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https://mathoverflow.net/questions/30647 | 4 | **Background**
Let $m,n$ be positive integers and consider the cyclic group $\mathbb{Z}\_{mn}$.
We have a natural epimorphism $\mathbb{Z}\_{mn} \to \mathbb{Z}\_n$ coming from the exact sequence
$$ 0 \to \mathbb{Z}\_m \to \mathbb{Z}\_{mn} \to \mathbb{Z}\_n \to 0 $$
where the homomorphism $\mathbb{Z}\_m \to \mathbb{Z}\_{mn}$ is multiplication by $n$.
Now let $\ell$ be another positive integer and consider the cyclic groups $\mathbb{Z}\_{\ell n}$ and $\mathbb{Z}\_{m n}$ with their respective epimorphisms to $\mathbb{Z}\_n$. Then the *fibered product* $\mathbb{Z}\_{\ell n} \times\_{\mathbb{Z}\_n} \mathbb{Z}\_{m n}$ is defined to be the subgroup of $\mathbb{Z}\_{\ell n} \times \mathbb{Z}\_{m n}$ consisting of elements $(a,b)$ with $a=b \mod n$.
(In less evil language, this is the categorical pullback of one of the two epimorphisms along the other.)
Now let $r \in \mathbb{Z}\_n^\times$ be a multiplicative unit, so that multiplication by $r$ defines an automorphism of $\mathbb{Z}\_n$, and consider modifying one of the epimorphisms above by post-composing with multiplication by $r$. This gives rise to a different (in general) fibered product
$$\mathbb{Z}\_{\ell n} \times\_{(\mathbb{Z}\_n,r)} \mathbb{Z}\_{m n} = \lbrace (a,b) \in \mathbb{Z}\_{\ell n} \times \mathbb{Z}\_{m n} \mid a = r b \mod n \rbrace,$$
which agrees with the one above when $r=1$.
**Question**
>
> Is there an easy way to determine the isomorphism type of $\mathbb{Z}\_{\ell n} \times\_{(\mathbb{Z}\_n,r)} \mathbb{Z}\_{m n}$, say as a product of cyclic groups?
>
>
>
In some special cases ($n=1$, $k = \ell$,...) one can do this without too much effort, but I was wondering whether there is a general expression. I would also be happy with any pointers to the literature. I'm also interested in fibered products of other finite groups (not necessarily cyclic) and was wondering whether it would be sensible to spend some time determining them up to isomorphism.
| https://mathoverflow.net/users/394 | Fibered products of cyclic groups | The group $G=\mathbb{Z}\_{\ell n} \times\_{(\mathbb{Z}\_n,r)} \mathbb{Z}\_{m n}$
is isomorphic to $\mathbb{Z}\_{an} \times\mathbb{Z}\_b$
where $a$ is the least common multiple of $\ell$ and $m$ and $b$ is
their highest common factor.
There is $r'\in\mathbb{Z}\_{\ell n}^\times$ with $r'\equiv r$ (mod $n$).
Then $g=(r',1)$ has order $an$ in $G$ which is then easily seen to have exponent
$an$. Then $\langle g\rangle$ is a direct summand of $G$.
As $G$ is a quotient of $\mathbb{Z}^2$ then it is the product of two
cyclic groups and so the other one must be $\mathbb{Z}\_b$ where $b=|G|/(an)
=\ell m/a$
is the highest common factor of $\ell$ and $m$.
| 6 | https://mathoverflow.net/users/4213 | 30656 | 19,955 |
https://mathoverflow.net/questions/30664 | 5 | Let $\phi\in C^\infty\_c(\mathbb R)$ be a smooth function with *compact support*.
For $h>0$ define the difference quotient $\phi\_h\in C^\infty\_c(\mathbb R)$ by $\phi\_h(t)=\dfrac{\phi(t+h)-\phi(t)}{h}$.
By definition, for fixed $t\in\mathbb R$, we have $\phi\_h(t)\to\phi'(t)$ as $h\to 0$.
>
> Question: Can we conclude, that $\phi\_h\to\phi'$ uniformly on $\mathbb R$ (as $h\to 0)$?
>
>
>
Motivation: This is used in a proof of Stone's Theorem on the existence of generators of operator groups I'm trying to understand.
| https://mathoverflow.net/users/1291 | Uniform convergence of difference quotient | By Taylor's theorem
$$\phi(t+h)=\phi(t)+h\phi'(t)+h^2\phi''(t+u(h,t)h)/2$$
where $0\le u(h,t)\le1$. So
$$\phi\_h(t)=\phi'(t)+h\phi''(t+u(h,t)h)/2.$$
As $\phi''$ is in $C^\infty\_c$ it's pretty clear that $\phi\_h\to\phi'$
uniformly.
| 4 | https://mathoverflow.net/users/4213 | 30667 | 19,962 |
https://mathoverflow.net/questions/30655 | 9 | There is a conjecture by Birkhoff which claims that for a simple closed $C^2$ plane curve $C$, if the billiard ball map is integrable then the curve is an ellipse.
Integrability here might be formulated as follows: there exists a neighbourhood of $C$ in the interior $Int(C)$ that is foliated by caustics (caustics being curves that are everywhere tangent to a given trajectory of the billiard ball).
I would be interested to know the current status (and progresses, if there are) of this conjecture.
| https://mathoverflow.net/users/7031 | Birkhoff conjecture about integrable billiards | I haven't heard of any recent breakthroughs. The strongest result that I know is due to [Misha Bialy](https://doi.org/10.1007/BF02572397):
>
> **Theorem.** If almost every phase point of the billiard ball map in a strictly convex billiard table belongs to an invariant circle, then the billiard table is a disc.
>
>
>
Stronger results are available for an outer version of the Birkhoff conjecture. Tabachnikov
proved that if the outer billiard map around a plane oval is algebraically integrable then the oval is an ellipse ([article](http://pjm.math.berkeley.edu/pjm/2008/235-1/pjm-v235-n1-p07-s.pdf), [arXiv version](https://arxiv.org/abs/0708.0255)).
| 5 | https://mathoverflow.net/users/5371 | 30668 | 19,963 |
https://mathoverflow.net/questions/30635 | 6 | All rings below are commutative with $1$.
Suppose $A\subset B$ is a subring and that $A\rightarrow A'$ is a faithfully flat ring homomorphism. [You may assume the rings are actually ${\mathbb C}$-algebras if it helps.] Suppose that $B' = A'\otimes\_A B$ is a localization of $A'$, i.e. there is a multiplicatively closed subset $S$ of $A'$ such that $B' = S^{-1}A'$. Must $B$ be a localization of $A$?
I find it hard to believe that the answer is "yes." But I'm having a mental block coming up with an example to show that it's "no."
| https://mathoverflow.net/users/2628 | Checking locally whether a homomorphism is a localization | Let $A$ be the coordinate ring of a smooth affine curve $X$ over $\mathbb C$, and let $p$ be a point of infinite order in the class group of $A$. Let $B$ be the coordinate ring of $X \smallsetminus \{p\}$, and let $C$ be the coordinate ring of an open subscheme $U$ of $X$ containing $p$ such that $p$ is principal in $U$. Set $A' = B \times C$. Then it it is easy to see that $A' \otimes\_A B$ is a localization of $A'$, while $B$ is not a localization of $A$.
| 5 | https://mathoverflow.net/users/4790 | 30670 | 19,964 |
https://mathoverflow.net/questions/30669 | 21 | Sometimes, dealing with the concrete and familiar Banach spaces of everyday life in maths, I happen nevertheless to ask myself about the generality of certain constructions. But, as I try to abstract such constructions up to the level of general Banach spaces, I immediately feel the cold air of wilderness coming from the less known side of this huge category. One typical difficulty is, that it's not at all obvious how to find, or to prove the existence, of (bounded linear) operators $T:E\to E.$
Certainly, there are always a lot of finite rank operators, furnished by the Hahn-Banach theorem, and their norm-limits. However, in an anonymous infinite dimensional Banach space, I do not see how to guarantee the existence of bounded linear operators different from a compact perturbation of $\lambda I:$ hence my questions:
Are there always other operators than $\lambda I+K $ on an infinite dimensional Banach space?
(in other words, can the Calkin algebra be reduced to just $\mathbb{C}\\ $)? More genarally, starting from a operator $T$, is there always a way to produce new operators different from a compact perturbations of the norm-closed algebra generated by $T$? Is there a class of Banach spaces that are rich of operators in some suitable sense (spaces with bases, for instance, but more in general?).
Thank you!
| https://mathoverflow.net/users/6101 | Banach spaces with few linear operators ? | Examples were constructed (about two years ago?) by [Argyros and Haydon](http://arxiv.org/abs/0903.3921). See this [blog post](http://gowers.wordpress.com/2009/02/07/a-remarkable-recent-result-in-banach-space-theory/) for some non-technical discussion. It seems worth noting, as one is almost obliged to, that the space originally constructed by A & H is an isomorphic predual of $\ell\_1$.
| 17 | https://mathoverflow.net/users/763 | 30671 | 19,965 |
https://mathoverflow.net/questions/30646 | 10 | If I'm not mistaken, it was in his seminal paper “An Essentially Undecidable Axiom System”, published in
Proceedings of the International Congress of Mathematics (1950), 729–730,
where R.M. Robinson proved that Gödel Incompleteness Theorem still applies to Peano Axioms if we drop the induction schema (hence showing that infinite axiomatization is not necessary for essential undecidability), in what we now call Robinson Arithmetic.
I would like to know:
* Is actually this paper what I should be looking for?
* Can it be found anywhere on the net? (I already tried on MathSciNet, SpringerLink, JSTOR and Google Scholar, without success)
* Can anyone pinpoint to closely related, or at least similar, accessible papers?
(Note: I already have the book "Undecidable theories", which he published in collaboration with Tarski, but I'd prefer to locate papers about 'Robinson theory', specifically).
| https://mathoverflow.net/users/1234 | How to locate the paper that established Robinson Arithmetic? | Hi Jose,
it's in the British library collection:
<http://snurl.com/z16ud>
Haven't checked what the fees are, but you could order it from there.
Alternatively, you could try the LMS:
<http://www.lms.ac.uk>
A good chance they will have the procs in their library, and you can get photocopies for a nominal fee.
Several other similar alternatives too.
| 3 | https://mathoverflow.net/users/7248 | 30686 | 19,976 |
https://mathoverflow.net/questions/30661 | 66 | What are nice examples of topological spaces $X$ and $Y$ such that $X$ and $Y$ are **not** homeomorphic but there do exist continuous bijections $f: X \to Y$ and $g: Y \to X$?
| https://mathoverflow.net/users/2060 | Non-homeomorphic spaces that have continuous bijections between them | Recycling an old (ca. 1998) sci.math post:
" Anyone know an example of two topological spaces $X$ and $Y$
with continuous bijections $f:X\to Y$ and $g:Y\to X$ such that
$f$ and $g$ are not homeomorphisms?
Let $X = Y = Z \times \{0,1\}$ as sets, where $Z$ is the set of integers.
We declare that the following subsets of $X$ are open for each $n>0$.
$$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$
This is a basis for a topology on $X$.
We declare that the following subsets of $Y$ are open for each $n>0$.
$$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$
This is a basis for a toplogy on $Y$.
Define $f:X\to Y$ and $g:Y\to X$ by $f((n,i))=(n,i)$ and $g((n,i))=(n+1,i).$
Then $f$ and $g$ are continuous bijections, but $X$ and $Y$ are not homeomorphic.
This example is due to G. Paseman.
David Radcliffe "
More generally, take a space X with three successively
finer topologies T, T' and T''. Form two spaces which have underlying
set ZxX, and "form the infinite sequences" .... T T T T' T'' T'' T'' ....
and ... T T T T T'' T'' T'' T'' .... The continuous maps will take a finer
topology in one sequence to a rougher topology in the other. You can
make them bijective, and show that they are obviously non-homeomorphic
for a judicious choice of X, T, T', and T''.
Gerhard "Ask Me About System Design" Paseman, 2010.07.05
| 31 | https://mathoverflow.net/users/3528 | 30695 | 19,982 |
https://mathoverflow.net/questions/30659 | 18 | I am a 19 yr old student new to all these ideas. I made the transformation $X(z)=\sum\_{n=1}^\infty z^n/n^2$. Therefore $X(1)=\pi^2/6$ as we all know (it is $\zeta(2)$). To calculate $X(1)$, I integrated
$$\frac{Y(z)}{z}=\sum\_{n=1}^\infty \frac{z^{n-1}}{n} $$
between 0 to 1;
We all know $Y(z)=-\log(1-z)$; so doing integral of $Y(z)/z$ from 0 to 1, I get
$$X(1)=\int\_0^1\ln\frac{z}{z-1}dz$$
that is $\pi^2/6$; Now integrating $X(z)/z$ between 0 to 1, I get $\sum\_{n=1}^\infty 1/n^3$;
therefore performing that integral I get $\sum1/n^3$ as
$$\int\_0^1 \frac{(\log x)^2 dx}{1-x};$$
which Iam unable to do.
So can anyone give an idea of how to find
$$\sum\_{n=1}^\infty\frac1{n^3}=\zeta(3)=\int\_0^1 \frac{(\log x)^2 dx}{1-x};$$
given
$$\zeta(2)=\int\_0^1 \frac{\log x\ dx}{1-x}=\frac{\pi^2}{6}.$$
Please take time to read all these things and help me out by giving some suggestion.
| https://mathoverflow.net/users/7346 | Establishing zeta(3) as a definite integral and its computation. | Dear Vamsi,
Unlike the special values $\zeta(2n)$ (for $n \geq 1$), which are known to be simple algebraic expressions in $\pi$ (in fact just rational multiples of $\pi^{2n}$), it is conjectured (but not known) that the values $\zeta(2n+1)$ are genuinely new irrationalities (and that in fact
each is genuinely different from the other); more precisely, they are conjectured to be algebraically independent of one another, and of $\pi$. There is no prior, classical, name
for these numbers, and in particular you should not expect to be able to evaluate your integral in terms of any numbers whose names you already know.
There is a theoretical basis for this conjecture: the kind of integrals that you are computing are callled "period integrals" (if you search, you will find a few other MO questions about periods, in this sense), and a general philosophy is that period integrals should have no more relations between them than those that are implied by elementary manipulations of integrals
(of the type that you made to compute $\zeta(2)$; and here I don't mean elementary in a disparaging sense, just in the sense of standard rules for computing integrals). In fact, period integrals are manifestations of underlying geometry (which I won't get into here; all I will say is that the geometry relevant to zeta values is the geometry of "mixed Tate motives"). One *can* show that the geometric objects underlying the odd zeta values are independent of one another, in a suitable sense, and of the geometric object underlying $\pi$ (which is basically the circle); what is missing is a proof that the period integrals faithfully reflect the underlying geometry (so that independence in geometry implies independence of period integrals). This is one of the big conjectures in contemporary arithmetic geometry and number theory, and so your question, which is a very nice one, is touching on some very fundamental (and difficult) mathematical issues.
Good luck as you continue your studies!
Best wishes,
Matthew Emerton
| 52 | https://mathoverflow.net/users/2874 | 30698 | 19,984 |
https://mathoverflow.net/questions/30662 | 9 | I am looking for a (Hausdorff or better) space $X$ and a subset $A$ of $X$ that is relatively countably compact (every sequence from $A$ has an accumulation point in $X$) such that the closure of $A$ is not countably compact.
It is known that in many "nice" spaces such examples do not exist (a classical case being normed spaces in their weak topology).
Edit: for $T\_4$ spaces this cannot happen, as the closure of a relatively countably compact subset is pseudocompact (Suppose A is relatively countably compact.
If f from cl(A) to R is unbounded then f|A is unbounded as well, as A is dense in cl(A).
So we can find {x\_n: n in N} in A such that |f(x)| >= n.
Let p in cl(A) be an accumulation point of this set, by A being relatively countably compact.
Then continuity at f implies that |f(x\_n)| <= |f(p)| + 1, for all but finitely many n.
This contradicts the choice of the x\_n, contradiction.)
So cl(A) is pseudocompact, and hence in a normal space, countably compact.
This explains the properties of the example given below.
| https://mathoverflow.net/users/2060 | Relatively countably compact subsets without countably compact closure. | Let $X$ be the space
$(\omega+1)\times(\omega\_1+1)-\{(\omega,\omega\_1)\}$, putting the transfinite order topology on each coordinate and the product topology on the whole space.
```
(0,omega_1) (1,omega_1) (2,omega_1) ---> O
: : :
: : : :
: : : :
: : : :
(0,alpha) (1,alpha) (2,alpha) ---> (omega,alpha)
: : : :
: : : :
(0,1) (1,1) (2,1) ---> (omega,1)
(0,0) (1,0) (2,0) ---> (omega,0)
```
Thus, the space consists of $\omega\_1+1$ many copies of a (horizontally) convergent sequence, stacked on top of each other, but with the limit of the final sequence omitted (at O in the diagram). Horizontally, the $\alpha$-th row has $(n,\alpha)$ converging to $(\omega,\alpha)$ as $n\to\omega$. Vertically, the $n$-th column has a transfinite sequence $(n,\alpha)$ converging to $(n,\omega\_1)$ as $\alpha\to\omega\_1$. This is a Hausdorff space, and comparatively nice in several ways.
Let $A=(\omega+1)\times\omega\_1\subset X$ be the lower portion of this diagram, without the top row. Any countable sequence from $A$ is bounded below a countable ordinal
and has accumulation points in $X$, and in fact in $A$ itself: either the sequence contains
infinitely many points from one coordinate, in which case
it accumulates on the limit supremum in that coordinate, or it
contains points from infinitely many coordinates, which
will accumulate on a point of the form $(\omega,\alpha)$ at
the right-most column in $X$. In particular, $A$ already contains all its limit points of sequences in $A$. Nevertheless, the closure of $A$ in $X$ is
all of $X$, and this includes the top row of points of the
form $(n,\omega\_1)$. But this sequence has no accumulation points
in $X$, as desired.
| 9 | https://mathoverflow.net/users/1946 | 30701 | 19,987 |
https://mathoverflow.net/questions/30611 | 3 | Let $E=\mathbb{C}/L$ be an elliptic curve. Then $\mathbb{C}$ is contractible, and $L$ is the fundamental group of $E$. What's interesting is that we can find the cohomology of $E$, which is the same as the sheaf cohomology $H^i(E,\mathbb{Z})$ for the constant sheaf $\mathbb{Z}$, by taking the group cohomology of $L$ with trivial action on $\mathbb{Z}$. What's more interesting is that if $\mathcal{O}^\times$ is the sheaf of invertible holomorphic functions on $E$, then we can find the sheaf cohomology of this by considering the group cohomology of the action of $L$ on the group of invertible holomorphic functions on $\mathbb{C}$ defined by $(\omega f)(z)=f(z+\omega)$ for $\omega \in L$. In the case of $H^0$, this is obvious, since the holomorphic functions which are fixed by $L$ ($H^0$ in group cohomology), which is the same as invertible holomorphic functions on $E$, which is the same as the global sections of the sheaf on $E$ (and this is true for many other sheaves). In the case of $H^1$, one can see that both are the group of invertible line bundles on $E$.
My question is: why is the group ($L$-module) of holomorphic functions on $\mathbb{C}$ the natural one to consider? In this case, it's clearer, since we know what holomorphic functions are. If in general we have a covering map $U \to X$ for some space $X$, with $U$ a contractible universal cover, then given a sheaf on $X$, how do we know what sheaf on $U$ to consider? Specifically, which sheaf on $U$ gives the sheaf cohomology when we take the group cohomology of its global sections?
| https://mathoverflow.net/users/1355 | Relation between sheaf and group cohomology | I doubt that in general one can construct a reasonable sheaf on $U$ with the required properties. To see what kind of bad things can happen, let us try to understand why this works for $X$ an elliptic curve and the sheaf $\cal{O}^{\times}$ on it.
We have the derived global sections functor from the $D^b$ of sheaves on $X$ to the $D^b$ of sheaves on a point, i.e. graded vector spaces. But given a sheaf $F$ on $X$ we can compute its global sections in a roundabout way: we can first take the pullback to $U$, then take global sections and then take the $G$-invariants where $G=\pi\_1(X)$. Passing to the derived categories we get $$R\Gamma (F)=R(R\Gamma f^{-1}(F))^G$$ where $F\in D^b(X)$, $f:U\to X$ is the projection, $R\Gamma f^{-1}$ is the right derived functor of the left exact functor $\Gamma f^{-1}$ and $R(\cdot)^G$ is the right derived functor of the functor of $G$-invariants (this functor goes from the $D^b$ of $G$-modules to graded vector spaces).
We have the Grothendieck spectral sequence that converges to $H^\ast(X,F)$ with the $E\_2$ sheet given by $$E\_2^{p,q}=H^p(G,H^q(U,f^{-1}(F)).$$
Now if $X$ is an elliptic curve, $F=\cal{O}^{\times}$ and $U=\mathbf{C}$, then it follows from the exponential exact sequence that $H^q(U,f^{-1}(F))=0$ for $q\neq 0$, so the above spectral sequence collapses and we get the required isomorphism $H^\ast (X,F)=H^\ast(G,H^0(U,f^{-1}(F))$. This also happens when say $F$ is locally constant and $U$ is contractible. But in general there seems no reason to expect the spectral sequence to collapse, let alone to be concentrated in one row only.
| 4 | https://mathoverflow.net/users/2349 | 30704 | 19,988 |
https://mathoverflow.net/questions/30696 | 5 | How can we prove that the moduli space,$M\_{g}(n)$, of genus $g$ Riemann surface with $n$ boundary components is homotopy equivalent to $M\_{g,n}$, that is ,the moduli space of genus $g$ Riemann surface with $n$ punctures? Thanks! (It is very intuitive, but it seems that I can't make it)
| https://mathoverflow.net/users/2391 | How can we show the spaces $M_{g}(n)$ and $M_{g, n}$ are homotopy equivalent? | A compact Riemann surface of genus $g$ with $n$ boundary components has a unique realization as a hyperbolic surface with geodesic boundary. One may see this by reflecting through the boundary and uniformizing. The uniqueness of the uniformization implies it is invariant under reflection, and therefore the fixed point set is geodesic.
Thus, the moduli space of genus $g$ Riemann surfaces with $n$ boundary components is equivalent to the space of hyperbolic surfaces with totally geodesic boundary. One may now insert a punctured disk into each boundary component, to obtain a Riemann surface with punctures. I don't know of a canonical way to do this, but for example for a boundary component of length $l$, one may attach isometrically the boundary of a punctured Euclidean disk of circumference $l$. The important thing is that this gluing only depends on $l$, and that it induces a conformal structure on the punctured surface. This gives a map between the spaces. Since the mapping class groups are the same, it induces a homotopy equivalence (in the category of orbifolds). Of course, there are some technical details one must carry out to make this argument rigorous. There are several other ways to fill in a punctured disk.
Another possible approach is to use the Weil-Petersson metric on moduli space. One can take the WP nearest point in the Deligne-Mumford compactification of moduli space, which is finite distance away since the WP metric is incomplete. Because the metric is CAT(0), a unique nearest point exists.
| 8 | https://mathoverflow.net/users/1345 | 30706 | 19,990 |
https://mathoverflow.net/questions/29880 | 1 | I would like to draw a curve between two points that minimizes the square of the second derivative integrated along the curve.
$J(y) = \int\_{1}^{0} {y}''^2 dx $
The first derivative for the start and end point are known and must be preserved, and all values on the curve between the start and end point must fall within some range l < e < u.
$y(0) = a, y(1) = b, y'(0) = c, y'(1) = d$
$y(t) = e, \forall t \in [0, 1] \implies l < e < u$
Additionally the curve should be continuous in the interval.
Sincere thanks to Wadim Zudilin for recommending posting an example, I know realize my first post was not even asking the right thing :(
| https://mathoverflow.net/users/7168 | Minimizing functionals constrained in a box | Just to elaborate a bit on what Rahul and I mentioned in the comments.
Take the action functional to be $\int\_0^1 (y'')^2 dt$, with prescribed boundary conditions $y(0) = a$, $y(1) = b$, $y'(0) = c$, $y'(1) = d$. For finding the free evolution, take the variation of the function relative to $y$ and set it to zero. Immediately this gives
$$\int\_0^1 y'' (\delta y)'' dt = 0 $$
for any perturbation. The boundary conditions prescribed implies that $\delta y(0) = \delta y(1) = \delta y'(0) = \delta y'(1) = 0$. So we are allowed to integrate by parts twice (assuming the solution is $C^4$) and obtain $y'''' = 0$, which implies that $y(t)$ is a cubic polynomial in time and thus has 4 free parameters, which we can fix by the boundary values.
The intuition for the bounded case is that, until the evolution hits the wall, locally the equation of motion should be identical to the free evolution. So the solution should be composed piecewise of cubic polynomials. Every time it hits the wall it should receive a hard impact, which suggests that $y''''(\tau\_i) = c\delta\_{\tau\_i}$, the Dirac delta. (The same way that for a hard billiard which away from the walls travel via $x'' = 0$ receive a delta function impact in the second derivative when it hits a wall.) This suggests that $y'''$ is a step function of $t$, and $y''$ is continuous.
For an example, let $l = -1, u = 1$, let the initial time be $t = 0$, and final time be $t = 2$. Let $a = b = 0$, and $c = d = 6$.
First solve the free problem: we want
$$ k\_3 t^3 + k\_2 t^2 + k\_1 t + k\_0 = y(t) $$
Plugging in the values for the four points we find by solving the linear system that the equation should be
$$ y(t) = 3t^3 - 9 t^2 + 6t $$
which achieves its local max and min in $[0,2]$ at $1\pm \sqrt{1/3}$, at which points $|y| = |\pm \sqrt{4/3}| > 1$. So the free evolution is no go.
There should be two break points, but for illustration we try first with one break point. Assume the break point is at $\tau$ where $y(\tau) = 1$. So we have a system of equations using the data at the points $2,0,\tau$ and the cubic ansatz
$$ y |\_{[0,\tau]} = \sum \alpha\_kt^k,\qquad y|\_{[\tau,2]} = \sum \beta\_k t^k$$
which leads to
$$\alpha\_0 = 0, \alpha\_1 = 6, \alpha\_3 \tau^3 + \alpha\_2\tau^2 + 6 \tau = 1, 3\alpha\_3\tau^2 + 2\alpha\_2\tau + 6 = 0$$
and
$$8\beta\_3 + 4\beta\_2 + 2\beta\_1 + \beta\_0 = 0, 12\beta\_3 + 4\beta\_2 + \beta\_1 = 6, \beta\_3\tau^3 + \beta\_2\tau^2 + \beta\_1\tau + \beta\_0 = 1, 3\beta\_3\tau^2 + 2\beta\_2\tau + \beta\_1 = 0$$
which gives
$$\alpha\_0 = 0, \alpha\_1 = 6, \alpha\_2 = -12 / \tau, \alpha\_3 = 6 / \tau^2$$
$$\beta\_3 = (6\tau - 2)/(\tau - 2)^3, \beta\_2 = -3(5\tau^2 - 3\tau + 6) / 2(\tau - 2)^3 $$
Continuity of $y''$ then implies
$$ 6\tau \alpha\_3 + 2\alpha\_2 = 6 \tau\beta\_3 + 2\beta\_2 $$
or
$$ 3\tau^2 + 23 \tau^2 - 42 \tau - 32 = 0$$
which has two negative roots and one positive one at $\tau = 2$ which we can throw out. And thus 1 break point it not enough.
A direct computation (if I did it right, which is not guaranteed) with 2 break points $\sigma\in [1,2]$ and $\tau \in[0,1]$ yields that $\tau = 1/2$ and $\sigma = 3/2$ an admissible pair. It is just linear algebra in the end.
| 0 | https://mathoverflow.net/users/3948 | 30713 | 19,996 |
https://mathoverflow.net/questions/25062 | 10 | So I understand in theory the definition of Ext and Tor, but when it comes to actually computing them, I'm stuck. For example, could someone show me how to compute $\text{Ext}(\mathbb{Z}/m\mathbb{Z}, \mathbb{Z}/n\mathbb{Z})$? I tried this by taking an injective resolution ($0 \to \mathbb{Z}/m\mathbb{Z} \to \mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z} \to 0$?) and I got an exact sequence with $\text{Ext}^1(\mathbb{Z}/m\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) = 0$, but I don't see what to do next?
P.S. No, this is not a homework question.
| https://mathoverflow.net/users/2503 | Examples of computing Ext and Tor functors? | $\mathbb{Q}/\mathbb{Z}$ is a pretty terrible abelian group, or a rather hard one, there may be better injective resolutions to work with. It would certainly be easier to do the projective resolution, use $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/n \to 0$. this will surely be easier to work through than the one involving $\mathbb{Q}/\mathbb{Z}$. Then compute the appropriate tensor product or hom group.
I started learning this stuff on more interesting modules as Schmidt suggests. For example, modules over the group ring of some cyclic group, or maybe an exterior algebra on two generators (if you make the generators of different gradings, in particular 1 and 3). This happens to be the category of modules you need to understand in order to compute complex connective k-theory!
this should help get you going. These computations are very fun!
| 11 | https://mathoverflow.net/users/3901 | 30714 | 19,997 |
https://mathoverflow.net/questions/30709 | 4 | Let $T:X\to Y$ be an operator between Banach spaces $X$ and $Y$. Assume that $X$ has a positive cone $C\subset X$, which generates $X$: every element of $X$ can be written as a difference of elements of the positive cone.
Assume now that we can show $$k\Vert x\Vert\le \Vert Tx\Vert$$ for some
$k>0$ and every $x\in C$.
Question: Are there any natural, general conditions on $T$, $X$ or both that allow to conclude that the image of $X$ under $T$ is closed in $Y$ (the above inequality holds for all $x\in X$, possibly with a different $k$)?
The motivating example is the following. Let $G=(V,E)$ be an infinite (oriented) graph and take $X=\ell\_1(V)$, $Y=\ell\_1(E)$. Let $T$ be the discrete gradient, $Tf(x,y)=f(y)-f(x)$. Then it is enough to check the inequality only for positive $f$, then use the triangle inequality.
