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https://mathoverflow.net/questions/25031
|
5
|
It is not difficult to see that any reduced fraction $\frac{p}{q}$
where $0 < p < q $ and both $p$ and $q$ have at most $N$
digits (where $N$ is a fixed integer) can be reconstructed
from its first $2N$ digits.
In other words, if we let
${\cal F}\_N= \lbrace (p,q) | 0 < p < q < {{10}^N} \rbrace $ and define
the mapping $ f : \ {\cal F}\_n \to { \mathbb N} $ by $ f(p,q)=$ integer\_part( $ \frac{10^{2N}p}{q} $) ,
then $f$ is injective. So there is a left inverse $g$, such that
$g(f(p,q))=(p,q)$ for any $(p,q) \in {\cal F}\_N$. What is the best way to compute
$g$ effectively ? There's always brute search, of course, but ...
|
https://mathoverflow.net/users/2389
|
Reconstructing a fraction from its first digits
|
Taking the continued fraction approximations of your decimal expansion until the denominators get larger than 10^N ought to work.
Edit: Let me add that you have to do a tiny bit more work to get the best rational approximants from the continued fraction, and that's probably the algorithm that should be used. See <http://en.wikipedia.org/wiki/Continued_fraction#Best_rational_approximations>
|
13
|
https://mathoverflow.net/users/353
|
25034
| 16,416 |
https://mathoverflow.net/questions/25009
|
7
|
After thinking about this [question](https://mathoverflow.net/questions/24932/counting-connected-manifolds/24938#24938) and reading this [one](https://mathoverflow.net/questions/4155/classification-problem-for-non-compact-manifolds/11259#11259) I am led to ask for an uncountable collection of homeomorphism types of boundaryless connected path-connected submanifolds of the plane.
My guess is that is suffices to consider complements of Cantor sets. However, I do not know how to distinguish ends (up to homeomorphism) sufficiently well to ensure that this works. Are there other, easier, invariants?
|
https://mathoverflow.net/users/1650
|
Counting submanifolds of the plane
|
See a [theorem of Richards](http://www.ams.org/mathscinet-getitem?mr=143186), which implies that homeomorphism types of planar surfaces are in 1-1 correspondence with homeo. types of compact subsets of the Cantor set. I think there should be uncountably many homeo. types of totally disconnected compactums, but I don't know a reference or an argument off the top of my head. I think this should be related to the ordinalities of the accumulation points, but I'm not sure which ordinals can occur.
Addendum: Googling, I found references to a result of Markiewicz-Sierpinski classifying
countable compact metric spaces up to homeomorphism by their [Cantor-Bendixson rank](http://en.wikipedia.org/wiki/Derived_set_%28mathematics%29#Cantor.E2.80.93Bendixson_rank) (see [section 3 of this paper](http://arxiv.org/abs/0710.1495) for a statement).
The CB-rank must be a countable [ordinal](http://en.wikipedia.org/wiki/Ordinal_number) $\zeta$, and the space is homeomorphic to the
ordinal $\omega^\zeta\cdot n+1$ with the order topology for some $n\in \mathbb{N}$. These may all be realized as
compact subsets of the line. This gives uncountably many non-homeomorphic compacta,
which by Richards' theorem implies that there are uncountably many planar surfaces.
|
9
|
https://mathoverflow.net/users/1345
|
25039
| 16,419 |
https://mathoverflow.net/questions/25020
|
13
|
Let ${\mathfrak g}$ be a Lie algebra in a symmetric monoidal category enriched over $K$-vector spaces, i.e., in particular, hom-s are $K$-vector spaces (where $K$ is a field of characteristic zero). What is its universal enveloping algebra?
As one can talk about associative and Lie algebras there, I can imagine the definition in terms of the universal property but I am interested in its existence, a construction, if you may. Completing the category appropriately (direct sums and direct summands) could give familiar tensor and symmetric algebras $T({\mathfrak g})$ and $S({\mathfrak g})$ (i.e. they are objects in a certain completion of the original category). Is there a way to quotient $T({\mathfrak g})$ or to deform $S({\mathfrak g})$ at this point?
|
https://mathoverflow.net/users/5301
|
What is the universal enveloping algebra?
|
I have now understood the situation better so my previous post has been replaced
by this. (The only thing that was in the original but will not be here are some
explicit formulas but Theo has given a reference for that.)
As I understand the question the poser wanted a construction of the enveloping
algebra of a Lie algebra in a symmetric monoidal pseudoabelian (i.e.,
idempotents have kernels) $K$-category $\mathcal C$ with arbitrary sums over a field
$K$ of characteristic zero. This means that for any $K[\Sigma\_n]$-module $M$ and
any object $V\in\mathcal C$ we can define $M\bigotimes\_{\Sigma\_n}V^{\otimes n}$ and for
any $\Sigma$-module $M\_\bullet$ (i.e., a collection $(M\_n)$ of
$K[\Sigma\_n]$-modules) we can define
$M(V):=\bigoplus\_nM\_n\bigotimes\_{\Sigma\_n}V^{\otimes n}$ which is an endofunctor
of $\mathcal C$. Furthermore, a map $M \to N$ of $\Sigma$-modules gives a natural
transformation of functors $M(V) \to N(V)$. In the particular case when $\mathcal C$ is
the category of $K$-vector spaces such a natural transformation comes from a
unique map of $\Sigma$-modules. The idea is to do what we know to do for $K$-vector
spaces, interpret it as a set of natural transformations, get the corresponding
maps of $\Sigma$-modules and use them to induce natural transformations for a general
$\mathcal C$.
Let thus $S(V)$ be the symmetric algebra on the $K$-vector space $V$, $T(V)$ the
tensor algebra on $V$ and $L(V)$ the free Lie algebra on $V$. Symmetrisation
gives an isomorphism $S(L(V)) \to U(L(V))=T(V)$, where $U(-)$ is the enveloping
algebra of Lie algebras. As $T(V)$ is a $T$-algebra (i.e., an associative
algebra) and we can use this isomorphism to give $S(L(V))$ a $T$-algebra
structure (i.e., a natural transformation $T(S(L(V))) \to S(L(V))$ fulfilling the
appropriate conditions with respect to the monad structure on
$T(-)$). Furthermore, if $\mathfrak g$ is a Lie algebra, then the $T$-algebra
structure on $S(\mathfrak g)$ induced by the isomorphism $S(\mathfrak g) \to U(\mathfrak
g)$ is given as the composite of $T(S(\mathfrak g)) \to T(S(L(\mathfrak g)))$ induced by the
inclusion $\mathfrak g \to L(\mathfrak g)$, the map $T(S(L(\mathfrak g))) \to S(L(\mathfrak
g))$ given by the $T$-module structure on $S(L(V))$ above and the map $S(L(\mathfrak
g)) S(\mathfrak g)$ induced by the structure map $L(\mathfrak
g) \to \mathfrak g$
Now, the functors $S(-)$, $L(-)$ and $T(-)$ are associated to $\Sigma$-modules which
will be denoted by the same letters (instead of the standard $Com$, $Lie$ and
$Ass$). Furthermore, composition of functors correspond to the plethysm
$\circ$. Hence we get that $S\circ L$ is a $T$-module, i.e., we have a map $T\circ S\circ L
\to S\circ L$ compatible with the operad structure on $T$. Consider now the case
of a general $\mathcal C$. Each of $S$, $L$ and $T$ give endofunctors on $\mathcal C$ and
$\circ$ again corresponds to composition. Let $\mathfrak g$ be a Lie algebra in
$\mathcal C$ and define a $T$-algebra structure (i.e., the structure of associative
algebra) on $S(\mathfrak g)$ as the composite
$$
T(S(\mathfrak g)) \to T(S(L(\mathfrak g))) \to S(L(\mathfrak g)) \to S(\mathfrak g)
$$
as above. The verification that this does indeed give a $T$-algebra structure is
just a question of unwinding the definitions. The fact that $S$ is an operad
gives us a natural transformation $V \to S(V)$ which applied to $\mathfrak g$ gives
a morphism $\mathfrak g \to S(\mathfrak g)$ which we now want to show is a Lie algebra
homomorphism. Here the Lie algebra structure on $S(\mathfrak g)$ is induced by its
$T$-algebra structure and the operad map $L \to T$. Again unwinding definitions
shows that it is indeed a Lie algebra morphism.
Finally assuming that $\mathfrak g \to A$ is a Lie algebra homomorphism where $A$ is
an associative algebra with $L \to T$ inducing its Lie algebra structure. Note
that we have an isomorphism (now going back to vector spaces) $S(L(V))\to T(V)$
and hence an isomorphism is $\Sigma$-modules $S\circ L=T$. This gives us a map $S(\mathfrak
g)\to S(L(\mathfrak g))=T(\mathfrak g) \to T(A) \to A$ and it is easy to see that this is
an algebra morphism.
|
6
|
https://mathoverflow.net/users/4008
|
25046
| 16,425 |
https://mathoverflow.net/questions/24103
|
10
|
Let *X* be a compact space.
Recall that its Čech cohomology $H^\bullet(X,\mathbb Z)$
is given by the colomit $\mathrm{colim}\_U\big(H^\*(C^\bullet(U;\mathbb Z),\delta)\big)$, where $U=(U\_i)$ runs over all open covers of *X*, ordered by refining.
For completeness, let us also recall that the *n*-cochains $C^n(U;\mathbb Z)$ are the group of continuous ℤ-valued functions on $\bigsqcup U\_{i\_1}\cap\ldots\cap U\_{i\_{n+1}}$.
Since *X* is compact, we may restrict ourselves to finite covers, without modifying the answer.
• **Definition:** A closed cover $V=(V\_i)$ of *X* is a finite collection of closed subsets $V\_i$ whose union is *X*.
We may now consider the modified Čech cohomology $\tilde H^\bullet(X,\mathbb Z)$, where we use closed covers instead of open covers.
• **Question:** Are $\tilde H^\bullet(X,\mathbb Z)$ and $H^\bullet(X,\mathbb Z)$ isomorphic?
---
*PS:* I know how to show that $\tilde H^1(X,\mathbb Z)$ and $H^1(X,\mathbb Z)$ are isomorphic, by using the fact that they both classify ℤ-principal bundles.
|
https://mathoverflow.net/users/5690
|
Čech cohomology of compact spaces via closed covers?
|
As Angelo says, there is a Mayer-Vietoris (spectral) sequence for closed covers. That comes from an exact sequence of sheaves, which also shows that closed covers are covers in the sense of a Grothendieck topology. Probably it's true for proper surjective maps in general.
I think that means that there is a geometric morphism from the topos of closed covers to the usual topos. This yields a comparison map $H^i\_{closed}\to H^i\_{open}$. If the space is Hausdorff, one can use partitions of unity to refine open covers by closed covers. That shows that the comparison map is surjective: any cohomology class is defined on an open cover, so one can refine it to a closed cover, so it comes from the closed cohomology. If one is careful about the bookkeeping, probably one can arrange a splitting of the Čech complexes, as in André's comment. But we can do better without mentioning this other topos.
We can compute the usual cohomology using the Čech process for open covers. But we could also do it for both open covers and closed covers. This won't change the answer, since closed covers are covers for the usual Grothendieck topology. But now we can get rid of the open covers, since the closed covers are cofinal.
I'm not sure what happens in the non-Hausdorff case. It is certainly not possible to refine open covers by closed covers, but it seems to me that the line with the doubled origin works anyway.
|
1
|
https://mathoverflow.net/users/4639
|
25057
| 16,431 |
https://mathoverflow.net/questions/25067
|
9
|
I'm looking for an answer to the following question. (An answer to a slightly different question would be good as well, since it could be useful for the same purpose.)
>
> Given a set *C* consisting of *n*
> subsets of {1, 2, ..., *n*}, each of
> size *k*, does there exist some small A
> $\subset$ {1, 2, ..., *n*} such that
> *A* intersects all (or all except a small number) of the sets in *C*?
>
>
>
Preferably, "small" will be $\epsilon$*n* where $\epsilon$ can be made arbitrarily small, as long as *n* and *k* are sufficiently large.
I'm hoping the answer is yes. Here is why some such *A* might exist: on average, each element of {1, 2, ..., *n*} intersects *k* sets in *C*, so one might hope to make do with *A* of size on the order of *n*/*k*.
This smells a bit like some version of Ramsey's theorem to me, or like the Erdős–Ko–Rado theorem, but it doesn't (as far as I can tell) follow directly from either.
|
https://mathoverflow.net/users/3410
|
Given n k-element subsets of n, is there a small subset A of n which intersects them all?
|
I believe, reading the abstract, that the paper "Transversal numbers of uniform hypergraphs", Graphs and Combinatorics 6, no. 1, 1990 by Noga Alon answers your question in the affirmative, for some definition of ``your question''. Namely, the worst case is that $A$ has to have size about $2\log k/k$ times $n$, and this multiplier tends to zero as $k$ tends to infinity.
Here's a free copy of the paper.
<http://www.cs.tau.ac.il/~nogaa/PDFS/Publications/Transversal%20numbers%20of%20uniform%20hypergraphs.pdf>
I'm certainly no expert on these matters and my advice would be to look at this and related literature on transversals of hypergraphs. Your collection $C$ of sets is the same thing as a $k$-uniform hypergraph, and the property that you want from $A$ is equivalent to it being a transversal.
Reading Alon's paper a little more I see that what you want is the easier direction of his argument (which gives a tight dependence on $k$). The basic idea is to choose your transversal randomly by picking elements of $\{1,\dots,n\}$ with an appropriate probability $p$. That way, with high probability, you'll hit most of the sets from your collection $C$, and then you just add in one extra element of $A$ for each un-hit set from $C$.
Reading a little further still, I see that the upper bound is probabilistic as well: that is, to make a collection $C$ which is ``bad'', the best plan is to choose sets in $C$ at random from amongst all $k$-element subsets of $\{1,\dots,n\}$.
There's probably literature on your ``almost transveral'' question, but I'll leave someone else to find it. My guess is that random does best in both directions there too.
|
12
|
https://mathoverflow.net/users/5575
|
25070
| 16,439 |
https://mathoverflow.net/questions/25072
|
3
|
Recall that in a triangulated category, all monomorphisms split (have a retraction). Let $F:C\to D$ be an exact functor between triangulated categories. It is an easy exercise to see that if $F$ is faithful then it detects monomorphisms: If $F(f)$ is a monomorphism then so is $f$. Same with epimorphisms of course, and with isomorphisms. But what about semi-simplicity in the following sense?
>
> **Definition**: Let us say that a morphism $f$ is *semi-simple* if there exists $g$ in the opposite direction such that $f=fgf$. (BTW, what is the right name for this?) Assuming $C$ idempotent-complete, $f$ is semi-simple $\iff$ $f$ is a composition of a split epimorphism, an isomorphism and a split monomorphism $\iff$ the exact triangle over $f$ is a sum of trivial triangles.
>
>
>
After trying for a while, I suspect that $F$ faithful is not enough to detect semi-simplicity, so:
>
> **Problem**: Find an exact faithful functor $F:C\to D$ between idempotent-complete triangulated categories and a morphism $f$ in $C$ which is not semi-simple but such that $F(f)$ is semi-simple in $D$.
>
>
>
Of course, it might also be true that faithfulness detects semi-simplicity. A proof of that would certainly count as an answer to the above problem!
What would make me really happy would be to have $F=(\ F:C\to C\ ,\ \mu:F^2\to F\ ,\ \eta:Id\to F\ )$ an exact faithful *monad* (a.k.a. triple) on the category $C=D$. In that case, $F$ faithful forces the unit $\eta$ to be objectwise a split monomorphism. If $\eta$ has moreover a natural retraction then $F$ detects semi-simplicity (easy) but this naturality of the retraction definitely fails for general monads.
|
https://mathoverflow.net/users/5417
|
Do exact faithful functors between triangulated categories detect semi-simplicity?
|
Let $A$ be a non semi-simple algebra and $C$ be the derived category of finite dimensional
$A-$modules. Let $D$ be the derived category of finite dimensional vector spaces and let
$F$ be the forgetful functor which maps $A-$module to its underlying vector space.
Then $F(f)$ is semi-simple for any $f$ (since any morphism in $D$ is semisimple).
UPDATE: as explained below this is not an answer since $F$ is not faithful. Here is an actual example: take for $C$ derived category of representations of an ADE quiver; let $D$ be as above and
let $F=RHom(T,?)$ where $T$ is a direct sum of all indecomposable representations of the quiver
(there are just finitely many of them). Then $F(f)$ is semi-simple for any $f$ as before; to check
that $F$ is faithful use the fact that any object of $C$ is isomorphic to direct sum of shifted
indecomposable representations of the quiver (since the category of representations of the quiver
has homological dimension 1).
|
4
|
https://mathoverflow.net/users/4158
|
25079
| 16,445 |
https://mathoverflow.net/questions/24905
|
2
|
When does the stationary density of an homogeneous Markov process exist?
|
https://mathoverflow.net/users/6127
|
Stationary Solutions of stochastic differential equations
|
It is hard to be brief here, but I will try.
One answer is: when the corresponding stationary Fokker-Planck equation (aka forward Kolmogorov equation) has a nonnegative integrable solution. The density is then obtained by normalization of that solution. This is not a very good answer because FP equations are often not so easy to analyze.
In fact, it is hard to guess what is the question you really wanted to ask. For example, one may say that your question is twofold:
1) when is there an invariant probability distribution?
2) If it exists, is it absolutely continuous w.r.t. Lebesgue?
Existence is guaranteed if the drift prevents the trajectories from going to infinity so that they spend a lot of time in a compact set. Some of the relevant keywords are: Lyapunov-Foster functions, Harris recurrence. Among weakest known conditions guaranteeing existence of invariant distributions are those due to Veretennikov.
Given part 1), absolute continuity of the stationary distributions can essentially be deduced from absolute continuity of transition probabilities. This part is easy if your equations are elliptic and not so easy if the noise excites only some directions.
The analysis can be quite nontrivial depending on how bad your equation is, as a look at a recent paper <http://arxiv.org/abs/0712.3884> might convince you.
|
6
|
https://mathoverflow.net/users/2968
|
25088
| 16,451 |
https://mathoverflow.net/questions/25089
|
10
|
When, if ever, can we view a differential form, e.g. like $dx \wedge dy$, as the similar looking expression used in physics to represent the product of "infinitesimals" e.g. $dx$ $dy$? In particular, I'm wondering why differential forms are anti-symmetric, e.g. $dx \wedge dy=-dy \wedge dx$, whereas in physics we often are happy to write $dx$ $dy=dy$ $dx$. Am I misunderstanding something basic?
|
https://mathoverflow.net/users/nan
|
Basic question about differential forms and physics
|
In both physics and mathematics, there are times when you want a signed multiple integral $dx \wedge dy$, and there are times when you want its unsigned counterpart $dx\;dy = |dx \wedge dy|$. The difference is that in physics, the notation $dx \wedge dy$ is typically paraphrased either with cross products or with antisymmetric indices. The exterior algebra of differential forms is a brilliant definition due to Elie Cartan. Physicists sometimes need ideas that are equivalent to Cartan's work in this topic, but in most areas of physics they simply didn't adopt his notation. One major exception is string theorists and certain gauge theorists, who by now understand Cartan perfectly well.
For example, the most elegant way to understand a surface integral, as you see it in Ampere's law or Stokes' theorem, is as a signed integral. It is the integral of a differential form
$$\omega(x(u,v),y(u,v),z(u,v)) = f(u,v) du \wedge dv$$
over a surface. But you can instead write it as the surface integral of a vector field $\vec{\omega} \cdot (\vec{du} \times \vec{dv})$. Any physicist can tell you that it's a signed integral; the only thing missing is Cartan's notation. A related example is Maxwell's equations. In low energy physics you write them as four equations with 3-vectors. In high-energy physics you write them as one or two equations with 4-vectors and 4-tensors with indices. You can also write the same equation using differential forms, but only gauge theorists and string theorists feel that they need that notation.
On the other hand, a mathematician who wants to use a probability density function or find an unsigned area or volume is perfectly happy to integrate with respect to $dx\;dy = |dx \wedge dy|$. Given other examples such as $ds = \sqrt{|dx|^2+|dy|^2}$ and $|dx \wedge dy|^p$, there is also a shift in emphasis: In more elementary use of Leibniz notation, the differentials are meant more as instructions for what kind of integral you are doing. In Cartan's notation, and in these other unsigned variations, the differentials become objects in their own right, basically what physicists would recognize as tensor fields with special transformation laws.
|
27
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https://mathoverflow.net/users/1450
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25098
| 16,459 |
https://mathoverflow.net/questions/25100
|
10
|
Suppose one has a set $S$ of positive real numbers, such that the usual numerical ordering on $S$ is a well-ordering. Is it possible for $S$ to have any countable ordinal as its order type, or are the order types that can be formed in this way more restricted than that?
|
https://mathoverflow.net/users/440
|
Order types of positive reals
|
Yes, one can have any countable ordering. Indeed any countable totally
ordered set can be embedded in $\mathbb{Q}$. Write your ordered set as
$ \lbrace a\_1,a\_2,\ldots \rbrace $
and define the embedding recursively: once you have placed $a\_1,\ldots,a\_{n-1}$
there will always be an interval to slot $a\_n$ into.
|
17
|
https://mathoverflow.net/users/4213
|
25101
| 16,461 |
https://mathoverflow.net/questions/25113
|
1
|
I hope this question is not too basic. I've asked various mathematicians in the past and had a good search through the Internet but with not a lot of luck.
I am interested in generalizations of Groups or Rings with more than the standard one or two operators. Perhaps one might say Sets with multiple (>2) ternary or even higher order operators.
I suspect someone may have at some point in the past proved that any generalisation can be reduced to a combination of Rings or Groups.
Another possibility is that this is something covered using Category Theory; then perhaps someone could point me to primer that covers the concept described.
|
https://mathoverflow.net/users/4563
|
Generalizations of Rings with multiple higher order Operators
|
As Robin says in his comment, the general framework of this is "universal algebra" or "Lawvere theories". Given the phrasing of the question, I shall refrain from directing you to the relevant nLab pages but instead point you to George Bergman's "An Invitation to General Algebra and Universal Constructions" which is available from his homepage [here](http://math.berkeley.edu/~gbergman/245/). I found this to be a very nice introduction to this subject.
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4
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https://mathoverflow.net/users/45
|
25116
| 16,471 |
https://mathoverflow.net/questions/25042
|
11
|
>
> Suppose that $v\_i$, for $i \in \{1, 2, \ldots 11, 12\}$, are twelve unit length vectors
> based at the origin in $R^3$. Suppose that $|v\_i - v\_j| \geq 1$ for all $i
> \neq j$. What arrangement of the $v\_i$ maximizes the number of pairs $\{i,j\}$
> so that $|v\_i - v\_j| = 1$?
>
>
>
If C is a cube of sidelength $\sqrt{2}$ centered at the origin then we can
place the $v\_i$ at the midpoints of the twelve edges. Taking the convex
hull of the $v\_i$ gives a cube-octahedron of edge-length one. See [here](https://en.wikipedia.org/wiki/Cuboctahedron)
for a picture. If you cut the cubeoctahedron along a hexagonal equator and
rotate the top half by sixty degrees you get another polyhedron. Both of
these have 24 edges. Are these the unique maximal solutions to the above
problem?
Notice that if you place the $v\_i$ at the arguably nicer vertices of a
icosahedron then the $v\_i$ become too widely separated. It is easy to
check this by making a physical model!
I spent some time thinking about areas of spherical polygons and restrictions on the graph of edges (and its dual graph) coming from the Euler characteristic. However, I don't think I got very far - in particular ruling out pentagons seems to be a crucial point that I couldn't deal with. Finally, to explain the problem title: instead of thinking of unit vectors with spacing restrictions, consider the (equivalent) problem of placing twelve identical spherical caps, of radius $\pi/12$, on the unit sphere with disjoint interiors in such a way as to maximize the number of points of tangency.
This question was asked of me by an applied mathematician. It comes from a problem involving packing balls in three-space, minimizing some quantity that is computed by knowing pairwise distances. The solution to the [kissing problem](https://en.wikipedia.org/wiki/Kissing_number_problem) thus justifies the "twelve" appearing in the problem statement. The projection of surrounding balls to a central one gives the spherical caps.
|
https://mathoverflow.net/users/1650
|
Packing twelve spherical caps to maximize tangencies
|
Interesting question. I can find answer using my program, which was made for solving Tammes problem for 13 points. But I need some time for answer.
UPD: I wrote program. Result: 24 is a maximal number of edges.
I did in three steps.
First, I enumerated planar graphs with 12 vertices with at least 25 edges, at most 5 edges in a vertex and at most hexagonal faces.
Total number of suc graphs is 67497.
Second, I eliminated by linear programming by considering values of face angles as variables.
My constrains was:
1. angle in triangle is ~1.2310
2. each angle no less than 1.2310
3. sum of angles around vertex is 2\*pi
4. opposite angles of rectangle are equal
5. sum of non-opposite angles in rectangle between 3.607 and 3.8213
I solve feasibility of this LP problem (with some tolerance)
After this step all graph were eliminated.
|
10
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https://mathoverflow.net/users/4311
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25124
| 16,476 |
https://mathoverflow.net/questions/25123
|
15
|
Long ago, manifolds were embedded subsets of euclidean space defined by polynomials. Later, using the gluing of open sets, people realized they could define manifolds intrinsically. And in certain cases, this lead to new manifolds which could not be realized as subsets of euclidean spaces. Eg non-orientable surfaces cannot be embedded in $R^3$, non-algebraic manifolds etc.
Nowadays, fractals are found as iterations of certain polynomial maps on euclidean space. (or some other ways, as wikipedia says, but still embdedded). Are we missing out on some fractals? Is there an intrinsic definition of fractals? (This would also have to come with an intrinsic definition of dimension. Haudsorff dimension is still an embedded dimensions, using covering of balls in euclidean space.)
|
https://mathoverflow.net/users/nan
|
Is there an intrinsic definition of fractal (i.e. not embedded in euclidean space)?
|
Topological dimension (say, covering dimension) $\dim\_\mathrm{T}$ and Hausdorff dimension $\dim\_\mathrm{H}$ both make sense for metric spaces. **Benoit Mandelbrot** defined $A$ to be a *fractal* iff $\dim\_\mathrm{T} A < \dim\_\mathrm{H} A$. The packing dimension $\dim\_\mathrm{P}$ also makes sense in metric space. **James Taylor** defined $A$ to be a *fractal* iff $\dim\_\mathrm{H} A = \dim\_\mathrm{P} A$. Also making sense for metric space is the definition of **Michael Barnsley** ... A *fractal* is an element of the hyperspace $\mathbb{H}(K)$ of a compact metric space $K$. Perhaps you have your own, different, definition?
Definitions for all these are in my book *Integral, Probability, and Fractal Measures*
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12
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https://mathoverflow.net/users/454
|
25128
| 16,479 |
https://mathoverflow.net/questions/25140
|
3
|
Recently, in a particular problem I was solving, I needed some kind of Radon-Nikodym theorem for measures where one of them is not necessarily absolutely continuous with respect to other.
My colleague informed that he believes that this slightly more general version is true:
Let $\mu$ and $\rho$ be two $\sigma$-finite measures defined one the same measurable space. Then there exists a nonnegative measurable function $f$ such that
$$\rho(A) = \int\_A f\ d\mu + \rho(A\cap B),$$
where $B$ is some measurable set such that $\mu(B)=0$.
However, I cannot prove it and thus convince myself that it is true (even in finite case), nor can I find a counterexample.
Of course, my approach is to use standard Radon-Nikodym theorem, and with its help hypothesis trivially reduces to:
Let $\mu$ and $\rho$ be two $\sigma$-finite measures defined one the same measurable space. Then there exists a measurable set $B$ such that $\mu(B)=0$ and $\rho(\ \cdot\ \cap B^c)$ is absolutely continuous with respect to $\mu$.
|
https://mathoverflow.net/users/6159
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"Radon-Nikodym theorem" for nonabsolute continuous measures
|
You looking for the [Lebesgue's decomposition theorem](http://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem).
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7
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https://mathoverflow.net/users/5542
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25143
| 16,485 |
https://mathoverflow.net/questions/24254
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15
|
Most mathematicians are aware that our species consists of two genders,
denoted for simplicity by the multisets $\lbrace X,X\rbrace$ and $\lbrace X,Y\rbrace$, with
offspring given by $\lbrace A,B\rbrace$ for $A\in\lbrace X,X\rbrace$
and $B\in \lbrace X,Y\rbrace$.
I am asking for the existence of combinatorial structures generalizing this construction.
More precisely, given a finite set $\mathcal C$ of $n$ distinct elements
($n=2$ and $\mathcal C=\lbrace X,Y\rbrace$ in the above example) and an integer $k$, we denote by $\mathcal M\_k$ the set of all multisets containing exactly
$k$ not necessarily distinct elements of $\mathcal C$.
A *gender partition* of a subset $\mathcal S\subset \mathcal M\_k$ is a partition of $\mathcal S$ into $k$ non-empty parts $\mathcal G\_1,\dots,\mathcal G\_k$ called genders such that the following two conditions hold:
(i) Given $(g\_1,\dots,g\_k)\in\mathcal G\_1\times \dots
\times \mathcal G\_k$, every element $(x\_1,\dots,x\_k)\in g\_1\times \dots\times g\_k$ gives rise to a multiset $\lbrace x\_1,\dots,x\_k\rbrace$ which is in $\mathcal S$.
(ii) Every multiset $g\in\mathcal S$ is of the form $\lbrace x\_1,\dots,x\_k\rbrace$
for $(x\_1,\dots,x\_k)\in g\_1\times \dots\times g\_k$ where $g\_i\in
\mathcal G\_i$ are suitable elements.
Examples with $\mathcal C=\lbrace X,Y\rbrace$ are:
(a) $\mathcal S= \lbrace X,X\rbrace\cup \lbrace X,Y\rbrace$ with $\mathcal G\_i$
given by singletons.
(b) $\mathcal S= \lbrace \lbrace X,X\rbrace,\lbrace Y,Y\rbrace\rbrace
\cup \lbrace X,Y\rbrace$ with $\mathcal G\_1=\lbrace \lbrace X,X\rbrace,\lbrace Y,Y\rbrace\rbrace$ and $\mathcal G\_2$ consisting of $\lbrace X,Y\rbrace$.
