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https://mathoverflow.net/questions/27657 | -4 | I am looking for factorization of polynomials of several variables in the way outlined below.
Consider a second degree polynomial of two variables over the complex numbers.
"P(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F" (see the edit below)
Experimenting with some polynomials of this sort showed me that factorization is possible in the following way.
```
"P(x,y) = (ax + by + c)(dx + ey + f)" (see the edit below) ,
```
the coefficients being over the complex numbers.
So, given an nth degree polynomial in n variables without a constant term, is it always possible to factorize it into n linear factors each having n variables in the above way? (This rings bells about the fundamental theorem of algebra).
Please suggest a reading material or journal, if any.
[EDIT: I am sorry, I erred. I have edited my question. The edit is that the polynomial has no constant term:
P(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey
and in the expected factorization, the last linear factor does not have a constant term, too:
P(x,y) = (ax + by + c)(dx + ey + f) ]
I kept the original question as it is for documentation purposes.
| https://mathoverflow.net/users/5627 | Factorizing polynomials of several variables (in a different perespective) | Let K be a field. The ring $K[X\_1,...,X\_n]$ is factorial, which means that any polynomial in n variables can be factored into a product of irreducible polynomials. But of course, these polynomials are not of degree 1 in general. If a polynomial can be factored as a product of terms of degree one, then its zero set is a finite union of hyperplanes. An interesting family of irreducible polynomials in C[X,Y] is given by $Y^2-X(X-1)(X-L)$, $L$ different from 0 and 1. The zero set in $P^2(C)$ is called an elliptic curve and it is diffeomorphic to a torus $S^1\times S^1$.
As a reference, I can point to Lang "Algebra", or Jacobson "Basic algebra".
| 3 | https://mathoverflow.net/users/6129 | 27659 | 18,074 |
https://mathoverflow.net/questions/27658 | 7 | Let $X$ be a topological space and $G$ be a topological group acting on $X$, both locally compact Hausdorff. Denote by $D^b(X)$ the derived category of sheaves (say of abelian groups) on $X$. We define a new category in the following way:
An object is an object $M$ of $D^b(X)$, together with an isomorphism
$$\pi^\*M\leftrightarrow \rho^\*M$$
that satisfies the cocycle condition, where $\pi,\rho\colon G\times X\to X$ are the projection and the action maps. A morphism is a morphism of objects in $D^b(X)$ which is compatible with the action isomorphisms.
Note that this category is a special case of the category of equivariant objects:
<http://ncatlab.org/nlab/show/equivariant+object>
>
> Is this category equivalent to the equivariant derived category? If not, why is the equivariant derived category a better construction in this case?
>
>
>
PS: Bernstein and Lunts give among others the following definition of the equivariant derived category in their book "Equivariant sheaves and functors" (p.32):
Denote by $[X/G]$ the "action simplex" $\Delta(n)=G^{n-1}\times X$.
By a simplicial sheaf we mean a collection of sheaves $F^n$ on $\Delta(n)$ together with maps
$\alpha\_h: h^\*F^m \rightarrow F^n$ for each map $h$ in $[X/G]$, which satisfy the cocycle condition:
$$\alpha\_{h' h}=\alpha\_h \circ h^\* \alpha\_{h'}$$
Denote by $Sh([X/G])$ the category of simplicial sheaves and by $Sh\_{eq}([X/G])$ the full subcategory where all $\alpha\_h$ are isomorphisms.
The derived equivariant category $D^b\_G(X)$ is then the full subcategory of $D^b(Sh([X/G]))$ consisting of objects, which have cohomology in $Sh\_{eq}([X/G])$.
| https://mathoverflow.net/users/2837 | Is this a definition of equivariant derived category? | The relation is easier to understand if $G$ is a discrete group. Then your
definition is equivalent to an action of $G$ on an object $M$ in $D^b(X)$. This
means that for each $g\in G$ we have a morphism $\phi\_g\colon M\rightarrow M$ in
$D^b(X)$ with, and this is the important part, $\phi\_{gh}=\phi\_g\phi\_h$
*in* $D^b(X)$. Seen from the point of view of complexes (i.e., before we
pass to $D^b(X)$) this means that we have a homotopy (or something slightly
worse depending on what kind of complex $M$ is) $\phi\_{gh}\sim\phi\_g\phi\_h$.
Experience tells us that such actions up to homotopy is too weak a notion to be
useful (things are a little bit tricky as there are non-trivial situations when
such an action can in fact be replaced up to homotopy by a true action) and in
general one should demand an actual action of $G$ on the complex $M$. The same
goes (and in fact even more so) for morphisms, there may be too many morphisms
if you just look at morphisms in $D^b(X)$ that commute *in* $D^b(X)$ with
the action of $G$.
**Addendum**: Donu makes an excellent point and I just want to elaborate. Take an additive $G$-action on some $(\mathbb Z/p)^n$ which does not lift to an additive action on $(\mathbb Z/p^2)^n$. Then we may look at the morphism $(\mathbb Z/p)^n \to (\mathbb Z/p)^n[1]$ whose mapping cone is $(\mathbb Z/p^2)^n$. It will be a $G$-map $D^b(pt)$ but there is no compatible action of $G$ on a mapping cone.
| 12 | https://mathoverflow.net/users/4008 | 27662 | 18,076 |
https://mathoverflow.net/questions/27621 | 10 | On [this](http://www.ms.u-tokyo.ac.jp/video/vgbook/idx_vguest2006.html) page, the interview is [here](http://www.ms.u-tokyo.ac.jp/video/vgbook/2006/0602kato_L.ram). Can someone provide an English translation?
| https://mathoverflow.net/users/436 | Transcription of an interview of Kazuya Kato | I can write a short summary of the nice interview, sacrificing lots of details, if that is really what you want. But, you can probably ask the video archive stuff [email protected]? They should be happy to receive your response.
| 14 | https://mathoverflow.net/users/36665 | 27669 | 18,082 |
https://mathoverflow.net/questions/27681 | 14 | How can the liar paradox be expressed concisely in symbols? In which formal languages?
| https://mathoverflow.net/users/3441 | Formulas for the liar paradox | The [Liar](http://en.wikipedia.org/wiki/Liar_paradox) is the statement "this sentence is false." It is expressible in any language able to perform self-reference and having a truth predicate. Thus, $L$ is a statement equivalent to $\neg T(L)$.
Goedel proved that the usual formal languages of mathematics, such as the language of arithmetic, are able to perform self-reference in the sense that for any assertion $\varphi(x)$ in the language of arithmetic, there is a sentence $\psi$ such that PA proves $\psi\iff\varphi(\langle\psi\rangle)$, where $\langle\psi\rangle$ denotes the Goedel code of $\psi$. Thus, the sentence $\psi$ asserts that "$\varphi$ holds of me".
Tarski observed that it follows from this that truth is not definable in arithmetic. Specifically, he proved that there can be no first order formula $T(x)$ such that $\psi\iff T(\langle\psi\rangle)$ holds for every sentence $\psi$. The reason is that the formula $\neg T(x)$ must have a fixed point, and so there will be a sentence $\psi$ for which PA proves $\psi\iff\neg T(\langle\psi\rangle)$, which would contradict the assumed property of T. The sentence $\psi$ is excactly the Liar.
Goedel observed that the concept of "provable", in contrast, is expressible, since a statement is provable (in PA) say, if and only if there is a finite sequence of symbols having the form of a proof. Thus, again by the fixed point lemma, there is a sentence $\psi$ such that PA proves $\psi\iff\neg\text{Prov}(\langle\psi\rangle)$. In other words, $\psi$ asserts "I am not provable". This statement is sufficiently close to the Liar paradox statement that one can fruitfully run the analysis, but instead of a contradiction, what one gets is that $\psi$ is true, but unprovable. This is how Goedel proved the Incompleteness Theorem.
| 23 | https://mathoverflow.net/users/1946 | 27686 | 18,095 |
https://mathoverflow.net/questions/27691 | 7 | Hanner's inequalities in the theory of $L^p$ spaces (see <http://en.wikipedia.org/wiki/Hanner>'s\_inequalities) look hard to come-up with at the first glance. Their proof (say, the one in Lieb & Loss "Analysis", Theorem 2.5.) gives no intuition (at least for me) how they come about. How does one see that these inequalities turn up naturally? Do you know a proof which at the same time hints to how one starts considering Hanner inequalities. I hope this is not too vague of a question. Both Wikipedia and Lieb & Loss mention that Hanner's inequalities have to do with uniform convexity of $L^p$ spaces, but from that alone I cannot see how they arise "naturally".
| https://mathoverflow.net/users/5498 | Hanner's inequalities: the intuition behind them | First, a probably unsatisfactory answer to your first question. I believe that Hanner proved his inequalities because he was investigating the modulus of convexity of $L^p$. So possibly he formed a guess on whatever basis, saw that guess would be correct if what have come to be known as Hanner's inequalities were true, then found that he could prove the inequalities. (I haven't actually looked at Hanner's paper so I might be wrong about the history, and in any case this is the kind of thought process that is seldom recorded in papers.)
A better answer is that Hanner's inequalities generalize the parallelogram identity in a very natural way.
As for your second question, I don't know that this really fits what you're asking for either, but [this paper](http://www.jstor.org/stable/2695768) contains the most natural-looking proof of Hanner's inequalities that I've seen.
**Added:** On further reflection, the best answer to the first question combines my first two paragraphs above. The parallelogram identity very precisely expresses the uniform convexity of $L^2$. So in investigating the modulus of convexity of $L^p$, it is natural to look for some inequality that generalizes the parallelogram identity to $L^p$.
| 7 | https://mathoverflow.net/users/1044 | 27692 | 18,099 |
https://mathoverflow.net/questions/27693 | 56 | I hope this question is suitable; this problem always bugs me. It is an issue of mathematical orthography.
It is good praxis, recommended in various essays on mathematical writing, to capitalize theorem names when recalling them: for instance one may write "thanks to Theorem 2.4" or "using ii) from Lemma 1.2.1" and so on.
Should these names be capitalized when they appear unnumbered? For instance which of the following is correct?
"Using the previous Lemma we deduce..." versus "Using the previous lemma we deduce..."
"The proof of Lemma 1.3 is postponed to next Section." versus "The proof of Lemma 1.3 is postponed to next section."
| https://mathoverflow.net/users/828 | Capitalization of theorem names | In English, proper nouns are capitalized. The numbered instances you mention are all usages as proper nouns, but merely refering to a lemma or corollary not by its name is not using a proper noun, and so is uncapitalized.
Thus, for example, one should write about the lemma before Theorem 1.2 having a proof similar to Lemma 5, while the main corollary of Section 2 does not.
---
**Edit.** Well, I've become conflicted. The Chicago Manual of Style, which I have always taken as my guide in such matters, asserts in item 7.136 that "the word *chapter* is lowercased and spelled out in text". And in 7.141 they favor *act 3* and *scene 5* in words denoting parts of poems and plays. This would seem to speak against *Section 2* and possibly against Theorem 1.2. In 7.135 they say that common titles such as foreward, preface, introduction, contents, etc. are lowercased, as in "Allan Nevins wrote the foreward to...". This may also be evidence against Theorem 1.2. But in 7.147 they favor Piano Sonata no. 2, which may be evidence in favor of Theorem 1.2. But they don't treat mathematical writing explicitly, and now I am less sure of what I have always believed, above. I do note that the CSM text itself refers to "fig. 1.2" and "figure 9.3", and not Figure 1.2, which would clearly speak against Theorem 1.2. So I am afraid that I may have to change my mind about this.
| 64 | https://mathoverflow.net/users/1946 | 27695 | 18,101 |
https://mathoverflow.net/questions/27723 | 2 | Let $A\_1,A\_2,\ldots,A\_k$ be finite sets. Furthermore, for each $i\in\{1,2,\ldots,k\}$, let $B\_i$ be a set whose elements are subsets of $A\_i$.
Is there any polynomial-time algorithm that decides whether there exists a choice of precisely one element $C\_i$ of each $B\_i$ such that for all $x\in (C\_1\cup C\_2\cup\ldots\cup C\_k)$ the following property is satisfied:
If $x\in A\_i$ then $x\in C\_i$ for each $i\in\{1,2,\ldots,k\}$?
Any pointer to a paper etc. would be greatly appreciated. Thanks.
| https://mathoverflow.net/users/6726 | poly-time algorithm to choose elements of sets | It seems to me that your problem is stronger than $\ell$-SAT.
In fact, let $A$ be the set of our literals. Assume that we have $p$ clauses. For each $i\in\left\lbrace 1,2,...,p\right\rbrace$, let $A\_i$ be the set of the literals occuring in the $i$-th clause, and let $B\_i$ be the set of all nonempty subsets of $A\_i$. Besides, add some more sets $A\_{p+1}$, $A\_{p+2}$, ..., $A\_i$ which are of the form {literal, its negation}, and for every such sets $A\_k$, let $B\_k$ be the set of its 1-element subsets. I think that a choice of $C\_i$ is the same as a satisfaction of all our clauses (the elements of $C\_1\cup C\_2\cup ...\cup C\_k$ corresponding to those literals that are satisfied).
| 3 | https://mathoverflow.net/users/2530 | 27731 | 18,122 |
https://mathoverflow.net/questions/27720 | 17 | As is well known, if $S$ is a semigroup in which the equations $a=bx$ and $a=yb$ have solutions for all $a$ and $b$, then $S$ is a group. This question arose when someone misunderstood the conditions as requiring that the solution to both equations be the same element of $S$. He suggested instead replacing one of the equations with a cancellation condition (he was thinking along the lines of trying to specify that the Cayley table would be a latin square). It is easy to see that there are semigroups that are not groups in which every equation of the form $a=xb$ has a solution and you can cancel on the right (the standard example that sets $ab=a$ for all $a,b$ works). What is not clear to me is what happens if the equations and cancellations are on the same side. That is:
>
> Suppose $S$ is a semigroup in which the following two conditions hold:
> 1. For all $a,b\in S$ there exists $x$ such that $a=xb$.
> - For all $a,b,c\in S$, if $ab=ac$ then $b=c$.
>
>
> Is $S$ a group?
>
It is easy to see that if $S$ contains an idempotent, then $S$ will be a group: if $e^2=e$, then for all $a\in S$ we have $e^2a=ea$, so $ea=a$ for all $a$; then solving $e=xa$ shows $S$ has a left identity and left inverses, hence is a group. In particular, $S$ will be a group if at least one cyclic subsemigroup of $S$ is finite, and also in particular if $S$ is finite.
I suspect that the answer will be "no" in full generality (that is, there are examples of semigroups $S$ that satisfy 1 and 2 above but are not groups), but I have not been able to construct one. Does any one have an example, a proof that $S$ will always be a group under these conditions, or a reference?
| https://mathoverflow.net/users/3959 | Do these conditions on a semigroup define a group? | I am a victim of timing... I had asked this of a colleague a few days ago and had received no answers, but today at lunch he gave me a counterexample and reference (Clifford and Preston's *The Algebraic Theory of Semigroups*, volume II, pp. 82-86). The example is the Baer-Levi semigroup: the semigroup of all one-to-one mappings $\alpha$ of a countable set $X$ into itself such that $X\setminus \alpha(X)$ is not finite.
Left cancellation follows trivially; if $a$ and $b$ are such mappings, then to construct $c\in S$ such that $a=cb$ proceed as follows: if $y=b(x)\in b(X)$, set $c(y)=c(b(x))=a(x)$. Now let $\delta$ be a one-to-one map from $X\setminus b(X)$ to $X\setminus a(X)$ with $(X\setminus a(X))\setminus\delta(X\setminus b(X))$ infinite. Then define $c(y)$ for $y\notin b(X)$ by $c(y)=\delta(y)$.
| 16 | https://mathoverflow.net/users/3959 | 27732 | 18,123 |
https://mathoverflow.net/questions/27722 | 4 | I'm wondering about a cross product for spectral sequences. I've got an idea, and I wonder if it is written up anywhere, or if it even holds water.
Let's start with three spectral sequences, $E, F$ and $G$. Assume that
$G\_1^{\*,\*} \cong E\_1^{\*,\*}\otimes F\_1^{\*,\*}$ as chain complexes.
Then the ordinary Künneth theorem gives us a map
$\times\_2: E\_2^{\*,\*} \otimes F\_2^{\*,\*} \to G\_2^{\*,\*}.$
Now $E\_2^{\*,\*} \otimes F\_2^{\*,\*}$ has a differential -- the standard one for the
tensor product of chain complexes, and I guess I have to hope that $\times\_2$
is a chain map. Given this, we apply Künneth again, and get
$\times\_3: E\_3^{\*,\*} \otimes F\_3^{\*,\*} \to
H^{\*,\*}( E\_2^{\*,\*} \otimes F\_2^{\*,\*}) \to
G\_3^{\*,\*}.$
Repeating the process leads to cross products
$\times\_r :E\_r^{\*,\*} \otimes F\_r^{\*,\*} \to
G\_r^{\*,\*}$
and presumably converging to the appropriate cross product at the end.
| https://mathoverflow.net/users/3634 | Tensor product of spectral sequences? | There is no reqson for $\times\_2$ to be a chain map. Pick for example a spectral sequence such that the differential in $E\_1$ is zero, such that the one on $d\_2$ is not, and pick a any product $\times\_1$ on $E\_1$ such that it is not a chain map on $E\_2=E\_1$.
You can get such structures, though. For example, suppose that $X$, $Y$ qnd $Z$ are filtered complexes and that $m:X\otimes Y\to Z$ is a chain map compatible with the filtrations. Then if $E\_X$, $E\_Y$ and $E\_Z$ are the spectral sequences gotten from $X$, $Y$ and $Z$, respectively, $m$ gives you maps $m\_r:E^X\_r\otimes E^Y\_r\to E^Z\_r$, all compatible with the differentials.
You'll find more details, for example, in McLeary's *User guide*.
| 9 | https://mathoverflow.net/users/1409 | 27746 | 18,132 |
https://mathoverflow.net/questions/27729 | 14 | One of the common definitions of homology using the singular chains, i.e. maps from the simplex into your space. The free abelian group on these can be made into a chain complex and one can take the homology of this. The result is usually called singular homology.
However, one can use a smaller chain complex instead, by taking the quotient with the degenerate chains (those that are the image of a degeneracy map $\sigma$). This will give the same result in homology.
In some articles, I've seen authors replace the singular chains with the normalized singular chains, often claiming that this is "for technical reasons" (a example is Costello's article on the Gromov-Witten potential associated to a TCFT). What are the important technical differences between these two functors? In which situations is there a preferred one?
| https://mathoverflow.net/users/798 | What are normalized singular chains good for? | Before going to linearity, note a technical advantage of normalized geometric realization is that it preserves products. That probably has linear consequences.
The simplest technical advantage of normalized chains is the [Dold-Kan theorem](http://ncatlab.org/nlab/show/Dold-Kan+correspondence) that the normalized chains give an equivalence of categories between simplicial abelian groups and chain complexes in positive degrees. But the tensor structures on the two categories (detailed in Tom's answer) [do not match](http://ncatlab.org/nlab/show/monoidal+Dold-Kan+correspondence) under this equivalence. The normalized functor is [both](http://golem.ph.utexas.edu/category/2009/11/doldkan_question.html#c029668) lax monoidal and oplax monoidal. I think the fat chain functor is still oplax monoidal. The technical advantage is that only the normalized functor is lax monoidal. My guess is that this is what Costello wants, in particular the consequence Tom mentions that a simplicial ring becomes a dga. Greg's parenthetical is another multiplicative property that I would expect to demonstrate the failure of the lax monoidal property, but I don't see that it does.
I have never seen an example where abnormal chains are *technically* advantageous; degenerate simplices tend only to get in the way of precise considerations. But if they don't get in the way, thinking about them may be a distraction, as Tom says.
| 10 | https://mathoverflow.net/users/4639 | 27752 | 18,136 |
https://mathoverflow.net/questions/27261 | 4 | I will begin by saying that $k=3$ might be a very specific case and this question is useless. Even if that is the case, I would like to know...
The sum of squares function $r\_k(n)$ is very famous. It counts the number of ways $n$ can be written as a sum of $k$ squares. In the case of $k=3$, when $n$ is squarefree and not $7\mod{8}$, $r\_k(n)$ is related to the class number of $\mathbb{Q}(\sqrt{-n})$. In the next (at least) two odd cases the function is still related to arithmetic constants of quadratic fields. C.f. "On the Representation of a Number as the Sum of any Number of Squares, and in Particular of five or seven", Hardy, 1918.
### Question
Are the numbers $r\_k(n)$ known to be related to special groups, like when $k=3$?
### Smaller Question
Is there a book with an in-depth account of these numbers and their arithmetic significance? (more than expressing them as coefficients of a modular form and proving bounds and (lots of) relations...)
| https://mathoverflow.net/users/2024 | Groups related to sum of squares function? | In my view, it depends a little what you mean by "related," but I don't see at first glance any natural group whose order is r\_k(n) for any k other than 3. Loosely speaking, representations of a form of rank m by the genus of a form of rank n are related to the set of double cosets
H(Q) \ H(A\_f) / H(Zhat)
where H is a form of SO\_{n-m} (this can be found e.g. in my paper with Venkatesh "Local-global principles..." but is certainly known to others). When H is abelian (i.e. when n-m = 2) this is naturally a group, otherwise not.
| 4 | https://mathoverflow.net/users/431 | 27759 | 18,142 |
https://mathoverflow.net/questions/27708 | 21 | When I study formal completion and formal schemes, on p.194 of Hartshorne's "Algebraic Geometry", he said "One sees easily that the stalks of the sheaf $\mathcal{O}\_{\hat{X}}$ are local rings."
Notice that here $\mathcal{O}\_{\hat{X}}$ is not the structure sheaf of X, there is a "hat" on the symbol $X$.
But I can't see the reason for that the stalks of the sheaf $\mathcal{O}\_{\hat{X}}$ are local rings.
Could someone explains this for me, thanks.
| https://mathoverflow.net/users/5482 | formal completion | Here is a self-contained explanation (hopefully without any blunder):
Locally, $\hat{X}$ is an affine formal scheme, so each point has a neighbourhood basis admitting of open sets $U$ admitting the following description:
there is a ring
$A$, with ideal $I$, such that the underlying topological space is $U\_0 :=$ Spec $A/I$, and the structure sheaf is the projective limit of the sheaves $\mathcal O\_{U\_n},$
where $U\_n :=$ Spec $A/I^{n+1}$.
(Note that the underlying topological space of all the $U\_n$ coincide, so as topological spaces $$U = U\_0 = \cdots = U\_n = \cdots .$$
The sheaves $\mathcal O\_{U\_n}$ are all sheaves on this same underlying topological space,
which form a projective system with obvious transition maps, corresponding to the surjections of rings $A/I^{n+1} \to A/I^n.$
Now if $x$ is a point of $\hat{X}$, and if we choose a neighbourhood $U$ of $x$ as above,
then the natural map on sheaves $\mathcal O\_{\hat{X}} \to \mathcal O\_{U\_0}$ induces
a natural map on stalks $\mathcal O\_{\hat{X},x} \to \mathcal O\_{U\_0,x}$.
The target is a local ring. Let $\mathfrak m\_x$ denote the preimage in $\mathcal O\_{\hat{X},x}$ of the maximal ideal
in $\mathcal O\_{U\_0,x}$; I claim that it is the unique maximal ideal of
$\mathcal O\_{\hat{X},x}$.
To see this, suppose that $f$ is an element of $\mathcal O\_{\hat{X},x}$ which does not
lie in $\mathfrak m\_x$. Then by definition of the stalk, $f$ extends to a section
of $\mathcal O\_{\hat{X}}$ over some neighbourhood of $x$, which (by shrinking $U$ as necessary) we may as well
assume is our affine neighbourhood $U$.
Thus we may think of $f$ as a section of the projective limit of $\mathcal O\_{U\_n}(U)$,
i.e. the projective limit of the rings $A/I^{n+1}$.
The assumption that $f$ is not in $\mathfrak m\_x$ says that its image in $\mathcal O\_{U\_0}(U)$ is not in the maximal ideal at $x$, and so shrinking $U$ further,
if necessary, we may assume that $f$ is a unit in $\mathcal O\_{U\_0}(U) = A/I$.
Thus $f$ is an element of the projective limit of $A/I^{n+1}$ which is a unit
in $A/I$. One easily verifies that $f$ is then a unit in every $A/I^{n+1}$,
and hence in the projective limit. Thus $f$ is a unit in the ring
$\mathcal O\_{\hat{X}}(U)$, and so in particular in the stalk $\mathcal O\_{\hat{X},x}$.
I've shown that every element of the stalk $\mathcal O\_{\hat{X},x}$ not lying
in $\mathfrak m\_x$ is a unit, which implies that $\mathcal O\_{\hat{X},x}$ is local
with maximal ideal $\mathfrak m\_x$.
| 30 | https://mathoverflow.net/users/2874 | 27780 | 18,158 |
https://mathoverflow.net/questions/27785 | 21 | If you have an infinite set X of cardinality k, then what is the cardinality of Sym(X) - the group of permutations of X ?
| https://mathoverflow.net/users/3537 | Cardinality of the permutations of an infinite set | $k^k$.
Easy that it's an upper bound. For lower bound split $X$ into two equinumerous
subsets; there are $\ge k^k$ permutations swapping the two subsets.
| 26 | https://mathoverflow.net/users/4213 | 27788 | 18,163 |
https://mathoverflow.net/questions/27747 | 5 | This question is motivated by some issue raised by David Speyer in [this question](https://mathoverflow.net/questions/27634/k-0r-mathbbz-but-some-f-g-projective-not-stably-free).
Let $R$ be a ring. Let $K\_0(R)$ and $G\_0(R)$ be the Grothendieck groups of f.g. projective modules and f.g. modules over $R$, respectively (you just kill all relations generated by short exact sequences). There is a natural map, called the *Cartan homomorphism* (see Serre's "Linear reps of finite groups", Chapter 15) $$c: K\_0(R) \to G\_0(R)$$ given by forgetting a module is projective.
In general, $c$ needs not be injective nor surjective. For non-surjectivity, take $R$ to be a local ring, then $K\_0(R)=\mathbb Z$ but $G\_0(R)$ can be huge (in particular, if $R$ is normal, $\mathbb Z\oplus \text{Cl}(R)$ is a quotient of $G\_0(R)$). Examples of non-injectivity can be found by taking $R$ to be some group rings, as the Cartan matrix is not always invertible, see for example Section 4 of [this paper](http://www.math.temple.edu/%7Elorenz/papers/reps.ps) by Martin Lorenz . But I don't know any commutative example of non-injectivity.
