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https://mathoverflow.net/questions/19160 | 1 | First of all excuse my ignorance in number theory, the following question might have a well-known solution or it might be an open problem, I just don't know enough in that area of mathematics (and many others).
Let $P\in \mathbb{Z}[X]$ irreducible and of degree at least 1. For $k\in \mathbb{N}, k\geq 2$, denote by $S\_k$ the set of integers $n$ such that there exists $m\in \mathbb{Z}$ and $P(n)=m^k$.
Is $S\_k$ finite?
| https://mathoverflow.net/users/3958 | Finite set of (perfect power) polynomial values? | As Qiaochu said in the comments, you must include Pell type equations as a special case, because they are the only counter example. At least for $k=2$, Siegel's theorem on integral points on algebraic curves implies that if your polynomial $P(x)$ has at least three distinct roots then $P(n)=m^2$ has only finitely many solutions. So your conjecture is, in particular, true for irreducible polynomials of degree higher than 2.
Anyway, for the general question for any exponent it's better that you read the full story in
"The diophantine equation f(x)=g(y)" by Y. Bilu and R.F. Tichy.
| 4 | https://mathoverflow.net/users/2384 | 19165 | 12,761 |
https://mathoverflow.net/questions/19148 | 21 | I always find the strong law of large numbers hard to motivate to students, especially non-mathematicians. The weak law (giving convergence in probability) is so much easier to prove; why is it worth so much trouble to upgrade the conclusion to almost sure convergence?
I think it comes down to not having a good sense of why, practically speaking, a.s. convergence is better than convergence i.p. Sure, I can prove that one implies the other and not conversely, but the counterexamples feel contrived. I understand the advantages of a.s. convergence on a technical level, but not on the level of everyday life.
So my question: how would you explain to, say, an engineer, the significance of having a.s. convergence as opposed to i.p.? Is there a "real-life" example of bad behavior that we're ruling out?
| https://mathoverflow.net/users/4832 | Motivation for strong law of large numbers | [Here](http://terrytao.wordpress.com/2008/06/18/the-strong-law-of-large-numbers/) is a nice post of T. Tao on SLLN. In the comments section he is asked a very similar question to which he answers the following: (I hope it's ok to reproduce it here, since it is buried down in the comments)
>
> Returning specifically to the question of finitary interpretations of the SLLN, these basically have to do with the situation in which one is simultaneously considering multiple averages $\overline{X}\_n$ of a single series of empirical samples, as opposed to considering just a single such average (which is basically the situation covered by the WLLN). For instance, if one had some random intensity field of grayscale pixels, and wanted to compare the average intensities at 10 x 10 blocks, 100 x 100 blocks, and 1000 x 1000 blocks, then the SLLN suggests that these intensities would be likely to be simultaneously close to the average intensity. (The WLLN only suggests that each of these spatial averages are individually likely to be close to the average intensity, but does not preclude the possibility that when one considers multiple such spatial averages at once, that a few outlying spatial averages will deviate from the average intensity. In my example with only three different averages, there isn’t much difference here, as the union bound only loses a factor of three at most for the failure probability, but the SLLN begins to show its strength over the WLLN when one is considering a very large number of averages at once.)
>
>
>
| 19 | https://mathoverflow.net/users/2384 | 19166 | 12,762 |
https://mathoverflow.net/questions/19149 | 6 | Let's say I have a linear regression model of the form $ y = B\_x x + I\_x + \epsilon $, where $B\_x$ is the beta coefficient of the $x$ term, $I\_x$ is the intercept term and $\epsilon$ is additive, normally distributed noise. If I have a dataset and perform linear regression, I get a value for $B\_x$, which indicates the slope of the relationship.
If I swap the roles of the $x$ and $y$ data, and try to fit a model of $x = B\_y y + I\_y + \epsilon$, I would expect intuitively that $B\_y = \frac{1}{B\_x}$. A simple geometric argument can be made to show that swapping the roles of $x$ and $y$ shouldn't change the position of the regression line w.r.t. any data point, and from here it seems like simple algebra that if $y = Bx + I$ then $x = \frac{1}{B} y + \frac{I}{B}$.
Where is this reasoning wrong? Can someone explain to me why $B\_x \neq \frac{1}{B\_y}$, preferably without resorting to tons of linear algebra or direct derivation from the normal equation?
| https://mathoverflow.net/users/4833 | Linear Regression Coefficients W/ X, Y swapped | Well, I think Mike McCoy's answer is "the right answer," but here's another way of thinking about it: the linear regression is looking for an approximation (up to the error $\epsilon$) for $y$ as a function of $x$. That is, we're given a non-noisy $x$ value, and from it we're computing a $y$ value, possibly with some noise. This situation is not symmetric in the variables -- in particular, flipping $x$ and $y$ means that the error is now in the independent variable, while our dependent variable is measured exactly.
One could, of course, find the equation of the line that minimizes the sum of the squares of the (perpendicular) distances from the data points. My guess is that the reason that this isn't done is related to my first paragraph and "physical" interpretations in which one of the variables is treated as dependent on the other.
Incidentally, it's not hard to think up silly examples for which $B\_x$ and $B\_y$ don't satisfy anything remotely like $B\_x \cdot B\_y = 1$. The first one that pops to mind is to consider the least-squares line for the points {(0, 1), (1, 0), (-1, 0), (0, -1)}. (Or fudge the positions of those points slightly to make it a shade less artificial.)
Another possible reason that the perpendicular distances method is nonstandard is that it doesn't guarantee a unique solution -- see for example the silly example in the preceding paragraph.
(**N.B.**: I don't actually know anything about statistics.)
| 3 | https://mathoverflow.net/users/4658 | 19171 | 12,764 |
https://mathoverflow.net/questions/19143 | 5 | Understanding adjoints has always been (and continues to be) a bit of a struggle for me.
Today I stumbled upon a property of adjoint functors which seemed extremely intuitive to me. I was wondering why this property isn't mentioned more often in introductory category theory literature, and whether or not it completely characterizes adjunctions.
If two functors $F:C\to D$ and $U:D\to C$ are adjoint $F\dashv U$, then for every $f:F(Y)\to X$ in $D$ there exists an $\hat f:Y\to U(X)$ in $C$ such that
$$ U(f)\circ \eta\_Y = \hat f$$
$$ \epsilon\_X\circ F(\hat f)=f$$
If we substitute the top equation into the bottom, we get
$$ \epsilon\_X\circ F(U(f)\circ \eta\_Y)=f$$
and by functoriality we get
$$ \epsilon\_X\circ F(U(f))\circ F(\eta\_Y)=f$$
$$ \epsilon\_X\circ (F\circ U)(f)\circ F(\eta\_Y)=f$$
What the last equation says is that we can recover any morphism $f$ from the action of the "round trip endofunctor" $F\circ U$ on it by pre-composing with $\epsilon\_X$ and post-composing with $F(\eta\_Y)$. These two morphisms are determined only by the domain and codomain of $f$ -- we only needed to know $X$ and $F(Y)$ in order to pick the two morphisms. We would have picked the same two morphisms for some $g\neq f$ as long as $g:F(Y)\to X$.
So, I believe it is correct to say that "if the domain of a morphism is within the range of a functor which has a right adjoint, then it can be recovered from the action of the composite endofunctor on it by pre-composition with some morphism and post-composition with some other morphism, where the choice of these two morphisms is completely determined by the domain and codomain of the original morphism". There is, of course, an equivalent statement for morphisms with a codomain in the range of a functor with a left adjoint.
So, my three questions are: (1) is this correct, (2) if so, why isn't it used to explain adjunctions to beginners (I certainly would have caught on quicker!) and (3) does the condition completely characterize adjoint functors?
Thanks,
| https://mathoverflow.net/users/2361 | "adjoint" =?= "inverse of composite endofunctor is uniform bi-composition" | (1) Yes. (2) Well, it doesn't give me any additional intuition. You didn't say why it helps you understand, so I can't judge what the advantage of it might be. I think this is really just a complicated way of giving the "bijection of hom-sets" condition.
(3) No, you need something more. For instance, let $r:B\to A$ be a surjection with section $s$, let $C$ have two objects $x$ and $y$ with $C(x,y)=B$, $C(y,x)=\emptyset$, and $C(x,x)=C(y,y)=1$ (only identities), let $D$ be similar using $A$ instead, and let $F:C\to D$ and $U:D\to C$ be the identity on objects and with action on arrows given by $r$ and $s$ respectively. Pick $\varepsilon$ and $\eta$ to be identities. Then every morphism in $D$ can be recovered, as you describe, but the components of $\eta$ are not natural, and the dual condition fails.
The "unknown (google)" comment above explained why if you additionally require the dual condition, plus naturality of $\eta$ and $\varepsilon$, then you do get an adjunction. (Although it's not clear to me from the condition you stated whether you wanted to require the morphism playing the role of $F(\eta)$ to actually be $F$ of something, which is also necessary for this argument to work.)
| 4 | https://mathoverflow.net/users/49 | 19176 | 12,768 |
https://mathoverflow.net/questions/19186 | 2 | Dear all
While studying the overlap distribution for two random Cantor sets (long story made short), I came across the following problem.
$G(k)$ is a complex valued function, and satisfy the following condition:
$G(k\mu) = G(k)^2+ \beta$
with $\beta,\mu$ constant (in my case $\beta=\frac{2}{9}, \mu = \frac{4}{3}$)
Is there a way to find the functional form of $G(k)$ which satisfy the condition?
Note that for $\beta = 0$, $G(k)=\exp\left(a k^{\log\_\mu 2}\right)$, ($a$ konstant) will satisfy the condition (easily verified), but I have no idea on how to find a solution for non-zero $\beta$. I'm a not a math student (I'm studying physics), but I have never seen problems like this before. Is there a way to find analytical expression for $G(k)$? Possible as an expansion?
I can generate a function which has this property on the computer. Writing $G(k)= x(k) + i y(k)$, with $x(k)=x(-k)$ and $y(k)=-y(-k)$ the function should look something like this:
<http://dl.dropbox.com/u/483049/xy.pdf>
--
jon
| https://mathoverflow.net/users/4626 | Finding Functional form for a given Scaling Condition | You do not give any smoothness requirement; I will look for an analytic $G$:
$$ G(k)=\sum\_{n=0}^\infty a\_nk^n.$$
In what follows, I assume also that $\mu=4/3$ and $\beta=2/9$. Expanding in a power series both sides of the equation and equating coefficients, we get that $a\_0=1/3$ or $a\_0=2/3$. In the first case we obtain the constant solution $G(k)=1/3$. But in the second case, we find a one parameter family of (formal) solutions, parametrized by the value of $a\_1$:
$$
a\_0=\frac23,\quad a\_1\in\mathbb{C},\quad a\_n=\frac{1}{\mu^n-\mu}\sum\_{i=1}^{n-1}a\_ia\_{n-i},\quad n>2.
$$
For $a\_1=0$ we obtain the constant solution $G(k)=2/3$. For other values of $a\_1$, one should check that the series has a positive radius of convergence.