(c) An example with $k=3$ (easily generalizable to arbitrary values of $k$)
is given by $\mathcal S=\lbrace \lbrace X,X,X\rbrace,\lbrace X,X,Y\rbrace,
\lbrace X,Y,Y\rbrace\rbrace$ with $\mathcal G\_i$ given by singletons.
More examples of gender partitions $\mathcal S=\mathcal G\_1\cup
\dots\cup \mathcal G\_k$ are fairly easy to construct. (And there are fairly easy
notions for "products", "quotients", one can split an element of $\mathcal C$
into several new elements, etc.)
The following additional condition is more difficult to satisfy:
Call a gender partition $\mathcal S=\mathcal G\_1\cup \dots\cup\mathcal G\_k$
*balanced* if $\mathcal S$ admits a stationary probability measure
$\mu$ giving equal weight $\frac{1}{k}$ to all genders $\mathcal G\_i$.
A probability measure $\mu$ on $\mathcal S$ is *stationary* if
the probabiliy $\mu(\lbrace x\_i,\dots,x\_k\rbrace)$ of every offspring
of $(g\_1,\dots,g\_k)$ (with respect to uniform choices for $x\_i\in g\_i$)
is proportional to $\prod\_{i=1}^k\mu(g\_i)$. (Stationary probability
measures exist always and are unique if $\mathcal S$ is
minimal in some sense.)
Example: The examples (a) and (b) above are balanced, (c) is not balanced.
Question: Produce other examples of balanced gender partitions.
Is there for example a balanced gender partition for $k=3$?
Remark: One can also consider probabilities on offsprings
which depend on the choice of $x\_i\in g\_i$. Example (c) is not balanced
even in this more general framework.
Variation: Instead of working with multisets, one can also work with
sequences of length $k$. An offspring of $k$ sequences
$g\_1=(g\_1(1),\dots,g\_1(k)),\dots,g\_k=(g\_k(1),\dots,g\_k(k))$ with
$g\_i\in\mathcal G\_i$ is then given by
$g\_1(\sigma(1)),\dots,g\_k(\sigma(k))$ where $\sigma$ is a (not necessarily arbitrary) permutation of
$\lbrace 1,\dots,k\rbrace$.
|
https://mathoverflow.net/users/4556
|
Procreation with several genders
|
So, I hope I understand the definitions correctly. Here's a way to construct an example with $k = 3$ genders (say A, B and C) using $n = 9$ sex chromosomes, which I will take to be the elements of $\mathbb{Z}/9\mathbb{Z}$: start with all 165 multisets of chromosomes unused. Choose any unused multiset $\{x, y, z\}$, and add it, together with the multisets $\{x + 3, y + 3, z + 3\}$ and $\{x + 6, y + 6, z + 6\}$, to gender A, add the multisets $\{x + 1, y + 1, z + 1\}, \{x + 4, y + 4, z + 4\}, \{x + 7, y + 7, z + 7\}$ to gender B and add the multisets $\{x + 2, y + 2, z + 2\}, \{x + 5, y + 5, z + 5\}, \{x + 8, y + 8, z + 8\}$ to gender C. Continue until every multiset of chromosomes is used up. The resulting genders (each consisting of 55 multisets of chromosomes) are symmetric and so (if I'm not mistaken) are balanced.
It's easy to see how to construct a wide variety of similar gender partitions for given $k$ if we may choose $n$ appropriately. These partitions have nice symmetry and use all possible multisets of chromosomes. This says nothing at all about, say, constructing gender partitions for $n = 2$ (though I believe that I've confirmed by case analysis that there are no balanced gender partitions for $k = 3$ and $n = 2$).
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6
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https://mathoverflow.net/users/4658
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25146
| 16,488 |
https://mathoverflow.net/questions/25141
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2
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This is an edited post of a post I made on sci.math (e.g. to fit MO markup) with
an elementary question on vector measures. Since it is almost a week and I have
received no answers, I am trying here. Below, I will "prove" a theorem that is
false (because it has a simple counter example) but I cannot find where the flaw
in the proof is. This is an elementary question, so I am not sure if it fits MO
standards. If it does not, feel free to close it down -- I will leave that
judgement to the moderators. I am also certain that once someone points out the
hole in my argument I am sure to slap myself on the forehead, cry out what a
complete idiot I am and vow not to show my face in public for the next few
years. But at this point, this problem is driving me bonkers, and I much prefer
my sanity over my reputation (if I have any).
WARNING: long post ahead.
SOME BACKGROUND: All Banach spaces are over the real field. Let $\Omega$ be a
Boolean algebra. In a harmless abuse of notation, the top element of $\Omega$
will be denoted by $\Omega$. By a measure on $\Omega$ I mean a *finitely
additive* map $u$ on $\Omega$ with values in a Banach space. If the codomain $B$
of $u$ is the real field I will call it a scalar measure. Note in particular
that (scalar) measures do *not* take infinite values. Neither
$\sigma$-completeness nor $\sigma$-additivity will be used anywhere (I told you
it was an elementary question).
Recall that if $u$ is a measure, then its *variation* $|u|$ is the (possibly
infinite) quantity,
$$|u|: E\mapsto \sup \{\sum\_{F\in \mathcal{E}}\|u(F)\|\}$$
where the supremum is taken over the set of all finite partitions $\mathcal{E}$
of $E$. Assuming $|u|$ is finite for every $E$, then $|u|$ is a scalar positive
measure (and therefore monotone and bounded).
On the other hand, the *semivariation* $\|u\|$ of $u$ is defined to be the
(possibly infinite) quantity,
$$\|u\|: E\mapsto \sup\{|b^{\ast}u|(E)\}$$
where the supremum is taken over all $b^{\ast}$ in the unit ball of the dual
space $B^{\ast}$. To avoid any possible misunderstandings, $|b^{\ast}u|$ is the
variation of the scalar measure $E\mapsto b^{\ast}(u(E))$.
Assuming $u$ has finite (or bounded) semivariation, that is, $\|u\|(E)$ is
finite for every $E$, then the semivariation is positive, monotone and
subadditive. An application of Hahn-Banach yields that for every $E$:
(A) $\|u(E)\| \leq \|u\|(E)$
Less trivial is the fact that $u$ has finite semivariation iff it is bounded.
This is a consequence of the fundamental inequality for the semivariation. Since
I will not need this inequality, I will just direct you to proposition 11, pg. 4
of the Diestel-Uhl monograph Vector Measures. Another elementary fact is that if
the codomain $B$ is finite-dimensional then the variation and the semivariation
agree (note: this fails in every infinite-dimensional space by
Dvoretzky-Rogers).
EXAMPLE: Let $\Omega$ be a Boolean algebra with an infinite partition of unity
(note: infinite cardinality of $\Omega$ is enough to guarantee this). Consider
the map $\chi:\Omega\to \mathbf{L}^{\infty}(\Omega)$ given by $E \mapsto
\chi(E)$, where $\chi(E)$ is the characteristic function of $E$. Then $\chi$ is
finitely additive and bounded but it is easy to see that its variation is
unbounded.
If you are wondering what is $\mathbf{L}^{\infty}(\Omega)$ for a general Boolean
algebra, suffice to say that such a space can indeed be constructed and with all
the right properties, but to not tarry too long, just take $\Omega$ to be the
power set of $\mathbb{N}$ and replace $\mathbf{L}^{\infty}(\Omega)$ with the
Banach space of bounded real-valued functions on $\mathbb{N}$ with the supremum
norm.
Now, I am going to "prove" that every bounded measure $v$ has bounded variation
in obvious contradiction with the above example. This will be done by
constructing a control measure for $v$ in a very special way. Since what I need
is for someone to tell me where and why I have gone astray, I am going to detail
the argument, even to the point of pedantry.
ARGUMENT: Denote by $\mathbf{BA}(\Omega)$ the space of bounded scalar measures
on $\Omega$. One can introduce a partial order on $\mathbf{BA}(\Omega)$ by
taking the pointwise order:
$$u \leq v \mbox{ iff } u(E) \leq v(E) \mbox{ for all } E \mbox{ in }\Omega$$
It is easy to see that with the pointwise order $\mathbf{BA}(\Omega)$ is a
partially ordered linear space.
Note: For partially ordered linear spaces, Banach lattices, etc. I will take as
my reference chapter 5, volume 3 of Fremlin's 5-volume work on measure theory.
It is available online, so googling will easily get you to it.
We can also put a norm on $\mathbf{BA}(\Omega)$ by taking the total variation,
that is, $\|u\| = |u|(\Omega)$.
Note: the context should make clear when $\|,\|$ denotes the norm of an element
of a Banach space or the semivariation.
Lemma 1: The space $\mathbf{BA}(\Omega)$ with the pointwise order and total
variation norm is a Banach lattice.
Proof: This can also be found in the books. One proof proceeds by noting that
$\mathbf{BA}(\Omega)$ is the dual of $\mathbf{L}^{\infty}(\Omega)$, which is a
Banach lattice. A (sketch of a) more direct proof goes as follows. Completeness
of $\mathbf{BA}(\Omega)$ is straightforward because only finite additivity is
involved. By elementary results on partially ordered linear spaces, to prove the
existence of arbitrary binary suprema and infima it suffices to prove the
existence of the supremum $\sup\{u, 0\}$ for every $u$ (the *positive part*
$u^{+}$ of $u$). This is given by $u^{+}: E \mapsto \sup\{u(F): F\leq E\}$. It
can also be seen that the *absolute value* $|u|$ of u defined by $\sup\{u, -u\}$
is just the variation of u, so that first, there is no ambiguity in my notation,
and second, the Banach lattice condition
(B) if $|u| \leq |v|$ then $\|u\| \leq \|v\|$
is trivially satisfied. Q. E. D.
The cone of positive elements of $\mathbf{BA}(\Omega)$ will be denoted simply by
$P$. Note that if $u$ is positive then it coincides with its variation.
Next, a lemma relating boundedness in the norm with boundedness for the
pointwise order.
Lemma 2: Let $A$ be a subset of $P$. Then $A$ is order-bounded iff it is
norm-bounded.
Proof: The direct implication follows from the Banach lattice condition (B).
For the converse implication, we prove the contrapositive. So suppose $A$ is
not order-bounded. Pick a non-zero positive $v$ and consider the sequence of
positive measures $v\_n = (n v)/\|v\|$. Since $A$ is not order-bounded there is
$u\_n$ in $A$ such that $u\_n \geq v\_n$ for every $n$. By the Banach lattice
condition (B) it follows that $\|u\_n\| \geq \|v\_n\| = n$, so that $A$ is not
norm-bounded. Q. E. D.
The next two lemmas amount to a proof that norm-bounded subsets of the positive
cone have a supremum. The structure of the proof is fairly standard and is
patterned after proofs of similar facts in other contexts (e.g. the fact that a
category has all coproducts if it has finite coproducts and filtered colimits).
Lemma 3: Let $(u\_i)$ be a norm-bounded, monotone net of positive measures. Then
the pointwise limit
(C) $E \mapsto \lim\_i u\_i(E) = \sup\{u\_i(E)\}$
exists and defines a map $u$ that is bounded, finitely additive and the supremum
of $(u\_i)$.
Proof: Since each $u\_i$ is positive (and therefore monotone) and $(u\_i)$ is
norm-bounded, by a constant $C$ say, then, $u\_i(E) \leq u\_i(\Omega) \leq C$ and
the supremum $\sup \{u\_i(E)\}$ exists. Since the net is monotone, this
supremum is just $\lim\_i u\_i(E)$ from which it follows that $u(E) = \lim\_i
u\_i(E)$ is finitely additive. Since it is positive, it is monotone and therefore
bounded by $u(\Omega) = \lim\_i u\_i(\Omega) \leq C$. Since the order is the
pointwise order, $u$ is the supremum of $(u\_i)$. Q. E. D.
Lemma 4: If $A$ is a norm-bounded subset of $P$ then it has a supremum.
Proof: By lemma 2, $A$ is order-bounded, by $v$ say. Consider the net $J
\mapsto \sup J$ where $J$ runs over the filtered partial order of the finite
subsets of $A$. The existence of $\sup J$ for every finite $J$ is guaranteed by
lemma 1. Since $v$ bounds $A$, it follows that $v$ bounds $\sup J$ for every
$J$. Since the net is monotone and norm-bounded by $\|v\|$, lemma 3 gives us its
supremum. A simple check proves that this supremum is precisely the supremum of
$A$. Q. E. D.
Now, let $v:\Omega\to B$ be a bounded measure. Since it is bounded, it has
finite semivariation. This implies that the set of positive measures
$\{|b^{\ast}v|\}$, where $b^{\ast}$ ranges over the unit ball of $B^{\ast}$,
is norm-bounded. By lemma 4, it has a supremum $u$. By the very definition of
the order structure, we have for every $E$, $|b^{\ast}v|(E) \leq u(E)$ and thus
by definition of the semivariation:
$$\|v\|(E) \leq u(E)$$
But by inequality (A) we have,
$$\|v(E)\| \leq u(E)$$
and this implies that $v$ has bounded variation!!!!!! For if $\{E\_n\}$ is a
finite partition of $E$ then
$$\sum\_n \|v(E\_n)\| \leq \sum\_n u(E\_n) = u(E) \leq u(\Omega)$$
Can anyone help me out here and point out where and why is my argument screwed?
Regards and thanks in advance,
G. Rodrigues
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https://mathoverflow.net/users/2562
|
Elementary vector measure question: what am I doing wrong?
|
Lemma 2 is false. Consider, e.g., a sequence of probability measures with mutually disjoint supports.
The problem with your argument is that just because $A$ is not ordered bounded you do not get elements of $A$ that are larger than a given measure. After all, $P$ is only partially ordered; not linearly ordered.
|
4
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https://mathoverflow.net/users/2554
|
25147
| 16,489 |
https://mathoverflow.net/questions/25132
|
8
|
A wikipedia page/paragraph on [ℵ₁](http://en.wikipedia.org/wiki/Aleph-null#Aleph-one) states:
1. "The definition of ℵ₁ implies (in
ZF, [Zermelo-Fraenkel set theory](http://en.wikipedia.org/wiki/Zermelo-Fraenkel_set_theory)
without the axiom of choice) that no
cardinal number is between ℵ₀ and
ℵ₁."
2. "If the [axiom of choice](http://en.wikipedia.org/wiki/Axiom_of_choice) (AC) is
used, it can be further proved that
the class of cardinal numbers is
totally ordered, and thus ℵ₁ is the
second-smallest infinite cardinal
number."
Can someone point me at a lay explanation of this please?
Is it simply saying that ℵ½'s existence is up to definition, or choice? And this has been shown by the [axiom of choice](http://en.wikipedia.org/wiki/Axiom_of_choice)?
|
https://mathoverflow.net/users/6156
|
Cardinality: Why is there no "ℵ½"?
|
The point is that without the Axiom of Choice, cardinalities are not linearly ordered, and it is possible under $\neg AC$ that there are additional cardinalities to the side of the $\aleph$'s. Thus, the issues is not additional cardinalities between $\aleph\_0$ and $\aleph\_1$, but rather additional cardinalities to the side, incomparable with these cardinalities.
Let me explain. We say that two sets $A$ and $B$ are *equinumerous* or *have the same cardinality* if there is a bijection $f:A\to B$. We say that $A$ has smaller-or-equal cardinality than $B$ if there is an injection $f:A\to B$. It is provable (without AC) that $A$ and $B$ have the same cardinality if and only if each is smaller-or-equal to the other (this is the Cantor-Shroeder-Bernstein theorem).
Under AC, every set is bijective with an ordinal, and so we may use these ordinals to select canonical representatives from the equinumerosity classes. Thus, under AC, the $\aleph\_\alpha$'s form all of the possible infinite cardinalities.
But when AC fails, the cardinalities are not linearly ordered (the linearity of cardinalities is equivalent to AC). Let me mention a few examples:
* It is a consequence of the Axiom of Determinacy that there is no $\omega\_1$ sequence of distinct reals. Thus, in any model of AD, the cardinality of the reals is uncountable, but incomparable to $\aleph\_1$. Thus, in such a model, it is no longer correct to say that $\aleph\_1$ is the smallest uncountable cardinal. One should say instead that $\aleph\_1$ is the smallest uncountable well-orderable cardinal.
* A more extreme example is provided by the Dedekind finite infinite sets. These sets are not finite, but also not bijective with any proper subset. It follows that they can have no countably infinite subsets. In particular, they are uncountable sets, but their cardinality is incomparable with $\omega$. Thus, in a model of $\neg AC$ having a Dedekind finite infinite set, it is no longer correct to say that $\aleph\_0$ is the smallest infinite cardinal.
Thus, the issue isn't whether there is something between $\aleph\_0$ and $\aleph\_1$, but rather, whether there are additional cardinalities to the side of these cardinalities.
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41
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https://mathoverflow.net/users/1946
|
25152
| 16,493 |
https://mathoverflow.net/questions/24813
|
5
|
Let $M$ be a compact, finite-dimensional Riemannian manifold, let $T: M \rightarrow M$ be an [Anosov diffeomorphism](http://en.wikipedia.org/wiki/Anosov_diffeomorphism), and let $\mu$ be a [Sinai-Ruelle-Bowen (probability) measure](http://www.scholarpedia.org/article/Hyperbolic_dynamics#Measure-theoretic_properties). Write $\mathcal{R} = \{ R\_1,\dots,R\_n \}$ for a [Markov partition](http://en.wikipedia.org/wiki/Markov_partition); write $p\_j^{(\mathcal{R})} := \mu(R\_j)$.
**Question:**
Does there ever/always exist $\mathcal{R}$ s.t. $p^{(\mathcal{R})}$ is a nontrivial uniform measure on $\{1,\dots,n\}$? If not, does there ever/always exist a sequence of partitions $\mathcal{R}\_m$ s.t. $p^{(\mathcal{R}\_m)}$ converges to a uniform measure in some nontrivial sense?
|
https://mathoverflow.net/users/1847
|
Do there exist Markov partitions with (nearly) uniform SRB measures?
|
Since (p\_1....,p\_n) is an eigenvector of the transition matrix, your uniformity assumption is satisfied iff the transition matrix associated to the partition is bistochastic. I guess this is rarely the case.
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2
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https://mathoverflow.net/users/6129
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25153
| 16,494 |
https://mathoverflow.net/questions/25071
|
2
|
I believe that "fewer than $\alpha$ variables" is equivalent to "every subformula has fewer than $\alpha$ free variables." The left-to-right implication is straightforward, and from right to left one can always rename variables. For example, the following trivial (and not very useful) theorem of ZFC has no more than two free variables in any subformula:
$$(\exists x) (\forall y) (y\notin x\ \&\ (\exists z) y\in z))$$
It uses three distinct variables (x, y, and z), but because no subformula contains more than two free variables, we can rewrite it to the equivalent:
$$(\exists x) (\forall y) (y\notin x\ \&\ (\exists x) y\in x))$$
So, are the two formulations of the restriction equivalent? If so, why isn't the latter formulation used more often in papers on bounded-variable logic? It seems to have the advantage that it doesn't impose a restriction on the (irrelevant) choice of *which* variables one uses, limiting only the number that appear free in any given subformula.
Thanks!
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https://mathoverflow.net/users/2361
|
Bounded-variable logic: "fewer than $\alpha$ variables" equivalent to "every subformula has fewer than $\alpha$ free variables"?
|
If you are speaking of infinitary logic, which your notation (and your other question) suggests, then the statement is not true. Take the case $\alpha=\omega$. Suppose that $\varphi\_n$ is a sentence that uses $n$ variables, and cannot be expressed equivalently with fewer than $n$ variables. But $\varphi\_n$ has no free variables. Consider the statement $\bigwedge\_n \varphi\_n$. This is a sentence, having no free variables, and it uses $\omega$ many variables. However, its subformulas are itself (with no free variables), plus the subformulas of any $\varphi\_n$, which each have at most $n$ free variables.
Thus, this assertion has the property that every subformula has fewer than $\omega$ free variables, but the assertion does not use fewer than $\omega$ variables. So it seems to be a negative example to your conjecture.
To take a specific example, let $\varphi\_n$ assert "there are at least $n$ distinct objects", which can be expressed in a finite formula using $n$ variables. The conjunction $\bigwedge\_n \varphi\_n$ asserts that the universe is infinite.
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2
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https://mathoverflow.net/users/1946
|
25159
| 16,497 |
https://mathoverflow.net/questions/25176
|
1
|
I'm modeling a game tech/build tree as a directed acyclic graph with a .dot file for visualization use in Graphviz.
Some of the dependencies discovered are redundant in the sense that while they are dependencies, they are satisfied via a longer yet required path.
```
a -> b
b -> c
a -> c // Unnecessary because we have to do b first.
```
And a longer example
```
a -> b
b -> c
c -> d
a -> d // Unnecessary between we have to do both b and c first.
```
Is there an algorithm to testing a graph for these unnecessary paths so that I could trim them from the .dot file? Perhaps this is more appropriately a programming question, but I'm guessing some use of graph theory applies here.
|
https://mathoverflow.net/users/6166
|
Remove unnecessary dependencies in a task graph?
|
AFAICT, what you want is called a [transitive reduction](http://en.wikipedia.org/wiki/Transitive_reduction) of the graph. La Wik claims that Graphviz can do the job somehow; consult its documentation.
|
2
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https://mathoverflow.net/users/6167
|
25177
| 16,508 |
https://mathoverflow.net/questions/25161
|
84
|
Background: When I first took measure theory/integration, I was bothered by the idea that the integral of a real-valued function w.r.t. a measure was defined first for nonnegative functions and only then for real-valued functions using the crutch of positive and negative parts (and only then for complex-valued functions using their real and imaginary parts). It seemed like a strange starting point to make the theory dependent on knowledge of the nonnegative function case when this certainly isn't necessary for Riemann integrals or infinite series: in those cases you just take the functions or sequences as they come to you and put no bias on positive or negative parts in making the definitions of integrating or summing.
Later on I learned about integration w.r.t a measure of Banach-space valued functions in Lang's Real and Functional Analysis. You can't break up a Banach-space valued function into positive and negative parts, so the whole positive/negative part business has to be tossed aside as a foundational concept. At the *end* of this development in the book Lang isolates the special aspects of integration for nonnegative real-valued functions (which potentially could take the value $\infty$). Overall it seemed like a more natural method.
Now I don't think a first course in integration theory has to start off with Banach-space valued functions, but there's no reason you couldn't take a cue from that future generalization by developing the real-valued case in the same way Banach-space valued functions are handled, thereby avoiding the positive/negative part business as part of the initial steps.
Finally my question: Why do analysts prefer the positive/negative part foundations for integration when there is a viable alternative that doesn't put any bias on which function values are above 0 or below 0 (which seems to me like an artificial distinction to make)?
Note: I know that the Lebesgue integral is an "absolute" integral, but I don't see that as a justification for making the very definition of the integral require treatment of nonnegative functions first. (Lang's book shows it is not necessary. I know analysts are not fond of his books, but I don't see a reason that the method he uses, which is just copying Bochner's development of the integral, should be so wildly unpopular.)
|
https://mathoverflow.net/users/3272
|
Why is Lebesgue integration taught using positive and negative parts of functions?
|
It's really the difference between two kinds of completions:
1. An order-theoretic completion. For this, it's easiest to start with non-negative functions, and have infinite values dealt with pretty naturally.
2. A metric completion. For this, it's more natural to start with finite-valued signed simple functions.
It's not exactly that simple -- historically, signed simple functions (well, actually, I think they used step functions) were used in an order-theoretic treatment by Riesz and Nagy. But I think this is a good way to look at the two ways of approaching this integral.
And needless to say, these two approaches generalize in two different contexts. They are both interesting and illuminate somewhat different aspects of the Lebesgue integral, even on the real line. For instance, the order-theoretic approach leads quickly to results such as the monotone convergence and bounded convergence theorems, while the metric approach leads naturally to the topology of convergence in measure and completeness of the $L\_p$ spaces.
|
51
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https://mathoverflow.net/users/6172
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25193
| 16,520 |
https://mathoverflow.net/questions/25170
|
2
|
It is not uncommon to describe interesting classes of field extensions by declaring that an extension $L|K$ belongs to that class if some type of problem with $K$-coefficiens has a property over $L$ if and only if it has the same property over $K$. I wonder about the following variant:
>
> **Question A**: For which field extensions $L|K$ the following is true?: Given finite dimensional $K$-vector spaces $U,V,W$, a $K$-bilinear form $\beta\_K:U\times V\to W$ is surjective if and only if the corresponding $L$-bilinear form $\beta\_L$ obtained by scalar extension is surjective.
>
>
>
Already in characteristic zero an answer to that would be nice. Also, I wonder for which $L|K$ surjectivity of $\beta\_K$ implies or is implied by surjectivity of $\beta\_L$.
One can take the geometric point of view: A bilinear form induces a map between associated projective spaces and one asks here for surjectivity of these maps on $K$ or $L$-rational points.
It is not hard to show that if $L$ is the reals or the $p$-adics, then surjectivity of a bilinear form over the rationals implies surjectivity with $L$-coefficients (the argument really uses both, density *and* local compactness). This is the setting in which the problem originally arised. I was also asking the following
>
> **Question B**: Given a bilinear form $\beta$ between finite dimensional $\mathbb Q$-vector spaces, is it true that $\beta$ is surjective if and only if for all primes $p$ (including $p=\infty$) the induced $\mathbb Q\_p$-bilinear form is surjective.
>
>
>
The answer to that is negative, see Poonen's explicit example below.
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https://mathoverflow.net/users/5952
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Surjectivity of bilinear forms.
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The answer to Question B is **no**, as I'll show below.
Let $U=V=\mathbf{Q}^3$ and $W=\mathbf{Q}^4$. Define
$$\beta((u\_1,u\_2,u\_3),(v\_1,v\_2,v\_3))=(u\_1 v\_1,u\_2 v\_2,u\_3 v\_3, (u\_1+u\_2+u\_3)(v\_1+v\_2+v\_3)).$$
**Claim 1:** $\beta$ is not surjective.
**Proof:** In fact, we will show that $(1,1,1,-1)$ is not in the image. If it were, then we would have a solution to
$$(u\_1+u\_2+u\_3)(u\_1^{-1}+u\_2^{-1}+u\_3^{-1})=-1.$$
Clearing denominators leads to an elliptic curve in $\mathbf{P}^2$, but MAGMA shows that all its rational points lie in the lines where some coordinate vanishes.
**Claim 2:** After base extension to any completion $k$ of $\mathbf{Q}$, the bilinear map $\beta$ becomes surjective.
**Proof:** Given $(a\_1,a\_2,a\_3,b) \in k^4$, we need to show that it is in the image. If $a\_1=0$, then set $u\_1=0$, $u\_2=1$, $u\_3=1$, $v\_2=a\_2$, $v\_3=a\_3$, and then solve for $v\_1$ in the remaining constraint. The same argument works if $a\_2=0$ or $a\_3=0$. If $a\_1,a\_2,a\_3$ are all nonzero, then we must find a solution to
$$(u\_1+u\_2+u\_3)(a\_1 u\_1^{-1}+a\_2 u\_2^{-1}+a\_3 u\_3^{-1})=b.$$
Clearing denominators leads to the equation of a projective plane curve
with a smooth $k$-point $(1:-1:0)$, so by the implicit function theorem
there exist nearby $k$-points with $u\_1,u\_2,u\_3$ all nonzero,
which gives us the solution we needed. $\square$
If you want a reasonable answer to Question A, I'd suggest that you try to make it more focused.
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Related MO questions: [What is the general opinion on the Generalized Continuum Hypothesis?](https://mathoverflow.net/questions/14338/what-is-the-general-opinion-on-the-generalized-continuum-hypothesis) ; [Completion of ZFC](https://mathoverflow.net/questions/46907/completion-of-zfc) ; [Complete resolutions of GCH](https://mathoverflow.net/questions/191139/complete-resolutions-of-gch;) [How far wrong could the Continuum Hypothesis be?](https://mathoverflow.net/questions/218407/how-far-wrong-could-the-continuum-hypothesis-be;) [When was the continuum hypothesis born?](https://mathoverflow.net/questions/135912/when-was-the-continuum-hypothesis-born)
Background
----------
The [Continuum Hypothesis](https://en.wikipedia.org/wiki/Continuum_hypothesis) (CH) posed by Cantor in 1890 asserts that $ \aleph\_1=2^{\aleph\_0}$. In other words, it asserts that every subset of the set of real numbers that contains the natural numbers has either the cardinality of the natural numbers or the cardinality of the real numbers. It was the first problem on the 1900 Hilbert's list of problems. The generalized continuum hypothesis asserts that there are no intermediate cardinals between every infinite set X and its power set.
Cohen proved that the CH is independent from the axioms of set theory. (Earlier Goedel showed that a positive answer is consistent with the axioms).
Several mathematicians proposed definite answers or approaches towards such answers regarding what the answer for the CH (and GCH) should be.
The question
------------
>
> My question asks for a description and explanation of the various approaches to the continuum hypothesis in a language which could be understood by non-professionals.
>
>
>
More background
---------------
I am aware of the existence of 2-3 approaches.
One is by Woodin described in two 2001 Notices of the AMS papers ([part 1](http://www.ams.org/notices/200106/fea-woodin.pdf), [part 2](http://www.ams.org/notices/200107/fea-woodin.pdf)).
Another by Shelah (perhaps in [this paper entitled "The Generalized Continuum Hypothesis revisited "](https://arxiv.org/abs/math/9809200)). See also the paper entitled "[You can enter Cantor paradise](https://arxiv.org/PS_cache/math/pdf/0102/0102056v1.pdf)" (Offered in Haim's answer.);
There is a [very nice presentation by Matt Foreman](http://www.math.helsinki.fi/logic/LC2003/presentations/foreman.pdf) discussing Woodin's approach and some other avenues. Another description of Woodin's answer [is by Lucca Belloti](http://www2.units.it/episteme/L&PS_Vol3No1/bellotti/bellotti-html/bellotti_L&PS_Vol3No1.html) (also [suggested by Haim](https://mathoverflow.net/questions/23829/solutions-to-the-continuum-hypothesis/23831#23831)).
The proposed answer $ 2^{\aleph\_0}=\aleph\_2$ goes back according to François to [Goedel](https://books.google.com/books?id=gDzbuUwma5MC&lpg=PP1&dq=collected%2520works%2520godel&pg=PA405#v=onepage&q=collected%2520works%2520godel&f=false). It is (perhaps) mentioned in Foreman's presentation. (I heard also from Menachem Magidor that this answer might have some advantages.)