Update: Given Andreas' response I realized I should probably ask not for general conditions but any sufficient condition that would give the above property, in particular in special cases like the motivating example above..
| https://mathoverflow.net/users/nan | When can closedness of the range of an operator be checked on a positive cone? | I'm reluctant to say no when the question is "Are there *any* natural, general conditions ..." but I'll take my chances. The following seems to be a counterexample with every nice property I can think of. Let X and Y both be the Hilbert space $\ell^2$, and label an orthonormal basis as $\{e\_n,f\_n:n\in\mathbf N\}$. Let $T$ be the (diagonal) operator that sends each $e\_n$ to itself and sends each $f\_n$ to $f\_n/n$. Note that this isn't closed; its range doesn't contain the vector $\sum\_0^\infty (f\_n/n)$ (because the pre-image would be $\sum\_0^\infty f\_n$ which isn't in $\ell^2$) but it does contain all the finite partial sums $\sum\_0^M (f\_n/n)$ that converge to it. Now let $C$ be the cone consisting of those vectors $\sum\_0^\infty (a\_ne\_n+b\_nf\_n)$ where $|b\_n|\leq a\_n$ for all $n$. On vectors of this form, the inequality $k\Vert x\Vert \leq \Vert Tx\Vert$ holds with $k=1/\sqrt 2$. (To prove it, consider each of the 2-dimensional spaces spanned by one $e\_n,f\_n$ pair.) Finally, each vector in $\ell^2$ is the difference of two vectors from $C$. (Again, to prove it, look in those 2-dimensional subspaces and check that there any vector $v$ is the difference of two vectors from $C$, each having at most twice the norm of $v$. The norm bound is used to assemble the 2-dimensional facts into the corresponding fact for $\ell^2$.)
| 2 | https://mathoverflow.net/users/6794 | 30717 | 19,998 |
https://mathoverflow.net/questions/30716 | 2 | I am interested in the possibility of generating probability distributions using inequality constraints. For instance assume that we have three urns with total of a 10 balls. Thus,
$a + b + c = 10$
Additionally it has the constraint
$a \geq b \geq c$
The allowable triples will correspond to the integer partitions of 10 with width 3, such as:
$(7, 2, 1)$
If you were given a ball and wanted to know the probability that it can from urn $a$, it seems reasonable to sum all the balls in the urn $a$ for all 14 partitions, divided by the total number of balls. In this particular case we would have,
$total \ ball \ count = balls \ per \ partition \* partition \ count = 10 \* 14 = 140$
$Prob(a) = 90 / 140 \approx .64$
So I have a few questions.
First, am I even on the mark? Is this a valid approach for calculating the probability that the ball comes from urn $a$? If so has this approached been discussed in literature, hopefully in more generality?
I am specifically interested in what the asymptotic behavior is as the number of balls and urns increase.
Here is the code I am using in Haskell to calculate probabilities:
<http://hpaste.org/fastcgi/hpaste.fcgi/view?id=27026#a27026>
| https://mathoverflow.net/users/7168 | Inequality constraints, probability distributions, and integer partitions | Whether this is a valid approach for calculating the probability will depend on what assumptions you are making on the probabilities of the various partitions. In particular, this calculation seems to require for its validity that all partitions be equally likely. Do you actually want (10,0,0) to be as likely as (4,3,3)?
Edit: Concerning asymptotic behavior, take the case where the number of balls equals the number of urns; call it $n$. One can ask for the proportion of balls in the first urn as a function of $n$, which is the same as asking for the average size of the biggest part in a partition of $n$ (which is also the same as asking for the average number of parts in a partition of $n$). This was studied by Kessler and Livingston, The expected number of parts in a partition of $n$, Monatshefte Math 81 (1976) 203-212. Let $P(n)$ be the total number of parts in all the partitions of $n$ (which is the same as the sum of the largest parts in all the partitions of $n$), and let $p(n)$ be the number of partitions of $n$. Then $${P(n)\over p(n)}=\sqrt{3n\over2\pi}(\log n+2\gamma-\log(\pi/2))+O(\log^3n)$$ where $\gamma$ is Euler's constant. More information at sequence A006128 at the Online Encyclopedia of Integer Sequences.
EDIT 15 July: Let $f\_{j,k}(n)$ be the sum of the $j$-th biggest parts over all partitions of $n$ into at most $k$ parts, $$f\_{j,k}(n)=\sum\_{\eqalign{a\_1\ge\dots\ge a\_k&\ge0\cr a\_1+\dots+a\_k&=n\cr}}a\_j,$$ and let $p\_k(n)$ be the number of partitions of $n$ into at most $k$ parts. You are asking for the asymptotic behavior of $f\_{j,k}(n)/np\_k(n)$ as $k$ and $n$ increase, but it is not clear how you want them to increase (that is, what relation they should bear to each other as they increase), nor is it clear what you want $j$ to be doing while $k$ and $n$ are increasing. Perhaps you are only interested in the case $j=1$, which is the case in your example.
In any event, you need asymptotics for $p\_k(n)$. From the generating function $$\sum\_np\_k(n)x^n={1\over(1-x)(1-x^2)\cdots(1-x^k)}$$ one can get, for fixed $k$, $p\_k(n)=C\_kn^{k-1}+O(n^{k-2})$, where $C\_k=(k!(k-1)!)^{-1}$ (I hope someone checks my calculations here).
Now I can tell you something about $f\_{k,k}(n)$ - this is the case when you're interested in the $\it last$ urn. Let's write $g\_k(n)=f\_{k,k}(n)$. It's not hard to prove that $g\_k(n+k)-g\_k(n)=p\_k(n)$, from which it follows that $$\sum\_ng\_k(n)x^n={x^k\over(1-x)\cdots(1-x^{k-1})(1-x^k)^2}$$ and then you get, again for fixed $k$, $g\_k(n)=D\_kn^k+O(n^{k-1})$, where $D\_k=k^{-1}(k!)^{-2}$. So, the probability that the ball is from the last urn approaches $k^{-2}$ as $n$ increases with $k$ fixed.
I didn't do so well with the first urn. Let's just take the case $k=3$. We want $$h\_3(n)=f\_{1,3}(n)=\sum\_{\eqalign{a\ge b\ge c&\ge0\cr a+b+c&=n\cr}}a.$$ Starting with $n=1$, this produces the sequence $1,3,6,11,17,27,37,52,69,90,113,144\dots$. I didn't find this sequence in the Online Encyclopedia of Integer Sequences. I think, but can't prove, that the generating function is $$\sum\_nh\_3(n)x^n={x+3x^2+4x^3+3x^4\over(1-x)^2(1-x^3)^2}.$$ This gives $h\_3(n)=(11/216)n^3+O(n^2)$. Together with $p\_3(n)=(1/12)n^2+O(n)$, you get the probability of the ball coming from the first of the three urns converging to 11/18, as $n$ increases (and conditional on my unproved formula for the generating function).
EDIT 8 February 2011: There's a typo in the last display, it should be $$\sum\_nh\_3(n)x^n={x+3x^2+4x^3+3x^4\over(1-x^2)^2(1-x^3)^2}.$$ George Andrews wrote out a proof of this formula (and his method should apply more generally to $\sum\_nf\_{j,k}(n)x^n$). Let $a=r+s+t$, $b=s+t$, $c=t$, then the generating function is $\sum\_{r,s,t}(r+s+t)x^{r+2s+3t}$, which can be split up into three pieces each of which is two geometric progressions times a sum of the shape $\sum\_uuy^u$. That last sum is well-known, and after some algebra the conjectured formula pops out.
| 3 | https://mathoverflow.net/users/3684 | 30722 | 20,001 |
https://mathoverflow.net/questions/30725 | 0 | I think it is possible to use only cosine function, but why the formula is used with sine?
I am trying to understand but don't know what to do after Fourier transform with imaginary part and real part.
| https://mathoverflow.net/users/3195 | Why do we use sine and cosine in fourier transform together? | I'm not sure which Fourier transform you mean, but I have only seen Fourier series written where the phase, e.g. $e^{2\pi i n x}$ for some integer n, is decomposed into its real and imaginary parts $cos(2\pi n x) +i sin(2\pi n x)$. Since sine and cosine are related to each other by translation, $sin (\pi/2 - x) = cos (x)$ and similarly $cos (\pi/2 - x) = sin (x)$ (this is a classic trig identity and is an instance of the sum angle formula where one of the angles is $\pi/2$ and one is -x). You can rewrite the above expression in terms of only sines or only cosines. The choice of sine or cosine is up to the author and purely a matter of preference. There is no "best way" to write the Fourier series.
Writing Fourier series in terms of sines and cosines seems to be antiquated within mathematics; I don't know if it is still popular in physics or engineering. Most modern books on Fourier series (e.g. any of Stein's books on Fourier analysis) write formulas in terms of the complex exponentials above. This is in part because the Fourier transform on the real line is written similarly, all the necessary information is encoded in the exponential like orthogonality relations and most likely because it's easier to write.
As far as what to do "after Fourier transform", I don't know what you mean. Could you please explain?
| 4 | https://mathoverflow.net/users/7361 | 30728 | 20,004 |
https://mathoverflow.net/questions/30723 | 6 | In some of my classes (e.g. graph theory, mechanics), the professors encourage the students to visualize solutions to problems; I do well in these classes. In other classes (e.g. linear algebra), we are encouraged instead to reason about abstract concepts; I usually do worse in these classes (relative to the same set of peers). My professors have corroborated my suspicion that the usefulness of visualization depends on the type of problem being solved.
Overall, it seems that I do well with visualization, but have a harder time reasoning about symbolic formulas or abstract entities.
Have people here found that certain areas of mathematics better lend themselves to specific types of mathematical reasoning? I am especially interested in identifying more subjects in math & science that emphasize visual reasoning since I very much enjoy these subjects.
| https://mathoverflow.net/users/4135 | What subfields of mathematics better lend themselves to visualization? | The classes that lend themselves to visualization certainly include graph theory
and mechanics, and I expect you would do well in classical geometry too. In
other cases, it may depend on which book you use. For example, I am sure
you would enjoy the approach to complex analysis in Tristan Needham's book
*Visual Complex Analysis* (Oxford University Press 1997), whereas you may
find other complex analysis books less friendly.
By all means use your strength in visualization to get a foothold in various
areas of math (this should also be possible in linear algebra), but try to develop other strengths too -- you will need them eventually.
| 9 | https://mathoverflow.net/users/1587 | 30733 | 20,009 |
https://mathoverflow.net/questions/30749 | 4 | Suppose I have a square n\*n, symmetric matrix with positive elements and zero diagonal.
For this to be considered a proper distance metric between n points, the triangle inequality needs to be satisfied (the other requirements follow from the definition).
Is there some standard measure that says to what extent this property is violated by a given matrix ?
In particular, is there a measure that is fast to compute and can such a thing be optimized for ? I.e. obtain solution X that is "as close to being a metric as possible".
| https://mathoverflow.net/users/7368 | Is there a standard measure for how close a matrix is to being a distance metric ? | There are a couple of plausible measures you could employ. One would be to minimize the Frobenius distance between the given matrix (call it $D$) and the target matrix $X$ . Since the space of all distance matrices that satisfy triangle inequality can be expressed using linear constraints, you end up with a least-squares problem that can be solved optimally.
Another measure that's more popular in the theoryCS community would be to find a matrix satisfying the triangle inequality where the worst-case ratio (the distortion) of distances was minimized. You could write this as "minimize $\lambda$ where $(1/\lambda)d\_{ij} \le x\_{ij} \le \lambda d\_{ij}$ for all (i,j) pairs", and again write the linear constraints ensuring that $X$ satisfies triangle inequality. this is a linear program.
It depends on whether you care about "worst-case" or "average-case" behaviour ultimately.
| 6 | https://mathoverflow.net/users/972 | 30754 | 20,023 |
https://mathoverflow.net/questions/30750 | 7 | This may turn out to be a bit embarassing, but here it is. It is probably not true that the ascending and descending central series (\*) of a nilpotent group have the same terms. (Or at least one of MacLane-Birkhoff, Rotman and Jacobson would have mentioned it.) However, I have been unable to find an example where they are different. I thought I had a sketch of proof that they are always equal, but there is a gap, of the kind where you feel it is not patchable.
I've proved it for a few nilpotent groups (dihedral of the square, any group of order p^3, the Heisenberg groups of dimension 3 and 4 over any ring -- I think the argument extends to any dimension), and checked a few more exotic examples in the excellent [Group Properties Wiki](http://groupprops.subwiki.org/wiki/Main_Page). So,
>
> What is the simplest (preferably finite) nilpotent group such that its a.c.s. and d.c.s. are different?
>
>
>
and
>
> Do the a.c.s. and d.c.s. coincide in some interesting general case?
>
>
>
(\*) For completeness, the ascending central series of a group G is defined by Z\_0 = 1, Z\_{i+1} = the pullback of Z(G/Z\_i(G)) along the projection, and the descending central series by G\_0 = G, G\_{i+1} = [G,G\_i]. The group G is nilpotent iff ever Z\_m = G or G\_m = 1. It turns out that if such m exists it is the same for both.
| https://mathoverflow.net/users/4367 | Nilpotent group with ascending and descending central series different? | It's false even if $m=2$. Try the product of a group of order $2$ and a dihedral group of order $8$. The center has order $4$ and the quotient by the center is abelian. The commutator subgroup has order $2$.
| 15 | https://mathoverflow.net/users/6666 | 30755 | 20,024 |
https://mathoverflow.net/questions/30643 | 12 | We are talking about ordinary reals in constructive mathematics.
1. Let represent each real number by infinite converging series:
$$r = [\;(a\_0,b\_0),(a\_1,b\_1),...,(a\_i,b\_i),...\;]$$
$$where\quad a\_i \leq b\_i\quad and \quad a\_i \leq a\_{i+1} \; and \; b\_{i+1} \leq b\_i$$
And interval $(a\_i,b\_i)$ converges: for any given rational $e > 0$ there is index $j$ such that $b\_k - a\_k < e$ for all $k \geq j$.
2. There are only one way to construct such a number: to build an algorithm that produces $ (a\_{i+1},b\_{i+1}) $ from (a,b) (or some nearly equivalent).
3. Let model algorithms by lambda terms (we are able to do so because lambda calculus is Turing complete).
4. It is easy to show that each lambda term may be represented by unique natural number (this is simple serialization/deserialization process, well known for every programmer).
5. So there is a one-to-one correspondence between real numbers and subset of natural numbers.
6. This imply that constructive reals and naturals are equipotent sets.
What are not ok with this reasoning and why?
| https://mathoverflow.net/users/7257 | Are real numbers countable in constructive mathematics? | I am going to attempt another answer which directly addresses what you wrote.
It **cannot** be shown constructively that every infinite subset of $\mathbb{N}$ is in bijective correspondence with $\mathbb{N}$. Thus your reasoning has a flaw when going from step 4 to step 5. By "infinite set" here I mean that there is an injective sequence from $\mathbb{N}$ into the set. "Bijective correspondence" means there is a bijection and its inverse between the sets. Concretely:
**Theorem:** Suppose every infinite subset of $\mathbb{N}$ is in bijective correspondence with $\mathbb{N}$. Then we can solve the Halting problem (assuming Markov principle).
*Proof*. The set
$$S = \lbrace n \mid \text{the $n$-th Turing machine diverges}\rbrace$$
is obviously infinite. Suppose we had a surjection $b : \mathbb{N} \to S$. But then we can semidecide whether a given Turing machine diverges, so we can decide whether a given Turing machien halts. (There is a hidden use of Markov principle here, can someone get rid of it?) QED.
Likewise, in your concrete example, it cannot be shown constructively that there is a bijection between $\mathbb{N}$ and constructive reals, or lambda terms representing them, even though it is the case that the lambda terms representing reals are in bijective correspondence with a subset of $\mathbb{N}$. That was precisely the point of my first answer: I gave a proof that any attempt at enumeration of the reals omits a real.
Furthermore, from discussion in my other answer it seems that you claim that in constructive mathematics all subsets of $\mathbb{N}$ are constructively enumerable. That is false. The set $S$ from the above proof is an example of a set which cannot be constructively enumerated. The set $S$ is of course constructively well defined (or "built" in your sense), as it is not too hard to write down a formula that describes it (using Kleene's predicate $T$).
| 9 | https://mathoverflow.net/users/1176 | 30757 | 20,026 |
https://mathoverflow.net/questions/30742 | 5 | You start with a bag of N recognizable balls. You pick them one by one and replace them until they have all been picked up at least once. So when you stop the ball you pick has not been picked before but all the others have been picked once or more.
Let $P\_N$ be the probability that all the others were actually picked twice or more.
Questions: does $P\_N$ have a limit as $N\to \infty$, is this limit 0 or can you compute it?
Note: This is a question that was asked at the Yahoo answers forum, by gianlino. Since no one has found an answer in that forum, I forwarded it here. Here is the original link. <http://answers.yahoo.com/question/index;_ylt=ApZ_Q6sS137DnEcNoUBqn.vty6IX;_ylv=3?qid=20100704050924AAfLYYJ>
| https://mathoverflow.net/users/7238 | Limit probability | One way to think about this sort of problem is to embed in continuous time. Take $N$ independent Poisson processes of rate 1. (Think of $N$ independent Geiger counters, each going off at rate 1, if you like). A point in the $i$th process corresponds to picking the $i$th ball. Since the processes are independent and all have the same rate, the sequence of ball selections is just a sequence of independent uniform choices, as we desire.
Let $M\_i(x)$ be the number of points in the $i$th Poisson process up to time $x$.
Then the distribution of $M\_i(x)$ is Poisson($x$). In particular,
$P(M\_i(x)\geq 2)= 1-(1+x)e^{-x}$. The time of the first point in such a process has exponential(1) distribution, so its probability density function is $e^{-x}$.
So fix one ball, say ball 1. Consider the event that when ball 1 is first chosen, all the other $N-1$ balls have each been chosen at least twice. To get the probability of this event, integrate over the
time that ball 1 is first chosen (i.e. the time of the first event in process 1):
$\int\_0^\infty e^{-x} P\big(M\_i(x)\geq 2 \text{ for } i=2,3,\dots,N\big) dx$
$=\int\_0^\infty e^{-x} \big(1-(1+x)e^{-x}\big)^{N-1} dx$.
The same applies for any ball, and the events are disjoint, so an exact answer to your question is
$N\int\_0^\infty e^{-x} \big(1-(1+x)e^{-x}\big)^{N-1} dx$.
I don't know if it's possible to get an exact expression for this integral, but it's easy enough to bound it. For any $K$
$N\int\_0^\infty e^{-x} \big(1-(1+x)e^{-x}\big)^{N-1} dx$
$\leq N \int\_0^K e^{-x} \big(1-e^{-x}\big)^{N-1} dx
+ N\int\_K^\infty e^{-x} \big(1-Ke^{-x}\big)^{N-1} dx$.
Take $K=\frac12 \log N$, for example; it's easy to evaluate both integrals
exactly (substitute $u=e^{-x}$) and to show that they both tend to 0 as $N\to\infty$.
| 12 | https://mathoverflow.net/users/5784 | 30760 | 20,028 |
https://mathoverflow.net/questions/30748 | 0 | Given this modified Dirichlet function: $f(x) = 0$ if $x$ is in $\mathbb{Q}$, else $f(x) = x$. I am wondering if this function is Darboux integrable on the interval $[0, 2]$.
I managed to show that every lower Darboux sum is equal to zero, therefore the lower Darboux integral is 0. My intuition tells me that the upper Darboux integral is also 0, but I can't think of how to show that.
I thought of using the fact that f(x) is Darboux integrable if and only if: for every epsilon there exists a partition of $[0, 2]$ such that $S(P) - s(P) < \epsilon$ where S, s are the lower Darboux sums. In our case $s(P)=0$ for every P so we have to show that for every $\epsilon>0$ there exists a partition such that $S(P) < \epsilon$, but didn't manage to come close to anything.
PS. If possible, please don't give a solution that uses Riemann's integral, I haven't studied it.
| https://mathoverflow.net/users/7369 | Modified Dirichlet function Darboux integrable on $[0,2]$? | The function is not integrable by Darboux (and equivalently by Riemann) as any point in (0,2] is a discontinuity point. (take a rational sequence approaching $x$, the values of $f(x\_n)$ are constantly $0$. If $x$ is irrational, then $\lim f(x\_n) \neq f(x)$)
If you want to stick to the Darboux integral definition, since the lower Darboux sum is always 0, and the upper Darboux sum of a given finite partition is $\sum\_{i = 1}^{n-1} (x\_{i+1}-x\_i)\cdot x\_{i+1}$.
In this case, the upper sum is just the sum of the rectangles created by the partition, so if we reduce each rectangle by the parts it exceeded the function $f(x)=x$
Take some finite partition $P = \langle 0 = x\_0,\ldots,x\_{n+1}=2 \rangle$ of $[0,2]$.
$\sum\_{i=0}^n (x\_{i+1} - x\_i)\cdot x\_{i+1} \ge $
$\sum\_{i=0}^n (x\_{i+1} - x\_i)\cdot x\_{i+1} - \frac{1}{2}\sum\_{i=0}^n(x\_{i+1} - x\_i)^2 =$
$\frac{1}{2}\sum\_{i=0}^n (2x\_{i+1}^2 - 2x\_{i+1}x\_i - x\_{i+1}^2 + 2x\_{i+1}x\_i - x\_i^2) =$
$\frac{1}{2}\sum\_{i=0}^n(x\_{i+1}^2 - x\_i^2) = 2$
and therefore we have that every upper sum is $\ge2$ for every partition.
| 2 | https://mathoverflow.net/users/7206 | 30764 | 20,032 |
https://mathoverflow.net/questions/30759 | 1 | Given a graph with a list of edges, is it possible to always construct a set of cycle bases for those edges, such that each and every edge is shared by at most 2 cycle bases?
The above question assumes that each and every edge must somehow belong to at least one cycle. IN other words, there is no vertex that is connected to one and only one edge.
| https://mathoverflow.net/users/807 | In a graph, is it always possible to construct a set of cycle bases, with each and every edge Is shared by at most 2 cycle bases? | Consider the complete graph on 7 vertices. It has 21 edges, so any set of cycles that utilizes each edge at most twice has size at most 42/3=14. But the cycle space of the graph has dimension 21-7+1=15, so you cannot have a basis with the requested property.
| 5 | https://mathoverflow.net/users/2368 | 30767 | 20,033 |
https://mathoverflow.net/questions/30765 | 16 | Is there a notion of an octonion prime?
A Gaussian integer $a + bi$ with $a$ and $b$ nonzero is prime if $a^2 + b^2$ is prime.
A quaternion $a + bi + cj + dk$ is prime iff $a^2 + b^2 +c^2 + d^2$ is prime.
I know there is an eight-square identity that underlies the octonions.
Is there a parallel statement, something like: an octonion is prime if its norm is prime?
I ask out of curiosity and ignorance.
**Addendum.**
The Conway-Smith book Bruce recommended is a great source on my question.
As there are several candidates for what constitutes an integral octonion,
the situation is complicated. But a short answer is that unique factorization
fails to hold, and so there is no clean notion of an octonion prime.
C.-S. select out and concentrate on what they dub the *octavian integers*, which,
as Bruce mentions, geometrically form the $E^8$ lattice.
Here is one pleasing result (p.113): If $\alpha \beta = \alpha' \beta'$, where
$\alpha, \alpha', \beta, \beta'$ are nonzero octavian integers, then the angle between $\alpha$ and $\alpha'$ is
the same as the angle between $\beta$ and $\beta'$.
---
A non-serious postscript:
Isn't it curious that
$\mathbb{N}$,
$\mathbb{C}$,
$\mathbb{H}$,
$\mathbb{O}$
correspond to N, C, H, O, the four atomic elements that comprise all proteins and much of organic
life? Water-space: $\mathbb{H}^2 \times \mathbb{O}$, methane-space: $\mathbb{C} \times \mathbb{H}^4$, ...
| https://mathoverflow.net/users/6094 | Gaussian primes, quaternion primes, ... octonions? | You should probably read "On Quaternions and Octonions" by J.H. Conway and D.A. Smith
P.S. It's "octonion" not "octonian"
Edit: The first thing you will find is a discussion of integral numbers. For the complex numbers you have $\mathbb{Z}[i]$ (aka Gaussian integers) which is an $A\_1\times A\_1$ lattice. You also have $\mathbb{Z}[\omega]$ (aka Eisenstein integers) which is an $A\_2$-lattice. For quaternions you have the integral quaternions (as above) which is an $A\_1^4$ lattice and you also have the Hurwitz integral quaternions (adjoin $(1+i+j+k)/2$) which as a lattice is $D\_4$. The Hurwitz numbers have "division with small remainder" property which makes them better. For the octonions you might start with doubling the Hurwitz integers to get a $D\_4\times D\_4$ lattice. Then you might add more and get $D\_8$. However the octavian integers are an $E\_8$ lattice. They are not unique (there are seven versions). These have several good properties:
Every left or right ideal is principal.
Every ideal is two-sided.
Then there is a discussion of prime factorisations.
Finally the automorphism group of the octaves has a simple subgroup of index 2 and order 12096. This group is $G\_2(2)$.
| 16 | https://mathoverflow.net/users/3992 | 30768 | 20,034 |
https://mathoverflow.net/questions/30771 | 1 | For all $n$, I need to find examples of rings $A\subset B$ such that:
i) $\dim A-\dim B\gt n$
ii) $\dim B-\dim A\gt n$
(where $\dim$ is the *Krull dimension*)
| https://mathoverflow.net/users/7332 | Krull Dimension | $\mathbb{Q} \subset \mathbb{Q}[x\_0, \dots, x\_n] \subset \mathbb{Q}(x\_0, \dots, x\_n)$.
| 14 | https://mathoverflow.net/users/460 | 30776 | 20,037 |
https://mathoverflow.net/questions/30481 | 9 | If I define an additive functor to be a functor on abelian categories such that the action of $F$ on ${\rm Hom}(A,B)$ is a group homomorphism, do I necessarily have that $F(\text{zero object}) = \text{zero object}$?
| https://mathoverflow.net/users/7313 | F(0) = 0? F: additive functor | Since the OP asked for a detailed answer:
Let $A$ be an object of an abelian (or additive) category. Then $A$ is a zero object if and only if the zero endomorphism is the identity endomorphism (and then $Hom(A,A)$ is the zero ring). If $F$ is any functor, it sends the identity morphism of $A$ to the identity morphism of $F(A)$. If in addition $F$ is additive (no pun intended), it also sends the zero morphism to the zero morphism. Thus $F(0)$ is a zero object.
Another exercise in the similar spirit: show that additive functor is additive on objects: it sends finite direct sums to direct sums. (Your question is the particular case of empty direct sum.)
| 18 | https://mathoverflow.net/users/2653 | 30779 | 20,040 |
https://mathoverflow.net/questions/30081 | 18 | I've never seen an authoritative explanation for the choice of the lower case letter $\ell$ or $l$ to denote an arbitrary prime different from a given prime $p$. This now has its own LaTeX command \ell, but has been in use at least since the old work of Taniyama and Weil involving *L* functions. That use of the upper case letter might have suggested the lower case here, I guess(?) The letter *q* would seem more natural in elementary number theory. The write-up of Serre's 1967 McGill lectures was published in 1968 by W.A. Benjamin under the title *Abelian l-adic representations and elliptic curves*. There his convention is to denote prime numbers by $\ell, \ell', p, \dots$, stating: "we mostly use the letter $\ell$ for $\ell$-adic representations and the letter $p$ for the residue characteristic of some valuation".
I've heard this question raised but not answered quite a few times. For instance, after a colloquium talk in Hamburg given by Bhama Srinivasan on Deligne-Lusztig characters, the elderly Ernst Witt asked the non-technical question I've just raised. (He had done impressive work in his youth but became a convert to the Nazi cause without apparently committing any war crimes. Possibly he was the young man reported to have shown up once at Emmy Noether's seminar wearing a pro-Nazi uniform. In old age he had retained some mental acuity but developed phobias about for example the flooring material in the math tower, which required talks like the ones Bhama and I gave to move to a remote building.)
[ADDED] Both Franz and quim point in the direction of how the symbol $l$ became common for prime numbers in Hilbert's development of Kummer's work. There he considers an $l$th root of unity ($l$ an odd prime) instead of $\lambda$ used earlier by Kummer. Later on I guess it became a default option for many people to use $l$ for a prime different from a given prime $p$, especially when $q$ became used commonly for a power of $p$.
| https://mathoverflow.net/users/4231 | Origin of symbol *l* for a prime different from a fixed prime? | This elaborates quim's answer. Kummer did indeed use $\lambda$ for denoting primes (in connection with cyclotomic fields); he borrowed the notation from Jacobi's articles on cyclotomy as well as from his notes of the number theory lectures in 1836/37. When Hilbert rewrote Kummer's contributions in his Zahlbericht, he started the chapter on cyclotomic fields with "Let $l$ denote an odd prime number". The reason for switching from the Greek to the Latin alphabet was Hilbert's custom to use Latin letters for rational numbers. Hilbert also used ${\mathfrak l}$ for prime ideals above $2$.
**Edit.** For what it's worth: Euler used primes $\lambda n + 1$ in art. 92 of his article [E449](http://www.math.dartmouth.edu/~euler/pages/E449.html).
| 10 | https://mathoverflow.net/users/3503 | 30799 | 20,053 |
https://mathoverflow.net/questions/30798 | 7 | i suppose it's fairly well known that if a (continuous, real-valued) function $f$ on the real line satisfies
$f(x-y)=f(x)-f(y)+const$
then it is necessarily linear.
are there any general characterizations of "approximate" linearity? for example, what can be said if $|f(x-y)-f(x)+f(y)-f(0)|$ is bounded by some small $\epsilon$? or, more relevant to what i need, if the $L^2$ norm of this difference is bounded by a small constant? in particular, suppose,
$$E[(f(X-Y)-f(Y)+f(Y)-f(0))^2]\leq\epsilon$$
for a pair of independent random variables $X,Y$. is there a sense in which $f$ is approximately linear?
| https://mathoverflow.net/users/7373 | approximately linear functions | Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality
$$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$
for all $x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers
([link](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=&s5=On%20the%20stability%20of%20the%20linear%20functional%20equation&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq)).