Another way of obtainig solutions is the following. Choose an arbitrary function $h\colon[1,\mu]\to\mathbb{C}$, and define $G(k)=h(k)$ if $1\le k<\mu$; for $\mu\le k<\mu^2$, let $G(k)=G(k/\mu)^2+\beta$; iterate this procedure to define $G$ on $[1,\infty)$. Now, for $1/\mu\le k<1$, let $G(k)=\pm\sqrt{G(\mu k)^2-\beta}$; iterate the procedure to define $G$ on $(0,1)$. Conditions can be impposed on the arbitrary function $h$ to make $G$ continuous, for instance ($G(\mu)=G(1)^2+\beta$).
| 3 | https://mathoverflow.net/users/1168 | 19196 | 12,778 |
https://mathoverflow.net/questions/19213 | 4 | Let the ring R be a MU`*`-module via a ring homomorphism φ and suppose it satisfies the condition of the Landweber exact functor theorem such that we obtain a cohomology theory $R^\*(-) := R \otimes\_{MU\_\*} MU^\*(-)$. If ω denotes the complex orientation class in $\widetilde{MU}^2(\mathbb{C}P^\infty)$, then R`*` is oriented by the class $\omega\_R := 1 \otimes \omega$.
Any other complex orientation of R`*` is obtainable by homogeneous power series θ with leading term x over R: θ(ω). These power series are in 1-1 correspondence with multiplicative natural transformations $t\_\theta\colon MU^\*(-) \to R^\*(-)$.
**Question**: Which tθ restrict to ring homomorphisms which satisfy the Landweber criterion on coefficients? For which theories is this true for any θ?
The place to start seems to be by noting that if the formal group law associated to R`*` (with the orientation given by ωR) is F, then tθ classifies the FGL $F^\theta(x,y) := \theta\big(F(\theta^{-1}(x),\theta^{-1}(y))\big)$ over R. Further, the p-series are related by $[p]\_{F^\theta}(x) = \theta\big([p]\_{F}(\theta^{-1}(x))\big)$, so it would suffice to show that the sequence of coefficients in the right degrees stay regular under this conjugation by θ.
This seems to be true for any θ as long as $[p]\_F(x)$ is of the form $\sum\_{n \geq1} a\_n x^{p^n}$ modulo p. In general, it is of the form $\sum\_{k\geq1} a\_kx^{kp^m}$, where m can be taken to be the height of the FGL (Ravenel's Green Book), but I don't see why it should be true in the general case.
I am sure this has been treated by someone, but have yet to see it on print. If anyone has seen question discussed somewhere, please let me know.
| https://mathoverflow.net/users/4877 | Changing the orientation of a Landweber exact cohomology theory | The property of being Landweber exact is independent of the orientation. In terms of Landweber's criterion, this is generally phrased as saying that the element vn is invariant modulo the ideal (p,v1,...,vn-1), and so any change-of-orientation (which induces a strict isomorphism on the formal group law) does not change the property of vn being or not being a zero divisor after modding out the previous terms.
This follows from Lemma A2.2.6 in Ravenel's green book, which implies that any endomorphism of the formal group law over an Fp-algebra R is of the form g(xph) for some h and some power series g. In particular, the p-series [p]F(x) over R/(p,v1,...,vn-1) has this property, and so the leading coefficient vn is invariant under strict isomorphisms.
It should be noted that vn is not invariant *before* taking this quotient, but that has no effect on whether these elements form a regular sequence in R.
| 6 | https://mathoverflow.net/users/360 | 19214 | 12,789 |
https://mathoverflow.net/questions/19215 | 7 | Let $(U\_i)\_{i\in I}$ be an open covering of a topological space $X$.
At <http://en.wikipedia.org/wiki/Nerve_of_an_open_covering>,
the nerve of the open covering is defined as follows:
>
> the nerve $N$ is the set of finite subsets of $I$ defined as follows:
>
>
> * the empty set belongs to $N$;
> * a finite set $J\subset I$ belongs to $N$ if and only if the intersection of the $U\_i$ whose subindices are in $J$ is non-empty.
>
>
>
On the other hand, <http://en.wikipedia.org/wiki/Nerve_(category_theory)> states:
>
> If $X$ is a topological space with open cover $U\_i$, the nerve of the cover is obtained from the above definitions by replacing the cover with the category obtained by regarding the cover as a partially ordered set with relation that of set inclusion.
>
>
>
Here, "the above definitions" refers to the usual construction of the nerve of a category: A vertex for each object, and a $k$-simplex for each $k$-tuple of composable morphisms.
My question is: Does this categorical construction really yield the previously defined nerve of the open covering?
For instance, cover the inverval by two intersecting invervals non of them containing the other one. Then it seems to me that the first construction yields two vertices connected by an edge, while the second construction yields to bare vertices.
What am i missing?
If the second definition is indeed wrong, what is the right way to obtain the nerve of an open covering as a special case of the nerve of a category?
| https://mathoverflow.net/users/1291 | Ambiguous definition of "nerve of an open covering" on wikipedia? | I think the second construction is not correct. If you replace the cover with the category whose objects are all **intersections of elements** of your original cover, then the two notions agree.
| 5 | https://mathoverflow.net/users/1231 | 19216 | 12,790 |
https://mathoverflow.net/questions/19218 | 9 | The following situation came up in my research:
Suppose two functions $f$ and $g$ map $[0,\infty)$ to (a subset of) itself. The function $f$ is linear and $g$ is quadratic, but $g$ is one-to-one on the interval $[0,\infty)$.
My conjecture/desired property: Any permutation of compositions of these two functions yields a unique polynomial.
The only progress I've made is the easy step of looking at degrees to argue that if two permutations are the same then they must have the same number of $g$'s.
For example, in my specific case $f(x)=x+1$ and $g(x)=x^2+x=x(x+1)$. Looking at length-3 compositions with two $g$'s gives the following different polynomials:
$f\circ g \circ g (x) = 1 + x + 2 x^2 + 2 x^3 + x^4$
$g\circ f \circ g (x) = 2 + 3 x + 4 x^2 + 2 x^3 + x^4$
$g\circ g \circ f (x) = 6 + 15 x + 14 x^2 + 6 x^3 + x^4$
I feel like this is probably easier than I'm making it, so please feel free to just give a reference for dealing with the semigroup(?) of polynomial composition.
| https://mathoverflow.net/users/3400 | Uniqueness in Composition of Polynomials | Your special case is right. More generally:
Let $f\left(x\right)=x+b$ with $b\neq 0$.
Let $g\left(x\right)=cx^2+dx+e$ with $c>0$, $d\in\mathbb R$ and $e\in\mathbb R$.
In fact, it is clear that every composition of $f$'s and $g$'s is a polynomial of positive degree and with positive leading coefficient (since $c>0$).
If we have a polynomial $P\in\mathbb R\left[X\right]$ which is a composition of $f$'s and $g$'s, we can always reconstruct the last step of the composition. Namely, we search for a nonnegative real $u$ such that $P-u=cQ^2+dQ+e$ for some polynomial $Q\in \mathbb R\left[X\right]$ of positive degree and with positive leading coefficient. If the last step has been a $g$, then $u=0$ must work; if the last step was an $f$, but some $g$ occured in the composition, then we must have a solution with $u\neq 0$ (in fact, if the last steps were $g$, $f$, $f$, ..., $f$ in this order, with $f$ occuring $k$ times, then $u$ must be $kb\neq 0$); if the composition consists of $f$'s only, then there is no solution (because $P$ must have degree $1$). The important thing is that the $u$, if it exists, is unique. In fact, if there would be two different $u$'s, then the two corresponding $Q$'s - let's call them $Q\_1$ and $Q\_2$ - would satisfy $\left(cQ\_1^2+dQ\_1+e\right)-\left(cQ\_2^2+dQ\_2+e\right)=w$ for some nonzero real $w$ (here, $w$ is the difference of the two $u$'s). This equation rewrites as $c\left(Q\_1-Q\_2\right)\left(Q\_1+Q\_2+1\right)=\left(c-d\right)Q\_1-\left(c-d\right)Q\_2+w$. Thus, (remembering that $c>0$) we conclude that
$\deg\left(Q\_1-Q\_2\right)+\deg\left(Q\_1+Q\_2+1\right)=\deg\left(c\left(Q\_1-Q\_2\right)\left(Q\_1+Q\_2+1\right)\right)$
$=\deg\left(\left(c-d\right)Q\_1-\left(c-d\right)Q\_2+w\right)\leq\max\left\lbrace \deg Q\_1,\deg Q\_2\right\rbrace$.
But at least one of the two degrees $\deg\left(Q\_1-Q\_2\right)$ and $\deg\left(Q\_1+Q\_2+1\right)$ must actually be equal to $\max\left\lbrace \deg Q\_1,\deg Q\_2\right\rbrace$ (because $Q\_1$ and $Q\_2$ are linear combinations of $Q\_1-Q\_2$ and $Q\_1+Q\_2$), and thus the other one must be zero or $-\infty$ (the degree of the zero polynomial). In other words, one of the polynomials $Q\_1-Q\_2$ and $Q\_1+Q\_2+1$ is constant. But the polynomial $Q\_1+Q\_2+1$ cannot be constant (because $Q\_1$ and $Q\_2$ have positive degree and positive leading coefficients). Hence, the polynomial $Q\_1-Q\_2$ is constant.
So let $Q\_1-Q\_2=k$ for $k\in\mathbb R$. Then, $\left(cQ\_1^2+dQ\_1+e\right)-\left(cQ\_2^2+dQ\_2+e\right)=w$ rewrites as $ck\left(Q\_1+Q\_2\right)+dk=0$ (since $Q\_1-Q\_2=k$). Hence, the polynomial $ck\left(Q\_1+Q\_2\right)$ also must be constant, so that $ck=0$ (since the polynomial $Q\_1+Q\_2$ is not constant, because $Q\_1$ and $Q\_2$ are two polynomials with positive degree and positive leading terms). Since $c>0$, this yields $k=0$, and thus $Q\_1-Q\_2=k=0$, so that $Q\_1=Q\_2$, and therefore $0=\left(cQ\_1^2+dQ\_1+e\right)-\left(cQ\_2^2+dQ\_2+e\right)=w$, contradicting $w\neq 0$.
| 7 | https://mathoverflow.net/users/2530 | 19221 | 12,792 |
https://mathoverflow.net/questions/19180 | 7 | Motivation: We have two examples:
(Abelian) Kummer theory (resp. Artin-Schreier theory) has a hidden cohomology theory given by Galois cohomology. The cocycle conditions become clear when you look at the multiplicative (resp. additive) form of Hlbert's theorem 90.