François G. Dorais mentioned an important paper by Todorcevic's entitled "[Comparing the Continuum with the First Two Uncountable Cardinals](http://www.math.toronto.edu/%7Estevo/continuum.ps)".
There is also a very rich theory (PCF theory) of cardinal arithmetic which deals with what can be proved in ZFC.
### Remark:
I included some information and links from the comments and answer in the body of question. What I would hope most from an answer is some friendly elementary descriptions of the proposed solutions.
---
There are by now a couple of long detailed excellent answers (that I still have to digest) by **Joel David Hamkins** and by **Andres Caicedo** and several other useful answers. (Unfortunately, I can accept only one answer.)
**Update** (February 2011): A new detailed answer was contributed by **Justin Moore**.
**Update** (Oct 2013) A user 'none' gave a link to [an article by Peter Koellner about the current status of CH](https://web.archive.org/web/20120403232634/http://logic.harvard.edu/EFI_CH.pdf):
**Update** (Jan 2014) A related popular article in "Quanta:" [To settle infinity dispute a new law of logic](https://www.simonsfoundation.org/quanta/20131126-to-settle-infinity-question-a-new-law-of-logic/)
**(belated) update**(Jan 2014) Joel David Hamkins links in a comment from 2012 a very interesting paper [Is the dream solution to the continuum hypothesis attainable](https://arxiv.org/abs/1203.4026) written by him about the possibility of a "dream solution to CH." A link to the paper and a short post [can be found here](http://jdh.hamkins.org/dream-solution-of-ch/).
**(belated) update** (Sept 2015) Here is a link to an interesting article: [Can the Continuum Hypothesis be Solved?](https://www.ias.edu/about/publications/ias-letter/articles/2011-fall/continuum-hypothesis-kennedy) By Juliette Kennedy
**Update** A videotaped lecture [The Continuum Hypothesis and the search for Mathematical Infinity](https://www.youtube.com/watch?v=nVF4N1Ix5WI) by Woodin from January 2015, with reference also to his changed opinion. (added May 2017)
**Update (Dec '15):** A very nice answer was added (but unfortunately deleted by owner, (2019) now replaced by a new answer) by Grigor. Let me quote its beginning (hopefully it will come back to life):
"One probably should add that the continuum hypothesis depends a lot on how you ask it.
1. $2^{\omega}=\omega\_1$
2. Every set of reals is either countable or has the same size as the continuum.
To me, 1 is a completely meaningless question, how do you even experiment it?
If I am not mistaken, Cantor actually asked 2..."
**Update** A 2011 videotaped lecture by Menachem Magidor: [Can the Continuum Problem be Solved?](https://youtu.be/3b_wbObaNgA) (I will try to add slides for more recent versions.)
**Update** (July 2019) Here are [slides of 2019 Woodin's lecture](https://www.uni-muenster.de/imperia/md/content/MathematicsMuenster/woodin-opening-colloquium.pdf) explaining his current view on the problem. (See the answer of Mohammad Golshani.)
**Update** (Sept 19, 2019) Here are [videos of the three 2016 Bernay's lectures by Hugh Woodin](https://gess.ethz.ch/en/news-and-events/paul-bernays-lectures/bernays-2016.html) on the continuum hypothesis and also the videos of the [three 2012 Bernay's lectures on the continuum hypothesis and related topics by Solomon Feferman](https://gess.ethz.ch/en/news-and-events/paul-bernays-lectures/bernays-2012.html).
**Update** (Sept '20) Here are [videos of the three 2020 Bernays' lectures](https://video.ethz.ch/speakers/bernays/2020.html) by Saharon Shelah on the continuum hypothesis.
**Update** (May '21) In a [new answer](https://mathoverflow.net/a/391766/1532), Ralf Schindler gave a link to his 2021 videotaped lecture in Wuhan, describing a result with David Asperó that shows a relation between two well-known axioms. It turns out that Martin's Maximum$^{++}$ implies Woodin's ℙ$\_{max}$ axiom. Both these axioms were known to imply the $\aleph\_2$ answer to CH. A link to the paper: <https://doi.org/10.4007/annals.2021.193.3.3>
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https://mathoverflow.net/users/1532
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Solutions to the Continuum Hypothesis
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Since you have already linked to some of the contemporary
primary sources, where of course the full accounts of those
views can be found, let me interpret your question as a
request for summary accounts of the various views on CH.
I'll just describe in a few sentences each of what I find
to be the main issues surrounding CH, beginning with some
historical views. Please forgive the necessary simplifications.
**Cantor.** Cantor introduced the Continuum Hypothesis
when he discovered the transfinite numbers and proved that
the reals are uncountable. It was quite natural to inquire
whether the continuum was the same as the first uncountable
cardinal. He became obsessed with this question, working on
it from various angles and sometimes switching opinion as
to the likely outcome. Giving birth to the field of
descriptive set theory, he settled the CH question for
closed sets of reals, by proving (the Cantor-Bendixon
theorem) that every closed set is the union of a countable
set and a perfect set. Sets with this perfect set property
cannot be counterexamples to CH, and Cantor hoped to extend
this method to additional larger classes of sets.
**Hilbert.** Hilbert thought the CH question so
important that he listed it as the first on his famous list
of problems at the opening of the 20th century.
**Goedel.** Goedel proved that CH holds in the
constructible universe $L$, and so is relatively consistent
with ZFC. Goedel viewed $L$ as a device for establishing
consistency, rather than as a description of our (Platonic)
mathematical world, and so he did not take this result to
settle CH. He hoped that the emerging large cardinal
concepts, such as measurable cardinals, would settle the CH
question, and as you mentioned, favored a solution of the form $2^\omega=\aleph\_2$.
**Cohen.** Cohen introduced the method of forcing and
used it to prove that $\neg$CH is relatively consistent
with ZFC. Every model of ZFC has a forcing extension with
$\neg$CH. Thus, the CH question is independent of ZFC,
neither provable nor refutable. Solovay observed that CH
also is forceable over any model of ZFC.
**Large cardinals.** Goedel's expectation that large
cardinals might settle CH was decisively refuted by the
Levy-Solovay theorem, which showed that one can force
either CH or $\neg$CH while preserving all known large
cardinals. Thus, there can be no direct implication from
large cardinals to either CH or $\neg$CH. At the same
time, Solovay extended Cantor's original strategy by
proving that if there are large cardinals, then increasing
levels of the projective hierarchy have the perfect set
property, and therefore do not admit counterexamples to CH.
All of the strongest large cardinal axioms considered today
imply that there are no projective counterexamples to CH. This can be seen as a complete affirmation of Cantor's original strategy.
**Basic Platonic position.** This is the realist view
that there is Platonic universe of sets that our axioms are
attempting to describe, in which every set-theoretic
question such as CH has a truth value. In my experience,
this is the most common or orthodox view in the
set-theoretic community. Several of the later more subtle
views rest solidly upon the idea that there is a fact of
the matter to be determined.
**Old-school dream solution of CH.** The hope was that
we might settle CH by finding a new set-theoretic principle
that we all agreed was obviously true for the intended
interpretation of sets (in the way that many find AC to be
obviously true, for example) and which also settled the CH
question. Then, we would extend ZFC to include this new
principle and thereby have an answer to CH. Unfortunately,
no such conclusive principles were found, although there
have been some proposals in this vein, such as [Freilings
axiom of symmetry](https://en.wikipedia.org/wiki/Freiling%27s_axiom_of_symmetry).
**Formalist view.** Rarely held by mathematicians,
although occasionally held by philosophers, this is the
anti-realist view that there is no truth of the matter of
CH, and that mathematics consists of (perhaps meaningless)
manipulations of strings of symbols in a formal system. The
formalist view can be taken to hold that the independence
result itself settles CH, since CH is neither provable nor
refutable in ZFC. One can have either CH or $\neg$CH as
axioms and form the new formal systems ZFC+CH or
ZFC+$\neg$CH. This view is often mocked in straw-man form,
suggesting that the formalist can have no preference for CH
or $\neg$CH, but philosophers defend more subtle versions,
where there can be reason to prefer one formal system to
another.
**Pragmatic view.** This is the view one finds in
practice, where mathematicians do not take a position on
CH, but feel free to use CH or $\neg$CH if it helps their
argument, keeping careful track of where it is used.
Usually, when either CH or $\neg$CH is used, then one
naturally inquires about the situation under the
alternative hypothesis, and this leads to numerous consistency or independence results.
**Cardinal invariants.** Exemplifying the pragmatic view, this is a very rich subject
studying various cardinal characteristics of the continuum,
such as the size of the smallest unbounded family of
functions $f:\omega\to\omega$, the additivity of the ideal
of measure-zero sets, or the smallest size family of
functions $f:\omega\to\omega$ that dominate all other such
functions. Since these characteristics are all uncountable
and at most the continuum, the entire theory trivializes
under CH, but under $\neg$CH is a rich, fascinating
subject.
**Canonical Inner models.** The paradigmatic canonical
inner model is Goedel's constructible universe $L$, which
satisfies CH and indeed, the Generalized Continuum
Hypothesis, as well as many other regularity properties.
Larger but still canonical inner models have been built by
Silver, Jensen, Mitchell, Steel and others that share the
GCH and these regularity properties, while also satisfying
larger large cardinal axioms than are possible in $L$. Most
set-theorists do not view these inner models as likely to
be the "real" universe, for similar reasons that they
reject $V=L$, but as the models accommodate larger and
larger large cardinals, it becomes increasingly difficult
to make this case. Even $V=L$ is compatible with the
existence of transitive set models of the very largest
large cardinals (since the assertion that such sets exist
is $\Sigma^1\_2$ and hence absolute to $L$). In this sense,
the canonical inner models are fundamentally compatible
with whatever kind of set theory we are imagining.
**Woodin.** In contrast to the Old-School Dream
Solution, Woodin has advanced a more technical argument in
favor of $\neg$CH. The main concepts include $\Omega$-logic
and the $\Omega$-conjecture, concerning the limits of
forcing-invariant assertions, particularly those
expressible in the structure $H\_{\omega\_2}$, where CH is
expressible. Woodin's is a decidedly Platonist position,
but from what I have seen, he has remained guarded in his
presentations, describing the argument as a proposal or
possible solution, despite the fact that others sometimes
characterize his position as more definitive.
**Foreman.** Foreman, who also comes from a strong
Platonist position, argues against Woodin's view. He writes
supremely well, and I recommend following the links to his
articles.
**Multiverse view.** This is the view, offered in
opposition to the Basic Platonist Position above, that we
do not have just one concept of set leading to a unique
set-theoretic universe, but rather a complex variety of set
concepts leading to many different set-theoretic worlds.
Indeed, the view is that much of set-theoretic research in
the past half-century has been about constructing these
various alternative worlds. Many of the alternative set
concepts, such as those arising by forcing or by large
cardinal embeddings are closely enough related to each
other that they can be compared from the perspective of
each other. The multiverse view of CH is that the CH
question is largely settled by the fact that we know
precisely how to build CH or $\neg$CH worlds close to any
given set-theoretic universe---the CH and $\neg$CH worlds
are in a sense dense among the set-theoretic universes. The
multiverse view is realist as opposed to formalist, since
it affirms the real nature of the set-theoretic worlds to
which the various set concepts give rise. On the Multiverse
view, the Old-School Dream Solution is impossible, since
our experience in the CH and $\neg$CH worlds will prevent
us from accepting any principle $\Phi$ that settles CH as
"obviously true". Rather, on the multiverse view we are to study all the possible set-theoretic worlds and especially how they relate to each other.
I should stop now, and I apologize for the length of this answer.
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https://mathoverflow.net/questions/25174
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I was curious if anyone has a reference for a formula giving the values of n and k so that $\binom{n}{k}<\binom{n+j}{k-1}$ for a fixed $j$.
Clearly this will be true if $k>\frac{n}{2}$ because then one will have that $\binom{n}{k}\le\binom{n}{k-1}<\binom{n+j}{k-1}$. One can improve on this result, and in the case where $j=1$ I have found precise conditions on $k$ in terms of $n$, but my approach is rather blunt and it seems like the general case will be quite tedious using my methods, even though this seems like a question that likely has an elegant combinatorial solution. I was wondering if anyone knows where a solution appears.
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https://mathoverflow.net/users/4535
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When is (n choose k) < (n+j choose k-1) for fixed j?
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It doesn't get that ugly, if you're mainly concerned with large $n$ and $k$. Simple order-of-magnitude stuff indicates that something like $k>\alpha n$ is true (where $\alpha$ depends on $j$).
The inequality $\binom{n}{k}<\binom{n+j}{k-1}$ is exactly equivalent to $$1 < \frac{k}{n-k+1} \prod\_{i=1}^j \frac{n+i}{n+i-k+1}.$$ Setting $k\approx \alpha n$ and letting $n\to \infty$, this implies the inequality $$\frac{\alpha}{(1-\alpha)^{j+1}} \geq 1.$$ This is easy to solve for $j$. To solve for $\alpha$ is rougher. Since $(1-m/j)^j\to e^{-m}$, one can see that for large $j$ we must have $\alpha>m/j$ (specifically, for each $m$, if $j$ is sufficiently large then $\alpha>m/j$).
It seems to me that $$\frac 1j < \alpha\_j < \frac{\log j}{j}$$ for $j\ge 5$ follows from calculus. So here's the result: Let $\alpha\_j$ be the unique real solution to $\alpha=(1-\alpha)^{j+1}$ with $0<\alpha<1$. If $n,k$ are sufficiently large and $k>\alpha\_j n$, then $\binom{n}{k}<\binom{n+j}{k-1}$. In particular, for each $j\ge 5$, if $n,k$ are sufficiently large with $k>\frac{\log(j)}{j} n $, then $\binom{n}{k}<\binom{n+j}{k-1}$.
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https://mathoverflow.net/users/935
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https://mathoverflow.net/questions/25211
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0
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Let's say I have a multiset of complex numbers $\lbrace a\_1,\cdots,a\_n\rbrace$ (so some of the elements may be repeated) and I would like to construct an entire function $p(z)$ with those numbers as zeroes. However, I also have a multiset of complex numbers $B = \lbrace b\_1,\cdots,b\_n \rbrace$ such that I wish $p(b\_i) = 1$ - **p is only 1 on the $b\_i$'s**.
It seems like trying to use Lagrange's polynomial interpolation formula gives you a polynomial with too high a degree (greater than $n$ and less than or equal to $2n$), and then there's the possibility that $p^{-1}(1) \nsubseteq B$.
I've been thinking about doing the following:
Let $g(z) = (x-a\_1) \cdots (x - a\_n)$, and then via Weierstrass construct an entire function $h(z)$ such that $e^{h(b\_i)} = 1/g(b\_i)$. Then it seems like the entire function $e^{h(z)}g(z)$ is getting somewhat closer to what I want - but then again I don't know if there are any other $\alpha$'s such that $e^{h(\alpha)}g(\alpha) = 1$ where $\alpha \notin B$.
The problem of polynomial interpolation and fitting seems very well studied; however, I can't seem to find a reference for this particular puzzle.
Thanks in advance!
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https://mathoverflow.net/users/5534
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Entire function interpolation with control over multiplicities/derivatives
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If I read you right,
you want an entire function that takes the values $0$ and $1$ at only
finitely many (specified) points. This implies that the function must be a polynomial,
by Picard's great theorem, since there will be deleted neighbourhoods of
infinity where the function misses two values.
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https://mathoverflow.net/users/4213
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25213
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https://mathoverflow.net/questions/25210
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7
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Recall that an integral domain $R$ is **atomic** if every nonzero nonunit admits at least one factorization into irreducible elements. (Indeed, hard-core factorization theorists have replaced the word "irreducible" by "atom".)
From prior reading, I happen to know that there exist atomic integral domains $R$ such that the univariate polynomial ring $R[t]$ is not atomic. This is a somewhat surprising pathology, because the implication is true if both instances of "atomic" are replaced by "UFD", "Noetherian" or "Ascending Chain Condition on Principal Ideals".
But I don't know a precise example or a reference, and I would like one for an expository article I'm writing. Of course, the chronologically earlier and logically simpler the example, the better.
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https://mathoverflow.net/users/1149
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Example sought of an atomic domain R such that R[t] is not atomic
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According to the book "Non-Noetherian commutative ring theory" by S.T. Chapman and S. Glaz the question was first asked in "Factorization of integral domains" by D.D. Anderson, D.F. Anderson, M. Zafrullah, Journal of Pure and Applied Algebra 69 (1990) 1-19 (question 1). An answer was given [here](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V0K-45FCX7K-1G&_user=10&_coverDate=07%2F02%2F1993&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1340527568&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=349d97d14c6369afb3bb28819cfd4e90) by M. Roitman.
There it was conjectured that $R[X]$ atomic $\implies$ $R[X,Y]$ is also atomic.
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https://mathoverflow.net/users/2384
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25214
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https://mathoverflow.net/questions/25122
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20
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Let me motivate my question a bit.
**Thm**. Let $X$ be a locally noetherian finite-dimensional regular scheme. If $X$ has enough locally frees, then the natural homomorphism $K^0(X)\longrightarrow K\_0(X)$ is an isomorphism.
A locally noetherian scheme has enough locally frees if every coherent sheaf is the quotient of a locally free coherent sheaf, $K^0(X)$ denotes the Grothendieck group of vector bundles on $X$ and $K\_0(X)$ denotes the Grothendieck group of coherent sheaves on $X$.
The above theorem is shown as follows.
By the regularity (and finite-dimensionality!) of $X$, we can construct a finite resolution by a standard procedure. (Surject onto the kernel at each stage with a vector bundle.) Then the "Euler characteristic" associated to this resolution is inverse to the natural morphism.
Now, I was looking through the literature (Weibel's book basically) and I saw that this theorem appears with the additional condition of separatedness. (Edit: This is not necessary. The point is that noetherian schemes that have enough locally frees are semi-separated.)
**Example**. Take the projective plane $X$ with a double origin. Then $K^0(X) \cong \mathbf{Z}^3$ whereas $K\_0(X) \cong \mathbf{Z}^4$.
**Example**. Take the affine plane $X$ with a double origin. Then $K^0(X) \cong \mathbf{Z}$, whereas $K\_0(X) \cong \mathbf{Z}\oplus \mathbf{Z}$.
So I figured I must be missing something...Thus, I ask:
**Q**. Are locally noetherian schemes that have enough locally frees separated?
**EDIT**.
The answer to the above question is "No" as the example by Antoine Chambert-Loir shows.
From Philipp Gross's answer, we conclude that a noetherian scheme which has enough locally frees is semi-separated. This means that, for every pair of affine open subsets $U,V\subset X$, it holds that $U\cap V$ is affine. Note that separated schemes are semi-separated and that Antoine's example is also semi-separated.
Taking a look at Totaro's article cited by Philipp Gross, we see that a regular noetherian scheme which is semi-separated has enough locally frees. (Do regular semi-separated and locally noetherian schemes have enough locally frees?)
This was (in a way) also remarked by Hailong Dao. He mentions the result of Kleiman and independently Illusie. Recently, Brenner and Schroer observed that their proof works also with $X$ noetherian semi-separated locally $\mathbf{Q}$-factorial. See page 4 of Totaro's paper. In short, separated is not really needed but semi-separated is.
Thus, we can conclude the following.
Suppose that $X$ is a regular and finite-dimensional scheme.
If $X$ has enough locally frees, then $K^0(X) \longrightarrow K\_0(X)$ is an isomorphism. For example, $X$ is noetherian and semi-separated.
Anyway, thanks to everybody for their answers. They helped me alot!
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https://mathoverflow.net/users/4333
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Are schemes that "have enough locally frees" necessarily separated
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The property that every coherent sheaf admits a surjection from a coherent locally free sheaf is also known as the *resolution property*.
The theorem can be refined as follows:
Every noetherian, locally $\mathbb Q$-factorial scheme with affine diagonal (equiv. semi-semiseparated) has the resolution property (where the resolving vector bundles are made up from line bundles).
This is Proposition 1.3 of the following paper:
*Brenner, Holger; Schröer, Stefan
Ample families, multihomogeneous spectra, and algebraization of formal schemes.
Pacific J. Math. 208 (2003), no. 2, 209--230.*
You will find a detailed discussion of the resolution property in
*Totaro, Burt.
The resolution property for schemes and stacks.
J. Reine Angew. Math. 577 (2004), 1--22*
Totaro proves in Proposition 3.1. that every scheme (or algebraic stack with affine stabilizers) has affine diagonal if it satisfies the resolution property.
The converse is also true for smooth schemes:
**Proposition 8.1** : Let $X$ be a smooth scheme of finite type over a field. Then the following are equivalent:
1. $X$ has affine diagonal.
2. X has the resolution property.
3. The natural map $K\_0^{naive} \to K\_0$ is surjective.
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15
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https://mathoverflow.net/users/4101
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25224
| 16,538 |
https://mathoverflow.net/questions/25220
|
14
|
Good Morning from Belgium,
I'm no stranger to the mantra that quiver-algebras are an extremely powerful tool (see for example the representation theory of finite dimensional algebras). But what is a bit unclear to me is what is known about what kind of algebras are quiver algebras. The first case I can prove is for graded rings with finite dimensional semisimple $A\_0$. But what about the ungraded case (I'm pretty sure that it is also true for finite-dimensional algebras over an algebraically closed field for example).
Is there a larger generality possible ?
Answers (and references) are as always much appreciated !
|
https://mathoverflow.net/users/4863
|
when are algebras quiver algebras ?
|
Louis, do you mean by a 'quiver-algebra' the path algebra of a quiver, or do you mean a quotient of a path algebra?
If the first, then I do not understand your comments. k[x,y] is graded with semi-simple part of degree zero but not a path algebra.
If you mean by quiver-algebra a quotient of a path algebra then the answer is simple : any finitely generated C-algebra will do as they are quotients of free algebras (path algebra of one vertex multiple loop quiver).
If you mean by quiver-algebra really the path algebra of a quiver, the answer is trickier.
If your algebra is finite dimensional (I'm always working over C) then the classification is : hereditary and basic (that is, all simples are one-dimensional). In that case, any hereditary is Morita equivalent to a path algebra. All this goes back to Gabriel.
If your algebra is infinite dimensional one has to be careful. Surely it must be formally smooth (that is, it has the lifting property for algebra maps through nilpotent ideals) and have a finite number of isoclasses of one-dimensional representations.
EDIT : Oops, if one has loops then there are of course infinitely many 1-dmls. I should have said that there are only finitely many components of 1-dml representations, all parametrized by affine spaces and such that one can pick one rep in each component to perform the trick with the structural morphism described below.
But that is not enough, take e.g. the groupalgebra of the modular group PSL(2,Z). It has 6 one-dimensionals but is not isomorphic to the 'obvious' quiver one would construct out of these 6 (arrows corresponding to extensions between the simples).
What one needs is that the structural morphism A --> End(SS) (where SS is the semi-simple on the finite number of one-dimls) splits so that A becomes a C^k-algebra. Then one can use formal smoothness of A and the path algebra with semi-simple part End(SS) and arrow part determined by M/M^2 (where M is the kernel of the structural morphism above) to prove that they are isomorphic. An argument like this appears in the paper by Cuntz and Quillen on noncommutative smoothness (they call a formally smooth algebra 'quasi-free').
|
13
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https://mathoverflow.net/users/2275
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25234
| 16,542 |
https://mathoverflow.net/questions/25238
|
4
|
Hi.
There is a really quick proof of the Nullstellensatz when the field is infinite (**edit : I meant uncountable**) (let's take $\mathbb{C}$ for example.)
It mainly uses the fact that $\mathbb{C}(x)$ is an extension of C of infinite and uncountable dimension.
I would like to know where (from who ? When ?) this idea came from ? I know that the well-known proof using entire rings and Noether normalisation came from Zariski, but I found nothing concerning this idea.
Thanks.
|
https://mathoverflow.net/users/6187
|
Origin of the elementary proof of the Nullstellensatz with an uncountable field
|
If this is the proof I think it is, in Exercise 4.31 of Eisenbud's book on commutative algebra he attributes it to Krull and van der Waerden.
|
6
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https://mathoverflow.net/users/317
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25241
| 16,548 |
https://mathoverflow.net/questions/25037
|
3
|
Say I have two heatmaps:
Each pixel of the heatmap represents a certain probability.
One heatmap is derived from empirical data, and the other heatmap is generated by an algorithm that is designed to simulate the natural process that underlies the empirical data.
I wish to tune the algorithm to make the generated heatmap match up as closely to the empirical heatmap as possible, but this is difficult without a proper metric to actually make a comparison. Thus, I wish to implement a metric that can return a value from 0 to 1 to make this comparison.
I am currently considering vector distance, mutual information, and KL-divergence. I am curious if anyone has experience or advice regarding this. --Thanks!
|
https://mathoverflow.net/users/6142
|
A metric for comparing two heatmaps
|
If this is a high-stake computation, on which you're willing to spend some comp. effort, you could use a $h\_{-1}$-Sobolev norm - in effect, compute Fourier coefficients of the difference of the heatmaps, and discount them by the wavenumber, before summing them up. I'm writing this from memory, so please check literature.
$$ d( A, B )^2 = \sum\_{k} \frac{ \mathcal{F}(A - B)^2\_k }{ 1 + (2\pi \vert k \vert)^2 }$$
Where $A,B$ are the heatmaps, $\mathcal{F}(A - B)\_k$ is the Fourier coefficient associated with the wavevector $k=(k\_x, k\_y)$, $\vert k \vert^2 = k\_x^2 + k\_y^2$.
This will have a "low-pass-filter" effect on the difference, so instead of just taking a plain $l\_2$ norm, which assigns equal weight to variations on large scale (coarse details), the above norm ( $h\_{-1}$-Sobolev norm) will discount variations in small-scale details and emphasize alignment of heat maps on the large scale level first. Due to the nature of your heatmaps, one computational, and the other being empirical, I believe you are bound to have small scale variations that you don't care about.
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3
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https://mathoverflow.net/users/992
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25255
| 16,555 |
https://mathoverflow.net/questions/25229
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3
|
For example, I take differentiability, analyticity, and algebraicity(of a function).
All(more or less) imply continuity. So when we define a differentiable function on $\mathbb R^n$ or an analytic function on $\mathbb C^n$, or a regular map on an affine space, we do not explicitly require that the functions are continuous. It follows automatically from the stronger condition.
But, when I look at the definitions in books of a global structure using sheaf theory, for a global definition of a morphism, ie on a differentiable manifold or an analytic space, or an abstract algebraic variety, the definition of a morphism requires a priori that the map be continuous, and then one requires that there is additionally a morphism of sheaves of algebras(of the suitable type of structure sheaves, depending on the local model used).
Why is this so? Is it something done for fancy, or is there a real need for the extra continuity assumption? I mean could things go wrong if this assumption is dropped?
|
https://mathoverflow.net/users/6031
|
Why is continuity required for sheaf-theoretic definitions of a structure on a space
|
As Andrea hints, if you start with sheaves then you need continuity to even begin talking about morphisms of sheaves.
However, if you're interesting in just defining, say, a smooth map between manifolds then you can simply write "$f \colon M \to N$ is smooth if, whenever $c \colon \mathbb{R} \to M$ is a smooth curve then $f \circ c \colon \mathbb{R} \to N$ is smooth". No assumption about continuity is needed there.
Indeed, once one gets to more exotic spaces, continuity becomes a hassle and is best left to one side. For example, the evaluation map $E \times E^\* \to \mathbb{R}$ is smooth for any locally convex topological vector space, $E$, but is only continuous for $E$ a normed vector space.
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3
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https://mathoverflow.net/users/45
|
25260
| 16,557 |
https://mathoverflow.net/questions/25256
|
9
|
On an elliptic curve given by a degree three equation y^2 = x(x - 1)(x - λ), we can define the group law in the following way (cf. Hartshorne):
1. We note that the map to its Jacobian given by $\mathcal{O}(p - p\_0)$ for a fixed point $p\_0$ is an isomorphism; ergo it inherits a group structure from the Jacobian.
2. In fact, if we embed it into $\mathbb{P}^2$ via the linear system $|3p\_0|$, then three colinear points have $p + q + r \sim 3p\_0$ and so this is in fact the group law inherited from Pic0.
Is there an analogous way to do this for Abelian varieties? In Lange & Birkenhake they simply define an Abelian variety to be $\mathbb{C}^n$ modulo a lattice, and so it automatically comes with a group structure. Still, this seems unsatisfying in comparison to the way we can do so for elliptic curves.
That being said, the previous method doesn't seem to work for Abelian varieties; divisors no longer correspond to formal sums of points and so any comparison with Pic0 wouldn't obviously yield a group structure on the points of X.
To make matters worse, the map that I tend to think of which takes X to Pic0(X) is given by $p \mapsto t\_p^\*L \otimes L^{-1}$ for a given line bundle $L$ on X, where $t\_p : X \to X$ is the map... defined by translation in X. So this map already requires the group structure on X to be defined.
So is there a way of defining the group law analogous to that of an elliptic curve?
NB: I do note that an elliptic curve can be defined as the zero locus of a cubic equation in $\mathbb{P}^2$, where I'm not sure how else we might define an Abelian variety other than as $\mathbb{C}^n$ modulo a lattice, and so perhaps the question is moot.
|
https://mathoverflow.net/users/1703
|
Is there an intrinsic way to define the group law on Abelian varieties?
|
Any torsor $V$ under an abelian variety over any field $K$ is caonically isomorphic to its degree $1$ Albanese variety $\operatorname{Alb}^1(V)$, which is itself a torsor under the degree $0$ Albanese variety $\operatorname{Alb}^0(V)$. Note that the $\operatorname{Alb}^0$ of any smooth projective variety is a complete, geometrically connected group variety.
Assuming there exists a $K$-rational point $O$, one can subtract $O$ to obtain an isomorphism $\operatorname{Alb}^1(V) \stackrel{\sim}{\rightarrow} \operatorname{Alb}^0(V)$. Pulling back via the composite of these isomorphisms puts a group structure on $V$ depending only on the chosen base point $O$.
|
7
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https://mathoverflow.net/users/1149
|
25262
| 16,559 |
https://mathoverflow.net/questions/25257
|
-2
|
If we roll 4 dices (fair), what is the probability of "sum of subset" being 5. e.g. 1432,1121, 2344, 2354 have a subset sum of 5. Can you illustrate how to calculate this.
|
https://mathoverflow.net/users/6191
|
probability of subset sum after rolling dice 4 times
|
The most straightforward way is also the most tedious: list all die rolls (1296 of them,
I assume), and count which ones have a subset sum of 5. You can save on some work by
looking at all rolls which have at least one 5 in them, and then all rolls which have no
5 but at least one 1 and one 4 or one 2 and one 3, and then enumerate the remaining cases.