>
> **Theorem.** Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$.
>
>
>
So in your case $E=L^2(\Omega)$, $E'=\mathbb R$.
**Edit.** As Yemon Choi indicated, finite dimensional versions of the result had been discovered earlier and independently. Check out, for instance, the Pólya and Szegö problem book (Ch 3, Problem 99):
>
> Assume that the terms of the sequence $a\_1,a\_2,a\_3,\dots$ satisfy the condition
> $$a\_m+a\_n-1 < a\_{m+n} < a\_m+a\_n+1.$$
> Then
> $$\lim\limits\_{n\to\infty}\frac{a\_n}{n}=\omega$$
> exists; $\omega$ is finite and we have
> $$\omega n-1 < a\_n < \omega n +1.$$
>
>
>
| 14 | https://mathoverflow.net/users/5371 | 30801 | 20,054 |
https://mathoverflow.net/questions/30800 | 2 | Say we have a branched $G$-Galois covering $X \rightarrow Y$ of surfaces, and assume that the branch locus (in $Y$) is a divisor with normal crossings. I will always assume that the ramification is tame, and you can pretend we're doing this with varieties over $\mathbb{C}$ if you prefer. This implies that the inertia groups of any two irreducible components of the ramification divisor commute, because they live in the inertia group of the point which is abelian. I've been told the following fact, and have been spending way too much time trying to understand why it's true:
Say the irreducible components $D\_1$ and $D\_2$ of the branch locus meet at a node. Let's say we pick a $P\_1$ and $P\_2$ irreducible preimages of $D\_1$ and $D\_2$ respectively, that meet.
If we blow up that node of the branch locus in $Y$ (and get the new scheme/variety $Y'$) and then normalize in $\kappa(X)$ (to get $X'$), we get a map $X' \rightarrow Y'$ where over $D\_1$, $D\_2$ and $E$ (the exceptional divisor) are $P\_1$, $P\_2$ and an irreducible divisor connecting them $P\_3$; such that if $I\_1$ was the inertia group of $P\_1$ and $I\_2$ the inertia group of $P\_2$ then $I\_1I\_2$ (which is defined because $I\_1$ and $I\_2$ commute) is the inertia group of $P\_3$.
Why is this true? I've been blowing things up and normalizing for way too long and it has been thoroughly uninsightful. I must be missing an important heuristic.
| https://mathoverflow.net/users/5309 | What happens to inertia groups after blow ups? | I'll do this over $\mathbb{C}$, since you'll let me get away with it.
The geometric fact you need to know is the following: If $D \subset Y$ is a branch divisor for the $G$ covering $X \to Y$, then the inertia group of $D$ is generated by the monodromy around a small loop $\gamma$ encircling $D$. In other words, near any smooth point of $D$, the divisor $D$ locally looks like $\{ z\_n=0 \} \subset \mathbb{C}^n$, and you are supposed to take the monodromy around $(0,0,\ldots, 0, r e^{i \theta} )$ for some small $r$. Well, that's not quite right because (1) the inertia group is only defined up to conjugacy and (2) I need to specify a path from my chosen basepoint in $Y$ to the loop I am computing monodromy around. Specifying the path in (2) lets me talk about inertia groups without saying "up to conjugacy". I'll ignore these issues.
So, suppose that $x$ is a point of $Y$ at which two components, $D\_1$ and $D\_2$, of the branch divisor cross normally. Passing to an analytic neighborhood of $x$, we may assume that $Y$ is the unit ball in $\mathbb{C}^2$, and that $D\_i = \{ z\_i =0 \}$. I'll use your notations $Y'$ and $E$. Inside $Y'$, consider small loops $\gamma\_1$, $\gamma\_2$ and $\delta$ around $D\_1$, $D\_2$ and $E$. The monodromy around these loops will generate the inertia groups at $D\_1$, $D\_2$ and $E$. The key fact is that,
>
> in $\pi\_1(Y' \setminus (D\_1 \cup D\_2 \cup E))$, we have $\delta= \gamma\_1 + \gamma\_2$.
>
>
>
Let's see why. $Y' \setminus (D\_1 \cup D\_2 \cup E) \cong Y \setminus (D\_1 \cup D\_2) \cong \{ (z\_1, z\_2) \in (\mathbb{C}^\*)^2 : |z\_1|^2+|z\_2|^2<1 \}$. This last space clearly retracts onto $(S^1)^2$, with fundamental group $\mathbb{Z}^2$. Explicitly, a small loop around $D\_1$ is given by $(r e^{i \theta}, s)$, for some small $r$ and $s$, and hence has class $(1,0)$. Similarly, a small loop around $D\_2$ has class $(0,1)$. A small loop around $E$ (this is the one you might have to think about) is given by $(r e^{i \theta}, s e^{i \theta})$ for $r$ and $s$ small, and hence has class $(1,1)$. As desired, $(1,1)=(1,0)+(0,1)$.
| 3 | https://mathoverflow.net/users/297 | 30803 | 20,055 |
https://mathoverflow.net/questions/30795 | 5 | Otal [Sur le nouage des géodésiques dans les variétés hyperboliques. C. R. Acad. Sci. Paris Sér. I Math. 320 (1995), no. 7, 847--852.] showed that "short" simple closed geodesics in 3-dimensional hyperbolic mapping tori are unknotted and unlinked with respect to the fiber, where "short" depends only on the genus of the fiber.
Another fact about short geodesics in hyperbolic mapping tori is the following: Given $\delta > 0$, the rotation part of a loxodromic isometry corresponding to a "short" simple closed geodesic is less than $\delta$, where "short" depends only on the genus of the fiber and $\delta$.
I haven't been able to find this written down. Does anyone know a reference for this fact?
| https://mathoverflow.net/users/4325 | Rotation part of short geodesics in hyperbolic mapping tori | This should follow from Minsky's paper *[The classification of Kleinian surface groups, I: Models and bounds](https://arxiv.org/abs/math/0302208)* on a priori bounds for surface groups, which is used in the proof of the ending lamination conjecture.
The punctured torus case (*[The classification of punctured-torus groups](http://www.ams.org/mathscinet-getitem?mr=1689341)*, Annals) is simpler and more explicit (see Theorem 4.1 and equations 4.4 and 4.5).
Addendum: Once I thought about it for a bit, I think it follows from much more elementary
considerations (in fact, I'm pretty sure someone explained this to me before, but I forgot the argument). Let $\Sigma$ be a surface. Suppose one has a very short geodesic $\gamma\subset M$, where $M\cong \Sigma\times \mathbb{R}$ is a hyperbolic manifold, then Otal's argument proves it is unknotted (this was actually known to Thurston, and generalized to multiple components by Otal). In fact, one may find a pleated surface $f:\Sigma \to M$ so that $\gamma$ is a closed geodesic on the image of this surface. Then the Margulis tube $V$ of $\gamma$ is of very large radius, and therefore its boundary $\partial V$ is very close to being a horosphere (i.e., its principle curvatures are very nearly $=1$) and is isometric to a Euclidean torus. The boundary slope $\gamma'\subset \partial V$ of the surface $\Sigma$ is a Euclidean geodesic of bounded length - this follows from an area estimate of a pleated annulus $A \subset \Sigma$ such that $f(A)$ cobounds $\gamma$ and $\gamma''$, where $\gamma''\sim \gamma'\subset \partial V$, which has $\mathrm{Area}(A) \approx \gamma''$ by a Gauss-Bonnet argument (if $V$ were a horocusp, then this would be an equality). But
$$
\mathrm{length}(\gamma')\leq \mathrm{length}(\gamma'')\approx \mathrm{Area}(A) \leq \mathrm{Area}(f^{-1}(\Sigma)) = -2\pi \chi(\Sigma).
$$
The meridian $\mu\subset \partial V$ is a curve intersecting $\gamma'$ once. We may assume that $\gamma',\mu\subset \partial V$ are chosen to be Euclidean geodesics. Then $\partial V \backslash (\gamma'\cup \mu)$ is a Euclidean parallelogram, with one pair of sides of bounded length corresponding to $\gamma'$. Since $V$ has very large radius, $\mu$ must be extremely long. The rotational part corresponds to the fraction of the offset between the two sides of the parallelogram corresponding to $\mu$.
But this implies that the rotational part of $\gamma$ is less than $$
2\pi \mathrm{length}(\gamma')/\mathrm{length}(\mu),
$$
which is very small, and approaches zero as $\mathrm{length}(\gamma)\to 0$.
| 8 | https://mathoverflow.net/users/1345 | 30816 | 20,060 |
https://mathoverflow.net/questions/30818 | 9 | The sum $\sum\_{n=1}^{\infty} 1/n^{s}$ is convergent for all real $s>1$ and diverges for all real $s \le 1$. The same holds for the sum $\sum\_{p \ prime} 1/p^{s}$. Thus, for the functions $f(n)= 1/n^s, s \in \mathbb{R}$ the sum $\sum\_{n=1}^{\infty}f(n)$ shows the same convergence behaviour as the sum $\sum\_{p \ prime}f(p)$.
The same holds, if I'm not mistaken, for the functions $f(n)= 1/(n (\ln n)^s), s \in \mathbb{R}$ (both for $n \in \mathbb{N}$ and for primes convergence iff $s>1$).
Question: Is there a real monotonic function $f$ such that $\sum\_{n=1}^{\infty}f(n)$ diverges whereas sum $\sum\_{p \ prime}f(p)$ converges? (The monotony requirement is for preventing 'artificial' solutions that single out the primes (as e.g. $f(n) = 2^n$ if $n$ is prime; $f(n)=n$ if $n$ is not prime)).
| https://mathoverflow.net/users/6415 | Sum f(p) over all primes convergent with sum f(n) over all natural numbers divergent? | I think, you are mistaken, sum $\sum 1/(p\log p)$ converges, since $p\_n\log p\_n$ behaves like $n(\log n)^2$
| 11 | https://mathoverflow.net/users/4312 | 30820 | 20,063 |
https://mathoverflow.net/questions/30828 | 8 | Given a natural number k, are there only finitely many finite simple groups with the property that all elements have order at most k?
This holds if I only look at the finite simple groups I understand
(e.g. alternating groups and SL(k,finite field)), but it's not clear to me whether this holds for all finite simple groups, even using their classification.
| https://mathoverflow.net/users/2985 | Finite simple groups with upper bound on order of elements | The classification of the finite simple groups implies that there are only finitely many finite simple groups of a given exponent $k$. To see this, first note that we can ignore the sporadic groups, as well as the cyclic groups of prime order. It is also also clear that there are only finitely many alternating groups of a given exponent. So we need only consider groups of (possibly twisted) Lie type over finite fields. Here we see that there are only finitely possibilities for the Lie type: otherwise, the Weyl groups would involve arbitrarily large alternating groups. Once the Lie type is fixed, there are only finitely many possibilities for the finite field: otherwise we would obtain semisimple/diagonal elements of arbitrarily large exponent.
There are currently no proofs of this result which do not use the classification of the finite simple groups.
| 14 | https://mathoverflow.net/users/4706 | 30832 | 20,070 |
https://mathoverflow.net/questions/29087 | 25 | For a given $n$, is there any characterization for the commutative subalgebras of $M\_n(\Bbb{C})$? I would like to know how many commutative subalgebras there are for each possible dimension.
In view of Chapman's answer, I am refining my previous question:
Given $k\leq n$, is there any way of describing the commutative subalgebras of $M\_n$ which are of dimension $k$.
| https://mathoverflow.net/users/6985 | Commutative subalgebras of M_n | If you are only concerned about commutative subalgebras of $M\_n(\mathbb{C})$ then there is a fairly easy characterization. So any self-adjoint abelian algebra is generated by a single self adjoint element (spectral theorem). Call this element T. Then T is diagonalizable and so the algebra it form will be the algebra of polynomials over it. Since it is diagonalizable that is a unitarty $u$ with $uTu^\*$ diagonal. And the algebra has dimension $k$ exactly when T has $k$ distinct non-zero eigenvalues.
Note: This is assuming that T is invertible. If T is not invertible then the polynomial algebra since it contains the constants will have dimension $k+1$.
So then we can view the algebra generated by T as an algebra of the form $u^\*Au$ where A is an algebra of diagonal matrices.
| 7 | https://mathoverflow.net/users/5732 | 30849 | 20,081 |
https://mathoverflow.net/questions/30463 | 2 | Suppose we have two finite groups given by presentations
$G=F/N\_1 , H=F/N\_2$
where $N\_1 \subset N\_2$ and $F$ is a free group of finite rank.
The canonical map $\pi: G \rightarrow H $ induces the inflation map between Schur Multipliers,
$Inf: M(H) \rightarrow M(G)$. (Take a cocyle in M(H) and compose with $\pi \times \pi$ to get a cocyle in M(G))
By Hops formula we have $M(G)\cong (N\_1\cap F')/[N\_1,F]$ and $M(H)\cong (N\_2\cap F')/[N\_2,F] $
What is the description of the inflation map in terms of Hopf's formula?
i.e does $$Inf: (N\_2\cap F')/[N\_2,F] \rightarrow (N\_1\cap F')/[N\_1,F] $$
have a nice description?
What can we say about its kernel?
Thanks in advance.
| https://mathoverflow.net/users/7307 | Hopf's formula and inflation map | I am very far from an expert, but I suspect you have written the map you want going in the wrong direction.
Have a look, for example, at the early part of the book review by Van der Kallen in The Bulletin of the AMS, Vol 10, Number 2 (1984), pages 330-333
He explains that "There is now some confusion as to what the term "Schur multipliers" refers to. The usual meaning is described by the formula $M(Q)=H\_2(Q,Z)$, which is also used when $Q$ is infinite. The competing candidate is its dual $Hom(H\_2(Q,Z),C^\*)= H^2(Q,C^\*)$..."
For finite groups these dual homology and cohomology groups are isomorphic, but not canonically isomorphic.
Your earler reference to cocycles suggests you begin with the cohomology version, but then the Hopf formula you write down is definitely a formula for $H\_2(G,Z)$
One version will be covariant, one contravariant.
To get from one to the other, you have to dualise, which changes the direction of the morphisms.
In cohomology, the inflation map goes from the cohomology of the quotient group to the cohomology of the original group.
In homology, the inflation map goes in the other direction - the opposite direction to the one you have written your arrow. in particular, if I understand it right, with Van der Kallen's "usual meaning" the inflation map should go from $M(G)$ to $M(H)$, not the way around you have it.
And there is certainly an obvious map in that direction.
.......................................
(Added after your comment). O.K. Let me re-express my answer in terms of the definition you are using.
With $M(G)=H^2(G,C^\*)$, I certainly agree that the inflation map goes from $M(H)$ to $M(G)$
Since $C^\*$ is divisible, the Universal Coefficient theorem shows that
$H^2(G,C^\*) \approx Hom(H\_2(G,Z),C^\*)$
And this isomorphism is natural.
We know that, for a finite group, $H\_2(G,Z)$ is a finite abelian group.
So $H\_2(G,Z)$ is isomorphic to $Hom(H\_2(G,Z),C^\*)$.
But this isomorphism is not natural- it depends on expressing $H\_2(G,Z)$as a direct sum of cyclic groups. A different sum will give rise to a different isomorphism. (Compare with the fact that there is no natural isomorphism between a finite-dimensional vector space and its dual).
Now the Hopf isomorphism
$H\_2(G,Z) \approx (N\_1 \cap F^')/[N\_1,F]$ depends only on $F$ and $N\_1$
So the moral of that is that you are using the wrong isomorphic copies to ask your
question. You don't have good enough control of the isomorphisms to the copies you are using.
If you want to relate inflation to Hopf's formula, you should be asking
What is the map from $Hom((N\_2 \cap F^')/[N \_2,F],C^\*)$ to
$Hom((N\_1 \cap F^')/[N \_1,F],C^\*)$ ?
And there is then an obvious candidate for the answer:
the dual of the obvious map from $(N\_1 \cap F^')/[N \_1,F]$ to $(N\_2 \cap F^')/[N \_2,F]$.
| 3 | https://mathoverflow.net/users/4648 | 30865 | 20,091 |
https://mathoverflow.net/questions/30868 | 18 | Is every finite codimensional subspace of a Banach space closed? Is it also complemented? I know how to answer the same questions for finite dimensional subspaces, but couldn't figure out the finite codimension case.
| https://mathoverflow.net/users/5498 | Subspaces of finite codimension in Banach spaces | It's a standard result that a linear functional from a Banach space to the underlying field (real or complex numbers) is continuous if and only if the its kernel is closed. Notice that its kernel is of codimension one. So, use the axiom of choice to find a discontinuous linear functional, and you have found a codimension one subspace which isn't closed. (As I was typing this, rpotrie got the same answer...)
As for complementation: well, this only makes sense for *closed* finite codimension subspaces. But then it's a perfectly reasonable question, and the answer is "yes". If F is of finite codimension in E, then by definition we can find a basis $\{x\_1,\cdots,x\_n\}$ for E/F. For each $k$ let $x\_k^\*$ be the linear functional on $E/F$ dual to $x\_k$, so $x\_k^\*(x\_j) = \delta\_{jk}$. Then let $\mu\_k$ be the composition of $E \rightarrow E/F$ with $x\_k^\*$. Finally, pick $y\_k\in E$ with $y\_k+F=x\_k$. Then the map
$$T:E\rightarrow E; x\mapsto \sum\_k \mu\_k(x) y\_k$$
is a projection of $E$ onto the span of the $y\_k$, and $I-T$ will be a projection onto $F$ (unless I've messed something up, which is possible).
| 14 | https://mathoverflow.net/users/406 | 30870 | 20,094 |
https://mathoverflow.net/questions/30788 | 15 | Inspired by [this](https://mathoverflow.net/questions/30345/what-is-an-example-of-a-compact-smooth-manifold-whose-k-theory-and-cech-cohomolog) question, I wonder if anyone can provide an example of a finite CW complex X for which the order of the torsion subgroup of $H^{even} (X; \mathbb{Z}) = \bigoplus\_{k=0}^\infty H^{2k} (X; \mathbb{Z})$ differs from the order of the torsion subgroup of $K^0 (X)$, where $K^0$ is complex topological K-theory. This is the same as asking for a non-zero differential in the Atiyah-Hirzebruch spectral sequence for some finite CW complex X, since this spectral sequence always collapses rationally.
Even better, is there an example in which X is a manifold? An orientable manifold?
Tom Goodwillie's answer to the question referenced above gave examples (real projective spaces) where the torsion subgroups are not isomorphic, but do have the same order.
It's interesting to note that the exponent of the images of these differentials is bounded by a universal constant, depending only on the starting page of the differential! This is a theorem of D. Arrlettaz (K-theory, 6: 347-361, 1992). You can even change the underlying spectrum (complex K-theory) without affecting the constant.
| https://mathoverflow.net/users/4042 | Torsion in K-theory versus torsion in cohomology | [This paper](http://arxiv.org/pdf/hep-th/0005103) by Volker Braun shows that the orientable 8-manifold $X=\mathbb{RP}^3\times \mathbb{RP}^5$ gives an example. One has $$K^0(X) \cong \mathbb{Z}^2 \oplus \mathbb{Z}/4\oplus (\mathbb{Z}/2)^2$$
and
$$H^{ev}(X) \cong \mathbb{Z}^2 \oplus (\mathbb{Z}/2)^5.$$
Braun does the calculations using the Künneth formulae for the two theories, with the discrepancy in size arising because the order of the tensor product of finite abelian groups is sensitive to their structure, not just their order.
One another remark is that Atiyah and Hirzebruch told us the $d\_3$-differential in their spectral sequence. It's the operation $Sq^3 \colon H^i(X;\mathbb{Z})\to H^{i+3}(X;\mathbb{Z})$ given by $Sq^3 := \beta\circ Sq^2 \circ r$, where $r$ is reduction mod 2 and $\beta$ the Bockstein. As you say, Dan, if this is non-vanishing, K-theory has smaller torsion. This happens iff there's a mod 2 cohomology class $u$ such that $u$ admits an integral lift but $Sq^2 (u) $ does not. Can someone think of a nice example where this occurs?
| 13 | https://mathoverflow.net/users/2356 | 30893 | 20,110 |
https://mathoverflow.net/questions/30907 | 11 | One way to define toric varieties is as quotients of affine $n-$space by the action of some torus. However, this is not strictly true as we need to throw away "bad points" which ruin this construction.
For example consider the construction of projective space as a toric variety. Let $\mathbb{G}\_m$ act on $\mathbb{A}^n$ in the obvious way. Then the quotient of $\mathbb{A}^n$ by this action is a single point, as "everything is rescaled to the origin". More rigously the only functions invariant under this action are the constants, thus the quotient is the spectrum of the ground field. To fix this we of course we remove the origin and then take the quotient and we get projective space as required.
So given an action of some torus on affine space, how do we know which points to remove before we take the quotient to make sure we get a toric variety?
My first naive guess is to remove the points which are fixed under the action, but Im wary it may be more subtle than that, as I know GIT can get quite technical.
| https://mathoverflow.net/users/5101 | Toric varieties as quotients of affine space | You want to read Section 2 of *[The homogenous ring of a toric variety](https://arxiv.org/abs/alg-geom/9210008)*, by David Cox. Nick Proudfoot has written an expository note, *[Geometric invariant theory and projective toric varieties](https://arxiv.org/abs/math/0502366)* on the projective case, which you might find helpful as well.
| 7 | https://mathoverflow.net/users/297 | 30908 | 20,119 |
https://mathoverflow.net/questions/30910 | 5 | Has the Robinson-Schensted correspondence, as explained by [Wikipedia](https://en.wikipedia.org/wiki/Robinson-Schensted_algorithm) or [Richard Stanley](http://math.mit.edu/%7Emusiker/rstan7-8.pdf), been implemented in any of the standard programming languages. I'm using Python, but I'm open to Java, C++, Mathematica, Matlab. On paper, the bumping is not so bad - I think 1364752 gives you a v-shaped tableau - but coding the algorithm may require linked lists.
The regular representation of a finite group can be decomposed into a direct sum of all the irreducible representations of G. The basis of the right-regular representation is the elements $g \in G$ and the group action is $\rho\_g(h) = hg$. Then every irreducible representation appears in the sum with multiplicity equal to its dimension
$$ |G| = \sum\_{\pi \in \text{Irr(G)}} (\dim \pi )^2$$
When G = S(n), the permutation group on n elements, the irreducible representations are indexed by Young-diagrams with n boxes and |G| = n!
The Robinson-Schensted correspondence takes this literally and bijectively takes in a permutation and spits out two pairs of (standard?) Young tableaux filled with numbers 1 thru n of the same shape.
| https://mathoverflow.net/users/1358 | Implementation of the Robinson-Schensted correspondence | It doesn't require linked lists, just arrays that can grow.
There's a [Java applet online](http://www.math.uconn.edu/~troby/Goggin/BumpingAlg.html) that implements it.
I'm sure there are other implementations online, but since I couldn't find any, as a start, here's a simple Python implementation. [Though it feels odd giving a programming answer here, and I'm sure several people here can write it much better!]
```
from bisect import bisect
def RSK(p):
'''Given a permutation p, spit out a pair of Young tableaux'''
P = []; Q = []
def insert(m, n=0):
'''Insert m into P, then place n in Q at the same place'''
for r in range(len(P)):
if m > P[r][-1]:
P[r].append(m); Q[r].append(n)
return
c = bisect(P[r], m)
P[r][c],m = m,P[r][c]
P.append([m])
Q.append([n])
for i in range(len(p)):
insert(int(p[i]), i+1)
return (P,Q)
print RSK('1364752')
```
Edit: Used binary search to improve from O(n3) to O(n2log n), which should matter only for very large permutations.
| 12 | https://mathoverflow.net/users/111 | 30914 | 20,122 |
https://mathoverflow.net/questions/30911 | 10 | Hartshorne's "Algebraic geometry" begins with the definition of (quasi-)affine and (quasi-)projective varieties over some fixed algebraically closed field. At a first glance, these seem to be quite different, so that I would have expected that one would pose questions *either* on quasi-affine *or* on quasi-projective varieties.
However, Hartshorne then defines a *variety* to be either a quasi-affine or a quasi-projective variety. These varieties (together with certain continuous and in some sense regular maps) then form the category of varieties.
>
> Here is my question: Is the above definition natural in the sense that we really want to compare quasi-affine and quasi-projective varieties or at least study them both at the same time?
>
>
>
For instance, is there a (non-trivial) example of a quasi-affine variety which is isomorphic in the above category to a quasi-projective variety? If not, isn't this "unifying" definition a bit artificial?
| https://mathoverflow.net/users/1291 | Is Hartshorne's definition of the category of varieties natural? | A quasiaffine variety IS quasi projective. Indeed it is an open set in an affine variety, which in turn is open in its projective closure. So one only considers quasiprojective varieties.
| 19 | https://mathoverflow.net/users/828 | 30916 | 20,124 |
https://mathoverflow.net/questions/30894 | 12 | I've been thinking about the [equiangular lines (or SIC-POVM) conjecture](http://en.wikipedia.org/wiki/SIC-POVM), and my conclusion is that the best means of attack would be through some kind of fixed point theorem -- I'm thinking specifically of geometric fixed point theorems, like Brouwer's. So my (rather vague) questions are:
1) is there some good survey article or classification for fixed point theorems?
2) are there fixed-point theorems which are related to actions of groups on geometric spaces?
3) has anybody tried this idea?
*Added:* In response to Joe's comment below, let me note that while the motivation is from quantum information theory, the equiangular lines conjecture is a purely classical geometry problem (see my comment below). The conjecture is really intriguing: numerical constructions of sets of equiangular lines have been found up to dimension 67, at which point the computer time required exceeded the patience of the investigators. However, only a handful of these numerical solutions have been shown to be rigorously correct by finding corresponding algebraic numbers. See [this recent paper](http://arxiv.org/abs/0910.5784).
| https://mathoverflow.net/users/2294 | Fixed point theorems and equiangular lines | The book "Fixed point theory" by Dugundji and Granas is a nice reference. The headers of the sections in the book give some kind of classification of fixed point theorems.
* results based on compactness
* order theoretic results
* results based on convexity
* Borsuk theorem and topological transitivity
* homology and fixed points
* Leray-Shauder degree and fixed point index
Part VI of the bibliography is really extensive and contains a finer classification of fixed point theorems.
| 7 | https://mathoverflow.net/users/6129 | 30933 | 20,135 |
https://mathoverflow.net/questions/30935 | 2 | If A and B are disjoint subsets of real numbers, and one of them is measurable can we say
m\*(A U B)=m\*(A)+m\*(B)?
I am unable to find counter example. I feel this is not true.
| https://mathoverflow.net/users/7401 | about measure theory | This is true. If for example A is measurable it is measurable in the sense of Caratheodory so that
For every set C we will have
$m\*(C) = m\*(C\cap A) + m\*(C \setminus A)$.
This with $C=A \cup B$ is your assertion
| 6 | https://mathoverflow.net/users/7402 | 30940 | 20,140 |
https://mathoverflow.net/questions/30938 | 6 | A lot is known about geometric and topological properties of diffeomorphism groups of surfaces (here, I am mainly thinking about the work of Smale and Eells-Elworthy). Is there anything known for orbisurfaces ? My first guess would be that many of these groups must have contractible components since singular points impose extra conditions somewhat similar to fixed points. Is there a good reference on this topic ?
| https://mathoverflow.net/users/7325 | Diffeomorphism groups of orbifolds | The result you want can be found in the following paper:
MR0955816 (89h:30028)
Earle, Clifford J.(1-CRNL); McMullen, Curt(1-MSRI)
Quasiconformal isotopies. Holomorphic functions and moduli, Vol. I (Berkeley, CA, 1986), 143--154,
Math. Sci. Res. Inst. Publ., 10, Springer, New York, 1988.
What they prove is actually pretty remarkable. Namely, let $S$ be a hyperbolic surface. Then there is a family $\phi\_t$ of self-maps of $\text{Diff}^{0}(S)$ such that $\phi\_0$ is the identity, such that $\phi\_1$ is the constant map taking each diffeomorphism to the identity diffeomorphism, and such if $f \in \text{Diff}^0(S)$ commutes with a finite order diffeomorphism $g$ of $S$, then $\phi\_t(f)$ also commutes with $g$ for all $t$. In other words, you can contract $\text{Diff}^0(S)$ in way that doesn't break any symmetries.
Now assume that $\Sigma = S / \Gamma$ is a good hyperbolic orbifold, where $\Gamma$ is a finite group of diffeomorphisms of $S$. The identity component of the orbifold diffeomorphism group of $\Sigma$ is then homeomorphic to
$$\text{Diff}^{0}(S,\Gamma) := \langle f \in \text{Diff}^{0}(S)\ |\ gfg^{-1}=f\ \text{for all}\ g \in \Gamma \rangle \subset \text{Diff}^{0}(S)$$
The null-homotopy $\phi\_t$ preserves $\text{Diff}^{0}(S,\Gamma)$, so it is contractible.
(EDIT : I made a slight fix to the definition of the orbifold diffeomorphism above. It doesn't change the argument. Thanks to Tom Church for pointing it out to me!).
| 7 | https://mathoverflow.net/users/317 | 30941 | 20,141 |
https://mathoverflow.net/questions/30715 | 12 | It is well known as Cohen's theorem that a commutative ring is Noetherian if all its
prime ideals are finitely generated. Is this statement true or false when prime ideals are replaced by maximal ideals?
| https://mathoverflow.net/users/5775 | A remark on Cohen's theorem | See the following paper (and search for its citations for related work)
Gilmer, R; Heinzer W.
A non-Noetherian two-dimensional Hilbert domain with principal maximal ideals,
Michigan J. Math. 23 (1976), 353-362
[Link](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-23/issue-4/A-non-Noetherian-two-dimensional-Hilbert-domain-with-principal-maximal/10.1307/mmj/1029001770.full)
| 7 | https://mathoverflow.net/users/6716 | 30942 | 20,142 |
https://mathoverflow.net/questions/29061 | 6 | C. Pugh and M. Shub showed in 1971 that, given an ergodic action of $G=\mathbb{R}^k$ on some separable finite measure space $(X,\mu)$, then all elements of $G$ , off a countable family of hyperplanes, are ergodic.