Descent theory for sheaves and stacks: In the case of sheaves, the cocycle condition is clear when we write $U\_{ij}:=U\_i \times\_U U\_j$ and look at the descent sequence. For stacks, the situation is even more obvious when you look at the coherence isomorphisms, which satisfy an explicit cocycle condition. The underlying cohomology theory here is Čech cohomology.
Question: Is this a general phenomenon in mathematics, that the presence of cocycle conditions is a good indicator that there is a cohomology theory determining things from behind the scenes, or is it just a coincidence and I just happened to see two very interesting and special cases?
| https://mathoverflow.net/users/1353 | Does the presence of cocycle conditions indicate the existence of an underlying cohomology theory? | I had lots of thoughts on that kind of question, and feel uneasy to speak as my answer can range from a tautology, through systematic and positive, but somewhat ignorant toward not-well understood cases, to mere impressions and (seeming?) "counterexample" oriented answer. The basic question is what you mean by a cocycle. Usually one talks on expressions of some higher categorical coherence, or about some notion of homotopy behind it. In such cases the answer is normally yes: the equivalent or homotopic cocycles will form cohomology classes and this can be in all understood cases done naturally and systematically. Higher [nonabelian cohomology](http://ncatlab.org/nlab/show/nonabelian+cohomology) can be done for all $n$, as now many frameworks know (Brown, Jardine, Toen, Street...) and [cohomology](http://ncatlab.org/nlab/show/cohomology) boils down to take homotopy classes into certain suspension of the coefficient object. For one recent framework we can advertise our own work [(pdf)](http://www.math.uni-hamburg.de/home/schreiber/nactwist.pdf).
I slightly believe anyway that some algebraic cases can be outside of the current homotopy categorical framework and I discussed that much on the n-category cafe, nforum and elsewhere. Namely model categories treat on equal footing homology and cohomology, while the minimal conditions on a setup to be able to do cohomology of homology is less than both simultaneously (cf. work of Rosenberg on "right exact structures" on a category, [pdf](http://www.mpim-bonn.mpg.de/preprints/send?bid=3623)).
Finally, we can imagine more complicated category-like structures where one can do much of the usual combinatorics but can not properly do the equivalence classes when needed for cohomology. There is one example which is maybe repairable, due Shahn Majid, namely he has a notion of bialgebra cocycles for a noncommutative and noncocommutative bialgebra. Now in special cocommutative or commutative cases like Lie algebras and/or abelian coefficients he recovers some known cohomology theories like Chevalley-Eilenberg cohomology for Lie algebras. In low dimensional cases he also gets some interesting nonabelian cocycles of much usage like Drinfel'd 2-twist and Drinfel'd 3-associator which are used in the study of monoidal categories, CFT, knot theory and quantum groups. In this example the differential and cocycles are defined for every $n$ but 17 years after the discovery, there is still no known way to define well the cohomology classes, for dimension 3 or more, for general bialgebra, despite the special cases and despite the cocycles and the differential. See the nlab page [bialgebra cocycle](http://ncatlab.org/nlab/show/bialgebra+cocycle) for the basics (and the references therein).
| 5 | https://mathoverflow.net/users/35833 | 19230 | 12,799 |
https://mathoverflow.net/questions/19174 | 36 | Does the following exist, and if not, does anyone besides me wish it did? A web site where a mathematician (say) could find other mathematicians who want to study the same book or paper, and arrange to meet via videoconference, and run their own informal seminar around that topic, and then disband when they're done.
The reason I ask is that I have plenty of material I'd like to learn this way. I have an interesting job in industry, with occasional mathematical challenges, but there are more topics I would like to learn outside of work, and that I didn't get a chance to cover in grad school. I live near two major universities, and they do offer some interesting courses I could audit, but I'd rather focus on the exact material I want to learn.
One reason such a site might be successful (or, more precisely, popular) is that I expect academic mathematicians and grad students would find great use for it as well, since it can be hard to branch out to new areas on your own, if your own department doesn't overlap enough. If you have a deep professional network in the math community, you are probably better off, but not everyone is in that position for one reason or another.
I expect such a site, if it doesn't exist today, wouldn't be hard to create. If the videoconferencing portion was done outside the site using Skype or another service, then the site would just have to match up the groups, and maybe offer other features to support a running seminar. It would basically be a specialized social networking site, or maybe even an application within an existing site like Facebook.
**UPDATE 1**: The more I think about it, the more I think meetup.com is a good fit too, since it already has the calendaring and rsvp-ing features. The downside is that each seminar would probably have to be a separate meetup, and meetup.com is sort of tied to geography.
**UPDATE 2**: I have spent a lot of time looking for good, free, non-sleazy video streaming services, and I ended up at [livestream.com](http://www.livestream.com/). They have a couple key features. They offer a free plan with a reasonable business model behind it (the "freemium" model with a paid tier and a free tier that shows ads to viewers, though I haven't seen what these ads look like yet). They have Mac and Windows client software that lets you use your webcam OR stream your desktop, the latter of which I think is very important as a replacement for a physical blackboard or whiteboard. The other piece I have been researching is drawing on the screen with a pen. The best solution I have found is slightly sub-optimal: [this pen](http://www.newegg.com/Product/Product.aspx?Item=N82E16838100017&cm_re=irisnotes-_-38-100-017-_-Product) or [this one](http://rads.stackoverflow.com/amzn/click/B0014BJIFM). They are similar in that they are standard ink pens with a wireless transmitter, and you clip a receiver to the paper and then write normally. The notes are either saved as images (the primary use case), or if you have a recent version of Windows, it can activate the pen features of the OS and you can draw directly into OneNote and other similar programs. This latter is what I want -- the seminar speaker can write on a paid of paper, and the remote attendees will see the writing rendered in real time on their screen. The problems are several: the pens cost at least $50-$60, I haven't tested them, reviews are scarce and Mac reviews are even scarcer. Most digital pen enthusiasm is focused on the Livescribe pen, which is strictly for saving notes and audio for later, and does not transmit in real time. Lastly, I also found [this chat software](http://thecoccinella.org/about) that supports audio and a whiteboard, and which is free. By the way, these pens all look really great for students, especially the Livescribe, because it syncs the audio with the notes, so you can play it back and know what was being said when something was written. And you can convert a session to Flash and upload it as a "pencast". I think pencasts could really be wonderful for mathematics presentations that are recorded offline and then uploaded for an audience to learn from asynchronously.
| https://mathoverflow.net/users/4837 | Informal online seminars or reading groups via videoconferencing? | I do not see much of a point in seeing people's faces and with full video either resolution is low, or jittering or the badnwidth is huge. So the solution is to have a simultanous voice and shared white board, which should be controlled by the individual elctronic tablet devices (mouse is not good for drawing). The electronic pen on the tablet is giving a one-dimensional (possibly with described path width, color etc.) trace, so if one just transmits the changes with the same resolution it amounts to many orders of magnitude lower data bandwidth.
It would be nice to have (and it is easy for professionals to create such a thing) a software for recording the lectures in a format which would include just the voice and the electronic pen on tablet path with the timing information on the movement of the pen path synchronous with the voice. I do not know which voice and whiteboard formats would be based on such synchonous timing information interspersed with audio and new path information. Of course one should be able to erase, and do the recording in chunks. Application which would be a viewer for such recording would be like a video but having a far higher resolution with smaller memory used as only 1-d information is added and most of the rest of the picture is static; and the voice is not that big of a deal. This kind of setup is a must for mathematical community. Many mathematical centers do big efforts to make videos where the focusing, jittering, shade and other problems as well as the mere size of video files make it impossible to be in good resolution and not hugely expensive or of big memory size. The sound + timed chalk/pen-path solution is much cheaper and scalable to hi resolution at present level of resources and technology.
| 13 | https://mathoverflow.net/users/35833 | 19231 | 12,800 |
https://mathoverflow.net/questions/19219 | 7 | When I first starting studying differential geometry, I asked my lecturer a question about smooth manifolds that didn't admit a partition of unity. He promptly told not to worry about such objects as they were only studied by the extremely eccentric. I would like to know if this is true, ie, does anyone study manifolds that don't admit a partition of unity (not whether such people are eccentric).
| https://mathoverflow.net/users/1977 | Smooth manifolds that don't admit a partition of unity | The answer to your stated question ("Does anyone study non-paracompact manifolds?") is certainly yes. Here are a few papers which do just this:
>
> Gauld, David.
> Manifolds at and beyond the limit of metrisability. (English summary) Proceedings of the Kirbyfest (Berkeley, CA, 1998), 125--133 (electronic),
> Geom. Topol. Monogr., 2, Geom. Topol. Publ., Coventry, 1999.
>
>
> <http://www.emis.de/journals/GT/ftp/main/m2/m2-7.pdf>
>
>
>
Among other things, Gauld references that there are two paracompact and two nonparacompact 1-manifolds, and $\aleph\_0$ paracompact and $2^{\aleph\_1}$ non-paracompact 2-manifolds. (That's a lot!)
>
> Foliations and foliated vector bundles, 1969 MIT lecture notes
>
>
> <http://www.foliations.org/surveys/FoliationLectNotes_Milnor.pdf>
>
>
>
Milnor entertains non-paracompact manifolds. In particular he constructs a (necessarily non-paracompact) surface with uncountable fundamental group. Milnor also says: "The main object of this exercise is to imbue the reader with suitable respect for non-paracompact manifolds."
>
> Balogh, Zoltan; Gruenhage, Gary.
> Two more perfectly normal non-metrizable manifolds. (English summary)
> Topology Appl. 151 (2005), no. 1-3, 260--272.
>
>
>
The existence of perfectly normal, non-metrizable (hence non-paracompact) manifolds is shown to depend upon one's set-theoretic assumptions.