A possibly quicker but still involved approach involves calculating the probabilities
that a sum appear on the kth roll but not the (k-1)th roll. Then one gets 1/6
of a probability of a sum on the first roll and 1/6 + 1/4 = 5/12 of a probability of
such a sum on the first or the second roll. The calculation for the third roll assumes
that no subset sum has occurred on the scone roll, and breaks down into 21 cases, starting
with 6 6 and going down to 1 1, if you like, but many of the cases can be treated the
same way, so that there are fewer types of cases to consider.
You can also try looking at rolls which miss having a subset sum of 5. Taking all even
die gets you a lower bound of 1/16 for not having a subset sum of 5, while looking
at all combinations of 3, 4, and 6, and removing the even combinations gives you (81-16)
more rolls which do not have a subset sum of 5, giving a lower bound of (1/8 -1/81) for
missing a subset sum of 5. There are still the extra cases to handle.
I do not see a simple characterization that allows a quick calculation of the
probability.
Gerhard "Ask Me About System Design" Paseman, 2010.05.19
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1
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https://mathoverflow.net/users/3568
|
25264
| 16,560 |
https://mathoverflow.net/questions/25259
|
11
|
Can anyone point me to an online database of Steiner triple systems?
My Google-fu is only getting me to descriptions of the few smallest ones, mostly Google book scans (which are rather useless to process using a computer...)
|
https://mathoverflow.net/users/1409
|
Database of Steiner triple systems
|
The first answer is not a database of the Steiner triple systems, but rather how many there are up to isomorphism. [It is known](http://oeis.org/A030129) that they there are 2 of order 13, 80 of order 15, and 11084874829 of order 19. The last number was computed by Kaski and Ostergard, and I suppose that the best approach is to ask them for their data. It's more or less the end of the story, because it is easy to make larger Steiner triple systems, but impossible to compile them into a complete database.
Actually [this paper](http://www.nbl.fi/~nbl4075/sts19cat.pdf) describes a compressed 39-gigabyte file with the Steiner triple systems of order 19, and says that it is available by e-mail request from three of the authors (including Kaski and Ostergard).
It looks like a number of people have the Steiner triple systems of order 15, but I didn't find a paper that simply lists them.
**Update 1:** It seems that everyone works from the paper "Small Steiner triple systems and their properties", by Mathon, Phelps, and Rosa. This paper is basically an encyclopedia of the 80 Steiner triple systems of order 15 and many of their properties. It also introduces a somewhat standard numbering. The thing to do at this point would be to transcribe the data in this widely cited paper into a file. Google seems to indicate that no such file has been posted to the web.
**Update 2:** It was done! See file data/steiner.tbl in [this GAP package](http://gap-system.org/Packages/loops.html) by Nagy and Vojtechovsky. For some reason, a Steiner triple system of order $n$ is also called a Steiner loop of order $n+1$, and that is the terminology that they use. They copied the data from Colbourn and Rosa, Triple Systems, which presumably is the same as in Mathon, Phelps, and Rosa.
How I found it: I Googled one of the hexadecimal strings used to describe one of the STS(15)s. None of the Google's heuristics worked for me, so instead I used the old-fashioned trick of searching for a very specific keyword. It also shows up in a LaTeX PHD thesis, but the GAP file, which has almost the same syntax as Python, is better.
|
18
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https://mathoverflow.net/users/1450
|
25266
| 16,562 |
https://mathoverflow.net/questions/25215
|
10
|
The standard reconstruction conjecture states that a graph is determined by its [**deck of vertex-deleted subgraphs**](http://en.wikipedia.org/wiki/Reconstruction_conjecture#Formal_statements).
>
> **Question**: Have other decks been investigated, finding out
> that only vertex-deleted subgraphs can do the job? If so: Which property of vertex-deleted subgraphs makes them exceptional?
>
>
>
I have three candidates in mind, others are conceivable. (For the sake of simplicity I consider only simple connected graphs $G$.)
1. the **deck of sub-maximal neighbourhoods**: Let the *sub-maximal neighbourhood of $v$* be the $v$-rooted graph constructed from $G$ by deleting all vertices with maximal distance from $v$.
2. the **deck of distinguishing neighbourhoods**: Let the *distinguishing neighbourhood of $v$* be the smallest $n$-neighbourhood of $v$ which distinguishes it from all vertices not conjugate to it ($n$-neighbourhood = the $v$-rooted induced subgraph containing all vertices $w$ with distance $d(v,w) \leq n$).
3. the **deck of crossref-deleted subgraphs**: Let the *crossref-deleted subgraph with respect to $v$* be the $v$-rooted graph constructed from $G$ by deleting all edges between vertices that have the same distance from $v$.
Note that the *vertex-deleted subgraph with respect to $v$* is nothing but the $v$-rooted graph constructed from $G$ by deleting all edges between $v$ and its neighbours.
I am not good in systematically constructing counterexamples, and I do not have very much intuition about general graphs. So, any counterexample to one of the candidates above would be very welcome.
What I *do* know is that a) trees are trivially reconstructible from their deck of crossref-deleted subgraphs, that b) graphs with one node of which the distinguishing neighbourhood is the whole graph are trivially reconstructible from their deck of distinguishing neighbourhoods, and that c) reconstructing (very) small graphs from one of the decks above is fun.
|
https://mathoverflow.net/users/2672
|
Reconstruction conjecture: Can other decks do the job?
|
I suspect that there are counterexamples to all three of your proposals. For example, you clarified that the distinguishing neighborhood of a vertex-transitive graph is just its 1-neighborhood. Then the dodecahedral graph and the Desargues graph will have the same decks; in each case you'll just have twenty copies of $K\_{1,3}$. Similarly, although I don't have an explicit counterexample offhand, I suspect that if you examine strongly regular graphs with the same degree and number of vertices, you will find plenty of pairs of graphs with the same decks of sub-maximal neighborhoods. (Strongly regular graphs have diameter 2 so again you're just looking at 1-neighborhoods.)
As for your question of what other reconstruction conjectures there are, the most famous is the edge reconstruction conjecture. There is also the vertex-switching reconstruction conjecture and the $k$-reconstruction conjecture (see Bondy's *Graph Reconstructor's Manual* for definitions). For more examples, you could try contacting Mark Ellingham at Vanderbilt, who keeps a list of papers on the reconstruction conjecture.
|
8
|
https://mathoverflow.net/users/3106
|
25267
| 16,563 |
https://mathoverflow.net/questions/25263
|
20
|
It is well known that if $p$ is an odd prime, exactly one half of the numbers $1, \dots, p-1$ are squares in $\mathbb{F}\_p$. What is less obvious is that among these $(p-1)/2$ squares, at least one half lie in the interval $[1, (p-1)/2]$.
I remember reading this fact many years ago on a very popular book in number theory, where it was claimed that this is an easy consequence of a more sophisticated formula of analytic number theory.
Sadly I forgot both the formula and the book. So the purpose of the question is double:
>
> 1) Has any simple way been found to derive the fact mentioned above?
>
>
> 2) Does anybody know a reference for the analytic number theory route to the proof?
>
>
>
|
https://mathoverflow.net/users/828
|
Most squares in the first half-interval
|
Nope! Amazingly enough, no elementary proof of this fact is yet known (Edit: See KConrad's answer). The difficulty is tied up in some pretty fantastic algebraic/analytic number theory, namely the analytic class number formula. But, without getting into that, here's the short form of the story: let $L(s)$ be the $L$-function attached to the character arising from the Kronecker symbol mod $q$. Then one can compute analytically via the Euler product formula for this $L$-function that (for an explicit positive constant $C$),
$$
L(1)=C\left[\sum\_{m=0}^{q/2}\left(\frac{m}{q}\right)-\sum\_{m=q/2}^{q}\left(\frac{m}{q}\right)\right]=\frac{\pi}{\left(2-\left(\frac{2}{q}\right)\right)q^{1/2}}\sum\_{m=0}^{q/2}\left(\frac{m}{q}\right),
$$
the positivity of which gives the desired statement about the distribution of quadratic residues.
For a reference (from whence I pulled this out of), Davenport's "Multiplicative Number Theory" is pretty fantastic. This is all done in the first 4-5 pages.
|
13
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https://mathoverflow.net/users/35575
|
25268
| 16,564 |
https://mathoverflow.net/questions/25269
|
10
|
Given a sequence $Y\_1, Y\_2, \dots$ of i.i.d. matrices in $\mathrm{GL}\_n(\mathbb R)$, there is a theorem of Furstenberg and Kesten which says that if $\mathbb E(\log\|Y\_1\|)$ is finite, there exists a constant $\gamma$ (the Lyapunov exponent) such that
$$\lim\_{n\rightarrow\infty}\frac{1}{n}\log\|Y\_n\dots Y\_1\| = \gamma$$
There are also versions of central limit theorems for this scenario. I'm pretty sure this is also known in a more general case (e.g. suppose we have a sequence of matrices $Y\_i$ of order 2, and I don't want to consider sequences of length $n$ in which $\dots Y\_i Y\_i\dots$ appears). I am wondering if anyone knows a good reference for theorems regarding Lyapunov exponents and central theorems in this case.
|
https://mathoverflow.net/users/6194
|
Random walks and Lyapunov exponents
|
*Random dynamical systems* by Ludwig Arnold contains a thorough discussion of various multiplicative ergodic theorems (including the Furstenberg-Kesten result), but not the central limit theorems. As far as I remember, the case of stationary sequences of linear stochastic iterations is also included there.
**Edit.** Concerning central limit theorems for products of random matrices, a quick search yields this [reference](https://projecteuclid.org/journals/annals-of-probability/volume-25/issue-4/Limit-theorems-for-products-of-positive-random-matrices/10.1214/aop/1023481103.full).
|
1
|
https://mathoverflow.net/users/5371
|
25271
| 16,565 |
https://mathoverflow.net/questions/25270
|
1
|
I have two circulant Cayley digraphs: that is, Cayley digraphs *X* = Cay(ℤ/*m*, *S*) and *Y* = Cay(ℤ/*n*, *T*), for odd integers *m* < *n*, and sets with sizes |*S*| = (*m* − 1)/2, and |*T*| = (*n* − 1)/2.
These digraphs are antisymmetric, in that *S* is disjoint from −*S*, and *T* is disjoint from −*T*. (It follows that for each distinct pair of vertices *a,b* in either graph, there is either an arc from *a* to *b*, or vice versa.)
**Question.** What conditions on *m*, *n*, *S*, and *T* must hold for *X* to be an induced directed subgraph of *Y*?
|
https://mathoverflow.net/users/3723
|
Conditions for subgraph relationship in circulant Cayley digraphs
|
I very much doubt that there is a nice answer for this.
I suspect that this question
is not essentially easier than the more general problem, where we allow $X$ to be any
tournament. If $n$ is a prime congruent to 3 mod 4 and $T$ is the set of non-zero
squares in $\mathbb{Z}/n$, the Cayley graph $Y$ is the Paley tournament. It follows from an
old result of Graham and Spencer that any smallish tournament is an induced directed
subgraph of $Y$. (Here ``smallish'' is technical term that means something like $\log(n)$,
or perhaps $\sqrt{(\log(n))}$.)
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3
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https://mathoverflow.net/users/1266
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25272
| 16,566 |
https://mathoverflow.net/questions/25151
|
3
|
In a paper that I am reading, the author is weighting edges in a graph using
$$w\_k \propto \det(D(p))$$
where $D(p)$ is the metric tensor (which if I understand correctly is a space-varying metric?). They say that $D(p) = 1$ is the Euclidean metric, $D(p) = \mbox{const}$ is a Riemannian metric. Can anyone explain how to interpret what the determinant of a metric means? And can you suggest (in layman's terms) a metric which is not constant in position?
I see $\det(D)$ so I think $D$ must be a matrix - is this correct? What is the matrix for a few simple metrics?
|
https://mathoverflow.net/users/5523
|
Determinant of a metric?
|
This is too long for a comment but still too short for an introductory course that you are asking for.
A Riemannian metric in the plane is (represented by) a matrix-valued function such that the value at every point $p$ is a symmetric matrix $g=g(p)$ of the form $g=\begin{pmatrix} E & F \\ F & G\end{pmatrix}$ which is positive definite as a quadratic form, that is, $E,G>0$ and $\det g >0$. Any matrix-valued function satisfying these properties is (represents) a Riemannian metric, by definition.
There are many examples. Usually introductions begin with metrics defined by parametric surfaces (that $r\_u$ and $r\_v$ stuff). This is only an example, one of the many, and it is irrelevant in the context of the paper. The paper considers Riemannian metrics not associated to any surface.
Let me try to explain what it is about. The above matrix $g$ defines a norm on $\mathbb R^2$ as follows: $\|(u,v)\|=\sqrt{Eu^2+2Fuv+Gv^2}$. This norm defines a metric $d$ by $d(x,y)=\|x-y\|$. This is a Riemannian metric associated with a constant metric tensor $g$. One can prove that this metric space is isometric to the standard Euclidean plane, and an isometry is given by a linear map (which can be written explicitly in terms of $E$, $F$ and $G$). So a constant metric tensor essentially produces a Euclidean metric written in some non-orthogonal coordinates.
What I said above is a reformulation of some bits from linear algebra. Riemannian geometry begins when $g$ is not constant. Then there is not a norm but a family of norms $\|\cdot\|\_p$, $p\in U$, where $U\subset\mathbb R^2$ is the domain where $g$ is defined. Every norm $\|\cdot\|\_p$ yields a metric $d\_p$ on its own private copy of $\mathbb R^2$ (this private copy is called the tangent space at $p$ and is usually denoted by $T\_pU$).
This family of metrics is "glued together" into one metric $d$ on $U$. It can be defined as the maximal metric satisfying the following condition: for every $p\in U$, one has
$$
\frac{d(x,y)}{d\_p(x,y)} \to 1 \ \ \ \text{as} \ \ x,y\to p .
$$
So, locally near a point $p$ the metric is approximately the same as the (essentially Euclidean) metric $d\_p$ defined by the matrix at $p$. This is sufficient for any first-order analysis.
The standard definition involves defining a length of a curve with respect to $g$ (integrate the length of the velocity vector, computed in that variable norm), and then defining the distance between points as the infimum of length of connecting curves. It is equivalent to the one I gave above.
|
6
|
https://mathoverflow.net/users/4354
|
25276
| 16,568 |
https://mathoverflow.net/questions/25275
|
6
|
Do you know how to construct a compact hyperbolic 3-manifold with three or four totally geodesic boundary components? The only constructions I could find have one boundary component. A reference would be appreciated.
|
https://mathoverflow.net/users/4325
|
Do you know how to construct a compact hyperbolic 3-manifold with three or four totally geodesic boundary components?
|
**Theorem**([Long and Niblo](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=long&s5=niblo&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq)): If $M$ is a 3-manifold and $S$ is an incompressible component of $\partial M$ then $\pi\_1 S$ is *separable* in $\pi\_1 M$ (pick a base point in $S$ to make sense of this).
This means that for any $\gamma\in \pi\_1M\smallsetminus \pi\_1 S$ there is a homomorphism $f$ from $\pi\_1M$ to a finite group such that $f(\gamma)\notin f(\pi\_1(S))$.
**Corollary:** If $|\pi\_1M:\pi\_1S|=\infty$ then $M$ had finite-sheeted coverings such that the preimage of $S$ has as many components as you like.
**Proof:** Choose a homomorphism $f$ such that $f(\pi\_1S)$ has large index in $f(\pi\_1M)$. Now your covering corresponds to $\ker f$. **QED**
So examples exists as finite covers of the example you already have!
|
11
|
https://mathoverflow.net/users/1463
|
25277
| 16,569 |
https://mathoverflow.net/questions/25282
|
13
|
It is described in many sources that algebraic topology had been a major source of innovation for algebraic geometry. It is said that the uses of cohomology, sheaves, spectral sequences etc. in algebraic geometry were motivated by algebraic topology. Moreover it is said that Weil conjectures arose out of inspiration from algebraic topology.
So it seems a very clear thing that algebraic topology tremendously influenced algebraic geometry, at least historically.
But are there influences in the other way? Did it ever happen that the modern developments in algebraic geometry were ever taken back to algebraic topology and led to developments over there?
Edit: Topological K-theory is one application. Are there more?
|
https://mathoverflow.net/users/6031
|
Are there applications of algebraic geometry into algebraic topology?
|
elliptic cohomology, topological modular forms, stacks, formal groups, genera, to name but a few.
|
15
|
https://mathoverflow.net/users/4183
|
25286
| 16,574 |
https://mathoverflow.net/questions/25288
|
3
|
I am looking for a good reference for Hilbert's irreducibility theorem, and ofproperties of Hilbert sets besides Serres Lectures on The Mordell-Weil Theorem. In particular, I am interested it to the following situation:
Assuming that a variety V is defined by a polynomial H(z,t) over a field k(s), where k has finite characteristic, and s is transcendental, I'm especially interested in how many elements of form a+bs are contained in the Hilbert set of H, defined as the set of {r in k(s) such that H(z,r) is irreducible over k(s)}.
Any answers would be much appreciated!
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https://mathoverflow.net/users/6198
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How many linear terms are in the Hilbert set of H(z,t), a polynomial in 2 variables over a field k(s) of transcendence degree one over a finite field?
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Try Serre, Topics in Galois Theory.
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https://mathoverflow.net/users/2938
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25290
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https://mathoverflow.net/questions/25190
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This should probably be an easy question, but I don't know how to answer it: Suppose *G* is a finitely generated presentable group. Suppose *a* is the absolute minimum of the sizes of all generating sets for *G* and *b* is the absolute minimum of the number of relations over all presentations of *G*. Question: Is it necessary that *G* has a presentation that simultaneously has *a* generators and *b* relations?
The case *b = 0* is just the fact that a free group cannot be generated by fewer elements than its free rank.
The problem could probably be interpreted in terms of CW-complexes (where the generators give rise to 1-cells and the relators give rise to 2-cells) but, because of my lack of familiarity with CW-complexes, I don't immediately see how to use these to solve the problem.
It also seems to be related to the notion of "deficiency" of a group, which is the (maximum possible over all presentations) difference #generators - #relations (under the opposite sign convention, the minimum possible difference #relations - #generators).
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https://mathoverflow.net/users/3040
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Does every finitely presentable group have a presentation that simultaneously minimizes the number of generators and number of relators?
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A stronger question, is the deficiency of $G$ realized for a presentation with
the minimal number of generators (rank($G$))? This question is asked in a [paper of Rapaport](http://www.ams.org/mathscinet-getitem?mr=308277), and proved to be true for nilpotent and 1-relator groups.
Addendum:
The question appears as Question 2, p. 2, of a [book by Gruenberg](http://books.google.com/books?id=1PQ1BVOl7WMC&lpg=PP7&ots=3Q_OnZqmeW&dq=gruenberg%2520relation%2520modules&lr&pg=PA2#v=onepage&q&f=false). Lubotzky
has answered the analogous question affirmatively in the category of profinite
groups ([Corollary 2.5](http://www.ams.org/mathscinet-getitem?mr=1848964)). (Note though that this is in the category of profinite presentations, so it does not imply an affirmative answer even for finite groups).
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https://mathoverflow.net/users/1345
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https://mathoverflow.net/questions/25252
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I have a (given) real-valued function defined over an area, $b(x)$, with $x\in \Omega \subset \mathbb{R}^2$.
I would like to find a smooth real-valued function $a(x)$ that maximises
$J = \int\_{x \in \Omega} a^3(x) + a(x)b(x)~dx$ with the constraint $\int\_{\Omega} a(x) ~dx=0$. If it helps prevent unbounded or irregular solutions I am happy to introduce further constraints or a higher-order penalty, eg,
$J = \int\_{x\in\Omega} a^3(x) + a(x)b(x) - \epsilon a^4(x)~dx$ (with the same constraint).
How would you solve this? Do I need to specify boundary conditions on $a$?
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https://mathoverflow.net/users/6190
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extremal problem (calculus of variations?)
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If we choose as a domain of the functional e.g. the space $L^3(\Omega)$ (with the constraint), and we assume $b\in L^{3/2}(\Omega),$ then $J$ is smooth, and if $\sup b(x)<\infty$ it has a bunch of critical points, all of the form described in the first answer, that is, all the $a(x)$ satisfying $3a(x)^2+b(x)=c$, where the constant $c\geq\sup b$ is arbitrary while the sign of $a(x)$ only has to satisfy the constraint condition $\int\_\Omega a(x)\ dx=0$. The second variation of $J$ at a critical point $a(x)$ is $D^2J(a)[h]^2= 6\int\_\Omega a(x)h^2(x)\ dx$. Thus, any critical point is neither a local minimizer nor a local maximizer (unless $a=0$, that only happens if $b$ is a constant and the problem is 1 dimensional). Moreover, $J$ is unbounded (choose a function $a(x)$ with $\int\_\Omega a(x)dx=0$ and
$\int\_\Omega a^3(x)dx\neq 0$, then for real $t$, $J(ta)$ is a third degree polynomial in $t$, thus unbounded both from above and from below).
If you take the perturbation with $\varepsilon>0$ the situation becomes more delicate . The natural domain is $L^4(\Omega)$ (with the constraint); assuming $b\in L^{4/3}(\Omega)$, $J$ is again smooth, unbounded from below, but now it is bounded from above; again it has a bunch of critical points, characterized by $3a(x)^2-4\varepsilon a^3(x)+b(x)=c$. There are no local minimizers (unless $b$ is constant) for the second variation
$6\int\_\Omega \left( a(x)-4\varepsilon a^2(x)\right) h^2(x)\ dx$ is not negative. It's not immediate to decide whether there are local or global maxima; certainly they can't be continuous since they have both to change sign and to verify $a(x)\leq 4\varepsilon a^2(x)$.
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https://mathoverflow.net/users/6101
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https://mathoverflow.net/questions/25289
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By explicit description I mean a description by explicit algebraic equations. What is the general method to find this in a situation in which you already know that the variety is projective by some other method, for instance by some major theorem whose internals you don't know, but took for granted?
I have two situations in mind:
$1$. The Grassmannian.
We have the Plucker embbedding:
$ \iota : \mathrm{Gr}\_{k}(K^n) \rightarrow \mathbb{P}(\wedge^k K^n) $
$\mathrm{span}( v\_1, \ldots, v\_k ) \mapsto K( v\_1 \wedge \cdots \wedge v\_k )$
By this we know that the Grassmannian is a closed analytic subspace of the projective space. Therefore we know that it is actually algebraic, by Chow's lemma. But how to explicitly describe the Grassmannian by equations?
$2$. Compact Riemann surfaces
We have some Riemann surface of arbitrary genus, for example given as a quotient of the upper half plane by some group. By some abstract theorem(Riemann-Roch??), we know that a Riemann surface can be embedded in the projective space as a closed subspace. And then again by Chow's lemma, it is actually algebraic. But the question remains how to compute explicit equations for this. Fortunately for elliptic curves we know the answer from Weierstrass theory. But what to do in the more general case?
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https://mathoverflow.net/users/6031
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How to determine explicit description for a projective variety?
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Your two questions are actually very different. If you already have a map to projective space, all you want is to find the relations among the functions giving the map. If the functions are given in some explicit way, e.g. power series, then you can turn finding all relations of a given degree into a linear algebra problem and from there find all relations by, e.g. dimension considerations.
The hard question is to exhibit functions on your variety. For Riemann surfaces, we know they are there by the Riemann existence theorem (and Riemann-Roch). Exhibiting them is not automatic. For the case of quotients of the upper-half plane, there is a huge classical literature for example for constructing a Hauptmodul (when the surface has genus zero). So, e.g. writing down the j-function from scratch is not obvious. There are other techniques (Poincare series, ...). Maybe somebody who knows more about this will give a better answer.
Finally, there is no magic bullet to constructing functions, unless you specify how you are given the variety in the first place. You need to make your question more precise.
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https://mathoverflow.net/users/2290
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https://mathoverflow.net/questions/25307
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Is there any sort of classification of (say finite) groups with the property that every subgroup is normal?
Of course, any abelian group has this property, but the quaternions show commutativity isn't necessary.
If there isn't a classification, can we at least say the group must be of prime power order, or even a power of two?
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https://mathoverflow.net/users/5513
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Groups with all subgroups normal
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These are called [Dedekind groups](https://en.wikipedia.org/wiki/Dedekind_group), and the non-abelian ones are called Hamiltonian groups. The finite ones were classified by Dedekind, and the classification extended to all groups by Baer. The non-abelian ones are a direct product of the quaternion group of order 8, an elementary abelian 2 group, and a periodic abelian group of odd order (or all of whose elements have odd order).
Periodic abelian groups all of whose elements have odd order can be quite complicated, but the finite ones are direct products of cyclic groups.
Your example does not have the property that all of its subgroups are normal when n ≥ 4. The subgroup generated by x1\*x2\*x3 is not normal, since (x1\*x2\*x3)^x4 = (a*x1)*(a*x2)*(a*x3) = a*x1\*x2\*x3, but x1\*x2\*x3 has order 2. For n = 3, your group is Q8 x 2, and so is Hamiltonian.
The cyclic group of order 6 and the direct product Q8 x 3 are two groups of non-(prime power) order with every subgroup normal.
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https://mathoverflow.net/users/3710
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A few years ago, somebody told me a lovely problem. I suspect there may be more to it (which I would be interested in learning), and would very much like to find a reference, it makes me uncomfortable to use it in class without being able to point to its source.
The problem is as follows. I'll post the solution I know, which is the reason I like it, as an answer, to give a bit of a chance to people who read it and want to think about it without being spoiled.
Assume the natural numbers are partitioned into finitely many arithmetic progressions. Then two of these progressions must have the same common difference.
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https://mathoverflow.net/users/6085
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Finitely many arithmetic progressions
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You're probably thinking of the proof, via generating functions, due to D J Newman. I don't have a reference to the first appearance in print, but it's in his book, A Problem Seminar, problem 90, on page 18, with solution on page 100.
I suppose that when you state the problem you must require finitely many but *at least two* arithmetic progressions.
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https://mathoverflow.net/questions/25249
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Let $a\_1,$ $a\_2,$ $\ldots,$ $a\_n$ be positive real numbers. Prove that
$$\sqrt{\frac{a\_1^2+\left( \frac{a\_1+a\_2}{2}\right)^2+\cdots +\left(\frac{a\_1+a\_2+\cdots +a\_n}{n}\right)^2}{n}} \le \frac{a\_1+\sqrt{\frac{a\_1^2+a\_2^2}{2}}+\cdots+\sqrt{\frac{a\_1^2+a\_2^2+\cdots +a\_n^2}{n}}}{n}.$$
I have proved this inequality for $n=2$ and $n=3.$ But I still cannot prove it for the general case. Can somone help me?
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https://mathoverflow.net/users/6180
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Another mixed mean inequality
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Mixed mean inequalities have been studied quite a bit, inspired mostly by the inequalities of [Carleman](http://en.wikipedia.org/wiki/Hardy%27s_inequality) and [Hardy](http://en.wikipedia.org/wiki/Hardy%27s_inequality), starting probably from [this](http://www.jstor.org/stable/2589560) article of K. Kedlaya. Note that the following holds if $r < s$ (in your case $r=1, s=2$):
$$\left(\frac{1}{n}\sum\_{k=1}^n \left(M\_k^{[r]}(\mathbf a)\right) ^s\right)^{1/s}\leq \left(\frac{1}{n}\sum\_{k=1}^n\left(M\_k^{[s]}(\mathbf a)\right) ^r \right)^{1/r}$$
Where $\mathbf a$ denotes a sequence of real numbers, and $M\_k^{[r]}$ denotes the $r$-th [power mean](http://en.wikipedia.org/wiki/Power_mean) of the first $k$ variables. You can find a proof of the general form in "Survey on classical inequalities" by T.M. Rassias (page 32, it is the original proof of B. Mond and J. Pecaric which was later extended to matrices and linear operators).
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https://mathoverflow.net/users/2384
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25319
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https://mathoverflow.net/questions/24923
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9
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Let me start with a simple observation. Suppose $f$ is a weight two newform of level $p^3$. Write $d$ for the size of the Galois orbit $f^\sigma, \sigma \in \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$. Then $d \geq (p-1)/2$. The proof is quite simple: associated to $f$ is an abelian variety $A\_f$ of dimension $d$ and conductor $p^{3d}$. If a prime $p$ divides the conductor of an abelian variety of dimension $d$, and $p>2d+1$, then the maximal power of $p$ dividing the conductor is $p^{2d}$ (I believe this is a theorem of Serre-Tate). Hence if we had $d < (p-1)/2$, this would immediately imply the absurd inequality $3d<2d$, so contradiction. Of course this also works for newforms of weight two and higher prime power level.
My questions:
1. Is this written down in the literature somewhere? It seems like such a simple observation that I want to guess that it is, but I cannot find a reference.
2. Is the same result true for higher weight newforms?
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https://mathoverflow.net/users/1464
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Galois orbits of newforms with prime power level
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See Section 4 of my paper [The Hecke algebra T\_k has large index](http://www.math.northwestern.edu/~emerton/pdffiles/index.pdf), joint with Frank Calegari, where this sort of argument is applied. Lemma 4.1 is a variant of the statement you prove; see also the last sentence.
In any case, the argument does generalize to higher weight newforms; it is just a statement
about $p$-torsion elements in $GL\_n(\mathbb Z\_p)$ (there are none when $p > n + 1$). This
in turn implies that a Galois rep. into $GL\_n(\mathbb Z\_p)$ must be tamely ramified at $p$
if $p > n + 1$, and hence has conductor at $p$ bounded by $p^n$ (because the exponent of the conductor is then just the codimension of the fixed subspace under tame inertia). Note also
(see our paper) that you can replace $\mathbb Z\_p$ by any unramified extension, and thus
that the argument controls not only the degree of the coefficient field of a newform, but also the ramification of $p$ in the coefficient field (the larger the degree of the coeff.
field, the more ramified $p$ has to be).
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https://mathoverflow.net/questions/25320
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Suppose the independent number of a graph is bounded. How small the clique number can be? linear?
It seems to be a natural problem to ask. but I could not find any reference.
Thanks.