*Is $c$ always injective if $R$ is commutative? How about if $R$ is commutative and Noetherian?*
(If this is true, one can prove the original question quoted above with the assumption $G\_0(R)=\mathbb Z$)
| https://mathoverflow.net/users/2083 | Seeking examples or proof: injectivity of Cartan homomorphism for commutative rings? | No, it is not injective in general, unless $R$ is regular notherian. There are many counterexamples; for a simple one you can take the ring $R := \mathbb C[t^2, t^3] \subseteq \mathbb C[t]$, compute that $G\_0(R) = \mathbb Z$, while $K\_0(R)$ maps onto the Picard group of $R$, which is the additive group $\mathbb C$.
| 9 | https://mathoverflow.net/users/4790 | 27790 | 18,164 |
https://mathoverflow.net/questions/27827 | 2 | The complex projective line is isomorphic to the 2-sphere, and so, has genus $0$. Does this result for all $CP^N$, that is, is the genus of $CP^N$ equal to $0$, for all $N$?
| https://mathoverflow.net/users/1977 | Genus of complex projective space | The geometric genus (the dimension of the space of global sections of the
canonical sheaf) of projective $n$-space is zero. See Hartshorne II.8.
| 6 | https://mathoverflow.net/users/4213 | 27829 | 18,187 |
https://mathoverflow.net/questions/27839 | 2 | In light of the answers given to [this question](https://mathoverflow.net/questions/27827/genus-of-complex-projective-space), I would like to pose a more general one: Do all Grassmannian spaces have genus 0? If so, do there exist any flag manifolds with non-zero genus?
| https://mathoverflow.net/users/1977 | Genus of Grassmannians and Flag Manifolds | EDIT: My first answer was confusing and not quite accurate. Let me try again.
In arbitrary characteristic, the structure sheaf of any homogeneous space $G/P$ (for $G$ a semisimple group) has no higher cohomology. This is an instance of Kempf's vanishing theorem. The space of sections is 1-dimensional, so this implies that the arithmetic genus is 0.
Now using Serre duality, we conclude that only the top cohomology of the canonical sheaf is nonzero, which means that the geometric genus is also 0.
| 8 | https://mathoverflow.net/users/321 | 27840 | 18,193 |
https://mathoverflow.net/questions/27825 | 2 | Hi folks, what is known about the $L^2$ space of holomorphic functions of 1 complex variable with the scalar product
$\langle f, g \rangle = \int dzd{\bar z} \frac{ {\bar f(z)} g(z) }{(1 + z{\bar z})^x}$
where $x > 2$ is a real number? The domain of integration is the entire complex plane. Poles are allowed in the functions so all possible powers in the Laurent expansion are allowed, $f(z) = \sum\_{n = -\infty}^\infty f\_n z^n$.
Is this a well-known space? Is an orthogonal basis readily available?
If $f(z)$ is a polynomial with sufficiently low degree then certainly it is in the above defined $L^2$ space. But there are much more functions that are okay, it seems, for instance $f(z) = \exp( -z )$. Or anything that falls off sufficiently fast.
The background is this: if $x=2j+2$ where $j$ is a half-integer and the holomorphic functions can only be at most $2j$ order polynomials, then the above defined space is the $2j+1$ dimensional irreducible unitary representation of $SU(2)$. The action of $g = [ [ a, b ], [ c, d ] ] \in SU(2)$ is
$(gf)(z) = (bz + d)^{2j} f\left( \frac{az+c}{bz+d} \right)$
Clearly, if $f(z)$ is a polynomial at most of order $2j$ then $(gf)(z)$ is also one. And the scalar product is the one I gave above, with $x=2j+2$.
Okay, this was the case for half-integer $j$. What is the deal with arbitrary $j$? Then I can still define the above scalar product with arbitrary $x$. The action above still preserves the scalar product. It is still a group action by $SU(2)$. Do I get an infinite dimensional representation of $SU(2)$? Is it reducible/irreducible?
| https://mathoverflow.net/users/4526 | L^2 space of holomorphic functions with given weight | Hi Daniel. As already said in my comment the space consists just of order polynomials of degree $\lfloor x - 1 \rfloor$. First, one can check that any function in the space must be holomorphic, since the weight doesn't help to integrate over poles. Then one gets from $\\| f \\| < \infty$ that $|f(x)| \leq |z|^{x-1 }$, so one has that $f$ is a degree $\lfloor x - 1 \rfloor$ polynomial by a consequence of Liouville's Theorem.
Helge
| 6 | https://mathoverflow.net/users/3983 | 27844 | 18,196 |
https://mathoverflow.net/questions/27836 | 9 | Let $G$ be a complex linear algebraic group, given to us as a closed subgroup of some $\mathrm{GL}(n,\mathbb{C})$. Suppose moreover that $G$ is semisimple. Then it's a fact that every finite-dimensional holomorphic representation of the complex Lie group $G(\mathbb{C})$ is actually an algebraic representation (i.e., given by polynomials in the matrix entries, together with $\mathrm{det}^{-1}$).
One can certainly deduce this from the highest-weight theory (that is, by provably constructing all the representations, and noting that everything you've constructed is in fact algebraic). But this isn't remotely satisfying.
In another direction, I was led by notes of Milne to a book and some papers by Dong Hoon Lee, and from those to a series of papers by Hochschild and Mostow. But those authors want to do something harder: classify the Lie groups (not necessarily semisimple!) that can be given an algebraic group structure such that all the holomorphic representations of the Lie group are algebraic representations of the algebraic group. It seems to me that if you start out with an algebraic group, and then assume that the group is semisimple, then most of the complications should go away.
So my question is, is there a satisfying and reasonably elementary proof of the fact in the first paragraph? I should say something about my motivation. I'll be teaching a Lie groups/algebras course next year, and when we talk about representation theory we'll observe this phenomenon; so it would be nice to explain it if there's a reasonable way to do so. Given that I can't expect my students to have had an algebraic geometry course, I'd want to minimize the algebraic geometry in the argument, possibly at the cost of making more serious use of the structure theory of Lie groups/algebras.
Here's an approach that I would find especially clarifying if it can be made to work. We're handed a faithful representation of $G(\mathbb{C})$ (the inclusion into $\mathrm{GL}(n,\mathbb{C})$) which is certainly algebraic. Its tensor powers are algebraic. Then the claim would be immediate by semisimplicity if one can show that every irreducible representation of $G(\mathbb{C})$ (or perhaps of Lie groups in some more general class than these) occurs as a subquotient of a tensor power of a faithful one. How might one prove the latter? (Can one prove the latter for compact real groups in a manner similar to the proof for finite groups, and then pass to semisimple complex groups by the unitary trick?)
| https://mathoverflow.net/users/379 | Algebraicity of holomorphic representations of a semisimple complex linear algebraic group |
>
> "Then the claim would be immediate by
> semisimplicity if one can show that
> every irreducible representation of
> $G(\mathbb{C})$ (or perhaps of Lie
> groups in some more general class than
> these) occurs as a subquotient of a
> tensor power of a faithful one. How
> might one prove the latter? (Can one
> prove the latter for compact real
> groups in a manner similar to the
> proof for finite groups, and then pass
> to semisimple complex groups by the
> unitary trick?)"
>
>
>
Yes. The analysis is not so pretty, but it is elementary. Let $K$ be a compact Lie group, $V$ a faithful representation, and $W$ any other representation. Just as in the finite group case, $\mathrm{Hom}\_K(W,V^{\otimes N}) \cong (W^\* \otimes V^{\otimes N})^K$, and the dimension of the latter is $\int\_K \overline{\chi\_W} \cdot \chi\_V^{N}$, where $\chi\_V$ and $\chi\_W$ are the characters of $V$ and $W$, and the integral is with respect to Haar measure. Let $d\_V$ and $d\_W$ be the dimensions of $V$ and $W$.
We now come to a technical nuisance. Let $Z$ be those elements of $K$ which are diagonal scalars in their action on $V$; this is a closed subgroup of $S^1$. For $g$ not in $Z$, we have $|\chi\_V(g)| < d\_V$. We first present the proof in the setting that $Z = \{ e \}$.
Choose a neighborhood $U$ of $\{ e \}$ small enough to be identified with an open disc in $\mathbb{R}^{\dim K}$. On $U$, we have the Taylor expansion $\chi\_V(g) = d\_V \exp(- Q(g-e) + O(g-e)^3)$, where $Q$ is a positive definite quadratic form; we also have $\chi\_W(g) = d\_W + O(g-e)$. Manipulating $\int\_U \overline{\chi\_W} \chi\_V^N$ should give you
$$\frac{d\_W \pi^{\dim K/2}}{\det Q} \cdot d\_V^N \cdot N^{-\dim K/2}(1+O(N^{-1/2}))$$
Meanwhile, there is some $D<d\_V$ such that $|\chi\_V(g)| < D$ for $g \in K \setminus U$.
So the integral of $\overline{\chi\_W} \chi\_V^N$ over $K \setminus U$ is $O(D^N)$, which is dominated by the $d\_V^N$ term in the $U$ integral.
We deduce that, unless $d\_W=0$, we have $\mathrm{Hom}\_K(W, V^{\otimes N})$ nonzero for $N$ sufficiently large.
If $Z$ is greater than $\{e \}$, then we can decompose $W$ into $Z$-isotypic pieces. Let $\tau$ be the identity character of the scalar diagonal matrices, and let the action of $Z$ on $W$ be by $\tau^k$. (If $\tau$ is finite, then $k$ is only defined modulo $|Z|$; just fix some choice of $k$). Then we want to consider maps from $W$ to $V\_N:=V^{k+Nd\_V} (\det \ )^{-N}$. $V\_N$ is constructed so that $\overline{\chi\_W} \chi\_{V\_N}$ is identically $d\_W$ on $Z$; one then uses the above argument with a neighborhood of $Z$ replacing a neighborhood of the origin.
| 1 | https://mathoverflow.net/users/297 | 27848 | 18,199 |
https://mathoverflow.net/questions/27830 | 0 | I am working on stability of nonlinear switched systems and recently, I have proven that switched systems with homogeneous, cooperative, Irreducible and commuting vector fields , i.e., vector fields with Lie bracket equal to 0, are D-stable under some condition. I was trying to find an example for such systems but surprisingly, I could not find any in the papers which have dealt with them.
Does anybody know a good example of commuting nonlinear vector fields? And are there any significance to such systems (from physical point of view)?
| https://mathoverflow.net/users/6748 | Commuting Nonlinear Vector Fields | Any set of nonlinear co-ordinates gives you a corresponding set of commuting vector fields. So any of the co-ordinates listed in the "See also" section of
<http://en.wikipedia.org/wiki/Elliptic_coordinates>
gives an example.
| 1 | https://mathoverflow.net/users/613 | 27852 | 18,201 |
https://mathoverflow.net/questions/27854 | 13 | I feel sort of silly asking this question. Unless I'm very much mistaken the paper I'm reading assumes the following statement:
Let $G$ be a finite group. We may embed it via the Cayley embedding into an ambient permutation group $G \leq S\_{|G|}$. Then any automorphism of $G$ comes from conjugation by an element in $N\_{S\_{|G|}}(G)$.
Is this statement true?
| https://mathoverflow.net/users/5756 | Does every automorphism of G come from an inner automorphism of S_G? | The statement is true. Let $g \in G$ and $\pi \in Aut(G)$. Let $\lambda\_{g}$ be the corresponding left translation by $g$. Regard $\pi$ and $\lambda\_{g}$ as elements of $Sym(G)$. Then for all $x \in G$,
$(\pi \lambda\_{g} \pi^{-1})(x) = (\pi \lambda\_{g}) ( \pi^{-1}(x)) =
\pi( g \pi^{-1}(x)) = \pi(g) x = \lambda\_{\pi(g)}(x)$.
Thus $\pi \lambda\_{g} \pi^{-1} = \lambda\_{\pi(g)}$.
| 16 | https://mathoverflow.net/users/4706 | 27856 | 18,202 |
https://mathoverflow.net/questions/27851 | 55 | Here is a question someone asked me a couple of years ago. I remember having spent a day or two thinking about it but did not manage to solve it. This may be an open problem, in which case I'd be interested to know the status of it.
Let $f$ be a one variable complex polynomial. Supposing $f$ has a common root with every $f^{(i)},i=1,\ldots,\deg f-1$, does it follow that $f$ is a power of a degree 1 polynomial?
upd: as pointed out by Pedro, this is indeed a conjecture (which makes me feel less badly about not being able to do it). But still the question about its status remains.
| https://mathoverflow.net/users/2349 | Polynomials having a common root with their derivatives | That is known as the Casas-Alvero conjecture. Check this out, for instance:
<https://arxiv.org/abs/math/0605090>
Not sure of its current status, though.
| 46 | https://mathoverflow.net/users/4290 | 27859 | 18,204 |
https://mathoverflow.net/questions/27853 | 5 | The finite dimensional irreducible unitary representations of $SU(2)$ are labelled by $j$ which needs to be half-integer, the dimension of the representation is $2j+1$. This is well-known, all is good.
If we do not require finite dimension for the representation, is it possible to make sense of representations with an arbitrary real number $j$? They will, presumably, be infinite dimensional but hopefully still unitary.
In the half-integer case when represented on at most $2j$ degree holomorphic polynomials, the 3 basis elements of the Lie-algebra in representation $j$ act as
$e\_1 = \frac{1-z^2}{2}\frac{d}{dz} + jz$
$e\_2 = \frac{1+z^2}{2i}\frac{d}{dz} + ijz$
$e\_3 = -z\frac{d}{dz} + j$
Clearly, if $j$ is not half-integer and we start from $f(z) = z$ and start acting on it with $e\_i$, it will generate an infinite dimensional space.
This kinda gives me the feeling that perhaps non-half-integer $j$ representations are still meaningful and are infinite dimensional. But I'm not sure, is this really the case or something will go wrong?
Basically what I'm asking is whether analytic continuation in $j$ makes any sense.
| https://mathoverflow.net/users/4526 | Infinite dimensional unitary representations of SU(2) for non-half-integer j? | I just wrote this answer on your last question. To summarize, you can't have an irreducible representation of a compact group that is infinite dimensional, unless the representation space is very exotic.
By the Peter-Weyl Theorem, all irreducible Hilbert space representations of a compact group (e.g. SU(2)) are finite dimensional. Thus, any infinite dimensional Hilbert space representation will be reducible.
What can be said for non-Hilbert space representations? Given a compact group G acting irreducibly (and continuously) on a locally convex topological vector space V, you can inject V into L2(G) by sending v in V to the function cv(g)=〈g⋅v,v'〉 where v' is basically any nonzero element of the continuous dual of V (this is where local convexity is used). (Note that the irreducibility of V implies the injectivity of the map.)
Thus V is finite dimensional: its image is not necessarily closed in L2(G), but, since V is irreducible, its image has to lie within a single irreducible component of L2(G), which are all finite dimensional.
So to find a representation that is infinite dimensional and irreducible, you'd have to look at non-locally convex vector spaces (actually, you need the dual space to not separate points). Like Lp with p<1.
| 11 | https://mathoverflow.net/users/6753 | 27869 | 18,210 |
https://mathoverflow.net/questions/27793 | 11 | I was reading Logicomix (a fictionalised account of logic from Frege to Gödel through Russell's eyes) and there was mention about two different versions of types developed by Russell and Whitehead for Principia Mathematica, unramified (first) and ramified. I don't expect that there is too much relation to the modern type theory (or is there?), but I'm curious to know what the difference is between them. In particular, it is stated that unramified types were not sufficient. How so?
I know the intuitive description of RW's types as a hierarchy of sets, but nothing beyond that.
| https://mathoverflow.net/users/4177 | Russell and Whitehead's types: ramified and unramified | Yes, this still occurs in modern type theory; in particular, you'll find it in the [calculus of constructions](http://en.wikipedia.org/wiki/Calculus_of_constructions) employed by the [Coq](http://www.lix.polytechnique.fr/coq/) language.
Consider the type called `Prop`, whose inhabitants are logical propositions (which are in turn inhabited by proofs). The type `Prop` does not belong to `Prop` -- this means that `Prop` exhibits **stratification**:
```
Check Prop.
Prop
: Type
```
However, note that `(forall a:Prop, a)` **does** have type `Prop`. So although `Prop` does not belong to `Prop`, things which *quantify over all of `Prop`* may still belong to `Prop`. So we can be more specific and say that `Prop` exhibits **unramified stratification**.
```
Check (forall a:Prop, a).
forall a : Prop, a
: Prop
```
By contrast, consider `Set`, whose inhabitants are datatypes (which are in turn inhabited by computations and the results of computations). `Set` does not belong to itself, so it too exhibits stratification:
```
Check Set.
Set
: Type
```
Unlike the previous example, things which *quantify over all of `Set`* do not belong to `Set`. This means that `Set` exhibits **ramified stratification**.
```
Check (forall a:Set, a).
forall a : Set, a
: Type
```
So, in short, "ramification" in Russell's type hierarchy is embodied today by what Coq calls "predicative" types -- that is, all types except `Prop`. If you quantify over a type, the resulting term no longer inhabits that type unless the type was impreciative (and Prop is the only impredicative type).
The higher levels of the Coq universe (`Type`) are also ramified, but Coq hides the ramification indices from you unless you ask to see them:
```
Set Printing Universes.
Check (forall a:Type, Type).
Type (* Top.15 *) -> Type (* Top.16 *)
: Type (* max((Top.15)+1, (Top.16)+1) *)
```
Think of `Top.15` as a variable, like $\alpha\_{15}$. Here, Coq is telling you that if you quantify over the $\alpha\_{15}^{th}$ universe to produce a result in the $\alpha\_{16}^{th}$ universe, the resulting term will fall in the $max(\alpha\_{15}+1, \alpha\_{16}+1)^{th}$ universe -- which is at least "one level up" from what you're quantifying over.
Just as it was later discovered that Russell's ramification was unnecessary (for logic), it turns out that predicativity is unnecessary for the purely logical portion of CiC (that is, `Prop`).
| 12 | https://mathoverflow.net/users/2361 | 27880 | 18,217 |
https://mathoverflow.net/questions/27884 | 6 | Consider the following linear Diophantine Equation::
```
ax + by + cz = d ------------ (1)
```
for all, a,b,c and d positive integers, and relatively prime, and assume a>b>c without loss of generality.
Can we find a lower bound on d which ensures at least one non-negative solution to this equation?
I know we can solve this problem easily for
```
ax+by = c. -------- (2)
```
The answer is
c>=ab, ------------(3)
which derives from the fact that the distance between two consecutive solutions of this equation is
```
$D = (\sqrt(a^2 + b^2))$ ----------(4)
```
and c>=ab ensures that the length of line in x-y plane is large enough to include at least one solution).
Since equation (1) is a plane in the xyz coordinate system, and the distance between consecutive solution can be shown to be DD = sqrt(b^2+c^2) (though this may not be smallest distance between solutions). I was thinking that if we can show that an inscribed circle with diameter DD can be enclosed within the triangle formed by x,y and z intercepts of Eq.(1), i.e. (c/a,0,0), (0,c/b,0) and (0,0,c/a), then we have at least one non-negative solution. But the in-circle radius has an inconvenient relationship with the original variables (a,b,c,d), and may not be a monotonic function of d.
Is there a smarter way to do this? and if such a bound exists, can it be extended to higher dimensions?
Thanks.
| https://mathoverflow.net/users/6759 | Non-negative integer solutions of a single Linear Diophantine Equation | The question of determining the lower bound on $d$ is called the Frobenius problem. For $2$ variables your bound can be improved: every integer starting from $(a-1)(b-1)$ is representable as a non-negative combination. Some general results on this problem are available in [this paper](http://www.jstor.org/pss/2371684), - even on the first page (which is open-access): it is stated that for $a>b>c$ the number $(c-1)(a-1)$ gives a bound. However, it is probably very far from optimal; there are various conjectures and partial results on the asymptotic behaviour of these numbers for big $a,b,c$ (and similarly for larger dimensions). In particular, some conjectures on Frobenius problem were formulated by late Vladimir Arnold in the past decade, see, e.g. [this article](http://www.springerlink.com/content/b373168270250837/) and [this article](http://www.springerlink.com/content/04tr526471584276/); some progress has been made in that direction, see e.g. like [this paper](http://www.turpion.org/php/paper.phtml?journal_id=sm&paper_id=4011).
| 8 | https://mathoverflow.net/users/1306 | 27886 | 18,221 |
https://mathoverflow.net/questions/27624 | 4 | I need some references for irreducible representations of the Heisenberg algebra with three generators or the category of its finite length modules.
Here, the Heisenberg algebra with three generators $x$, $y$ and $z$ is defined to be the Lie algebra whose underlying vector space is generated by $x$, $y$, and $z$, and the commutator relations are give as follows: $[x,y]=-[y,x]=z$, and all other commuator relations are zero. So it is a nilpotent (and hence solvable) Lie algebra with a one dimensional center $Kz$, where $K$ is the ground field of zero characteristic.
(The Heisenberg algbera in $2n+1$ vectors is defined in same way, for more details I would suggest J. Diximier, "Enveloping algberas".)
| https://mathoverflow.net/users/5604 | Irreducible representations of Heisenberg algebra | If we just consider central representations, i.e., those for which $z$ acts by a nonzero scalar, then up to a certain kind of equivalence (given by conjugation with algebra isomorphisms) there is a unique irreducible representation of the algebra. It is infinite dimensional, given by polynomials in one variable. If the action of $z$ is multiplication by $\lambda \in K^\times$, then we can write the representation as $K[x]$, where the action of $x$ is multiplication by $x$, and the action of $y$ is $\lambda \frac{\partial}{\partial x}$.
If you want to rigidify the equivalence to isomorphism, (i.e., you want to remember the specific actions of the elements $x,y,z$ instead of just the algebra structure), then there is a parameter space of irreducible representations, with one coordinate describing the action of $z$ (i.e., taking values in nonzero elements of $K$), and the rest of the coordinates describing a line in the span of $x,y$. The construction generalizes to the $2n+1$ dimensional setting, where the representation is given by polynomial functions on a Lagrangian subspace of a $2n$ dimensional symplectic vector space, but you lose uniqueness. The parameter space of these representations involves a torus and the Lagrangian Grassmannian, but I don't remember if it has the geometric structure of a product or some kind of fibration.
The finite length modules are in bijection with finitely generated [holonomic D-modules](http://en.wikipedia.org/wiki/D-module) on the affine line, since the central condition endows the modules with an action of the [Weyl algebra](http://en.wikipedia.org/wiki/Weyl_algebra), which is a quotient of the universal enveloping algebra. As zamanjan notes in a comment here, when $n > 1$, this fails rather spectacularly. Stafford (1983) showed that there are irreducible modules with characteristic cycle of dimension $2n-1 > n$, and Bernstein-Lunts showed that in the $n=2$ case, this is in some sense a property of the generic irreducible module.
For non-central representations, things are considerably messier. When $z$ acts trivially, you're asking for pairs (or $2n$-tuples) of commuting matrices, and when $z$ is arbitrary, you can have finite dimensional representations look a lot like representations of arbitrary nilpotent lie algebras.
Regarding references, I guess anything about D-modules should work, but depending on your background, they may be hard to read. Howe has a paper called "A century of Lie theory", and Rosenberg has a paper called "A selective history of the Stone-von-Neumann theorem", both of which could be useful.
| 5 | https://mathoverflow.net/users/121 | 27897 | 18,229 |
https://mathoverflow.net/questions/27713 | 1 | I'm trying to solve the BVLS problem for huge (2e6x2e6) matrices which are very sparse (4 elements per row). Does anybody have a recommendation for a free solver (preferably a library of routines)?
The BVLS problem is defined as:
$\underset{l \le x \le u}{\min} \lVert Ax - b \rVert\_2^2$
| https://mathoverflow.net/users/1899 | Recommendations for a large scale bounded variable least squares (BVLS) solver for sparse matrices | This is such a well-solved problem that there are many software packages that have built in functions for this.
Here are a selection of built-in functions in different software packages that can be used:
In Matlab: lsqlin (type help lsqlin into Matlab and it tells you exactly what to type. I have just (approximately) solved your problem with random sparse matrices and it works great.)
KNITRO for Mathematica this package also solves this exact problem but I don't have this software so I can't tell you which exact function.
For a free solver I have found this: <http://sourceforge.net/projects/quadprog/>
However it assumes that $A$ has full column rank. This is just because this algorithm uses the dual problem which exists when the Hessian $A^TA$ is positive definite.
| 2 | https://mathoverflow.net/users/2011 | 27902 | 18,233 |
https://mathoverflow.net/questions/27901 | 1 | It is well known that if $X$ is a first countable topological space and $Y$ is a topological space, then $f : X \rightarrow Y$ is continuous iff
$$\forall x \in {\rm map}(\mathbb{N},X),\forall p \in X \quad x\_{n} \rightarrow p \Rightarrow f(x\_{n}) \rightarrow f(p)$$
It is also well known that if $X$ and $Y$ are metric spaces and $f : X \rightarrow Y$ is uniformly continuous, then $f$ maps Cauchy sequences to Cauchy sequences.
By analogy it seems plausible that if a function between metric spaces maps Cauchy sequences to Cauchy sequences then it must be uniformly continuous. However mimicking the proof of the analogous result for continuous maps doesn't work, which makes me think the result if false. Does anyone know any counterexamples?
Also on the uniform continuity wikipedia page, it says that the result is true if $X$ and $Y$ are subsets of $\mathbb{R}^{n}$. EDIT: It actually doesn't say this, I misread the page.
| https://mathoverflow.net/users/4002 | Does Cauchy continuity imply uniform continuity? [No.] | No it's not true.
f(x) = x^2 on whole real line.
It maps Cauchy sequences to Cauchy sequences but it's not uniformly continuous on the whole real line.
| 12 | https://mathoverflow.net/users/3124 | 27918 | 18,247 |
https://mathoverflow.net/questions/27912 | 1 | How can one show $\displaystyle\frac{(m+n-1)!}{m ! (n-1)!}\leq \left[\frac{e (m+n-1)}{n}\right]^{n-1}$ ?
| https://mathoverflow.net/users/6766 | bound for binomial coefficients | Denote the quotient of the right and left hand sides,
$$
f(m,n)=\biggl(\frac{e(m+n-1)}n\biggr)^{n-1}\bigg/\binom{m+n-1}m.
$$
Then $f(m,1)=1$ for all $m\in\mathbb N$ and
$$
\frac{f(m,n+1)}{f(m,n)}
=\frac{e}{\biggl(1+\dfrac1n\biggr)^n}\cdot\biggl(1+\frac1{m+n-1}\biggr)^{n-1} > 1,
$$
that is,
$$
f(m,n+1)> f(m,n)>\dots> f(m,2)> f(m,1)=1.
$$
This proves the required inequality.
| 4 | https://mathoverflow.net/users/4953 | 27923 | 18,251 |
https://mathoverflow.net/questions/27896 | 14 | According to the OEIS ([A002966](http://oeis.org/A002966)) there are 294314 solutions in positive integers to the equation
$$\sum\_{i=1}^7\frac{1}{x\_i}=1$$ assuming $x\_1\leq x\_2\leq\cdots\leq x\_7$.