Is there an analogous statement in the topological setting, with ergodicity replaced by minimality (i.e. all orbits are dense) and $X$ assumed to be compact ?
| https://mathoverflow.net/users/6129 | Minimal elements of minimal R^k actions | A colleague pointed out the following counterexample.
Let $h\_t$ be the horocyclic flow on a negatively curved compact surface S. This R action is known to be minimal. Now Consider the $R^2$ action on $S\times S$ given by $(s,t)\rightarrow (h\_s,h\_t)$. This action is again minimal.
The action of the diagonal $\{(s,s), s\in R\}$ is not minimal since the orbit of any point (x,x) stays in the diagonal.
Let $\theta\in R$. The action of the line $\{(s,\theta s), s\in R\}$ is not minimal because it is conjugated to the diagonal action. This comes from the fact that the two actions $h\_{\theta s}$ and $h\_s$ are conjugated by the geodesic flow.
As a result, there are no elements in $R^2$ acting minimally, although $R^2$ itself acts minimally.
| 4 | https://mathoverflow.net/users/6129 | 30945 | 20,144 |
https://mathoverflow.net/questions/30925 | 8 | Famous Robinson Schensted Knuth correspondence gives a correspondence between the matrices with non-negative integer entries and pair of semi standard tableaux. The proof that I have seen is highly combinatorial e.g. in Knuth's paper [Permutations, matrices, and generalized young tableaux]. Does there exist a geometrical proof of this correspondence?
| https://mathoverflow.net/users/7386 | Geometric proof of Robinson-Schensted-Knuth correspondence? | This depends on the meaning of the word "geometric". If you are thinking of RSK and want the geometry in the way the algorithm is *presented* (in the case of permutations only), Viennot's paper [Une forme geometrique de la correspondance de Robinson–Schensted](https://doi.org/10.1007/BFb0090011) [mentioned](https://mathoverflow.net/a/30932) by PeterR is your answer. If you are thinking of geometry of flag varieties, you might like Steinberg's theorem (see van Leeuwen's thesis [A Robinson–Schensted algorithm in the geometry of flags for Classical Groups](http://www-math.univ-poitiers.fr/~maavl/pdf/thesis.pdf) and his *J. Algebra* paper [An Application of Hopf Algebra techniques to Representations of Finite Classical Groups](http://www-math.univ-poitiers.fr/%7Emaavl/pdf/Hopf.pdf). Finally, if you want RSK to be a map between integer points in polytopes, there are several versions of that, going back to Gansner in 1981 (you can find a description of that and references in the second half of the paper [Hook length formula and geometric combinatorics](https://www.mat.univie.ac.at/~slc/wpapers/s46pak.pdf) of mine).
| 18 | https://mathoverflow.net/users/4040 | 30950 | 20,146 |
https://mathoverflow.net/questions/30948 | 8 | I'm trying to read [a paper](http://arxiv.org/abs/0808.0350) of Boris Kalinin on the cohomology of dynamical systems for a project. The material is geared towards topologically transitive Anosov diffeomorphisms (which is how the initial (abelian) results were proved by Livsic). However, he axiomatizes things for homeomorphisms of metric spaces. It is necessary in the proof to use a version of the Anosov closing lemma, but it looks stronger than the one I've seen. I'd like to know whether elementary techniques will suffice to prove it.
**Background:** The statement of the closing lemma that I learned initially uis as follows. Let $M$ be a compact manifold, $f$ an Anosov diffeomorphism. Put a metric $d$ on $M$, and fix $\epsilon>0$. Then there is $\delta$ such that if $n \in \mathbb{N}$ and $d(f^n(x), x)<\delta$, then there is $p \in X$ with $f^n(p)=p$ and $d(p,x)<\epsilon$. In other words, "approximately periodic points" can be approximately closely by actual ones.
The property Kalinin stipulates in section 1 of his paper is that there is a type of *exponential* closeness. In other words, Kalinin wants that $c,\delta, \gamma>0$ exist such that any $x \in X$ with $d(f^n(x),x)<\delta$ can have the whole orbit be exponentially approximated by a periodic orbit (of $p$ with $f^n(p)=p$). More precisely, one has
$d(f^i(p), f^i(x))< c d(f^i(x),x) e^{-\gamma \min(i,n-i)}$ for each $i=0, \dots, n-1$. This means that the orbits of $p$ and $x$ get even closer in the middle, and this is a strenghtening of the usual closing condition.
One can motivate this fact for Anosov diffeomorphisms geometrically by considering hyperbolic linear maps and drawing a picture of the stable and unstable subspaces, and I am told that it is a straightforward (and "effective") generalization of the usual statement of the closing lemma. This seems more like an intuitive aid rather than a rigorous proof for general Anosov diffeomorphisms, though.
However, Kalinin goes on to say more.
He assumes that there exists $y \in X$ such that $d(f^i(x), f^i(y)) \leq \delta e^{-\gamma i}$, $d(f^i(y)), f^i(p)) \leq \delta e^{-\gamma(n-i)}$. Immediately thereafter, he states that this is true for Anosov diffeomorphisms in view of the closing lemma.
**Questions:**
1) Can one prove the (stronger) version of the closing lemma in Kalinin's paper using the usual statement itself standard techniques (i.e. successive approximation, basic linear algebra for hyperbolic maps, or lemmas like this one)? The books I have seen do not mention it, and certainly say nothing about a point $y$ as in the statement.
2) Does anyone know a good reference for this material (or for the general theory of Anosov diffeomorphisms, for that matter)?
| https://mathoverflow.net/users/344 | Kalinin's formulation of the Anosov closing lemma | The closing lemma as stated by Kalinin can be found in many textbooks e.g. Katok-Hasselblatt "Introduction to the modern theory of dynamical systems", corollary 6.4.17.
The closing lemma really gives a periodic point close to x,
with iterates also close to the iterates of x until the orbit of x returns. That's not just the fact that periodic points are close to non-wandering points.
The point y is obtained by taking the intersection of the local stable set of x with the nth pull-back of the local unstable set of $f^n(p)$. Draw a picture to understand what's going on.
There are "geometric" proofs of the closing lemma that build y before p. And of course the original article of Livsic contains such a proof.
| 3 | https://mathoverflow.net/users/6129 | 30954 | 20,149 |
https://mathoverflow.net/questions/30917 | 5 | What's the relation between Voevodsky's motivic cohomology and Quillen K-theory of a scheme?
| https://mathoverflow.net/users/nan | Relation between motivic cohomology and Quillen K-theory | You should look at Marc Levine's preprint "K-theory and motivic cohomology of schemes, I". The version on the UIUC K-theory server seems to be older than the version on his [website](http://www.uni-due.de/~bm0032/publ/Publ.html) .
Roughly speaking, the motivic spectral sequence starts from motivic cohomology and converges to algebraic K-theory. This spectral sequence was conjectured by Beilinson and first written down for nice fields by Bloch and Lichtenbaum (pre-print on the UIUC server), and was extended to more general schemes by Friedlander and Suslin (Ann. Sci. École Norm. Sup. (4) 35 (2002), no. 6, 773--875). I think these papers were written before the bulk of Voevodsky's work, and the E\_2 term was described in terms of Bloch's higher Chow groups. The correspondence between motivic cohomology and higher Chow groups is Theorem 1.2 in Levine's paper, and is due to Voevodsky ("Motivic cohomology groups are isomorphic to higher Chow groups in any characteristic," Int. Math. Res. Not. 2002, no. 7, 351--355).
| 7 | https://mathoverflow.net/users/4042 | 30955 | 20,150 |
https://mathoverflow.net/questions/30501 | 8 | Define 2 power series over the field $\mathbb Z/2\mathbb Z$ by $f=1+x+x^3+x^6+\dots$, the exponents being the triangular numbers, and $g=1+x+x^4+x^9+\dots$, the exponents being the squares. Write $f/g$ as $c\_0+c\_1x+c\_2x^2+\dots$ with each $c\_n$ in $\mathbb Z/2\mathbb Z$.
**Question.** Is it true that when $n$ is even then $c\_n$ is 1 precisely when $n$ is in the set of
even triangular numbers $\lbrace 0,6,10,28,36,\dots\rbrace$? [Kevin O'Bryant](https://mathoverflow.net/users/935/kevin-obryant) has verified that this holds when $n$ is 512 or less.
**Remark.** If one writes $1/g$ as $b\_0+b\_1x+b\_2x^2+\dots$, then $n\mapsto b\_n$ is the characteristic
function $\bmod 2$ of the set $B$ studied by O'Bryant, Cooper and Eichhorn (see [this](https://mathoverflow.net/questions/26839/) and [this](https://mathoverflow.net/questions/28462/) questions
of O'Bryant on MO); they show that when $n$ is even then $b\_n$ is 1 precisely when $n$ is twice
a square. A positive answer to my question would give a nice characterization of those
elements of $B$ that are congruent to $7 \bmod 16$.
(I've used the modular forms tag because of the formal similarity of $f$ and $g$ to Jacobi
theta functions, and the motivation of O'Bryant, Cooper and Eichhorn in looking at $B$).
| https://mathoverflow.net/users/6214 | Variations on a theme of O'Bryant, Cooper and Eichhorn concerning power series over $\mathbb Z/2\mathbb Z$ | (Part 1)--My argument uses the following curious fact about ideals in $Z[i]$ and $Z[\sqrt{-2}].$
Suppose $n=8m+1$. Let $I=I(n)$ and $J=J(n)$ be the number of ideals of norm $n$ in $Z[i]$ and $Z[\sqrt{-2}]$. Then $I\equiv J$ (4) except when $m$ is odd triangular, in which case $I\equiv J+2$ (4).
As a corollary we find that in $Z/2[[x]]$, $fg^2-fg$ is $x+x^3+x^{15}+x^{21}+\cdots$, the exponents being the odd triangular numbers. (I wonder if this is previously observed, and if it's
related to congruence relations for modular forms). To prove the corollary it suffices to
show: Let $I\_1=I\_1(m)$ be the number of solutions of $m=t+2s$ and $J\_1=J\_1(m)$ be the number of solutions of $m=t+s$ with $t$ triangular, $s$ a square. Then when $m$ is odd triangular, $I\_1-J\_1$ is
odd; otherwise it is even.
We compare $I\_1(m)$ with $I(n)$. Suppose $m=t+2s$. Then $n=(8t+1)+16s$ and $8t+1=x^2$, $16s=y^2$ with
$x$ odd and $x$, $y$ in $N$. $x+iy$ and $x-iy$ generate ideals of norm $n$ in $Z[i]$. These 2 ideals are
distinct except when $m$ is triangular and $s=y=0$. Using the fact that $Z[i]$ is a PID we find
that every ideal of norm $n$ comes from a decomposition $m=t+2s$, and that $I=2I\_1$, except when
$m$ is triangular in which case $I=2I\_1-1$.
Suppose $m=t+s$. Then $n=(8t+1)+8s$, and $8t+1=x^2$, $4s=y^2$ with $x$ odd and $x$, $y$ in $N$. $x+y\sqrt{-2}$ and $x-y\sqrt{-2}$ generate ideals of norm $n$ in $Z[\sqrt{-2}]$. As above, we find
that $J=2J\_1$, except when $m$ is triangular in which case $J$ is $2J\_1-1$. Combining the
result of this paragraph and the last with the curious fact we get the corollary.
One now derives the answer to my question. Let $R=Z/2[[x^2]]$. As R-module, $A$ is the direct sum of $R$ and $xR.$ Let $pr$ be the $R$-linear map $A\to R$ which is id on $R$ and 0 on $xR$.
Since $fg^2-fg=x+x^3+x^{15}+\cdots$, $pr(fg^2)=pr(fg)$. Since $pr$ is R linear and $1/g^2$ is in $R$.
$pr(f)=pr(f/g)$. So for even $m$, the coefficient of $x^m$ in $f/g$ is the coefficient of $x^m$ in $f$,
answering my question. (In his answer Wadim saw the projection argument but missed the
implications)---To be continued
| 1 | https://mathoverflow.net/users/6214 | 30958 | 20,153 |
https://mathoverflow.net/questions/30891 | 6 | A binary quartic form
$aX^4+bX^3Y+cX^2Y^2+dXY^3+Y^4$
decomposes as a product of linear factors $Y-t\_jX$, $j=1,...,4$.
I would like to have an explicit formula for symmetrization of the crossratio of $t\_j$.
| https://mathoverflow.net/users/7393 | explicit formula for the j-invariant of binary quartic form | The $j$ invariant is
$j=\frac{S^3}{S^3-27T^2}$
where
$S=a-\frac{bd}{4}+\frac{c^2}{12}$
and
$T=\frac{ac}{6}+\frac{bcd}{48}-\frac{c^3}{216}-\frac{ad^2}{16}-\frac{b^2}{16}$
for more details see my article ["A computational solution to a question by Beauville on the invariants of the binary quintic"](https://www.sciencedirect.com/science/article/pii/S0021869306000287), J. Algebra **303** (2006) 771-788. The preprint version is [here](https://arxiv.org/abs/math/0508421).
| 11 | https://mathoverflow.net/users/7410 | 30967 | 20,159 |
https://mathoverflow.net/questions/30890 | 6 | Hello,
Maybe it is too vague a question, but I would like to ask if anybody could say some explanatory words about the importance (for infinity category study) of studying all the kinds of fibrations there are for simplicial sets (there are more than eight: right, left, mid, Kan, trivial, and so on..). In general, it might be believed that there are few "important" notions, so why do we have so many fibrations, all of which are studied?
Thank you
| https://mathoverflow.net/users/2095 | Fibrations of Simplicial sets | I'm not an expert, but, here is my understanding. Right-fibrations are important because they are the infinity-version of a category fibered in groupoids (that is an infinity-category fibered in infinity-groupoids). In particular, given an infinity-category $C$,there is a model structure on $sSet/C$, called the contravariant model structure, such that the fibrant and cofibrant objects are precisely the right-fibrations over $S$, and this model structure is Quillen-equivalent (through a generalization of the Grothendieck construction) to the projective model-structure of simplicial presheaves over $w(C)$, where $w(C)$ is a simplicial category and $w$ is the left-adjoint to the homotopy-coherent nerve. Both of these (simplicial) model categories model $Fun(C^{op},\infty-Gpd)$- the infinity-categeory of "weak presheaves in infinity groupoids". So, the upshot is, right fibrations are the infinity-analogue of Grothendieck fibrations in groupoids and provide a model for weak presheaves. This presheaf infinity-topos is the starting point for higher topos theory; infinity topoi are just left-exact (accessible) localizations of such presheaf-infinity categories.
Now, dually, left-fibrations are a model for "infinity-categories COfibered in infinity-groupoids".
The next step, is Cartesian-fibrations. Cartesian-fibrations are the infinity-version of categories fibered in categories (not necessarily groupoids), i.e. they are "infinity categories fibered in infinity-categories". Nearly everything above goes through again, except we need to work with marked-simplicial sets, where we "mark the cartesian-edges".
Again, dually, CoCartesian fibrations model "infinity categories cofibered in infinity-categories". You may wonder why we need both notions? In fact, we need both notions TOGETHER in order to define adjunctions. An adjunction between two infinity-categories $C$ and $D$ is a functor $K \to \Delta[1]$ which is simultaneously a Cartesian-fibration and a CoCartesian fibration, together with Joyal-equivalences $K\_{0} \cong C$ and $K\_{1} \cong D$. This definition is a generalization to the infinity-world of a characterization of adjunctions using cographs.
Now, Kan-fibrations are a relic of homotopy theory. They are the fibrations on the Quillen-model structure on simplicial sets. In a similar spirit, categorical fibrations are the fibrations in the Joyal-model structure on simplicial sets. Other than that, they are not that well behaved; they don't really play a role in infinity-category theory.
Finally, inner fibrations, as far as I know, are only used in defining Cartesian fibrations. That is, a Cartesian fibration is defined to be an inner fibration satisfying extra properties.
I hope this helps.
| 8 | https://mathoverflow.net/users/4528 | 30984 | 20,171 |
https://mathoverflow.net/questions/30904 | 7 | It is known that to prove completeness of first-order logic for countable languages WKL0 is enough. But, is it the weakest subsystem where one can prove it?
What about the incompleteness theorems? Is it known which are the weakest subsystems of second order arithmetic where one would be able to prove each of them?
| https://mathoverflow.net/users/6466 | Weakest subsystems of second order arithmetic for mathematical logic | In fact, the incompleteness and completeness theorems can be proven in subsystems of second-order arithmetic weaker than RCA-0: incompleteness can be proven in EFA (first-order elementary arithmetic), which proves exponentiation total, but cannot prove iterated exponentiation to be total. In fact, systems much weaker than EFA can prove incompleteness: Solovay has shown that any sane system of arithmetic (more or less, any first-order equational logic where there are reasonable definitions of zero and successor) strong enough to prove that multiplication is total can prove incompleteness. But EFA is interesting because "Exponential Function Arithmetic is the weakest system in use for which the coding of finite objects by nonnegative integers is worry free" (Friedman 2010): EFA is a reasonable first-order base upon which to build reverse mathematics.
EFA can be usefully extended to the language of second-order arithmetic using the comprehension scheme ∀x (φ(x) ↔ ψ(x)) → **∃Y** ∀x (x ∈ **Y** ↔ φ(x)), where where φ and ψ are Σ-0-1 and Π-0-1 predicates which may have free second-order variables (this definition is from Avigad 2003). This language, call it ERCA-0, is then an analog of RCA-0-like that is a conservative extension of EFA. Avigad shows how this base system can be considered as a weaker base theory for reverse mathematics, with a series of weaker analogs to other fixtures of the reverse mathematics landscape: in particular, EWKL-0, that analog of WKL-0, can prove the completeness theorem.
To summarise: ERCA-0 is weaker than RCA-0 and can prove the incompleteness theorems; EWKL-0 is weaker than WKL-0 and can prove the completeness theorem. We can hope for weaker systems still, but Friedman's remark suggests that such systems will be more complex, and less suitable for reverse mathematics: there's a sense in which we might expect this to be around the best "weak" base system.
**References**
1. Avigad, 2003, [Number theory and elementary arithmetic](http://www.andrew.cmu.edu/user/avigad/Papers/elementary.pdf). NB. Avigad calls elementary arithmetic, EA.
2. Friedman, 2010, [Concrete Incompleteness from EFA through Large Cardinals](http://www.math.ohio-state.edu/~friedman/pdf/ConIncompAmst051010.pdf).
| 7 | https://mathoverflow.net/users/3154 | 31008 | 20,182 |
https://mathoverflow.net/questions/30998 | 9 | While giving the first of eight lectures on introductory model theory and its applications yesterday, I stated Hilbert's 17th problem (or rather, Artin's Theorem): if $f \in \mathbb{R}[t\_1,\ldots,t\_n]$ is positive semidefinite -- i.e., non-negative when evaluated at every $x = (x\_1,\ldots,x\_n) \in \mathbb{R}^n$ -- then it is a sum of squares of rational functions. One naturally asks (i) must $f$ be a sum of squares of polynomials, and (ii) do we know how many rational functions are necessary? The general answers here are no (Motzkin) and no more than $2^n$ (Pfister). Then I mentioned that the case of $n=1$ is a very nice exercise, because one can prove in this case that indeed $f(t)$ is positive semidefinite iff it is a sum of two (and not necessarily one, clearly) squares of polynomials. Finally I muttered that this was a sort of function field analogue of Fermat's Two Squares Theorem (F2ST).
So I thought about how to prove this result, and I was able to come up with a proof that follows the same recipe as the Gaussian integers proof of F2ST. Then I realized that the key step of the proof was that a monic irreducible quadratic polynomial over $\mathbb{R}$ is a sum of two squares, which can be shown by...completing the square.
But then today I went back to the general setup of a "Gaussian integers" proof, and I came up with the following definition and theorem.
**Definition**: An integral domain $R$ is *imaginary* if $-1$ is a square in its fraction field; otherwise it is *nonimaginary*. (In fact I will mostly be considering Dedekind domains, hence integrally closed, and in this case if $-1$ is a square in the fraction field it's already a square in $R$, so no need to worry much about that distinction.) Note that nonimaginary is a much weaker condition than the fraction field being formally real.
(Definition: An element $f$ in a domain $R$ is a sum of two squares up to a unit if there exist $a,b \in R$ and $u \in R^{\times}$ such that $f = u(a^2+b^2)$.)
**Theorem**: Let $R$ be a nonimaginary domain such that $R[i]$ ($= R[t]/(t^2+1)$) is a PID.
a) Let $p$ be a prime element of $R$ (i.e., $pR$ is a prime ideal). Then $p$ is a sum of two squares up to a unit iff the residue field $R/(p)$ is imaginary.
b) Suppose moreover that $R$ is a PID. Then a nonzero element $f$ of $R$ is a sum of two squares up to a unit iff $\operatorname{ord}\_p(f)$ is even for each prime element $p$ of $R$ such that $R/(p)$ is nonimaginary.
[**Proof**: Introduce the "Gaussian" ring $R[i]$ and the norm map $N: R[i] \rightarrow R$. Follow your nose, referring back to the proof of F2ST as needed.]
**Corollaries**: 1) F2ST. 2) Artin-Pfister for $n = 1$. 3) A characterization of sums of two squares in a polynomial ring over a nonimaginary finite field (a 1967 theorem of Leahey).
4) Let $p \equiv 3,7 \pmod{20}$ be a prime number. Then $p$ *is* a sum of two squares up to a unit in $\mathbb{Z}[\sqrt{-5}]$ but is not (by F2ST) a sum of two squares in $\mathbb{Z}$.
Finally the questions:
>
> Have you seen anything like this result before?
>
>
>
I haven't, explicitly, but somehow I feel subconsciously that I may have. It's hard to believe that this is something new under the sun.
>
> What do you make of the strange situation in which $R$ is not a PID but $R[i]$ is?
>
>
>
Note that one might think this impossible, but $R = \mathbb{R}[x,y]/(x^2+y^2-1)$ is an example. [Reference: Theorem 12 of [Elliptic Dedekind domains revisited](http://alpha.math.uga.edu/%7Epete/ellipticded.pdf).] Do you have any idea about how one might go about producing more such examples, e.g. with $R$ the ring of integers of a number field (or a localization thereof)?
---
**Addendum**: As I commented on below, a good answer to the first question seems to be the paper
>
> MR0578805 (81h:10028)
> Choi, M. D.; Lam, T. Y.; Reznick, B.; Rosenberg, A.
> Sums of squares in some integral domains.
> J. Algebra 65 (1980), no. 1, 234--256.
>
>
>
In this paper, they prove the theorem above with slightly different hypotheses: $R$ is a nonimaginary UFD such that $R[i]$ is also a UFD. Looking back at my proof, the only reason I assumed PID was not to worry about the distinction between $R/pR$ and its fraction field. Just now I went back to check that everything works okay with PID replaced by UFD. So the second question becomes more important: what are some examples to exploit the fact that $R[i]$, but not $R$, needs to be a UFD?
| https://mathoverflow.net/users/1149 | Sums of two squares in (certain) integral domains | Let $K$ be a complex quadratic number field such that $K(i)$ has class number $1$. If $K$ has class number $\ne 1$, then $K(i)$ must be the Hilbert class field of $K$, which, in this case, coincides with the genus class field of $K$. By genus theory, the discriminant of $K$ must have the form $d = -4p$ for a prime number $p \equiv 1 \bmod 4$.
In these cases, the ring of integers in $K$ is $R = {\mathbb Z}[\sqrt{-p}]$, and the ring of integers in $L$ is ${\mathbb Z}[i, (1 + \sqrt{p})/2] \ne R[i]$.
Finding number fields of higher degree with this property seems to be an interesting problem;
I can't think of an obvious approach in general, but if I find anything, I'll let you know.
**Edit 1**. The argument works for all imaginary number fields: if $R$ is the ring of integers in a number field $K$, and if $S = R[i]$ is the ring of integers in the extension $L = K(i)$, then disc$(L) = \pm 4$ disc$(K)^2$ (this is a simple determinant calculation: take an integral basis
$\{\alpha\_1, \ldots, \alpha\_n\}$ for $K$; then $\{\alpha\_1, \ldots, \alpha\_n, i\alpha\_1, \ldots, i\alpha\_n\}$ is an integral basis for $L$).
On the other hand, if $L$ has class number $1$ and $K$ is not a PID, then $K$ has class number $2$ and $L$ is the Hilbert class field of $K$. This implies disc$(L) = \pm$ disc$(K)^2$.
The problem in the non-imaginary case is that $K(i)$ might be unramified at all finite primes, but not at infinity; in this case, $K$ has class number $2$ in the strict sense, yet its ring of integers is a UFD.
Remark 2. By looking at $p = \alpha^2 + \beta^2$ modulo $4$ it follows (unless I did something stupid) that primes $p \equiv 3, 7 \bmod 20$ are not sums of two squares in
${\mathbb Z}[\sqrt{-5}]$.
**Edit 2**. Your suggestion to look at rings $R[\frac12]$, where $R$ is the ring of integers of a quadratic number field, seems to work for $K = {\mathbb Q}(\sqrt{-17})$, which has a cyclic class group of order $4$. The ring $R[\frac12]$ has class number $2$ because $2$ is ramified and so generates a class of order $2$, and $S = R[\frac12,i]$ is the integral closure of $R[\frac12]$ in the extension $L = K(i)$. The ring $R[\frac12]$ has class number $2$, and $S$ is a UFD since $L$ has class number $2$, and its class group is generated by one of the prime ideals above $2$ (the ramified prime above $2$ in $K$ splits in $L$).
**Edit 3**. The corollary concerning sums of two squares in $R[\frac12]$ shows that primes $p \equiv 3 \bmod 4$ splitting in $K$ are sums of two squares up to units. In fact it follows from genus theory that the prime ideals above such $p$ are not in the principal genus, hence lie in the same class as the prime above $2$ or in its inverse. This implies that either $2p = x^2 + 17y^2$ or $8p = x^2 + 17y^2$, giving a representation of $p$ as a sum of two squares up to a unit (remember $2$ is invertible). Thus everything is working fine.
| 9 | https://mathoverflow.net/users/3503 | 31010 | 20,183 |
https://mathoverflow.net/questions/31009 | 6 | Hi guys,
I am able to prove that any symmetric manifold is complete (Consider a local geodesic and use the symmetry to flip it, effectively doubling the length of the geodesic, ad infinitum). I want to use a similar procedure to prove that a manifold whose isometries act transitively is complete, i.e there is always an isometry which maps the start point of a local geodesic to its end point, preserving the geodesic. I am, however, unable to ensure that it is not `rotated' in the process, i.e I want the pushforward of the initial tangent, by the isometry, to be the final tangent, ensuring the resultant doubled geodesic is smooth.
My Lie group theory is a bit scratchy but I assume there is a method which allows me to construct the correct pushforward using only transitivity.
Any ideas would be great,
regards,
MK
| https://mathoverflow.net/users/4890 | Action of the group of isometries on a manifold | By the Hopf-Rinow theorem, you only have to prove that the manifold is a complete metric space. By homogeneity, the injectivity radius is bounded from below by a uniform positive constant. Using this and the compacity of balls whose radius is smaller than the injectivity radius of their center it is easy to check the convergence of Cauchy sequences.
Another way to do this is to interpret the bound on injectivity radius, $r$ say, in term of geodesic extension: a geodesic $\gamma$ defined on $[a,b]$ can be extended to a geodesic defined on $(a-r,b+r)$. From this the conclusion follows.
| 14 | https://mathoverflow.net/users/4961 | 31012 | 20,184 |
https://mathoverflow.net/questions/31006 | 1 | Let us suppose the the group $G:=\mathbb{Z}/2\mathbb{Z}=(1,i)$ freely act on a smooth projective variety/$k$ $X$ and denote by $Y$ the G.I.T. quotient $X/G$. Let $\pi:X\longrightarrow Y$ the quotinet map. Now take a $G$-linearised coherent sheaf $(\mathcal{F}, \lambda)$, one can construct the sheaf of invariants of $\mathcal{F}$,the sheaf $\pi\_\ast(\mathcal{F})^G$, that is a sheaf on $Y$. Then given $\overline{\alpha}=\pi(\alpha)$ a point of $Y$ it is possible to consider the fiber of $\pi\_\*(\mathcal{F})^G$ at $\overline{\alpha}$, $\pi\_\ast(\mathcal{F})^G(\overline{\alpha})$. On the other and one can consider the fiber of $\pi\_\ast(\mathcal{F})(\overline\alpha)$ that is (I think) isomorphic to the direct sum
$$\mathcal{F}(\alpha)\oplus\mathcal F(i\cdot\alpha)\simeq \mathcal F\otimes(k(\alpha)\oplus k(i\cdot\alpha))$$
Now this vector space admit a $G$-action induced by $\lambda$, given by $\lambda\otimes\sigma$ where sigma is the action on $k(\alpha)\oplus k(i\cdot\alpha)$ given by permutation.
My question is: the fiber of the sheaf of invariants is isomorphic to the invariant subspace of the fiber (in the given action)?
I think the answer is no (the first is a subspace of the latter), but I would prefer it to be yes...
If the answer is yes, how do I prove it? could you give me some references?
If the answer is no, could you give me an explicit counterexample?
Thank you very much for the time you dedicated to me and a special thanks to every one who will answer me
best regards
Stgermain
| https://mathoverflow.net/users/6949 | The fiber of the sheaf of invariants | It all works out as well as you could want in every possible sense because of the freeness of the action. As you know, you can make a cover by $G$-stable affine opens, so the real work is in that case. So we focus on the affine case, and then all hypotheses on the affine can be removed: let $A$ be any ring whatsoever and $G$ a finite group acting freely on $X = $ Spec($A$) in the sense of acting freely on the set of geometric points valued in any algebraically closed field, which forces the strongest sense of acting freely on points valued in any ring at all. Thus, $G \times X := \coprod\_{g \in G} X \rightarrow X \times X$
as functors via $(g,x) \mapsto (x, g.x)$ is a subfunctor inclusion and thus an equivalence relation on $X$ in the sense of functors. Now for the real content: that equivalence relation condition, coupled with $G \times X \rightrightarrows X$ being finite locally free (even finite etale) implies that SGA3, Expose V, 4.1(iv) applies, so $X := {\rm{Spec}}(A)$ is a finite etale cover of $Y := {\rm{Spec}}(A^G)$ and the natural map $$\coprod\_ {g \in G} X \rightarrow X \times\_ Y X$$ via $x\_g \mapsto (x, g.x)$ on the $g$th copy of $X$ is an isomorphism. This is a deep result of Grothendieck in such generality. In particular, if $A^G$ is a $B$-algebra and $B \rightarrow B'$ is any map of rings then the natural map $$B' \otimes\_B A^G \rightarrow (B' \otimes\_B A)^G$$ is an isomorphism.