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https://mathoverflow.net/users/6200
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Suppose the independent number of a graph is bounded. How small the clique number can be?
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This is basically a question in Ramsey theory. The Ramsey number $R(s, t)$ is the minimum integer $n$ for which every red-blue coloring of the edges of a complete $n$-vertex graph induces either a red complete graph of order $s$ or a blue complete graph of order $t$. So for example Kim (*Random Structures and Algorithms* 7 (1995), 173-207) showed that $R(3,t)\asymp t^2/\log t$. Roughly speaking this means that a graph with $t^2/\log t$ vertices with independence number at most 2 must have a clique of size $t$. So in this case the clique number grows like the square root of the number of vertices of the graph. I'm not sure what the best known results are in general but this should point you to the right literature.
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https://mathoverflow.net/questions/25285
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I am just looking for a basic introduction to the Podles sphere and its topology. All I know is that it's a q-deformation of $S^2$.
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https://mathoverflow.net/users/1358
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Introduction to the Podles Sphere
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First of all some terminology. One usually talks about Podles spheres, since they are a one parameter family. If you say **the** Podles sphere you probably mean the one that is often referred to as the **standard** one. My indications will refer to the whole family.
You do not clarify your background and your directions. The family of Podles spheres has been used as an example in all possible meaning of the word *quantization*, I guess you could easily find a list of hundred of references about it. I'll stick to some paper that can be considered as foundational in various aspects.
First: as NC C\*-algebra the first reading should be Podles original paper, which is simple and well written P.Podles Quantum spheres, Lett.Math.Phys. 14, 193-202 (1987).
From the point of view of quantum homogeneous spaces, i.e. *-coideal subalgebras in Hopf-*-algebras I strongly suggest M.Dijkhuizen and T. Koornwinder Quantum homogeneous spaces, duality and quantum 2-spheres, Geom.Dedicata 52, 291-315 (1994).
In both these approaches, more or less evidently, q-special functions pop up at some point.
M. Noumi and Mimachi K., Quantum 2-spheres and big q-Jacobi polynomials, Comm. Math. Phys. 128, 521-531 (1990) is the one not to avoid.
Last but not least you may be interested in understanding Podles spheres as deformations of Poisson structures (say à la Rieffel), which is beautifully explained in A.Sheu, Quantization of the Poisson SU(2) and its Poisson homogeneous space - the 2 sphere, Comm. Math. Phys. 135, 217-232 (1991). I suggest that here you first read the appendix by Lu and Weinstein where the Poisson structures are explained neatly and simply, and then go the quantization part (some of which may result rather technical at first).
Then, of course, as mathphysicist mentioned, the whole issue of putting spectral triples opens up; the literature there is much more scattered (several attempts and several choices as well) and I guess one should just simply dive into open sea and see what happens...
ADDED:
Personally I would start with the paper by Dijkhuizen and Koornwinder that settles the algebraic part (generators and relations) and has a down-to-earth approach, without technicalities (if you know a little about Hopf algebras). I would not dismiss the paper by Noumi and Mimachi if you're interested in spectral triples. q-special functions means harmonic analysis on the sphere: it shouldn't surprise it is important if you look for Dirac operators satisfying some invariance condition.
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https://mathoverflow.net/users/6032
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https://mathoverflow.net/questions/25344
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Is there any "well-known" algebraically closed field that is uncountable other than $\mathbb{C}$ ?
The algebraic closure of $\mathbb{C}(X)$ would work, but is it meaningful, is this field used in some topics ?
Have you other examples ?
Thank you.
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https://mathoverflow.net/users/6187
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uncountable algebraically closed field other than C ?
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The algebraic closure of $\mathbb{F}\_p((t))$ is uncountable of characteristic $p$. It comes up naturally in number theory and algebraic geometry.
For every characteristic $p \geq 0$ and uncountable cardinal $\kappa$, there is up to isomorphism exactly one algebraically closed field of characteristic $p$ and cardinality $\kappa$. The examples of $\mathbb{C}$ and closures of Laurent series fields as above give you the ones of continuum cardinality and all characteristics. Indeed I do not know any specific reason to consider algebraically closed fields of larger than continuum cardinality.
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If you want to compute crystalline cohomology of a smooth proper variety $X$ over a perfect field $k$ of characteristic $p$, the first thing you might want to try is to lift $X$ to the Witt ring $W\_k$ of $k$. If that succeeds, compute de Rham cohomology of the lift over $W\_k$ instead, which in general will be much easier to do. Neglecting torsion, this de Rham cohomology is the same as the crystalline cohomology of $X$.
I would like to have an example at hand where this approach fails: Can you give an example for
>
> A smooth proper variety $X$ over the finite field with $p$ elements, such that there is no smooth proper scheme of finite type over $\mathbb Z\_p$ whose special fibre is $X$.
>
>
>
The reason why such examples *have* to exist is metamathematical: If there werent any, the pain one undergoes constructing crystalline cohomology would be unnecessary.
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https://mathoverflow.net/users/5952
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Lifting varieties to characteristic zero.
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This paper of Serre gives an example (I've justed pasted I. Barsotti's math-sci review).
(The paper can be found in Serre's "Collected Works vol. II 1960-1971)
>
> Serre, Jean-Pierre Exemples de
> variétés projectives en
> caractéristique $p$ non relevables en
> caractéristique zéro. (French) Proc.
> Nat. Acad. Sci. U.S.A. 47 1961
> 108--109.
>
>
> An example of a non-singular
> projective variety $X\_0$, over an
> algebraically closed field $k$ of
> characteristic $p$, which is not the
> image, $\text{mod}\,p$, of any variety
> $X$ over a complete local ring of
> characteristic 0 with $k$ as residue
> field. The variety $X\_0$ is obtained
> by selecting, in a 5-dimensional
> projective space $S$, and for $p>5$, a
> non-singular variety $Y\_0$ which has
> no fixed point for an abelian finite
> subgroup $G$ with at least 5
> generators of period $p$, of the group
> $\Pi(k)$ of projective transformations
> of $S$, but which is transformed into
> itself by $G$; then $X\_0=Y\_0/G$. The
> reason for the impossibility is that
> $\Pi(K)$, for a $K$ of characteristic
> 0, does not contain a subgroup
> isomorphic to $G$. {Misprint: on the
> last line on p. 108 one should read
> $s(\sigma)=\exp(h(\sigma)N)$.}
>
>
>
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https://mathoverflow.net/questions/25354
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Let me first run through the setting of my question in an example I understand well; that of modular curves. If $Y\_1(N)$ denotes the usual modular curve over the complexes, the quotient of the upper half plane by the congruence subgroup $\Gamma=\Gamma\_1(N)$, then there are two kinds of sheaves that one often sees showing up in the theory of automorphic forms in this setting:
1) Locally constant sheaves. The ones showing up typically come from representations of $\Gamma$ of the form $Symm^{k-2}(\mathbf{C}^2)$, with $\Gamma$ acting in the obvious way on $\mathbf{C}^2$. These sheaves---call them $V\_k$---are related to classical modular forms of weight $k$ via the Eichler-Shimura correspondence. They only exist for $k\geq2$ (weight 1 forms are not cohomological) and representation-theoreticically the sheaves are associated to representations of the algebraic group $SL(2)$ (the reason one starts in weight 2 rather than weight 0 is that there is a correction factor of "half the sum of the positive roots").
2) Coherent sheaves. The ones showing up here are powers $\omega^k$ of a canonical line bundle $\omega$ coming from the universal elliptic curve. The global sections of $\omega^k$ (which are bounded at the cusps) are classical modular forms of weight $k$. Although there are no classical modular forms of negative weight, the sheaf $\omega^k$ still makes sense for $k<0$ (in contrast to case 1 above). I am much vaguer about what is conceptually going on here. I have it in my mind that here $k$ is somehow a representation of the group $SO(2,\mathbf{R})$.
Now my question: what is the generalisation of this to arbitrary, say, PEL Shimura varieties? Part (1) I understand: I can consider algebraic representations of the reductive group I'm working with and for each such gadget I can make a locally constant sheaf. But Part (2) I understand less. I am guessing I can construct a big vector bundle on my moduli space coming from the abelian variety. Now, given some representation of some group or other, I can build coherent sheaves somehow, possibly by "changing the structure group" somehow. For which representations of which group does this give me a coherent sheaf on the moduli space?
Basically---what is the general yoga for supplying natural *coherent* sheaves on Shimura varieties, which specialises to the construction of $\omega^k$ in the modular curve case, and which explains why $\omega^k$ exists even for $k<0$?
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https://mathoverflow.net/users/1384
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Constructing coherent sheaves on Shimura varieties.
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If the Shimura variety is attached to the Shimura data $h:\mathbb S \to G\_{/\mathbb R}$,
and if as usual $K$ denotes the centralizer in $G\_{/\mathbb R}$ of $h$,
then the automorphic vector bundles are $G(\mathbb R)$-equivariant bundles on $X:= G(\mathbb R)/K(\mathbb R)$ attached to representations of the algebraic
group $K$.
Specifically, if $V$ is the representation (over $\mathbb C$, say), then the associated
vector bundle $\mathcal V$ is given by $(G(\mathbb R)\times V)/K(\mathbb R).$
Note that if $G = GL\_2$ and we are in the modular curve case, then $K$ is equal to
$\mathbb C^{\times}$ (thought of as an algebraic group over $\mathbb R$), its complexification is $\mathbb G\_m\times \mathbb G\_m$, and so there are two integral
parameters describing its irreps (which are just one-dimensional in this case). One of them is the usual $k$ of $\omega^k$; the other can be chosen so as to correspond to twisting by a power of the determinant, which does not change the underlying
line bundle, but changes the equivariant structure (which we don't think about so explicitly in the classical language).
Now $K$ is the Levi in a parabolic $P$ (defined over $\mathbb R$), and there is an open embedding (of complex analytic spaces)
$X := G(\mathbb R)/K(\mathbb R) \hookrightarrow G(\mathbb C)/P(\mathbb C) =: D$.
(If you look at Deligne's Corvalis article and unravel things, you'll see that this is
how the complex structure on $X$ is defined: in the Hodge-theoretic formulation given there, $P$ is the parabolic preserving the Hodge filtration on the Hodge-structure corresponding to the base point of $X$.)
The automorphic vector bundles are naturally defined on the flag variety $D$
(as $\mathcal V := (G(\mathbb C)\times V)/P(\mathbb C)$), and then restricted to $X$.
(So in the modular curve case, $D$ is the projective line, and $\omega^k$ comes from
$\mathcal O(k)$.)
Now the automorphic bundles, being $G$-equivariant, descend to bundles on each
Shimura variety quotient $Sh(X,K\_f)$ of $X \times G(\mathbb A\_f)$, and have canonical models defined over the reflex field. In the PEL case, one will be able to construct these canonical models using data from the abelian varieties (think about how the abelian varieties give rise to Hodge structures with Mumford--Tate group equal to $G$ in the first place). In general, this result is due to Milne (see his answer).
Added: A colleague pointed out to me, regarding Kevin's initial question,
that it can be understood much more generally. Namely, all automorphic
vector bundles are locally free coherent sheaves, equivariant under the
action of the finite adeles, but not all coherent sheaves necessarily
have these properties! Practically nothing is known about the $K$-theory of
Shimura varieties, although the question is obviously of fundamental
interest.
The very deep current work on cycles on Shimura varieties, from various
points of view, must be the beginning of a substantial theory whose ultimate
shape no one is in a position to imagine. So Kevin's question should be
seen
as a program for future collaboration between number theorists and algebraic
geometers (at least).
Also, let me remark that I'm told that the term "automorphic vector bundle" was invented in
a conversation between Michael Harris and Jim Milne in the second half of the 80s.
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https://mathoverflow.net/questions/25323
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14
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First, yes, I've seen Mumford's paper of this title. I'm actually interested in specific ones, and looking for really the most elementary/elegant proof possible.
I'm told that for $g\geq 2$ it is known that the Picard groups of $\mathcal{M}\_g$ and $\mathcal{A}\_g$ (the moduli spaces of curves of genus $g$ and abelian varieties of dimension $g$) are both isomorphic to $\mathbb{Z}$ (at least, over $\mathbb{C}$). What's the most efficient way to compute this? In fact, for $\mathcal{M}\_g$, it's even generated by the Hodge bundle, I'm told. Ideally I want to avoid using stacks (though if stacks give an elegant proof, I'm open to them) and also would like to be able to calculate the degrees of some natural bundles, though I get that that's going to be a bit harder, so I want to focus this question on the computation of the Picard group.
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https://mathoverflow.net/users/622
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Picard Groups of Moduli Problems
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I'll just talk about the calculation of $\text{Pic}(\mathcal{M}\_g)$ as a group (showing that it is generated by the Hodge bundle is then a calculation).
I think the most elementary way to view this problem is to think in terms of orbifolds rather than stacks. Recall that $\mathcal{M}\_g$ is the quotient of Tecichmuller space $\mathcal{T}\_g$ by the mapping class group $\text{Mod}\_g$ (this is the curves analogue of $\mathcal{A}\_g$ being the quotient of the Siegel upper half plane by the symplectic group). This action is properly discontinuous but not free (that's why we have an orbifold/stack rather than an honest space). A line bundle on $\mathcal{M}\_g$ is then a $\text{Mod}\_g$-equivariant line bundle on $\mathcal{T}\_g$. There is an equivariant first Chern class homomorphism $c\_1 : \text{Pic}(\mathcal{M}\_g) \rightarrow H^2(\text{Mod}\_g;\mathbb{Z})$. Mumford showed that $H^1(\text{Mod}\_g;\mathbb{Z})=0$, so $\text{Pic}(\mathcal{M}\_g)$ cannot vary continuously. This implies that $c\_1$ is injective. Later, Harer proved that $H^2(\text{Mod}\_g;\mathbb{Z}) \cong \mathbb{Z}$ for $g$ large. Since the Hodge bundle is nontrivial, $c\_1$ cannot be the zero map, so we conclude that $\text{Pic}(\mathcal{M}\_g) \cong \mathbb{Z}$.
Let me now recommend three places that contain more details about the above point of view. First, Hain has a survey entitled "Moduli of Riemann Surfaces, Trancendental Aspects", a large portion of which is devoted to the calculation of the Picard group. He gives many more details of the above sketch. He also shows how to show that the Hodge bundle generates the Picard group. Second, in the first couple of sections of my paper "The Picard Group of the Moduli Space of Curves With Level Structures" I give some extra details about things like Chern classes of orbifold line bundles. Finally, Hain has another survey "Lectures on Moduli Spaces of Elliptic Curves" in which he works all the above out for the moduli space of elliptic curves, where things are a little more concrete.
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16
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https://mathoverflow.net/users/317
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25359
| 16,624 |
https://mathoverflow.net/questions/25227
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9
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Recently, in response to deciding the Continuum Hypothesis $CH$, Hamkins and Gitman have proposed consider a multiverse of set-theoretic universes, some in which $CH$ is true, some in which $\neg CH$ is true (and some in which $CH$ is not a relevant hypothesis?).
In formulating logical languages, there has been an outstanding problem: that of deciding the Law of the Excluded Middle $p\vee\neg p$. In classical logic $p\vee\neg p$ is true, but in intuitionistic logic this is not the case. With $CH$, the pragmatic mathematician tries to avoid invoking $CH$; if he assumes $CH$ or $\neg CH$, he will state it clearly. In everyday mathematics practice, the mathematician does use $p\vee\neg p$, but we do see an effort to give *effective constructions*, and *hard estimates in analysis.*
It seems that people are using classical logic basically because it's a core logic, in analogy to the constructible universe $L$ in which $CH$ is true. It is the logic first discovered (and often wrongly attributed as the Platonic choice) and the consistency and strength of other languages proven in terms of this core logic. Afterwards people come up with other logics, like Brouwer coming up with intuitionistic logic, in which a fundamental principle, the Law of the Excluded Middle, does not hold. It seems to me that this debate regarding the Law of Excluded Middle can be formalized using the multiverse approach.
So has anyone tried to use the multiverse approach towards considering these plurality of languages? Perhaps by considering a multiverse of topoi?
**PS.** The difficulty for arithemetizable syntax to describe continuous properties of geometric/measure-theoretic space is central (may I say?) to the difficulty of deciding $CH$. In a like fashion, the difference between classical and intuitionistic logic plays up in comparing the Dedekind-reals and the Cauchy-reals. In other words, continuous properties have different descriptions in these two logics. I'm hoping that work on the Law of the Excluded Middle will shed some light on trying to use discrete languages to model continuous properties. There's more where this comes from, but it is enough for now...
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https://mathoverflow.net/users/nan
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Using the multiverse approach to decide the law of the exluded middle?
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I am not sure I understand all remarks that Colin made, and I disagree with some of them, but I can comment on the idea of "multiverse". Let us consider the following positions:
1. There is a standard mathematical universe.
2. There are many mathematical universes.
3. There is a multiverse of mathematics.
These are supposed to be informal statements. You may interpret "universe" as "model of set theory ZF(C)" or "topos" if you like, but I don't want to fix a particular interpretation.
I would call the first position "naive absolutism". While many mathematicians subscribe to it on some level, logicians have known for a long time that this is not a very useful thing to do. For example, someone who seriously believes that there is just one mathematical universe would reject model theory as fiction, or at least brand it as a technical device without real mathematical content. Such opinions can actually slow down mathematical progress, as is historically witnessed by the obstruction of "Euclidean absolutism" in the development of geometry.
The second position is what you get if you take model theory seriously. Set theorists have produced many different kinds of models of set theory. Why should we pretend that one of them is the best one? You might be tempted to say that "the best mathematical universe is the one I am in" but this leads to an unbearably subjective position and strange questions such as "how do you know which one you are in?" At any rate, it is *boring* to stay in one universe all the time, so I don't understand why some people want to stick to having just one cozy universe.
I need to explain the difference between the second and the third position. By "multiverse" I mean an ambient of universes which form a structure, rather than just a bare collection of separate universes. The difference between studying a "collection of universes" and studying a "multiverse" is roughly the same as the difference between Euclid's geometry and [Erlangen program](http://en.wikipedia.org/wiki/Erlangen_program) -- both study points and lines but conceptual understanding is at different levels. Likewise, a meta-mathematician might prove interesting theorems about models of set theory, or he might consider the overall structure of set-theoretic models.
It should be obvious at this point that there cannot be just one notion of "multiverse". I can think of at least two:
* Set theory studies the multiverse of models of ZF. The structure is studied via notions of forcing, and probably other things I am not aware of.
* Topos theory studies the multiverse of toposes. The structure of the multiverse is expressed as a 2-category of toposes (and geometric morphisms).
I do not mean to belittle set theory, but in a certain sense topos theory is more advanced than set theory because it uses *algebraic methods* to study the multiverse (I consider category theory to be an extension of algebra). In this sense the formulation of forcing in terms of complete Boolean algebras by Scott and Solovay was a step in the right direction because it brought set theory closer to algebra. Set theorists should learn from topos theorists that *transformations* between set-theoretic models are far more interesting than the models themselves.
In the present context the question "classical or intuitionistic logic" becomes "what kind of multiverse". If multiverse is "an ambient of universes, each of which supports the development of mathematics" then taking our multiverse to be either too small or to big will cause trouble:
* if the multiverse is too small, we will be puzzled by its ad hoc properties and we will look in vain for overall structure (imagine doing analysis with only rational numbers),
* if the multiverse is too big, its overall structure will be poor and it will include universes whose internal mathematics is too far removed from our own mathematical experience (imagine doing analysis on arbitrary rings--I am sure it's possible but it's unlike classical analysis).
Topos theory gains little by restricting to Boolean toposes. I have never heard a topos-theorist say "I wish all toposes were Boolean". Also, toposes occurring "in nature" (sheaves on a site) typically are not Boolean, which speaks in favor of intuitionistic mathematics.
An example of an ad hoc property in too small a multiverse occurs in set theory. We construct models of set theory by forming Boolean-valued sets which are then quotiented by an ultrafilter. What is the ultrafilter quotient for? The algebraic properties of Boolean-valued sets are hardly improved when we pass to the quotient, not to mention that it stands no chance of having an explicit description. A possible explanation is this: we are looking only at one part of the set-theoretic multiverse, namely the part encompassed by Tarskian model theory. Our limited view makes us think that the ultraquotient is a necessity, but the construction of Boolean-valued models exposes the ultraquotient as a combination of two standard operations (product followed by a quotient). We draw the natural conclusion: a model of classical first-order theory should be a structure that measures validity of sentences in a general complete Boolean algebra. The Boolean algebra $\lbrace 0,1 \rbrace$ must give up its primacy. What shall we gain? Presumably a more reasonable overall structure. At first sight I can tell that it will be easy to form products of models, and that these products will have the standard universal property (contrast this with ultraproducts which lack a reasonable universal property because they are a combination of a categorical limit and colimit). Of course, there must be much more.
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16
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https://mathoverflow.net/users/1176
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25366
| 16,628 |
https://mathoverflow.net/questions/25348
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6
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Let $R$ be a ring. Consider the inclusion functor from the category of finitely generated, projective $R$-modules to the category of all finitely generated $R$-modules. For which rings does it have a left adjoint.
For example for any principal ideal domain $R$, we have the structure theorem of f.g. $R$-modules. The upper left adjoint is given by $M\mapsto M/torsion$.
In general there is no reason, why such a construction should exist. I tried to check the assumptions for Freyds adjoint functor theorem. But it seems quite hard to check whether, they are satisfied for a given ring $R$.
So does anyone know a counterexample, where such a adjoint doesn't exist or a weaker property than PID, that implies the existence ?
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https://mathoverflow.net/users/3969
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For which rings does a projectivization of modules exist?
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A necessary condition is that, for any finitely generated R-module M, the Hom-object $Hom\_R(M,R)$ is a finitely generated projective right R-module. This is because if a left adjoint existed, this would be isomorphic to the set $Hom\_R("Proj(M)",R)$, and this is a summand of a free right module.
Consider the case $R = \mathbb{Z}/p^2$, $M = \mathbb{Z}/p$, $P = R$. Then $$Hom\_{\mathbb{Z}/p^2}(\mathbb{Z}/p, \mathbb{Z}/p^2) \cong \mathbb{Z}/p$$
and so this ring admits no projectivization.
Conversely, if this condition is satisfied then we can define $D\_R M = Hom\_R(M,R)$, and $DM$ is always finitely generated projective with a natural double-duality map $M \to D\_{R^{op}} D\_R M$ which is an isomorphism for projective modules. This provides the desired adjoint, because any map $M \to P$ produces a natural factorization
$$M \to DDM \to DDP \leftarrow^{\sim} P$$
which is the desired adjoint.
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6
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https://mathoverflow.net/users/360
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25372
| 16,631 |
https://mathoverflow.net/questions/25375
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12
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Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of these two vector spaces of cardinalities $c$ and $c\times c = c$, so they are isomorphic as vector spaces over $\mathbb{Q}$.
Is there a way to prove that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups which does not use AC? Are there models of set theory in which these groups are not isomorphic? I'm also curious whether there is a proof (perhaps using AC) which does not make use of this vector space machinery, though I'm fairly sure the above proof is the simplest.
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https://mathoverflow.net/users/5963
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AC in group isomorphism between R and R^2
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You cannot prove that $\mathbb{R}$ and $\mathbb{R}^{2}$ are isomorphic in $ZF$. To see this, note that the map $(x,y) \mapsto (x,0)$ is a nontrivial non-surjective additive endomorphism of $\mathbb{R}^{2}$. Assuming the existence of suitable large cardinals, every map $f: \mathbb{R} \to \mathbb{R}$ is measurable in $L(\mathbb{R})$ and it follows that the only additive endomorphisms of $\mathbb{R}$ in $L(\mathbb{R})$ are the maps $x \mapsto rx$ for some $r \in \mathbb{R}$. Thus $\mathbb{R} \not \cong \mathbb{R}^{2}$ in $L(\mathbb{R})$.
There is an interesting open problem related to your question of whether an isomorphism can be found which does not make use of the vector space machinery.
If we add a Ramsey ultrafilter $\mathcal{U}$ to $L(\mathbb{R})$, then we obtain an interesting model $L(\mathbb{R})[\mathcal{U}]$ which contains the ultrafilter $\mathcal{U}$ while still retaining some of the "nice" properties of $L(\mathbb{R})$. In particular, neither $\mathbb{R}$ nor $\mathbb{R}^{2}$ has a basis as a vector space over $\mathbb{Q}$ in $L(\mathbb{R})[\mathcal{U}]$ . However, it is unknown whether or not $\mathbb{R} \cong \mathbb{R}^{2}$ in $L(\mathbb{R})[\mathcal{U}]$. In other words, does an ultrafilter help in trying to construct such an isomorphism?
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12
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https://mathoverflow.net/users/4706
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25380
| 16,634 |
https://mathoverflow.net/questions/25374
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17
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[Restated from stackoverflow](https://stackoverflow.com/questions/2872959/algorithm-how-to-tell-if-an-array-is-a-permutation-in-on/2874631#2874631):
Given:
* array a[1:N] consisting of integer elements in the range 1:N
Is there a way to detect whether the array is a permutation (no duplicates) or whether it has any duplicate elements, in O(N) steps and O(1) space without modifying the original array?
**clarification:** the array takes space of size N but is a given input, you are allowed a fixed amount of additional space to use.
---
The stackoverflow crowd has dithered around enough to make me think this is nontrivial. I did find a few papers on it citing a problem originally stated by Berlekamp and Buhler (see ["The Duplicate Detection Problem", S. Kamal Abdali, 2003](https://stackoverflow.com/questions/2872959/algorithm-how-to-tell-if-an-array-is-a-permutation-in-on/2874631#2874631))
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https://mathoverflow.net/users/1305
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duplicate detection problem
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It's at least possible to test whether the input is a permutation with a randomized algorithm that uses O(1) space, always answers "yes" when it is a permutation, and answers "yes" incorrectly when it is not a permutation only with very small probability.
Simply pick a hash function $h(x)$, compute $\sum\_{i=1}^n h(i)$, compute $\sum\_{i=1}^n h(a[i])$, and compare the two sums.
Ok, some care needs to be used in defining and choosing among an appropriate family of hash functions if you want a rigorous solution (and I suppose we do want one, since we're on mathoverflow not stackoverflow). Probably the simplest way is just to fill another array $H$ with random numbers and let $h(x)=H[x]$, but that is unacceptable because it uses too much space. I'll leave this part as unsolved and state this as a partial answer rather than claiming full rigor at this point.
See also my paper [Space-Efficient Straggler Identification in Round-Trip Data Streams via Newton's Identitities and Invertible Bloom Filters](http://arxiv.org/abs/0704.3313) which solves a more general problem (if there are O(1) duplicates, say which ones are duplicated, using only O(1) space) with the same lacuna in how the hash functions are defined. It also contains a proof that an algorithm that makes only a single pass over the data cannot solve the problem exactly and deterministically, but of course that doesn't apply to algorithms with random access to the input array.
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16
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https://mathoverflow.net/users/440
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25384
| 16,637 |
https://mathoverflow.net/questions/25379
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7
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The Cauhy-Crofton formula relates the length of a curve to an integral over lines:
$$
\text{length} (\gamma) = \frac14\iint n\_\gamma(\varphi, p)\; d\varphi\; dp,
$$
where $\gamma$ is a curve and $n\_\gamma(\varphi, p)$ is the number of times the line defined by the angle $\varphi$ and the distance to the origin $p$ intersects the curve.
This formula has had a major impact on image analysis in the last decade. It gives a method to discretize functionals used in e.g. image segmentation and solve them with graph methods.
My question is: is there a similar formula, but for curvature instead? I.e.
$$
\int\_\gamma |\kappa(x)|^p dx = ?
$$
Now using lines would not be enough, bur perhaps circles?
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https://mathoverflow.net/users/818
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Cauchy-Crofton formula for curvature
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I doubt it. Let's first dissect your Cauchy-Crofton formula. Up to minor technical assumptions, what we have is that
$$\mbox{length}(\gamma) = \int\_{\mathbb{R}^2} \delta\_\gamma(x) d^2x$$
where $\delta\_\gamma(x)d^2x$ represents the measure concentrated on $\gamma$. Locally you can think of its as the pull back of the Dirac delta via the charaterization function of $\gamma$ as a submanifold.
Next we do a change of coordinates. Observe that (not being too precise here) $\mathbb{R}^2 = TS^1 \times \mathbb{R} / S^1$ (basically you can rotate a standard coordinate system; in other words, the set of all lines on the plane can be identified with $TS^1$, and we take its Cartesian product with $\mathbb{R}$ to measure along each of the lines). So let $\pi$ be the canonical projection map from $TS^1\times \mathbb{R}$ to $\mathbb{R}^2$. Then your integral can be re-written as
$$ \mbox{length}(\gamma) = \frac{1}{|S^1|}\int\_{TS^1\times \mathbb{R}} \pi^\*\delta\_\gamma(y) dy$$
where $\pi^\*\delta\_\gamma(y) dy$ is the pull back measure. Now you integrate out the fibers $\mathbb{R}$ first and by simple geometry you see that
$$ \int\_{TS^1\times\mathbb{R}}\pi^\*\delta\_\gamma(y) dy = \int\_{TS^1} n(s) ds $$
where $n(s)$ is the number of times the line indexed by the coordinate $s$ intersects $\gamma$. So up to some normalization constants we recover the Cauchy-Crofton formula.
Now, if you want to get an integral of the curvature $k$ along the curve, you can write it as
$$ \mbox{curvature integral} = \int\_{\mathbb{R}^2} |k(x)|^p \delta\_\gamma(x) d^2x$$
the procedure of lifting to the space of lines is no problem, so you can again get an integral over $TS^1\times \mathbb{R}$. The problem is that I can't see any obvious way of integrating out the additional factor of $\mathbb{R}$. In the Cauchy-Crofton case, each time the line hits your curve it picks up a unit bundle of mass. In the curvature case, you pick up some number which depends on the curvature at the point of intersection.
Using circles won't help either. You can consider a foliation (or almost a foliation) of $\mathbb{R}^2$ by some family of curves, each of which can be identified with some curve $m$. Then you can do the same thing as above: let $M$ be the parameter space of this foliation (the foliation can be moved under rigid motion transformations to get a new foliation) then $\mathbb{R}^2$ can be identified as $M\times m / G$ by $G$ being some subgroup of the Euclidean symmetries. So formally your integral can be re-written as
$$ \frac{1}{|G|}\int\_{M\times m} \mbox{something} dy $$
but to simplify down to just an integral over the parameter space $M$, you need to integrate out along the fibre $m$, and to get a simple expression at the end you almost certainly need that the family of curves $m$ in your foliation must be very special. Unless $m$ is adapted to $\gamma$ such that the integral along $m$ of $|k|^p\delta\_\gamma$ can be easily evaluated, you have no hope of arriving at a simple integral expression.