And so forth. I could find 10 more papers without much effort. I'm not sure I could find 100. (A MathSciNet search with "manifold" and ("nonmetrizable" or "non-paracompact") in the anywhere fields doesn't return many hits.) So some serious mathematicians take non-paracompact manifolds seriously enough to write some papers about them. On the other hand, although one could use more complimentary language than "extremely eccentric", your lecturer's take on non-paracompact manifolds seems to be an accurate reflection of how most geometric topologists feel: they seem mostly to be used as a source of counterexamples and to be of interest to general and set-theoretic topologists.
| 13 | https://mathoverflow.net/users/1149 | 19237 | 12,803 |
https://mathoverflow.net/questions/19243 | 5 | My question is motivated by the previous discussion 'Why is a topology made of open sets?'. While the axioms for arbitrary unions and finite intersections are without doubt essential to the concept of a topological space, the 1st axiom (the set itself and the empty set are open) seems rather technical. So, do we really need these conditions in order to build most (if not all) of the point-set-topology without significant changes? In other words, if we leave out the 1st axiom, can point-set-topology stil remain as useful and powerful in terms of what we actually need to do analysis and geometry?
EDIT: Expanded the topic name in accordance with my question.
| https://mathoverflow.net/users/1849 | Do the empty set AND the entire set really need to be open? | Here's a boring reason, and it may or may not convince you: any function $f : X \to Y$ between topological spaces has the property that the preimage of the entire space $Y$ is the entire space $X$, and the preimage of the empty subset of $Y$ is the empty subset of $X$. So if you allow topological spaces in which either the entire set or the empty set is not open, there are **no continuous functions** from these spaces to "classical" topological spaces! Given that you agree with me that this is undesirable behavior, I think you are forced to make the entire set and/or the empty set either always open or always not open, and I think if you pick the second option then nothing changes except that, as KConrad says, it becomes unnecessarily harder to say things.
Actually, the situation is even worse: if the empty set isn't allowed to be open in $X$ then the continuous functions $X \to Y$ cannot miss any open set in $Y$, and if the entire set isn't allowed to be open in $X$ then the continuous functions $X \to Y$ cannot take values entirely in a proper open subset of $Y$. I think these are both much more unnatural than allowing the entire set and the empty set to be open. This is assuming you agree that the standard definition of continuity is natural.
| 19 | https://mathoverflow.net/users/290 | 19248 | 12,811 |
https://mathoverflow.net/questions/19238 | 5 | For some reason my thinking is *very* fuzzy today, so I apologize for the following rather silly question below...
Let $T$ be an ergodic transformation of $(X,\Omega, \mathbb{P})$ and let $X$ be partitioned into $n < \infty$ disjoint sets $R\_j$ of positive measure. For $x \in R\_k$ define $\tau(x) := \inf \{\ell>0:T^\ell x \in R\_k\}$. The Kac lemma (see, e.g. <http://arxiv.org/abs/math/0505625>) gives that $\int\_{R\_k} \tau(x) \ d\mathbb{P}(x) = 1$.
Now $\int\_X \tau(x) \ d\mathbb{P}(x) = \sum\_k \int\_{R\_k} \tau(x) \ d\mathbb{P}(x) = n$, or equivalently $\mathbb{E}\tau = n$.
**Can anyone provide a sanity check on the above assertion that the expected return time is just the size of the partition?** I've never seen this explicitly stated as a corollary of the Kac lemma, which seems odd.
| https://mathoverflow.net/users/1847 | Is the average first return time of a partitioned ergodic transformation just the number of elements in the partition? | I found a 2003 paper of Choe containing your sanity check, called "[A universal law of logarithm of the recurrence time](http://iopscience.iop.org/0951-7715/16/3/306?ejredirect=migration)". See the first few lines of section 3 on page 888. The "$K\_n$" used there is essentially your $\tau$, but corresponding to a partition $P\_n$ in a sequence of partitions. See definition 1.7 on page 885 for what the notation means.
| 3 | https://mathoverflow.net/users/1119 | 19249 | 12,812 |
https://mathoverflow.net/questions/19264 | 12 | This is related to the previous question of how to define a conductor of an elliptic curve or a Galois representation.
>
> What motivated the use of the word "conductor" in the first place?
>
>
>
A friend of mine once pointed out the amusing idea that one can think of the conductor of an elliptic curves as "someone" driving a train which lets you off at the level of the associated modular form.
A similar statement can be made concerning Szpiro's conjecture, which provides asymptotic bounds on several invariants of an elliptic curve in terms of its conductor. Here one might think of the conductor as "someone" who controls this symphony of invariants consisting of the minimal discriminant, the real period, the modular degree, and the order of the Shafarevich-Tate group (assuming BSD).
Was there some statement of this sort which motivated Artin's original definition of the conductor?
---
Does anyone have a reference for the first appearance of the word conductor in this context?
---
I apologize if this question is inappropriate for MO.
| https://mathoverflow.net/users/4872 | What is the etymology for the term conductor? | It is a translation from the German Führer (which also is the reason that
in older literature, as well as a fair bit of current literature, the conductor
is denoted as f in various fonts). Originally the term conductor appeared in
complex multiplication and class field theory: the conductor of an abelian extension is a certain ideal that controls the situation. Then it drifted off into other
areas of number theory to describe parameters that control other situations.
Of course in English we tend not to think of conductor as a leader
in the strong sense of Führer, but more in a musical sense, so it seems like a weird translation. But back in the
1930s the English translation was leader rather than conductor, at least once: see the review of Fueter’s book on complex multiplication in the 1931 Bulletin of the Amer. Math.
Society, page 655. The reviewer writes in the second paragraph "First there is a careful treatment of those ray class fields whose leaders are multiples of the ideal..."
You can find the review yourself at
<http://www.ams.org/bull/1931-37-09/S0002-9904-1931-05214-9/S0002-9904-1931-05214-9.pdf>.
I stumbled onto that reference quite by chance (a couple of years ago). If anyone knows other places in older papers in English where conductors were called leaders, please post them as comments below. Thanks!
Concerning Artin's conductor, he was generalizing to non-abelian Galois extensions the parameter already defined for abelian extensions and called the conductor. So it was natural to use the same name for it in the general case.
Edit: I just did a google search on "leader conductor abelian" and the first hit
is this answer. Incredible: it was posted less than 15 minutes ago!
| 24 | https://mathoverflow.net/users/3272 | 19265 | 12,822 |
https://mathoverflow.net/questions/19266 | 1 | Here by $P^n$ I mean $CP^n$, and what I want to do is to calculate the number of global sections of the holomorphic tangent bundle of $CP^n$.
If $n=1$, it is well known that $h^0(P^1, TP^1)=h^o(P^1,\mathcal{O}\_{P^1}(2))=3$.
If $n>1$, I did some calculation in local coordinates, and find out that
$h^0(P^n, TP^n) = n(n+1)$.
I am not sure if this is the correct answer and wonder if anyone else has calculated this before.
Besides, does anybody know the value of $h^1(P^n, TP^n)$? Even the $n=2$ case is enough for me. Many thanks!
| https://mathoverflow.net/users/3569 | Computing the dimension of the module of global holomorphic vector fields for complex projective n-space | The dimension of $H^0(\mathbb P^n,T\mathbb P^n)$ is $(n+1)^2-1$ and
$h^1(\mathbb P^n, T \mathbb P^n)=0$. Using the Euler sequence (see for instance
Griffiths-Harris, Principles of Algebraic Geometry) you can reduce the computation of these guys to the computation of
the comology of $\mathcal O\_{\mathbb P^n}$ and $\mathcal O\_{\mathbb P^n}(1)$.
For the dimension of the space of holomorphic vector fields on $\mathbb P^n$ it is perhaps easier
to realize that $$Aut(\mathbb P^n) = PSL(n+1, \mathbb C)$$ and its Lie algebra is
$$\mathfrak{sl}(n+1,\mathbb C)$$.
| 6 | https://mathoverflow.net/users/605 | 19268 | 12,823 |
https://mathoverflow.net/questions/19258 | 5 | Hi people,
I'm interested in results, such as the Gauß-Bonnet theorem, Fàry-Milnor theorem or classification theorems for manifolds, which give topological properties from geometric considerations. Can anyone recommend some good texts? In particular I'd like to see a nice proof of Fàry-Milnor and of the theorem of turning tangents (total curvature of an imbedded plane curve is $2\pi$).
thanks
| https://mathoverflow.net/users/4890 | Topological results from geometry | About the Fary–Milnor theorem. Milnor's original proof is already very nice (see [here](http://www.jstor.org/stable/1969467)). I also very much like [this proof](http://www.jstor.org/stable/119165) by Alexander & Bishop (see also a version of this proof in [my book](http://www.math.ucla.edu/~pak/book.htm)).
| 5 | https://mathoverflow.net/users/4040 | 19271 | 12,825 |
https://mathoverflow.net/questions/19240 | 48 | **4-colour Theorem.** Every planar graph is 4-colourable.
This theorem of course has a well-known history. It was first proven by Appel and Haken in 1976, but their proof was met with skepticism because it heavily relied on the use of computers. The situation was partially remedied 20 years later, when Robertson, Sanders, Seymour, and Thomas published a new proof of the theorem. This new proof still relied on computer analysis, but to such a lower extent that their proof was actually verifiable. Finally, in 2005, Gonthier and Werner used the [Coq](http://en.wikipedia.org/wiki/Coq) proof assistant to formalize a proof, so I suppose only the most die hard skeptics remain.
My question stems from reading [An Update on the Four-Color Theorem](https://www.ams.org/notices/199807/thomas.pdf) by Robin Thomas. The paper describes several interesting reformulations of the 4-colour theorem. Here is one:
Note that the cross-product on vectors in $\mathbb{R}^3$ is not an associative operation. We therefore define a *bracketing* of a cross-product $v\_1 \times \dots v\_n$ to be a set of brackets which makes the product well-defined.
**Theorem.** Let $i, j, k$ be the standard unit vectors in $\mathbb{R}^3$. For any two different bracketings of the product $v\_1 \times \dots \times v\_n$, there is an assignment of $i,j,k$ to $v\_1, \dots, v\_n$ such that the two products are equal and non-zero.
The surprising fact is that this innocent looking theorem implies the 4-colour theorem.
**Question.** Is anyone working on an algebraic proof of the 4-colour theorem (say by trying to prove the above theorem)? If so, what techniques are involved? What partial progress has been made? Or do most people consider the effort/reward ratio of such an endeavor to be too high?