Similarly for 8 summands there are 159330691 solutions.
My question: What are they? Is there a way of counting them without knowing them?
The bound for $x\_n$ for $n$ summands is double exponential and I could only compute the solutions up to $n=6$ with Maple.
| https://mathoverflow.net/users/6355 | Diophantine equation: Egyptian fraction representations of 1 | As far as I know, the only significant result to speed up these calculations is that $E\_2(\frac{p}{q}) = \frac{1}{2}|\lbrace d: d | q^2, d \equiv -q (mod p) \rbrace|$, where $E\_2(p/q)$ represents the number of decompositions into 2 unit fractions, and each matching $d$ represents the decomposition $\frac{p}{q} = \frac{qp}{q(q+d)} + \frac{dp}{q(q+d)}$. (Take floor() or ceil() depending on whether you want to allow repeats.)
When I've coded this in the past, I called one of 4 different functions depending on a) whether $p=1$ or not, and b) whether $q/p \ge min$ or not, where $min$ is the greatest denominator I'm already using. When $p=1$ and $q \ge min$, in particular, we can just calculate $\tau(q^2)/2$ from the factorisation of $q$; in the other cases I actually walked the factors from $q/p$ to $\sqrt{q}$.
So: yes, you can count the number of matching sets without generating the 7 elements of each set, but computationally the elements are just a whisker away.
| 12 | https://mathoverflow.net/users/6089 | 27925 | 18,252 |
https://mathoverflow.net/questions/27922 | 2 | For introduction, Ethiointegers are integers which get reversed when multiplied by another number.
For instance,
2178 \* 4 = 8712
1089 \* 9 = 9801
I couldn't find such numbers, even by another name anywhere else except in one journal which I think is not electronically accessible.
**My question**: Has numbers like this been treated anywhere? If so, what interesting properties do they have?
[E.g. One property I consider interesting: sumofdivisors(1089) = 1729 = 1728 + 1 (Hardy's taxi number 12^3 + 1^3, [more about this sum](http://www.mathpages.com/home/kmath028.htm) ), also 1728 is a rearrangement of 2178!]
[Added later] Eulertotient(2178) = Eulertotient(1089)
| https://mathoverflow.net/users/5627 | Ethio Integers? | These unnamed numbers were famous enough to make it to [A Mathematician's Apology](http://en.wikipedia.org/wiki/A_Mathematician%27s_Apology). After mentioning what you wrote above Hardy writes:
>
> These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals much to a mathematician. The proofs are neither difficult nor interesting—merely a little tiresome. The theorems are not serious; and it is plain that one reason (though perhaps not the most important) is the extreme speciality of both the enunciations and the proofs, which are not capable of any significant generalization.
>
>
>
These numbers are listed in [OEIS](http://www2.research.att.com/~njas/sequences/A008919), and there it is mentioned that
>
> Theorem (David W. Wilson): If reverse(n) = k\*n in base 10, then k = 1, 4 or 9.
>
>
>
Off the top of my head I can make the conjecture that none of these numbers contain the digit 3, for example, but I don't know of any interesting mathematical ideas that could be involved in the proof (this doesn't mean anything though, and is of course very subjective).
Edit: To answer your question about places where these numbers have been considered, I will mention [this](http://arxiv.org/abs/math/0511366) article by Lara Pudwell. The author's views are opposite to Hardy's, and she describes a few related problems and generalizations.
| 11 | https://mathoverflow.net/users/2384 | 27927 | 18,253 |
https://mathoverflow.net/questions/27931 | 38 | The $j$-function and the fact that 163 and 67 have class number 1 explain why:
$\operatorname{exp}(\pi\cdot \sqrt{163}) = 262537412640768743.99999999999925$,
$\operatorname{exp}(\pi\cdot \sqrt{67}) = 147197952743.9999987$.
But is there any explanation for these?:
$\frac{163}{\operatorname{ln}(163)} = 31.9999987 \approx 2^5$,
$\frac{67}{\operatorname{ln}(67)} = 15.93 \approx 2^4$,
$\frac{17}{\operatorname{ln}(17)} = 6.00025$.
These numbers seem too close to integers to occur by chance.
| https://mathoverflow.net/users/6769 | Why Is $\frac{163}{\operatorname{ln}(163)}$ a Near-Integer? | On the other hand, *Mathematica* gives LogIntegral[163]=43.075210908806756346563... and LogIntegral[67]=22.6520420103880266691324... so this does not appear to be connected to x/Ln[x] in the context of the Prime Number Theorem
| 4 | https://mathoverflow.net/users/6756 | 27932 | 18,255 |
https://mathoverflow.net/questions/27878 | -1 | [Akaike's information criterion](http://en.wikipedia.org/wiki/Akaike_information_criterion) is a measure of the goodness of fit of an estimated statistical model that accounts for both the fit quality and model complexity. One way to calculate AIC is as follows:
$\mathit{AIC}=2k + n[\ln(\mathit{RSS})]\,$
, where $k$ is the number of parameters in the model and $\mathit{RSS}$ is the residual sum of squares.
Assume two models:
$f\_1(x) = ax$
and
$f\_2(x) = \frac{ax}{b + cx} + d \sin(x)$
.
Is the number of parameters, $k$, in both the models equal 1?
**EDIT** In the equations above only $x$ is a variable, while $a$, $b$, $c$ ... are constant parameters
| https://mathoverflow.net/users/5823 | Number of parameters in Akaike's information criterion | The question should be: how many of the parameters need to get estimated based on the data? If three of them were somehow known independently of the data, then I say for present purposes there's only one.
The nonlinearity in $x$ in the $\sin x$ term is not a concern because $f\_2(x)$ is linear in $\sin x$ and the parameter is multiplied by that. But I'm wondering whether fitting by minimizing the residual sum of square makes sense when you have nonlinearity in $x$ in the other term.
Akaike's information criterion is one of many used for this purpose that I've heard of, and I'm really not familiar with any of them except to a minor extent that of Mallows.
| 0 | https://mathoverflow.net/users/6316 | 27966 | 18,278 |
https://mathoverflow.net/questions/27941 | 1 | The Fourier series of a function (B-spline) is given by:
$$s(x)=\sum\_{j=-\infty}^{\infty}\operatorname{sinc}\Bigl[\pi\frac{j}{K}\Bigr]^{p}\exp[2\pi ijx]$$
But the B-spline has only finite support. How can one see this using its Fourier series representation?
| https://mathoverflow.net/users/3589 | Fourier series of B-spline | A function is a piecewise polynomial if and only f it is a linear combination of functions, each of them having some derivative equal to a finite sum of Dirac measures.
The jth Fourier coefficient of the Dirac at a is $e^{2\pi ija}$, and integrating amounts to multiplying the coefficient by some power of $1\over j$.
As a result, a function is a spline with finite support if and only if its Fourier coefficients $c\_j$ can be written as a finite linear combination of terms of the form $e^{i\pi ja}\over j^k$, $k\in \mathbb{N},a\in \mathbb{R}$.
You can see that this is the case in your formula by expanding the coefficient $c\_j=({{e^{i\pi j/K}-e^{i\pi j/K}}\over j/K})^p$.
| 4 | https://mathoverflow.net/users/6129 | 27969 | 18,280 |
https://mathoverflow.net/questions/27971 | 60 | Cayley's theorem makes groups nice: a closed set of bijections is a group and a group is a closed set of bijections- beautiful, natural and understandable canonically as symmetry. It is not so much a technical theorem as a glorious wellspring of intuition- something, at least from my perspective, that rings are missing; and I want to know why.
Certainly the axiom system is more complicated- so there is no way you're going to get as simple a characterisation as you do with groups- but surely there must be some sort of universal object for rings of a given cardinality, analogous to the symmetric group in group theory. I would be surprised if it was a ring- the multiplicative and additive properties of a ring could be changed (somewhat) independently of one another- but perhaps a fibration of automorphisms over a group? If so is there a natural(ish) way of interpreting it?
Perhaps it's possible for a certain subclass of rings, perhaps it's possible but useless, perhaps it's impossible for specific reasons, in which case: the more specific the better.
**Edit:** So Jack's answer seems to have covered it (and quickly!): endomorphisms of abelian groups is nice! But can we do better? Is there a chance that 'abelian' can be unwound to the extent we can make this about sets again- or is that too much to hope for?
| https://mathoverflow.net/users/5869 | Why is there no Cayley's Theorem for rings? | Every (associative, unital) ring is a subring of the endomorphism ring of its underlying additive group. Rings act on abelian groups; groups act on sets. The universal action on an abelian group is its endomorphism ring; the universal action on a set is the symmetric group. Modules are rings that remember their action on an abelian group; permutation groups are groups that remember their action on a set.
A set is determined by its cardinality, but for abelian groups cardinality is not a very useful invariant. Rather than "order" of a ring, consider the isomorphism class of its underlying additive group. This is even commonly done in the finite ring case, where the order still has some mild control, but not as much as the isomorphism type of the additive group.
| 84 | https://mathoverflow.net/users/3710 | 27974 | 18,282 |
https://mathoverflow.net/questions/20265 | 2 | Here is my question: how to define global section functor from D-module on affine flag variety to representation of affine Lie algebra?
Let's me explain the difficulty: it seems there doesn't exist global definition of D-module on ind-scheme. For affine flag variety, it is a union of finite dimensional subvarieties, and usually we can't make them smooth. We should think of a D-module on a singular variety as a usual D-module on big smooth space which supports on this singular variety. On the other hand, the global sections of D-module depends on the embedding of singular variety to the other smooth One.
I really don't know how to think of global section functor of D-module on affine flag variety, so I don't know how to formulate the localization theorem.
Maybe I should look at Frenkel-Gaitsgory's paper, but I'm afraid it is a question before reading their papers.
Moreover, I would like to know what is the status of localization theorem for affine Lie algebra?
1. at Critical level
2. at noncritical level
| https://mathoverflow.net/users/5082 | About localization theorem for affine Lie algebra? | The main problem seems to be that you think the global section functor for (twisted) D-modules on a singular variety depends on a choice of embedding into a smooth variety. This is not true - D-modules can be defined on singular spaces using the infinitesimal site, and you can define global sections without any choice of embedding. Beilinson and Drinfeld describe the characteristic zero theory in section 7.10 of their unfinished book on Hitchin's integrable system, available from [this page](http://www.math.harvard.edu/~gaitsgde/grad_2009/)
Also, here are [notes on D-modules on ind-schemes](http://www.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/Apr15%28LimsofCats%29.pdf), from Dennis Gaitsgory's seminar.
The derived global section functor used in localization is constructed in section 23.5 of the [Frenkel-Gaitsgory paper](http://arxiv.org/abs/0712.0788). If you read the introduction of the paper, you will find a statement of their results, and you will find a claim that much less is known away from the critical level.
| 2 | https://mathoverflow.net/users/121 | 27977 | 18,285 |
https://mathoverflow.net/questions/27509 | 5 | Let $\gamma$ be a simple loop in a spine of a strongly irreducible Heegaard splitting of a closed 3-manifold $M$ with torsion-free fundamental group. Does $\gamma$ necessarily correspond to a primitive element of the fundamental group of $M$, or is it possible for $\gamma$ to be a power of some other element?
I suspect that $\gamma$ is not necessarily primitive in the fundamental group of $M$, but I do not know of any examples.
Edit: I added torsion-free fundamental group after Charlie's comment. In particular, I am most interested in this question for hyperbolic 3-manifolds.
| https://mathoverflow.net/users/4325 | Is a simple loop in a spine of a strongly irreducible Heegaard splitting primitive in the fundamental group? | **New Answer:** Take a 2-bridge knot, and perform hyperbolic
Dehn filling (so that the core of the Dehn filling is geodesic),
and so that the filling slope has intersection number $>1$ with
the meridian. Then the meridian will not be primitive, since it
will be a multiple of the core of the Dehn filling. 2-bridge
knots have a genus two Heegaard splitting, which has a spine
for the handlebody which is a [wedge of two meridians
at the bottom](http://www.math.uic.edu/~agol/parabolic/parabolic02.html). This remains a spine in the
Dehn filling, so the loops represented by the meridians are not
primitive. This also works for the (2,n) torus knots (which are
2-bridge), so I think Charlie's answer is right (at least for
many small Seifert fibered-spaces).
**Old (non)Answer:**
Here's almost an example. All punctured torus bundles have
[Heegaard genus $\leq 3$](http://www.ams.org/mathscinet-getitem?mr=1701636), and many have Heegaard genus 3 (I discussed
this [once in my defunct blog](http://www2.math.uic.edu/~agol/blog/040628.pdf)). One may find a genus 3 Heegaard splitting
of any once punctured torus bundle by taking two copies of a
fiber, tubing them together along the boundary on one side,
and adding a handle to the other side (drill out discs from
both fibers, and glue an annulus in). By a [theorem of
Moriah-Rubinstein](http://www.ams.org/mathscinet-getitem?mr=1487722), most Dehn fillings will also have
Heegaard genus 3 if the punctured torus bundle does.
The Heegaard splitting of the punctured torus bundle
has one side which is a handlebody,
and the other side a compression body. We may think of the
handlebody as a product neighborhood of the fiber (which is
a punctured torus) with a 1-handle attached. We may find a
spine for the handlebody which consists of a wedge of two
loops which is a spine for the punctured torus, together with
another loop going through the 1-handle.
Now, the peripheral curve of the punctured torus is not primitive
in Dehn fillings along curves which intersect the longitude
multiple times. This curve is represented in the spine not
as an embedded curve, but has multiplicity two (since it is
a commutator of the generators). If we choose a [small punctured torus bundle](http://www.ams.org/mathscinet-getitem?mr=675414)
of genus three, most Dehn fillings will be small (non-Haken) of
Heegaard genus 3, and so this Heegaard surface will be strongly
irreducible. But the peripheral curve will not be primitive,
since it will be a multiple of the core of the Dehn filling. However,
it is not embedded in the spine.
Even though this doesn't answer the question, it gives a strategy
for trying to find an example. Namely, if one can find a 1-cusped
hyperbolic 3-manifold which is small, and contains an incompressible
surface with boundary, such that the boundary slope is a generator in
the surface,
and such that a tubular neighborhood of the surface (or a slight modification by
drilling a hole in a tubular neighborhood)
is a minimal genus Heegaard splitting, then many Dehn fillings will
have the desired property.
| 4 | https://mathoverflow.net/users/1345 | 27983 | 18,291 |
https://mathoverflow.net/questions/27984 | 5 | "Random" modules of the same size over a polynomial ring seem to always have the same Betti table. By a "random" module I mean the cokernel of a matrix whose entries are random forms of a fixed degree.
For example if we take the quotient of the polynomial ring in three variables by five random cubics:
$S = \mathbb{Q}[x1,x2,x3]$
$M$ = coker random( S^1, S^{5:-3} )
then Macaulay2 "always" (e.g. 1000 out of 1000 times) gives the following Betti table
total: 1 5 9 5
0: 1 . . .
1: . . . .
2: . 5 . .
3: . . 9 5
It seems that the behavior can be explained by the fact that we can resolve the cokernel of a generic matrix of the given form and this resolution remains exact when specializing to any point in a Zariski open subset of some affine space. My question is whether anyone knows a slick proof of this fact.
To elaborate: we can adjoin a new variable to our original ring for each coefficient appearing in each entry of the matrix. So in the above example we would adjoin 10\*5 = 50 new variables to $S$, say $y1..y50$. Call the new ring $T$. Consider the $1x3$ matrix $N$ over $T$ whose entries are cubic in the $x\_i$ and linear in the $y\_i$. Resolve the cokernel of $N$ over $T$ to get a complex $F$. We can then substitute any point in $\mathbb{Q}^{50}$ into the maps of $F$ to get a complex over the original ring $S$. The claim is that this complex is exact on a Zariski open set of the affine space $\mathbb{Q}^{50}$.
It seems like this must be well known but I'm having trouble finding references.
| https://mathoverflow.net/users/3293 | Does the free resolution of the cokernel of a generic matrix remain exact on a Zariski open set? | It sounds like the question you mean to ask is the following: if $X$ is an integral noetherian scheme with generic point $\eta$ and $C^{\bullet}$ is a finite complex of coherent sheaves on $X$ such that $C^{\bullet}\_ {\eta}$ is exact, then does there exist a dense open $U$ in $X$ such that $C^{\bullet}|\_U$ is exact and has exact fibers? If that is your question, then the answer is yes. It is instructive to make your own proof by using generic flatness considerations for finite modules over noetherian domains (applied to various kernels, cokernels, images, etc.). Alternatively, look at EGA IV$\_3$, 9.4.2 and 9.4.3 for vast generalizations without noetherian hypotheses (e.g., 9.4.3 concerns compatibility of formation of homology sheaves with respect to passage to fibers over some dense open, not assuming exactness at the generic fiber). Even in this hyper-generality, the principle of using "generic flatness" remains the same.
| 6 | https://mathoverflow.net/users/6773 | 27986 | 18,292 |
https://mathoverflow.net/questions/27989 | 8 | We know that a countably additive translation invariant measure with $\mu([0,1]) = 1$ cannot be defined on the power set of $\mathbb R$. This is because $[0,1]$ can be partitioned into countably many congruent sets, with the help of the axiom of choice.
But I was wondering whether a *finitely additive* measure with these properties would be possible? I know it wouldn't be possible for dimension $n>2$ because of the Banach-Tarski paradox, but I am curious about $n=1$. If such a measure can be constructed on $\mathcal P(\mathbb R)$, would that be unique?
| https://mathoverflow.net/users/1229 | Finitely additive translation invariant measure on $\mathcal P(\mathbb R)$ | Banach-Tarski poses a problem for existence of measures that are invariant under all rigid motions, not just translation. The existence of finitely additive translation-invariant measures that agree with Lebesgue measure on Lebesgue-measurable sets is a consequence of the Hahn-Banach theorem. This is exercise 21 in chapter 10 of Royden's *Real Analysis*. The extensions given by Hahn-Banach don't seem to have any uniqueness properties, so I doubt this measure is unique.
| 7 | https://mathoverflow.net/users/121 | 27994 | 18,297 |
https://mathoverflow.net/questions/27965 | 1 | For me the definition of amenability of an at most countable discrete group (with counting measure) is existence of a Folner sequence. Assuming this, why is every countable discrete abelian group amenable? What is the Folner sequence that does the job?
| https://mathoverflow.net/users/5498 | Countable discrete abelian group amenable | The direct limit $G = \bigcup\_n G\_n$ of a nested sequence of countable amenable groups $G\_n$ is still amenable, since every finite set $S$ in $G$ will lie in one of the $G\_n$ and thus there must exist some finite set $F\_S$ which is not shifted very much by $S$. Since there are only a countable number of $S$, one can diagonalise and obtain a Folner sequence for $G$.
Since every countable abelian group is the direct limit of finitely generated abelian groups, which have polynomial growth and are thus amenable, every countable abelian group is amenable.
An instructive example is the free group $\bigcup\_n Z^n$ on countably many generators. Here, the sets $\{-N\_n,\ldots,N\_n\}^n$ will form a Folner sequence if $N\_n$ grows sufficiently rapidly in n.
I have some notes on amenability that cover these topics at
<http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/>
| 5 | https://mathoverflow.net/users/766 | 27999 | 18,301 |
https://mathoverflow.net/questions/28000 | 13 | There are (at least) two ways of writing down the Dirichlet L-function associated to a given character χ: as a Dirichlet series
$$\sum\_{n=1}^\infty \frac{\chi(n)}{n^s}$$
or as an Euler product
$$\prod\_{p\mbox{ prime}} \left(1-\frac{\chi(p)}{p^s}\right)^{-1}.$$
Correspondingly, this gives two ways of restricting a Dirichlet L-function to an arithmetic progression, by considering either
$$A(s)=\sum\_{n\equiv a\pmod{q}} \frac{\chi(n)}{n^s}$$
or
$$B(s)=\prod\_{p\equiv a\pmod{q}} \left(1-\frac{\chi(p)}{p^s}\right)^{-1}.$$
I am assuming here that *a* and *q* are relatively prime. The function *A(s)* is a fairly classical object and has been studied extensively.
I am interested in finding out some analytic information on the function *B(s)*: can it be meromorphically continued to the complex plane? What are its poles, if any? What are the singular parts corresponding to these poles? If that makes things any easier, I would even be happy to know about this in the special cases (*a*, *q*)=(1, 3) and (*a*, *q*)=(2, 3).
I have had no success in tracking this down, so I am hoping that somebody will either know of some references where this is worked out, or some hints on how I could go about doing it myself.
| https://mathoverflow.net/users/6243 | What are the analytic properties of Dirichlet Euler products restricted to arithmetic progressions? | It's best to split this up into two cases.
Case 1: $\chi(a) = 1$. Then for $\Re(s) > 1$,
$$\prod\_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod\_{p \equiv a \pmod{q}} \left(1 - \frac{1}{p^s}\right)^{-1} = \sum\_{n \in \left\langle \mathcal{P} \right\rangle} \frac{1}{n^s},$$
where $\left\langle \mathcal{P} \right\rangle$ is the multiplicative semigroup generated by the set of primes $\mathcal{P}$ consisting of all $p \equiv a \pmod{q}$. So this is just the Burgess zeta function $\zeta\_{\mathcal{P}}(s)$. Now there exist Burgess zeta functions that cannot be holomorphically extended to $1 + it$ for any $t \in \mathbb{R}$ (this is mentioned for example in Terry Tao's paper "A Remark on Partial Sums Involving the Mobius Functions", which I'm pretty sure is available somewhere on the arxiv). In this case, however, I have no idea; perhaps some of the relevant literature discusses it.
Case 2: $\chi(a) = -1$. Then
$$\prod\_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod\_{p \equiv a \pmod{q}} \left(1 + \frac{1}{p^s}\right)^{-1} = \sum\_{n \in \left\langle \mathcal{P} \right\rangle} \frac{\lambda(n)}{n^s},$$
where $\lambda(n)$ is the Liouville function. Equivalently,
$$\prod\_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \frac{\zeta\_{\mathcal{P}}(2s)}{\zeta\_{\mathcal{P}}(s)},$$
so it comes down to the same thing; determining whether $\zeta\_{\mathcal{P}}(s)$ extends holomorphically to the line $\Re(s) = 1$ and beyond.
EDIT: Recall that the prime number theorem for arithmetic progressions says that
$$\pi(x;q,a) = \frac{1}{\varphi(q)} \mathrm{li}(x) + O\_A(x \exp(-c\_1 (\log x)^{1/2})$$
for fixed $A > 0$ with $q \leq (\log x)^A$. An application of a result of Diamond (cf. *Asymptotic Distribution of Beurling's Generalized Integers*) then implies that
$$N\_{\mathcal{P}}(x) = \sum\_{n \in \left\langle \mathcal{P} \right\rangle, \; n \leq x}{1} = a x + O\_A(x \exp(-c\_2 (\log x \log \log x)^{1/3})$$
for some particular $a > 0$. By partial summation, we have that for $\Re(s) > 1$,
$$\zeta\_{\mathcal{P}}(s) = \frac{as}{s-1} + s \int\_{1}^{\infty} \frac{N\_{\mathcal{P}}(x) - ax}{x^{s+1}} \: dx .$$
Diamond's result implies that this integral is uniformly convergent for $\Re(s) \geq 1$, and so it is continuous in this half-plane. Thus $\zeta\_{\mathcal{P}}(s) = c/(s-1) + r\_0(s)$ with $r\_0(s)$ continuous for $\Re(s) \geq 1$, and so $\zeta\_{\mathcal{P}}(s)$ extends to $\Re(s) \geq 1$ with a singularity at $s = 1$. Moreover, it is not difficult to show that $\zeta\_{\mathcal{P}}(1+it) \neq 0$ for all $t \in \mathbb{R}$; a version of this is shown in Montgomery and Vaughan's *Multiplicative Number Theory I: Classical Theory* section 8.4.
Note also that assuming the generalised Riemann Hypothesis, it is possible to strengthen this meromorphic extension of $\zeta\_{\mathcal{P}}(s)$ to $\Re(s) > 1/2$ with $\zeta\_{\mathcal{P}}(s)$ nonvanishing in this open half-plane; see Titus W. Hilberdink and Michel L. Lapidus, *Beurling Zeta Functions, Generalised Primes and Fractal Membranes*.
| 11 | https://mathoverflow.net/users/3803 | 28004 | 18,305 |
https://mathoverflow.net/questions/27997 | 2 | Let $X = \mathbb{R}^n$, and consider a nondegenerate representation $\rho: C\_0(X) \to B(H)$ where $B(H)$ is the algebra of bounded operators on a separable Hilbert space. The support of a vector $v \in H$ is defined to be the complement in $X$ of the union of all open sets $U$ such that $\rho(f)v = 0$ for every $f \in C\_0(U)$.
Suppose $v$ has compact support. My intuition is that any function $g \in C\_0(X)$ which restricts to the zero function on $supp(v)$ should satisfy $\rho(g)v = 0$, but I can't quite prove it. Here is what I have so far.
Let $\mathcal{F}$ denote the collection of open sets $U$ such that $\rho(f)v = 0$ for $f \in C\_0(U)$. By definition $V = supp(v)^c$ is the union of the open sets in $\mathcal{F}$, and moreover a simple partition of unity argument shows that $\mathcal{F}$ is closed under finite unions. If we can show that $V \in \mathcal{F}$ then we are done because by the hypotheses $g \in C\_0(V)$. I'm sure I'm just missing something simple; can anyone help?
This result should be true if $X$ is any separable metric space equipped with a proper coarse structure, so I suppose the proper setting for this question is metric geometry. That should explain the title of the question and the tags.
| https://mathoverflow.net/users/4362 | Is this a correct interpretation of support in coarse geometry? | **Edit** I have amended the proof to cover the general case following a suggestion of Matthew Daws.