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5
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https://mathoverflow.net/users/3948
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25387
| 16,639 |
https://mathoverflow.net/questions/25386
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9
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Is it true that the intersection of a maximal ideal in $A[x]$ with $A$ is a maximal ideal in $A$?
Let's say A is Noetherian. I would be surprised if it isn't true but somehow I can't seem to show it. Any help or tip will be appreciated. Thanks!
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https://mathoverflow.net/users/5292
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Maximal ideal in polynomial ring
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No, it's not true in general. E.g. the pricipal ideal generated by $px - 1$ is maximal
in $\mathbb Z\_p[x]$ (for any prime $p$); the quotient $\mathbb Z\_p[x]/(p x - 1)$ is precisely the field $\mathbb Q\_p$. However, the intersection of this ideal with $\mathbb Z\_p$ is equal to the zero
ideal, which is not maximal.
If the ring $A$ is Jacobson, then the result you want is true. (For in this case, if $\mathfrak m$ is a maximal ideal in $A[x]$, then
$A[x]/\mathfrak m$ is a finite type $A$-algebra which is a field, therefore is finite over $A$
by the Jacobson hypothesis, and so the image of $A$ in $A[x]/\mathfrak m$ (which is equal to
$A/A\cap \mathfrak m$) is itself a field, and so $A\cap \mathfrak m$ is maximal in $A$.)
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14
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https://mathoverflow.net/users/2874
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25388
| 16,640 |
https://mathoverflow.net/questions/25389
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16
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Let $M$-be a differentiable manifold. Then, suppose to capture the underlying geometry we apply the singular homology theory. In the singular co-chain, there is geometry in every dimension. We look at the maps from simplexes, look at the cycles and go modulo the boundaries. This has a satisfying geometric feel, though I need to internalize it a bit more(which matter I tried to address in other questions).
Now on the other hand, let $\Omega^1$ be the space of $1$-forms on the space. The rest of the de Rham complex comes out of this object, wholly through algebraic processes, ie by taking the exterior powers and also the exterior derivative. After getting this object in hand, the journey upto getting the de Rham cohomology ring is entirely algebraic.
And by the de Rham theorem, this second method is equally as good as the more geometric first method. In the second method no geometry is explicitly involved anywhere in any terms after the first term. So the module of $1$-forms somehow magically capture all the geometry of the space $M$ without need of any explicit geometry. This is amazing from an algebraic point of view since we have less geometric stuff to understand.
This makes me wonder for the conceptual reason why this is true. I know that one should not look a gift horse in the mouth. But there is the need to understand why there is such a marked difference in the two approaches to capturing the geometry in a manifold, viz, through de Rham cohomology in differential topology, and through singular homology in algebraic topology. I would be grateful for any explanations why merely looking at all the $1$-forms is so informative.
Edited in response to comments: I meant, the de Rham cohomology is as good as singular homology for differentiable manifolds. What is "geometry of rational homotopy type"? And what is "geometry of real homotopy type"?
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https://mathoverflow.net/users/6031
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Is a conceptual explanation possible for why the space of 1-forms on a manifold captures all its geometry?
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When you say that the geometry of the whole space 'just comes out of' the space of 1-forms, there is naturally a huge amount of machinery hiding underneath the surface: $\Omega^n$ does not just drop out under successive exterior derivatives $d^n$: $d^2$ is the zero map- in fact the whole point of DeRham cohomology is that the exterior derivative is not surjective.
What's really going on is with the ring structure of $\Omega^\*$- the wedge product of differential forms is designed to mimic the determinant for subspaces not neccessarily equal the whole space, so that we can calculate eg. signed 2-volume in a 4-dimensional space with ease. (This follows from the fact that the n-volume of an n-parallellopiped is [up to scalar] the unique skew symmetric n-linear form on an n-dimensional subspace).
Wedged asortments of 1 forms added together give different weightings of n-volume to different orientations of n-planes in the tangent space- giving a way of measuring n-volume locally, knitting together into a global picture across a manifold of how to measure for example an n-dimensional embedded submanifold.
Edit: okay so the real question here is why can De Rham cohomology do what homology does (ie count the k-dimensional holes in a space)? What follows will be hands-wavy, but less so than my original finale.
Suppose we have a k dimensional hole in our manifold, with no intention of providing proof or even consistent definitions I claim the following moral argument:
1. Let's pretend that our hole looks normalish, let's say we have a neighbourhood of the hole 'homotopic' to $\mathbb{R}^k\setminus0$
2. The volume form of $S^{k-1}$, '$\omega$' is a k-1-form in $\mathbb{R}^k\setminus0$ (though not yet a smooth k-1-form when considered in the ambient space)
3. $\alpha:=\omega(\frac{x}{||x||})$ is a smooth k-1 form $\in \Omega^{k-1}(\mathbb{R}^k\setminus0)$
4. We can see because of the $||x||$ bit that if we have a copy of $S^{k-1}$ which we measure with this form, expanding it or contracting it will not change a thing, homotopies that cause overlap will be sorted by the volume form bit (negative and positive orientations will cancel each other out)
5. Thus, morally, $d\alpha=0$
6. Now suppose we had a $\beta$ with $d\beta=\alpha$, well then we could use it to measure half of an ebbedded $S^{k-2}: S^{k-2}\_+ $ so let $\beta(S^{k-2}\_+)=b$
7. By stokes', as we rotated our $S^{k-2}\_+$ around through the spare dimension giving say $r(S^{k-2}\_+)$ then $\beta(r(S^{k-2}\_+))-b$ would be the k-1 spherical volume in between them
8. Rotating the full way around would give a value, simultaneously of $vol\_{k-1} (S^{k-1})$ and 0
9. so there is no $\beta$, and $\alpha \in H^{k-1}(\mathbb{R}^k \setminus 0)$
10. Bump it, extend it by zero, add dimensions by the pullback of the projection, pull it back through a chart and you've probably got an element of your original cohomology group
11. Question:So we can count the holes- do we count anything else and ruin our counting? Answer: nope- to get that kind of homotopy invariance needs a singularity- and smooth sections of the cotangent bundle don't take too kindly to singularities unless they're in a hole where the manifold isn't.
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19
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https://mathoverflow.net/users/5869
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25393
| 16,644 |
https://mathoverflow.net/questions/25371
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6
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Suppose I have a simple, simply connected (linear) algebraic group $\mathcal{G}$ over an algebraically-closed field $k$, which could have any characteristic. In fact, to keep things simple, let's imagine $\mathcal{G}$ is simply-laced. Let $\mathcal{B}$ be a choice of Borel, and $\mathcal{T}$ be a maximal torus for $\mathcal{B}$ (and hence for $\mathcal{G}$). Our $\mathcal{B}$ gives a choice of simple positive roots. I'll write $\mathbb{G}\_M$ for the multiplicative group of $k$.
Now, suppose we look at the Dynkin diagram for $\mathcal{G}$, and imagine removing a single node from the diagram; for simplicity let's imagine this doesn't disconnect the diagram. Then I believe we should be able to find a subgroup $\mathbb{G}\_M\times\mathcal{G}'$ in $\mathcal{G}$ and a Borel subgroup $\mathcal{B}'$ in $\mathcal{G}'$, such that the inclusion $\mathbb{G}\_M\times\mathcal{B}'\hookrightarrow \mathcal{G}$ factors through $\mathcal{B}$; such that the Dynkin diagram for $\mathcal{G}'$ is just the Dynkin diagram for $\mathcal{G}$ with our chosen node removed; and indeed such that when we use the inclusion $\mathcal{G'}\hookrightarrow\mathcal{G}$ to map the simple roots of $\mathcal{G}'$ determined by $\mathcal{B'}$ into roots of $\mathcal{G}$, and hence map the Dynkin diagram of $\mathcal{G}'$ into the Dynkin diagram of $\mathcal{G}$, we get the obvious inclusion of 'the diagram with the node removed' into 'the original diagram'.
Moreover, if instead I did disconnect the diagram, I should get a similar picture, with a subgroup $\mathbb{G}\_M\times\mathcal{G}\_1\times\dots\times\mathcal{G}\_k$ where the groups $\mathcal{G}\_k$ have Dynkin diagrams corresponding to the components of the picture I get by removing the node from the original Dynkin diagram.
My question is: a) am I right that this kind of operation is all legal and above board, b) are there any caveats one should be aware of, and c) what are good references to appeal to to make this kind of thing rigorous.
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https://mathoverflow.net/users/3513
|
Chopping up Dynkin diagrams
|
Brian's comment does what you want, and describes the *almost* direct product caveat.
A standard and excellent reference for all things of this nature is Demazure's "Sous-groupes Paraboliques des groupes réductifs," Exposè XXVI of SGA3. I believe that Borel and Tits earlier work "Groupes Réductifs" has similar results covered, in the case of a base field instead of base scheme. Maybe this would be better for your question.
There's no need for the assumptions simple, simply-connected, simply-laced, nor the assumption that you're working over an algebraically closed field. The general setting is a reductive group $G$ over a base scheme $S$.
For a (connected) reductive group $G$ over an algebraically closed field, once you fix $T \subset B \subset G$ as you wish, there's a bijection between subsets of the vertex-set of the Dynkin diagram and parabolic subgroups $P$ of $G$ containing $B$ (so-called standard parabolics). Each parabolic subgroup has a standard Levi decomposition $P = LU$, where $U$ is the unipotent radical of $P$, and $L$ is a reductive group containing $T$.
When $G$ is a simply-connected group, the group $L$ has simply-connected semisimple derived subgroup $L'$. The Dynkin diagram of $L'$ is the subdiagram induced by subset of the vertex-set of the Dynkin diagram of $G$.
Getting the isogeny class right -- e.g. identifying a precise surjective map from $G\_m^r \times L' \rightarrow L$ with finite kernel -- is kind of a pain, in my experience. Typically, I just "know the answer" for classical groups (e.g., Jim's example, in $SL\_3$, one finds a Levi $L \cong GL\_2$) and for certain exceptional groups. Occasionally, I have to do the tedious work of determining the root datum (in the sense of Borel) of the Levi to figure it out (or more commonly looking through the literature to find someone else who's done it). Oh how I wish that this were implemented in SAGE.
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https://mathoverflow.net/users/3545
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25394
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https://mathoverflow.net/questions/25332
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4
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I am certainly sure that any one who has read Gil Kalai's witty community wiki has benefited a lot.
Here I follow a similar track in asking this question.
So let's compose a list of fundamental theorems in mathematics which may not even have the tag "fundamental" but have serious wight in the respective branch of math.
I will start with the elementary and very popular ones.(Please add a description if the theorem is fundamental but still not so well-known)
Thanks for all your effort.
1. FTA: The Fundamental Theorem of Arithmetic (or Unique-Prime-Factorization Theorem):
->Any integer greater than 1 can be written as a unique product (up to ordering of the factors) of prime numbers.
2. FTA: The Fundamental theorem of Algebra:
-> The field of complex numbers is algebraically closed
3. FTC: The fundamental theorem of calculus:
-> Has two parts and specifies the relationship between the two central operations of calculus: differentiation and integration.
4. FTLP: The fundamental theorem of linear programming:
-> In a weak formulation, states that the maxima and minima of a linear function over a convex polygonal region occur at the region's corners.
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https://mathoverflow.net/users/5627
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Fundamental theorems
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In his book *Topics in Geometric Group Theory*, Pierre de la Harpe calls the following result the **Fundamental Observation of Geometric Group Theory** (though he also calls it a theorem!). It is also often called the Svarc--Milnor Lemma. Roughly speaking, it asserts that the coarse geometry of a group is captured by any suitably nice action of that group by isometries on a metric space.
**Theorem.** Let $X$ be a metric space that is geodesic and proper, let $\Gamma$ be a group and let $\Gamma$ act properly discontinuously and cocompactly by isometries on $X$. Then $\Gamma$ is finitely generated, and furthermore for any $x\_0\in X$ the map $\Gamma\to X$ given by
$\gamma\mapsto\gamma x\_0$
is a quasi-isometry.
**Remarks.**
1. $\Gamma$ is endowed with the word metric (with respect to some choice of finite generating set).
2. A map of metric space $f:Y\to X$ is a *quasi-isometric embedding* if there are constants $\lambda\geq 1$, $\mu\geq 0$ such that
$\lambda d\_Y(y\_1,y\_2)+\mu\geq d\_X(f(y\_1),f(y\_2))\geq \frac{1}{\lambda} d\_Y(y\_1,y\_2)-\mu$
for all $y\_1,y\_2\in Y$. It is a *quasi-isometry* if, furthermore, for every $x\in X$ there is $y\in Y$ such that $d(x,f(y))\leq \mu$.
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5
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https://mathoverflow.net/users/1463
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25397
| 16,648 |
https://mathoverflow.net/questions/25365
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1
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The Wikipedia article on coherent sheaves makes the following claim (without any reference), which I had trouble proving or finding a reference for: on an algebraic variety *X* (or I guess possibly even on a locally noetherian scheme), the coherent sheaves can be defined as the smallest class of sheaves of $\mathcal{O}\_X$-modules with the following two properties:
i) the sheaf $\mathcal{O}\_X$ is itself coherent;
ii) if, in a short exact sequence of sheaves, two of the sheaves are coherent, then so is the third.
I'm skeptical, but I would still like to know if this is true. If so, does anyone know a reference?
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https://mathoverflow.net/users/1310
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Characterisation of coherent sheaves on an algebraic variety
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This is false. Every sheaf in that class would have zero first Chern class, since $c\_1$ is additive over short exact sequences.
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https://mathoverflow.net/users/4716
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25401
| 16,650 |
https://mathoverflow.net/questions/25402
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57
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Ben Green and Terrence Tao proved that there are arbitrary length arithmetic progressions among the primes.
Now, consider an arithmetic progression with starting term $a$ and common difference $d$. According to Dirichlet's theorem(suitably strengthened), the primes are "equally distributed" in each residue class modulo $d$. Therefore we imagine that the Green-Tao theorem should still be true if instead of primes we consider only those positive primes that are congruent to $a$ modulo $d$. That is, Green-Tao theorem is true for primes within a given arithmetic progression.
Question: Is something known about this stronger statement?
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https://mathoverflow.net/users/6031
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Is the Green-Tao theorem true for primes within a given arithmetic progression?
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The Green-Tao is true for any subset of the primes of positive relative density; the primes in a fixed arithmetic progression to modulus $d$ have relative density $1/\phi(d)$.
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137
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https://mathoverflow.net/users/1464
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25403
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https://mathoverflow.net/questions/25383
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4
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I'm trying to understand Proposition 2.9 of this [paper](http://arxiv.org/abs/0809.3031v2) on weakly group theoretical fusion categories.
First of all I have a problem with understanding the settings for de-equivariantization process. It is written on page 5 of the paper that one needs $\mathcal{E}=\mathrm{Rep}(G)\subset Z(\mathcal{C})$ such that $\mathcal{E}$ embeds into $\mathcal{C}$ via the forgetful functor $Z(\mathcal{C})\rightarrow \mathcal{C}$.
**Question 1.**
I would like to know what "embeds" means exactly.
Rephrasing when a functor is an embedding functor is? Is it enough to be injective on objects and to send simple objects into simple objects. Obviously this is necessary.
**Question 2.** In a concrete example where $\mathcal{C}=\mathrm{Rep}(A)$ for a Hopf algebra $A$ what would be the conditions to be imposed for the corresponding composition to be an embedding.
Is the corresponding category $\mathcal{C}\_G$ the category of representations of a Hopf algebra, i.e does it posses a fiber functor?
**Question 3.** The third question I have is regarding the de-equivariantization $\mathcal{E}'\_G $ from the second part of the proof proposition.
I understood the construction of $\mathcal{E}$ and clearly $\mathcal{E}\subset \mathcal{E}'$ since $\mathcal{E}$ is symmetric. The question I have is why after composing to the restriction functor $Z(\mathcal{E}') \rightarrow \mathcal{E}'$ one still has an inclusion.
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https://mathoverflow.net/users/2805
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De-equivariantization by Rep(G)
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Sebastian, sorry for confusion. Here are the answers:
Question 1. "embeds" means that the functor ${\mathcal E}\to {\mathcal C}$ is fully faithful.
Equivalently, this functor sends simple objects to simple ones and non-isomorphic simple objects to non-isomorphic ones (so your guess is correct).
Question 2: In the situation when ${\mathcal C}=Rep(A)$ this means that the group algebra of $G$ is a Hopf quotient of the Hopf algebra $A$ and the functor ${\mathcal E}\to {\mathcal C}$ is isomorphic to the pullback functor $Rep(G)\to Rep(A)$. We do not expect that the category ${\mathcal C}\_G$ has a fiber functor in this situation (It should not be too difficult to
construct an explicit counterexample but I do not remember it right now).
Question 3: Here you compose embedding ${\mathcal E}\to {\mathcal E}'$ with functor
${\mathcal E}'\to Z({\mathcal E}')$ (which comes from the braiding on ${\mathcal E}'$) and functor $Z({\mathcal E}')\to {\mathcal E}'$. But notice that the composition ${\mathcal E}'\to Z({\mathcal E}')\to {\mathcal E}'$ is just the identity functor, so in the end of the day you compose embedding ${\mathcal E}\to {\mathcal E}'$ with the identity functor.
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9
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https://mathoverflow.net/users/4158
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25406
| 16,653 |
https://mathoverflow.net/questions/25399
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4
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In complexity theory, when a uniform family of circuits recognises a language is it the case that each of the input gates is on a path to the output gate?
That is, there are no input gates with wires connected to gates that do not eventually connect to the output gate.
The definitions found in the literature are general enough for sections of the circuit to not connect to the output gate.
However, it seems to me that without assuming that all input gates have a path to the output graph some things, for example reductions, get needlessly tricky.
Has anyone seen this mentioned before?
Is it a well known assumption?
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https://mathoverflow.net/users/2644
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Is every input gate of a Boolean Circuit (to decide a language) on a path to the output gate?
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It's not necessarily the case that each input gate is on a path to the output gate. Determining if there is a path in a directed (acyclic) graph from one node s to another node t is **NLOGSPACE**-complete, so it is not a condition that you can arbitrarily enforce on (say) **LOGSPACE**-uniform circuits. It is easy to enforce this condition without loss of generality on **P**-uniform circuits: if you only require that the circuit be constructed in polynomial time, then after building the circuit it is easy to compute the strongly connected components of the graph representing the circuit, and throw out all gates that do not have a path to the output gate. Hence the inputs that do not eventually connect to the output would simply not be present. (Note I am not sure if this is precisely what you want.)
Still, I can't think of any typical usages of circuits where some input gates may not have a path to the output. Most reductions involving circuits are from machines to circuits, and we typically assume that a machine reads all its input bits. And when one is trying to prove a circuit lower bound, one always considers functions that depend on every bit.
**UPDATE:** Here's a silly way to enforce the condition you want, but I don't know if it will help you in your particular case. Suppose for simplicity that the number of inputs $n$ is a power of $2$. Make a complete binary tree of $2n-1$ new gates (where all edges lead towards the root of the tree), so we have $n$ leaves. Label all nodes in the tree as $AND$ gates. For the $i$th leaf in the tree, lead in the inputs $x\_i$ and $\neg x\_i$. Now $OR$ the root of this binary tree with the output gate of your original circuit, and make this $OR$ your new output gate. Clearly every input now has a path to an output gate and the functionality of the circuit has not changed. Moreover this transformation is extremely uniform; I am pretty sure it can be made **DLOGTIME**-uniform. But is this really what you need for your problem?
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https://mathoverflow.net/users/2618
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25412
| 16,656 |
https://mathoverflow.net/questions/25360
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I remember to have read that the L-function of an elliptic curve, which a priori only converges for $\Re s > \frac{3}{2}$ also converges at $s=1$ provided that the $L$-function
satisfies the functional equation.
I always thought that this is due to the fact that in this case the L-function is also the L-function of a modular form and in this case we have better convergence. However, the modular forms which correspond to these curves are cusp forms of weight 2 and so have a priori even worse convergence properties, namely convergence for $\Re s > 2$.
So I now wonder, whether I remember correctly and the the claim above is indeed correct?
If so, I would like to see a reference for the proof.
What is the reason for this fact. Does some non-trivial fact about elliptic curves play a role. It will (almost surely) not hold for arbitrary modular forms or cusp forms, since they always satisfy the functional equation. What do we need? An Euler product?
EDIT: The $L$-function extends to an entire function. But what I am interested in is the original series representation. Is it true that the series representation for the $L$-function of an elliptic curve which converges a priori for $\Re s > \frac{3}{2}$, is valid also for a bigger strip $\Re s > c$ with $c<\frac{3}{2}$.
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https://mathoverflow.net/users/3757
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Convergence of L-series
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Wood, see the article by Kumar Murty in Seminar on Fermat's Last Theorem. He shows how the L-series converges (conditionally!) on Re(s) > 5/6, thus in particular at s = 1. You can find the book on Google books and do a search on "5/6" to find the page. OK, I just did that and will tell you: it's on page 15. He proves the theorem for any Dirichlet series converging abs. for Re(s) > 3/2 and having an analytic continuation and suitable functional equation relating values at s and 2-s. (Thus in practice it is a theorem about L-functions of suitable weight 2 modular forms, certainly nothing *directly* about elliptic curves!) He also says that what he describes is a special case of a more general result, with citation.
Where the Dirichlet series converges, it still represents the L-function that may have been analytically continued to the wider region by some other method, since Dirichlet series are analytic on the half-plane to the right of any point where they converge and moreover at any point where they converge the value is the limit of the function taken along the line to the right of the point (Abel's theorem for power series on discs works for Dirichlet series on right half-planes). So we are assured that if you happen to know the series itself converges somewhere new it still equals the orthodox analytic continuation (whatever that means).
On the other hand, the Euler product has surprises. Goldfeld discovered that if the Euler product for an ell. curve over Q converges at s = 1, in the natural sense of partial Euler products over primes up to x as x goes to $\infty$, and the value of the Euler product is nonzero, then this value is *not* L(1) but rather is off from this by a factor of $\sqrt{2}$. Of course there was no real input about elliptic curves directly: Goldfeld was *assuming* the ell. curve was modular (he was writing in the 1980s) and used that right away.
It turns out exactly the same thing happens for *quadratic* Dirichlet L-functions at s = 1/2: if the partial Euler product at s = 1/2 converges to a nonzero value then again you're off by a factor of $\sqrt{2}$ from L(1/2), but in one setting the factor is $\sqrt{2}$ and in the other it's $1/\sqrt{2}$. For non-quadratic Dirichlet L-functions there's no funny business: if the Euler product converges at s = 1/2 to a nonzero value (which, by the way, it always should since nobody expects Dirichlet L-functions vanish at 1/2) then the value will be L(1/2). I first heard about Goldfeld's result at a talk by Karl Rubin (he gave the usual heuristic for BSD conjecture by looking at the Euler product at s = 1 and some wiseguy in the audience asked if it really did converge at s = 1 and Rubin mentioned there was a paper of Goldfeld on that), but when I read Goldfeld's paper I was confused by part of it, so in trying to work it out in the simpler example of Dirichlet L-functions I wound up seeing I could prove the same kind of theorem for any Euler product over a global field having the properties everyone expects it should have. This turns out to be related to properties of symmetric and exterior square Euler products and is morally the same quadratic bias (called Chebyshev's bias) that Sarnak and Rubinstein found when they worked out comparative statistics on the number of primes up to x in different congruence class mod $m$: classes of squares or non-squares exhibit different fine growth rates compared to one another. For more details on the partial Euler products, including some numerical examples, see my paper <http://www.math.uconn.edu/~kconrad/articles/eulerprod.pdf>.
By the way, one can definitely observe this $\sqrt{2}$ business happening numerically but we'll never expect to *prove* it happens since it actually implies the Riemann hypothesis for the relevant L-function. If the product converges to a nonzero value at a point on the critical line it is no real surprise that you could prove the Dirichlet series for the log of the Euler product converges everywhere to the right of the critical line, which implies the Euler product itself converges to a nonzero value everywhere to the right of the critical line, hence Riemann hypothesis. In fact, as I show in that paper, this (suitable) Euler product convergence at a point on the critical line is actually equivalent to something which at present appears to lies deeper than the Riemann hypothesis but is still plausible. There's no lack of results which imply RH but are themselves false, you see. This probably isn't one of them.
In summary, the Dirichlet series for ell. curve L-functions (over $\mathbf Q$) can provably be shown to converge on a wider region than they are usually said to converge and still equal the L-function there, including s = 1, while the Euler product probably does converge at s = 1 too but if the value is not 0 then it's not going to converge to what you expect... unless you think the Riemann hypothesis is false.
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https://mathoverflow.net/users/3272
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https://mathoverflow.net/questions/25431
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3
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A friend of mine in the department needs to know if the following PDE has been extensively studied
$$ u\_t = (u^2)\_{xx}$$
Or more generally, replacing the square by any function of $u$. One would like to know uniqueness and existence of its solution, and smoothness property, for example.
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https://mathoverflow.net/users/4923
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Identifying the nonlinear parabolic PDE $u_t = (u^2)_{xx}$.
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Since $(u^2)\_{xx} = (2uu\_x)\_x$ this is (up to a rescaling) the [porous medium equation](http://mathworld.wolfram.com/PorousMediumEquation.html) for $m=1$.
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5
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https://mathoverflow.net/users/1847
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25433
| 16,664 |
https://mathoverflow.net/questions/25428
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30
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I have heard that [Polylogarithms](http://en.wikipedia.org/wiki/Polylogarithm) are very interesting things. The wikipedia page shows a lot of interesting identities. These functions are indeed supposed to have caught the attention of Ramanujan. Moreover, they seem to be important in physics for various purposes like Bose-Einstein integrals, which I am not really knowledgeable enough to understand. These are all things I have heard from people after I queried "why polylogarithms are interesting".
So this function
$$Li\_s (z) = \sum\_{k=1}^\infty \frac{z^k}{k^s}$$
is very interesting and has a lot of useful properties as I can see. Especially for integral values of $s$.
What is bothering me is the following.
Let $f(z) = \sum a\_n z^n$ be analytic inside the unit disc. That is, the radius of convergence at least $1$. For simplicity, we assume it is $1$, i.e.,
$$r = \frac{1}{\limsup\_{n\rightarrow\infty}\sqrt[n]{|a\_n|}} = 1$$
implying that inside the unit disc, term-by-term differentiation and integration are possible. Since $\left( a\_n \right)^{1/n}$ goes to $1$ as $n$ goes to $\infty$, it is also true that $\left(na\_n \right)^{1/n}$ goes to $1$. Similarly, $\left(\frac{a\_n}{n} \right)^{1/n}$ also goes to $1$. Now, defining
$$f(z,s) = \sum\_{n=1}^\infty \frac{a\_n z^n}{n^s}$$
it is the case that $f\_{2}(z) = f(z,2)$ and in general iterating, $f\_n(z) = f(z, n)$ are analytic inside the unit disc. In fact, uniform bounds hold true for compact sets contained in the domain $|z| \leq 1$ and $|s| \leq K$ for arbitrary $K$, and so $f(z,s)$ is analytic in the domain $D^\circ \times \mathbb C$, where $z \in D^\circ$, the open unit disc, and $s \in \mathbb C$.
So any analytic function in the unit disc can be extended just like the logarithm is extended to polylogarithms. But there is a rich theory about polylogarithms, and I haven't heard of any such theories about other functions analytic inside the unit disc.
So, what makes polylogarithm so amenable to an extended theory of polylogarithms yielding so many results? Why is this not giving such an interesting theory for just any other complex functions?
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https://mathoverflow.net/users/6031
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What is special about polylogarithms that leads to so many interesting identities and applications?
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The reason why polylogarithm are so important/interesting/ubiquitous is they are the simplest non trivial examples of analytical functions underlying variations of mixed Hodge structure. This goes back to Beilinson and Deligne.
A variation of mixed Hodge structure is a very sophisticated gadget. You can think of it as
1. a nice differential equation (the underlying connexion)
2. its solutions (the underlying local system of horizontal sections) with a $\mathbb{Q}$-structure
3. some additional data that make the structure very rigid.
Typical examples of VMHS on $X$ come from the cohomology of families of varieties parametrized by $X$. They can be used to encode the interaction between between topological and arithmetical properties of $X$.
For example, there is a rank 2 variation of mixed Hodge structure $K \in Ext^1\_{VMHS(\mathbb{C}^\times)}(\mathbb{Q},\mathbb{Q}(1))$ over $\mathbb{C}^\times$ known as the Kummer sheaf. The underlying local system $K\_{\mathbb{Q}}$ has fiber $H\_1(\mathbb{C}^\times,\{1,z\};\mathbb{Q})$ at $z$. The underlying connexion is a trivial vector bundle of rank 2 with nilpotent connexion
$$
\nabla = d -
\begin{pmatrix} 0 & 0 \cr %
\frac{dt}{t} & 0
\end{pmatrix}
$$
The "periods" are obtained by integrating the coefficient of the matrix over the paths $\gamma \in H\_1(\mathbb{C}^\times,\{1,z\};\mathbb{Q})$. So we get a the non trivial period by integrating $\frac{dt}{t}$ over paths $[1,z]$, i.e. we get determinations of $\log(z)$. Conclusion: we have an object $K$ in $VMHS(\mathbb{C}^\times)$ "categorizing" the classical logarithm function. On the arithmetic side, the transcendance of $\log(z)$ for generic $z$ mirrors the fact that $H\_1(\mathbb{C}^\times) = \mathbb{Z}$.
The same can be done for polylogarithm functions. The Logarithmic sheaf is the symmetric algebra $Log := Sym(K)$ (it corresponds to the whole family of $\log^n(z)$, $n\in \mathbb{N}$). The Polylogarithm sheaf is a canonical extension $Pol$ of $\mathbb{Q}$ by the restriction of $Log(1)$ to $\mathbb{P}^1\setminus \{0,1,\infty\}$. Its periods encodes the monodromy of the polylogarithm functions in the same way the Kummer sheaf does for the logarithm function. These are the most elementary unipotent variations of mixed Hodge structures.