I think it would be interesting to have an algebraic proof, even a very long one, particularly if the algebraic proof does not use computers. Given its connection to many other areas (Temperley-Lieb Algebras), the problem seems to be amenable to other forms of attack.
| https://mathoverflow.net/users/2233 | Algebraic proof of 4-colour theorem? | There is a classical approach by Birkhoff and Lewis, which remained dormant for decades. It was recently revived by Cautis and Jackson (start [here](https://core.ac.uk/download/pdf/82080353.pdf) [“The matrix of chromatic joins and the Temperley-Lieb algebra”, *J. Combin. Theory* **89** (2003), 109–155] and proceed [here](https://www.ams.org/tran/2010-362-01/S0002-9947-09-04836-3/S0002-9947-09-04836-3.pdf) [“On Tutte's chromatic invariant”, *Trans. Amer. Math. Soc.* **362** (2010), 491–507]), using the Temperley-Lieb algebra.
| 16 | https://mathoverflow.net/users/4040 | 19274 | 12,827 |
https://mathoverflow.net/questions/19276 | 4 | Let $X$ be a finite CW complex. Is there an integer $N,$ such that $H^i(X,F)=0$ for all $i>N$ and all abelian sheaves $F$ on $X?$ The cohomology is defined to be the derived functor of the global section.
| https://mathoverflow.net/users/370 | singular cohomological dimension | You can take $N=\dim X$, according to Proposition 3.1.5 in Dimca, [Alexandru. Sheaves in topology. Universitext. Springer-Verlag, Berlin, 2004. xvi+236 pp. [MR2050072](http://www.ams.org/mathscinet-getitem?mr=MR2050072)]
| 4 | https://mathoverflow.net/users/1409 | 19278 | 12,830 |
https://mathoverflow.net/questions/19193 | 6 | I have been perusing Harthorne for some time, and I noticed something: it is well known that the class group on $\mathbb{P}^n\_k$ is $\mathbb{Z}$. But as I look at Harthorne's proof it seems to me that it works in much greater generality. Namely if I consider any projective scheme $X=\operatorname{Proj}(A)$, where $A$ is a graded UFD in such a way that there exists an irreducible element of degree $1$, then the exact same reasoning shows that the class group of $X$ is also $\mathbb{Z}$, and generated by the prime divisor $(a)$. Is this true?
| https://mathoverflow.net/users/4863 | Divisors on Proj(UFD) | Well, if you read on to Chapter 2, exercise 6.3, then it is stated that:
$$Cl(A) \cong Cl(X)/\mathbb Z[H]$$
here $[H]$ represents the hyperplane section. So the answer is yes.
There is a less well-known but very nice generalization. Suppose that $X$ is smooth. Let $R=A\_m$ be the local ring of A at the irrelevant ideal. Then one has a (graded) isomorphism of $\mathbb Q$- vector spaces:
$$CH(X)\_{\mathbb Q}/[H]CH(X)\_{\mathbb Q} \cong A\_\*(R)\_{\mathbb Q}$$
Here $CH(X)$ is the Chow *ring* of $X$ and $A\_\*(R)$ is the total Chow group of $R$. Details can be found in this [paper](https://projecteuclid.org/journals/tohoku-mathematical-journal/volume-48/issue-1/A-remark-on-the-Riemann-Roch-formula-on-affine-schemes/10.2748/tmj/1178225414.full) by Kurano.
| 3 | https://mathoverflow.net/users/2083 | 19283 | 12,834 |
https://mathoverflow.net/questions/19285 | 30 | Let $X$ be a set and let ${\mathcal N}$ be a collection of nets on $X.$
I've been told by several different people that ${\mathcal N}$ is the collection of convergent nets on $X$ with respect to some topology if and only if it satisfies some axioms. I've also been told these axioms are not very pretty.
Once or twice I've tried to figure out what these axioms might be but never came up with anything very satisfying. Of course one could just recode the usual axioms regarding open sets as statements about nets and then claim to have done the job. But, come on, that's nothing to be proud of.
Has anyone seen topology axiomatized this way? Does anyone remember the rules?
| https://mathoverflow.net/users/4783 | How do you axiomatize topology via nets? | Yes. This is given in Kelley's *General Topology*. (Kelley was one of the main mathematicians who developed the theory of nets so that it would be useful in topology generally rather than just certain applications in analysis.)
In the section "Convergence Classes" at the end of Chapter 2 of his book, Kelley lists the following axioms for convergent nets in a topological space $X$
a) If $S$ is a net such that $S\_n = s$ for each $n$ [i.e., a constant net], then $S$ converges to $s$.
b) If $S$ converges to $s$, so does each subnet.
c) If $S$ does not converge to $s$, then there is a subnet of $S$, no subnet of which converges to $s$.
d) (Theorem on iterated limits): Let $D$ be a directed set. For each $m \in D$, let $E\_m$ be a directed set, let $F$ be the product $D \times \prod\_{m \in D} E\_m$ and for $(m,f)$ in $F$ let $R(m,f) = (m,f(m))$. If $S(m,n)$ is an element of $X$ for each $m \in D$ and $n \in E\_m$ and $\lim\_m \lim\_n S(m,n) = s$, then $S \circ R$ converges to $s$.
He has previously shown that in any topological space, convergence of nets satisfies a) through d). (The first three are easy; part d) is, I believe, an original result of his.) In this section he proves the converse: given a set $S$ and a set $\mathcal{C}$ of pairs (net,point) satisfying the four axioms above, there exists a unique topology on $S$ such that a net $N$ converges to $s \in X$ iff $(N,s) \in \mathcal{C}$.
I have always found property d) to be unappealing bordering on completely opaque, but that's a purely personal statement.
**Addendum**: I would be very interested to know if anyone has ever put this characterization to any useful purpose. A couple of years ago I decided to relearn general topology and write notes this time. The flower of my efforts was an essay on convergence in topological spaces that seems to cover all the bases (especially, comparing nets and filters) more solidly than in any text I have seen.
[http://alpha.math.uga.edu/~pete/convergence.pdf](http://alpha.math.uga.edu/%7Epete/convergence.pdf)
But "even" in these notes I didn't talk about either the theorem on iterated limits or (consequently) Kelley's theorem above: I honestly just couldn't internalize it without putting a lot more thought into it. But I've always felt/worried that there must be some insight and content there...
| 30 | https://mathoverflow.net/users/1149 | 19288 | 12,837 |
https://mathoverflow.net/questions/19255 | 5 | I've been running into the following type of partition problem.
>
> Given positive integers *h*, *r*, *k*, and a real number ε ∈ (0,1), find *n* such that if every (unordered) *r*-tuple from an *n* element set *X* is assigned a set of at least ε*k* 'valid' colors out of a total of *k* possible colors, then you can find *H* ⊆ *X* of size *h* and a single color which is 'valid' for all *r*-tuples from *H*.
>
>
>
Lower bounds on the smallest such *n* can be obtained from lower bounds for Ramsey's Theorem. If *k* is sufficiently large, then partition the set of colors into [1/ε] pairwise disjoint sets of approximately equal size to emulate a proper [1/ε]-coloring of *r*-tuples. A simple pigeonhole argument shows that this is essentially sharp when *r* = 1 and *k* is large enough, i.e. one color must be 'valid' for at least *n*ε points.
Is the Ramsey bound more or less sharp for *r* > 1 or are there better lower bounds? The interesting case is when *k* is large since the proposed Ramsey lower bound is (surprisingly?) independent of *k*.
| https://mathoverflow.net/users/2000 | Bounds on a partition theorem with ambivalent colors | I do not think that the lower bound could depend only on epsilon. Below is the sketch of my argument.
Fix h=3, r=2, eps=1/4, thus we color the edges of a graph, each with 25% of all the colors and we are looking for a "monochromatic" triangle. Let us take k random bipartitions of the vertices and color the corresponding edges of the bipartite graph with one color. Using [Hoeffding](http://en.wikipedia.org/wiki/Hoeffding%2527s_inequality) or some similar inequality we get that for big enough k every edge is colored at least k/4 times if n is at most exp(ck), where c is some fixed constant with some positive probability. Therefore the bound must depend on k and not only on epsilon.
| 4 | https://mathoverflow.net/users/955 | 19289 | 12,838 |
https://mathoverflow.net/questions/19303 | 7 | What's a good example to illustrate the fact that a function all of whose partial derivatives exist may not be continuous?
| https://mathoverflow.net/users/2612 | Good example of a non-continuous function all of whose partial derivatives exist | The standard example I have seen is: $f(x,y)=\frac{2xy}{x^2+y^2}$.
| 10 | https://mathoverflow.net/users/4500 | 19304 | 12,850 |
https://mathoverflow.net/questions/19309 | 10 | I am looking for a reference for the following fact:
The orthogonal group $O\_{2n}$ over an algebraically closed field of characteristic 2
has exactly two connected components.
To be more precise, let $O\_q$ denote the orthogonal group of the quadratic form $q(x)=x\_1 x\_2 +x\_3 x\_4+\cdots +x\_{2n-1}x\_{2n}$
over an algebraically closed field $k$.
In characteristic $p\neq 2$ the determinant takes two values on $O\_q$, 1 and $-1$,
and therefore the subgroup $SO\_q:=O\_q\cap SL\_{2n}$ is of index 2 in $O\_q$; it is known that $O\_q\cap SL\_{2n}$ is connected.
In characteristic 2 the determinant takes only one value 1 on $O\_q$ (because $-1=1$), and therefore $O\_q\cap SL\_{2n}=O\_q$.
Still there is a homomorphism $D\colon O\_q\to \mathbf{Z}/2\mathbf{Z}$ given by a polynomial $D$ called the Dickson invariant,
see J.A.~Dieudonn\'e, Pseudo-discriminant and Dickson invariant, Pacific. J. Math. 5 (1955), 907--910.
This homomorphism $D$ indeed takes both values 0 and 1 on $O\_q$, and therefore its kernel ker $D$
is a closed subgroup of index 2 in $O\_q$. I would like to know that ker $D$ is connected.
In other words, I am looking for a reference to the assertion that the orthogonal group $O\_q$ has at most two connected components.
This is proved in Brian Conrad's handout "Properties of orthogonal groups" to his course Math 252 "Algebraic groups",
see <http://math.stanford.edu/~conrad/252Page/handouts/O(q).pdf> . Is there any other reference for this fact?
I will be grateful to any references, comments, etc.
Mikhail Borovoi
| https://mathoverflow.net/users/4149 | Connected components of the orthogonal group O(2n) in characteristic 2. | Presumably this is treated in detail in chapter 7 of the book
The Classical Groups and K-Theory, by A.J.Hahn and O.T.O'Meara.
On page 424 it says in theorem 7.2.23 that the elementary subgroup has index two.
And elementary matrices are in the connected component of 1.