By the definition of $supp(v)$, for any $x$ in $supp(v)^c$ there exists an open set $U(x)\subset supp(v)^c$ containing $x$ such that $\rho(f)v=0$ for all $f\in C\_0(U(x)).$ If $g$ has compact support $K\subset supp(v)^c,$ there is a finite subset $U\_1,\ldots,U\_m$ of $\{U(x)\}$ covering $K.$ Using a partition of unity, $g=g\_1+\ldots+g\_M$ where $g\_i\in C\_0(U\_{k(i)})$. Therefore, $\rho(g)v=\sum\_i \rho(g\_i)v=0.$ In general, by the lemma below, we can approximate $g$ by a sequence $\{g\_n\}$ of continuous functions with compact support disjoint from $supp(v),$ so $\rho(g)v=\rho(\lim g\_n)v=\lim \rho(g\_n)v=0.\square$
**Lemma** Suppose that $L\subset X$ is compact and $g\in C\_0(X)$ restricts to zero function on $L.$ Then $g=\lim g\_n,$ where $g\_n\in C\_0(X)$ has compact support disjoint from $L.$
**Proof** The function $g\_n$ is obtained from $g$ by a smooth cutoff at distance $1/n$ from $L.$ The approximation property follows from the fact that $g$ vanishes on $L$.
More formally, let $h:\mathbb{R}\to [0,1]$ be a continuous function such that $h(y)=0$ for $y\leq 1$, $h(y)=1$ for $y\geq 2.$
```
___
/
/
/ h(x)
____/
1 2
```
Since $L$ is compact, the distance function $d\_X(\cdot,L)$ is well-defined and continuous. Let $L\_n$ be the open $1/n$-neighborhood of $L$ in $X$. Set
$$g\_n(x)=h(nd\_X(x,L))g(x).$$
By construction, $g\_n$ vanishes on $L\_n$ and coincides with $g$ on $L\_{2n}^c$. Its support is compact and is contained in $L\_n^c.$ Moreover,
$$ \|g-g\_n\|\leq \sup\_{x\in {L}\_{2n}} |g(x)|.$$
The right hand side is a non-negative monotone decreasing sequence. Suppose that there exists a sequence of points $x\_n\in L\_{2n}$ such that $g(x\_n)$ is bounded away from $0.$ Since $g$ is compactly supported, this sequence has an accumulation point $x.$ Then $x\in L$ and so $g(x)=0,$ which is a contradiction. $\square$
| 4 | https://mathoverflow.net/users/5740 | 28005 | 18,306 |
https://mathoverflow.net/questions/27972 | 18 | Is it true that a line bundle is relatively ample iff its restsriction to fibers is? If so, what would be the reference?
| https://mathoverflow.net/users/6772 | Relatively ample line bundles | If you admit the map to be proper and the schemes to be reasonably good it is true.
A reference I know is Lazarsfeld's book "Positivity in algebraic geometry", paragraph 1.7.
| 5 | https://mathoverflow.net/users/6430 | 28006 | 18,307 |
https://mathoverflow.net/questions/8269 | 11 | This is a question about forcing. I have seen the following fact mentioned in multiple places, but have not been able to find a proof: if a random real is added to a transitive model of ZFC, then in the generic extension the set of reals in the ground model becomes meager.
My guess is that one should be able to, in some natural way, directly construct from a random real a countable sequence of nowhere dense sets covering the ground model reals, but I am not sure.
| https://mathoverflow.net/users/2436 | Adding a random real makes the set of ground model reals meager | The proof is based on the fact that there is a decomposition ${\bf R}=A\cup B$ of the reals such that $A$, $B$ are (very simple) Borel sets, $A$ is meager, $B$ is of measure zero, and ${\bf R}=A\cup B$ even holds if after forcing we reinterpret the sets. Nos let $s$ be a random real. If $r\in {\bf R}$ is an old real, then $s\notin r+B$, so $s\in r+A$, that is, the meager
$s-A$ contains all old reals.
| 8 | https://mathoverflow.net/users/6647 | 28024 | 18,319 |
https://mathoverflow.net/questions/28028 | 33 | It is well known how to construct a Laplacian on a fractal using the Dirichlet forms (see e.g.
[the survey article](http://www.ams.org/notices/199910/fea-strichartz.pdf) by Strichartz). This implies, in particular, that a fractal can be "heated", i.e. one can write (and solve) the heat equation on the fractal.
**The question is**, can one run a fluid flow through a fractal set? In other words, is there
a proper way to write the Navier-Stokes equations on a fractal? In order to do this, it seems that we need a "correct" notion of *divergence* at least.
More generally, *is there a "correct" way to define a differential form on a fractal?*
| https://mathoverflow.net/users/5371 | How to define a differential form on a fractal? | Take a look at [Jenny Harrison, "Flux across nonsmooth boundaries and fractal Gauss/Green/Stokes' theorem,"](http://iopscience.iop.org/0305-4470/32/28/310/) which should at least answer your question about a "correct" notion of flux and divergence for a fractal domain -- here's the abstract:
>
> By replacing the parametrization of a
> domain with polyhedral approximations
> we give optimal extensions of theorems
> of Gauss, Green and Stokes'. Permitted
> domains of integration range from
> smooth submanifolds to structures that
> may not be locally Euclidean and have
> no tangent vectors defined anywhere.
> One may still calculate divergence and
> curl over a domain, and flux across
> its boundary which itself may have no
> normal vectors defined anywhere.
>
>
>
| 15 | https://mathoverflow.net/users/1557 | 28029 | 18,321 |
https://mathoverflow.net/questions/28025 | 16 | I am sorry for the following question, because the actual answer to this question is in the beautiful works of Feferman and Jeroslow, but, unfortunately, I havn't any time to go into that specific field right now and maybe some of you is aware of the answer.
The situation with Gödel's second incompleteness theorem is quite delicate. Let Pf(a, b) means that a is the Gödel number of proof of statement with Gödel number b and Neg(a, b) means that b is the Gödel number of negation of statement with Gödel number a. The main result of Gödel's work is representability of predicates Pf and Neg in formal arithmetic. Knowing that, we can define the Consistency of formal arithmetic in formal arithmetic as follows:
" it is not the case that there exist a statement A such that A and (not) A are provable in formal arithmetic " or in the language of formal arithmetic:
$$\forall x\_1, x\_2, x\_3, x\_4 [\neg (\operatorname{Pf}(x\_1, x\_3) \wedge \operatorname{Pf}(x\_2, x\_4) \wedge \operatorname{Neg}(x\_3, x\_4))].$$
Let's denote last proposition by W. Gödel's second incompleteness theorem says that W is not provable in formal arithmetic, i. e. the consistency of formal arithmetic is not provable in formal arithmetic.
But Solomon Feferman in his remarkable paper of 1960 "Arithmetization of metamathematics in general setting" have found other formalization of consistency of formal arithmetic W' that is provable in formal arithmetic (see notes by E. Mendelson to the second theorem in formal arithmetic chapter of his "Intoduction to Mathematical logic"). Jeroslow ("Consistency statements in formal theories"), thereafter, studied consistency statements in a broader way.
Can someone explain the nature of W'?
(obviously, $W\Leftrightarrow W'$ can not be proven in PA).
Can we interpret the provability of W' as the proof of consistency of formal arithmetic in formal arithmetic?
How well are various consistency statements grasped today?
Thanks in advance.
| https://mathoverflow.net/users/6307 | Clarification of Gödel's second incompleteness theorem | The key idea Feferman is exploiting is that there can be two different enumerations of the axioms of a theory, so that the theory does not prove that the two enumerations give the same theory.
Here is an example. Let $A$ be a set of the axioms defined by some formula $\phi(n)$ (that is, $\phi(x)$ holds for exactly those $x$ that are in $A$). Define a new formula $\psi(n)$ like so:
$\psi(n) \equiv \phi(n) \land \text{Con}( \langle x \leq n : \phi(x)\rangle)$
Where Con(σ) is a formula which says that no contradiction is provable from the axioms listed in the sequence σ.
In the case where $A$ is the set of axioms for a suitable consistent theory $T$ that satisfies the second incompleteness theorem, the following hold:
(1) In the standard model, we have $\phi(n) \Leftrightarrow \psi(n)$ for all $n$, because $T$ really is consistent.
(2) T does not prove that $\phi(n) \Leftrightarrow \psi(n)$ for all $n$, because this equivalence implies that T is consistent.
(3) If we use $\psi$ to define a formula Conψ(T), then T will prove (under the assumption that the empty theory is consistent, if this is not provable in T) that the theory defined by ψ is consistent. However, T will not prove Conφ(T), which is the usual consistency sentence for T.
This kind of example is presented in a more precise way in Feferman's 1960 paper that you mentioned, along with precise hypotheses on the theory and sharper results.
My opinion is that we cannot regard a proof of Conψ(T) as a proof of the consistency of T, because although φ and ψ are extensionally identical, they do not intensionally represent the same theory. Feferman expresses a similar idea on his p. 69. Of course, this is a matter of philosophy or interpretation rather than a formal mathematical question.
---
***Addendum***
The difference between extensional and intensional equality is easiest to explain by example. Let A be the empty set and let B be the set of integers larger than 1 that do not have a unique prime factorization. Then we know B is also the empty set: so A and B are *extensionally* equal. But the definition of B is much different than the definition of A: so A and B are *intensionally* different.
This distinction is often important in contexts like computability and formal logic where the things that you work with are actually definitions (also called codes, indices, Gödel numbers, or notations) rather than the objects being defined. In many cases, extensional equality is problematic, because of computability or effectiveness problems. For example, in my answer above, we know that φ and ψ define the same set in the real world, but this fact requires proof, and that proof may be impossible in the object theory we are dealing with. On the other hand, intensional equality is easy to decide, provided you are working directly with definitions.
| 21 | https://mathoverflow.net/users/5442 | 28031 | 18,322 |
https://mathoverflow.net/questions/28008 | 12 | Disclaimer:
-----------
I am asking this question as an improvement to [this question](https://mathoverflow.net/questions/27881/who-is-the-last-mathematician-that-understood-all-of-mathematics), which should be community wiki. This is in line with the actions taken by Andy Putman in a similar case (cf. [meta](http://mathoverflow.tqft.net/discussion/399/i-hope-that-people-will-do-this-in-the-future/)).
See the relevant [meta thread](http://mathoverflow.tqft.net/discussion/439/who-is-the-last-mathematician-that-understood-all-of-mathematics/) about the previous question.
Edit: If it wasn't already obvious, I only asked this question to prevent the other one (which was not made community wiki) from being reopened.
Question:
---------
The scope of mathematics has grown immensely since ancient times. At what point in time did it become impossible for a single person to understand the majority of mathematics enough to keep current with contemporary research?
Edit: Clarified the wording.
| https://mathoverflow.net/users/1353 | At what point in history did it become impossible for a person to understand most of mathematics? | The world's output of scientific papers increased exponentially from 1700 to 1950.
One online source is [this article](http://www.its.caltech.edu/~dg/crunch_art.html) (which is concerned with what has happened since then). The author displays a graph (whose source is a 1961 book entitled "Science since Babylon" by Derek da Solla Price) showing exponential increase in the cumulative number of scientific journals founded; an increase by a factor of 10 every 50 years or so, with around 10 journals recorded in 1750.
Perhaps someone can locate similar statistics specific to mathematics, but it's reasonable to expect the same pattern. If so, it is a long time since any individual could follow the primary mathematical literature in anything close to its entirety.
But then, gobbling papers is not how leading mathematicians (or scientists) actually operate.
By making judicious choices of what to pursue when, and with sufficient brilliance and vision, it is possible even today to make decisive contributions to many fields. Serre has done so in, and between, algebraic topology, complex analytic geometry, algebraic geometry, commutative algebra and group theory, and continues to do so in algebraic number theory/representation theory/modular forms.
| 11 | https://mathoverflow.net/users/2356 | 28033 | 18,324 |
https://mathoverflow.net/questions/25691 | 25 | G(n,p)
------
We are familiar with the standard notion of random graphs where you fixed the number n of vertices and choose every edge to belong to the graph with probability 1/2 (or p) independently. This model is referred to as **G(n,1/2)** or more generally **G(n,p).**
Random graphs with prescribed marginal behavior
-----------------------------------------------
Now, suppose that rather than prescribe the probability for every edge, you presecribe the marginal probability for every induced subgraph H on r vertices. (So r should be a small integer, 3,4,5, etc.) You make the additional assumptions that
a) the probability $p\_H$ does not depend on the identity of the r vertices,
and
b) It depends only on the isomorphism type of $H$.
So, for example: for r=3 you can think about the case that $p\_H = 1/9$ if H is a triangle or an empty graph and $p\_H=7/54$ otherwise.
Once these $p\_H$ are assigned you consider among all the probability distributions with these marginal behavior (if there are any) the one with maximal entropy. (But this choice **is negotiable**; if there is something different worth doing this is fine too.)
My questions:
-------------
1) Are these models studied in the literature?
2) When are such $p\_H$'s feasible?
3) Given such feasible $p\_H$'s say on graphs with 4 vertices, is there a quick algorithm to sample from the maximal-Entropy distribution which will allow to experiment with this model?
### Background
This question is motivated by a recent talk by Nati Linial in our "basic notion" seminar on extremal graph theory. (Maybe these are well studied models that I simply forgot, but I don't recall it now.)
| https://mathoverflow.net/users/1532 | Some models for random graphs that I am curious about | The Lovasz-Szegedy theory of *graphons* is likely to be relevant. Every measurable symmetric function $p: [0,1] \times [0,1] \to [0,1]$ (otherwise known as a graphon) determines a random graph model, in which every vertex v is assigned a colour c(v) uniformly at random from the unit interval [0,1], and then any two vertices v, w are connected by an edge with an independent probability of p(c(v),c(w)). These are in some sense the only models of large graphs in the sense that any sequence of increasingly large graphs has a subsequence that converges to a graphon (in the sense that the $p\_H$ statistics converge).
Each $p\_H$ (in the asymptotic limit $n \to \infty$) can be read off from the graphon as an integral. For instance, the density of triangles is $\int\int\int\_{[0,1]^3} p(x,y) p(y,z) p(z,x)\ dx dy dz$.
If the $p\_H$ are specified for *all* finite graphs H, then this determines p up to change of variables (measure-preserving bijections on [0,1]) outside of a set of measure zero. But if one only specifies the $p\_H$ for a finite number of H then there are multiple choices for p (and in some cases, no choices at all) and it is not obvious to me how to find a solution or even to determine whether when a solution exists. (Note that even for just two choices of H, one being an edge and the other being a bipartite graph, the question of determining the possible values of $p\_H$ is essentially Sidorenko's conjecture, which is still not fully resolved.) But perhaps numerical methods (annealing, gradient descent, etc.) may be able to find solutions some of the time (though they will hardly be "canonical").
| 16 | https://mathoverflow.net/users/766 | 28045 | 18,331 |
https://mathoverflow.net/questions/28027 | 4 | Let $L/K$ be a finite Galois extension with Galois group $G$ and $V$ a $L$-vector space, on which $G$ acts by $K$-automorphisms satisfying $g(\lambda v)=g(\lambda) g(v)$. It is known that the canonical map
$V^G \otimes\_K L \to V$
is an isomorphism. However, I can't find any short and nice proof for that. Actually I'm wondering if it is possible to construct an explicit inverse map, by choosing a basis of $L/K$ and taking some average with respect to $G$. Any ideas? I'm interested in the case of positive characteristic.
| https://mathoverflow.net/users/2841 | Galois descent, explicit inverse map | Martin, <http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf>
is a handout on this kind of stuff
and Theorem 2.14 there gives a proof of the bijection between $K$-forms of $V$ and $G$-structures on $V$. It doesn't qualify as "short", and whether it's "nice" or not is too
subjective. I wrote it for a target audience that knows only Galois theory and tensor products.
As for an explicit inverse map, see the top of page 6.
Let $Tr\_G \colon V \rightarrow V^G$ by $Tr\_G(v) = \sum\_{\sigma \in G} \sigma(v)$. If $d = [L:K]$ and $\alpha\_1,\dots,\alpha\_d$ is a $K$-basis of $L$, there exist
$\beta\_1,\dots,\beta\_d$ in $L$ such that
$$
v = \sum\_{j=1}^d \alpha\_jTr\_G(\beta\_j v)
$$
for all $v$ in $V$. The right side provides a decomposition coming from $L \otimes\_K V^G$.
| 6 | https://mathoverflow.net/users/3272 | 28049 | 18,333 |
https://mathoverflow.net/questions/27584 | 5 | Hi,
Having a random variable $X$ I am trying to find a stochastic process $Z\_t$ such that:
$$P[Z\_t>T] = P[X > T | X > t]$$
for all $T>t$, or a proof that such a process does not exist.
Please note that this question is not related to any homework and that I actually need this result for my research in financial maths.
**Edit**
I haven't really mentioned it, but what I am really after is some sort of closed form formula for $Z\_t$, ideally as a function of $X$ and $t$.
| https://mathoverflow.net/users/3160 | Process equivalent to conditional probability | Cool problem. The process you are after is certainly not unique, but here is a reasonably explicit construction of an increasing jump process $Z\_t$ with the property you want. (Under a couple of assumptions which I think are implicit in your statement).
The assumptions: 1) $X$ is a positive random variable, $P(X>0)=1$. 2) $X$ is unbounded, $P(X>T)\neq 0$. (In fact the construction works without the second assumption, but then $Z\_t$ stops after some random time.)
Let $X\_0, X\_1,X\_2,\ldots$ be the following Markov chain:
1.) $X\_0=0$ with probability one.
2.) Given $X\_0,\ldots,X\_{j-1}$ let the distribution of $X\_j$ be
$$P(X\_j >T|X\_1,\ldots,X\_{j-1}) = P(X >T|X>X\_{j-1}).$$
(In particular $X\_1$ is a "copy" of $X$: $P(X\_1 >T)=P(X>T)$. The distributions of $X\_2,\ldots$ are more complicated.)
Clearly $X\_j$ is a strictly increasing sequence with probability one. One can also show that $\lim\_n X\_n =\infty$ with probability one. Since $X\_0=0$ it follows that for any $t\ge 0$ we can find a unique (random) $j\_t$ such that
$$X\_{j\_t-1} \le t< X\_{j\_t}.$$
Let
$$Z\_t = X\_{j\_t},$$
so $Z\_t$ is piecewise constant and increasing.
To see that $Z\_t$ has the property you want, note that
$$P(Z\_t >T)=\sum\_{j=1}^\infty P( (X\_j>T) \& (j\_t =j)).$$
Let $\nu$ be the probability measure for the distribution of $X\_{j-1}$. Then by the definitions of $j\_t$ and of $X\_j$,
$$P( (X\_j>T) \& (j\_t =j)) = \int\_{(0,t]} P((X >T) \& (X>t) |X>x) d \nu(x).$$
Since the even $(X >t) \subset (X>x)$ for $x \le t$ in the domain of integration we have
$$P((X>T) \& (X>t) |X>x) = \frac{P((X>T)\& (X>t))}{P(X>t)} \frac{P(X>t)}{P(X>x)} =P(X>T|X>t) P(X>t|X>x).$$
Thus
$$P( (X\_j>T) \& (j\_t =j)) = P(X>T|X>t) P((X\_j >t) \& (X\_{j-1}\le t)) = P(X>T|X>t)P(j\_t=j),$$
and so
$$P(Z\_t >T) = P(X>T | X>t) \sum\_{j=1}^\infty P(j\_t=j)= P(X>T | X>t)$$
as desired!
(By the way, you can construct the sequence $X\_1,\ldots$ as follows. Let $Y\_1,\ldots$ be a sequence of independent identically distributed random variables, each with the distribution of $X$. We will take $X\_j=Y\_{n\_j}$ with $n\_j$ a certain random sequence that depends on $Y\_1,\ldots$. Let $X\_1=Y\_1$ and given $X\_j=Y\_{n\_j}$ let $n\_{j+1}$ be the first index $n$ larger than $n\_j$ such that $Y\_{n\_j} < Y\_{n}$. So long as $P(X>T)\neq 0$ for all $T>0$ we produce in this way an infinite sequence $X\_1,\ldots$.)
| 6 | https://mathoverflow.net/users/6781 | 28050 | 18,334 |
https://mathoverflow.net/questions/28053 | 7 | This question probably has a simple and immediate answer which escapes me now. (And, I should admit, it's more my curiosity than anything else.) The only natural way to construct a group structure on the cartesian product $G\times H$ of two groups $G$ and $H$ (in particular, ``natural'' to me means that on each factor the group product should be the original one) is the semi-direct product in the case when one group acts on another one by automorphisms. Are there any natural constructions of a group structure on $G\times H$ where neither factor is a normal subgroup?
Update: I was pointed out that the notion of Zappa-Szep product that appears in the answer given by Steven Gubkin is also mentioned in [an earlier MO discussion](https://mathoverflow.net/questions/5528/when-does-a-subgroup-h-of-a-group-g-have-a-complement-in-g); I thought I'd link it here for some sort of connectivity.
| https://mathoverflow.net/users/1306 | Group structures on the cartesian product of two groups | Wikipedia to the rescue!
<http://en.wikipedia.org/wiki/Zappa-Szep_product>
| 10 | https://mathoverflow.net/users/1106 | 28058 | 18,338 |
https://mathoverflow.net/questions/28056 | 31 | **Question.** Given a Turing-machine program $e$, which
is guaranteed to run in polynomial time, can we computably
find such a polynomial?
In other words, is there a
computable function $e\mapsto p\_e$, such that whenever $e$
is a Turing-machine program that runs in polynomial time,
then $p\_e$ is such a polynomial time bound? That is, $p\_e$
is a polynomial over the integers in one variable and
program $e$ on every input $n$ runs in time at most
$p\_e(|n|)$, where $|n|$ is the length of the input $n$.
(Note that I impose no requirement on $p\_e$ when $e$ is not
a polynomial-time program, and I am not asking whether the
function $e\mapsto p\_e$ is polynomial-time computable, but
rather, just whether it is computable at all.)
In the field of complexity theory, it is common to treat
polynomial-time algorithms as coming equipped with an
explicit polynomial clock, that counts steps during the
computation and forces a halt when expired. This convention
allows for certain conveniences in the theory. In the field
of computability theory, however, one does not usually
assume that a polynomial-time algorithm comes equipped with
such a counter. My question is whether we can computably
produce such a counter just from the Turing machine
program.
*I expect a negative answer.* I think there is no such
computable function $e\mapsto p\_e$, and the question is
really about how we are to prove this. But I don't know...
Of course, given a program $e$, we can get finitely many
sample points for a lower bound on the polynomial, but this
doesn't seem helpful. Furthermore, it seems that the lesson
of [Rice's Theorem](http://en.wikipedia.org/wiki/Rice%27s_theorem) is
that we cannot expect to compute nontrivial information by
actually looking at the program itself, and I take this as
evidence against an affirmative answer. At the same time,
Rice's theorem does not directly apply, since the
polynomial $p\_e$ is not dependent on the set or function
that $e$ computes, but rather on the way that it computes
it. So I'm not sure.
Finally, let me mention that this question is related to and
inspired by [this recent interesting MO question about the
impossibility of converting NP algorithms to P
algorithms](https://mathoverflow.net/questions/27867/non-existence-of-algorithm-converting-np-algorithm-to-p-algorithm).
Several of the proposed answers there hinged critically on
whether the polynomial-time counter was part of the input
or not. In particular, an affirmative answer to the present
question leads to a solution of that question by those
answers. My expectation, however, is for a negative answer
here and an answer there ruling out a computable
transformation.
| https://mathoverflow.net/users/1946 | Given a polynomial-time algorithm, can we compute an explicit polynomial time bound just from the program? | [Edit: A bug in the original proof has been fixed, thanks to a comment by Francois Dorais.]
The answer is no. This kind of thing can be proved by what I call a "gas tank" argument. First enumerate all Turing machines in some manner $N\_1, N\_2, N\_3, \ldots$ Then construct a sequence of Turing machines $M\_1, M\_2, M\_3, \ldots$ as follows. On an input of length $n$, $M\_i$ simulates $N\_i$ (on empty input) for up to $n$ steps. If $N\_i$ does not halt within that time, then $M\_i$ halts immediately after the $n$th step. However, if $N\_i$ halts within the first $n$ steps, then $M\_i$ "runs out of gas" and starts behaving erratically, which in this context means (say) that it continues running for $n^e$ steps before halting where $e$ is the number of steps that $N\_i$ took to halt.
Now if we had a program $P$ that could compute a polynomial upper bound on any polytime machine, then we could determine whether $N\_i$ halts by calling $P$ on $M\_i$, reading off the exponent $e$, and simulating $N\_i$ for (at most) $e$ steps. If $N\_i$ doesn't halt by then, then we know it will never halt.
Of course this proof technique is very general; for example, $M\_i$ can be made to simulate any fixed $M$ as long as it has gas but then do something else when it runs out of gas, proving that it will be undecidable whether an arbitrary given machine behaves like $M$.
| 41 | https://mathoverflow.net/users/3106 | 28060 | 18,340 |
https://mathoverflow.net/questions/28062 | 6 | Let $M$ be a compact Riemannian manifold and let $T : M \rightarrow M$ be Anosov. I have read [here](http://books.google.com/books?id=hFN6oiecbrYC&pg=PA7) that it is an open problem to prove that $T$ is topologically mixing if $M$ is connected. [Katok and Hasselblatt](http://books.google.com/books?id=9nL7ZX8Djp4C&dq&pg=PA151) point out that if $(T, \mu)$ is mixing, then the restriction of $T$ to the support of $\mu$ is topologically mixing. Yet I have also read [here](http://www.jstor.org/pss/2373810) (among other places) that Axiom A diffeomorphisms (and hence Anosov diffeomorphisms) satisfy exponential decay of correlations, which seems like it implies mixing (take characteristic functions). Still, I see references in the physics literature (which is where I am coming from) to "mixing Anosov maps".
While the Anosov alternative for *flows* is clear enough, I haven't found a characterization of mixing properties for Anosov diffeomorphisms that sorts out these apparently contradictory statements.
With that in mind, my question is (sort of): under what circumstances is an Anosov diffeomorphism guaranteed to be mixing?
| https://mathoverflow.net/users/1847 | When is an Anosov diffeomorphism mixing? | I don't have a proper answer to your main question beyond pointing to the list of equivalent properties in Pesin's book (your first reference), which you've obviously seen already. However, I'll point out that in Ruelle's paper (your third reference), the first main theorem (on page 3), which contains a statement on exponential decay of correlations, is proved under the hypothesis that the unstable manifolds are dense in the attractor. As in Pesin's book, that will imply mixing (on the attractor), so there's no mystery in the coexistence of the result stated by Ruelle and the open problem stated by Pesin.