Now we have "categorized" the classical polylogarithm functions. In fact, this can actually be done on a more fundamental level using only algebraic cycles defined over $\mathbb{Z}$ (this is the motivic story, the variation of mixed Hodge structure being just a realization of the motivic object). This has very interisting arithmetic consequences. For example, specializing to 1, this implies that we have motivic cohomology classes in $H^1(\mathbb{Z},\mathbb{Q}(n))$ whose images under the Hodge regulator corresponds to $\zeta(n)$. Using this picture, the period conjecture then implies that the $\zeta(2n+1)$ are algebraicly independent over $\mathbb{Q}$. To give you an idea of how powerful this intuition is: we can't even prove $\zeta(5)$ is irrational!
In conclusion, the polylogarithm functions are interesting because they correspond to non trivial algebraic cycles. This leads to interactions between the analytical properties of the functions, arithmetics of special values and algebraic geometry. Lots of classical functions should have similar interpretations. For example, there is a similar picture for Euler's Beta and Gamma functions.
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https://mathoverflow.net/users/1985
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25446
| 16,672 |
https://mathoverflow.net/questions/25439
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18
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I am in the (slow) process of editing my [notes on Lie Groups and Quantum Groups (V Serganova, Math 261B, UC Berkeley, Spring 2010](http://math.berkeley.edu/~theojf/QuantumGroups10.pdf). Mostly I can fill in gaps to arguments, but I have found myself completely stuck in one step of one proof. One possibility that would get me unstuck is a positive answer to the following (which may be obviously false or trivial, but I'm not thinking well):
>
> **Question:** Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb K$, and if necessary you may assume that $\mathbb K = \mathbb C$ and that $\mathfrak g$ is semisimple. Then $\mathfrak g$ acts on itself by the adjoint action, and on polynomial functions $f : \mathfrak g \to \mathbb K$ via derivations. A polynomial $f: \mathfrak g \to \mathbb K\,$ is *$\mathfrak g$-invariant* if $\mathfrak g \cdot f = 0$. For example, let $\pi: \mathfrak g \to \mathfrak{gl}(V)$ be any finite-dimensional reprensentation. Then $x \mapsto \operatorname{tr}\_V \bigl(\pi(x)^n\bigr)$ is $\mathfrak g$-invariant for any $n\in \mathbb N$. Is every $\mathfrak g$-invariant function of this form? Or at least a sum of products of functions on this form?
>
>
>
When $\mathfrak g$ is one of the classical groups $\mathfrak{sl},\mathfrak{so},\mathfrak{sp}$, or the exceptional group $G\_2$ the answer is yes, because we did those examples in the aforementioned class notes. But I have no good grasp for the $E$ series, and I don't know if the statement holds for non-semisimples.
What I'm actually trying to prove is a weaker statement, but I figured I'd ask the stronger question, because to me the answer is not obviously "no". The weaker statement:
>
> **Claim:** Let $\mathfrak g$ be a finite-dimensional semisimple Lie algebra over $\mathbb C$. Then every $\mathfrak g$-invariant function is constant on nilpotent elements of $\mathfrak g$. (Recall that $x\in \mathfrak g$ is *nilpotent* if $\operatorname{ad}(x) = [x,] \in \mathfrak{gl}(\mathfrak g)$ is a nilpotent matrix — some power of it vanishes.)
>
>
>
It's clear that the spectrum of any nilpotent matrix is $\{0\}$, and for a semisimple Lie algebra, any nilpotent element acts nilpotently in all representations. For the classical groups, in the notes we exhibited generators for the rings of $\mathfrak g$-invariant functions as traces of representations, and so we can just check the above claim. But we did not do the $E$ series or $F\_4$.
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https://mathoverflow.net/users/78
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Is every G-invariant function on a Lie algebra a trace?
|
The answer to the general question is "no":
>
> If $\mathfrak{g}$ is solvable, by Lie's theorem its commutant $\mathfrak{g}^{\prime}=[\mathfrak{g},\mathfrak{g}]$ is represented by strictly upper triangular matrices in a suitable basis in any finite-dimensional module. Hence all "trace generated" polynomials are zero on $\mathfrak{g}^{\prime}$; in other words, they factor through the abelianization $\mathfrak{g}/\mathfrak{g}^{\prime}$ and are generated by *linear* invariant polynomials. Unless the adjoint action of $G$ with Lie algebra $\mathfrak{g}$ on $\mathfrak{g}^{\prime}$ has a Zariski dense orbit, there are invariant polynomials that cannot be obtained in this way.
>
The answer to the claim is "yes", this is Kostant's theorem from his celebrated paper:
>
> If $G$ is a complex semisimple group then its nullcone $\mathcal{N}\subset\mathfrak{g}$ is the Zariski closure of a single adjoint orbit consisting of regular nilpotent elements.
>
Kostant actually proved that the nullcone is the scheme-theoretic complete intersection defined by $rk\;G$ homogeneous positive degree algebra generators of $\mathbb{C}[\mathfrak{g}]^G$ — this is the connection with the Chevalley theorem mentioned by others. But for the present purpose, it is enough to show that regular nilpotents are Zariski open and dense in $\mathcal{N}\cap\mathfrak{n},$ and a good way of doing it was indicated by David Speyer.
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https://mathoverflow.net/users/5740
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25450
| 16,675 |
https://mathoverflow.net/questions/25473
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8
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For prime *p* sufficiently large, there is always an integer *q* such that *q* is a residue mod *p*, but neither *q*−1 nor *q*+1 are; the number of such residues scales like *p*/8 (and similarly for any sequence of residues/non-residues in three consecutive integers).
What are the best lower bounds on primes *p*, for which such "isolated" residues are guaranteed to exist? Do they exist, for instance, for all *p* ≡ 3 (mod 4) aside from *p* = 3?
(If this is a typical homework problem, please point me to a textbook for which it is an exercise.)
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https://mathoverflow.net/users/3723
|
Isolated quadratic residues in integers mod p
|
I'll write $\chi(x)$ for the Legendre symbol modulo $p$.
Consider
$$f(x)=(\chi(x)+1)(\chi(x-1)-1)(\chi(x+1)-1).$$
Then $f(x)=8$ if $x$ is an isolated quadratic residue and $0$ otherwise
(unless $x$ is $0$, $\pm1$ which are exceptional cases that have to be factored
in to the bookkeeping eventually). Thus $S=\sum\_{x=1}^p f(x)$
is eight times the number of isolated primes plus a fudge factor. But expanding out
$S$ gives sums such as $\sum\chi(x)$ and $\sum\chi(x(x-1))$ which are easy to deal
with, and also $T=\sum\chi(x^3-x)$. This final sum is related to the number of
points on the elliptic curve $y^2=x^3-x$ and so is bounded by $2\sqrt p$ by Hasse's
bound. But if $p\equiv3$ (mod $4$), $T=0$ as the $x$ and $-x$ terms cancel. In this
case one gets an exact formula for the number of isolated quadratic residues.
One can extend this ideal to bigger patterns of residues and nonresidues.
As the polynomials become larger
they are related to hyperelliptic curves, but from the Riemann hypothesis
for curve one can still get bounds.
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https://mathoverflow.net/users/4213
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25480
| 16,694 |
https://mathoverflow.net/questions/25443
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3
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I was told recently that if I take a 2-torus (genus 2) and remove 1 point, then this is homotopy equivalent to a torus with 3 points removed. This may be really easy but I don't see it.
Thank you!
|
https://mathoverflow.net/users/18
|
Homotopy Equivalence of Punctured Tori
|
A surface minus a finite number of points is homotopy equivalent to a bouquet of circles, and two bouquets of circles are homotopy equivalent iff they have the same number of circles.
This two observations and a little picture to see how many circles are involved in your example should do it :)
|
7
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https://mathoverflow.net/users/1409
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25487
| 16,699 |
https://mathoverflow.net/questions/25470
|
19
|
If we let $\Omega\subset\mathbb{R}^d$ with $d=1,2,3$ and define $\mathcal{H}^1(\Omega)=(w\in L\_2(\Omega): \frac{\partial w}{\partial x\_i}\in L\_2(\Omega), i=1,...,d)$. My tutor has repeated several times:
1. If $d=1$ then $\mathcal{H}^1(\Omega)\subset\mathcal{C}^0(\Omega)$.
2. If $d=2$ then $\mathcal{H}^2(\Omega)\subset\mathcal{C}^0(\Omega)$ but $\mathcal{H}^1(\Omega)\not\subset\mathcal{C}^0(\Omega)$.
3. If $d=3$ then $\mathcal{H}^3(\Omega)\subset\mathcal{C}^0(\Omega)$ but $\mathcal{H}^2(\Omega)\not\subset\mathcal{C}^0(\Omega)$.
I was interested in trying to show these relationships. Does anyone know any references that would be useful.
Thanks in advance.
|
https://mathoverflow.net/users/2011
|
When is Sobolev space a subset of the continuous functions?
|
I understand from your post that you'd like to show those facts by yourself first, and not necessarily to approach the whole theory now (I like your approach). Trivial hint: start with smooth functions with compact support in $\Omega$, and try to bound their $L^\infty$ norm in terms of the $H^d$ norm. Also, I suggest that you try building counter-examples by yourself for the case of non-inclusions.
Reference: Brezis' book of Functional Analysis may give you nice hints.
|
16
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https://mathoverflow.net/users/6101
|
25492
| 16,703 |
https://mathoverflow.net/questions/25461
|
10
|
Can you give a non-trivial example of an integer weight cusp form which does not lie in the old subspace and it has $a\_p=0$ for all primes $p$?
If such a form cannot exist then why?
|
https://mathoverflow.net/users/2344
|
Modular forms with prime Fourier coefficients zero
|
Write $f=\sum c\_i f\_i$ as a sum over new eigenforms. Your condition is thus equivalent to $\sum c\_i \lambda\_i(p)=0$ for all $p$. Taking the absolute value squared of this and summing over $p\leq X$ gives
$0=\sum\_{i,j}c\_i \overline{c\_j} \sum\_{p\leq X} \lambda\_i(p)\overline{\lambda\_j(p)}$.
By the pnt for Rankin-Selberg L-functions, the inner sum over primes is $\sim X (\log{X})^{-1}$ if $i=j$, and is $o(X (\log{X})^{-1})$ otherwise. Taking $X$ very large we obtain $0=cX(\log{X})^{-1}+o(X(\log{X})^{-1})$, so contradiction.
|
13
|
https://mathoverflow.net/users/1464
|
25495
| 16,705 |
https://mathoverflow.net/questions/25491
|
27
|
A wonderful piece of classic mathematics, well-known especially to combinatorialists and to complex analysis people, and that, in my opinion, deserves more popularity even in elementary mathematics, is the Lagrange inversion theorem for power series (even in the formal context).
Starting from the exact sequence
$0 \rightarrow \mathbb{C} \rightarrow \mathbb{C}((z)) \xrightarrow{D} \mathbb{C}((z)) \;\xrightarrow{ \mathrm{Res} }\; \mathbb{C} \rightarrow 0,$
and using the simple rules of the formal derivative **D**, and of the formal residue **Res** (that by definition is the linear form taking the formal Laurent series $f(z)=\sum\_{k=m}^{\infty}f\_k z^k \in \mathbb{C}((z))$ to the coefficient $f\_{-1}$ of $z^{-1}$) one easily proves:
>
> (**Lagrange inversion formula**): If $f(z):=\sum\_{k=1}^{\infty}f\_k z^k\in \mathbb{C}[[z]]$ and $g(z):=\sum\_{k=1}^{\infty}g\_k z^k\in\mathbb{C}[[z]]$ are composition inverse of each other, the
> coefficients of the (multiplicative) powers of $f$ and $g$ are linked
> by the formula $$n[z^n]g^k=k[z^{-k}]f^{-n},$$ and in particular (for
> $k=1$), $$[z^n]g=\frac{1}{n} \mathrm{Res}( f^{-n} ).$$
>
>
>
(to whom didn't know it: enjoy computing the power series expansion of the composition inverse of $f(z):=z+z^m$,
or of $f(z):=z\exp(z),$ and of their powers).
**My question**: what are the generalization of this theorem in wider context. I mean, in the same way that, just to make one example, the archetypal ODE $u'=\lambda u$ procreates the theory of semigroups of evolution in Banach spaces.
Also, I'd be grateful to learn some new nice application of this classic theorem.
(notation: for $f=\sum\_{k=m}^{\infty}f\_k z^k \in \mathbb{C}((z))$ the symbol $[z^k]f$ stands, of course, for the coefficient $f\_{k}$)
|
https://mathoverflow.net/users/6101
|
Genealogy of the Lagrange inversion theorem
|
The Lagrange inversion formula allows to give a combinatorial interpretation of the Jacobian conjecture. See for example the classic paper of Bass, Connell and Write, The Jacobian conjecture: reduction of degree and formal expansion of the inverse, <http://www.ams.org/journals/bull/1982-07-02/S0273-0979-1982-15032-7/S0273-0979-1982-15032-7.pdf>.
|
9
|
https://mathoverflow.net/users/4790
|
25496
| 16,706 |
https://mathoverflow.net/questions/25499
|
8
|
Suppose $\alpha > 1$ is irrational. Are there infinitely many primes of the form $\left\lfloor \alpha n \right\rfloor$? Is the number of $p \leq X$ of this form $\sim \alpha^{-1} X (\log{X})^{-1}$? I know this is the kind of thing the circle method was born to do, but I cannot for the life of me find a reference for this!
|
https://mathoverflow.net/users/1464
|
Primes in quasi-arithmetic progressions?
|
I think the uniform distribution mod1 of $\{p/\alpha\}$ is due to Vinogradov, and the asmptotic for primes in a Beatty sequence $\sim \frac{\pi(x)}{\alpha}$ is an immediate consequence. Indeed for $p$ to be equal to some $\lfloor k\alpha\rfloor$ it is equivalent to $1-\frac{1}{\alpha}<\frac{p}{\alpha}-\lfloor \frac{p}{\alpha}\rfloor<1$. So you just need the fractional part of $p/\alpha$ to be on a fixed interval of length $\alpha$ mod1.
On a related note [this](http://arxiv.org/abs/0708.1015) paper discusses the general sequence $q\lfloor \alpha n+\beta\rfloor +a$.
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8
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https://mathoverflow.net/users/2384
|
25510
| 16,712 |
https://mathoverflow.net/questions/25413
|
4
|
Is there an algorithm which, given two polynomials in $n$ variables with real coefficients, $p(x)$, and $q(x)$, will determine whether the zero sets $p^{-1}(0), q^{-1}(0)\subset R^n$, are homeomorphic to each other?
(also same question for polynomials over $C$ with $R^n$ replaced by $C^n$).
|
https://mathoverflow.net/users/5365
|
Determining if two algebraic sets are homeomorphic
|
I think the answer to the "real" version of the question is no. Here are some remarks.
1. One can realize each smooth manifold as a real algebraic variety in a Euclidean space. So one can realize each smooth compact manifold as the zero set of a single polynomial, by taking the sum of the squares of the polynomials that generate the ideal of the corresponding algebraic variety.
2. In dimensions $\leq 7$ every PL-manifold admits a smooth structure. So given a PL-manifold one can "smoothen" it and construct a homeomorphic real algebraic variety. The question is whether this can be done constructively (and moreover so that the resulting variety is defined over the rationals, if one considers the "rational" version). This looks plausible. If it is true then one can use the fact that the homeomorphism problem for PL-manifolds of dimension 4 is undecidable.
|
3
|
https://mathoverflow.net/users/2349
|
25518
| 16,718 |
https://mathoverflow.net/questions/25513
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14
|
This is related to [another question](https://mathoverflow.net/questions/15569) in which it is proved that Zariski open sets are dense in analytic topology.
But it is intuitive that something more is true. Namely, that they are the sets where some polynomials vanish, and consideration of a few examples in $\mathbb R^n$ where they are of Lebesgue measure $0$, suggest strongly that the Zariski-closed sets(except the whole affine space) are of measure $0$ in $\mathbb C^n$ as well. This should be quite simple; but I am unable to prove it due to inexperience in measure theory.
The nice thing about proving this is that once this is done, then we are able to claim safely that so-and-so statement is true almost everywhere, if it is true on a Zariski-open set.
So, in a more measure theoretic formulation:
>
> Let $X$ be a set in $\mathbb C^n$ contained in the zero locus of some collection of polynomials. How to show that $X$ is of measure $0$?
>
>
>
In fact my feeling is that more should be true, ie, we can replace polynomials by analytic functions at least, and get the same result.
|
https://mathoverflow.net/users/6031
|
Zariski closed sets in C^n are of measure 0
|
If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.
Indeed, almost every point of $Z$ is a [density point](http://en.wikipedia.org/wiki/Lebesgue%2527s_density_theorem). It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Thus $f$ has zero Taylor expansion at some point, hence $f\equiv 0$.
|
26
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https://mathoverflow.net/users/4354
|
25519
| 16,719 |
https://mathoverflow.net/questions/25482
|
15
|
One of my friends asked me whether or not the inclusion of the category of Grothendieck toposes into elementary toposes has a left adjoint. We are looking at the categories of geometric morphisms. I am not really sure how to start but nothing seems to rule it out immediately.
|
https://mathoverflow.net/users/1106
|
Is there a "Grothendieckification" functor from elementary toposes to Grothendieck toposes?
|
No, it doesn't. If it did, then it would preserve limits. But the category of Grothendieck toposes and geometric morphisms has a terminal object, namely the category of sets, while there are elementary toposes not admitting any geometric morphism to Set (for instance, any small elementary topos).
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23
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https://mathoverflow.net/users/49
|
25523
| 16,722 |
https://mathoverflow.net/questions/25527
|
13
|
I'm helping to teach an undergraduate algebraic geometry course (out of Hatcher's textbook). We have recently defined the degree of a map of spheres using homology, and the professor and I thought it would be nice if we could give some kind of argument that such a map is determined up to homotopy by its degree. I know two proofs of this: one using the Freudenthal suspension theorem, and the other using the Pontryagin correspondence between homotopy classes of [smooth] maps and framed cobordism classes of framed submanifolds (see Milnor, Topology from a Differential Viewpoint). Unfortunately, neither of these arguments would be accessible to our students, who have only seen the fundamental group and homology (no higher homotopy theory) and who are not necessarily expected to know any differential topology.
Thus, I ask the following question:
>
> Is there an elementary argument (i.e., that can be understood by someone who only knows about homology and the fundamental group) that the degree of a map of spheres determines its homotopy type?
>
>
>
More precisely, what we have (or will have) available is most of the material in the first two chapters of Hatcher, not including the "additional topics."
If necessary, I'm willing to make plausible assumptions that the students may not know how to prove, such as
-Replacing $f \colon S^n \to S^n$ by a homotopic map if necessary, we may assume that there exists points with only finitely many preimages, such that $f$ is a homeomorphism locally about each preimage. (i.e., regular values)
-Every map of CW complexes is homotopic to a cellular map.
-The degree map $\pi\_n(S^n) \to \mathbb{Z}$ is a group homomorphism. [This reduces us to showing that a degree-0 map is nullhomotopic.]
|
https://mathoverflow.net/users/5094
|
An elementary proof that the degree of a map of spheres determines its homotopy type
|
Take a look at Exercise 15 in Section 4.1, page 359 of the book you're referring to. This outlines an argument that should be the sort of thing you're looking for. The main step is to deform a given map to be linear in a neighborhood of the preimage of a point, using either simplicial approximation or the argument that proves the cellular approximation theorem. Once this is done, the rest is essentially the Pontryagin-Thom argument (in a very simple setting), plus the fact that $GL(n,\mathbb R)$ has just two path-components.
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19
|
https://mathoverflow.net/users/23571
|
25533
| 16,728 |
https://mathoverflow.net/questions/25535
|
5
|
The responses to [another question](https://mathoverflow.net/questions/17732/) clarifies that the best known examples of distributions that are not measures, are the derivatives of the delta and such. What I want to know is: Is that the only way a distribution is not a measure?
>
> Are there distributions that are not measures, their derivatives, and their derivatives, and so on?
>
>
>
|
https://mathoverflow.net/users/6031
|
Distributions more complicated than the Dirac δ and derivatives
|
There is a structure theorem for distributions that shows that they are all (possibly infinite, but locally finite) sums of derivatives.
Here's a proper statement. Let $T$ be a distribution on $\mathbb{R}^n$. Then there exists continuous functions $f\_{\alpha}$ such that $T = \sum\_{\alpha} (\frac{\partial}{\partial x})^{\alpha} f\_{\alpha}$, where for each bounded open set $\Omega$, all but a finite number of the distributions $(\frac{\partial}{\partial x})^{\alpha} f\_{\alpha}$ vanish identically on $\Omega$. Here $\alpha$ is a multiindex.
I recommend perusing the beautiful little book "A Guide to Distribution Theory and Fourier Transforms" by Strichartz. The above theorem is discussed in section 6.2.
|
20
|
https://mathoverflow.net/users/317
|
25540
| 16,732 |
https://mathoverflow.net/questions/25532
|
7
|
**Question:**
What does an element of $\mathcal F \big( L^\infty(\mathbb R)\big)$ look like locally?
As formulated, the question might be a bit difficult to answer since the Fourier transform of a function *f* ∈ *L*∞(ℝ) is a distribution, and it is not easy to "write down" a distribution.
So let me first illustrate the situation at hand with an easy example:
>
> ***Example:***
> The Fourier transform of the Heaviside function *H*(x) (i.e. the characteristic function of the positive reals) is given by
> a linear combination of the function 1/x and of the Dirac delta function
> (see [this](http://en.wikipedia.org/wiki/Heaviside_step_function#Fourier_transform) Wikipaedia entry for the exact formula, as well as for the meaning of the distribution "1/x").
>
>
>
The formalism of distributions is bit overkill for talking about measures, and things that look like 1/x.
For example, the primitive of an element of $\mathcal F \big( L^\infty(\mathbb R)\big)$ is
always a function (well defined outside of a set of measure zero).
Using the above observation, we get the following
**Reformulation of the question:**
Let *f*(x) ∈ *L*∞(ℝ) be a function, and let *g*(x) be a primitive of its Fourier transform.
• What can *g*(x) look like locally?
• What local conditions must *g* satisfy in order to have a chance of coming from some
*f* ∈ *L*∞(ℝ)?
• On what kind of sets can *g* fail to be continuous?
|
https://mathoverflow.net/users/5690
|
What does the Fourier transform of an L-infinity function look like locally?
|
It is pretty much the same as to describe the class $G$ of functions $g$ on the circle whose Fourier coefficients decay as $O(|k|^{-1})$. There is no nice "space side" property $P$ that would characterize them but for every nice "space side" property $P$ one can figure out in finite time if it holds for all such functions or not.
As to your particular questions, the answers are
1) On the circle being in this class it is a local property (this needs compactness of the circle) because if $g\in G$, then the product of $g$ and any sufficiently smooth function is in $G$ and you can do partitions of unity.
2) There are obvious inclusions $BV\subset G\subset BMO$ ("bounded variation" and "bounded mean oscillation"). If you need something tighter than that, tell the family of comparison spaces you want to use.
3) Since $\sum\_{k\ge 1} \frac 1kz^k$ is unbounded at $1$ and continuous everywhere else, we can move such spikes around to create a function that is locally unbounded on any closed set we want and discontinuous on any $F\_\sigma$ set we want including the entire circle.
|
8
|
https://mathoverflow.net/users/1131
|
25544
| 16,734 |
https://mathoverflow.net/questions/25541
|
6
|
Let $X$ be a topological space and let $|Sing(X)|$ be the geometric realization of the total singular complex of $X$.
Then $|Sing(X)|$ is a CW complex with one cell for each non-degenerate singular simplex. There's a natural map $f:|Sing(X)|\to X$ and there's a theorem that says that $f$ is a weak homotopy equivalence. That is, $f$ induces isomorphisms of homotopy groups.
Then it seems that Whitehead theorem applies and gives that $f$ is homotopy equivalence as long as $X$ is homotopy equivalent to a CW complex (i.e. $X$ is m-cofibrant). Is that correct?
Is there an example when $f$ is not homotopy equivalence? Any examples that come up in "real life"?
|
https://mathoverflow.net/users/3375
|
Is geometric realization of the total singular complex of a space homotopy equivalent to the space?
|
The map from the (realization of the) singular complex of a space $X$ to $X$ is a homotopy equivalence if and only if $X$ is homotopy equivalent to a CW complex, so to get examples where the map is not a homotopy equivalence you just need spaces that are not homotopy equivalent to CW complexes. There are plenty of these that come up all the time in various places outside algebraic topology. For example, a compact space with infinitely many path-components isn't homotopy equivalent to a CW complex. For an explicit example, take the Cantor set, or more simply just a convergent sequence with its limit point. Suspending these examples gives path-connected examples, and iterated suspensions give examples with higher connectivity. More generally, any compact space having a nonfinitely generated homology group is an example. There are lots of other ways that spaces not homotopy equivalent to CW complexes arise too.
|
15
|
https://mathoverflow.net/users/23571
|
25547
| 16,737 |
https://mathoverflow.net/questions/25557
|
4
|
I'm looking for a proof of that the only spheres with almost complex structure are $S^2$ and $S^6$. I've googled "almost complex structure sphere", but all I get is comments saying that "this fact is well-known".
Are there good write-ups on this topic? Thanks in advance.
|
https://mathoverflow.net/users/nan
|
References on almost complex structures on spheres
|
I think this "well known fact" was proved first by Borel and Serre,
Borel, A., Serre, J. P.: Groupes de Lie et puissances réduites de Steenrod. Amer. J. Math.75, 409–448 (1953)
For a more detailed timeline, see [Differential Geometry: Geometry in mathematical physics and related topics](http://books.google.de/books?id=1s-UePH5KfAC&pg=PA100&lpg=PA100&dq=almost+complex+structure+on+spheres+S%5E2+S%5E6&source=bl&ots=HZDRuFRPB6&sig=T_jsFRQlEIhPPBTvfGHOZp5426I&hl=de&ei=9qn3S9ShO6KemwPUgenHAg&sa=X&oi=book_result&ct=result&resnum=6&ved=0CC8Q6AEwBTgU#v=onepage&q&f=false) by Greene and Yau (p.100).
Or as BCnrd suggested [Almost complex and complex structures](http://books.google.de/books?id=-yz03cskUOgC&pg=PA233&lpg=PA233&dq=almost+complex+structure+on+spheres+S%5E2+S%5E6&source=bl&ots=KxGtjBRHCP&sig=AANscUt-KM5KILqIaPMJ1zRId94&hl=de&ei=9qn3S9ShO6KemwPUgenHAg&sa=X&oi=book_result&ct=result&resnum=3&ved=0CCEQ6AEwAjgU#v=onepage&q&f=false) by C. C. Hsiung (Chapter VI)
|
6
|
https://mathoverflow.net/users/675
|
25566
| 16,752 |
https://mathoverflow.net/questions/25576
|
3
|
I've come across a reference in a paper to the
>
> Hecke relation for the universal R-matrix of a quasi-triangular Hopf algebra.
>
>
>
I've looked around, standard references, online etc, but can't seem to find what this Hecke relation is. Can anyone point me in the right direction, or even better, tell me what it is?
|
https://mathoverflow.net/users/2612
|
Reference for the Hecke relation for the universal R-matrix
|
For the Drinfeld-Jimbo quantum universal enveloping algebras, see Proposition 24 of Chapter 8 in the book Quantum Groups and Their Representations, by Klimyk and Schmudgen. This relation is just in the type A situation, for $\mathfrak{gl}\_n$ or $\mathfrak{sl}\_n$. The relation they get is
$$
(\hat{R} - q)(\hat{R} + q^{-1}) = 0,
$$
where $R$ is the R-matrix for the vector representation of $U\_q(\mathfrak{g})$, $\hat{R} = \tau \circ R$, where $\tau$ is the tensor flip, and all the conventions are those of Klimyk and Schmudgen.
In other situations, the map $\hat{R}$ has more than just two eigenvalues, so the spectral decomposition is more complicated, and the characteristic polynomial (i.e. the Hecke relation) is more complicated.
|
8
|
https://mathoverflow.net/users/703
|
25579
| 16,760 |
https://mathoverflow.net/questions/25572
|
2
|
Given a smooth projective variety $X$ over some algebraically closed field $k$
and a locally free sheaf $R$ of $O\_X$-algebras, e.g. central simple algebras or orders.
If $M$ is a left $R$-module which is locally free over $O\_X$, is it true that $M$ is locally projective over $R$? For example if $X$ is a curve a torsion free $O\_X$-module would be locally projective over $R$. Or do we need more conditions for $R$ and $M$ for this to be true?
Why can one compute $Ext\_R^1(M,M)$ via $H^1(\mathcal{H}om\_R(M,M))$ only in the case $M$ is locally projective over $R$?
|
https://mathoverflow.net/users/3233
|
Connection: locally free - locally projective
|
It is true for locally free sheaves of algebras that are central simple algebras at every point, though. These are known as sheaves of Azumaya algebras; I mention them since they were brought up in the question. I don't have a reference here, but the proof is not hard.
|
2
|
https://mathoverflow.net/users/4790
|
25580
| 16,761 |
https://mathoverflow.net/questions/25505
|
3
|
The ring Zn:={0,1,..,n-1} under addition and multiplication modulo n.
Suppose a,b,c,x $\in$ Zn are nonzero and the cyclic order R(a,b,c) holds, then under what conditions does R(ax,bx,cx) hold ?
|
https://mathoverflow.net/users/3537
|
Cyclic order relation in Zn
|
**Answer.** Under the condition
$$
\left\lbrace\frac{x(b-a)}n\right\rbrace<\left\lbrace\frac{x(c-a)}n\right\rbrace
\qquad\qquad(\*)
$$
where $\lbrace\cdot\rbrace$ denotes the fractional part of a real number.
The condition $R(a,b,c)$ is equivalent to $R(0,b-a,c-a)$ and means that the least residues of $b-a$ and $c-a$ modulo $n$ are ordered in the increasing order as integers. Note that the least residue of an integer $m$ modulo $n$ can be expressed by means of $n\cdot\lbrace m/n\rbrace$.