So the $G$-invariant map $\pi:X \rightarrow Y$ is a $G$-torsor for the etale topology and hence $Y$ is a quotient of $X$ by the $G$-action in all good senses (quotient sheaf, good behavior under base change, universal mapping property, etc.). In particular, by etale descent theory (see the example of "Galois covers" in section 6.2 or so of the book "Neron Models") it follows that $\pi\_{\ast}^G$ and $\pi^{\ast}$ induce inverse equivalence between the category of quasi-coherent $O\_Y$-modules and $G$-equivariant quasi-coherent $O\_X$-modules.
In particular, for $G$-equivariant quasi-coherent $O\_X$-modules $F$, the formation of $\pi\_{\ast}(F)$ and $\pi\_{\ast}(F)^G$ commute with any base change on the quotient (e.g., passing to a fiber there). Hence, the geometric fibers of $\pi\_{\ast}(F)$ are the $G$-invariants on the fibers, as you desired, and more vividly the fiber of $\pi\_{\ast}(F)$ at a geometric point $y$ of $Y$ is the direct sum of the fibers at the points in the $G$-orbit fiber of geometric points on $X$ over $y$, with $G$ acting simply transitively on this collection of fibers at points of $X\_y$.
Edit: As t3suji points out, the situation is much simpler (without needing freeness conditions) if $|G|$ is a unit on the schemes in question.
| 3 | https://mathoverflow.net/users/6773 | 31029 | 20,188 |
https://mathoverflow.net/questions/31004 | 80 | At various times I've heard the statement that computing the group structure of $\pi\_k S^n$ is *algorithmic*. But I've never come across a reference claiming this.
Is there a precise algorithm written down anywhere in the literature? Is there one in folklore, and if so what are the run-time estimates? Presumably they're pretty bad since nobody seems to ever mention them.
Are there any families for which there are better algorithms, say for the stable homotopy groups of spheres? or $\pi\_k S^2$ ?
edit: I asked Francis Sergeraert a few questions related to his project. Apparently it's still an open question as to whether or not there is an exponential run-time algorithm to compute $\pi\_k S^2$.
| https://mathoverflow.net/users/1465 | Computational complexity of computing homotopy groups of spheres | [Francis Sergeraert](http://www-fourier.ujf-grenoble.fr/~sergerar/) and his coworkers have implemented his effective algebraic topology theory in a program named Kenzo. It seems capable of computing any $\pi\_n(S^k)$ (in fact homotopy groups of any simply connected finite CW complex), although I don't know how far it is feasible. For instance $\pi\_6 S^3$ is [computed](http://www-fourier.ujf-grenoble.fr/~sergerar/Papers/Genova-1.txt) in about 30 seconds. In a 2002 [paper](http://www-fourier.ujf-grenoble.fr/~sergerar/Papers/Constructive-AT.pdf), they mention other algorithms by Rolf Schön and by Justin Smith, not implemented at that time.
| 44 | https://mathoverflow.net/users/6451 | 31042 | 20,192 |
https://mathoverflow.net/questions/30989 | 4 | Some NP-complete optimization problems, like the knapsack problem, have a solution reachable in polynomial time that is guaranteed to be within arbitrary ε of the optimum answer. (aka PTAS - polynomial time approximation scheme)
Some decision problems, like testing primes, have probabilistic solutions (like Rabin's) where you can get to arbitrary ε certainty of having the right answer. (aka BPP - bounded error, probabilistic, polynomial time)
I'm aware these are very different things theoretically, but I'm going to lump them together and call them "ε-P" - i.e. problems that have 'approximate' (in certainty or optimality) solutions in polynomial time, to within whatever ε one wants.
My question is, how many NP problems are "ε-P", like the above?
---
Answer as I understand it:
Certain problems that are "MAX SNP-hard" have no PTAS. These include: metric traveling salesman, maximum bounded common induced subgraph, three dimensional matching, maximum H-matching, MAX-3SAT, MAX-CUT, vertex cover, and independent set.
NP-complete problems probably don't have BPPs.
However, there's no clear *positive* answer (i.e. what NP problems *do* have a PTAS/BPP). Brownie points if you can supply one.
---
FYI: I am not a mathematician. (My areas are social neuroscience, computer hacking, etc.)
So this is probably not nearly precisely characterized enough to answer precisely, and I am not able to do so. I'm going to give a motivated explanation; please fill in the gaps and correct my errors as you see fit. My boyfriend is a mathematician (algebraic combinatorics) and can translate stuff that's over my head, so don't feel obliged to talk down to me.
This is a pragmatic rather than theoretical question (motivated purely by curiosity), so 'good-enough' answers are good enough. ;-)
| https://mathoverflow.net/users/7418 | Does NP = "epsilon-P" (PTAS / BPP)? | The answer to this question is essentially given in previous answers, but I'll try to state it more completely. It really depends on the problem. All NP-complete problems are equivalent in how hard it is to find their exact solution, but they vary widely in how hard it is to approximate them. Many of them can be shown hard to approximate by using the PCP theorem. A few were known to be hard to approximate before the PCP theorem. There are many which have a polynomial time approximation scheme (PTAS), and so are "easy" to approximate (for some meaning of "easy"). A few have a fully polynomial time approximation scheme (FPTAS), and so are easy to approximate (for a much more satisfying meaning of "easy").
There are no known NP-complete problems which have probabilistic algorithms (like primality testing does) -- this would imply BPP=NP, which is something that computer scientists think is very unlikely.
| 10 | https://mathoverflow.net/users/2294 | 31046 | 20,195 |
https://mathoverflow.net/questions/31035 | 16 | I just want to ask if there is any deeper motivation or clear geometric "sense" behind the barycentric subdivision. Some friend asked me about this a few months ago, looking back the section at Hatcher, I still feel quite confused. I remember one friend told me combinatorically one can do this from posets back to posets, but this does not give me any way to "understand" it properly. In some books (Bredon, for example), the author use excision property as one of the axioms, I'm wondering "where they came from, why they make any sense?".
| https://mathoverflow.net/users/5175 | Deeper meanings of barycentric subdivision | There can be many reasons for subdividing simplices, barycentrically or otherwise.
For a simplicial complex (triangulated space) there are the simplicial homology groups. These are known to be isomorphic to the singular homology groups, therefore (1) invariant under homeomorphism, and in particular (2) invariant under (not necessarily barycentric) subdivision. Before the invention of singular homology, I believe that (1) was unknown. Fact (2) was a key part of the theory. Subdivision is important simply because even if your space is made out of simplices you will sometimes care about subsets which are only unions of simplices after you cut the space up finer. In simplicial homology, excision is an easy algebraic fact, stemming from the fact that when a complex is a union of two subcomplexes then every simplex is in one or the other (or both).
In singular theory, as you know, invariance under homeomorphism is a triviality but excision requires some work. The point is that when a space is a union of two open sets then (bad news) not every singular simplex is in one or the other but (good news) simplices can be systematically replaced by combinations of smaller simplices to show that this does not matter. This is where subdivision is used, and there is no reason it has to be barycentric. It's like with the fundamental group: you might explore a space by using maps of a standard unit interval into it, but in proving the Seifert-Van-Kampen Theorem you might want to subdivide that interval into little pieces.
Barycentric subdivision also rises in PL (piecewise linear) topology in one other specific technical way that has nothing much to do with homology: regular neighborhoods. In a finite simplicial complex $K$, the smallest neighborhood of a given subcomplex $L$ that is itself a subcomplex does not in general have $L$ as a deformation retract, but this becomes true if you first barycentrically subdivide *twice*.
And in the interplay between categories and simplicial constructions barycentric subdivision turns up in various ways.
ADDED in response to Hatcher's answer and its comment thread:
Yes, there is a way of extending to all $n$ the pattern that begins: cut a segment in half, cut a triangle into four equal pieces using midpoints of edges ... It is sometimes called "edgewise subdivision", I believe. It may be realized for simplicial sets as follows: A simplicial set is a functor $\Delta^{op}\to Set$ where $\Delta$ is the category of standard nonempty ordered finite sets; its subdivision is obtained by composing with (the opposite of) the functor $\Delta\to\Delta$ which takes an ordered set to two copies of that set laid end to end. This leaves the realization unchanged. Applied to a standard $n$-simplex, it gives a certain subdivision with $2^n$ pieces. If $n>2$ then the pieces are not all the same shape. If $n=3$ you get a tetrahedron cut into four scaled-down models of itself sitting in the corners and four more whose union is an octahedron; these four all share an edge, the only internal edge that there is. It's not immediately clear to me what diameter estimate is available for the pieces.
This can be generalized so that you now cut an edge into $k$ equal pieces and a triangle into $k^2$ congruent pieces (almost half of which are upside down) and in general cut an $n$-simplex into $k^n$ pieces. This $k$-fold edgewise subdivision plays a role in the area of cyclic homology and related things: when a simplicial set $X$ has the kind of extra structure that makes it a cyclic set (a suitable action of a cyclic group of order $m$ on the set $X\_{m-1}$ for all $m>0$) then its realization has an action of the circle group, and to make the action of the subgroup of order $k$ appear as a simplicial action you can do the $k$-fold edgewise subdivision described above.
There is also another edgewise subdivision. In this one the $1$-simplex is cut in half as before and the $2$-simplex is cut into four pieces in the following way: join the middle vertex to the midpoint of the opposite side, and join that midpoint to the midpoints of both of the other sides. This construction corresponds to the functor $\Delta\to\Delta$ that takes an ordered set to two copies of the same laid end to end but with the order reversed in one copy.
See also my answer to the recent MO question "Endofunctors of the Simplex Category".
The second edgewise subdivision that I described can be used to analyze the relationship between two definitions of algebraic $K$-theory: Quillen's $Q$-construction is essentially a subdivision of Waldhausen's $S$-construction.
| 17 | https://mathoverflow.net/users/6666 | 31048 | 20,196 |
https://mathoverflow.net/questions/31020 | 1 | Which method would you recommend for error estimation of the following approximation?
$$\frac{1}{K} \sum\_{j=0}^{K-1}\frac{cos(2\pi\frac{j}{K}u)}{P\_{n}(\cos[\pi\frac{j}{K}])}\approx\int\_{0}^{1}\frac{cos(2\pi xu)}{P\_{n}(\cos[\pi x])}dx$$
Here $P\_{n}$ some polynomial
$u=1,2...K/2$
$\frac{1}{12k^2}f''(\psi)$ is a very bad estimator
| https://mathoverflow.net/users/3589 | trapezoidal rule error approximation. What if f''(x)/12n^2 doesn't work? | My first guess would be to use the Euler-Maclaurin summation formula ([Wikipedia article](http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_summation_formula)). This proves, amongst other things, that the error goes down exponentially if the integrand is a periodic function on [0,1].
**Added:** After thinking about it a bit more, I'm wondering about some things. Firstly, the formula given in the question is not the trapezoidal rule (as promised in the title and suggested by the result for the error), but it is the rectangle rule which is only first order. Secondly, if the integrand has poles in [0,1] (that is, if $P\_n(\cos(\pi x))=0$ for some $x\in[0,1]$), then the error estimate becomes meaningless; in this case you probably need different techniques like complex analysis to prove anything. A final remark: perhaps you can use the elementary techniques explained in: Weideman, "Numerical integration of periodic functions: a few examples", *Amer. Math. Monthly* **109** (2002), no. 1, 21-36 ([MathSciNet](http://www.ams.org/mathscinet-getitem?mr=1903510)).
I think I need some more background in order to have further help. In particular, do you know anything about the polynomials $P\_n$, and what kind of result do you hope to get?
| 3 | https://mathoverflow.net/users/2610 | 31053 | 20,199 |
https://mathoverflow.net/questions/31050 | 8 |
>
> Let $G$ be the group of orientation-preserving homeomorphisms (or, if you prefer, diffeomorphisms) of the real line. Does there exist a natural way to associate, to each function $f \in G$, a homomorphism $\phi\_f \colon \mathbb{R} \to G$ such that $\phi\_f(1) = f$?
>
>
>
Motivation: Many of us, in high school or before, have wondered whether for a given function $f$, it is possible to find a function $f^{1/2}$ such that $f^{1/2} \circ f^{1/2} = f$. I have never seen a satisfactory answer. This is intended as a more "grown-up" version: Is there some natural definition of $f^r$ ($= \phi\_f(r)$), for $r \in \mathbb{R}$, such that $f^1 = f$ and $f^r \circ f^{-s} = f^{r-s}$?
| https://mathoverflow.net/users/5094 | Homomorphisms from $\mathbb{R}$ to $\mathrm{Homeo}^+(\mathbb{R})$, or "fractional iterations" | This has some relation with [this question](https://mathoverflow.net/questions/26462/noninteger-iterates-of-functions-how-to-get-ode-from-flow-at-a-given-time/26472#26472), but it is obviously different.
In general, a homeomorphism of $\mathbb{R}$ which preserves the orientation may have or not fixed points.
If it has no fixed points, then it is conjugated to a translation and thus, one can easily construct such $\phi\_f$.
The other case is not much more difficult, since one can consider the set of fixed points of $f$ (that is, such that $f(x)=x$) which is closed and then do the trick in the complement of that set and leave fixed the set of fixed points for every $t$ (when defining $\phi\_f(t)$). Notice that in either case, there is in general no unique way to do this.
It is also interesting that the diffeomorphism case is quite different, in particular, one can easyly construct a diffeomorphism $f:\mathbb{R}\to \mathbb{R}$ such that there is no diffeomorphism $g$ such that $g\circ g =f$. This can be seen in the paper provided by Helge (in fact it has to do with distortion and the fact that if you take one contracting point, there are restrictions to construct, for example a square root, see Section 1 of [this paper](http://arxiv.org/PS_cache/arxiv/pdf/0710/0710.3989v1.pdf)).
ADDED RELATED REFERENCE: In [this paper](http://www.ams.org/journals/bull/1974-80-03/S0002-9904-1974-13470-1/S0002-9904-1974-13470-1.pdf), Palis gives a not so difficult proof that $C^1$-generic diffeomorphisms (which belong to a $G\_\delta$-dense subset of $Diff^1(M)$) of a compact manifold, the diffeomorphisms are not the time one map of a flow.
| 8 | https://mathoverflow.net/users/5753 | 31056 | 20,202 |
https://mathoverflow.net/questions/31001 | 5 | In fact, it is a simple problem. I just want to know whether there are some interesting proof.
$Z[x\_1, x\_2, ......, x\_{n^2-1}]$ and $Z[y\_{11}, ......, y\_{1n}, y\_{21}, ......, y\_{nn}]/(det(y\_{ij})-1))$, where $Z$ is integer.
One way to prove is select a prime number,say $p=2$,then localize these two rings, one can count the number of elements in both rings and they are NOT equal.
Question:
Is there any other geometric way to "see" they are obviously not isomorphic to each other?
Any related comments are welcome.
Thanks
The reason I want to ask is some one argued that the proof I gave above is not natural. He thought this is not a "Grothendieck style proof"
| https://mathoverflow.net/users/1851 | How to prove these two rings are not isomorphic | Does your critic dislike that the argument seems not applicable over general rings? But it is: if there's an isomorphism over some ring $R$ then we can descend to a finitely generated subring and pass to the quotient by a maximal ideal to get such an isomorphism over a finite field, and then count points.
Or maybe your critic would prefer to invoke the fact that any group variety structure on an affine space over a field $k$ is unipotent, which ${\rm{SL}}\_n$ is not? Here is a Grothedieck-style proof of this fact about affine space, exactly in the same spirit as the preceding argument: if an affine space over $k$ has a non-unipotent $k$-group structure then by increasing $k$ to an algebraic closure it would (by virtue of being smooth, connected, and affine) contain a nontrivial $k$-split $k$-torus as closed $k$-subgroup. We can then once again descend this property to a subring of $k$ finitely generated over $\mathbf{Z}$ and specialize to a finite field and conclude by counting points: the size of affine space over $k$ of size $q$ is a power of $q$, whereas the nontrivial $k$-split $k$-torus subgroup forces the total number of $k$-points to be divisible by $q-1$, so we get a contradiction as long as $q \ne 2$, and we can certainly always arrange that by increasing the finite field before "counting" anyway.
(There is a more "direct" proof of this general fact about group structures on affine spaces in Springer's book on algebraic groups, but it is kind of complicated. The specialization trick sure makes it easier, at the cost of better algebro-geometric technique to work over a base that is not a field during the middle of that process.)
| 7 | https://mathoverflow.net/users/3927 | 31057 | 20,203 |
https://mathoverflow.net/questions/31058 | 16 | This is a problem I had a look at some years ago but always had the feeling that I was missing something behind its motivation.
D.H. Lehmer says in his 1947 paper, “The Vanishing of Ramanujan's Function τ(n),” that it is natural to ask whether τ(n)=0 for any n>0.
My question is: Why is it natural to wonder whether τ(n)=0 any n>0?
Are there any particular arithmetic properties among the many satisfied by τ(n) that would lead one to ponder its vanishing? The problem is mentioned [here](http://mathworld.wolfram.com/TauFunction.html), where it's stated that it was a conjecture of Lehmer, although it's not actually presented as a conjecture in his paper, more a curiosity.
Maybe there is no deep reason to ponder the vanishing of τ(n), in which case that would be a satisfactory answer too.
| https://mathoverflow.net/users/7330 | The vanishing of Ramanujan's Function tau(n) | The key to your question is [lacunarity](https://en.wikipedia.org/wiki/Lacunary_function) in modular functions.
The tau function, as we know, occurs as the coefficient of the [Discriminant function](https://planetmath.org/ModularDiscriminant), which in turn is the 24th power of the [Eta function](https://mathworld.wolfram.com/DedekindEtaFunction.html). The Eta function was known to be lacunary (having gaps or zero coefficients). Therefore it was natural for Lehmer in 1947 to wonder if coefficients of powers of eta are also zero. See the opening passage of the following paper
MR0021027 (9,12b) Lehmer, D. H. The vanishing of Ramanujan's function $\tau(n)$. Duke Math. J. 14, (1947). 429--433. <http://projecteuclid.org/euclid.dmj/1077474140>
| 10 | https://mathoverflow.net/users/5372 | 31070 | 20,209 |
https://mathoverflow.net/questions/30975 | 14 | Let $F(s)=\sum\_{n\geq 1}\frac{a\_n}{n^s}$ be a Dirichlet series with (finite) abscissa of absolute convergence $\sigma\_a$. It can be shown that $\forall \sigma >\sigma\_a:$
$$\lim\_{T\to\infty}\frac{1}{2T}\int\_{-T}^{T}F(\sigma+ it)n^{it}\mathrm{d}t=\frac{a\_n}{n^{\sigma}}.$$
The natural question arises, given some function $F$ holomorphic in some half-plane, under what conditions does it have a representation as a Dirichlet series. I believe this is a very broad question, so I would actually like to make things a bit more specific.
For fixed $c\geq 0$ let $H:=H\_c:=\{z\in\mathbb{C}:\Re(z)>c\}$ be some half-plane and let $f\in\mathcal{O}(H)$. For $\sigma> c$ define the linear functional
$$\Phi\_{n,\sigma}: f\mapsto\lim\_{T\to\infty}\frac{1}{2T}\int\_{-T}^{T}f(\sigma + it)n^{it}\mathrm{d}t$$
I have intentionally left out the actual domain of $\Phi\_{n,\sigma}$ in $\mathcal{O}(H)$ (since it is rather part of the general question than a known fact). It can be easily seen though that $\Phi\_{n,\sigma}$ is not well-defined on the whole $\mathcal{O}(H)$. Let $\sigma\_0>c$ be fixed real number.
(Q1) Provided $\{\Phi\_{n,\sigma\_0}(f)\} \_{n\in\mathbb{N}}$ exists, does it follow that $\{ \Phi\_{n,\sigma}(f)\} \_{n\in\mathbb{N}}$ exists for all $\sigma>\sigma\_0$?
(Q2) Provided $\{\Phi\_{n,\sigma\_0}(f)\} \_{n\in I}$ exists, where $I\subset\mathbb{N}$ is some infinite subset, does it follow that $\Phi\_{n,\sigma\_0}(f)$ exists for all $n\in\mathbb{N}$? How about sufficiently large finite subset $I\subset\mathbb{C}$?
(Q3) Provided that $n^{\sigma\_0}\Phi\_{n,\sigma\_0}(f)=:a\_n$ exists for all $n\in\mathbb{N}$. Does it follow that the Dirichlet series $\sum\_{n\geq 1}\frac{a\_n}{n^s}$ is absolute(?) convergent in some half-plane? If it is convergent, does it represent $f$ in that half-plane?
(Q4) And the more general question: Are there any known conditions when an analytic function admits expansion as an ordinary Dirichlet series?
I am also aware of the existence of a series of papers of A.F. Leont'ev on the representations of analytic functions as Dirichlet series, e.g. "On the representation of analytic functions by Dirichlet series", A. F. Leont'ev 1969 Math. USSR Sb. 9 111 and "On conditions of expandibility of analytic functions in Dirichlet series", A. F. Leont'ev 1972 Math. USSR Izv. 6 1265, etc. English translations as well as some of the original are available at [iopsciences](http://iopscience.iop.org/search?filtersubmit=FILTER+NOW&f=titleabs&fieldedquery=&fullsearch=&searchRefinement=true&searchType=fullText&time=all&previousquery=Dirichlet+series&pacsDisplay=&catDisplay=&nameDisplay=&yearDisplay=&authorDisplay=&filterDisplay=&pacsCloudDisplay=¤t_refinement_all_authors=A+F++Leont%27ev¤t_refinement_issn=0025-5726¤t_refinement_issn=0025-5734&sk=creation_date&query=Dirichlet&navsubmit=FILTER+NOW). Unfortunately for me, I don´t have institutional access to those :-(
However, while Leontev´s papers appear to be fundamental for the subject, they all date back to the period 1969-1975. So I was hoping that there might be some good serveys or other types of good references summarizing the recent developments, respectively the most general results in the subject so far and that would be also "easier to have" than the aforementioned papers. Also, per Andrey Rekalo´s comment it seems that Leont'ev´s work is not really applicable to the more specific case of representation by ordinary Dirichlet series.
Thank you in advance for any input!
| https://mathoverflow.net/users/1849 | Dirichlet series expansion of an analytic function | A.F. Leont'ev continued to work on [general Dirichlet series](http://en.wikipedia.org/wiki/General_Dirichlet_series) well into 1980s (until his death in 1987). Actually, he published three monographs on the subject from 1976 to 1983! He made a short summary of his earlier results for the 1974 ICM in Vancouver (a free preview of the lecture is available [here](http://books.google.co.uk/books?hl=en&lr=&id=9mFsIglfGbQC&oi=fnd&pg=PA1&dq=%2522On+the+representation+of+analytic+functions+by+Dirichlet+series%2522&ots=uDlLE7BbYG&sig=J8YlBMGGKyX2XvW3y3FFpoTNrSU#v=onepage&q=%2522On%2520the%2520representation%2520of%2520analytic%2520functions%2520by%2520Dirichlet%2520series%2522&f=false)).
A.F. Leont'ev obtained in some sense final results on the representation of analytic functions by general Dirichlet series of the form
$$f(s)=\sum\limits\_{n=1}^{\infty}a\_n e^{-\lambda\_n s},\quad s\in D\subset \mathbb C.$$
He studied Dirichlet series in bounded and unbounded convex domains (including half-planes). The problem is that his results may not be directly applicable to the `ordinary' Dirichlet series with $\lambda\_n=\ln n$. A typical Leont'ev's theorem for half-planes is as follows ([link to the original article in Russian](http://www.mathnet.ru/php/getFT.phtml?jrnid=sm&paperid=3278&what=fullt&option_lang=rus)).
>
> **Theorem.** For every $\rho>1$, there is a sequence $\lambda\_n>0$, $n\in\mathbb N$, satisfying the condition
> $$\lim\limits\_{n\to\infty}\frac{n}{\lambda\_n^\rho}=\tau,\quad 0<\tau<\infty,$$
> such that any function $f$, which is analytic in the right half-plane $\Re z > 0$ , can be represented in the form
> $$f(z)=\sum\limits\_{n=1}^{\infty}a\_n e^{-\lambda\_n z}+\Phi(z),\qquad \Re z > 0,$$
> where $\Phi$ is entire.
>
>
>
This obviously doesn't cover the case $\lambda\_n=\ln n$, $n\in\mathbb N$.
Anyway, if you're interested and if you have a colleague who speaks Russian I can send you a couple of original articles by Leont'ev (PDF files). (**Edit:** sent.) By the way, the papers you've mentioned both deal with the case of convergence in a bounded domain $D$.
| 8 | https://mathoverflow.net/users/5371 | 31075 | 20,211 |
https://mathoverflow.net/questions/31077 | 1 | Define
g(x) = (1+x) ln(1+x) - x.
One can check that g is strictly monotonically increasing for x>=0 by checking its derivative is ln(1+x). So g is invertible and its inverse is also strictly monotically increasing.
Is there an explicit closed form for its inverse?
With a page of calculations I can prove that
(1/2) x/ln(1+sqrt(x)) <= g^{-1}(x) <= 2 x/ln(1+sqrt(x))
for all x>=0. Is this obvious? Can estimates like this be found in the literature?
| https://mathoverflow.net/users/7438 | Inverse of (1+x) ln(1+x) - x | Just to expand on Qiaochu's comment: let $W$ stand for the [Lambert W-function](http://en.wikipedia.org/wiki/Lambert_W_function), then if $g(x)=z$, we readily find that
$$
x=\exp\big(W((z-1)/e)+1\big)-1.
$$
| 6 | https://mathoverflow.net/users/2149 | 31080 | 20,213 |
https://mathoverflow.net/questions/30589 | 3 | Let $G=GL(2,p)$ be the group of linear transformations. It acts on the set $X=F\_p^2$.
Then $C^X$ is a linear representation of $G$. What is the decomposition of this representation into irreducible representations?
*Note*: Let $d\_i$ denote the number of isomorphic irreducible representations. Then $\sum d\_i^2$ is equal to the number of orbits of $G$ when $G$ acts on $X\times X$. The number of orbits of $G$ in $X\times X$ is equal to $p+1$.
| https://mathoverflow.net/users/4246 | Decomposition of GL(2,p) into irreducible representations | **Note:** *This was written up concurrently with David's answer, but wasn't proofread and didn't get past the captcha stage due to technical problems.*
It is more common to consider the representation of $G=GL(2,\mathbb{F}\_p)$ on $\mathbb{C}^Y$, where $Y=X\setminus 0$. This is a direct sum over all multiplicative characters $\chi$ of $\mathbb{F}\_p$ of the induced representations of $G$ in the homogeneous functions of homogeneity degree $\chi:$
$$\mathbb{C}^Y=\bigoplus\_{\chi\in \mathbb{F}\_p^\ast}(\mathbb{C}^Y)\_{\chi}, \quad (\mathbb{C}^X)\_{\chi}=\{f:Y\to \mathbb{C}: f(\lambda a)=\chi(\lambda)f(a)\ \text{for all}\ \lambda\in \mathbb{F}\_p \},$$
and $\mathbb{C}^X=\mathbb{C}\oplus\mathbb{C}^Y,$ where the first summand corresponds to the functions supported at $0\in\mathbb{F}\_p^2.$
Each induced representation has dimension $p+1.$ For the trivial character, the induced representation contains a trivial subrepresentation of $GL(2,\mathbb{F}\_p)$ (constants), with irreducible $p$-dimensional quotient (Steinberg representation). All other induced representations are irreducible and pairwise non-isomorphic (they all have different central characters). The situation is a bit more complicated for restrictions to $H=SL(2,\mathbb{F}\_p)$ (denoted the same by abuse of notation): for any $\chi,\ (\mathbb{C}^Y)\_{\chi}\simeq (\mathbb{C}^Y)\_{\chi^{-1}}$ and if $p$ is odd and $\psi$ is a character of order $2$ then the representation $(\mathbb{C}^Y)\_{\psi}$ is a direct sum of two non-isomorphic $(p+1)/2$-dimensional representations, all other representations remain irreducible and pairwise non-isomorphic. These facts can be verified *ad hoc* by the "sum of squares" calculation or, more systematically, by using the Mackey theory for finite groups. This accounts for close to the half of irreducible representations of $H$; to get the rest, one needs to induce from the non-split torus $\simeq \mathbb{F}\_{p^2}^\ast/\mathbb{F}\_{p}^\ast. $
| 3 | https://mathoverflow.net/users/5740 | 31083 | 20,216 |
https://mathoverflow.net/questions/31100 | 1 | In need for something equivalent to the continuity-definition of real functions I use the following definition of "coarse-continuity" for sequences. Has it been known already? Has it even got a name?
Definition: A function $f(x)$ with $x \in \mathbb{N}$ is called coarsely continuous if and only if there exists a fixed positive constant $C$ such that
${\forall}$ $x, y \in \mathbb{N}$, $|y-x| \ge 1$ : $\dfrac{|f(y) – f(x)|}{|y-x|} < C$.
| https://mathoverflow.net/users/7441 | Has coarse continuity been known already? | Functions like this are called [Lipschitz](http://en.wikipedia.org/wiki/Lipschitz_continuity). The definition works for maps between any two metric spaces. There is also the notion of being coarse lipschitz:
If you have a function $f : X \to Y$ between two metric spaces, and constants $K \geq 1$ and $C \geq 0$, then $f$ is $(K,C)$--coarse lipschitz if
$d\_Y(f(x),f(y)) \leq K \ d\_X(x,y) + C$ for any $x$ and $y$ in $X$.
| 6 | https://mathoverflow.net/users/1335 | 31101 | 20,228 |
https://mathoverflow.net/questions/14918 | 17 | Are there examples of sets of natural numbers that are proven to be decidable but by non-constructive proofs only?
| https://mathoverflow.net/users/2672 | Non-constructive proofs of decidability? | When I teach computability, I usually use the following example to illustrate the point.