Wilberd
| 8 | https://mathoverflow.net/users/4794 | 19327 | 12,862 |
https://mathoverflow.net/questions/19269 | 42 | What are the examples of some mathematicians coming very close to a very promising theory or a correct proof of a big conjecture but not making or missing the last step?
| https://mathoverflow.net/users/1851 | What are some examples of narrowly missed discoveries in the history of mathematics? | Freeman Dyson discusses a few examples of this in his article [*Missed Opportunities*](http://www.ams.org/bull/1972-78-05/S0002-9904-1972-12971-9/S0002-9904-1972-12971-9.pdf). One that I thought was particularly striking was that mathematicians could have discovered special relativity decades before Einstein just by staring at Maxwell's equations hard enough, and also on the basis that the representation theory of the Poincare group is simpler than the representation theory of the Galilean group.
| 35 | https://mathoverflow.net/users/290 | 19332 | 12,866 |
https://mathoverflow.net/questions/19348 | 15 | Space forms are complete (connected) Riemannian manifolds of constant sectional curvature.
These fall into three classes: Euclidean, with universal covering isometric to $\mathbb{R}^n$,
spherical, with universal covering isometric to $S^n$, and hyperbolic, with universal covering isometric to $\mathbb{H}^n$.
Does there exist compact space forms of the same dimension from two different classes having isomorphic fundamental groups?
This cannot happen for $n = 2$, since the Gauss-Bonnet theorem
$$
\int\_M K{ \ }\mathrm{vol}\_M = 2{\pi}\chi(M)
$$
shows that the Euler characteristic $\chi(M)$ is positive, zero, or negative when the space form $M$ is Euclidean, spherical or hyperbolic respectively. But for (closed) surfaces the fundamental group determines the Euler characteristic.
It is essential for the question to require compactness, otherwise there are trivial examples. Dividing out $\mathbb{R}^2$ by the group of isometries generated by
$(x,y) \mapsto (x+1,y)$ yields a Euclidean space form with fundamental group isomorphic to $\mathbb{Z}$, while dividing out $\mathbb{H}^2$ by the group of isometries generated by
the hyperbolic isometry $(x,y) \mapsto (x+1,y)$ yields a hyperbolic space form also with fundamental group isomorphic to $\mathbb{Z}$.
The standard reference on space forms is Spaces of Constant Curvature by Joseph A. Wolf.
The classification of Euclidean space forms is given in Chapter 3, and of spherical ones in Chapter 7. Wolf does not treat hyperbolic space forms, possibly because not much was known about them in 1967. Unfortunately, the fundamental groups are infinite for the compact Euclidean space forms, and finite for the spherical space forms (which are necessarily compact, being quotients of $S^n$). So a hypothetical example has to involve a hyperbolic space form.
An example might drop out of the theory of three-manifolds. In dimension three the space forms belong to three of the eight Thurston model geometries. A pair yielding an example would have to be one Euclidean and one hyperbolic, since it follows from Perelman's geometrization theorem that the spherical ones are precisely those with finite fundamental group.
| https://mathoverflow.net/users/3304 | Fundamental group of a compact space form. | The fundamental group of a compact hyperbolic space form has exponential growth, according to a well-known theorem of Milnor [Milnor, J. A note on curvature and fundamental group. J. Differential Geometry 2 1968 1--7. [MR0232311](http://www.ams.org/mathscinet-getitem?mr=MR0232311)]. Bieberbach groups are, on the other hand, polynomial: indeed, their translations subgroups have finite index, so by Gromov's theorem they have polynomial growth.
| 15 | https://mathoverflow.net/users/1409 | 19352 | 12,879 |
https://mathoverflow.net/questions/19345 | 3 | Let $X$ be a scheme. Suppose you are given a sheaf of Abelian groups $\mathcal{A}$ over $X$. How can you determine if $\mathcal{A}$ is the underlying Abelian sheaf of a sheaf of $O\_X$-modules? In other words, is it possible to characterize in some (interesting) way the essential image of the forgetful functor from $Mod(O\_X)$ to $Ab(X)$?
I'm not sure a non-tautological answer exists...
| https://mathoverflow.net/users/4721 | How to characterize Abelian sheaves that are quasi-coherent? | 1) There is a very simple example that shows that it is impossible to answer the question of whether $\mathcal{A}$ comes from a quasi-coherent sheaf $\mathcal{F}$ on $X$ if all one is given is the underlying topological space $|X|$ and $\mathcal{A}$ as a sheaf on $|X|$. Namely, if $|X|$ is a point and $\mathcal{A}$ is such that $\mathcal{A}(|X|)=\mathbf{Q}$, then either outcome is possible: the answer is YES if $X=\operatorname{Spec} \mathbf{Q}$, but NO if $X=\operatorname{Spec} \mathbf{F}\_p$.
2) There are some nontrivial *necessary* conditions that one can state in terms of the topological space and the sheaf of abelian groups alone. For example, in order for $\mathcal{A}$ to come from a quasi-coherent sheaf, there must exist an open covering $(U\_i)$ of $|X|$ such that the sheaf $\mathcal{A}|\_{U\_i}$ on $U\_i$ is acyclic for every $i$.
3) The condition in 2) is definitely not sufficient, even if the scheme structure on $|X|$ is not specified in advance. For instance the constant sheaf $\mathbf{Z}/6\mathbf{Z}$ on a point is acyclic, but it cannot be a quasi-coherent sheaf for any scheme structure on the point.
| 13 | https://mathoverflow.net/users/2757 | 19361 | 12,885 |
https://mathoverflow.net/questions/19377 | 5 | It might be well-known (and sorry if it is), but a quick search did not return the answer.
Consider prime numbers $p \neq q$.
Are $\displaystyle \frac{p^q-1}{p-1}$ and $\displaystyle \frac{q^p-1}{q-1}$ relatively prime?
| https://mathoverflow.net/users/3958 | Prime numbers that lead to relatively prime | The answer is no. As the Wikipedia article in my comment states, the counterexample $p = 17, q = 3313$ was found by Stephens in 1971, but the stronger question of whether one can ever divide the other is a famous open problem because its solution would greatly simplify a step in the proof of the Feit-Thompson theorem.
| 18 | https://mathoverflow.net/users/290 | 19387 | 12,903 |
https://mathoverflow.net/questions/19388 | 5 | I've always thought of the degree of a subvariety of projective space as the degree of the divisor that defines the (given) embedding into projective space. It's been pointed out to me that this works only for curves. Now I'm confused: is there a similar characterization of the degree of a general subvariety of some projective space?
| https://mathoverflow.net/users/3238 | Degrees of subvarieties of projective space | If $X\subset \mathbb P^n$ is a subvariety of dimension $m$ embedded by a linear system $V \subset H^0(X,\mathcal O\_X(D))$ then the degree of $X$ is equal to $D^m$.
| 7 | https://mathoverflow.net/users/605 | 19389 | 12,904 |
https://mathoverflow.net/questions/19390 | 41 | The modular curve $X\_0(N)$ has good reduction at all primes $p$ not dividing $N$. At such a prime, the Eichler-Shimura relation expresses the Hecke operator $T\_p$ (as an element of the ring of correspondences on the points of $X\_0(N)$ in $\overline{ \mathbb{F}\_p }$) in terms of the geometric Frobenius map. This is already weird enough; the definitions of the Hecke operators with which I'm familiar give no indication that such a relation should be true, and the sources I've read so far give no intuition as to why such a relation should be true. (In fact, I still don't feel very comfortable with the Hecke operators themselves; I like the definition given in Milne the best so far, but any enlightening alternate definitions are welcome if they shed light on the question.)
What's much weirder, to me anyway, is an important corollary of the Eichler-Shimura relation, which says that given a cusp eigenform $f$ of weight $2$ with respect to $\Gamma\_0(N)$ it is possible to construct an elliptic curve $E\_f$ whose L-function is (more or less) the Mellin transform of $f$.
There are several reasons this is weird to me, but here's a specific one. The modular form $f$ satisfies a functional equation more or less by definition. Its Mellin transform therefore satisfies a corresponding functional equation (part of which has been absorbed into the fact that $f$ is written in terms of its Fourier coefficients), again more or less by definition. The zeta function of an elliptic curve, however, satisfies a functional equation because ([or so I'm told](https://mathoverflow.net/questions/2040/why-are-functional-equations-important)) of Poincaré duality in the étale cohomology. So:
> What on Earth does Poincaré duality have to do with modular symmetry?
(Ignore the above; I seem to have mixed up the local and global functional equations again.)
One of the many things that's weird about the above is that the L-function of an elliptic curve naturally has an Euler product, but for modular forms the Euler product for the Mellin transform comes about because of certain properties of the Hecke operators (which, again, I don't really understand conceptually). What do these properties have to do with multiplying local zeta functions together?
---
I guess I should also clarify what I mean by "intuition":
For the first part of the question, if something in the definition of the Hecke operators suggests that they should be related to the Frobenius map if certain natural things were true, and the proof of Eichler-Shimura (which I haven't really looked at yet...) consists of verifying these natural things are true, that would be great intuition. I would appreciate an answer telling me whether or not this was the case in terms of "first principles."
For the second part of the question, intuition might more naturally come from a more sophisticated perspective. Here I would appreciate an answer about the "big picture" instead, giving some vague high-level sense of how this all fits into more general things people know about automorphic forms and so forth.
| https://mathoverflow.net/users/290 | Intuition behind the Eichler-Shimura relation? | (1) Short answer to first question: $T\_p$ is about $p$-isogenies, and in char. $p$ there is a canonical $p$-isogeny, namely Frobenius.
Details:
The Hecke correspondence $T\_p$ has the following definition, in modular terms:
Let $(E,C)$ be a point of $X\_0(N)$, i.e. a modular curve together with a cyclic subgroup
of order $N$. Now $T\_p$ (for $p$ not dividing $N$) is a correspondence (multi-valued function) which maps
$(E,C)$ to $\sum\_D (E/D, (C+D)/D)$, where $D$ runs over all subgroups of $E$ of degree $p$.
(There are $p+1$ of these.)
Here is another way to write this, which will work better in char. $p$:
map $(E,C)$ to $\sum\_{\phi:E \rightarrow E'}(E',\phi(C)),$
where the sum is over all degree $p$ isogenies $\phi:E\rightarrow E'.$ Giving a degree
$p$ isogeny in char. 0 is the same as choosing a subgroup of order $D$ of $E$ (its kernel),
but in char. $p$ the kernel of an isogeny can be a subgroup scheme which is non-reduced,
and so has no points, and hence can't be described simply in terms of subgroups of points.
Thus this latter description is the better one to use to compute the reduction of the
correspondence $T\_p$ mod $p$.
Now if $E$ is an elliptic curve in char. $p$, any $p$-isogeny $E \to E'$ is either
Frobenius $Fr$, or the dual isogeny to Frobenius (often called Vershiebung).