As far as I know the best that you can say in general is that Anosov diffeos are mixing on the non-wandering set, or on the closure of an unstable manifold. So an equivalent question is, "Under what circumstances is every point non-wandering for an Anosov diffeo?" (Or, "Under what circumstances are unstable manifolds dense?")
| 4 | https://mathoverflow.net/users/5701 | 28065 | 18,343 |
https://mathoverflow.net/questions/28063 | 14 | Here are four remarks about the homology and homotopy type of a compact, complex manifold $M$:
1. If $M$ is Kähler, then it is symplectic and thus $H^2(M,\mathbb{R}) \ne 0$. (Also, as explained in a [blog posting by David Speyer](http://sbseminar.wordpress.com/2008/02/14/complex-manifolds-which-are-not-algebraic/), you still have $H^2(M,\mathbb{R}) \ne 0$ even if $M$ is non-projective but algebraic.)
2. An interesting first example of a non-Kähler manifold is a Hopf manifold, by definition $(\mathbb{C}^n\setminus 0)/\Gamma\_r$, where $\Gamma\_r$ is a rescaling by $r$ with $|r| \ne 0,1$. This example has $H^1(M,\mathbb{R}) \ne 0$.
3. On the other hand, even-dimensional, compact Lie groups have left-invariant complex structures. If $M$ is such a manifold and is simply connected, then it is also 2-connected. $H^1(M,\mathbb{Z}) = H^2(M,\mathbb{Z}) = 0$ and $M$ is manifestly not Kähler. On the other hand, no such example is 3-connected and you always have $H^3(M,\mathbb{R}) \ne 0$.
4. There is (or was) a long-standing conjecture that no even-dimensional sphere other than $S^2$ has a complex structure.
So, question: Is there for each $n$, a compact, complex manifold $M$ which is $n$-connected?
| https://mathoverflow.net/users/1450 | Highly connected, compact complex manifolds | E. Calabi, B. Eckmann, *A class of compact, complex manifolds which are not algebraic.*
Ann. of Math. (2) 58, (1953). 494–500.
From Chern's MR review (MR0057539):
>
> This paper defines on the topological product $S^{2p+1} \times S^{2q+1}$ of two spheres of dimensions $2p+1$ and $2q+1$ respectively, $p$ > 0, a complex analytic structure. The complex manifold so obtained ... admits a complex analytic fibering, with two-dimensional tori as fibers and having as base space the product $\mathbb{P}^p \times \mathbb{P}^q$ of complex projective spaces of (complex) dimensions $p$ and $q$ respectively.
>
| 20 | https://mathoverflow.net/users/2356 | 28067 | 18,344 |
https://mathoverflow.net/questions/28047 | 15 | Some time ago I asked a question on consecutive numbers represented integrally by an integral positive binary quadratic form. It has occurred to me that, instead, the Green-Tao theorem may include a result on arithmetic progressions represented by a positive binary form. So my question is whether that is the case, do we already know that a positive binary form represents arbitrarily long arithmetic progressions? These would be primes in this general setting, thus quite different from consecutive integers of course.
My main reference is
David A. Cox,
Primes of the form $x^2 + n y^2.$ He defines the Dirichlet density on page 169. Then he states the
Chebotarev Density Theorem (8.17) on page 170. Finally he gives the Dirichlet density of primes represented by a positive binary form on page 188, Theorem 9.12. EDIT::: Not difficult to state: with discriminant $ \Delta < 0$ and class number $ h(\Delta),$ if the form is ambiguous (such as the principal form) the Dirichlet density of the set of primes it represents is $$ \frac{1}{2 h(\Delta)},$$
while if the form is not ambiguous the Dirichlet density is
$$ \frac{1}{h(\Delta)}.$$ On page 190 he does the example $ \Delta = -56.$ Here $x^2 + 14 y^2$ represents a set of primes with Dirichlet density $1/8,$ while $2 x^2 + 7 y^2$ also gets density $1/8,$ but in the other genus $ 3 x^2 + 2 x y + 5 y^2 $ and $ 3 x^2 - 2 x y + 5 y^2 $ each represent the same set of primes with density $1/4.$ Note on page 195 we have Exercise 9.17, that the sum of these densities for any discriminant must be $1/2.$ A little fiddling, not mentioned in the book, shows that each genus (of a fixed discriminant $\Delta$) represents the same total density, something we really want because of the relationship between genera and arithmetic progressions of primes.
From an earlier answer by David Hansen it would appear that the only missing ingredient is a comparison between Dirichlet density and "relative density." I have never been sure on this point, does a positive binary form represent the same relative density of primes as its Dirichlet density? Anyway, see:
[Is the Green-Tao theorem true for primes within a given arithmetic progression?](https://mathoverflow.net/questions/25402/is-the-green-tao-theorem-true-for-primes-within-a-given-arithmetic-progression/25403#25403)
My earlier question, about which I should say that I have come to believe there is no upper bound on the length of intervals represented, despite the great difficulty finding examples:
[Can a positive binary quadratic form represent 14 consecutive numbers?](https://mathoverflow.net/questions/23943/can-a-positive-binary-quadratic-form-represent-14-consecutive-numbers)
On the Green-Tao theorem itself:
<http://arxiv.org/abs/math.NT/0404188>
<http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem>
same-day EDIT: I looked up Dirichlet density on wikipedia. As relates to the earlier David Hansen answer I liked, wikipedia stops short of saying that the Dirichlet density of primes in an arithmetic progression is identical to the relative density. Personally, I cannot see how the relative density could be anything else, but that is just my opinion. So I think I am also asking for references that prove the relative density is equal to the Dirichlet density in some naturally-occurring situations.
<http://en.wikipedia.org/wiki/Dirichlet_density>
Later on the same day EDIT: There is stronger language in this next wikipedia page, so I think we can conclude that for primes in an arithmetic progression the Dirichlet density and the relative density are equal, but I would still like a more substantial reference. I'm the nervous type. I worry.
<http://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem>
| https://mathoverflow.net/users/3324 | The Green-Tao theorem and positive binary quadratic forms | Edit: {The answer to your question, "...do we already know that a positive binary form represents arbitrarily long arithmetic progressions?" is yes. See the second paragraph below.}
If the relative density exists, so does the Dirichlet density and they are equal. The converse is not true in general. For primes in a given arithmetic progression, both densities exist. See Lang's Algebraic Number Theory, Ch. VIII.4 and XV. Given those facts, one approach to the problem would be trying to show that the relative density of the set of primes represented by a positive binary quadratic form actually exists (I have no idea how hard this might be).
On the other hand, if you only want to know about a.p.'s of primes represented by a positive quadratic form, a better approach might be answering the question, "Does Green-Tao still hold for sets of primes with positive Dirichlet density?" The answer is yes since G-T only requires that the limsup of the relative density be positive, and positive Dirichlet density implies positive limsup (if the limsup were 0, the lim would be 0).
| 15 | https://mathoverflow.net/users/5513 | 28069 | 18,346 |
https://mathoverflow.net/questions/28054 | 22 | The group $Diff(S^n)$ ($C^\infty$-smooth diffeomorphisms of the $n$-sphere) has many interesting subgroups. But one question I've never seen explored is what are its "big" finite-dimensional subgroups?
For example, $Diff(S^n)$ contains a finite-dimensional Lie subgroup of dimension $n+2 \choose 2$, the subgroup of conformal automorphisms of $S^n$. Similarly it contains a compact Lie subgroup of dimension $n+1 \choose 2$, the isometry group of $S^n$.
Is it known that:
1) A finite-dimensional Lie subgroup of $Diff(S^n)$ having dimension at least $n+2 \choose 2$ is conjugate to a subgroup of the conformal automorphism group of $S^n$ ? (Answer, no, see Algori's answer below). Modified question: As Algori notes, $GL\_{n+1}(\mathbb R) / \mathbb R\_{>0}$ acts on $S^n$ and has dimension $n^2+2n$. So is a finite-dimensional Lie subgroup of $Diff(S^n)$ of dimension $n^2+2n$ (or larger) conjugate to a subgroup of this group?
2) A compact Lie subgroup of $Diff(S^n)$ having dimension at least $n+1 \choose 2$ is conjugate to a subgroup of the isometry group of $S^n$ ? (Answer: Yes, see Torsten Ekedahl's post below)
For example, arbitrary compact subgroups of $Diff(S^n)$ do not have to be conjugate to subgroups of the above two groups -- perhaps the earliest examples of these came from exotic projective and lens spaces. But I have little sense for how high-dimensional these "exotic" subgroups of $Diff(S^n)$ can be.
| https://mathoverflow.net/users/1465 | "Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known? | You can make big Lie groups act effectively on small manifolds by cheating: make the group a product of groups, with each factor acting by compactly supported diffeomorphisms on a different disjoint open subset. So the additive group $\mathbb R^N$ becomes a subgroup of $Diff(S^1)$ by flowing along $N$ commuting vector fields supported in $N$ disjoint arcs.
(added later) A similar cheat: let $a\_1,\dots ,a\_N$ be linearly independent functions of one variable. Then $a\_1(x)\frac{\partial}{\partial y},\dots ,a\_N(x)\frac{\partial}{\partial y}$ are independent commuting vector fields in the $x,y$ plane. Modify this example to make it compactly supported if you like.
(added still later) My proposed extension of the first cheat to a semisimple group (comment thread of Torsten's answer) is doomed: Choose a point on the circle and choose an element of SL\_2(R) that fixes this point and acts on the tangent space there with eigenvalue c>1. Lifting the group element to the universal covering group in the right way, you get an element g of the latter group that fixes all the points above the given point in the universal covering space of the circle, in each case with eigenvalue c. But now if this line with this action could be embedded in a longer line with trivial action outside then there would be a sequence of fixed points of g with eigenvalue c converging to a fixed point of g with eigenvalue 1, contradiction.
| 33 | https://mathoverflow.net/users/6666 | 28081 | 18,354 |
https://mathoverflow.net/questions/28088 | 21 | As usual I expect to be critisised for "duplicating"
[this question](https://mathoverflow.net/questions/27931/). But I do not! As Gjergji immediately
notified, that question was from numerology. The one I ask you here
(after putting it in my [response](https://mathoverflow.net/questions/27931/why-is-163-ln163-a-near-integer/28002#28002)) is a mathematics
question motivated by Kevin's (O'Bryant) comment to the earlier post.
**Problem.**
For any $\epsilon>0$, there exists an $n$ such that
$\|n/\log(n)\|<\epsilon$ where $\|\ \cdot\ \|$ denotes the
distance to the nearest integer.
In spite of the simple formulation, it is likely that
the diophantine problem is open. I wonder whether it follows
from some known conjectures (for example, Schanuel's conjecture).
| https://mathoverflow.net/users/4953 | When is $n/\ln(n)$ close to an integer? | If $f(x)=\frac{x}{\log x}$, then $f'(x)=\frac{1}{\log x} - \frac{1}{(\log x)^2}$, which tends to zero as $x\rightarrow \infty$. Choose some large real number $x$ for which $f(x)$ is integral. Then the value of $f$ on any integer near $x$ must be very close to integral.
| 42 | https://mathoverflow.net/users/5513 | 28090 | 18,359 |
https://mathoverflow.net/questions/28103 | 2 | I have recursive polynomials
$$Q\_{n}(t)=tQ\_{n-1}(t)+\frac{1-t^{2}}{n+1}Q'\_{n-1}(t)$$
and
$$Q\_{0}(t)=1$$
Is there a theory for finding a factorisation of recursive polynomials?
It is possible to show that
$$\sum\_{s=-\infty}^{\infty}sinc(\pi(x+s))^{p+1}=Q\_{p-1}[cos(\pi x)]$$
where $$sinc(x)=sin(x)/x$$
Maybe you have some ideas for my case?
| https://mathoverflow.net/users/3589 | roots of recursive polynomials | I have realised that your recurrent relation is exactly the one which appears in Eq. (2.4) in [[*Izvestiya: Mathematics* **66**:3 (2002) 489--542]](http://dx.doi.org/10.1070/IM2002v066n03ABEH000387) (see
also [here](http://wain.mi.ras.ru/PS/zete_main.pdf)). The properties of the corresponding polynomials are expressed in Lemma 2.2 and around (they seem to be exactly the ones you are trying to establish).
| 3 | https://mathoverflow.net/users/4953 | 28105 | 18,365 |
https://mathoverflow.net/questions/28093 | 12 | Let $\delta$ denote a non-zero complex algebraic differential operator in a single variable x. That is, it can be written as a sum
$$ \delta = \sum\_i f\_i\partial\_x^i$$
where there $f\_i$ are complex polynomials in x.
Let $R=\mathbb{C}[x]$, and consider the image of $\delta$ as a map on R. As a subspace of $R$, does $Im(\delta)$ always contain an non-trivial ideal?
It does in every case I can think of where there is some trick I can use to understand the image better:
* When $\delta$ is a function.
* When $\delta$ is a constant coefficient differential operator.
* When $\delta$ has order 1.
* When $\delta$ is homogeneous for the Euler grading; that is, it takes monomials to monomials.
It seems like it should be related to the simpler fact that $\delta$ is zero if $\delta$ kills functions of unboundedly high degree, which can be shown from the Formal Continuity of differential operators.
**Remark.** For more than one variable, the above question is false. If $\delta=x\partial\_x-y\partial\_y$, then $\delta$ is homogeneous for the Euler bigrading (it takes monomials to monomials), but it kills all monomials of the form $x^iy^i$. Since any monomial ideal in $\mathbb{C}[x,y]$ must contain some monomial of this form, the image of this $\delta$ contains no ideal.
| https://mathoverflow.net/users/750 | Does the image of a differential operator always contain an ideal? | No. Let $\delta=x-\partial$ and $L=Im(\delta).$ I claim that $L$ does not contain any non-zero ideal of $\mathbb{C}[x].$ Indeed, $x^k\equiv (k-1)x^{k-2}\ (\mod L)$ and, by induction,
$$x^{2n+1}\equiv (2n)!!x\equiv 0(\mod L),\ x^{2n}\equiv (2n-1)!!\ (\mod L).$$
Thus $L$ contains all odd powers of $x$ and has codimension 1 in $R.$
Suppose that $L$ contains a principal ideal $(f)$. Let $f=f\_0+f\_1$ be the decomposition of $f$ into the even and odd parts ($f\_0$ is the span of the even degree monomials of $f$). Then $x^{2N}f\_0\in L$ and $x^{2N+1}f\_1\in L$ for any $N\geq 0.$
At least one of $f\_0$ and $xf\_1$ is non-zero and has the form $g=\sum\_n a\_{n}x^{2n}.$
Then for any $N\geq 0,$
$$x^{2N}g=\sum\_n a\_nx^{2(n+N)}\equiv \sum\_n (2n+2N)!!a\_n\ (\mod L)\quad
\text{ and }\quad \sum\_n (2n+2N)!!a\_n=0.$$
However, this is impossible: for sufficiently large $N,$ the term involving $a\_n\ne 0$ with the largest $n$ clearly dominates the rest of the sum.
| 14 | https://mathoverflow.net/users/5740 | 28107 | 18,367 |
https://mathoverflow.net/questions/1388 | 39 | Given a set Ω and a σ-algebra F of subsets, is there some natural way to assign something like a "uniform" measure on the space of all measurable functions on this space? (I suppose first it would be useful to know if there's even a natural σ-algebra to use on this space.)
The reason I'm asking is because I'd like to do the following. Let Ω be the (2-dimensional) surface of a sphere, with the uniform probability distribution. Let F be the Borel σ-algebra, and let G be the sub-algebra consisting of all measurable sets composed of lines of longitude. (That is, S is in G iff S is measurable and for all x in S, S contains all points with the same longitude as x.) Let A be the set of all points with latitude 60 degrees north or higher (a disc around the north pole).
Let f be a G-measurable function defined on Ω such that the integral of f over any G-measurable set B equals the measure of (A\cap B). (This is a standard tool in defining the conditional probability of A given G-measurable sets.) It's not hard to show that for any such function f, for almost-all x, f(x) will equal the unconditional measure of A.
What I'd like to be able to say is that for any x, for almost-all such functions f, f(x) will equal the unconditional measure of A. However, I can't say "almost-all" on the functions unless I have some measure on the space of functions.
Clearly I can do this by concentrating all the measure on the single constant function in this set. But I'd like to be able to pick out this most "generic" such function even in cases where A isn't so nice and symmetric.
Maybe there's some other, simpler question I should be asking first?
| https://mathoverflow.net/users/445 | Is there a natural measures on the space of measurable functions? | Let I be the unit interval with the Borel $\sigma$-algebra. There is no $\sigma$-algebra on the set of measurable functions from I to I such that the evaluation functional $e:I^I\times I\to I$ given by $e(f,x)=f(x)$ is measurable, as shown by Robert Aumann [here](http://www.ma.huji.ac.il/raumann/pdf/66.pdf), so even finding useful $\sigma$-algebras is a problem.
However, t is possible to talk about "almost all" functions in a function space even when it is not possible to have an appropriate measure. The trick is to find a characterization of a set having full (or zero) measure that can be applied to function spaces. There is a generalization of Lebesgue measure zero, independently found by various authors and known as *Haar measure zero* or *shyness* that should be applicable to your problem. A nice survey of the theory and some of its extensions can be found [here](http://www.ams.org/journals/bull/2005-42-03/S0273-0979-05-01060-8/).
| 42 | https://mathoverflow.net/users/35357 | 28114 | 18,370 |
https://mathoverflow.net/questions/28113 | 2 | **Q** is the rational number field.
p is a prime number.
q is a prime number other than p.
$k\_{p^r}$ is a cyclotomic field.
$k\_{p^r}$=**Q**(x) where x is exp(2$\pi$i/$p^r$).
[$k\_{p^r}$:**Q**]=$p^{r-1}(p-1)$.
Question: Does q remain a prime in the integer ring of $k\_{p^r}$?
| https://mathoverflow.net/users/2666 | Cyclotomic Fields over Q and prime ideals | Theorem I.2.13 of Washington's book on cyclotomic fields says the following: $K$ is the $n$th cyclotomic field and $p\nmid n$, let $f$ be the smallest positive integer such that $p^f\equiv 1 (\mathrm{mod}~n)$. Then $p$ splits into $\phi(n)/f$ distinct primes in $K$.
| 10 | https://mathoverflow.net/users/1021 | 28121 | 18,375 |
https://mathoverflow.net/questions/28010 | 1 | Hi
Just a short question. How are the IQR of the boxplot related to the confidence interval of a sample? Is the IQR actually the 50% confidence interval?
| https://mathoverflow.net/users/5357 | Boxplot IQR and confidence interval | My answer didn't seem to score any points with anyone, and rdchat's answer is lousy, so let's look more closely.
Suppose $X\_1,\dots,X\_n$ are an i.i.d. sample from a normally distributed population with unknown mean $\mu$ and unknown variance $\sigma^2$, and we seek a confidence interval for the population mean. As usual, let $\overline{X} = \left(X\_1 + \cdots + X\_n\right)/n$ be the sample mean and $S^2 = 1/(n-1)\sum\_{i=1}^n (X\_i - \overline{X})^2$ be the (unbiased) sample variance. (BTW, unbiasedness is overrated. Everybody knows that, except some non-statisticians. Apparently there are lots of those.)
Now
$$
\frac{\overline{X} - \mu}{\sigma/\sqrt{n}}
$$
is normally distributed with mean 0 and variance 1, and
$$
\frac{\overline{X} - \mu}{S/\sqrt{n}}
$$
has a Student's t-distribution with $n-1$ degrees of freedom (called "Student's" because it's named after William Sealey Gosset, of course---and some of the content of this forum makes one suspect that a lot of people here don't know that standard bit of folklore). Go to the table and look up the value of $A\_n$ for which
$$
\Pr\left( -A\_n < \frac{\overline{X} - \mu}{S/\sqrt{n}} < A\_n \right) = \frac{1}{2},
$$
or in other words
$$
\Pr\left(\overline{X} - A\_n \frac{S}{\sqrt{n}} < \mu < \overline{X} + A\_n \frac{S}{\sqrt{n}}\right) = \frac{1}{2}.
$$
Then
$$
\overline{X} \pm A\_n \frac{S}{\sqrt{n}}
$$
are the endpoints of a 50% confidence interval for $\mu$.
Important point: The length of the confidence interval goes to 0 as the sample size increases, since $\sqrt{n}$ is in the denominator (and $A\_n$ approaches the value one would get for the normal distribution rather than for Student's distribution). But the sample quartiles do not get closer together in the limit as $n$ grows, since ("almost surely") they approach the population quartiles.
So the answer is NO.
| 1 | https://mathoverflow.net/users/6316 | 28126 | 18,376 |
https://mathoverflow.net/questions/28119 | 11 | Hello,
Joel Dogde, in a comment on his question "Roots of unity in different completions of a number field", says the following, about the analogy between number fields and function fields :
*Number of roots of unity in number fields is something like the size of the constant field for function fields.*
Could anyone explain that ?
Thanks.
| https://mathoverflow.net/users/2330 | Why the roots of unity are the analogs of constants ? | One answer: the roots of unity in $K$ are [precisely the elements](https://mathoverflow.net/questions/10911/english-reference-for-a-result-of-kronecker) of $K$ which have absolute value $1$ for every absolute value on $K$; the elements of the constant field have this property for function fields.
| 13 | https://mathoverflow.net/users/297 | 28129 | 18,378 |
https://mathoverflow.net/questions/26040 | 23 | Given a positive integer $a$, the *Ramsey number* $R(a)$ is the least $n$ such that whenever the edges of the complete graph $K\_n$ are colored using only two colors, we necessarily have a copy of $K\_a$ with all its edges of the same color.
For example, $R(3)= 6$, which is usually stated by saying that in a party of 6 people, necessarily there are 3 that know each other, or 3 that don't know one another; but there is a party of 5 people without this property. This is probably known to everybody.
Slightly less known is the fact that any such coloring of $K\_6$ in fact contains **2** monochromatic triangles.
The *Ramsey multiplicity* $m(a)$ (there does not seem to be a standard notation) is the largest number $m$ of monochromatic copies of $K\_a$ that we can guarantee in any 2-coloring of $K\_{R(a)}$. For example, $m(3)=2$, and Piwakowski and Radziszowski showed around 1999 that $m(4)=9$.
I have a couple of questions (please forgive me if they are trivial, I'm just beginning to form my intuitions in this field) :
1. Is it known that $m(n)$ is monotonically increasing?
2. Do we know anything about the rate of growth of the function $m(n)$?
I suspect that the answer to both questions is yes and that reasonable bounds for $m(n)$ are known, but haven't been able to locate any references. The best I know is that $$ m(n)\le \frac{\binom{r(n)}n}{2^{\binom n2-1}}, $$
(proved by Burr and Rosta in 1980), which is probably too high, and a recent result of Conlon suggests that
$$ m(n)\ge C\frac{\binom{r(n)}n}{2^{n(3n-1)/2}} $$
for some appropriate $C$. I say "suggests" because Conlon's results carry some additional implicit constants that I haven't checked can be absorbed this way. (Please let me know if I am completely off the mark here.) [**Edit:** Unfortunately, Conlon's bounds (in his paper "On the Ramsey Multiplicity of Complete Graphs") do not apply here. No lower bound beyond $m(n)\ge1$ seems known.]
---
"Update": William Gasarch's Open Problems column in the June 2020 issue of the *ACM SIGACT News* is devoted to [Ramsey multiplicity](https://doi.org/10.1145/3406678.3406685).
| https://mathoverflow.net/users/6085 | Ramsey multiplicity | I emailed David Conlon about this question. He agreed to let me share his answer. In short, the problem very much seems to be open (I've added the relevant tag). As Thomas mentions, the upper bound I cite is straightforward. *And nothing better is known!*
If one looks for papers on Ramsey multiplicity, a few come up, but they deal with a different concept, that I explain below. The quotes are from Conlon's emails.
>
> Unfortunately, the concept you're talking about is also known as the Ramsey
> multiplicity! There are very few references as far as I know. The only one I
> can think of offhand is the Piwakowski and Radziszowski paper which you
> quoted. Perhaps there's something in the references to that paper, but I
> doubt it somehow.
>
>
>
Indeed, in the papers I have seen (included P-R, where $m(4)=9$ is proved), there are no arguments about $m(n)$ for general $n$ (or even $n=5$).
>
> The function you're interested in is rather amorphous, I'm afraid. My result
> will imply that if $n \ge 4^t$ you must have at least $n^t/2^{3t^2/2}$ copies of
> $K\_t$ or thereabouts. But when your number is below $4^t$ it implies nothing.
>
>
> In general, because we don't understand the Ramsey function, I find it hard
> to imagine how we might be able to say anything at all about $m(n)$. Unless
> there's an elementary argument which gives something. It reminds me of
> estimating the difference between successive Ramsey numbers like $r(n,n)$ and
> $r(n,n+1)$, where, though the difference is almost certainly exponential, the
> largest difference that can be guaranteed is tiny (I think linear or
> quadratic even, though I can't remember exactly).
>
>
>
Here, $r(m,n)$ are the usual Ramsey numbers (what I called $R(n)$ in the question, is $r(n,n)$ in this notation). In general, $r(m,n)$ is the smallest $k$ such that any coloring of the edges of $K\_k$ with blue and red either contains a blue copy of $K\_m$ or a red copy of $K\_n$.
>
> The only thing that appears clear to me is an upper bound following from the
> probabilistic method, namely $\displaystyle \frac{\binom{r(n)}{n}}{2^{\binom n2}}$. It's not even
> obvious how one would approach showing that the multiplicity is at least 2!
>
>
>
Finally, as to the question of how to call this concept:
>
> I'd suggest that this be called the **critical multiplicity** or something like
> that, just to distinguish it from the usual multiplicity function.
>
>
>
The usual Ramsey multiplicity is defined as follows. It is significantly better understood than $m(n)$.
Let $k\_t(n)$ be the minimum number of monochromatic copies of $K\_t$ within a two-coloring of the
edges of $K\_n$, and let $$ c\_t(n)= \frac{k\_t(n)}{\binom nt} $$ be the minimum proportion of monochromatic copies of $K\_t$ in such a two-coloring.
It is known that the numbers $c\_t(n)$ increase with $n$. The *Ramsey multiplicity* of $t$ (or of $K\_t$) is $\displaystyle c\_t\lim\_{n\to\infty} c\_t(n)$.
(Relevant references can be found in Conlon's paper mentioned in the question.)
| 12 | https://mathoverflow.net/users/6085 | 28132 | 18,381 |
https://mathoverflow.net/questions/28104 | 16 | It occured to me that the Sieve of Eratosthenes eventually generates the same prime numbers, independently of the ones chosen at the beginning. (We generally start with 2, but we could chose any finite set of integers >= 2, and it would still end up generating the same primes, after a "recalibrating" phase).