Let $f(n)=1$, if there are $n$ consecutive $1$s somewhere in the decimal expansion of $\pi$, and $f(n)=0$ otherwise. Is this a computable function?
Some students might try naively to compute it like this: on input $n$, start to enumerate the digits of $\pi$, and look for $n$ consecutive $1$s. If found, then output $1$. But then they realize: what if on a particular input, you have searched for 10 years, and still not found the instance? You don't seem justified in outputting $0$ quite yet, since perhaps you might find the consecutive $1$s by searching a bit more.
Nevertheless, we can prove that the function is computable as follows. Either there are arbitrarily long strings of $1$ in $\pi$ or there is a longest string of $1$s of some length $N$. In the former case, the function $f$ is the constant $1$ function, which is definitely computable. In the latter case, $f$ is the function with value $1$ for all input $n\lt N$ and value $0$ for $n\geq N$, which for any fixed $N$ is also a computable function.
So we have proved that $f$ is computable in effect by providing an infinite list of programs and proving that *one* of them computes $f$, but we don't know which one exactly. Indeed, I believe it is an open question of number theory which case is the right one. In this sense, this example has a resemblence to Gerhard's examples.
| 22 | https://mathoverflow.net/users/1946 | 31111 | 20,235 |
https://mathoverflow.net/questions/31038 | 5 | Let A and B be C\*-algebras, and let $\phi:A\rightarrow B$ be a *surjective* \*-homomorphism. Then $\phi$ is non-degenerate, and so we can extend it to \*-homomorphism between the multiplier algebras: $\tilde\phi: M(A)\rightarrow M(B)$. It's rather tempting to believe that then, surely, $\tilde\phi$ is also surjective. But I cannot for the life of me think of a proof. Any ideas...?
**Background:** The multiplier algebra $M(A)$ is the largest C\*-algebra containing A as an essential ideal. Concretely, pick some "large enough" representation of A (either $A\rightarrow B(H)$ a non-degenerate \*-representation, or $A\rightarrow A^{\*\*}$ say) and then $M(A) = \{ x : xa,ax\in A \ (a\in A)\}$ the idealiser of $A$ in our large ambient algebra. As $\phi$ surjects, it's very easy to define $\tilde\phi$: we simply have that $$\tilde\phi(x) \phi(a) = \phi(xa), \quad \phi(a) \tilde\phi(x) = \phi(ax).$$
This is well-defined, for if $\phi(a)=0$, then given an approximate identity $(e\_i)$ for A, we have that $\phi(xa) = \lim\_i \phi(xe\_i a) = \lim\_i \phi(xe\_i) \phi(a) = 0$, and so forth. Indeed, if $B\subseteq B(K)$ say, then $\tilde\phi(x)$ is the limit (strong operator topology say) of the net $\phi(xe\_i)$ in $B(K)$; then clearly this is in the idealiser of $B$, and so does define a member of $M(B)$.
| https://mathoverflow.net/users/406 | Surjective *-homs between multiplier algebras | This is true if $A$ is $\sigma$-unital, and is sometimes called the "noncommutative Tietze extension theorem". A good reference is Proposition 6.8 in Lance's *Hilbert $C^\*$-modules*. Proposition 3.12.10 in Pedersen's *$C^\*$-algebras and their automorphism groups* covers the separable case, which was first proved by Akemann, Pedersen, and Tomiyama in a 1973 paper called "[Multipliers of C\*-algebras](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WJJ-4CRHY72-43&_user=440026&_coverDate=07%252F31%252F1973&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1394971669&_rerunOrigin=google&_acct=C000020939&_version=1&_urlVersion=0&_userid=440026&md5=9ed3e6657608f669d2437593c461f3f6)".
Pedersen points out in Section 3.12.11 that you can get counterexamples in the commutative case by considering non-normal locally compact Hausdorff spaces, so that Tietze's extension theorem doesn't apply.
---
Akemann, Pedersen, and Tomiyama are more explicit:
>
> In fact let $X$ be a locally compact Hausdorff
> space which is not normal, and consider two disjoint closed sets $Y\_1$
> and $Y\_2$ such that the function $b$ which is one on $Y\_1$ and zero on $Y\_2$ has
> no continuous extension to $X$. The restriction map of $C\_0(X)$ to
> $C\_0(Y\_2\cup Y\_2)$ is surjective, and $b\in M(C\_0(Y\_1\cup Y\_2))$, but $b$ is not the
> image of a multiplier of $C\_0(X)$.
>
>
>
| 6 | https://mathoverflow.net/users/1119 | 31117 | 20,239 |
https://mathoverflow.net/questions/31113 | 113 | Zagier has a very short proof ([MR1041893](https://mathscinet.ams.org/mathscinet-getitem?mr=1041893), [JSTOR](https://www.jstor.org/stable/2323918)) for the fact that every prime number $p$ of the form $4k+1$ is the sum of two squares.
The proof defines an involution of the set $S= \lbrace (x,y,z) \in N^3: x^2+4yz=p \rbrace $ which is easily seen to have exactly one fixed point. This shows that the involution that swaps $y$ and $z$ has a fixed point too, implying the theorem.
This simple proof has always been quite mysterious to me. Looking at a precursor of this proof by Heath-Brown did not make it easier to see what, if anything, is going behind the scenes.
There are similar proofs for the representation of primes using some other quadratic forms, with much more involved involutions.
Now, my question is: is there any way to see where these involutions come from and to what extent they can be used to prove similar statements?
| https://mathoverflow.net/users/3635 | Zagier's one-sentence proof of a theorem of Fermat | [This paper](http://www.math.tugraz.at/~elsholtz/WWW/papers/zagierenglish9thjuly2002.ps) by Christian Elsholtz seems to be exactly what you're looking for. It motivates the Zagier/Liouville/Heath-Brown proof and uses the method to prove some other similar statements. Here is a [German version](http://www.math.tugraz.at/~elsholtz/WWW/papers/papers10zmasem060.pdf), with slightly different content.
Essentially, Elsholtz takes the idea of using a group action and examining orbits as given (and why not -- it's relatively common) and writes down the axioms such a group action would have to fulfill to be useful in a proof of the two-squares theorem. He then **algorithmically** determines that there is a unique group action satisfying his axioms -- that is, the one in the Zagier proof. The important thing is that having written down these (fairly natural) axioms, there's no cleverness required; finding the involution in Zagier's proof boils down to solving a system of equations.
| 95 | https://mathoverflow.net/users/6950 | 31121 | 20,242 |
https://mathoverflow.net/questions/31118 | 5 | Is there a way to determine a formula giving all *integer* values of $x$ for which the value of a polynomial $P(x)$ with *integer coefficients* is a square?
That is, is there a closed formula for:
$X = \{ x \in \mathbb{N} : \exists \ n \in \mathbb{N} : P(x) = n^2 \}$ ?
I'm interested in particular in $P'(x) = 8x^2-8x+1$, but am wondering about the general case as well.
For $P'(x)$ a sample of $X$ is $\{ 1, 3, 15, 85, 493, 2871, 16731, 97513, \ldots \}$.
| https://mathoverflow.net/users/7258 | Integer polynomials taking square values | There's a fairly detailed explanation of the solution to a similar equation [here](https://web.archive.org/web/20150908090650/http://mathforum.org/library/drmath/view/73118.html). See also [this page](https://www.alpertron.com.ar/QUAD.HTM), which can give you an automated step-by-step solution to such quadratic diophantine equations.
I'll also add that the command Reduce[8 x^2 - 8 x + 1 - y^2 == 0 && Element[x | y, Integers], {x, y}] will produce the answer to your particular problem in Mathematica fairly quickly. I'm making this an answer because the output is too huge to fit into the comments.
```
(C[1] \[Element] Integers && C[1] >= 0 &&
x == 1/32 (16 +
4 (-2 (17 - 12 Sqrt[2])^C[1] +
Sqrt[2] (17 - 12 Sqrt[2])^C[1] - 2 (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) &&
y == 1/2 ((17 - 12 Sqrt[2])^C[1] -
Sqrt[2] (17 - 12 Sqrt[2])^C[1] + (17 + 12 Sqrt[2])^C[1] +
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) || (C[1] \[Element] Integers &&
C[1] >= 0 &&
x == 1/32 (16 +
4 (-2 (17 - 12 Sqrt[2])^C[1] +
Sqrt[2] (17 - 12 Sqrt[2])^C[1] - 2 (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) &&
y == 1/2 (-(17 - 12 Sqrt[2])^C[1] +
Sqrt[2] (17 - 12 Sqrt[2])^C[1] - (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) || (C[1] \[Element] Integers &&
C[1] >= 0 &&
x == 1/32 (16 -
4 (-2 (17 - 12 Sqrt[2])^C[1] +
Sqrt[2] (17 - 12 Sqrt[2])^C[1] - 2 (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) &&
y == 1/2 ((17 - 12 Sqrt[2])^C[1] -
Sqrt[2] (17 - 12 Sqrt[2])^C[1] + (17 + 12 Sqrt[2])^C[1] +
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) || (C[1] \[Element] Integers &&
C[1] >= 0 &&
x == 1/32 (16 -
4 (-2 (17 - 12 Sqrt[2])^C[1] +
Sqrt[2] (17 - 12 Sqrt[2])^C[1] - 2 (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) &&
y == 1/2 (-(17 - 12 Sqrt[2])^C[1] +
Sqrt[2] (17 - 12 Sqrt[2])^C[1] - (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) || (C[1] \[Element] Integers &&
C[1] >= 0 &&
x == 1/32 (16 +
4 (2 (17 - 12 Sqrt[2])^C[1] + Sqrt[2] (17 - 12 Sqrt[2])^C[1] +
2 (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) &&
y == 1/2 (-(17 - 12 Sqrt[2])^C[1] -
Sqrt[2] (17 - 12 Sqrt[2])^C[1] - (17 + 12 Sqrt[2])^C[1] +
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) || (C[1] \[Element] Integers &&
C[1] >= 0 &&
x == 1/32 (16 +
4 (2 (17 - 12 Sqrt[2])^C[1] + Sqrt[2] (17 - 12 Sqrt[2])^C[1] +
2 (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) &&
y == 1/2 ((17 - 12 Sqrt[2])^C[1] +
Sqrt[2] (17 - 12 Sqrt[2])^C[1] + (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) || (C[1] \[Element] Integers &&
C[1] >= 0 &&
x == 1/32 (16 -
4 (2 (17 - 12 Sqrt[2])^C[1] + Sqrt[2] (17 - 12 Sqrt[2])^C[1] +
2 (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) &&
y == 1/2 (-(17 - 12 Sqrt[2])^C[1] -
Sqrt[2] (17 - 12 Sqrt[2])^C[1] - (17 + 12 Sqrt[2])^C[1] +
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) || (C[1] \[Element] Integers &&
C[1] >= 0 &&
x == 1/32 (16 -
4 (2 (17 - 12 Sqrt[2])^C[1] + Sqrt[2] (17 - 12 Sqrt[2])^C[1] +
2 (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1])) &&
y == 1/2 ((17 - 12 Sqrt[2])^C[1] +
Sqrt[2] (17 - 12 Sqrt[2])^C[1] + (17 + 12 Sqrt[2])^C[1] -
Sqrt[2] (17 + 12 Sqrt[2])^C[1]))
```
| 5 | https://mathoverflow.net/users/353 | 31125 | 20,244 |
https://mathoverflow.net/questions/31109 | 38 | I am interested in the differences between algebras and coalgebras. Naively, it does not seem as though there is much difference: after all, all you have done is to reverse the arrows in the definitions. There are some simple differences:
The dual of a coalgebra is naturally an algebra but the dual of an algebra need not be naturally a coalgebra.
There is the Artin-Wedderburn classification of semisimple algebras. I am not aware of a classification even of simple, semisimple coalgebras.
More surprising is: a finitely generated comodule is finite dimensional.
This question is about a more striking difference. The free algebra on a vector space $V$ is $T(V)$, the tensor algebra on $V$. I have been told that there is no explicit construction of the free coalgebra on a vector space. However these discussions took place following the consumption of alcohol. What is known about free coalgebras?
| https://mathoverflow.net/users/3992 | Is there an explicit construction of a free coalgebra? | There cannot be a "free coalgebra" functor, at least in what I think is the standard usage. Namely, suppose that "orange" is a type of algebraic object, for which there is a natural "forgetful" functor from "orange" objects to "blue" objects. Then the "free orange" functor from Blue to Orange is the *left* adjoint, if it exists, to the forgetful map from Orange to Blue.
Suppose that the forgetful map from coalgebras to vector spaces had a left adjoint; then it would itself be a right adjoint, and so would preserve products. Now the product in the category of coalgebras is something huge — think about the coproduct in the category of algebras, which is some sort of [free product](http://en.wikipedia.org/wiki/Free_product) — and it's clear that the forgetful map does not preserve products.
On the other hand, the coproduct in the category of coalgebras is given by the direct sum of underlying vector spaces, and so the forgetful map does preserve coproducts. This suggests that it may itself be a left adjoint; i.e. it may have a *right* adjoint from vector spaces to coalgebras, which should be called the "cofree coalgebra" on a vector space.
Let me assume axiom of choice, so that I can present the construction in terms of a basis. Then I believe that the cofree coalgebra on the vector space with basis $L$ (for "letters") is the graded vector space whose basis consists of all words in $L$, with the comultiplication given by $\Delta(w) = \sum\_{a,b| ab = w} a \otimes b$, where $a,b,w$ are words in $L$. I.e. the cofree coalgebra has the same underlying vector space as the free algebra, with the dual multiplication. It's clear that for finite-dimensional vector spaces, the cofree coalgebra on a vector space is (canonically isomorphic to) the graded dual of the free algebra on the dual vector space. Anyway, this is clearly a coalgebra, and the map to the vector space is zero on all words that are not singletons and identity on the singletons. I haven't checked the universal property, though.
**Edit:** The description above of the cofree coalgebra is incorrect. I learned the correct version from Alex Chirvasitu. The description is as follows. Let $V$ be a vector space, and write $\mathcal T(V)$ for the tensor algebra of $V$, i.e. for the free associative algebra generated by $V$. Then the cofree coassociative algebra cogenerated by $V$ is constructed as follows. First, construct $\mathcal T(V^\ast)$, and second construct its *finite dual* $\mathcal T(V^\ast)^\circ$, which is the direct limit of duals to finite-dimensional quotients of $\mathcal T(V^\ast)$. There is a natural inclusion $\mathcal T(V^\ast)^\circ \hookrightarrow \mathcal T(V^\ast)^\ast$, and a natural map $\mathcal T(V^\ast)^\ast \to V^{\ast\ast}$ dual to the inclusion $V^\ast \to \mathcal T(V^\ast)$. Finally, construct $\operatorname{Cofree}(V)$ **as the union of all subcoalgebras of $\mathcal T(V^\ast)^\circ$ that map to $V \subseteq V^{\ast\ast}$ under the map $\mathcal T(V^\ast)^\circ \hookrightarrow \mathcal T(V^\ast)^\ast \to V^{\ast\ast}$. Details are in section 6.4 (and specifically 6.4.2) of the book *Hopf Algebras* by Moss E. Sweedler.**
In any case, $\operatorname{Cofree}(V)$ is something like the coalgebra of "finitely supported distributions on $V$" (or, anyway, that's is how to think of it in the cocomutative version). For example, when $V = \mathbb k$ is one-dimensional, and $\mathbb k = \bar{\mathbb k}$ is algebraically closed, then $\operatorname{Cofree}(V) = \bigoplus\_{\kappa \in \mathbb k} \mathcal T(\mathbb k)$. I should emphasize that now when I write $\mathcal T(\mathbb k)$, in characteristic non-zero I do not mean to give it the Hopf algebra structure. Rather, $\mathcal T(\mathbb k)$ has a basis $\lbrace x^{(n)}\rbrace$, and the comultiplication is $x^{(n)} \mapsto \sum x^{(k)} \otimes x^{(n-k)}$. Identifying $x^{(n)} = x^n/n!$, this is the comultiplication on the "divided power" algebra. It's reasonable to think of the $\kappa$th summand as consisting of (divided power) polynomials times $\exp(\kappa x)$, but maybe better to think of it as the algebra of descendants of $\delta(x - \kappa)$ — but this is just some Fourier duality.
In the non-algebraically-closed case, there are also summands corresponding to other closed points in the affine line. **end edit**
---
I should mention that in my mind the largest difference between algebras and coalgebras (by which I mean, and I assume you mean also, "associative unital algebras in Vect" and "coassociative counital coalgebras in Vect", respectively) is one of finiteness. You hinted at the difference in your answer: if $A$ is a (coassociative counital) coalgebra (in Vect), then it is a colimit (sum) of its finite-dimensional subcoalgebras, and moreover if $X$ is any $A$-comodule, then $X$ is a colimit of its finite-dimensional sub-A-comodules. This is absolutely not true for algebras. It's just not the case that every algebra is a limit of its finite-dimensional quotient algebras. A good example is any field of infinite-dimension.
It follows from the finiteness of the corepresentation theory that a coalgebra can be reconstructed from its category of *finite-dimensional* corepresentations. Let $A$ be a coalgebra, $\text{f.d.comod}\_A$ its category of finite-dimensional right comodules, and $F : \text{f.d.comod}\_A \to \text{f.d.Vect}$ the obvious forgetful map. Then there is a coalgebra $\operatorname{End}^\vee(F)$, defined as some natural coequalizer in the same way that the algebra of natural transformations $F\to F$ is defines as some equalizer, and there is a canonical coalgebra isomorphism $A \cong \operatorname{End}^\vee(F)$. (Proof: see André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, Category theory (Como, 1990), Lecture Notes in Math., vol. 1488, Springer, Berlin, 1991, pp. 413–492. MR1173027 (93f:18015).)
For an algebra, on the other hand, knowing its finite-dimensional representation theory is not nearly enough to determine the algebra. Again, the example is of an infinite-dimensional field (e.g. the field of rational functions). On the other hand, it is true that knowing the *full* representation theory of an algebra determines the algebra. Namely, if $A$ is an (associcative, unital) algebra (in Vect), $\text{mod}\_A$ its category of all right modules, and $F: \text{mod}\_A \to \text{Vect}$ the forgetful map, then there is a canonical isomorphism $A \cong \operatorname{End}(F)$. (Proof: $F$ has a left adjoint, $V \mapsto V\otimes A$. But $V \mapsto V\otimes \operatorname{End}(F)$ is also left-adjoint to $F$. The algebra structure comes from the adjunction: the $\text{mod}\_A$ map $A\otimes A \to A$ corresponds to the vector space map $\operatorname{id}: A\to A$.) ((Note that you don't actually need the *full* representation theory, which probably doesn't exist foundationally, but you do need modules at least as large as $A$.))
All this all means is that if you believe that almost everything is finite-dimensional, you should reject algebras as "wrong" and coalgebras as "right", whereas if you like infinite-dimensional objects, algebras are the way to go.
| 25 | https://mathoverflow.net/users/78 | 31126 | 20,245 |
https://mathoverflow.net/questions/30898 | 8 | It seems that there are three basic ways to prove an inequality eg $x>0$.
1. Show that x is a sum of squares.
2. Use an entropy argument. (Entropy always increases)
3. Convexity.
Are there other means?
Edit: I was looking for something fundamental. For instance Lagrange multipliers reduce to convexity. I have not read Steele's book, but is there a way to prove monotonicity that doesn't reduce to entropy? And what is the meaning of positivity?
Also, I would not consider the bootstraping method, normalization to change additive to multiplicative inequalities, and changing equalities to inequalities as methods to prove inequalities. These method only change the form of the inequality, replacing the original inequality by an (or a class of) equivalent ones. Further, the proof of the equivalence follows elementarily from the definition of real numbers.
As for proofs of fundamental theorem of algebra, the question again is, what do they reduce too? These arguments are high level concepts mainly involving arithmetic, topology or geometry, but what do they reduce to at the level of the inequality?
Further edit: Perhaps I was looking too narrowly at first. Thank you to all contributions for opening to my eyes to the myriad possibilities of proving and interpreting inequalities in other contexts!!
| https://mathoverflow.net/users/nan | Ways to prove an inequality | I don't think your question is a mathematical one, for the question about what do all inequalities eventually reduce to has a simple answer: axioms. I interpret it as a metamathematical question and still I believe the closest answer is the suggestion above about using everything you know.
An inequality is a fairly general mathematical term, which can be attributed to any comparison. One example is complexity hierarchies where you compare which of two problems has the highest complexity, can be solved faster etc. Another one is studying convergence of series, that is comparing a quantity and infinity, here you find Tauberian theory etc. Even though you did not specify in your question which kind of inequalities are you interested in primarily, I am assuming that you are talking about comparing two functions of several real/complex variables. I would be surprised if there is a list of exclusive methods that inequalities of this sort follow from. It is my impression that there is a plethora of theorems/principles/tricks available and the proof of an inequality is usually a combination of some of these. I will list a few things that come to my mind when I'm trying to prove an inequality, I hope it helps a bit.
First I try to see if the inequality will follow from an equality. That is to recognize the terms in your expression as part of some identity you are already familiar with. I disagree with you when you say this shouldn't be counted as a method to prove inequalities. Say you want to prove that $A\geq B$, and you can prove $A=B+C^2$, then, sure, the inequality follows from using "squares are nonnegative", but most of the time it is the identity that proves to be the hardest step. Here's an example, given reals $a\_1,a\_2,\dots, a\_n$, you want to prove that $$\sum\_{i,j=1}^n \frac{a\_ia\_j}{1+|i-j|} \geq 0.$$
After you realize that sum is just equal to $$\frac{1}{2\pi}\cdot\int\_{0}^{2\pi}{\int\_{0}^{1}{\frac{1-r^{2}}{1-2r\cos(x)+r^{2}}\cdot |\sum\_{k=1}^{n}{a\_{k}e^{-ikx}}|^{2}dx dr}}$$ then, yes, everything is obvious, but spotting the equality is clearly the nontrivial step in the proof.
In some instances it might be helpful to think about combinatorics, probability, algebra or geometry. Is the quantity $x$ enumerating objects you are familiar with, the probability of an event, the dimension of a vector space, or the area/volume of a region? There is plenty of inequalities that follow this way. Think of Littlewood-Richardson coeficients for example.
Another helpful factor is symmetry. Is your inequality invariant under permuting some of its variables? While I don't remember right now the paper, Polya has an article where he talks about the "principle of nonsufficient reason", which basically boils down to the strategy that if your function is symmetric enough, then so are it's extremal points (there is no sufficient reason to expect assymetry in the maximal/minimal points, is how he puts it). This is similar in vein to using Langrange multipliers. Note however that sometimes it is the oposite of this that comes in handy. [Schur's inequality](http://en.wikipedia.org/wiki/Schur%27s_inequality), for example is known to be impossible to prove using "symmetric methods", one must break the symmetry by assuming an arbitrary ordering on the variables. (I think it was sent by Schur to Hardy as an example of a symmetric polynomial inequality that doesn't follow from Muirhead's theorem, see below.)
Majorization theory is yet another powerful tool. The best reference that comes to mind is Marshall and Olkin's book "Inequalities: Theory of Majorization and Its Applications". This is related to what you call convexity and some other notions. Note that there is a lot of literature devoted to inequalities involving "almost convex" functions, where a weaker notion than convexity is usually used. Also note the concepts of Schur-convexity, quasiconvexity, pseudoconvexity etc. One of the simplest applications of majorization theory is [Muirhead's inequality](http://en.wikipedia.org/wiki/Muirhead%27s_inequality) which generalizes already a lot of classical inequalities and inequalities such as the ones that appear in competitions.
Sometimes you might want to take advantage of the duality between discrete and continuous. So depending on which tools you have at your disposal you may choose to prove, say the inequality $$\sum\_{n=1}^{\infty}\left(\frac{a\_1+\cdots+a\_n}{n}\right)^p\le \left(\frac{p}{p-1}\right)^p \sum\_{n=1}^{\infty}a\_n^p$$ or it's continuous/integral version
$$\int\_{0}^{\infty}\left(\frac{1}{x}\int\_0^x f(t)dt\right)^p dx \le \left(\frac{p}{p-1}\right)^p \int\_{0}^{\infty} f(x)^p dx$$ I've found this useful in different occasions (in both directions).
Other things that come to mind but that I'm too lazy to describe are "integration preserves positivity", [uncertainity principle](http://en.wikipedia.org/wiki/Uncertainty_principle#Uncertainty_theorems_in_harmonic_analysis), using the mean value theorem to reduce the number of variables etc. What also comes in handy, sometimes, is searching if others have considered your inequality before. This might prevent you from spending too much time on an inequality like $$\sum\_{d|n}d \le H\_n+e^{H\_n}\log H\_n$$ where $H\_n=\sum\_{k=1}^n \frac{1}{k}$.
| 9 | https://mathoverflow.net/users/2384 | 31135 | 20,250 |
https://mathoverflow.net/questions/31072 | 2 | As a phd-student I've wandered into a question of colourings of graphs and wondered what was known about them.
Given a finite graph G, where the maximum degree of a vertex is d, I'm interested in colourings where not only are no adjacent vertices the same colour, but also that no vertex has two neighbours of the same colour. [in other words, a vertex and all vertices adjacent to it, are all coloured distinctly]
It's easy to see that the number of colours is at least d + 1, and I'm interested in when this is this actual number of colours needed. (although information on the general case is also of interest)
Also within my work I am mainly looking at regular graphs. But again, that is merely the cases I'm working with and information, thoughts, references on the general case would be muchly appreciated.
To pose it as specific questions:
When can a graph G be coloured, as above, with only d + 1 colours?
When can a regular graph G be coloured, as above, with only d + 1 colours?
Is it NP-Complete to find such colourings? (I feel it is because it seems similar to edge colourings)
| https://mathoverflow.net/users/5965 | Colourings of Graphs with extra conditions | Qiaochu Yuan commented that your problem is equivalent to coloring what is known as the **square** $G^{2}$ of the graph $G$. For more details on coloring the square of a graph, see "The chromatic number of graph powers", N. Alon and B. Mohar, Combinatorics, Probability and Computing (1993) 11, 1-10. On-line at <http://www.math.tau.ac.il/~nogaa/PDFS/am8.pdf>
| 2 | https://mathoverflow.net/users/5883 | 31139 | 20,253 |
https://mathoverflow.net/questions/31147 | 15 | I'm looking for a good reference that has a detailed treatment of obstruction theory in the case where the target space is not simple. The specific situation I am interested in involves lifting a map of 3-skeletons from a $K(G, 1)$ to an arbitrary homotopy 3-type $X$ to the 4-skeletons (and hence to a true map between the spaces); the obstruction to this "should" live in $H^4(G, \pi\_3(X))$ (with the appropriate action of $G$ on $\pi\_3(X)$) via a local coefficient system, but I've been having trouble hashing out the details. Experts I've asked have given answers ranging from saying that it's impossible to saying that they're certain that it's possible but they don't know a reference that does it. Books I've looked at tend to gloss over the details, but they seem to indicate that this should work. Can anybody set me straight on this?
| https://mathoverflow.net/users/396 | Obstruction theory for non-simple spaces | Paul Olum developed some obstruction theory for maps into non-simple spaces back in the 1940-ies and 50-ies. You may want to check out his paper "Obstructions to extensions and homotopies", Annals of Mathematics, Vol 52, 1950, pp 1-50, if you have not looked at it yet.
| 14 | https://mathoverflow.net/users/6668 | 31149 | 20,257 |
https://mathoverflow.net/questions/31150 | 4 | The Hardy-Littlewood Conjecture F [1] involves the infinite product
$$\prod\left(1-\frac{1}{\varpi-1}\left(\frac D\varpi\right)\right)$$
where $\varpi$ ranges over the odd primes and $\left(\frac D\varpi\right)$ is the Legendre symbol.
Is there a good way to calculate this? The product converges very slowly, and none of the standard methods (Cohen-Villegas-Zagier, Wynn, etc.) seem to work because of the unpredictable sign changes.
Given that *D* is fixed, it suffices to calculate the partial products in various congruence classes; I don't know if this is a viable approach.
Another possibility: I've seen almost magical series acceleration with the zeta function, it may work here.
[1] G. H. Hardy, J. E. Littlewood. "Some of the problems of partitio numerorum III: On the expression of a large number as a sum of primes". *Acta Mathematica* **44** (1923), pp. 1-70.
| https://mathoverflow.net/users/6043 | Calculating the infinite product from the Hardy-Littlewood Conjecture F | This problem was studied by a few, and the ideas involve too much latex to write here. Mainly there are ideas of transforming to crazy weighted sums and then use ERH to bound errors from crazier integrals. It suffices to say that the culmination of this research is the freely available paper:
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> [New Quadratic Polynomials With High Densities Of Prime Values](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.46.3281), Jacobson and Williams 1999
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In that year, they used their ideas to compute the constant for discriminants of up to 72 digits! (assuming ERH) The bottleneck of the process seems to be the calculation of the algebraic invariants class number and regulator. Over a decade has passed, the technology today should be able use the same methods to get up to 100 digits, within reasonable time, and up to 110 digits with a bit more time (apparently, two weeks on a cluster):
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> [Practical Improvements to Class Group and Regulator Computation of Real Quadratic Fields](http://arxiv.org/abs/1005.0205), Biasse and Jacobson 2010
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> [Improvements in the Computation of Ideal Class Groups of Imaginary Quadratic Number Fields](http://hal.inria.fr/docs/00/39/74/08/PDF/biasse.pdf), Biasse 2009
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| 2 | https://mathoverflow.net/users/2024 | 31161 | 20,264 |
https://mathoverflow.net/questions/31153 | 10 | I'm inspired by the polymath project. It might be great for few undergraduates to work together on a research topic.
What are some research problems with the following properties(Experimental mathematics is a field containing problems with the criteria below):
1. Accessible to undergraduates
2. There can be many reasonable approaches to the problem
3. People with computer science, applied math or other related backgrounds can also contribute
| https://mathoverflow.net/users/6886 | Problem suggestions for polymath for undergraduates research | Pick any of the problems in the archives of [Al Zimmermann's Programming Contests](http://www.azspcs.net/), and make progress either on the theoretic side (tighter upper bounds / lower bounds / asymptotics) or the computational side.