Now Frobenius takes an elliptic curve $E$ with $j$-invariant $j$ to the elliptic curve
$E^{(p)}$ with $j$-invariant
$j^p$. So the correspondence on $X\_0(N)$ in char. $p$ which maps $(E,C)$ to $(E^{(p)},
Fr(C))$ is itself the Frobenius correspondence on $X\_0(N)$. And the correspondence
which maps $(E,C)$ to its image under the dual to Frobenius is the transpose
to Frobenius (domain and codomain are switched). Since there are no other $p$-isogenies in char. $p$
we see that $T\_p$ mod $p = Fr + Fr'$ as correspondences on $X\_0(N)$ in char. $p$;
this is the Eichler--Shimura relation.
(2) Note that only weight 2 eigenforms with rational Hecke eigenvalues give elliptic curves;
more general eigenforms give abelian varieties.
An easy computation shows that if $f$ is a Hecke eigenform, than the $L$-funcion
$L(f,s)$, obtained by Mellin transform, has a degree 2 Euler product. A more conceptual answer would probably involve describing how automorphic representations
factor as a tensor product of local factors, but that it a very different topic from Eichler--Shimura, and I won't say more here.
| 54 | https://mathoverflow.net/users/2874 | 19399 | 12,909 |
https://mathoverflow.net/questions/19414 | 19 | Let $S$ be an uncountable set. Does there exist a probability measure which is defined on *all* subsets of $S$, with $P({x}) = 0$ for any element $x$ of S ?
If I remove the condition $P({x}) = 0$, then I can trivially get a measure defined on all subsets as follows:
Fix some $a \in S$. For any subset $U \subset S$, define
$$
P(U) =
\begin{cases}
1, \quad \text{if} \quad a \in U \\
0, \quad \text{otherwise}
\end{cases}
$$
But what happens if I am not allowed to put nonzero probability on individual points ?
| https://mathoverflow.net/users/4279 | Existence of probability measure defined on all subsets | The existence of such a measure is equiconsistent to the existence of a [measurable cardinal](https://en.wikipedia.org/wiki/Measurable_cardinal), one of the large cardinal notions, and if ZFC is consistent, cannot be proved in ZFC. (See the notion of real-valued measurable cardinal on the Wikipedia page.)
| 17 | https://mathoverflow.net/users/1946 | 19415 | 12,916 |
https://mathoverflow.net/questions/19363 | 8 | Background
----------
---
Let $X$ and $S$ be simplicial sets, i.e. presheaves on $\Delta$, the so-called *topologist's simplex category*, which is the category of finite nonempty ordinals with morphisms given by order preserving maps.
How can we derive the structure of the face and degeneracy maps of the join from either of the two equivalent formulas for it below:
The *Day Convolution*, which extends the monoidal product to the presheaf category:
$$(X\star S)\_{n}:=\int^{[c],[c^\prime] \in \Delta\_a}X\_{c}\times S\_{c^\prime}\times Hom\_{\Delta\_a}([n],[c]\boxplus[c^\prime])$$
Where $\Delta\_a$ is the augmented simplex category, and $\boxplus$ denotes the ordinal sum. The augmented simplex category is the category of all finite ordinals (note that this includes the empty ordinal, written $[-1]:=\emptyset$.
The join formula (for $J$ a finite nonempty linearly ordered set):
$$(X\star S)(J)=\coprod\_{I\cup I=J}X(I) \times S(I')$$
Where $\forall (i \in I \text{ and } i' \in I'),$ $i < i'$, which implies that $I$ and $I'$ are disjoint.
Then we would like to derive the following formulas for the face maps (and implicitly the degeneracy maps):
>
> The $i$-th face map $d\_i : (S\star T)\_n \to (S\star T)\_{n-1}$ is defined on $S\_n$ and $T\_n$ using the $i$-th face map on $S$ and $T$. Given $\sigma \in S\_j\text{ and }\tau\in T\_k$ , we have:
>
>
> $$d\_i (\sigma, \tau) = (d\_i \sigma,\tau)\text{ if } i \leq j, j ≠ 0.$$
> $$d\_i (\sigma, \tau) = (\sigma,d\_{i-j-1} \tau) \text{ if } i > j, k ≠ 0.$$
> $$d\_0(\sigma, \tau) = \tau \in T\_{n-1} \subseteq (S\star T)\_{n-1} \text{ if } j = 0$$
> $$d\_n(\sigma, \tau) = \sigma \in S\_{n-1} \subset (S\star T)\_{n-1}\text{ if } k = 0$$
>
>
>
We note that the special cases at the bottom come directly from the inclusion of augmentation in the formula for the join.
Edit: Another note here: I got these formulas from a different source, so the indexing may be off by a factor of -1.
Question
--------
---
How can we derive the concrete formulas for the face and degeneracy maps from the definition of the join (I don't want a geometric explanation. There should be a precise algebraic or combinatorial reason why this is the case.)?
Less importantly, how can we show that the two definitions of the join are in fact equivalent?
**Edit:**
Ideally, an answer would show how to induce one of the maps by a universal property.
Note also that in the second formula, we allow $I$ or $I'$ to be empty, and we extend the definition of a simplicial set to an augmented simplicial set such that $X([-1])=\*$, i.e. the set with one element.
A further note about the first formula for the join: $\boxplus$ denotes the ordinal sum. That is, $[n]\boxplus [m]\cong [n+m+1]$. However, it is important to notice that there is no natural isomorphism $[n]\boxplus [m]\to [m]\boxplus [n]$. That is, there is no way to construct this morphism in a way that is natural in both coordinates of the bifunctor. This is important to note, because without it, it's not clear that the ordinal sum is asymmetrical.
| https://mathoverflow.net/users/1353 | The Join of Simplicial Sets ~Finale~ | (A note: I am going to regard simplicial sets as also defined on the empty ordinal as well, with $X(\emptyset) = \*$, which is required for the join formula. This is implicit in your first definition and will remove the need for two extra cases for $d\_i$ at the end.)
Regarding the "minor" question. The short explanation is that this follows by decomposing the Hom-set according to the preimage of $c$ and $c'$ in $n$, and observing that each decomposition of $n$ provides an initial choice.
In more category-theoretic language, one way to rewrite the convolution is using the "over" category:
$$
(X \star S)\_n = \int^{[n] \to [c] \boxplus [c']} X\_c \times S\_{c'}
$$
where now the coend is taken over the comma category $n \downarrow \boxplus$ whose objects are triples $([c],[c'],f)$ of a pair of objects of $\Delta$ and a morphism from $[n]$ to their ordinal sum. We note that this comma category decomposes as a disjoint union of categories: each $([c],[c'],f)$ determines a decomposition $[n] = f^{-1} [c] \cup f^{-1} [c']$ into a disjoint union, and morphisms preserve such a decomposition. Therefore,
$$
[n] \downarrow \boxplus \simeq \coprod\_{[n] = I \cup I'} (I \downarrow \Delta) \times (I' \downarrow \Delta)
$$
This makes the coend decompose:
$$
(X \star S)\_n = \coprod\_{[n] = I \cup I'} \int^{I \to [c], I' \to [c']} X\_c \times S\_{c'}
$$
However, the comma category $(I \downarrow \Delta)\times(I' \downarrow \Delta)$ has an initial object: $I \times I'$ itself. Thus, the coend degenerates down to simply being the value:
$$
(X \star S)\_n = \coprod\_{[n] = I \cup I'} X\_{|I|} \times S\_{|I'|}
$$
This is slightly different notation for the second definition of the join that you gave.
Now, as for the boundary formulas.
The definition of $d\_i$ is as follows. For each $0 \leq i \leq n$, there is a unique map $d^i:[n-1] \to [n]$ in $\Delta$ whose image is $[n] \setminus \{i\}$: $d^i(x) = x$ for $x < i$, and $d^i(x) = x+1$ for $x \geq i$. The induced map $(X \star S)\_n \to (X \star S)\_{n-1}$ is the map induced by applying the contravariant functor to $d^i$.
Since $(X \star S)\_n$ is a disjoint union of sets, it suffices to show that the formula is correct on $X(I) \times S(I')$ for all decompositions of $[n]$ into $I \cup I'$, where $|I| = j+1$ and $|I'| = k+1$. There are two possibilities: either $i \in I$ when $0 \leq i \leq j$, or $i \in I'$ when $j < i \leq n$.
In either case, the map $[n-1] \to [n] = I \cup I'$ induces, by taking preimages, a unique ordered decomposition $[n-1] = J \cup J'$ of $[n-1]$. If $i \in I$, then $J$ has size $|I| - 1$ and $J'$ is mapped isomorphically to $I'$ by $d^i$. In this case, the map $d^i$ is isomorphic to the map $d^i \boxplus id$ on $[j-1] \boxplus [k] \to [j] \boxplus [k]$. If $i \in I'$, we have the reverse possibility, with $d^i$ isomorphic to $id \boxplus d^{i-j-1}$ (the upper index necessary because inserting the identity at the beginning adds $j+1$ elements to the ordered set at the beginning).
In the case $0 \leq i \leq j$, the induced map
$$
d\_i: X(I) \times S(I') \to \coprod\_{[n-1] = K \cup K'} X(K) \times S(K')
$$
is therefore the map $X(d^i) \times id: X(I) \times S(I') \to X(J) \times S(J')$, followed by the inclusion into the coproduct. In the case $j < i \leq n$, the map is $id \times S(d^{i-n-1})$ followed by inclusion.
This recovers the formula for $d\_i$ that you have written down, up to inserting copies of a point $\*$ as in the remark at the beginning.
| 12 | https://mathoverflow.net/users/360 | 19419 | 12,919 |
https://mathoverflow.net/questions/19420 | 133 | It's "well-known" that the 19th century [Italian school of algebraic geometry](https://en.wikipedia.org/wiki/Italian_school_of_algebraic_geometry) made great progress but also started to flounder due to lack of rigour, possibly in part due to the fact that foundations (comm alg etc) were only just being laid, and possibly (as far as I know) due to the fact that in the 19th century not everyone had come round to the axiomatic way of doing things (perhaps in those days one could use geometric plausibility arguments and they would not be shouted down as non-rigorous and hence invalid? I have no real idea about how maths was done then).