The relations ($\*$) are hardly related for different values of $x=1,2,\dots$: one can always construct $n$ and residues $a$ and $b$ in such a way that for a given range of $x\in X\_1\cup X\_2$ ($X\_1$ and $X\_2$ are two finite disjoint sets of $\mathbb Z\_{>0}$) the inequality ($\*$) hold for $x\in X\_1$ and does not hold for $x\in X\_2$.
|
4
|
https://mathoverflow.net/users/4953
|
25582
| 16,762 |
https://mathoverflow.net/questions/25472
|
15
|
In the simplest cases, the fundamental group serves as a measure of the number of 2-dimensional "holes" in a space. It is interesting to know whether they capture the following type of "hole".
This example may look pathological, but one must understand where one gets stuck, when one tries to study pathological spaces. It helps one in understanding where exactly all the extra nice conditions are used, and hopefully this type of approach will help in minimizing the number of false beliefs we unconsciously have.
The line with the double origin is the following space. In the union $\{0\} \times \mathbb R \cup \{ 1 \} \times \mathbb R$, impose the equivalence relation $(0, x) \sim (1, x) $ iff $x \neq 0$.
This space is locally like the real line, ie a $1$-manifold in everything except the Hausdorff condition. It is connected, path-connected and semilocally simply connected. Just the sort of nice space you study in the theory of fundamental group and covering spaces, except for the (significant) pathology that it has one inconvenient extra point violating the Hausdorff-ness.
It seems that the usual methods of computing the fundamental group are not working for this space. Van Kampen's theorem in particular does not apply. Also the covering spaces are weird, just like this space. In fact this space would have been a covering of $\mathbb R$, were it not for the condition that the preimage of every point is a *disjoint* union of open sets.
So, what if we try to compute the fundamental group of this space? I would be satisfied to know whether it trivial or not. Say, is the collection of homotopy classes of loops based at $1$ nontrivial? It is possible to speculate that a certain loop based at $1$ which passes through both the origins on this special line, in such a way that it passes through the "upper" origin, ie $(1,0)$ on the way left, and it passes through the lower origin, ie $(0,0)$, ought not to be homotopic to the constant loop based at $1$. But how to go about proving/disproving this statement?
|
https://mathoverflow.net/users/6031
|
Fundamental group of the line with the double origin.
|
The earlier answers showing that the fundamental group of this space is infinite cyclic by determining its universal cover or by constructing a fiber bundle over it with contractible fibers are very nice, but it's also possible to compute $\pi\_1(X)$ by applying the classical van Kampen theorem not to $X$ itself but to the mapping cylinder of a map from the circle to $X$ representing the supposed generator of $\pi\_1(X)$, namely the map that sends the upper and lower halves of $S^1$ to arcs in $X$ from $+1$ to $-1$ in the two copies of $\mathbb R$ in $X$. Decompose the mapping cylinder into the two open sets $A$ and $B$ which are the complements of the two "bad" points in $X$ (regarding $X$ as a subspace of the mapping cylinder). Taking a little care with the point-set topology, one can check that $A$, $B$ and $A\cap B$ each deformation retract onto the circle end of the mapping cylinder. Then van Kampen's theorem says that $\pi\_1$ of the mapping cylinder, which is isomorphic to $\pi\_1(X)$, is isomorphic to the free product of two copies of $\mathbb Z$ amalgamated into a single $\mathbb Z$.
An interesting fact about $X$ is that it is not homotopy equivalent to a CW complex, or in fact to any Hausdorff space. For if one had a homotopy equivalence $f:X \to Y$ with $Y$ Hausdorff then $f$ would send the two bad points of $X$ to the same point of $Y$ so $f$ would factor through the quotient space of $X$ obtained by identifying these two bad points. This quotient is just $\mathbb R$ and the quotient map $X \to \mathbb R$ is not injective on $\pi\_1$, so the same is true for $f$ and $f$ can't be a homotopy equivalence.
|
30
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https://mathoverflow.net/users/23571
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25584
| 16,764 |
https://mathoverflow.net/questions/25578
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8
|
Suppose (Ln) is a sequence of loops in a torus S1 × S1 converging in the Hausdorff metric to some set L in the torus. Suppose also that for each loop Ln the projection map p:S1 × S1 -> S1 defined by p(x,y) =x when restricted to Ln is not null-homotopic, then can we conclude that the restriction of the projection map p to L is also not null homotopic? Or are there counter examples?
|
https://mathoverflow.net/users/996
|
limiting behaviour of converging loops on a torus
|
Your intuition is correct: the map p : L → *S*1 is not nullhomotopic.
If p : L → *S*1 was nullhomotopic, then it would factor through the universal cover ℝ of *S*1.
The inclusion ι : L → *S*1 × *S*1 would therefore factor through ℝ × *S*1.
Let ι' : L → ℝ × *S*1 denote a lift of ι,
and let L(0) := ι'(L).
The subset L(0) ⊂ ℝ × *S*1 is then one of the connected components of the preimage of L in ℝ × *S*1.
Let us call the other components L(*n*) for *n* ∈ ℤ.
Let *C*0 ⊂ ℝ × *S*1 be a loop in that
separates L(0) from L(1).
We may also assume that *C*0 is not nullhomotopic in ℝ × *S*1 [this still needs a small argument...].
The projection *C* ⊂ *S*1 × *S*1 of *C*0 is then a loop in the complement of L,
and
represents the element (0,1) of ℤ×ℤ = π1(*S*1 × *S*1).
Since L lies in the complement of *C* and L*n* → L in the Hausdorff metric,
there exists an *n* such that L*n* lies in the complement of *C*.
This contradicts your assumtion that p : L*n* → *S*1 is nullhomotopic.
|
4
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https://mathoverflow.net/users/5690
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25589
| 16,768 |
https://mathoverflow.net/questions/25223
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1
|
Considering a complex algebraic group G defined over the reals, one knows from an article of Borel and Harish-Chandra (Arithmetic subgroups of algebraic groups, Annals of Mathematics 75 (1962)) that G is reductive (as a complex group) if and only if G(R), the subgroup of its real points, is reductive (as a real group). One natural question is whether the same is true replacing the reals by an arbitrary field (say of characteristic 0) and the complexes by its algebraic closure?
Clarifying:
If $G$ is an affine reductive algebraic group defined over $k$ we know that $G$ can be seen as a subgroup of $Gl(n,\bar{k})$, with $\bar{k}$ the algebraic closure of k. Let $G(k)=G\cap Gl(n,k)$. My question is if $G(k)$ is also reductive? Meaning reductive when the unipotent radical of the group is trivial.
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https://mathoverflow.net/users/6179
|
Reductive groups question
|
I'm a little hesitant to say anything in the face of all the comments above, but I think Ana is asking for an answer to the following question: If $G$ is a reductive algebraic $k$-group, and we choose:
1. an algebraic closure $\overline{k}$ of $k$,
2. a $\overline{k}$-group embedding $G\_{\overline{k}} \hookrightarrow GL\_n$,
3. a set-theoretic normal subgroup $U \trianglelefteq G(k)$ such that any element $g \in U$ has unipotent image in $GL\_n(\overline{k})$ under the composition $\operatorname{Spec} \overline{k} \to \operatorname{Spec} k \overset{g}{\to} G \to GL\_n$,
then is $U$ necessarily trivial?
I think the answer is yes, but I don't know a complete proof. When $k$ is infinite, the Zariski closure of $U$ in $G$ is a normal unipotent subgroup (at least, it sounds plausible). When $k$ is finite, you may have to eliminate small order cases by hand.
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0
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https://mathoverflow.net/users/121
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25590
| 16,769 |
https://mathoverflow.net/questions/25597
|
5
|
I was reading this question:
[limiting behaviour of converging loops on a torus](https://mathoverflow.net/questions/25578/limiting-behaviour-of-converging-loops-on-a-torus)
And I wanted to be able to give an argument along the lines of: "If your loops are converging in your torus, their projections must converge in your $S^1$", but a quick google search gives me no results along these lines- do they exist? If not why not?
I am aware that if either of your spaces are unbounded then a sensible topology isn't particularly forthcoming, but is there a situation in which a result of this form can make sense? As a starting point let's set the bar at:
>
> Do compact fibrations induce maps on their subsets that are continuous wrt Hausdorff distance?
>
>
>
Can we do better? Can we do a little worse? Or does none of this make sense?
|
https://mathoverflow.net/users/5869
|
Do continuous maps give continuity in the 'topology' of Hausdorff distance?
|
Any uniformly continuous map $f$:X→Y between metric spaces induces a uniformly continuous map $C\mapsto \overline{f(C)}\ $ between the spaces of closed subsets wrto the Hausdorff distances; in fact with the same modulus of continuity. (Just recall that the Hausdorff distance between A and B is less than δ if and only if for any a∈A there is some b∈B with d(a,b)<δ and for any b∈B, there is some a∈A with d(a,b)<δ).
PS: As to the topologic side of the question (stability of topological or homotopical properties of subsets of a space under perturbations in Hausdorff distance). As far as I know these things usually work well in an ANR metric space X, because of the homotopy extension property. For instance, any closed subset of X that is contractible in X (meaning that the inclusion map $C\to X$ is null homotopic) has a contractible nbd in X. If X is also compact, the nbd is a uniform nbd, so if a sequence of closed sets $C\_n$ converges in the Hausdorff sense to C, the $C\_n$ are eventually contractible in X, too. Generalizing a bit, we may also say that the Lusternik-Schnirelman category of C (minimum cardinality of a contractible closed covering of C) is lower semicontinuous wrt the Hausdorff distance (the category of the limit is larger or equal to the limit of the category along the sequence). Reference (for ANR's and homotopy): e.g. the first 2 chapters of Spanier book.
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5
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https://mathoverflow.net/users/6101
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25601
| 16,775 |
https://mathoverflow.net/questions/25423
|
13
|
I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}\_q({\mathfrak sl}\_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}\_N$, and let $R$ be a universal R-matrix for ${\cal U}\_q({\mathfrak sl}\_N)$ . If we denote the usual dual pairing of ${\cal U}\_q({\mathfrak sl\_N})$ with
$SU\_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$R^{ir}\_{js} = \langle R, u^i\_j \otimes u^r\_s \rangle = q^{\delta\_{ir}}\delta\_{ij}\delta\_{rs} + (q-q^{-1})\theta (i-r)\delta\_{is}\delta\_{jr}.$$ A natural question to ask is whether such a formula exists for $(R^{-1})\_{js}^{ir}=\langle R^{-1}, u^i\_j \otimes u^r\_s \rangle$. An obvious guess would be to take the inverse of the matrix $[R\_{js}^{ir}]\_{i,r,j,s}$.
That is, to guess that $$[(R^{-1})\_{js}^{ir}] \_{i,r,j,s}$$
is equal to $$([R\_{js}^{ir}]\_{i,r,j,s})^{-1}.$$ This guess is confirmed by the fact that
$$
\delta\_{ij}\delta\_{rs} = \langle R R^{-1},u^i\_j \otimes u^r\_s\rangle = \sum\_{k,l} \langle R,u^i\_k \otimes u^r\_l \rangle \langle R^{-1},u^k\_j \otimes u^l\_s \rangle= \sum\_{k,l} R\_{kl}^{ir} (R^{-1})^{kl}\_{js}.
$$
The matrix inverse is easy to calculate and gives us the formula
$$\langle R, u^i\_j \otimes u^r\_s \rangle = q^{-\delta\_{ir}}\delta\_{ij}\delta\_{rs} - (q-q^{-1})\theta (i-r)\delta\_{is}\delta\_{jr}.
$$
Now let's try and test this result: As is very well known $(S \otimes $id)$R = R^{-1}$. Thus, $$\langle R^{-1}, u^i\_j \otimes u^r\_s \rangle = \langle R, S(u^i\_j) \otimes u^r\_s \rangle.$$ In the case of $N=2$, $i=j=r=s=1$, we have $S(u^1\_1) = u^2\_2$, and so,
$$
(R^{-1})^{11}\_{11}=\langle R^{-1},u^1\_1 \otimes u^1\_1 \rangle = \langle R,S(u^1\_1) \otimes u^1\_1 \rangle = \langle R,u^2\_2 \otimes u^1\_1 \rangle = R^{21}\_{21} = 1.
$$
But the formula above gives us that $(R^{-1})^{11}\_{11}$ is equal to
$$ q^{-\delta\_{11}}\delta\_{11}\delta\_{11} - (q-q^{-1})\theta (1-1)\delta\_{11}\delta\_{11} = q^{-1},$$
Moreover, performing the analogous calculations for the other possible values of $i,j,r,s$ we get two different matrices: Using the general formula we get
$$
\left(
\begin{array} {cccc}
q^{-1} & 0 & 0 & 0 \\\
0 & 1 & 0 & 0 \\\
0 & -\lambda & 1 & 0 \\\
0 & 0 & 0 & q^{-1} \\\
\end{array}
\right);
$$
and using the equality $(S \otimes$id$)(R) = R^{-1}$ we get
$$
\left(
\begin{array} {cccc}
1 & 0 & 0 & -\lambda \\\
0 & q & 0 & 0 \\\
0 & 0 & q & 0 \\\
0 & 0 & 0 & 1 \\\
\end{array}
\right),
$$
where $\lambda = (q-q^{-1})$.
I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text
|
https://mathoverflow.net/users/1867
|
The Inverse of a Universal R-Matrix for Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2)
|
I think the only issue here is a harmless error in your calculation and that there is a normalization of the $R$-matrix for $U\_q(sl\_N)$ by a factor of $q^{1/2}$ which you have omitted (See 8.4.2 of Klymik Schmudgen).
First, I get $(R^{-1})^{21}\_{12} = -q^{-1} R^{21}\_{12}$,
because $\langle(S\otimes id)(R),a^2\_1\otimes a^1\_2\rangle = \langle R,S(a^2\_1)\otimes a^1\_2 \rangle = -q^{-1} \langle R,a^2\_1\otimes a^1\_2\rangle$, using that $S(b)=-q^{-1} b$ from Proposition 4.1.2.3 of Klymik Schmudgen. So the actual matrix you should get should just be $q^{-1}$ times what you had expected to get.
Now the factor of q^{-1} here is because you had multiplied the actual universal R matrix by $q^{1/2}$ and so $(\lambda A)^{-1} = \lambda^{-1} A^{-1}$, so there's a factor of $\lambda^2$ as a discrepancy.
I hope this helps!
|
9
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https://mathoverflow.net/users/1040
|
25604
| 16,776 |
https://mathoverflow.net/questions/25602
|
2
|
Motivated by [this question](https://mathoverflow.net/questions/25423/the-inverse-of-a-universal-r-matrix-for-quantized-universal-enveloping-algebra-of), I'm going to ask a much more direct one: Let $R$ be a universal R-matrix for the quasitriangular Hopf algebra ${\cal U}\_q ({\mathfrak sl}\_N)$, $~$ $R ^{-1}$ its inverse, and $\langle \cdot,\cdot \rangle$ the dual pairing between ${\cal U}\_q ({\mathfrak sl}\_N)$ and $SU\_q[N]$, do we have a formula for
$$
\langle R^{-1},u^i\_j \otimes u^r\_s\rangle
$$
analogous to the formula
$$
\langle R,u^i\_j \otimes u^r\_s\rangle = q^{\delta\_{ir}}\delta\_{ij}\delta\_{rs} + (q-q^{-1})\theta (i-r)\delta\_{is}\delta\_{jr},
$$
where $\theta$ denotes the Heaviside symbol.
|
https://mathoverflow.net/users/1095
|
Formula for the Matrix Elements of the Inverse of special linear Universal R-Matrix of Uq(sln)
|
John, see my answer there. It is as the OP thought. The formula for $R^{-1}$ on the vector represenation can only be the inverse of the formula for $R$, namely
$\langle R^{-1},u^i\_j\otimes u^r\_s \rangle = q^{-1}(q^{-\delta\_{ir}} \delta\_{ij}\delta\_{rs} + (q^{-1}-q)\theta(i-r)\delta(is)\delta(jr))$.
As I explained, the formula you wrote there is really for $q^{\frac 12} \langle R,u^i\_j\otimes u^r\_s \rangle$.
|
2
|
https://mathoverflow.net/users/1040
|
25606
| 16,777 |
https://mathoverflow.net/questions/25610
|
5
|
Can anybody please give me an example of a binary operation under which N forms a group? More generally, how to find some operations to make possibly any set a group?
|
https://mathoverflow.net/users/6258
|
Making N (set of all positive integers) a group
|
As explained by Arturo, a simple example of a group structure on **N** is the operation a ⊕ b = f-1(f(a) + f(b)) where f:**N**→**Z** is defined by f(2n) = n and f(2n+1) = -n.
The statement that any nonempty set admits a group structure is equivalent to the Axiom of Choice! This is explained in [this answer](https://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf/12988#12988).
|
18
|
https://mathoverflow.net/users/2000
|
25613
| 16,782 |
https://mathoverflow.net/questions/16994
|
63
|
Can anyone suggest a relatively gentle linear algebra text that integrates vector spaces and matrix algebra right from the start? I've found in the past that students react in very negative ways to the introduction of abstract vector spaces mid-way through a course. Sometimes it feels as though I've walked into class and said "Forget math. Let's learn ancient Greek instead." Sometimes the students realize that Greek is interesting too, but it can take a lot of convincing! Hence I would really like to let students know, right from the start, what they're getting themselves into.
Does anyone know of a text that might help me do this in a not-too-advanced manner? One possibility, I guess, is Linear Algebra Done Right by Axler, but are there others? Axler's book might be too advanced.
Or would anyone caution me against trying this, based on past experience?
|
https://mathoverflow.net/users/4042
|
Linear Algebra Texts?
|
For teaching the type of course that Dan described, I'd like to recommend David Lay's "Linear algebra". It is very thoroughly thought out and well written, with uniform difficulty level, some applications, and several possible routes/courses that he explains in the instructor's edition. Vector spaces are introduced in Chapter 4, following the chapters on linear systems, matrices, and determinants. Due to built-in redundancy, you can get there earlier, but I don't see any advantage to that. The chapter on matrices has a couple of sections that "preview" abstract linear algebra by studying the subspaces of $\mathbb{R}^n$.
|
24
|
https://mathoverflow.net/users/5740
|
25614
| 16,783 |
https://mathoverflow.net/questions/25509
|
15
|
I've recently been interested in the following type of functions. A total computable function *f*:**N**→**N** is *effectively closed* if there is a computable function *p* such that *f*[**N** \ We] = **N** \ W*p*(e), where We is the e-th c.e. set.
>
> Have effectively closed functions been studied? If so, what are they normally called?
>
>
>
I would also appreciate pointers to some uses and/or alternative characterizations of effectively closed functions.
**Motivation.** It is well-known that there is a near-perfect analogy between the adjectives *computable* and *continuous*. For example, a total function *f*:**N**→**N** is computable if and only if it is effectively continuous, i.e. there is a computable function *p* such that *f*-1[We] = W*p*(e). [For the backward implication, let *q* be a computable function such that W*q*(n) = {n} and use the composite *p*∘*q* to enumerate the graph of *f*.] A similar trick shows that a total function is effectively open if and only if it is computable. However, a total computable function is not necessarily effectively closed since that entails that the range of *f* is computable and, indeed, that *f* maps every computable set onto a computable set. Also, the notion is nontrivial since non-constant polynomials and increasing functions are effectively closed.
---
**Update.** Joel David Hamkins gave the following characterization of effectively closed computable functions: they are the computable functions *f*:**N**→**N** for which there is a computable *b*:**N**→**N** such that *f*-1(n) ⊆ {0,1,...,b(n)} for every n ∈ **N**. Although I accepted Joel's answer, the main question is still open.
|
https://mathoverflow.net/users/2000
|
Effectively closed computable functions
|
I like your concept a lot, and have been able to find a characterization.
Suppose that $f:N\to N$ is effectively closed in your sense.
First, as you mentioned, it is easy to see that $\text{ran}(f)$ is
computable, since by taking $W\_e$ to be empty your equation shows that
$\text{ran}(f)$ is both c.e. and co-c.e.
Second, I claim that $f$ is finite-to-one. To see this,
suppose that $f^{-1}(k)$ is infinite for some $k$. Define
a c.e. set $W\_e$ as follows: At stage $s$, if we see that
$k$ is still not in $W\_{\rho(e),s}$, the state-$s$ approximation
to $W\_{\rho(e)}$, then enumerate the next element of
$f^{-1}(k)$ into $W\_e$. (Although this definition may look
circular, since I am defining $W\_e$ by reference to
$W\_{\rho(e)}$, the definition is legitimate by an
application of the Recursion Theorem. That is, I really
define $W\_{r(e)}$, and then find $e$ such that $W\_e=W\_{r(e)}$.)
Note that if $k$ is never enumerated into $W\_{\rho(e)}$,
then I will eventually put all of $f^{-1}(k)$ into $W\_e$,
which will result in $k\notin f[N-W\_e]$, but $k\in
N-W\_{\rho(e)}$, a contradiction. Alternatively, if $k\in
W\_{\rho(e),s}$, then $f^{-1}(k)\cap W\_e$ has at most
$s$ members, and so there are $a\in N-W\_e$ with $f(a)=k$,
placing $k$ into $f[N-W\_e]$ but not in $N-W\_{\rho(e)}$,
again a contradiction.
A similar argument shows actually that the
function $k\mapsto f^{-1}(k)$ is computable. Namely,
define the set $W\_e$ by the following procedure. At stage
$s$, look at every $k\leq s$, and if $k\notin
W\_{\rho(e),s}$, then enumerate all of $f^{-1}(k)\cap s$ into
$W\_e$. (Again, appeal to Recursion Theorem to get such an
$e$.) In other words, as long as $k$ is not in
$W\_{\rho(e),s}$, then we put all elements of $f^{-1}(k)$
below $s$ into $W\_e$.
If $k\notin W\_{\rho(e)}$, then $f^{-1}(k)\subset W\_e$, and
so $k\notin f[N-W\_e]$, contradicting $k\in N-W\_{\rho(e)}$.
Thus, $W\_{\rho(e)}=N$. From this, it follows that $W\_e=N$.
Now, note that $k\in W\_{\rho(e)}$ implies $k\in W\_{\rho(e),s\_k}$
for some stage $s\_k$, and so $f^{-1}(k)$ is a subset of $s\_k$.
By applying $f$ to each value below $s\_k$, we see that the
map $k\mapsto f^{-1}(k)$ is a computable function.
This means that $f$ has a particularly simple form.
Namely, there is a computable partition $N=\bigsqcup\_k
B\_k$, with each $B\_k$ finite, such that $f$ maps elements of
$B\_k$ to $k$. (Note that some $B\_k$ may be empty.)
Conversely, every function with such a form is
computably closed in your sense. Suppose that $f$ arises
from such a computable partition of $N$ into finite sets
$B\_k$. Given any program $e$, enumerate $k$ into $W\_{\rho(e)}$ when
all of $B\_k$ gets enumerated into $W\_e$. It follows that
$f[N-W\_e]=N-W\_{\rho(e)}$, as desired.
This provides a characterization of the effectively closed
computable functions:
**Theorem.** A computable function $f:N\to N$ is
effectively closed if and only if $f$ is finite-to-one and
the map $k\mapsto f^{-1}(k)$ is computable.
|
7
|
https://mathoverflow.net/users/1946
|
25619
| 16,787 |
https://mathoverflow.net/questions/25603
|
21
|
This concerns a number of basic questions about ample line bundles on a variety $X$
and maps to projective space. I have searched related questions and not found answers, but I apologize if I missed something. I'll work with schemes of finite type over a field $k$ for simplicity.
**Background**
A quasi-coherent sheaf $F$ on a $k$-scheme $X$ is *globally generated* if the natural map $H^{0}(X,F)\otimes \mathcal{O}\_{X} \rightarrow F$ is a surjection of sheaves. Basically, this says that for any point on $X$, there is at least one section of $F$ that doesn't vanish at that point, so there are enough sections of $F$ to see all the points of $X$. (EDIT: As pointed out in the comments below, this last sentence does not describe a situation equivalent to being globally generated. Perhaps it is better to say that
globally generated means that for each point $x \in X$, $F$ has some rank $r$ at $x$ and globally generated means that there are at least $r$ sections of $F$ that are linearly independent over $x$.)
The notion of globally generated is especially useful when $F=L$ is a line bundle on $X$.
If $V$ is a finite dimensional subspace of $H^{0}(X,L)$ such that $V \otimes \mathcal{O} \rightarrow L$ is surjective, then we get a morphism $\varphi\_{V}:X \rightarrow \mathbb{P}(V)$
by the universal property of the projective space $\mathbb{P}(V)$ of hyperplanes in $V$.
Essentially, given a point $x \in X$, we look at the fibre over $x$ of the surjection
$V \otimes \mathcal{O} \rightarrow L$ to get a quotient $V \rightarrow L\_{x}$. The kernel
is a hyperplane in $V$, and the morphism $\varphi\_{V}$ sends $x$ to that hyperplane as a point in $\mathbb{P}(V)$.
So how to build globally generated sheaves? A line bundle $L$ is called *ample*
if for every coherent sheaf $F$, $F \otimes L^{\otimes n}$ is globally generated for all large $n$. The smallest $n$ after which this becomes true can depend on $F$.
Finally, a line bundle is called *very ample* if $L$ is globally generated and $\varphi\_{V}$ is an embedding
for some subspace of sections $V$.
There are various properties of and criteria for ample line bundles, which can be found in Hartshorne, for example. What we need for the questions below are the following: $L$ is ample if and only if $L^{m}$ is ample for some $m$ if and only if $L^{n}$
is very ample for some $n$; if $L$ is ample, eventually $L^{k}$ will have sections, be globally generated, be very ample, and have no higher cohomology.
**Questions**
1. Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally generated but not very ample, and of very ample line bundles with higher cohomology?
2. Given an ample line bundle $L$, what is the minimal number $k$ so that I can be sure $L^{k}$ has sections, is globally generated, is very ample? Is $k$ related to the dimension of $X$?
3. If $L$ is very ample, I can use it to embed $X$ into some projective space. Then by projecting from points off of $X \subset \mathbb{P}^{N}$, I can eventually get a finite morphism $X \rightarrow \mathbb{P}^{d}$, where $d$ is the dimension of $X$. But what if
I just know that $L$ is ample and globally generated? Can I also use it to get such a finite morphism to $\mathbb{P}^{d}$?
|
https://mathoverflow.net/users/6254
|
Ample line bundles, sections, morphisms to projective space
|
*1. Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally generated but not very ample, and of very ample line bundles with higher cohomology?*
On a curve of genus $g$, a general divisor of degree $d \le g-1$ has no sections. Of course, if $d>0$ then it is ample.
$K\_X$ on a hyperelliptic curve is globally generated but not very ample.
Look at $L=\mathcal O(1)$ on a plane curve of genus $d$. Then from
$$ 0\to \mathcal O\_{\mathbb P^2}(1-d) \to \mathcal O\_{\mathbb P^2}(1) \to \mathcal O\_C(1)\to 0$$
you see that $H^1(\mathcal O\_C(1))=H^2(\mathcal O\_{\mathbb P^2}(1-d))$ which is dual to $H^0(\mathcal O\_{\mathbb P^2}(d-4))$. So that's nonzero for $\ge4$.
*2. Given an ample line bundle $L$, what is the minimal number $k$ so that I can be sure $L^k$ has sections, is globally generated, is very ample? Is $k$ related to the dimension of $X$?*
Again, just look at the divisor of a degree 1 on a curve of genus $g$. You need $k\ge g$, so you see that there is no bound in terms of the dimension.
It turns out that a better right question to ask is about the *adjoint* line bundles $\omega\_X\otimes L^k$ ($K\_X+kL$ written additively). Then the basic guiding conjecture is by Fujita, and which says that for $k\ge \dim X+1$ the sheaf is globally generated, and for $k\ge \dim X+2$ it is very ample. This is proved for $\dim X=2$, proved with slightly worse bounds for $\dim X=3$. For higher dimensions the best result is due to Angehrn-Siu who gave a quadratic bound on $k$ instead of linear. There are some small improvements for some special cases.
*3. If $L$ is very ample, I can use it to embed $X$ into some projective space. Then by projecting from points off of $X\subset \mathbb P^N$, I can eventually get a finite morphism $X\to \mathbb{P}^d$, where $d$ is the dimension of $X$. But what if I just know that $L$ is ample and globally generated? Can I also use it to get such a finite morphism to $\mathbb P^d$?*
But of course $L$ gives a morphism $f$, and it follows that $f$ is finite: $f$ contracts no curve so $f$ is quasifinite, and $f$ is projective (since $X$ was assumed to be projective). And quasifinite + proper = finite.
|
25
|
https://mathoverflow.net/users/1784
|
25622
| 16,789 |
https://mathoverflow.net/questions/25623
|
3
|
The cardinality of the set of all root paths in the infinite complete binary tree is equal to the cardinality of the Continuum. The same holds true for k-ary trees for any finite k. But what is the case for k infinite?
|
https://mathoverflow.net/users/1320
|
Cardinality of the set of all paths in the infinite complete infinitary tree
|
Assuming your path has countable length, the set of all paths in a $k$-ary tree will have cardinality $k^{\aleph\_0}$. Indeed, at each step you have $k$ choices, and there are $\aleph\_0$ steps (think of a path as a function from $\mathbb{N}$ to $[k]$).
|
6
|
https://mathoverflow.net/users/5513
|
25628
| 16,792 |
https://mathoverflow.net/questions/25474
|
8
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Recall that a commutative ring is Jacobson if every prime ideal is the intersection of the maximal ideals that contain it.
In the exercises of a commutative algebra course I gave I asked the students to show that a commutative ring is Jacobson if and only if every non-maximal prime ideal is the intersection of the prime ideals that strictly contain it. I now suspect that somewhere in the back of my mind I had imposed the condition that the ring should be Noetherian without actually saying this. Of course, Jacobson rings will always have this other property, and the converse is straightforward to prove if there is no strictly ascending chains of prime ideals. But is the result true in general?
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https://mathoverflow.net/users/345
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Primes in a (commutative) Jacobson ring
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The result is true in general.
We may assume a counterexample is given in the form of a domain $R$ satisfying the second property but with nontrivial Jacobson radical, i.e. the closed points of Spec $R$ are not dense. Let D be an affine open neighborhood of $(0)$ in Spec $R$ which contains no closed points. Since D is affine, there exists some $x\in$ D which is closed in D. That is,
$\overline{\lbrace x\rbrace}\setminus x\subset\text{Spec }R\setminus D$.
Since $D$ is open, this implies
$x\not\in \overline{\overline{\lbrace x\rbrace}\setminus x}$,
but this contradicts the requirements of the second property.
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11
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https://mathoverflow.net/users/5513
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25633
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