A specific nice example could be [Point Packing](http://www.azspcs.net/Contest/PointPacking).
| 2 | https://mathoverflow.net/users/25 | 31162 | 20,265 |
https://mathoverflow.net/questions/31163 | 15 | Let $l^2$ be a Hilbert space of infinite sequences $(z\_0, z\_1, \cdots)$ with finite $\sum\_{i=0}^{\infty} |z\_i|^2$.
Are there any simple example of unbounded linear opearator $T: l^2 \to l^2$ with $D(T)=l^2$?
| https://mathoverflow.net/users/7079 | Unbounded linear operator defined on $l^2$ | No there aren't any simple, or even any constructive, examples of everywhere defined unbounded operators. The only way to obtain such a thing is to use Zorn's Lemma to extend a densely defined unbounded operator. Densely defined unbounded operators are easy to find.
Zorn's lemma is applied as follows. Let $A$ be an operator on a domain $\mathcal D$. Consider the set $E$ of all extensions of $A$, that is the collection of operators $A'$ on domains $\mathcal D' \supset \mathcal D$ that agree with $A$ when restricted to $\mathcal D$. Then $E$ is partially ordered by inclusion on domains. Furthermore, any linear chain has an upper bound, by taking unions of domains. So there is a maximal element by Zorn. Finally, suppose the maximal element $A$ is defined on a domain $\mathcal D'$ that is not all of $\ell^2$. Let $v$ be any vector in the complement of $\mathcal D'$. Define an extension of $A'$ on $\mathcal D'+\{a v\}$ by, say, mapping $v$ to zero. This contradicts maximality, so any maximal element is globally defined.
| 15 | https://mathoverflow.net/users/6781 | 31166 | 20,267 |
https://mathoverflow.net/questions/31165 | 6 | Let $K$ be a algebraic number field of degree $n$ over $\mathbb{Q}$, and $O$ its ring of integers. Let $P$ be a prime ideal of $O$ and $(p)=P \cap \mathbb{Z}$.
Is it true that the localization $O\_{P}$ is a rank $n$ free module over $\mathbb{Z}\_{(p)}$ (the localization of $\mathbb{Z}$ at $(p)$) if and only if $P$ is the only prime above $(p)$?
| https://mathoverflow.net/users/7456 | Localizations as free, finite rank modules | Well, if $P$ is not the only prime above $p$, then $O\_P$ cannot be a finitely-generated $\mathbb{Z}\_{(p)}$-module for the following reason. Suppose $Q$ is another prime ideal above $p$ and select $\beta\in Q\setminus P$. Then $\beta^{-1}\in O\_P$. If $O\_P$ were finitely-generated as a module over $\mathbb{Z}\_{(p)}$, then it would be integral over $\mathbb{Z}\_{(p)}$, and hence would be contained in the integral closure of $\mathbb{Z}\_{(p)}$ in $K$, which is $O\_p$. But then $\beta^{-1}\in O\_p$, so $1/\beta=\alpha/m$ for some integer $m$ not divisible by $p$. This means that $m=\alpha\beta\in Q$, whence $m\in Q\cap\mathbb{Z}=(p)$, a contradiction.
| 6 | https://mathoverflow.net/users/4351 | 31182 | 20,276 |
https://mathoverflow.net/questions/31154 | 24 | Recall that a **$(k,k+1,\dots,k+n)$-TQFT** is (supposed to be) a functor from the $n$-category whose $j$-morphisms are (isomorphism classes of) compact $(k+j)$-dimensional manifolds with boundary to some target category, usually your favorite version of $n$-Vect. When $k=0$, a full "classification" of TQFTs with a given target category is given in:
* Lurie, Jacob. On the classification of topological field theories. *Current developments in mathematics, 2008*, 129--280, Int. Press, Somerville, MA, 2009. 58Jxx (57Rxx) [MR2555928](http://www.ams.org/mathscinet-getitem?mr=2555928). [arXiv:0905.0465](http://arxiv.org/abs/0905.0465).
Or, rather, Lurie first provides reasonable definitions for a number of things, end then proves that there is an equivalence of $n$-categories between the $(0,\dots,n)$-TQFTs with target $\mathcal V$ and the $n$-groupoid of ("fully") dualizable objects in $\mathcal V$. (The classification is not particularly effective in two ways: given a dualizable object, which is the value the TQFT assigns to a point, it can be still very hard to understand the functor on complicated manifolds; and given a category, it can be still very hard to classify its dualizable objects.) For a review, see [nLab: cobordism hypothesis](http://ncatlab.org/nlab/show/cobordism+hypothesis).
But Lurie's result does not describe all gadgets that deserve to be called "TQFT"s. For example, it is a classical folk theorem that $(1,2)$-TQFTs are the same as commutative cocommutative [Frobenius algebras](http://en.wikipedia.org/wiki/Frobenius_algebra). I think that there are other similar results of this nature, but I don't know of any theory that puts them all into a single framework. Hence:
>
> **Question:** Is there a classification, similar to Lurie's, for $(k,\dots,k+n)$-TQFTs with a give target $n$-category?
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| https://mathoverflow.net/users/78 | What's the current state of the classification of not-fully-extended TQFTs? | Moore and Seiberg's result (Phys. Lett. **212B** (1988) p.451) on classifying modular functors can be thought of as classification of (1,2,3) theories. (M&S only do the 1 and 2 of (1,2,3), but it's not hard to extend to 3 as well; see "On Witten's 3-manifold Invariants" [here](http://canyon23.net/math/).)
My guess is that extending this style of classification to any of the adjacent slots (1,2,3,4), (2,3,4) or (2,3) would be very difficult. For (1,2,3,4) one would need to start by describing a categorified action of mapping class groups of surfaces in terms of local data; the uncategorified version is already long and messy (see refs above). For (2,3,4) one would need to characterize mapping class groups of 3-manifolds in terms of local data (Hatcher-Thurston for 3-manifolds).
| 21 | https://mathoverflow.net/users/284 | 31184 | 20,277 |
https://mathoverflow.net/questions/31179 | 3 | Let $S$ be an irregular surface of general type over $\mathbb{C}$ and $a \colon S \to A:=\textrm{Alb}(S)$ be its Albanese map. Let Def($S$) and Def($A$) be the bases of the Kuranishi family of $S$ and $A$, respectively. Then $a$ induces a map $f \colon \textrm{Def}(S) \to \textrm{Def}(A)$, whose differential is $f\_\* \colon H^1(S, T\_S) \to
H^1(A, T\_A)$.
Since every small deformation of $S$ is again a surface of general type, if Def(S) is generically smooth then the image of $f\_\*$ is contained in the subspace of dimension $\frac{g(g+1)}{2}$ of $H^1(A, T\_A)$ corresponding to the algebraic deformations of $A$ (here $g := \dim(A)$ ).
Is this still true when Def($S$) is not generically smooth, i.e. everywhere non-reduced?
I suspect that the answer should be "yes", but I would like to see a rigorous proof (or a counterexample, if my guess is wrong).
| https://mathoverflow.net/users/7460 | Variation of the Albanese map | Yes, this is true. The point is that the Albanese is well defined in families, as a family of abelian varieties; that is, given a flat family of smooth projective varieties, there is a projective smooth family of Albanese varieties over the same base. This family is the dual of the family of Pic^0, which is well known to be smooth and projective.
| 4 | https://mathoverflow.net/users/4790 | 31190 | 20,281 |
https://mathoverflow.net/questions/31091 | 0 | In Lambek and Scott's "Introduction to higher order categorical logic" (1988), they state that every Heyting Algebra can be understood as a bicartesian closed category.
On the other hand, fixing a bicartesian closed category, and using $A \cong B$ to denote that morphisms1 exists between $A$ and $B$, we can see that every bicartesian closed category exhibits the [intuitionistic equational axiomatization](http://en.wikipedia.org/wiki/Heyting_algebra#Characterization_using_the_axioms_of_intuitionistic_logic) of a Heyting algebra. Specifically, we can observe that:
1. If $X \to Y \cong 1$ and $Y \to X \cong 1$ then $X \cong Y$
2. If $1 \to X \cong 1$ then $X \cong 1$
3. $X \to (Y \to X) \cong 1$
4. $(X \to (Y \to Z)) \to (X \to Y) \to (X \to Z) \cong 1$
5. $X \times Y \to X \cong 1$
6. $X \times Y \to Y \cong 1$
7. $X \to Y \to X \times Y \cong 1$
8. $X \to X + Y \cong 1$
9. $Y \to X + Y \cong 1$
10. $(X \to Z) \to (Y \to Z) \to (X + Y \to Z) \cong 1$
11. $0 \to X \cong 1$
Here $\to$ is an exponential, $\times$ is a product, and $+$ is a co-product, $1$ is a final object and $0$ is an initial object.
I cannot find the statement of this in Lambek & Scott, however. So I have two questions:
(A) Does this follow from some general theorem regarding bicartesian closed categories?
(B) Is this a folk theorem, or is there a place in the literature where this is established?
---
I originally wrote isomorphism here, but as Andreas Blass notes this is not true (for instance, in the category of sets). However, as noted below, this is true if we weaken the statement to *equimorphic*.
| https://mathoverflow.net/users/7348 | Bicartesian closed categories and Heyting algebras | As Andreas Blass observed, those identities do not hold in all bicartesian closed categories. However, they are true if "isomorphism" is replaced by "equimorphism." In a poset category, equimorphism and isomorphism are the same and thus these equations do verify that a bicartesian closed poset category is a Heyting algebra.
That said, I suppose that the answer to your underlying question is the [Curry–Howard isomorphism](http://en.wikipedia.org/wiki/Curry%E2%80%93Howard_correspondence). Under this interpretation, an intuitionistic proof of a proposition like $X \land Y \to X$ can be interpreted as a morphism $X \times Y \to X$ or $1 \to X^{X \times Y}$. In this particular case, the obvious proof gives the projection $\pi\_1:X \times Y \to X$ but this is by no means the only possible morphism $X \times Y \to X$.
| 2 | https://mathoverflow.net/users/2000 | 31192 | 20,282 |
https://mathoverflow.net/questions/30168 | 4 | Suppose a $\mathcal{H}^{1}$ measurable set $A\subset \mathbb{R}^{n}$ has positive Hausdorff density $\Theta^{1}(\mathcal{H}^{1},A,x)=c>0$ in a point $x\in A$.
If we have a decomposition $A=B\cup C$ with disjoint $B$ and $C$, can it happen that $\Theta^{1}(\mathcal{H}^{1},B,x)$ and $\Theta^{1}(\mathcal{H}^{1},C,x)$ do not
exists? Can it even happen that the lower densities of these sets are both zero? Are there conditions on $A$ to prevent this?
More specifically:
Let $C\_{s,\epsilon}(x)$ denote the double cone in direction $s\in\mathbb{S}^{n-1}$ with opening angle $\epsilon$ at the point $x$.
Is it possible that $\Theta^{1}(\mathcal{H}^{1},A\cap C\_{s,\epsilon}(x),x)$ does not exists for all $s\in \mathbb{S}^{n-1}$ and all $\epsilon\in (0,\epsilon\_{s})$?
($\epsilon\_{s}$ is some maximal angle depending on the direction $s$) Is it possible that the lower Hausdorff density vanishes for all these double cones?
| https://mathoverflow.net/users/1272 | Does positive density imply existence of the density for some part of a decomposition? | Today at our problem coffee someone (I don't know if he wants his name mentioned) showed me a counterexample to the first part of my question, in case anyone else is interested.
We take $X:= [0,1]\subset \mathbb{R}^{2}$ - for $\mathbb{R}^{n}$ the same argument should work, only with slightly more complicated notation - and $x=0$.
Let $\lambda\_{k}\in (0,1)$
be a positive sequence converging monotonically to zero. We define the monotonically decreasing sequence
$a\_{n}=\prod\_{i=1}^{n}\lambda\_{i}$ and for $n\in\mathbb{N}$ we set
$A\_{0}:=[1,a\_{1}], A\_{n}:=[a\_{2n+1},a\_{2n}], B\_{n}:=(a\_{2n},a\_{2n-1})$
and finally
$A:=\bigcup\_{n\in\mathbb{N}\cup\lbrace 0\rbrace} A\_{n}\cup\lbrace 0\rbrace$, $B:=\bigcup\_{n\in\mathbb{N}}B\_{n}$.
Since $a\_{n}\leq \lambda\_{1}^{n}\rightarrow 0$ we have $A\cup B=X$.
If we choose $r\_{n}=a\_{2n}$ and $R\_{n}=a\_{2n+1}$ we get
$\mathcal{H}^{1}(A\cap \overline B\_{R\_{n}}(x))\leq a\_{2n+2}=\prod\_{i=1}^{2n+2}\lambda\_{i}\leq \lambda\_{2n+2}R\_{n}$
and
$\mathcal{H}^{1}(B\cap \overline B\_{r\_{n}}(x))\leq a\_{2n+1}=\prod\_{i=1}^{2n+1}\lambda\_{i}\leq \lambda\_{2n+1}r\_{n}$, which shows that both lower densities are $0$.
| 1 | https://mathoverflow.net/users/1272 | 31196 | 20,283 |
https://mathoverflow.net/questions/31198 | 20 | Since a lie group is a manifold with the structure of a continuous group, then each point of the manifold [Edit: provided we fix a metric, for example an invariant or bi-invariant one] has some scalar curvature R.
>
> **Question** [Edited] Is there a nice formula which expresses the scalar curvature at a point of the manifold in terms of the lie algebra of the group?
>
>
>
| https://mathoverflow.net/users/7466 | Curvature of a Lie group | See Exercice 1 in Chapter 4 of Do Carmo's "Riemannian Geometry".
The formula is $R(X,Y)Z = \frac 1 4 [[X,Y], Z]$.
In particular, if $X$ and $Y$ are orthonormal, the sectional curvature of the generated plane is
$K(\sigma)= \frac 1 4 \|[X,Y]\|^2$
Which is always $\geq 0$.
EDIT: In view of the comments, it is important to add that this is for a bi-invariant metric.
| 23 | https://mathoverflow.net/users/5753 | 31200 | 20,285 |
https://mathoverflow.net/questions/26839 | 26 | There is a unique nonempty set $B$ of nonnegative integers such that
every positive integer can be written in the form
$$b + s^2, b\in B, s\ge0$$
in an even number of ways.
$B = \{0, 1, 2, 3, 5, 7, 8, 9, 13, 17, 18, 23, 27, 29, 31, 32, 35, 37,$
$ 39, 41, 45, 47, 49, 50, 53, 55, 59, 61, 63, 71, 72,$
$ 73, 79, 81, 83, 87, 89, 91, 97, 98, 101, 103, 107,$
$ 109, 113, 115, 117, 121, 127, 128, 137, 139, 149,$
$ 151, 153, 157, 159, 162, 167, 171, 173, 181, 183,$
$ 191, 193, 197,\dots\}$
Does the set $B$ have positive density?
Now for some context. Every set $A$ of nonnegative integers that contains 0 has a unique set $B$ of nonnegative integers so that
$$\left( \sum\_{a\in A} q^a \right) \, \left( \sum\_{b\in B} q^b \right) = 1$$
in the ring ${\mathbb F}\_2[[q]]$ of binary power series. We call $B$ the reciprocal of $A$.
As a consequence of a Euler's pentagonal number theorem, the reciprocal of the set $\{n(3n+1)/2 \colon n \in \mathbb{Z}\}$ is the set $\{ n \colon p(n)\equiv 1 \bmod 2\}$, where $p(n)$ is ordinary partition function. Almost nothing interesting is known about the parity of the partition function, but computationally it seems to be even and odd with equal frequency. This question arises out of an effort to put the parity of the partition function into some context.
In this article ([arxiv](http://arxiv.org/abs/math/0506496), [Int. J. Number Theory 2 (2006), no. 4, 499--522](http://www.worldscinet.com/ijnt/02/0204/S1793042106000693.html)), Josh Cooper, Dennis Eichhorn and I investigated the properties of $A$ that lead to $B$ having positive density, and all of our data and partial results can be summed up in the following conjecture:
>
> **Conjecture**: If $A$ contains 0,
> is not periodic, and is uniformly
> distributed in every congruence class
> modulo every power of 2, then $B$ has
> positive density.
>
>
>
Letting $A$ be the set of squares, we were able to prove that the even numbers in $B$ are exactly $\{2k^2 \colon k\ge 0\}$, and we were able to classify the $1\mod 4$ elements of $B$.
**Update** Greg Kuperberg's answer concerning the conjecture displayed above is, while not quite a disproof, utterly convincing. So convincing, I can no longer understand how I thought the conjecture could plausibly be true. In [our paper](http://arxiv.org/abs/math/0506496), we described it as "the strongest conjecture that is consistent with our theorems, our experiments, and Conjecture 1.1", so I see we weren't too enthusiastic about its truth. We should have been even less so!
The question directly asked, the density of the reciprocal of the squares, remains unanswered. Paul Monsky has introduced a new (to me, at least) approach, and has made striking progress both in the answer below and in his answer to [this question](https://mathoverflow.net/questions/28462/why-are-there-usually-an-even-number-of-representations-as-a-sum-of-11-squares).
I love Greg's answer to the question I didn't dare ask, and want to accept it, but Paul's is more directly relevant to the question I did ask.
Here are some computational counts of the number of elements of $B\cap[0,2^{23}]$ in particular congruence classes.
```
(1 mod 4, 371867), (3 mod 4, 760697)
(1 mod 8, 185336), (5 mod 8, 186531), (3 mod 8, 294045), (7 mod 8, 466652)
(1 mod 16, 92703), (5 mod 16, 93236), (9 mod 16, 92633), (13 mod 16, 93295),
(3 mod 16, 147232), (11 mod 16, 146813),
(7 mod 16, 204808), (15 mod 16, 261844)
(7 mod 32, 102487), (23 mod 32, 102321),
(15 mod 32, 130895), (31 mod 32, 130949)
```
Since there was a specific request for 15 mod 32 data, here are the first 10 such numbers in $B$: (47,79,271,559,623,687,719,815,879,911). Here are the last 10 that I've computed: (8388539, 8388551, 8388559, 8388563, 8388567, 8388571, 8388581,
8388591, 8388593, 8388603, 8388607)
| https://mathoverflow.net/users/935 | How thick is the reciprocal of the squares | In a related question, (Why are there usually..), O'Bryant characterized the elements of B that = 3 mod 8, and asked why the number of such that are at most X appears to be small; my
answer to his question showed that the number is O(Xloglog(X)/log(X)). In this answer I'll
sketch a proof that the same result holds for elements of B that =7 mod 16. (The remaining
case of elements that =15 mod 16 looks much harder). We need a characterization result:
Lemma : Let g in Z/2[[x]] be 1+x+x^4+x^9+.., the exponents being the squares. If n=16m+1,
the coefficient of x^n in 1/g^7 is 1 precisely when n is a square.
To see this let f=1+x+x^3+x^6+.., the exponents being the triangular numbers. Then xf^8+g=
g^4, so 1/g^7=(xf^8/g^8)+(1/g^4). Since n is odd, the coefficient in question is that of x^n
in xf^8/g^8, which is the coefficient of x^2m in f/g. My answer to my MO question, "Variations..", shows that this is 1 precisely when 2m is triangular, i.e. when n is a square.
Characterization result: Suppose k=7 (16). Then k is in B precisely when the number of ways of writing k as 2(square)+4(square)+square is odd.
Proof: 1/g=(g^2)*(g^4)*(1/g^7). So the coefficient of x^k in 1/g is the mod 2 reduction of
the number of ways to write k as 2(square)+4(square)+(the exponent,c, of a monomial
appearing non-trivially in 1/g^7). c must be odd, and since 1/g^7=g/g^8, c=1 (8). If
c=9 (16), then mod 16, 7=2(square)+4(square)+9, which is impossible. So c=1 (16), and by the lemma c is a square. Conversely if k=2(square) +4(square) +a square s, then s=1 (16), and the lemma shows that x^(s) appears non-trivially in 1/g^7.
Theorem: Suppose k in B=7(16). Then there are at most 2 primes occurring to odd exponent
in the factorization of k. So the number of such k in B that are at most X is O(Xloglog(X)/
log(X)).
(The proof is along the same lines as my answer to O'Bryants 11 square problem, but now one has to work with the ternary form x^2+y^2+2\*z^2 rather than Gauss' x^2+y^2+z^2).
| 6 | https://mathoverflow.net/users/6214 | 31204 | 20,289 |
https://mathoverflow.net/questions/31173 | 3 | This can be seen as a follow up my question here:
[Is there a notion of "fibered category with boxproducts"?](https://mathoverflow.net/questions/28152/is-there-a-notion-of-fibered-category-with-boxproducts)
Given a monoidal fibration $f:E\rightarrow B$
(i.e. a strict monoidal functor between monoidal categories which is a fibration of ordianary categories)
where the base is a cartesian monoidal category endowed with a grothendieck topology.
What are the right conditions for such a fibration be called a stack?
I guess it is not enough to ask that $E(X)\rightarrow Desc(X,U)$ is an equivalence of ordinary categories. Insted one should need some further condition that ensures the following:
"if $(\phi\_i)$ can be glued to $\phi$ and $(\psi\_j)$ can be glued to $\psi$ than $(\phi\_i\boxtimes \psi\_j)$ can be glued to $\phi\boxtimes \psi$"
Does this notion exist yet? What would be the right condition?
Examples I have in mind are
$B$=geometric objects for example smooth varieties and
$E$=sheaves for example $\mathcal{D}\_X$-modules
| https://mathoverflow.net/users/2837 | Notion of stack fibered in monoidal categories? | I would take the view point that a monoidal category is a bicategory with one object. (Then a category fibered in monoidal categories should be the same thing as a weak functor into bicategories that "only hits monoidal categories".) In other words, what you should have is that this fibration is a 2-stack when viewed as a fibration in bicategories. The descent condition should then be that the canonical map $E|\_X \to Desc(X,U)$ be an equivalence of bicategories, where each of these monoidal categories is viewed as a bicategory, which is equivalent to Jeff's comment; this is just saying that $E|\_X \to Desc(X,U)$ is a monoidal equivalence.
| 4 | https://mathoverflow.net/users/4528 | 31206 | 20,291 |
https://mathoverflow.net/questions/31212 | 5 | Let $g, r, a, b$ be positive integers. In Friedman's urn model we have an urn with $r$ red and $g$ green balls in it. In each step we take one ball out of urn, register its color and return it to the urn. Additionally, we put $a$ more balls of this color and $b$ more balls of the other color.
Let $X\_n$ be the relative amount of green balls in the box after $n$-th step. It can be proven (e.g. Richard Durrett, *Probability: Theory and Examples*, pages 254-255) that
$$\lim\_{n\to\infty}X\_n = \frac12\quad\text{a.e.}$$
If one thinks about this statement for a second it would most probably strike him as extremely counter-intuitive. The proof cited above make use of square integrable martingales, and unfortunately doesn't seem to give *intuitive* explanation of this phenomena.
I'm looking for an explanation which would explain on some heuristic level why this result is in some sense logical. An idea for more intuitive proof would also definitely be helpful.
I would also like to note that if we take $b=0$ the model becomes well-known Polya-Eggenberger urn model for which we have
$$\lim\_{n\to\infty}X\_n\sim B\left(\frac{g}{a}, \frac{r}{a}\right).$$
| https://mathoverflow.net/users/6159 | Intuitive "proof" or explanation of a result in Friedman's urn | You might be interested in [the article](http://www.jstor.org/stable/2238205?seq=2) by David A. Freedman on Friedman's urn. He reports a simple and intuitive proof due to Ornstein, which only uses the strong law
of large numbers.
In his notation the urn contains $W\_n$ white balls and $B\_n$ black balls at time $n$, $a$ and $b$ have the same meaning as in the question.
>
> D. Ornstein has obtained this very intuitive proof that $(W\_n + B\_n)^{-1}W\_n$ converges
> to $1/2$ with probability 1 for $b > 0$. Suppose first $a > b$. If $0 \leq x \leq 1$ and
> $$\mathbb P\left\{\limsup \frac{W\_n}{W\_n + B\_n} \leq x\right\} = 1,$$
> by an easy variation of the Strong Law, with probability 1, in $N$ trials there will be at most $Nx + o(N)$ drawings of a white ball; so at least $N(1 - x) - o(N)$ drawings of black. Therefore, with probability 1, $\limsup (W\_n+ B\_n)^{-1}B\_n$ is bounded above by
> $$\lim\limits\_{N\to\infty}\frac{a[Nx + o(N)] + b[N(1 - x) - o(N)]}{N(a + b)}=\frac{b+(a-b)x}{a+b}.$$
> Starting with $x = 1$ and iterating,
> $$\mathbb P\left\{\limsup \frac{W\_n}{W\_n + B\_n} \leq \frac{1}{2}\right\} = 1$$
> follows. Interchange white and black to complete the proof for $a > b$. If $a < b$, and
> $$\mathbb P\{\limsup (W\_n + B\_n)^{-1}W\_n \leq x\} = 1,$$
> then a similar argument shows
> $$\mathbb P\left\{\limsup\frac{B\_n}{W\_n + B\_n} < \frac{a+(b-a)x}{a + b}\right\}=1$$
> The argument proceeds as before, except both colors must be considered simultaneously.
>
>
>
| 4 | https://mathoverflow.net/users/5371 | 31218 | 20,295 |
https://mathoverflow.net/questions/30238 | 13 | Reading Princeton Companion I found out that every finitely presented group can be realized as the fundamental group of a 4-manifold.
When starting to write this answer I found this related [MO question](https://mathoverflow.net/questions/15411/finite-generated-group-realized-as-fundamental-group-of-manifolds/15421#15421). However, my question has to do with one of its answers (which is similar to the hint given in the Princeton Companion).
The proceedure for constructing this manifold from a given presentation is first to construct a CW-complex with that fundamental group (by wedge sum of circles puting 2-cells to cover the relations) embedding it in $R^5$ and considering the frontier of a tubular neighborhood.
Two questions come up to me (which are maybe trivial):
1- Why this cannot be done in $R^4$? Or can it be but the result is not the same?
2- Why the resulting manifold has the desired fundamental group?
| https://mathoverflow.net/users/5753 | Constructing 4-manifolds with fundamental group with a given presentation. | Related to question (1), suppose you wanted to get a 4-manifold with boundary as a submanifold of $\mathbb{R}^4$ with fundamental group a finitely presented group, with presentation complex $K$. It is a [result of Curtis](http://www.ams.org/mathscinet-getitem?mr=140114) that any 2-complex $K$ is
homotopy equivalent to a 2-complex which embeds in $\mathbb{R}^4$ (see also the Stallings
reference in the comment on this [MO question](https://mathoverflow.net/questions/19618/when-does-a-cw-complex-of-dimension-2-embedd-in-r4) and [Dranishnikov-Repovs](http://www.ams.org/mathscinet-getitem?mr=1201430) - in fact this works if the 2-complex $K$ has only one vertex). However, to thicken up $K\subset \mathbb{R}^4$ to get a tubular neighborhood which is a 4-manifold (with boundary) in $\mathbb{R}^4$ which retracts to $K$, $K$ must be nicely embedded. For example, the 2-cells should be locally flat. I'm not sure if this holds true for the embeddings of Curtis or for the other constructions.
To answer (2), think about what happens when you thicken up the complex. Thickening some points in $\mathbb{R}^5$ gives some 5-balls, whose boundary is a union of 4-spheres. Thickening an interval attached to some points, one gets a 1-handle $D^1\times D^4$, whose boundary removes two 4-balls from the 4-spheres and attaches in $D^1 \times S^3$, giving a 4-manifold with free fundamental group (a connect sum of $S^3\times S^1$'s). The 2-cells thicken up to 2-handles $D^2\times D^3$, which remove $S^1\times D^3$ from the 4-manifold (which doesn't change the fundamental group, since $S^1$ is codimension 3), and replaces it with $D^2 \times S^2$, which has the effect of killing the element in the free group corresponding to the circle by Van Kampen. So you see that this gives the same thing as the construction in Henry Wilton's answer to the other question.
| 7 | https://mathoverflow.net/users/1345 | 31226 | 20,299 |
https://mathoverflow.net/questions/31223 | 7 | In BBD mixed sheaves and weights for them were only defined for ($\overline{\mathbb{Q}\_l}$-)sheaves over a variety $X\_0$ defined over a finite field $F$. Weights start to behave better when one extends coefficients from $F$ to its algebraic closure i.e. passes from $X\_0$ to $X$.
Now, BBD was published in 1982. Are any significant improvements and/or generalizations known in this field now? There is a paper by Huber and a book by Jannsen where mixed sheaves and weights for them are mentioned. Yet these authors were not able to generalize the results of BBD (in fact, an example of Jannsen shows that this is probably impossible). They also didn't extend scalars. So, are there any other papers on this subject?
Upd. There seems to be two basic ways to define weights (for sheaves) explicitly. The first method uses weights of Hodge structures. It seems that this method can work only for something like the category of mixed Hodge modules. Possibly, I will study these categories in the future. Yet at the moment I study motives, and it seems that 'motivic' people usually do not understand mixed Hodge modules (and so did not relate them with motives).
So, I am currently interested in the second method. It uses the eigenvalues of the Frobenius action. So, was anything interesting done using THIS approach after 1982?
| https://mathoverflow.net/users/2191 | In what setting does one usually define mixed sheaves and weights for them? | Admittedly, I'm almost completely ignorant about the $\ell$-adic setting. But, in case the following at least gets at the spirit of your question: in the de Rham setting, I found an old (1990s) preprint of Saito ("On the formalism of mixed sheaves," now TeXed up and available [on the arxiv](http://arxiv.org/abs/math/0611597)) useful in understanding the formal structure of (the system of categories of) mixed sheaves. The book by Peters and Steenbrink ("Mixed Hodge Structures") also has a nice section (14.1, "An axiomatic introduction") in the chapter (14) on mixed Hodge modules that explains the picture well.
| 7 | https://mathoverflow.net/users/2628 | 31229 | 20,301 |
https://mathoverflow.net/questions/31178 | 12 | I've asked a question like this before, but now I'm more interested in counting the number of covers.