If I take 3 and 4 as my first primes, starting at 5:
* 5 is prime,
* 6 is not (twice 3),
* 7 is prime,
* 8 is not,
* 9 is not,
* 10 is not,
* etc.
Eventually, I will find all the primes as if I had started with 2 only.
To me, it means that we can generate big prime numbers without any knowledge of their predecessors (until a certain point). If I want primes higher than 1000000, then I can generate them without any knowledge of primes under, say, 1000. (It may not be as effective computationnally, but I find this philosophically interesting.)
Is this result already known ?
Are there any known implications ?
**Edit**
The number from which we are assured to get the right primes again is after the square of the last missing natural prime.
That is, if I start with a seed set containing only the number 12, the highest missing natural prime is 11, so I'll end up having 121 as my last not-natural prime.
Non-natural primes found are 12,14,15,16,18,20,21,22,25,27,33,35,49,55,77 and 121.
This is a bit better than what I thought at first (namely, as stated below, somewhere under the square of the highest seed).
| https://mathoverflow.net/users/2446 | Sieve of Eratosthenes - eventual independence from initial values | So, let's see if I can precisify your claim. We start with a finite "seed set" that we assert to be Ghi-Om-prime (the seed set must not contain 1). Numbers smaller than the largest seed we completely ignore. Now for numbers larger than any seed prime, we run the Seive. You claim that there is some cut-off, depending of course on the seed set but hopefully not growing too quickly, so that after that cut-off, the primes match the primes in the seed set.
Recall how the Seive works. A number is GO-prime iff it is not divisible by any smaller GO-prime. Conversely, the GO-composites are precisely those numbers that are divisible by some smaller number that is either in your seed set or larger than the largest number in the seed set. In particular, every GO-composite is a true composite. So suppose there is a true composite that is not GO-composite. Then all its proper divisors are less than the largest seed. But every non-prime has a proper divisor that is at least as large as its square root. So the largest true composite that is not GO-composite is strictly less than the square of the largest seed.
This proves your claim, with the bound you suggested in the comment to Noldorin.
In any case, some remarks. First, if we state the rules for the Sieve correctly, and in particular disallow 1 as either prime or composite, then the true primes are precisely the GO primes where the seed set is empty.
Second, although I haven't seen this particular variation of the Sieve before, as I said in the comments to the question, you should check out [Sanjoy Mahajan's thesis (PDF)](http://www.inference.phy.cam.ac.uk/sanjoy/thesis/thesis-letter.pdf). In the last chapter, he proposes the following probabilistic variation of the Sieve. Namely, let's define the probability that $p$ SM-divides $n$ to be $1/p$. Then the probability that $n$ is SM-prime is
$$ \prod\_{p < \sqrt n} (\text{probability that $p$ is SM-prime})\times (\text{probability that $p$ does not SM-divide $n$})$$
The point is that SM-primality is nicely smooth, rather than exhibiting the strange jumps in the true prime spectrum, but still approximates the spectrum well for certain types of estimates. Sanjoy also considers a few other probabilistic models (whether to stop the product at $\sqrt n$ or $n$, for example), including some where he seeds the model with some early assertions of SM-primality. Sanjoy is a physicist, and so does not work at mathematicians' level of rigor, and says as much: the point of his model is to lead to new conjectures about distribution of true primes.
| 15 | https://mathoverflow.net/users/78 | 28142 | 18,389 |
https://mathoverflow.net/questions/28135 | 4 | I have a {0,1}, invertible, triangular matrix, that I would like to show is totally unimodular. Are there any known results on the total unimodularity of classes of triangular matrices?
| https://mathoverflow.net/users/19029 | When is a triangular matrix totally unimodular? | [Seymour's decomposition theorem](http://en.wikipedia.org/wiki/Matroid#Regular_matroids) for regular matroids yields a polynomial-time algorithm for testing if any {0,1,-1} matrix is totally unimodular. Unfortunately, due to the sound of paint drying on this [question](https://mathoverflow.net/questions/27346/has-anyone-implemented-a-recognition-algorithm-for-totally-unimodular-matrices) of Gordon Royle, it seems as if no one has yet implemented it.
| 5 | https://mathoverflow.net/users/2233 | 28144 | 18,390 |
https://mathoverflow.net/questions/28092 | 5 | It's known that finding the intersection of n halfplanes in 2-d takes $\Omega(n\log n)$ time. Does the lower bound apply if we change the question to *deciding* whether the intersection is non-empty?
| https://mathoverflow.net/users/6645 | Feasibility of linear programs | It seems this can be done in linear time. Algorithms that solve linear programs are also capable of deciding whether the LP is feasible or not and 2-d linear programs can be solved in linear time (linear in terms of the number of constraints). So to decide whether a set of n halfplanes is non-empty or not, just solve the LP that has those halfplanes as its constraints with any objective function.
| 2 | https://mathoverflow.net/users/6645 | 28146 | 18,392 |
https://mathoverflow.net/questions/28147 | 150 | I was helping a student study for a functional analysis exam and the question came up as to when, in practice, one needs to consider the Banach space $L^p$ for some value of $p$ other than the obvious ones of $p=1$, $p=2$, and $p=\infty$. I don't know much analysis and the best thing I could think of was Littlewood's 4/3 inequality. In its most elementary form, this inequality states that if $A = (a\_{ij})$ is an $m\times n$ matrix with real entries, and we define the norm
$$\|A\| = \sup\biggl(\left|\sum\_{i=1}^m \sum\_{j=1}^n a\_{ij}s\_it\_j\right| : |s\_i| \le 1, |t\_j| \le 1\biggr)$$
then
$$\biggl(\sum\_{i,j} |a\_{ij}|^{4/3}\biggr)^{3/4} \le \sqrt{2} \|A\|.$$
Are there more convincing examples of the importance of "exotic" values of $p$? I remember wondering about this as an undergraduate but never pursued it. As I think about it now, it does seem a bit odd from a pedagogical point of view that none of the textbooks I've seen give any applications involving specific values of $p$. I didn't run into Littlewood's 4/3 inequality until later in life.
[Edit: Thanks for the many responses, which exceeded my expectations! Perhaps I should have anticipated that this question would generate a big list; at any rate, I have added the big-list tag. My choice of which answer to accept was necessarily somewhat arbitrary; all the top responses are excellent.]
| https://mathoverflow.net/users/3106 | Why do we care about $L^p$ spaces besides $p = 1$, $p = 2$, and $p = \infty$? | Huge chunks of the theory of nonlinear PDEs rely critically on analysis in $L^p$-spaces.
* Let's take the 3D Navier-Stokes equations for example. Leray proved in 1933 existence of *a weak* solution to the corresponding Cauchy problem with initial data from the space $L^2(\mathbb R^3)$. Unfortunately, it is still a major open problem whether the Leray weak solution is unique. But if one chooses the initial data from $L^3(\mathbb R^3)$, then Kato showed that there is *a unique strong* solution to the Navier-Stokes equations (which is known to exist locally in time). $L^3$ is the "weakest" $L^p$-space of initial data which is known to give rise to unique solutions of the 3D Navier-Stokes.
* In some cases the structure of the equations suggests the choice of $L^p$ as the most natural space to work in. For instance, many equations stemming from non-Newtonian fluid dynamics and image processing involve the $p$-Laplacian
$\nabla\left(|\nabla u|^{p-2}\nabla u\right)$ with $1 < p < \infty.$ Here the $L^p$-space
and $L^p$-based Sobolev spaces provide a natural framework to study well-posedness and regularity issues.
---
* Yet another example from harmonic analysis (which goes back to Paley and Zigmund, I think). Let
$$F(x,\omega)=\sum\limits\_{n\in\mathbb Z^d} g\_n(\omega)c\_ne^{inx},\quad x\in \mathbb T^d,$$
where $g\_n$ is a sequence of independent normalized Gaussians and $(c\_n)$ is a non-random element of $l^2(\mathbb Z^d)$. Then the function $F$ *belongs almost surely to any* $L^p(\mathbb T^d)$, $2\leq p <\infty$ and it *does not* belong almost surely to $L^{\infty}(\mathbb T^d)$.
There have been very recent applications of this resut to the existence of solutions to the nonlinear Schrodinger equations with random initial data (due to Burq, Gérard, Tzvetkov et al).
| 69 | https://mathoverflow.net/users/5371 | 28150 | 18,395 |
https://mathoverflow.net/questions/28152 | 4 | Is there a notion of fibered category with box products?
By this I roughly mean a fibration $C\rightarrow B$ where $B$ has finite products,
along with functors
$$\boxtimes: C(X)\times C(Y)\rightarrow C(X\times Y)$$
and some coherent isomorphisms, for example:
$$(f\times g)^\* (M\boxtimes N) \leftrightarrow (f^\* M) \boxtimes (g^\* N)$$
and
$$(M\boxtimes N)\boxtimes L \leftrightarrow M\boxtimes (N \boxtimes L) $$
This situation occurs often in geometry, for example:
B=Varieties, C=quasi coherent sheaves, D-modules
| https://mathoverflow.net/users/2837 | Is there a notion of "fibered category with boxproducts"? | Yes, there is. In [this paper](http://www.tac.mta.ca/tac/volumes/20/18/20-18abs.html) I called it a *monoidal fibration* (with cartesian monoidal base), but I'm sure that other people had thought about it before. There are some nice things you can say especially in the case when the base is cartesian; for instance you automatically get a monoidal structure on each fiber defined by $M\otimes N = \Delta^\*(M\boxtimes N)$, and you can recover the box-product from the fiberwise monoidal structure via $M\boxtimes N = \mathrm{pr}\_1^\* M \otimes \mathrm{pr}\_2^\* N$.
| 4 | https://mathoverflow.net/users/49 | 28163 | 18,401 |
https://mathoverflow.net/questions/28168 | 0 | Let δ is a proximity.
I will call a set A connected iff for every partition {X,Y} of the set A holds X δ Y.
Question: Let A and B are sets with non-empty intersection. Let both A and B are connected. Prove or give a counter-example that A∪B is also connected.
(This question arouse as a special example of a more general theorem. I spend may be half of hour attempting to prove it and after these my efforts failed, I desire to share this question.)
| https://mathoverflow.net/users/4086 | Connectedness of a union regading a proximity | Consider $X\cap A$ and $Y\cap A$, starting from a partition $\lbrace X,Y\rbrace$ of $A\cup B$. If both intersections are nonempty we are done, as $(X\cap A)\delta(Y\cap A)$. Otherwise, $A\subseteq X$, say, but then $X\cap B$ and $Y\cap B$ are nonempty and we find $(X\cap B)\delta(Y\cap B)$.
In either case $X\delta Y$.
| 3 | https://mathoverflow.net/users/5903 | 28173 | 18,410 |
https://mathoverflow.net/questions/28186 | 4 | Can we find a sequence $u\_n$ of positive real numbers such that
$\sum\_{n=1}^\infty u\_n^2$ is finite, yet $\sum\_{n=1}^\infty ({u\_1+u\_2+...+u\_n\over n})^2$ is infinite ?
After several attempts, I think this is not possible, but I can't prove that the finiteness of the first sum implies the finiteness of the second sum.
| https://mathoverflow.net/users/6129 | Convergence of the sum of squares of averages of a sequence whose sum of squares is convergent | Hardy's inequality says
$$\sum\_{n=1}^{\infty}\left(\frac{a\_1+\cdots+a\_n}{n}\right)^p\le \left(\frac{p}{p-1}\right)^{p}\sum\_{n=1}^{\infty}a\_n^p$$
for any $p>1$.
| 10 | https://mathoverflow.net/users/2384 | 28187 | 18,422 |
https://mathoverflow.net/questions/28162 | 2 | When some papers say"XXX fibration", I see it seems that it is just that the surjective map f: X ---> Y, such that the fiber is XXX,but it is really not "fibration",I didn't see it prove that it is a fibration. So could you tell me if I am wrong? Thanks!
| https://mathoverflow.net/users/2391 | Is an algebraic geometer's fibration also an algebraic topologist's fibration? | Now that the intent of the question has become clear, I'll attempt to take it out of limbo by transferring the content of the comments - my own (TP) and Boyarsky's - into a community wiki answer.
In algebraic or complex analytic geometry, a fibration is a map from a variety to a lower-dimensional variety having some reasonable properties (proper, surjective, flat). I'm not sure if there's a generally accepted, precise definition in this generality, but for instance, a *Lefschetz fibration* on a connected complex manifold $M$ is a proper, surjective holomorphic map $M\to C$ to a Riemann surface, with non-degenerate critical points and distinct critical values.
This usage is *in*consistent with that of algebraic topologists.
A Lefschetz fibration, unless it happens to be a submersion (hence a smooth fibre bundle, by Ehresmann's theorem), is not a fibration in the senses of algebraic topology (Serre or Hurewicz). A singular fibre has the homotopy-type of a regular fibre with a middle-dimensional cell attached. So, the topological Euler characteristics of regular and singular fibres differ by 1.
| 5 | https://mathoverflow.net/users/2356 | 28190 | 18,424 |
https://mathoverflow.net/questions/14212 | 35 | Here's a a famous problem:
If a rectangle $R$ is tiled by rectangles $T\_i$ each of which has at least one integer side length, then the tiled rectangle $R$ has at least one integer side length.
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------
There are a number of proofs of this result (14 proofs in this particular [paper](http://www.jstor.org/pss/2322213) by Stan Wagon). One would think this problem is a tedious exercise in combinatorics, but the broad range of solutions which do not rely on combinatorial methods makes me wonder what deeper principles are at work here. In particular, my question is about the proof using double integrals which I sketch out below:
Suppose the given rectangle $R$ has dimensions $a\times b$ and without loss of generality suppose $R$ has a corner at coordinate $(0,0)$. Notice that $\int\_m^n\sin(2\pi x)dx=0$ iff $m\pm n$ is an integer. Thus, for any tile rectangle $T\_i$, we have that:
$$\int\int\_{T\_i}\sin(2\pi x)\sin(2\pi y)dA=0$$
If we sum over all tile rectangles $T\_i$, we get that the area integral over $R$ is also zero:
$$\int\int\_{R}\sin(2\pi x)\sin(2\pi y)dA=\sum\_i\int\int\_{T\_i}\sin(2\pi x)\sin(2\pi y)dA=0$$
Since the corner of the rectangle is at $(0,0)$, it follows that either $a$ or $b$ must be an integer.
**My question is as follows: where exactly does such a proof come from and how does it generalize to other questions concerning tiling? There is obviously a deeper principle at work here. What exactly is that principle?**
One can pick other functions to integrate over such as $x-[x]-1/2$ and the result will follow. *It just seems like black magic that this works*. It's as if the functions you are integrating over tease out the geometric properties of your shape in an effortless way.
EDIT: It's likely that one doesn't necessary need integrals to think in the same flavor as this solution. You're essentially looking at both side lengths in parallel with linear test functions on individual tiles. However, this doesn't really explain the deeper principles here, in particular how one could generalize this method to more difficult questions by choosing appropriate "test functions."
| https://mathoverflow.net/users/934 | Tiling a rectangle with a hint of magic | It is not at all obvious to me that there is any deep principle at work in the double-integral proof. In my mind, the double-integral proof is really the same as the checkerboard proof. You're just trying to come up with a translation-invariant function on rectangles that is (a) additive and (b) zero if and only if the rectangle has the property of interest. All that matters is that you pick a function that cancels itself out in the same way as the checkerboard pattern does. The choice of the sine function for this purpose is amusing but not deep.
If you're looking for deeper principles then I would recommend Rick Kenyon's paper "A note on tiling with integer-sided rectangles," *J. Combin. Theory Ser. A* 74 (1996), no. 2, 321-332. Using this problem as an example, Kenyon demonstrates the concept of the Conway-Lagarias *tiling group*, a powerful tool for studying tiling problems.
| 15 | https://mathoverflow.net/users/3106 | 28191 | 18,425 |
https://mathoverflow.net/questions/26582 | 2 | I am looking for a reference to a proof of the following well-know fact (cited for example by
B.Farb in ``Relatively hyperbolic groups'', Geom. Funct. Anal. 8 (1998), no. 5, 810--840); [MR1650094](https://mathscinet.ams.org/mathscinet-getitem?mr=1650094),
[DOI:10.1007/s000390050075](https://doi.org/10.1007/s000390050075).
Suppose $X$ is the universal covering of a negatively curved Riemannian manifold, let $O$ be an open horoball in $X$ and let $H=\partial O$ be the horospherical boundary of $O$.
Also suppose that $\gamma\colon [0,1]\to X\setminus O$ is a rectifiable path such that $d(\gamma (t), H)\geq k>0$ for every $t\in [0,1]$, and let $\pi\colon X\setminus O\to H$
the (well-defined) nearest-point projection. Then, there exists $\alpha>0$ (only depending on the curvature of $X$) such that the length $L(\pi\circ\gamma)$ of $\pi\circ\gamma$ is bounded above by $e^{-\alpha k} L(\gamma)$.
Of course, this fact can be reduced to the computation of the Lipschitz constant of the projection of a horosphere onto another horosphere having the same basepoint.
When $X$ is the real hyperbolic $n$-space, such a computation is very easy, and it is likely that the variable curvature case can be reduced to the hyperbolic case via some comparison theorem. However, I was wondering if there is some standard reference I could rely on.
| https://mathoverflow.net/users/6206 | Reference for the geometry of horospheres | Try [Geometry of horospheres](http://www.google.com/url?sa=t&source=web&cd=4&ved=0CCEQFjAD&url=http%3A%2F%2Fwww.intlpress.com%2FJDG%2Farchive%2F1977%2F12-4-481.pdf&ei=8-EWTOknwYHyBp6y3PMI&usg=AFQjCNFd1SZWgJqlUWCA-8lYqLHdWUtP3w) by
Heintze and Im Hof.
| 6 | https://mathoverflow.net/users/1573 | 28194 | 18,428 |
https://mathoverflow.net/questions/28157 | 4 | Some days ago, I posted a question about [models of arithmetic and incompleteness](https://mathoverflow.net/questions/26676/incompleteness-and-nonstandard-models-of-arithmetic). I then made a mixture of too many scattered ideas. Thinking again about such matters, I realize that what really annoyed me was the assertion by Ken Kunen that the circularity in the informal definition of natural number (what one gets starting from 0 by iterating the successor operation a finite number of times) is broken “by formalizing the properties of the order relation on ω” ( page 23 of his “The Foundations of Mathematics”). What does actually “breaking the circularity” mean? Is there a precise model theoretic statement that expresses this meaning? And what about proving that statement? Is that possible?
| https://mathoverflow.net/users/6466 | Breaking the circularity in the definition of N | Looking at the draft that was linked above, it's more clear what Kunen means. He is just saying that the informal "definition" of the natural numbers that you might think of in school is circular when examined closely. And it is, in the sense that you have to start with some undefined concept, be it "natural number", "finite set", "proof", etc., to capture finiteness.
However, Kunen does not dwell on that sort of philosohical point. He is simply saying that there *is* a formal and non-circular definition of ω in set theory, as the smallest infinite ordinal. This does give a rigorous definition, but it doesn't ensure that "finite" in an aribitrary model corresponds to our actual notion of finite. That is something that cannot be ensured in first-order logic.
| 6 | https://mathoverflow.net/users/5442 | 28240 | 18,455 |
https://mathoverflow.net/questions/28237 | 17 | If $F$ is a global field then a standard exact sequence relating the Brauer groups of $F$ and its completions is the following:
$$0\to Br(F)\to\oplus\_v Br(F\_v)\to\mathbf{Q}/\mathbf{Z}\to 0.$$
The last non-trivial map here is "sum", with each local $Br(F\_v)$ canonically injecting into $\mathbf{Q}/\mathbf{Z}$ by local class field theory.
In particular I can build a class of $Br(F)$ by writing down a finite number of elements $c\_v\in Br(F\_v)\subseteq \mathbf{Q}/\mathbf{Z}$, one for each element of a finite set $S$ of places of $v$, rigging it so that the sum $\sum\_vc\_v$ is zero in $\mathbf{Q}/\mathbf{Z}$.
This element of the global Brauer group gives rise to an equivalence class of central simple algebras over $F$, and if my understanding is correct this equivalence class will contain precisely one division algebra $D$ (and all the other elements of the equiv class will be $M\_n(D)$ for $n=1,2,3,\ldots$).
My naive question: is the dimension of $D$ equal to $m^2$, with $m$ the lcm of the denominators of the $c\_v$? I just realised that I've always assumed that this was the case, and I'd also always assumed in the local case that the dimension of the division algebra $D\_v$ associated to $c\_v$ was the square of the denominator of $c\_v$. But it's only now, in writing notes on this stuff, that I realise I have no reference for it. Is it true??
| https://mathoverflow.net/users/1384 | Dimension of central simple algebra over a global field "built using class field theory". | To paraphrase Igor Pak: OK, this I know. It is remarkable how difficult it is to track down a reference which gives an actual proof for this fact (moreover applicable to all global fields). The notes of Pete Clark don't give a proof or a reference for a proof, and its omission in Cassels-Frohlich is an uncorrected error. :)
But here is a reference: Theorem 3.6 in the notes on Honda-Tate theory on Kirsten Eisentraeger's webpage. The assertion is even stronger: one can find a *cyclic* splitting field of the expected minimal degree. A moment's reflection leads one to realize what is actually going on: in the non-archimedean local theory we know that one can always arrange the splitting field to be the *unramified* one of the expected minimal degree (already in Serre's Local Fields, and part of the story of the "local invariant"), so in particular it is cyclic in that case. Taking into account the real case, and using the exactness at the left of the global-to-local sequence for Brauer groups, the global problem reduces to making a global cyclic extension inducing specified local ones at finitely many places and having a predicted degree which is lcm of local degrees (in the local theory the degree is actually all that really matters, not the unramifiedness).
Enter Grunwald-Wang... and since all that matters locally is the degree, if we don't care about global cyclicity but just global degree and some local degrees then weak approximation & Krasner's Lemma suffice to do the job (so for the question as asked, in which there's no cyclicity, the global problem is actually very elementary once the local case is settled!). Note that in Cassels-Frohlich the global cyclic splitting field is addressed, but not its degree (since Grunwald-Wang is not addressed in Cassels-Frohlich).
Historically the existence of a global *cyclic* splitting field, moreover of the expected degree, was regarded as one of the real triumphs of global class field theory, and the early attempts at class field theory by the German school were intimately tied up with this problem of the cyclic splitting field. This is why it was such a shock to Artin when Wang discovered that Grunwald's proof of local-to-global for cyclic extensions was not true (but fortunately Wang's fix was sufficient); see Roquette's historical notes on CFT.
Finally, to put this in perspective, it should be noted (as remarked in Eisentraeger's notes) that there are examples of complex function fields in transcendence degree 3 admitting nontrivial 2-torsion Brauer classes not represented by a quaternion division algebra! (The appearance of trdeg 3 is reasonable, as the period-index problem for surfaces over an algebraically closed field was proved by deJong.)
| 12 | https://mathoverflow.net/users/6773 | 28245 | 18,458 |
https://mathoverflow.net/questions/28248 | 8 | Given an operation of say a topological group on a topological space, one can form the quotient stack X//G:
the stack associated to the action groupoid.
Does this stack satisfy some kind of universal property?
| https://mathoverflow.net/users/2837 | Universal property of X//G? | Probably the simplest example is when the space $X$ is a single point. Then $pt//G$ classifies principal $G$-bundles. Here, the action groupoid is just $G$ considered as a one-object groupoid. The one object, $pt$ becomes an atlas for the stack, so we have a representable epimorphism $pt \to pt//G$. This is in fact the universal principal $G$-bundle! Given any map $X \to pt//G$, the 2-fibered product $X \times\_{pt//G} pt\to X$ is a principal $G$-bundle. What we have in general is, given any topological groupoid $H$, its associated stack (stackification of $Hom(blank,H)$ ) is equivalent to $Bun\_H:T \mapsto Bun\_H(T)$, where $Bun\_H(T)$ is the groupoid of principal $H$-bundles over $T$. So, by Yoneda, we have for all spaces $T$, an equivalence of groupoids $Hom(T,Bun\_H) \cong Bun\_H(T)$. Moreover, you can show that for any map $T \to Bun\_H$, the principal bundle $P$ to which the map corresponds is the 2-fibered product $T \times\_{Bun\_H} H\_0 \to T$. Applying this to the case when $H$ is the group $G$, we get what I claimed. Now, if instead $H$ is the action groupoid of $G$ on a space $X$, we get that $X//G$ classifies principal $G \ltimes X$-bundles. This answers the question as far as what maps INTO the stack yield.
As Chris suggested, for any topological groupoid, you can take its enriched-nerve to get a simplicial space. Applying Yoneda on each component, you get a simplicial stack. The weak 2-colimit of this, is equivalent to the stack associated to the groupoid. (In fact, you only need to consider the 2-truncation of this diagram). Spelling this out, you get Proposition 3.19 out of Noohi's Foundations of Topological Stacks I. This roughly says that maps from the stack associated to a groupoid $H$ into another stack $Y$ are the same as maps $f:H\_0 \to Y$ (or by Yoneda elements $f \in Y(H\_0)\_0$) together with an isomorphism $f \circ s \to f \circ t$ satisfying some obvious coherencies, where $s$ and $t$ are the source and target map of the groupoid $H\_1 \rightrightarrows H\_0$. In the case where $H$ is the action groupoid $G \ltimes X$, you get maps $f:X \to Y$, together with isomorphisms $f \circ proj \to f \circ \rho$, where $\rho:G \times X \to X$ is the action. This becomes more concrete, if for example, $Y$ is not some arbitrary stack, but instead the stack of principal $K$-bundles for some group $K$. Then $Hom(X//G,Bun\_K)$ is the groupoid with objects principal $G$-bundles $P$ over $X$ together with an isomorphism $proj^\*P \cong \rho^\*P$ (satisfying a cocycle condition), with the obvious arrows.
If you were wondering why $X//G$ was the "like the ordinary quotient but better", then the following heuristics should help. The 2-truncated guy whose colimit equals $X//G$ is "like the colimit expressing $X/G$ but with the action built in more". In fact, we we took the UNtruncated guy, so, the simplicial space associated to the action groupoid, then it "IS" the Borel construction (or rather its geometric realization). In particular, this implies the homotopy type of $X//G$ is the same as the homotopy quotient $EG \times\_G X$. Now consider one more thing- if $G$ is acting on $X$ we have the commutative square with $G \times X$ on the upper left, $X$ on bottom left and on the upper right, and $X/G$ (the coarse quotient) on the bottom right, with map up top being the action, and the map down below being the projection, then this is a pullback diagram if and only if the action is faithful. If you replace $X/G$ with $X//G$, then this is ALWAYS a (2-)pullback diagram. Hence, going to stacks "makes all actions faithful up to homotopy" (more accurately, stacks allow us to encode the isotropy data that would otherwise be lost in such a way that any action becomes as good as a faithful one).
| 9 | https://mathoverflow.net/users/4528 | 28251 | 18,462 |
https://mathoverflow.net/questions/28044 | 8 | The osculating circle at a point of a smooth plane curve can be obtained by considering three points on the curve and the circle defined by them. When the three points approach $P$, the circle becomes the osculating circle at $P$.
