) \to A\_1 \otimes B\_2 \otimes C\_2$.
I hope these hints convey the idea, because actually drawing the diagrams here would be a somewhat time-consuming job. :-)
| 7 | https://mathoverflow.net/users/2926 | 43453 | 27,627 |
https://mathoverflow.net/questions/43434 | 10 | I'm asking the question on a bit of whim, but I do wonder what answers I would get. The Goldbach conjecture is usually discussed in the realm of the distribution of primes and/or probability. Methods I've seen in the past are mostly analytic.
Have there been methods of attack on this problems that are at their core not about the distribution of primes, and whose methodology steers away from hard analysis? Have there been methods of attack that are completely in the realm of algebra? How have they fared? (for example: was a version of it stated and proven over function fields?)
| https://mathoverflow.net/users/5756 | Algebraic aspects of the Goldbach conjecture | You might be interested in this article [on Goldbach over function fields](http://arxiv.org/abs/0912.1702). The approach is rather geometric/algebraic, so it does pass your "steers away from hard analysis" test.
| 6 | https://mathoverflow.net/users/35416 | 43455 | 27,628 |
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