We suppose given the following data.
1. A positive integer $d$
2. A finite set of closed points $B= ( b\_1,\ldots,b\_n )$ in $\mathbf{P}^1\_\mathbf{C}$
3. Branch types $T\_1,\ldots, T\_n$.
**Question.** How many branched covers of $\mathbf{P}^1\_\mathbf{C}$ exist which are branched only over $b\_i$ (with branch type $T\_i$ over each $b\_i$)?
The answer lies within the Hurwitz number for $(T\_1,\ldots,T\_n)$. This translates the problem to combinatorial group theory.
Now, for my main question:
**Q1.** Can one ``count'' covers of $\textrm{Spec} \mathbf{Z}$ as above? That is, can one count the number of finite field extensions $$\mathbf{Q}\subset K$$ of given degree $d=[K:\mathbf{Q}]$ which are unramified outside a given set of prime numbers $p\_1,\ldots,p\_n$ with ramification types $T\_1,\ldots,T\_n$?
I know that one can use Minkowski's Geometry of Numbers to give some nontrivial bounds on the discriminant. Is this the best we can do?
| https://mathoverflow.net/users/4333 | Counting branched covers of the projective line and Spec Z | One thing to keep in mind is that the analogue of Spec Z is really P^1 over a finite field k, not P^1/C. And here already one does *not* have a simple "Hurwitz-type formula" for the number of G-covers with given branching which are defined over k.
Just to give an example which may be illustrative; suppose that G = S\_3, and you require that the inertia at the primes p\_1, ... p\_n is tame and maps to a transposition in G. The extensions of Q of this kind are more or less in bijection with the etale Z/3Z covers of the quadratic field K = Q(sqrt(N)) where N = p\_1....p\_n. (I am being careless about the real place here.) In any event, to "count" the number of covers is in effect to compute the size of the 3-part of the class group of K. There is not going to be a nice formula for this, and in particular it will depend unpredictably on the primes in question. On the other hand, you can compute the *average* of this quantity over squarefree integers N, by Davenport-Heilbronn.
So I would say:
"No" to your Q1 as stated.
"Yes, for some choices of G," to your Q1 *on average* -- e.g. for G = S\_3 (by Davenport-Heilbronn), for G = D\_4 (Cohen-Diaz y Diaz-Olivier), for G = S\_4, S\_5 (by Bhargava, though perhaps some slight and presumably true refinement of Bhargava to squarefree discriminants is needed), for G = D\_p when K is F\_ell(T) by work of myself, Venkatesh and Westerland.
| 9 | https://mathoverflow.net/users/431 | 31233 | 20,303 |
https://mathoverflow.net/questions/31243 | 5 | What is the entropy of a normal distribution with mean 0 and variance \sigma?
Thanks!
| https://mathoverflow.net/users/7475 | entropy of normal distribution | I found [here](http://www.cis.hut.fi/ahonkela/dippa/node94.html) that "the negative differential entropy of the normal distribution"
(which may not be what you are asking for?) is:
$$-\frac{1}{2} [ \log (2 \pi \sigma^2 ) + 1 ] ,$$
independent of $\mu$.
| 6 | https://mathoverflow.net/users/6094 | 31245 | 20,309 |
https://mathoverflow.net/questions/31208 | 6 | Let $f:[n]\times [n] \rightarrow [0,1]$ be a function from pair of integers to the real interval [0.1]. I would like to find sets of complex vectors
$X= \{x\_i\}$ and $Y=\{y\_j\}$ satisfying $x\_i\cdot y\_j=f(i,j)$, in such a way that
the vectors in $X$ and in $Y$ are as small as possible. More precisely,
set $m= \max\_{i,j} f(i,j)$ and $N=\max\_{i,j}[\|x\_i\|,\|y\_j\|]$.
1. What is the minimal $N$ such that $x\_i\cdot y\_j = f(i,j)$ for all $i,j\in [n]$?
2. Is there an upper bound on $N$ purely in function of $m$, i.e., with no dependence
on $n$?
3. If the answer to question two is no, what is the best upper bound that we can
give for $N$ in function of $n$ and $m$? A trivial upper bound is
$$N \leq \max\_i{\sum\_{j} f(i,j) } \leq mn.$$
but I believe that the dependence on $n$ might be lowered.
cordially,
mateus
| https://mathoverflow.net/users/6442 | Inner products and norms | Such questions have been dealt with. Note first that your $m$ is just the norm of the matrix $F$ (see Robin's comment) as an operator from $\ell\_1^n$ to $\ell\_\infty^n$. $M$ also has a name, it is the $\gamma\_2$ norm of this operator (This is the minimal product of
$$\|Y\|\_{1\to 2}\|X\|\_{2\to\infty}$$ over all factorizations $F=XY$ . $\|Z\|\_{p\to q}$ denotes the norm of Z as an operator from $\ell\_p$ to $\ell\_q$.)
It is not hard to see that $M=\gamma\_2(F)\le \sqrt n \|F\|\_{1\to\infty}=\sqrt n m$.
For a random $0,1$ matrix $F$ one gets that this estimate is tight, up to a universal constant.
You can look here [Link](https://doi.org/10.1007/s00493-007-2160-5) for details.
In particular Cor 5.2 there (it deals with random $\pm 1$ matrices but it is easy to go between those and random $0,1$ matrices).
| 10 | https://mathoverflow.net/users/6921 | 31246 | 20,310 |
https://mathoverflow.net/questions/31250 | 53 | A naive and idle number theory question from a topologist (but not a knot theorist):
I have heard it said (and this came up recently at MO) that there is a fruitful analogy between Spec $\mathbb Z$ and the $3$-sphere. I gather that from an etale point of view the former is $3$-dimensional and simply connected; from the same point of view the subschemes Spec $\mathbb Z/p$ are $1$-dimensional and very much like circles; and the Legendre symbols for two odd primes that figure in quadratic reciprocity are said to be analogous to linking numbers of knots. So, prompted by a recent MO question, I started thinking:
The abelianized fundamental group of the complement of Spec $\mathbb Z/p$ (the group of $p$-adic units) is not terribly different from the abelianized fundamental group of a knot complement (an infinite cyclic group). For nontrivial knots, there is a lot more to the fundamental group of a knot complement than its abelianization. The next little bit, the abelianization of the commutator subgroup (or $H\_1$ of the infinite cyclic cover) has an action of that infinite cyclic group, and I recall that the Alexander polynomial of the knot may be created out of this action.
So there must be some analogue of that in number theory, right? Like, some construction involving ideal class groups or idele class groups of $p$-power cyclotomic fields can be interpreted as the Alexander polynomial of a prime number?
| https://mathoverflow.net/users/6666 | Prime numbers as knots: Alexander polynomial | [This article](http://arxiv.org/abs/0904.3399v1) appears to discuss the relationship between Alexander polynomials in knot theory and Iwasawa polynomials in number theory, although I haven't looked at it in detail. I discovered this paper in [This Week's Finds 257](http://math.ucr.edu/home/baez/week257.html), which gives a number of references for this sort of thing.
| 20 | https://mathoverflow.net/users/396 | 31253 | 20,312 |
https://mathoverflow.net/questions/31131 | 11 | Given a CM field we can use its maximal order (and a choice of CM type) to construct an abelian variety $\mathbb{C}^g/\Lambda$ with complex multiplication by the maximal order.
How do I (or where can I find information on) explicitly write down equations for a projective embedding of this variety, and the action of the CM order on points? Is this implemented anywhere?
For genus one we can use the Eisenstein series to find the coefficients of a Weierstrass model for the elliptic curve. So I'm looking for a generalization of this.
| https://mathoverflow.net/users/2024 | CM field to Torus to Abelian Variety? | This is not the answer. I am adding some relevant papers (some possessing good examples) which won't fit in the comments section.
I have been wanting to know the answer to this question as well. It seems one has to find an ample line bundle, and then calculate the Riemann theta relations which define the projective embedding. The equation(s) of the embedding is decided by relation between the theta functions. The wikipedia page on [equations for abelian varieties](http://en.wikipedia.org/wiki/Equations_defining_abelian_varieties) was written Charles Matthews. Perhaps he might have more to say on this question...
MR0946234 (89i:14038) Barth, Wolf . Abelian surfaces with $(1,2)$-polarization.
Algebraic geometry, Sendai, 1985, 41--84, Adv. Stud. Pure Math., 10, North-Holland, Amsterdam, 1987.
MR1257320 (95e:14033) Barth, W. ; Nieto, I. Abelian surfaces of type $(1,3)$ and quartic surfaces with $16$ skew lines. J. Algebraic Geom. 3 (1994), no. 2, 173--222.
MR1336597 (96h:14064) Barth, W. Quadratic equations for level-$3$ abelian surfaces.
Abelian varieties (Egloffstein, 1993),
1--18, de Gruyter, Berlin, 1995.
MR1602020 (99d:14046) Gross, Mark ; Popescu, Sorin . Equations of $(1,d)$-polarized abelian surfaces.
Math. Ann. 310 (1998), no. 2, 333--377.
MR2194379 (2006m:14054) Gunji, Keiichi . Defining equations of the universal abelian surfaces with level three
structure.
Manuscripta Math. 119 (2006), no. 1, 61--96.
MR0611469 (82h:14028) Sasaki, Ryuji . Some remarks on the equations defining abelian varieties.
Math. Z. 177 (1981), no. 1, 49--60.
MR1048533 (91e:14043) Birkenhake, Ch. ; Lange, H. Cubic theta relations. J. Reine Angew. Math. 407 (1990), 167--177.
| 3 | https://mathoverflow.net/users/5372 | 31255 | 20,313 |
https://mathoverflow.net/questions/31248 | 20 | I have sitting in front of me two 2-categories. On the left, I have the 2-category GPOID, whose:
* objects are groupoids;
* 1-morphisms are (left-principal?) bibundles;
* 2-morphisms are bibundle homomorphisms.
On the right, I have the 2-category ALG, whose:
* objects are algebras (over $\mathbb C$, say);
* 1-morphisms are (adjectives?) bimodules;
* 2-morphisms are bimodule homomorphisms.
And probably I should go through and add "in TOP" to every word on the left and "C-star" to every word on the right.
(I have the impression that ALG is two-equivalent to another category, which I will put on the far right, whose:
* objects are cocomplete VECT-enriched categories (with extra adjectives?);
* 1-morphisms are cocontinuous VECT-enriched functors;
* 2-morphisms are VECT-enriched natural transformations.
In one direction, the functor takes an algebra to its full category of modules. In the other direction, there might be extra adjectives needed, and I mean to appeal to the Mitchell embedding theorem; but I'm pretty sure an equivalence exists if I insist that every object on the far right comes with a cocontinuous faithful VECT-enriched functor to VECT, and my idea is that Mitchell says that every category admits such a functor.
So anyway, the point is that either category on the right or far right is a sort of "algebraic" category, as opposed to the more "geometric" category on the left.)
Then I've been told on numerous occasions that there is a close relationship between GPOID and ALG. See, for example, the discussion at [Geometric interpretation of group rings?](https://mathoverflow.net/questions/27375/) — in fact, it's reasonable to think of the present question as a follow-up on that one.
The relationship is something like the following. To each (locally compact Hausdorff) topological, say, groupoid, we can associate a C-star, say, algebra — the construction restricts in various special cases to: a (locally compact Hausdorff) topological space $X$ going to its algebra of continuous vanishing-at-infinity functions $C\_0(X)$; a finite group $G$ going to its group ring $\mathbb C G$; etc. The construction extends to 1- and 2-morphisms to build a (contra)functor. At least if I get all the adjectives right, the functor should be a two-equivalence.
>
> **Question:** What's the precisification of what I have said above? What exactly is the two-functor from groupoids to algebras, and which adjectives make it into an equivalence of two-categories? Groupoids have natural "disjoint union" and "product" constructions; these presumably correspond to Cartesian product and tensor product (?!? that's not the coproduct in the category of algebras, but maybe in this two-category it is?) on the algebraic side?
>
>
>
Let me end with an example to illustrate my confusion, which I brought up in [Op. cit.](https://mathoverflow.net/questions/27375/). Let $G$ be a finite abelian group; then it has a Pontryagin dual $\hat G$. Now, there is a canonical way to think of $G$ as a groupoid: it is the groupoid $\{\text{pt}\}//G$ with only one object and with $G$ many morphisms. If I'm not mistaken, the corresponding algebra should be the group algebra $\mathbb C G$. But $\mathbb C G$ is also the algebra $C\_0[\hat G]$ of functions on the *space* $\hat G$. And if there is one thing I am certain of, it is that the underlying space of $\hat G$ (a groupoid with no non-identity morphisms) and the one-object groupoid $\{\text{pt}\}//G$ are not equivalent as groupoids. And yet their "function algebras" are the same. So clearly I am confused.
| https://mathoverflow.net/users/78 | What is the precise relationship between groupoid language and noncommutative algebra language? | I think the problem is that the left hand side is part of commutative geometry, while the right hand side is part of noncommutative geometry (regardless of whatever vague claims I may have made in my previous response). Groupoids (or stacks) certainly have interesting noncommutative aspects that are captured by the construction you discuss, but it's not an equivalence. More precisely, (an algebro-geometric version of) your construction attaches to a stack the category of (quasi)coherent sheaves (which is given by replacing algebras by their categories of modules as you suggest). However, one cannot hope to recover the stack from this category, as your finite group example shows: representations of a finite group are the same as (algebraic) vector bundles on BG, or on $\widehat G$, and these are certainly nonisomorphic.
In order to get something that can be an equivalence with appropriate adjectives, we need to change the right hand side into its commutative version: replace categories by tensor (symmetric monoidal) categories - ie remember that vector bundles/coherent sheaves have a tensor product. Now we're in a good situation: Tannakian theory tells us that in a variety of settings this construction is an equivalence. For example tensor of representations corresponds to tensor of vector bundles on BG, but has nothing to do with tensor of vector bundles on $\widehat G$. This is of course anathema to noncommutative geometers - vector bundles on a noncommutative space (modules for noncommutative rings) don't have tensor products. It's in this sense that the LHS is part of commutative geometry.
A nice formal way to say this is by looking at the adjoint functor to the functor from stacks to categories, or to symmetric monoidal categories, given by considering bundles/sheaves with or without tensor product. The first adjoint takes a category to the moduli stack of objects in it; the second is the Tannakian reconstruction functor, sending a tensor category to its "spectrum" (fiber functors over various rings). The first functor is very far from recovering a stack - the moduli stack of objects in a category of modules for a ring is much much bigger than the spectrum of the ring!
| 11 | https://mathoverflow.net/users/582 | 31256 | 20,314 |
https://mathoverflow.net/questions/31254 | 0 | Any hints how to compute this sum
$$\sum\_{i=1}^{N-1}\left[i\frac{K}{N}\right]^{p}?$$
where K < N , $\left[\cdot\right]$ denotes fractional part,
$p\in N$
| https://mathoverflow.net/users/3589 | sum of fractional parts | The article [On Certain Sums of Fractional Parts](https://doi.org/10.1007/BF01238638 "Arch. Math 25, 41–44 (1974). zbMATH review at https://zbmath.org/?q=an:0277.10005") by Gandhi and Williams answers your question for $p=1$; it's likely that since 1974 this result has been generalized, but I wasn't able to find a reference.
| 3 | https://mathoverflow.net/users/35336 | 31258 | 20,316 |
https://mathoverflow.net/questions/31278 | 7 | Let $\mathcal{R}$ be the hyperfinite factor of type $\rm{II}\_1$ and let $\mathbb{F}\_n$ be a free group with $n$ generators. Let $\alpha$ be an action of $\mathbb{F}\_n$ on $\mathcal{R}$.
Does the von Neumann crossed product $\mathcal{R}\rtimes\_{\alpha}\mathbb{F}\_n$ have the QWEP?
Remarks: Since $\mathbb{F}\_n$ is a residually finite group, the group von Neumann algebra $\rm{VN}(\mathbb{F}\_n)$ has the QWEP. Moreover $\mathcal{R}$ has the QWEP.
| https://mathoverflow.net/users/5210 | Does a crossed product R⋊_α F_n of the hyperfinite factor of type II_1 and a free group have the QWEP? | Yes. If $a$ and $b$ are generators of $\mathbb F\_2$ then $\mathcal R \rtimes\_\alpha \mathbb F\_2$ decomposes as an amalgamated free product of $(\mathcal R \rtimes\_\alpha \langle a \rangle)$ and $(\mathcal R \rtimes\_\alpha \langle b \rangle)$ over $\mathcal R$, where each of these are hyperfinite. Brown, Dykema, and Jung showed in <http://arxiv.org/abs/math/0609080> that for separable finite von Neumann algebras being embeddable into $\mathcal R^\omega$ is stable under amalgamated free products over a hyperfinite von Neumann algebra. Thus $\mathcal R \rtimes\_\alpha \mathbb F\_2$ is embeddable into $\mathcal R^\omega$, which is equivalent to QWEP. Induction then gives the case when $2 \leq n < \infty$, and the case $n = \infty$ then follows since QWEP is preserved under (the weak-closure of) increasing unions.
Related to this, Collins and Dykema in <http://arxiv.org/abs/1003.1675> have recently shown that the class of Sophic groups is stable under taking amalgamated free products over amenable groups.
I believe this is an open problem however if we consider arbitrary residually finite groups instead of only $\mathbb F\_n$.
| 9 | https://mathoverflow.net/users/6460 | 31282 | 20,329 |
https://mathoverflow.net/questions/31270 | 19 | Hello, I would like to ask you if there is a mathematical theory, that is complete (in the sense of Goedel's theorem) but practically applicable. I know about Robinson arithmetic that is very limited but incomplete already. So, I would like to know if there is some mathematics that could be practically used (expressiveness) and reduced to logics (completeness).
I'm very new to the site and to the maths as well, so please tell me if that's a silly question.
| https://mathoverflow.net/users/6702 | Complete mathematics | You probably intended to restrict the question to effectively axiomatizable theories. Otherwise, for example, the first-order theory of the standard model of arithmetic is a complete theory, as is the theory of the standard model of ZFC.
Gödel's incompleteness theorem establishes some limitations on which effective theories can be complete. It shows that no effective, complete, consistent theory can interpret even weak theories of arithmetic such as Robinson arithmetic. However, there are many mathematically interesting theories that do not interpret the natural numbers.
Examples of complete, consistent, effectively axiomatizable theories include:
* For any prime $p$, the theory of algebraically closed fields of characteristic $p$
* The theory of real closed ordered fields, mentioned by Ricky Demer
* The theory of dense linear orderings without endpoints
* Many axiomatizations of Euclidean geometry
| 27 | https://mathoverflow.net/users/5442 | 31286 | 20,331 |
https://mathoverflow.net/questions/31275 | 28 | Let $R$ be a nonzero commutative ring with $1$, such that all finite matrices over $R$ have a *Smith normal form*. Does it follow that $R$ is a principal ideal domain?
If this fails, suppose we additionally suppose that $R$ is an integral domain?
What can we say if we impose the additional condition that the diagonal entries be unique up to associates?
| https://mathoverflow.net/users/nan | Does Smith normal form imply PID? | The implication is false without the assumption that R is Noetherian, because finite matrices don't detect enough information about infinitely generated ideals.
For example, let R be the ring
$$
\bigcup\_{n \geq 0} k[[t^{1/n}]]
$$
where $k$ is a field (an indiscrete valuation ring). Any finite matrix with coefficients in R comes from a subring $k[[t^{1/N}]]$ for some large $N$, and hence can be reduced to Smith normal form within this smaller PID.
However, the ideal $\cup (t^{1/N})$ is not principal.
| 33 | https://mathoverflow.net/users/360 | 31287 | 20,332 |
https://mathoverflow.net/questions/31288 | 0 |
>
> **Possible Duplicate:**
>
> [AC in group isomorphism between R and R^2](https://mathoverflow.net/questions/25375/ac-in-group-isomorphism-between-r-and-r2)
>
>
>
Somewhere, I recall being told that there is an isomorphism between $\mathbb{R}$ and $\mathbb{C}$ under addition. However, despite a rather lengthy search, I have been unable to find anything to support this fact, although Paul Yale of Pomona College, in his paper, "Automorphisms of the Complex Numbers," showed that there are "wild" automorphisms of $\mathbb{C}$ that require the axiom of choice to construct. Given that rather surprising fact, it does not seem too unlikely that there could be an isomorphism between $\mathbb{R}$ and $\mathbb{C}$. So, the question is: Is it possible for there to be an isomorphism between $\mathbb{R}$ and $\mathbb{C}$, and if so, what is is?
| https://mathoverflow.net/users/6856 | Is there an Isomorphism between R and C under addition? | As vector spaces over the rationals, they have the same dimension, so the only tricky part is the difficulty in finding a basis.
| 5 | https://mathoverflow.net/users/3684 | 31289 | 20,333 |
https://mathoverflow.net/questions/26079 | -2 | Given a directed acyclic graph `G` and a path made up from its set of nodes `N`, what is the closest approximate match to N, equipped with an intuitive notion of distance?
A path in a directed acyclic graph is essentially a string (that uses the set of nodes as the alphabet), so when comparing paths one can make use of edit distances developed for approximate string matching:
There are many algorithms for approximate string matching:
<http://en.wikipedia.org/wiki/Approximate_string_matching>
This string matching answers the question when the `G` itself is also a path.. then we're merely asking to compare two strings.
Asking if an arbitrary graph `A` built from the same set of nodes is a sub-graph of `G` is the general case of the problem, but I'm only interested in the case where `A` is a path and `G` is directed & acyclic.
Any general pointers are also welcome.
--
Edit: I ended up using a dynamic programming algorithm, independently also suggested in the accepted answer. Good call! It is probably the most accurate option as well, and barely more "complex" than the string-to-string case when the average # of edges per node is low.
| https://mathoverflow.net/users/6321 | Algorithm for determining if a path exists in a graph or if not, the closest edit distance. | If graph is acyclic you can use some sort of dynamic programming.
Let $a\_{u,k}$ be the best distance you can get if you start from vertex $u \in G$ and consider only $k$ last vertices of your given path.
It's quite straightforward how to calculate all $a\_{u,k}$ based on all values of $a\_{u',k'}$ with $u'$ "after" $u$ and $k' < k$: you just iterate over all edges going from vertex $u$.
This approach works in $O(|G| \times strlen)$ time.
| 1 | https://mathoverflow.net/users/7079 | 31296 | 20,337 |
https://mathoverflow.net/questions/31295 | 17 | Throughout, let $f$ be a Lebesgue measurable function (or continuous if you wish, but this is probably no easier). (*Questions with distributions etc. are possible also but I want to keep things simple here*).
### FINAL CLARIFICATION/REWRITE!!
Thanks to all who have commented so far. I will need some more time to digest it properly. The original forms of the questions are at the end; here I have rewritten the questions, hopefully more clearly; sorry for my poor explanation before!
**Definition** $\mathcal{M}\_a$, the space of *absolutely null moment functions*, is the set of all measurable $f$ satisfying
$$
\int\_0^\infty t^n |f(t)| dt < \infty,
\quad \int\_0^\infty t^n f(t) dt = 0, \qquad \forall \, n=0,1,2,\ldots.
$$
$\mathcal{M}$, the space of *null moment functions*, is the set of all measurable $f$ satisfying
$$
\int\_0^T |f(t)| dt < \infty \quad \forall \, T>0,
\qquad
\lim\_{R \to \infty} \int\_0^R t^n f(t) dt = 0, \quad \forall \, n=0,1,2,\ldots
$$
Thus, trivially $0 \in \mathcal{M}\_a \subseteq \mathcal{M}$, but $\mathcal{M}\_a$ contains many other non-trivial functions. It seems certain that $\mathcal{M}\_a \ne \mathcal{M}$ (I would be amazed if the spaces were equal), although constructing an explicit example seems tricky.
**Definition** Given a function $\psi \geq 0$, let $G(\psi)$ be the set of all $f$ such that $|f| \leq \psi$.
(*Of course we're identifying functions equal a.e., so really we should consider equivalence classes etc. just as for* $L^p$ *spaces*).
***Rephrased Question 1*** Find general simple necessary and/or sufficient conditions on $\psi$ with the property
$$
G(\psi) \cap \mathcal{M}\_a = \{ 0 \}.
$$
***Rephrased Question 2*** The same as Question 1, but with $G(\psi) \cap \mathcal{M}$ instead.
Or, if $G(\psi)$ is not the appropriate space for these problems, consider $\int\_T^{2T} |f| dt \leq g(T) $ or $\int\_n^{n+1} |f| dt \leq A\_n$ or something similar instead. Finding the correct kind of restrictions on $f$ is part of the problem.
Thus, $G(\psi) \cap \mathcal{M}\_a = \{ 0 \}$ for $\psi(t) = \exp(-\delta t)$, by the discussion below; and also for any compactly supported $\psi \in L^1$.
But $G(\psi) \cap \mathcal{M}\_a \ne \{ 0 \}$ for $\psi(t) = \exp(-t^{1/4}) |\sin(t^{1/4})|$.
---
### Original QUESTION 1
If
$$
\int\_0^\infty t^n |f(t)| dt < \infty,
\quad \int\_0^\infty t^n f(t) dt = 0, \qquad \forall \, n=0,1,2,\ldots,
$$
when must $f \equiv 0$ almost everywhere?
---
***EDIT: CLARIFICATION***: this is really about *classes of functions, expressed in terms of a growth/decay rate function* $\phi$, *which give* **unique** *solutions to the moment problem*. ***I am NOT asking*** how to solve the moment problem itself!
If possible, find *necessary and sufficient conditions* on functions $\phi \searrow 0$ with the property that
$$
\int\_R^\infty |f(t)| dt \leq \phi(R) \qquad \Rightarrow \qquad f \equiv 0.
$$
So, $\phi(R) = \int\_R^\infty \exp(-t^{1/4}) |\sin(t^{1/4})| dt$ is *not enough* (by Example 2 below); nor is $\phi(R) = \int\_R^\infty |f(t)| dt$ with f given by "coudy" in his/her answer below.
But $\phi = \chi\_{[0,b]}$ is enough (by Example 1 below); moreover $\phi(R) = \exp(-\delta R)$ would be enough for any fixed $\delta > 0$, by my discussion of Example 1 below, because the relevant Laplace transform $F = \mathcal{L}f$ is analytic on the half-plane $\{ \mathrm{Re}(z) > -\delta \}$. So we want to know about the gap *between* $\exp(-\delta R)$ and functions like that given by "coudy" below.
***FURTHER CLARIFICATION:*** by **analogy**, maybe an example from PDEs will explain better (I'm not saying this is related to my problem; I'm saying that this is *the same kind of result* as what I want):
**Definition** A *null temperature function* is a continuous function $u = u(x,t) : \mathbb{R} \times [0, \infty) \to \mathbb{R}$ such that the *heat equation* is satisfied, i.e. $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ in $\{ t>0 \}$, and $u(x, 0) = 0$ for all $x$.
**Theorem** There exists a null temperature function satisfying $|u(x,t)| \leq \exp(A/t)$ with $A>0$, such that $u(x,t) \not\equiv 0$ for all $t>0$.
**Theorem** Let $u$ be a null temperature function satisfying $|u(x,t)| \leq A \exp(B t^{-\delta})$, for some $A,B>0$ and $\delta<1$. Then $u \equiv 0$.
So here the "critical" growth rate for null temperature functions is roughly $\exp(A/t)$. I am looking for a similar thing with "null moment functions".
Note that this is a ***totally different problem*** to: *given* $v$, *find some* $u$ satisfying the heat equation such that $u(x,0)=v(x)$.
---
### Original QUESTION 2
If instead we have only $\int\_0^R |f(t)| dt < \infty$ for each $R>0$, and
$$
\lim\_{R \to \infty} \int\_0^R t^n f(t) dt = 0, \qquad n=0,1,2,\ldots
$$
when must $f \equiv 0$ almost everywhere? *(I have very little idea about this)*.
---
I think these questions are clearly very natural, interesting, and important, but Googling etc. didn't work well (I tried "vanishing moments" and other phrases, but there's just too much stuff out there). Standard known examples/methods follow.
***Example 1:*** if $f$ is compactly supported on $[a,b]$, say, then $f \equiv 0$ a.e. because polynomials are dense in $C[a,b]$.
***Example 2:*** by taking imaginary parts of $\int\_0^\infty t^{4n+3} \exp(-(1+i)t) dt \in \mathbb{R}$, we get
$$
\int\_0^\infty t^{4n+3} e^{-t} \sin t \, dt = 0,
$$
and so by substituting $t = x^{1/4}$,
$$
\int\_0^\infty x^n \exp(-x^{1/4}) \sin(x^{1/4}) dx = 0, \qquad n=0,1,2,\ldots
$$
*Alternative method for Example 1:* consider the Laplace transform $F(z) = \int\_0^\infty e^{-zt} f(t) dt$. In Example 1, $F$ is an entire function such that $F^{(n)}(0) = 0$ for all $n$, so $F \equiv 0$ and thus $f \equiv 0$ a.e. as required.
So, any condition on $f$ forcing $F$ to be analytic on some disc with centre $0$ is enough; but can we do better?
In Example 2, $f \in L^1(0,\infty)$ and so $F$ is bounded and analytic on $\{ \mathrm{Re}(z) > 0 \}$, and continuous on the boundary, with $\lim\_{z \to 0} F^{(n)}(z) = 0$ for all $n$. But this is still not enough to force $F \equiv 0$.
| https://mathoverflow.net/users/6651 | Let a function f have all moments zero. What conditions force f to be identically zero? | The answer to the first question is no. The following example is standard in probability theory, see e.g. Billingsley "probability and measure", example 30.2.
$$f(x) = {1\over \sqrt{2\pi}\ x}\ e^{-{(\ln x)^2\over 2}}\ \sin(2\pi \ln x) \ {\bf 1}\_{[0,\infty[}(x)$$
You can check that all the moments are zero using the change of variable $\ln x = s+k$.
This example is used to show that a probability measure is not always determined by its moments.
| 19 | https://mathoverflow.net/users/6129 | 31299 | 20,339 |
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