Generalizing this method to a conic defined by five points on a curve, one obtains an 'osculating conic' at each point of a smooth plane curve. It is pretty straightforward to show that this osculating conic is an ellipse (resp. parabola, hyperbola) if $y'' y^{(4)} > \frac{5}{3}y'''^2$ (resp. =, <).
Thus one can classify points on a plane curve, e.g. all points on $y=e^x$ are hyperbolic (which is not obvious to the 'naked eye'). The aforementioned criterion is an affine differential invariant.
I'm wondering why these 'osculating conics' seem to be relatively unknown (they would give a nice textbook example connecting elementary calculus and linear algebra - you can find the criterion above in one handwritten page, starting with the Taylor expansion) and whether there are any applications (generalization of evolute; can you get the curve back from the focal curve of the osculating conics)?
Furthermore it would be interesting to know whether there are any results for 'osculating cubics' (nine points define a cubic plane curve) or 'osculating quartics' (fourteen points define a quartic plane curve) or for 'osculating quadrics' (would in general yield another classification of points on a surface beyond the usual elliptic/parabolic/hyperbolic one).
| https://mathoverflow.net/users/6415 | Osculating conics and cubics and beyond | These highly osculating curves were studied, in particular by V.I. Arnol'd. One of the important references will be:
Topological invariants of plane curves and caustics.
Dean Jacqueline B. Lewis Memorial Lectures presented at Rutgers University, New Brunswick, New Jersey. University Lecture Series, 5. American Mathematical Society, Providence, RI, 1994.
More precisely, what was studied are the points of the curve, where the level of its tangency with (say) conics is higher than expected. I guess these are exactly the points that (using your terminology) separate the elliptic part of the curve from the hyperbolic part.
The key words for this research are *Extactic points* (terminology proposed by D. Eisenbud). Using google scholar you can find a complete text of Arnol'd, called:
Remarks on the extatic points of plane curves, V.I. Arnold - The Gelfand Mathematical Seminars, 1993-1995.
This article contains some generalisations of the *four vertex* theorem. <http://en.wikipedia.org/wiki/Four-vertex_theorem>
One more nice reference is a paper of Tabachnikov and Timorin. <https://arxiv.org/abs/math/0602317>
| 10 | https://mathoverflow.net/users/943 | 28256 | 18,466 |
https://mathoverflow.net/questions/22263 | 4 | We are interested in the following question (definitions and references are given below):
Main Question: Given an upper-semicontinuous polyhedral multifunction $F:R^n \rightarrow R^m$, is there always a Lipschitz continuous function $g:R^n \rightarrow R^m$ such that $g(x) \in F(x)$ for all $x \in R^n$ ?
In general, *upper* semicontinuity is probably not the right property to guarantee continuous selections (see [HP-1], page 89). On the other hand, polyhedrality of the multifunction may help.
Related Question: Under what *mild* conditions is a positive answer to the above question guaranteed?
Motivation: Solutions of finite-dimensional variational inequalities (VI) or complementarity problems (CP) (see [FP-1]) typically have upper semicontinuous solution maps. Furthermore, if the functions and sets defining the VI or CP have affine or linear (or polyhedral) structure, their solution maps are polyhedral multifunctions. The main question posed above is then a natural question to ask in the study of differential inclusions $\dot{x} \in G(x)$ that have these solution maps appearing in the definition of $G(x)$.
Definitions:
1. A *multifunction* is simply a *set-valued* map.
2. A multifunction $F:R^n \rightarrow R^m$ is said to be *upper semicontinuous* at a point $\bar{x}$ if for every open set $\mathcal{V}$ containing $F(\bar{x})$, there exists an open neighbourhood $\mathcal{U}$ of $\bar{x}$ such that, for each $x \in \mathcal{U}$, $\mathcal{V}$ contains $F(x)$.
3. A multifunction is said to be *polyhedral* if its graph is a polyhedral subset of $R^{n + m}$.
Referenecs:
1. [FP-1] F. Facchinei and J-S Pang, Finite-Dimensional Variational Inequalities and Complementarity Problems (vol. 1), pp. 138-139.
2. [HP-1] S. Hu and N.S. Papageorgiou, Handbook of Multivalued Analysis (vol. 1), p. 36 and p. 89.
3. [M-1] Ernest Michael, Continuous Selections I, The Annals of Mathematics, Vol. 63, (1956), pp. 361-382.
4. [M-2] Ernest Michael, Continuous Selections II, The Annals of Mathematics, Vol. 64, (1956), pp. 562-580.
5. [M-2] Ernest Michael, Continuous Selections III, The Annals of Mathematics, Vol. 65, (1957), pp. 375-390.
| https://mathoverflow.net/users/5526 | Do upper-semicontinuous polyhedral multifunctions have Lipschitz continuous selections? | Hi,
I happen to be working on semi-algebraic set-valued maps, and I might have a partial answer in [1]. I guess when you say polyhedral, you mean that the graph of the set-valued map is a union of finitely many polyhedrons. If that is the case, polyhedral set-valued maps are semi-algebraic. Semi-algebraic set-valued maps are strictly continuous (A generalization of Lipschitz continuity for set-valued maps: See [1]) except on a set of smaller dimension. I hope this means that there is a selection that is Lipschitz except on a set of smaller dimension.
[1] A. Daniilidis and C.H.J. Pang, Continuity and Differentiability of set-valued maps revisited in the light of tame geometry. <http://arxiv.org/abs/0905.0373>
[2] R.T. Rockafellar and R.J.-B. Wets, Variational Analysis.
| 4 | https://mathoverflow.net/users/6821 | 28258 | 18,468 |
https://mathoverflow.net/questions/28233 | 5 | is there any algorithm known for computing (middle perversity)intersection homology of complex toric varieties based on their combinatorial data? I'm not looking for a computer program.
Regards,
Peter
| https://mathoverflow.net/users/6398 | Intersection homology for toric varieties | See
Braden, Tom and MacPherson, Robert, From moment graphs to intersection cohomology, Math. Ann. 321 (2001), no. 3, 533--551.
| 5 | https://mathoverflow.net/users/3077 | 28260 | 18,470 |
https://mathoverflow.net/questions/28263 | 1 | This must be a well-known exercise with spectral sequences, but I don't know a reference for it. I'm trying to figure out when does $Tot$ commute with colimits.
More precisely, let $X$ be a double cochain complex of, say, $R$-modules, $R$ a commutative ring with unit, or, more generally, a double complex in an abelian category. Let $\cal{C}$ denote the category of these double cochain complexes.
We have two different *total functors*, $\mbox{Tot}^\prod$ and $\mbox{Tot}^{\bigoplus}$, from the category of double complexes to the category of cochain complexes:
$$
\mbox{Tot}^{\prod}(X)^n = \prod\_{p+q=n}X^{p,q} \qquad \mbox{and} \qquad \mbox{Tot}^{\bigoplus}(X)^n = \bigoplus\_{p+q=n}X^{p,q} \quad .
$$
Let $\mbox{Tot}$ denote anyone of them and let $I$ be a (filtered) category, and $X: I \longrightarrow \cal{C}$ a functor. We have a natural morphism
$$
\theta: \varinjlim\_i \mbox{Tot} (X\_i) \longrightarrow \mbox{Tot} (\varinjlim\_i X\_i) \quad .
$$
When dealing with $\mbox{Tot}^\bigoplus$, this $\theta$ is an isomorphism, because a direct sum is a colimit and colimits commute with colimits.
What happens when we take $\mbox{Tot}^\prod$? Is $\theta$ at least a quasi-isomorphism (a morphism inducing an isomorphism in cohomology)? In which cases? Do we need some extra hypothesis on the abelian category (AB...)? Is the hypothesis "filtered" really needed, or we can deal with arbitrary colimits in general?
Of course, if our double complex has finite diagonals, then $\mbox{Tot}^\prod = \mbox{Tot}^\bigoplus$, and we are done. But what happens without this hypothesis?
I'm mainly interested in the case of a right half-plane double complex, that is $X^{p,q} = 0$ if $p<0$, but I'll be glad to learn about all possible cases.
Any references or hints will be welcome.
| https://mathoverflow.net/users/1246 | Tot and colimits | Imagine that all double complexes in the image of your functor X: I → C have both differentials equal to zero. Moreover, all terms of these bicomplexes outside of a fixed diagonal are also zero. Then you are asking, quite simply, whether colimits commute with countable products. If they don't, your morphism θ cannot be a quasi-isomorphism (being a non-isomorphism of complexes with zero differentials). And of course, if in a certain abelian category countable filtered colimits commute with countable products (and both exist), then all objects of this category are zero.
| 8 | https://mathoverflow.net/users/2106 | 28273 | 18,476 |
https://mathoverflow.net/questions/28271 | 8 | Suppose we have a smooth dynamical system on $R^n$ (defined by a system of ODEs). Assuming that the system has a finite set of fixed points, I am interested in knowing (or obtaining references about) what is the behaviour of its fixed point structure under perturbations of the ODEs. More specifically, i would like to know under which conditions the total number of fixed points remains the same. By perturbations I mean generic changes in the system of equations... the more general the better.
I am sorry if the question is too basic, my interest comes from the study of the so-called renormalization flow in field theories. In particular, it would be important for me to generate an intuition about the conditions under which the approximations performed over a dynamical system alters its fixed point structure.
| https://mathoverflow.net/users/6091 | Persistence of fixed points under perturbation in dynamical systems | A good reference for this sort of thing is Guckenheimer and Holmes, *Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields*, Springer-Verlag, 1983, if you're not already familiar with it (and even if you are, for that matter). Chapter 3 in particular is relevant to your question.
| 3 | https://mathoverflow.net/users/5701 | 28285 | 18,483 |
https://mathoverflow.net/questions/28280 | 9 | In his answer to my question
[The Green-Tao theorem and positive binary quadratic forms](https://mathoverflow.net/questions/28047/the-green-tao-theorem-and-positive-binary-quadratic-forms)
Kevin Ventullo answers my initial question in the affirmative. What remains is the title question here, of separate interest to me.
Any integral positive binary quadratic form integrally represents a set of primes with known Dirichlet density. This is an application of the Cebotarev density theorem (or Chebotarev or Tchebotarev), in particular it is Theorem 9.12 in David A. Cox, Primes of the form $x^2 + n y^2,$ with the example $\Delta = -56$ on page 190. I typed this out in the previous question.
Now, Jurgen Neukirch "Class Field Theory" points to Serre "A Course in Arithmetic," and on page 76 Serre says that the set of primes p such that a fixed polynomial has a root $\pmod p$ has a natural density, and refers to K. Prachar "Primzahlverteilung" chapter 5 section 7. By results (theorem 9.2, page 180) in the Cox book, this means that the principal form
$x^2+ny^2$ or $x^2+xy+ky^2$ does represent a set of primes with a natural density, therefore equalling the Dirichlet density. And by the result on arithmetic progressions, a full genus of forms has a natural density.
Combining observations, the principal form always has a natural density of primes, any full genus does, therefore we are done for one class per genus, and in the case of two classes per genus we are done with the principal genus and any genus with two distinct opposite forms. So we have natural densities for Cox's $\Delta = -56$ example. We are also done with the principal genus when it has three classes.
So, (and I would love a reference), does every positive binary quadratic form represent a set of primes for which the natural density exists?
| https://mathoverflow.net/users/3324 | Does a positive binary quadratic form represent a set of primes possessing a natural density | Accoring to [H. Lenstra](http://websites.math.leidenuniv.nl/algebra/Lenstra-Chebotarev.pdf), Chebotarev's theorem holds both for Dirichlet and
for natural density (but he doesn't give a reference in this document).
Applying Chebotarev to the extension $H/\mathbb{Q}$ where $H$ is the Hilbert
class field of $\mathbb{Q}(\sqrt{-n})$ gives the result you want.
(At least for primitive discriminants; for non-primitive discriminants
you need an appropriate generalization of the Hilbert class field).
**Added in response to Will's comment**
There is always a suitable field. Let $-D$ be a primitive negative
discriminant and let $d=-m^2D$ be a general discriminant. Let $H\_n$
be the abelian extension of $K=\mathbb{Q}{\sqrt{-D}}$ where a non-ramified
prime splits iff its is principal and generated by an
element of $\mathbb{Z}+m\mathcal{O}\_K$. Such a field $H\_n$ is
called a *ring class field* and exists by class field theory. It also
is an extension of $K$ by a singular value of the $j$-function.
Then $G\_m=\mathrm{Gal}(H\_n/\mathbb{Q})$ is a generalized dihedral group.
There is a correspondence between conjugacy classes in $G\_m$ and pairs
of equivalence classes of $ax^2\pm bxy+cy^2$ of discriminant $d$, such that
an unramified prime has its Frobenius in a conjugacy class iff it it
represented by the corresponding form. This is why we can apply Chebotarev.
**Even more added**
A good reference for ring class fields is Cox's book *Primes of the Form
$x^2+ny^2$*.
| 8 | https://mathoverflow.net/users/4213 | 28287 | 18,485 |
https://mathoverflow.net/questions/28070 | 4 | Hello all. This is probably a simple problem for you guys, but my geometry is a bit rusty and I am hoping that you can help.
I am trying to arrange an arbitrary number of objects around the circumference of an ellipse. My first stab at the problem resulted in the use of a simple rotational matrix like this (note that I am using graphical Cartesian coordinates where y increases down from the top of the screen).
```
def rotate(x, y, theta)
x_p = (x * Math.cos(theta)) - (y * Math.sin(theta))
y_p = (y * Math.cos(theta)) + (x * Math.sin(theta))
return [x_p, y_p]
end
```
That is obviously a circle, so I use the x coordinate and feed it into the equation of my ellipse to get the y(s).
This works and does what it is supposed to, but the problem is that I am only incrementing theta to find my points, so obviously they are not equidistant around the circumference of the ellipse. That is what I am after, and any help would be greatly appreciated.
You can find a description of the problem with pictures in this thread:
<http://www.bigresource.com/Tracker/Track-flash-DO1WzX6KNq/>
| https://mathoverflow.net/users/2416 | Finding n points that are equidistant around the circumference of an ellipse | This answer assumes you are interested in finding $n$ points on an ellipse such that the arc lengths between successive points are equal.
As others mentioned, this problem involves an elliptic integral, which has no elementary expression. However, many scientific computation libraries are able to compute this function numerically, and you can take advantage of that. It looks like you're using Python; one possibility here would be the GNU Scientific Library (gsl) for which Python bindings are available.
Say your ellipse is parametrized by $x(t) = a \cos t$, $y(t) = b \sin t$ where $a$ is the semimajor and $b$ the semiminor axis. Let $e = \sqrt{1-b^2/a^2}$ be the eccentricity. The arclength from $(x(0), y(0)) = (a,0)$ counterclockwise to $(x(t), y(t))$ is given by $s(t)=a E(t,e)$, where $E(t,e) = \int\_0^t \sqrt{1-e^2 \sin^2\theta}d\theta$ is the "incomplete elliptic integral of the second kind". gsl implements this function as `gsl_sf_ellint_E`. The total arc length of your ellipse is $S = s(2\pi)$, which you can compute, so you need to find the value $t\_k$ such that $s(t\_k) = kS/n$, for each $k=0,\dots,n$. That is, you must solve for $t$ in the equation $s(t) = kS/n$.
But you are able to compute $s(t)$ numerically, and you also know the derivative $s'(t) = a \sqrt{1 - e^2 \sin^2 t}$ thanks to the fundamental theorem of calculus. Using this, it is a standard procedure to solve the equation iteratively using something like Newton's method. You could implement it yourself with a bit of reading, or use one of gsl's more sophisticated implementations, e.g. `gsl_root_fdfsolver_steffenson`.
| 6 | https://mathoverflow.net/users/4832 | 28290 | 18,488 |
https://mathoverflow.net/questions/28275 | 3 | I'm looking for a small research problem an undergraduate would be capable of after taking just an abstract algebra course, introductory algebraic geometry (at level of Miles Reid's book and Ideals, Varieties & Algorithms), and a course in number theory. Is there a website that would have a decent listing, or possibly a book one can recommend that may have small open research problems?
| https://mathoverflow.net/users/6824 | Looking for an undergraduate research problem in algebraic geometry or algebraic number theory | Your question is missing a crucial word in the first sentence (capable of ...). Is the missing word "understanding" or "solving"?
Anyway, here is a problem: Find the maximum number of points of a curve of genus $g$ over $\mathbb{F}\_q$, for some values of $g,q$ for which this number is not known (check for values at <http://www.manypoints.org/> )
| 2 | https://mathoverflow.net/users/2290 | 28291 | 18,489 |
https://mathoverflow.net/questions/26312 | 5 | Recently Francisco Santos has announced that he has a counterexample to the Hirsch conjecture. The last I heard it was circulating among several people and there would be a public version of it available soon. I am curious how close it is to release. Also has there been any progress in the attempt to find the vertices of the counterexample. The last I heard to find the vertices a series of steps had to be done and each step increased the complexity of the problem by a geometric factor making it difficult to complete the computation.
| https://mathoverflow.net/users/1098 | A Counterexample to the HIrsch Conjecture | The public version is now out. It is available [here](http://arxiv.org/abs/1006.2814)
| 3 | https://mathoverflow.net/users/1098 | 28294 | 18,492 |
https://mathoverflow.net/questions/28282 | 5 | Let $G$ be a simple algebraic group over a field $k$, and let $U$ be the unipotent radical of a Borel subgroup $B$. Because $B$ normalises $U$, the group $H = B/U$ acts on the coordinate ring $\mathcal{O} = k[X]$ of the basic affine space $X = G/U$ via $(h.f)(x) = f(xh)$. We get a decomposition of $\mathcal{O}$ into a direct sum
$\mathcal{O} = \oplus\_{\lambda \in \Lambda^+} \mathcal{O}^\lambda$
where the Weyl module $\mathcal{O}^\lambda$ is the set of all $f \in \mathcal{O}$ such that $h.f = \lambda(h)f$ for all $h \in H$. Because the action of $H$ commutes with the action of $G$ on $k[X]$ given by $g.f(x) = f(g^{-1}x)$, each $\mathcal{O}^\lambda$ is a $G$-submodule of $\mathcal{O}$. We can also identify $\mathcal{O}^\lambda$ with the space of global sections $H^0(G/B, \lambda)$.
Next, multiplication in $\mathcal{O}$ induces a $G$-module map $\mathcal{O}^\lambda \otimes \mathcal{O}^\mu \to \mathcal{O}^{\lambda + \mu}$ for any $\lambda, \mu \in \Lambda^+$. Since $\mathcal{O}$ has no zero-divisors, this map is non-zero.
Now if the base field $k$ has characteristic zero, it is well-known that the $G$-modules $\mathcal{O}^\lambda$ for $\lambda \in \Lambda^+$ are irreducible, so the multiplication map above must be surjective. Does this remain true when the characteristic of $k$ is positive, when the Weyl modules $\mathcal{O}^\lambda$ are no longer irreducible in general?
| https://mathoverflow.net/users/6827 | Tensor products of Weyl modules in positive characteristic | The question has an affirmative answer and a fairly long history as well, but the
proof uses some nontrivial ideas. The notation used here is nonstandard relative to that found in Jantzen's book *Representations of Algebraic Groups* (second edition, AMS, 2003). Also, a "Weyl module" (in the usual sense) of a given highest weight is the dual of the module of global sections for a related line bundle on the flag variety, using Kempf's vanishing theorem (1976). The term "Weyl module" was coined by Carter and Lusztig in their paper on special linear groups, partly because the formal character is given by Weyl's formula.
A Weyl module has a unique simple quotient, while the corresponding module has this module as its unique simple submodule.
There was a series of papers by Lakshmibai-Musili-Seshadri on the geometry of flag varieties in prime characteristic, in which they stated along the way that the tensor product of
these dual Weyl modules maps onto the one specified by the sum of highest weights. (Their proof may not be rigorous. In any case, Kempf's theorem comes into play here.) A focused reference is the paper in J. Algebra 27 (1982) by Jian-pan Wang, "Sheaf cohomology on $G/B$ and tensor product of Weyl modules". That paper followed up a suggestion of mine that such a tensor product should have a filtration with appropriate Weyl modules as subquotients. The paper by Olivier Matthieu in Duke Math. J. 59 (1989) used Frobenius splitting techniques to prove this in full generality after the partial results by Wang and then by Steve Donkin in Springer Lecture Notes 1140 (1985). Eventually all of this gets folded into the general theory of "tilting modules" for reductive algebraic groups (Chapter G in Jantzen).
[ADDED] As Ekedahl just pointed out, a treatment is given in the more recent and more extensive book by Brion and Kumar along with history.
| 7 | https://mathoverflow.net/users/4231 | 28301 | 18,495 |
https://mathoverflow.net/questions/28265 | 37 | In the common Hodge theory books, the authors usually cite other sources for the theory of elliptic operators, like in the Book about Hodge Theory of Claire Voisin, where you find on page 128 the Theorem 5.22:
>
> Let P : E → F be an elliptic differential operator on a compact manifold. Assume that E and F are of the same rank, and are equipped with metrics. Then Ker P is a finite-dimensional subspace of the smooth sections of E and Im P is a closed subspace of finite codimension of the smooth sections of F and we have an L²-orthogonal direct sum decomposition:
>
>
> smooth sections of E = Ker P ⊕ Im P\*
>
>
> (where P\* is the formal L²-adjoint of P)
>
>
>
In the case of Hodge Theory, we consider the elliptic self-adjoint operator Δ, the Laplacian (or: Hodge-Laplacian, or Laplace-Beltrami operator).
A proof for this theorem is in Wells' "Differential Analysis on Complex Manifolds", Chapter IV - but it takes about 40 pages, which is quite some effort!
---
Now that I'm learning the theory of elliptic operators (in part, because I want to patch this gap in my understanding of Hodge Theory), **I wonder if this "functional analysis" is really always necessary**.
**Do you know of any class of complex manifolds** (most likely some restricted class of complex projective varieties) **where you can get the theorem above without using the theory of elliptic operators** (or at least, where you can simplify the proofs that much that you don't notice you're working with elliptic operators)**?** Maybe the general theorem really requires functional analysis (I think so), but the **Hodge decomposition might follow from other arguments**.
I would be very happy to see some arguments proving special cases of Hodge decomposition on, say, Riemann surfaces. I would be even happier to hear why this is implausible (this would motivate me to learn more about these fascinating elliptic differential operators).
If this ends up being argumentative and subjective, feel free to use the community wiki hammer.
| https://mathoverflow.net/users/956 | Proving Hodge decomposition without using the theory of elliptic operators? | The hard part of the proof of the Hodge decomposition (which is where the serious functional analysis is used) is the construction of the Green's operator. In Section 1.4 of Lange and Birkenhake's "Complex Abelian Varieties", they prove the Hodge decomposition for complex tori using an easy Fourier series argument to construct the Green's operator.
| 23 | https://mathoverflow.net/users/317 | 28307 | 18,497 |
https://mathoverflow.net/questions/28250 | 9 | I have a specific problem, but would also like to know how to tackle the general case. I will first state the genral question. Let $M$ be an embedded submanifold of $\mathbb{R}^n$ and let $F: \mathbb{R}^n \to \mathbb{R}^n $ be a smooth map. How do I go about checking whether $F(M)$ is a smooth embedded submanifold of $\mathbb{R}^n$ or not? The specific problem I have is the following :- Let $F:\mathbb{C}^2\to \mathbb{C}^2$ the map $(z\_1,z\_2) \mapsto (z\_1+z\_2,z\_1z\_2)$ and let $M$ be the unit sphere in $\mathbb{C}^2$, i.e., $\lbrace (z\_1,z\_2) : |z\_1|^2 + |z|^2 = 1 \rbrace $. Is $F(M)$ an embedded submanifold of $\mathbb{C}^2$ (considered as $\mathbb{R}^4$)?
| https://mathoverflow.net/users/36038 | Checking whether the image of a smooth map is a manifold | The specific $F(M)$ is not a smooth submanifold. Here is an argument.
To simplify formulas, I renormalize the sphere: let it be the set of $(z\_1,z\_2)\in\mathbb C^2$ such that $|z\_1|^2+|z\_2^2|=2$ rather than 1. Then, as Gregory Arone pointed out, $F(M)$ is the set of $(b,c)\in\mathbb C^2$ such that the roots $z\_1,z\_2$ of the equation $z^2-bz+c$ satisfy $|z\_1|^2+|z\_2^2|=2$. I claim that it is not a smooth manifold near the point $p:=(2,1)\in F(M)$.
Let us intersect $F(M)$ with two planes: $\{b=2\}$ and $\{c=1\}$. If it is is a smooth submanifold, at least one of the intersections should be locally (near $p$) a 1-dimensional smooth submanifold of the respective plane (otherwise both planes are tangent to $F(M)$ at $p$, but this is impossible since they span the whole space). But this is not the case:
If $b=2$, the equation is $z^2-2z+c=0$, hence $z\_1+z\_2=2$, then $|z\_1|+|z\_2|\ge 2$ and therefore $|z\_1|^2+|z\_2|^2\ge 2$. Equality is attained only for $z\_1=z\_2=1$, thus the intersection is a single point $c=1$, not a 1-dimensional submanifold.
If $c=1$, the equation is $z^2-bz+1$, hence $z\_1z\_2=1$, then $|z\_1|\cdot|z\_2|=1$ and therefore $|z\_1|^2+|z\_2|^2\ge 2$. The equality is attained if and only if $|z\_1|=|z\_2|=1$, so $z\_1$ and $z\_2$ are conjugate to each other. The set of $b$'s for which this happens is the real line segment $[-2,2]$ which is not a submanifold near 2.
| 9 | https://mathoverflow.net/users/4354 | 28308 | 18,498 |
https://mathoverflow.net/questions/28305 | 4 | I just read for the first time the definition of an internally approachable set, which says:
A set $N$ is internally approachable (I.A.) of length $\mu$ iff there is a sequence $(N\_{\alpha} : \alpha < \mu)$ for which the following holds: $N=\bigcup\_{\alpha< \mu} N\_{\alpha}$ and for all $\beta < \mu$ $( N\_{\alpha} : \alpha < \beta ) \in N$.
Now if $N \prec (H(\theta), \in, < )$ is I.A. of length $\mu$. Is it true that
(a) If $\alpha < \mu$ then $\alpha \in N$
(b) If $\alpha < \mu$ then $N\_{\alpha} \in N$ ?
This is trivial if $N$ is transitive, and I'm quite sure that both (a) and (b) hold but I need a good argument.
| https://mathoverflow.net/users/4753 | Some consequences of internally approachable structures | Yes, both (a) and (b) follow from your definition without assuming that $N$ is transitive. You say that for every $\beta\lt\mu$ the sequence $\langle N\_\alpha | \alpha\lt \beta\rangle$ is in $N$. This implies that $\beta$ is in $N$, since $\beta$ is the length of this sequence and $N$ computes this length correctly by elementarity. Thus, every $\beta\lt\mu$ is in $N$ and so (a) holds (renaming $\alpha$ to $\beta$). It now follows that (b) also holds, assuming that $\mu$ is a limit ordinal, since once we have the sequence of length $\alpha+1$ in $N$, we may evaluate this sequence at $\alpha$ to deduce that $N\_\alpha$ is in $N$.
| 4 | https://mathoverflow.net/users/1946 | 28310 | 18,499 |
https://mathoverflow.net/questions/28112 | 6 | Let $R$ be a complete discrete valuation ring with field of fractions $K$ and algebraically closed residue field $k$, and let $X$ be a proper, smooth, geometrically connected curve over $K$. Take a finite extension $L/K$ and a regular proper model $\widetilde{X}$ of $X$ over the ring of integers of $L$ whose special fibre $\widetilde X\_k$ is semi-stable. Let $e$ be the ramification index of $L/K$. The reduction graph of $X$ is a metrised graph obtained as follows. Take a set of intervals of lengh $1/e$ indexed by the singular points of $\widetilde X\_k$. For every singular point $x$ of $\widetilde X\_k$, label the endpoints of the corresponding edge by the two irreducible components (possibly the same) on which $x$ lies. For every irreducible component $C$ of $\widetilde X\_k$, identify all the endpoints labelled $C$. The result (as a metric space) is independent of the choice of $L$.
**Question:** Suppose we have a regular proper model of $X$ over $R$ whose special fibre $X\_k$ is reduced, but not necessarily semi-stable. Suppose furthermore that we know the irreducible components and singular points of $X\_k$ and their intersection multiplicities in this model. What can be said about the reduction graph of $X$?
It seems reasonable to ask this question in this generality, but I am actually interested in the modular curves X1(*n*) over W*p*[*ζ**p*2], where *p* is a prime number dividing *n* exactly twice and W*p* is the ring of Witt vectors of an algebraic closure of **F***p*. In this situation non-semi-stable models as above were found by Katz and Mazur. What I would like specifically is an upper bound on the diameter of the reduction graph for these curves.
| https://mathoverflow.net/users/5944 | Diameter of reduction graph of a curve over a complete discrete valuation ring | This is a sequel to the above comments.
Consider an elliptic curve $E$ over $K$ with additive reduction over $K$ and multiplicative reduction over some extension $L/K$. Then we can find a quadratic extension $L/K$, and the Kodaira symbole of $E$ over $K$ is $I^\*\_m$ for some $m$ (see below), and the group of components of $E\_L$ is $Z/nZ$, where $n=[L:K]\nu\_K(\Delta)$ and $\Delta$ is the minimal discriminant of $E$ over $K$.
Now how to compute $n$ from data over $K$ ? I mean from data that you can read from the special fiber of the minimal regular model over $K$ ? The link between $n$ and $m$ is $n/2=(2+\delta) + (m+4)-1$ (Ogg's formula), and $\delta$ is the Swan conductor of $E$. If the residue characteristic is different from $2$, then $\delta=0$ and $n=2(m+5)$. Perfect.
But if the residue characteristic is 2 (so wild ramification happens), then $\delta$ can be arbitrarily big (if the absolute ramification index of $K$ is big). So $n$ is not a function of data from the special fiber over $K$.
| 6 | https://mathoverflow.net/users/3485 | 28315 | 18,502 |
https://mathoverflow.net/questions/26313 | 1 | In page 21 of *A Problem seminar*, D. J. Newman presents a novel way (at least for me) to determine the expectation of a discrete random variable. He refers to this expression as the **failure probability formula**. His formula goes like this
$f\_{0}+f\_{1}+f\_{2}+\ldots$
where $f\_{n}$ is the probability that the experiment fails to produce the desired outcome for $n$ steps.
He goes on to outlining two ways to establish the validity of this formula, but I'm having a hard time to unravel his second proof. I'll insert it here in order for my inquiry to be self-contained:
>
> Since the expected amount of time spent on the $n$ trial is equal to the probability that the $n$ trial occurs, and since this equal to $f\_{n-1}$, we do obtain the net expectation of $f\_{0}+f\_{1}+f\_{2}+\ldots$ as asserted.
>
>
>
Can any of you guys explain in greater detail this Newmanian argument? I'd also like to know whether the denomination **failure probability formula** is a standard one in the mainstream of Probability literature.
Thank you very much for your continued support.
| https://mathoverflow.net/users/1593 | Failure probability formula | Well, this is certainly a known idea, but I suspect it's not important enough to have its own name. For example, I believe it is used without special mention in Hammersley's "A Few Seedlings of Research" (Proc. 6th Berkeley Symp. on Math. Stat. and Prob., 1971), which has as its target audience a graduate student just beginning research in mathematical statistics.
Here's one way of thinking about the argument: imagine the countably infinite sequence of experiments starting all lined up and ready to go; each experiment will run if the preceding one runs but is unsuccessful. The expected number of steps is the same as the expected number of these experiments that actually run. Thus, the expected number of steps is the sum over $n$ of the probability that the $n$th experiment runs. But this is precisely $f\_{n - 1}$ (the probability that all the experiments before it fail).
---
An alternative perspective: let $p\_n$ be the probability that it requires exactly $n$ steps. Then summing
$$\begin{array}{ccccc}
p\_1 & & & & \\\\
p\_2 & p\_2 & & & \\\\
p\_3 & p\_3 & p\_3 & & \\\\
p\_4 & p\_4 & p\_4 & p\_4 & \\\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}$$
first by rows gives the "usual" expression for expected value while summing first by columns gives this expression you're dealing with. (If everything converges then it converges absolutely, so no need to worry about rearranging.)
Basically the same phenomenon is going on in the continuous version Michael Hardy mentions -- in that case we are calculating the area of one region in the $xy$-plane in two different ways, integrating with respect to $x$ in one and with respect to $y$ in the other.
| 3 | https://mathoverflow.net/users/4658 | 28316 | 18,503 |
https://mathoverflow.net/questions/28313 | 5 | Let $B$ be the closed unit ball in $\mathbb R^n$ and $f\colon B\to B$ a continuous map.
Consider the infinite product $B^{\mathbb Z}$ equipped with the product topology. Consider the solenoid
$$
S\_f=\{\{x\_n; n\in\mathbb Z\}: x\_{n+1}=f(x\_n)\}
$$
equipped with the induced topology.
Question: Is $S\_f$ contractible? If yes, is there a deformation retraction?
Motivation is here: [Two commuting mappings in the disk](https://mathoverflow.net/questions/3332/two-commuting-mappings-in-the-disk/28003#28003)
| https://mathoverflow.net/users/2029 | Solenoid of a continuous map of a ball, is it contractible? | No, it is not even path-connected in general, already for $n=1$.
Consider the folding map $f:[0,1]\to[0,1]$, namely $f(t)=2t$ for $t\le 1/2$ and $f(t)=2(1-t)$ for $t\ge 1/2$. There is no path connecting the orbits of the two fixed points: 0 and 2/3.
Indeed, suppose there is a continuous path $t\mapsto \{x\_n(t):n\in\mathbb Z\}$ in $S\_f$ such that $x\_n(0)=0$ and $x\_n(1)=2/3$ for all $n\in\mathbb Z$. Consider $x\_0(t)$ and $x\_{-n}(t)$ where $n>0$. While $x\_{-n}(t)$ travels from 0 to 1/2, $x\_0(t)$ will take the value 0 at $t\_0=0$, then the value 1 at some $t\_1>t\_0$, then 0 at some $t\_2>t\_1$, then 1 again, and so on, $2^n$ times. Since $t\mapsto x\_0(t)$ is a continuous function, it can have only finite number of alternating 0 and 1 values, but $n$ can be arbitrarily large, a contradiction.
| 9 | https://mathoverflow.net/users/4354 | 28318 | 18,505 |
https://mathoverflow.net/questions/28235 | 2 | Except the original Grönwall's theorem that $$\limsup\_{n \to \infty} \frac{\sigma(n)}{n \log \log n} = e^{\gamma},$$ and the two variants $$\limsup\_{\begin{smallmatrix} n\to\infty\cr n\ \text{is square free}\end{smallmatrix}} \frac{\sigma(n)}{n \log \log n} = \frac{6e^{\gamma}}{\pi^2}$$ and $$\limsup\_{\begin{smallmatrix} n\to\infty\cr n\ \text{is odd}\end{smallmatrix}} \frac{\sigma(n)}{n \log \log n} = \frac{e^{\gamma}}{2}$$ that have been proven [here](http://www.mpim-bonn.mpg.de/preprints/send?bid=2960), are there any other similar statements known?
| https://mathoverflow.net/users/2525 | Variants of Grönwall's theorem | One example possessing a limit is the colossally abundant numbers of Alaoglu and Erdos,
<http://en.wikipedia.org/wiki/Colossally_abundant_number>
where the limit of the Choie, Lichiardopol, Moree and Sole's
$$f\_1(a\_n) = \frac{\sigma(a\_n)}{a\_n \log \log a\_n}$$
is the same
$$ e^\gamma .$$
That is, the limit for these numbers is the lim sup for all numbers.
These are more natural than people realize. There is a simple recipe that takes some $ \epsilon > 0$ and gives an explicit factorization for the best value $n\_\epsilon;$ see page 7 in the Briggs pdf
"Notes on the Riemann hypothesis and abundant numbers" at the bottom of the Wikipedia entry. The exponent of a prime $p$ in the factorization of $n\_\epsilon$ is
$$ \left\lfloor \log\_p \left( \frac{p^{1 + \epsilon} - 1}{p^\epsilon -1} \right) \right\rfloor - 1 $$
The process of making a sequence of "champion" numbers this way was invented by Ramanujan.
| 2 | https://mathoverflow.net/users/3324 | 28328 | 18,512 |
https://mathoverflow.net/questions/28347 | 2 | Roth's theorem states that for every real algebraic $\alpha$ and $\epsilon>0$, there is a $c>0$ such that
$$|\alpha -\frac{p}{q}| > \frac{c}{q^{2+\epsilon}}.$$
Lang's conjecture strengthened this to
$$|\alpha -\frac{p}{q}| > \frac{c}{q^2 (\log q)^{1+\epsilon}}.$$
A naive further strengthening would be to ask for
$$|\alpha -\frac{p}{q}| > \frac{c}{q^2},$$
and this is satisfied by all $\alpha$ with bounded
partial quotients $a\_n$ (Khinchin Thm 23).
Life would be nice and simple if it were true that all algebraic $\alpha$ has bounded $a\_n$. This seems too much to be asking for but I am ignorant of any counterexample. Does anyone knows a counter example or even some heuristic reasoning why this is unlikely to be true ? Thanks in advance.
| https://mathoverflow.net/users/1894 | Is it possile for all real algebraic numbers to have continued fractions with bounded partial quotients ? | There is no example of an algebraic number of degree $> 2$ for which the boundedness or not of the entries of the continued fraction has been determined. In a 1976 Annals of Math paper, Baum & Sweet treat the analogous problem with the ring of rational integers replaced by $\mathbb{F}\_ 2[x]$ (and its infinite place). They found algebraic quantities over $\mathbb{F}\_ 2(x)$ contained in $\mathbb{F}\_ 2((x))$ of degree $> 2$ whose "continued fraction" entries have bounded degree, and others with unbounded degree. So it all looks a bit puzzling.
| 10 | https://mathoverflow.net/users/6773 | 28348 | 18,524 |
https://mathoverflow.net/questions/28345 | 6 | Iwaniec and Friedlander wrote a short survey article for the notices of the AMS, entitled "What is the Parity Phenomenon?"
<http://www.ams.org/notices/200907/rtx090700817p.pdf>
At the end of the article they refer to a young mathematician:
"Sometimes it almost seems as though there is a
ghost in the House of Prime Numbers. Perhaps that
will be ruled out some day. There are suggestions
of a youngster who might do this, one who will
come from the Automorphic Room of the house."
Does anyone know who this mathematician is?
Thanks!
| https://mathoverflow.net/users/2547 | Who is the Youngster in the Automorphic Room? | They wrote a mixed technical summary and allegory, something of a prose poem. The allegorical part is concentrated in three paragraphs. These are the second paragraph, the last paragraph, and one in the middle in which people in the "Analytic Room" regard their methods as recent in that Euler is only about three hundred years old. The youngster is evidently anyone from the "Automorphic Room." So the suggestion is that progress is likely to come from automorphic methods, and no specific mathematician is indicated. My best guess on the "ghost" is the possibility of an "exceptional" character whose $L$-function could possess a bad zero. This is from the next to last paragraph, after mention of the "Algebraic Room."
Overnight my impression of the piece clarified a bit, from a pretty vague sense that the inhabitants in the Rooms were not specific people, even from the distant past. I am close to Unreasonable Sin's comment of just an hour ago. I think the inhabitants of the Rooms are ideas more than specific theorems, techniques, particular papers. So the Youngster is an idea. Note Victor Protsak's comment of ten hours ago, I think he got it exactly right. Victor is a smart man. Heed him.
For examples of this style, see the fantasy short story collections of Lord Dunsany,
<http://en.wikipedia.org/wiki/Edward_Plunkett,_18th_Baron_of_Dunsany>
especially, from 1905, The Gods of Pegana
<http://en.wikipedia.org/wiki/The_Gods_of_Peg%C4%81na>
evidently available online in full, see for instance
<http://en.wikisource.org/wiki/The_Gods_of_Peg%C4%81na#The_Chaunt_of_the_Priests>
| 13 | https://mathoverflow.net/users/3324 | 28352 | 18,528 |
https://mathoverflow.net/questions/28295 | 8 | I have come across this inequality in the paper "Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type" <http://www.math.msu.edu/~fedja/Published/paper.ps> by Nazarov and he calls it by the name of Salem Inequality (which according to him is well known, but I can't find a reference).
If I have understood it correctly, the Inequality says that if $p$ is an exponential polynomial whose exponents are well separated, then the average value of square of the modulus of $p$ over a sufficiently large interval dominates the sum of the square of the modulus of its coefficients.
Let $p(t) = \Sigma\_{k=1}^n c\_k e^{ i \lambda\_k t}$, where $ \lambda\_1<\lambda\_2\dots<\lambda\_n \in \mathbb R$ and $\lambda\_k$'s satisfies a separation condition i.e., $\lambda\_{k+1}-\lambda\_k \geq \Delta >0$. Let $I$ be an interval of length bigger than $4\pi / \Delta$, then $$\sum\_{k=1}^{n} |c\_k|^2 \leq \frac{4}{|I|} \int\_I |p(t)|^2 dt. $$
How can one prove this Inequality? This surely would have a lot of applications (and as he says must be well known – maybe under a different name?). I would appreciate some references to such inequalities in general. Also I find it curious that the length of the interval does not seem to depend on $n$ and depends only on $\Delta$.
| https://mathoverflow.net/users/6766 | Salem Inequality | Following a cue from Wadim, this inequality is Theorem 9.1 in Chapter 5 of Zygmund's *Trigonometric series*, vol 1. Note that although the book is mostly dealing with trigonometric series, the proof is given for general lacunary $\lambda\_k.$ (Salem was a good friend of Zygmund's; see the preface to the book.)
| 8 | https://mathoverflow.net/users/5740 | 28353 | 18,529 |
https://mathoverflow.net/questions/28354 | 5 | There is a nice tool in representation theory, the Howe duality, which as I know works for certain pairs of classical Lie algebras (the reference to the complete list of Howe dual pairs is appreciated very much). Is there any extension of the Howe duality for exceptional algebras?
| https://mathoverflow.net/users/2052 | Howe duality for exceptional algebras | There is a theory of dual reductive pairs and examples for exceptional Lie algebras.
For example, in $E\_8$ we have dual reductive pairs $(A\_1,E\_7)$, $(A\_2,E\_6)$, $(G\_2,F\_4)$, $(D\_4,D\_4)$. These are used implicitly in constructions of $E\_8$; for example $(G\_2,F\_4)$
corresponds to the Freudenthal-Tits construction and $(D\_4,D\_4)$ corresponds to the triality construction.
A reference is:
MR1264015 (95b:17010) Rubenthaler, Hubert
Les paires duales dans les algèbres de Lie réductives.
(French) [Dual pairs in reductive Lie algebras]
Astérisque No. 219 (1994), 121 pp.
These seem to be of interest from the point of view of automorphic forms which I know nothing about.
| 3 | https://mathoverflow.net/users/3992 | 28359 | 18,532 |
https://mathoverflow.net/questions/28299 | 9 | Consider simple, undirected [Erdős–Rényi](http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93R%C3%A9nyi_model) graphs $G(n,p)$, where $n$ is the number of vertices and $p$ is the probability for each pair of vertices to form an edge. Many properties of these graphs are known - in particular, $G(n,p)$ is almost surely connected when $p \gt (1 + \epsilon)\frac{\log(n)}{n}$, and the largest clique in $G(n, \frac{1}{2})$ is almost surely about $2\log\_2(n)$.
**What is known about the vertex connectivity number $\kappa(G)$, $G\in G(n,p)$, the minimum number of vertices that one must remove in order to disconnect the graph?**
It is known that for fixed $k$ and fixed $p\in (0,1)$, almost every graph in $G(n,p)$ is $k$-connected, but what is the expected connectivity as a function of $p$ and $n$?
| https://mathoverflow.net/users/3920 | Vertex connectivity of random graphs? | The expected connectivity cannot be higher than the expected minimal degree, which jumps to roughly $pn$ after getting into the range $p\gg\frac{\log n}{n}$. On the other hand, sloppily counting potential clusters of size $m < n/2$ that have boundaries of less than $k$ vertices gives a probability of $\binom{n}{m}\binom{n-m}{k}(1-p)^{m(n-m-k)}$, which is for $k \ll n$ decreasing in $m$ up to $m\approx \frac{n-k}{2}$ and increasing after that value, so we can get an estimate by considering only $m=1$ (checking for vertices with at most $k$ neighbours) and $m=\frac{n}{2}$:
$$
\binom{n}{n/2}\binom{n/2}{k}(1-p)^{n(n-2k)/4} < \exp(n \log 2+k \log n - pn(n-2k)/4) <
$$
$$
< \exp(n \log 2 - pn(\frac{n}{4}-\frac{k}{2}-\log n)) < \exp(-\frac{n \log n}{4} + n \log 2 +2(\log n)^2),
$$
this latter number tending to $0$ fast enough to ignore it. So, the expected connectivity is the expected minimal degree and is roughly $pn$ once $p$ exceeds $\log n/n$. Do you need the behaviour of expected connectivity specifically in this region?
| 5 | https://mathoverflow.net/users/2368 | 28366 | 18,536 |
https://mathoverflow.net/questions/28362 | 1 | The starting point is that it is known that the Hankel determinants for the Catalan sequence give the number of nested sequences of Dyck paths. I would like to promote this to symmetric functions.
This is motivated by some representation theory.
The naive idea is to start with the sequence of symmetric functions $s\_{n,n}$ and take the Hankel determinants using the inner product (that is product in the group ring of $S(2n)$) instead of the usual outer product. However this doesn't make sense.
Take the $2 \times 2$ case. Then the naive determinant is
$$ \left|\begin{array}{cc} s\_{n-1,n-1} & s\_{n,n} \\\ s\_{n,n} & s\_{n+1,n+1}\end{array}\right|$$
The inner product of the two diagonal terms is defined but the inner product of the two off-diagonal terms is not.
The idea that I want to test is that this is $\sum\_\lambda s\_\lambda$ where the sum is over the conjugates of the partitions $4^a2^{n-2a}$.
Any suggestions on how to fix this? If this does get fixed then I would like to know how to calculate the result. The difficulty is that I have not seen an implementation of the inner product in the computer algebra systems I use, Magma and Sage (which I think both use the same source for symmetric functions).
| https://mathoverflow.net/users/3992 | Hankel determinants of symmetric functions | It is unlikely to obtain for such a determinant the sum of all Schur functions indexed by partitions of $2n$ with four parts all even or all odd. Indeed, this sum is already equal to the inner product $s\_{n,n}\ast s\_{n,n}$ (see [arXiv:0809.3469](https://arxiv.org/abs/0809.3469)).
About your second question: you can compute the inner product of symmetric functions in SAGE using the "kronecker\_product" command. For instance, compute the inner product of Schur functions $s\_{6,2}\ast s\_{5,3}$ as follows:
s=SymmetricFunctionAlgebra(QQ,basis='schur')
s([6,2]).kronecker\_product(s([5,3]))
You may also use Maple with John Stembridge's package SF and the command "itensor".
| 1 | https://mathoverflow.net/users/6768 | 28368 | 18,538 |
https://mathoverflow.net/questions/28361 | 9 | My motivation is to understand the following situation: Given absolutely and almost simple algebraic group $G$ defined over a number field $k$ and a finite valuation $v$ on $k$, when $G(k\_v)$ can be compact (with respect to the $p$-adic topology)?
I more or less understand that if $G=SL\_1(D)$ where $D$ is a division ring of dimension $n^2$ and of order $n$ in the Brauer group over $Q\_p$ then $G(Q\_p)$ is compact. I also understand that $Spin(q)$ when $q$ has more than $5$ variable cannot be compact over local non-archimedean fields.
Are there more examples? I think that one can classify all the examples but I don't manage to do it or find a reference for it.
Can someone outline a route to take in order to understand it thoroughly (for someone with basic understanding of algebraic groups and Galois cohomology)?
**Thanks a lot!**
| https://mathoverflow.net/users/6836 | Compact simple simply connected algebraic groups over $Q_p$ or other local non-archimedean fields | Here are some remarks on the answers of Charles Matthews, Kevin Buzzard, and Victor Protsak. For justification of Kevin Buzzard's claim, see G. Prasad, "An elementary proof of a theorem of Bruhat-Tits-Rousseau and of a theorem of Tits" from Bull. Soc. Math. France 110 (1982), pp. 197--202, for an incredibly elegant and short proof that over any henselian valued field $F$, a connected reductive $F$-group $G$ is $F$-anisotropic if and only if $G(F)$ is "bounded" (a property defined in terms of a choice of closed immersion of $G$ into an affine space over $F$, the choice of which doesn't matter; this is meaningful for any affine $F$-scheme of finite type and equivalent to compactness when $F$ is locally compact). Platanov-Rapinchuk has a universal assumption that all fields of characteristic 0 (except when they're finite), so unfortunately that reference is insufficient for uniform arguments over all non-archimedean local fields. I suppose (near?-)circularity (suggested by Victor Protsak) is a more serious issue. :)
There remains the matter of determining, for locally compact non-archimedean $F$ and connected reductive $F$-groups, precisely when the $F$-anisotropic case can actually occur. As Victor Protsak mentioned, via Bruhat-Tits theory one sees that for connected semisimple $F$-groups which are *absolutely simple* and *simply connected* over a non-archimedean local field $F$ (i.e., $G$ a simply connected $F$-form of a Chevalley group), such forms never exist away from type A, and in type A the $F$-anisotropic examples are precisely the $F$-groups of norm-1 units of central simple algebras over $F$. (Note the contrast with the case $F = \mathbb{R}$, for which there's always a "compact form" of any Chevalley type.)
Let me now briefly explain why this handles the general connected reductive case, by a standard kind of argument with central isogenies and separable Weil restriction. (This is explained also in the article [2] of Tits referenced in Charles Matthews' answer.) If $f:G' \rightarrow G$ is a (possibly inseparable) central $F$-isogeny between connected reductive $F$-groups then the preimage of an $F$-torus of $G$ is an $F$-torus of $G'$ (since maximal tori in $G'$ are their own functorial centralizers, so $\ker f$ is of multiplicative type, nothing funny happens when $f$ is not separable). Since $F$-anisotropcity of an $F$-torus is invariant under $F$-isogenies (as we see using the $F$-rational character group, or more direct arguments), it follows that $G$ is $F$-anisotropic if and only if $G'$ is. (This argument has the advantage of working over any field $F$, in contrast with a direct attack on the topology of rational points by using finiteness theorems for Galois cohomology of connected reductive groups.) Thus, by considering an arbitrary connected reductive $F$-group $G$ and letting $G'$ denote the product of its maximal central $F$-torus and the simply connected central cover of the derived group $D(G)$, we see that the problem comes down to the simply connected case.
But in the *simply connected* semisimple case, the general structure of connected semisimple groups over fields (in terms of central isogenous quotient of direct product of commuting simple "factors") implies that $G$ is uniquely a direct product of commuting $F$-simple connected semisimple $F$-groups, each of which is simply connected, so we may assume $G$ is $F$-simple. Then by an elementary result of Borel and Tits (6.21 in "Groupes reductif", IHES), $G = {\rm{Res}}\_ {F'/F}(G')$ for a finite separable extension $F'/F$ and a connected semisimple $F'$-group $G'$ that is *absolutely* simple and simply connected. By the good behavior of Weil restriction with respect to the formation of the *topological* group of rational points, it follows that the equality ${\rm{Res}}\_ {F'/F}(G')(F) = G'(F')$ of abstract groups is a homeomorphism, so we can replace $(G,F)$ with $(G',F')$ to reduce to the case when $G$ is also absolutely simple, the case addressed by Victor Protsak above.
(A more algebraic argument with Galois descent relating maximal $F'$-tori in $G'$ and maximal $F$-tori in its Weil restriction to $F$ shows the equivalence of anisotropicity for $G'$ and its Weil restriction through the finite separable $F'/F$, where $F$ can be taken to be any field at all.)
Conclusion: for non-archimedean local $F$, the $F$-anisotropic connected reductive $F$-groups are precisely the central quotients of products of an $F$-anisotropic torus and groups of norm-1 units of central division algebras over finite separable extensions of $F$.
| 13 | https://mathoverflow.net/users/6773 | 28377 | 18,543 |
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