But someone asked me for an explicit example of a false result "proved" by this school, and I was at a loss. Can anyone point me to an explicit example? Preferably a published paper that contained arguments which were at the time at least partially accepted by the community as being OK but in fact have holes in? Actually, to be honest I'd probably prefer some sort of English historical summary of such things, but I do have access to (living and rigorous) Italian algebraic geometers if necessary ;-)
EDIT: A few people have posted solutions which hang upon the Italian-ness or otherwise of the person making the mathematical mistake. It was not my intention to bring the Italian-ness or otherwise of mathematicians into the question! Let me clarify the underlying issue: a friend of mine, interested in logic, asked me about (a) Grothendieck's point of view of set theory and (b) a precise way that one could formulate the statement that he "made algebraic geometry rigorous". My question stemmed from a desire to answer his.
| https://mathoverflow.net/users/1384 | what mistakes did the Italian algebraic geometers actually make? | As for a result that was not simply incorrectly proved, but actually false, there is the case of the [Severi bound](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=MR&pg8=ET&r=1&review_format=html&s4=severi&s5=massimo&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq)(\*) for the maximum number of singular double points of a surface in P^3. The prediction implies that there are no surfaces in P^3 **of degree 6** with more than 52 nodes, but in fact there are such surfaces in P^3 with 65 nodes such as the Barth sextic (and this is optimal by [Jaffe--Ruberman](http://www.ams.org/mathscinet-getitem?mr=1486992)).
(\*) Francesco Severi; "Sul massimo numero di nodi di una superficie di dato ordine dello spazio ordinario o di una forma di un iperspazio." Ann. Mat. Pura Appl. (4) 25, (1946). 1--41, [MR0025179](https://mathscinet.ams.org/mathscinet-getitem?mr=0025179), [doi:10.1007/bf02418077](https://doi.org/10.1007/bf02418077).
| 74 | https://mathoverflow.net/users/4344 | 19428 | 12,923 |
https://mathoverflow.net/questions/19339 | 29 | I think I am confused about some terminology in algebraic geometry, specifically the meaning of the term "torsor". Suppose that I fix a scheme S. I want to work with torsors over S. Let $\mu$ be a sheaf of abelian groups over S. Then my understanding is that a $\mu$-torsor, what ever that is, should be classified by the cohomology gorup $H^1(X; \mu) \cong \check H^1(X; \mu)$.
Now suppose that $\mu$ is representable in the category of schemes over S, i.e. there is a group object $$\mathbb{G} \to S$$
in the category of schemes over $S$, and maps (over S) to $\mathbb{G}$ is the same as $\mu$. Lots of interesting example arise this way.
I also thought that in this case a torsor over S can be defined as a scheme $P \to S$ over S with an action of the group $\mathbb{G}$ such that locally in S it is trivial. I.e. there exists a cover $U \to S$ such that
$$ P \times\_S U \cong \mathbb{G} \times\_S U $$
as spaces over S with a $\mathbb{G}$-action (or rather as spaces over U with a $\mathbb{G} \times\_S U$-action).
The part that confuses me is that these two notions don't seem to agree. Here is an example that I think shows the difference. Let $S= \mathbb{A}^1$ be the affine line (over a field k) and let $x\_1$ and $x\_2$ by two distinct points in $S$. Consider the subscheme $Y = x\_1 \cup x\_2$, and let $C\_Y$ be the complement of Y in S. Let $A$ be your favorite finite abelian group which we consider as a constant sheaf over S. Then we have an exact sequence of sheaves over S,
$$0 \to A\_{C\_Y} \to A \to i\_\*A \to 0$$
Where $i\_\*A(U) = A(U \cap Y)$. I believe the first two are representable by schemes over S, namely $$C\_Y \times A \cup S \times 0$$
and $S \times A$, where we are viewing the finite set $A$ as a scheme over $k$ (and these products are scheme-theoretic products of schemes over $spec \; k$).
In any event, the long exact sequence in sheaf cohomology shows that
$$H^1(S; A\_{C\_Y}) \cong \check H^1(S; A\_{C\_Y}) \cong A$$
and it is easy to build an explicit C$\check{\text{e}}$ch cocycle using the covering given by the two opens consisting of the subschemes $U\_i = S \setminus x\_i$, for $i = 1,2$.
Now the problem comes when I try to glue these together to get a representable object over S, i.e. a torsor in the second sense. Then I am looking at the coequalizer of
$$C\_Y \times A \rightrightarrows (C\_Y\cup C\_Y) \times A$$
where the first map is the inclusion and the second is the usual inclusion together with addition by a given fixed element $a \in A$. This seems to just gives back the trivial "torsor" $C\_Y \times A$.
Am I doing this calculation wrong, or is there really a difference between these two notions of torsor?
| https://mathoverflow.net/users/184 | Torsors in Algebraic Geometry? | As remarked by Brian Conrad above, there is an excellent explanation of all this in Milne's book *Étale cohomology*, Section III.4. There wouldn't be much point in reproducing the details here, but the main issues are:
* You need to decide whether a torsor is going to be a scheme over *S* which locally looks like a trivial torsor, or merely a sheaf of sets over *S* which locally looks like a trivial torsor. What people mean by "torsor" can be either of these things. As Milne says, "The question of which sheaf torsors arise from schemes is, in general, quite delicate". If you go for the sheaf definition, then isomorphism classes of torsors are indeed classified by $\check H^1(S,G)$. Beware that if *G* is not commutative then you need to define $\check H^1(S,G)$ appropriately as a pointed set.
* You need to decide which topology all this is happening in; the usual definition of torsor uses the flat topology, though if *G* is smooth over *S* then you can use the étale topology instead.
* Depending on what topology you're using, and what *S* and *G* look like, there may be issues about whether $\check H^1(S,G)$ and $H^1(S,G)$ are isomorphic.
| 20 | https://mathoverflow.net/users/3753 | 19432 | 12,924 |
https://mathoverflow.net/questions/19431 | 5 | Take a cusp form $f$ and let $f(q) = q + a\_2q^2 + q\_3q^3 + \ldots$" denote its $q$-expansion (assume that the $a\_k$ are integers, and that $f$ comes from an elliptic curve $E$). Of course the series $f(1) = 1 + a\_2 + a\_3 + \ldots$ diverges, but I wonder whether there is any work on evaluating $f(1)$ via some regularization method (I do not even know whether $\lim\_{q \to 1-0} f(q)$ exists or not). I am kind of hoping that $f(1)$ is connected with the order of the Tate-Shafarevich group of the associated elliptic curve $E$ and perhaps the order of its torsion group (the actual expression would have to be invariant under isogenies).
Does this ring a bell with anyone?
| https://mathoverflow.net/users/3503 | Values of cusp forms at q = 1 ? | I'll give a slightly uncertain answer, based somewhat on my recollection of conversations with Zagier a month ago about similar questions.
If we were to imitate Euler, we might consider $f(1)$ as
$$f(1) = \sum\_{n \geq 1} a\_n = \sum\_{n \geq 1} a\_n n^{-0} = L(f,0).$$
So the analytic continuation of the L-function suggests that $f(1)$ should be identified with the value of the L-function at zero. By the functional equation, this relates to the L-function at the right edge of the critical strip.
So, for a cusp form of weight two, arising from an elliptic curve $E$ over $Q$, the value $L(f,0)$ is related to $L(E,2)$. An interpretation of this L-value, conjectured by Zagier, was proven by Goncharov and Levin, in "Zagier's conjecture on $L(E,2)$", Invent. Math. 132 (1998).
As for the analytic question, you are considering the "value" of a cusp form $f$ on the real axis, which bounds the upper half-plane. Almost by definition, there is a Sato hyperfunction $f\_{bdr}$ on the real axis, which describes this boundary behavior of the holomorphic function $f$ on the upper half-plane. I am not sure if the following is published, but I have the impression that there might be a preprint now or soon which proves the following result:
At every (positive? I don't recall) rational number $q$, the hyperfunction $f\_{bdr}$ is $C^\infty$ at $q$. Its value at $1$ is $L(f,0)$ as described above.
I think that saying "a hyperfunction is $C^\infty$ at $q$" means that the hyperfunction can be expressed as the distributional derivative of a continuous function -- $f = g^{(k)}$ for some $k \geq 0$ -- and $g$ happens to be $C^\infty$ at $q$. But I'm not much of an analyst.
I think that the value $f(1)$ also exists as $\lim\_{z \rightarrow 1} f(z)$ limit, if $z$ approaches $1$ via a geodesic in the upper half-plane.
I don't think you'll see Sha or the torsion directly, as these appear at the central value $L(f,1)$. On the other hand, I do think you'll find $L(f,-n)$ for all $n \geq 0$ (or equivalently, $L(f,2+n)$ ), by looking at the derivatives $f^{(n)}(1)$ of the boundary hyperfunction of $f$ at $1$.
| 2 | https://mathoverflow.net/users/3545 | 19433 | 12,925 |
https://mathoverflow.net/questions/19406 | 11 | For which $n \times n$ correlation matrix $C$ can one construct Bernoulli random variables $(B\_1, \ldots, B\_n)$ with correlation $C$ ?
Following the approach described in this MO [thread](https://mathoverflow.net/questions/18268/discrete-stochastic-process-exponentially-correlated-bernoulli), one can think of the following construction. Define independent Bernoulli random variables $B\_{k\_1, \ldots, k\_n}$ for $(k\_1, \ldots, k\_n) \in \mathbb{Z}^n$ and another independent $\mathbb{Z}^k$-valued random variable $I=(I\_1, \ldots, I\_n)$. Then $(B\_{I\_1}, \ldots, B\_{I\_n})$ is a correlated Bernoulli vector.
1: Is there any example of correlation structure that cannot be obtained this way ?
2: Any easy example of correlation matrix $C$ that cannot be the correlation matrix of a $\{0,1\}^n$ valued random vector ?
| https://mathoverflow.net/users/1590 | Constructing Bernoulli random variables with prescribed correlation | Here's a pretty general construction. Take unit vectors $v\_1,\dots,v\_n$ in $\mathbb{R}^n$ and let $u$ be a random unit vector (chosen with the uniform probability measure on the unit sphere). Define $B\_i$ to be 1 if the inner product of $u$ and $v\_i$ is positive and -1 otherwise. Then the correlation between $B\_i$ and $B\_j$ is the inner product of $v\_i$ and $v\_j$. (I haven't checked that carefully but I think it's true.)
Added in the light of the comment below: OK, I should have checked. It's actually not the inner product but π minus twice the angle between the two vectors all over π. I.e., it depends linearly on the angle between the two vectors, is 1 when that angle is zero and -1 when it is π. The angle is the inverse cos of the inner product, which gives us a formula.
So it gives us a fairly big supply of matrices -- I can't quite decide whether it gives us all (or rather all for which the variables take two values, each with probability 1/2).
A more general-looking construction is this. Take any probability space and let $A\_1,...,A\_n$ be sets of measure 1/2. Pick a random point x and let $B\_i$ be 1 if x is in $A\_i$ and -1 otherwise. But that becomes trivial, because if you have any set of Bernoulli variables taking the values $\pm 1$ with probability 1/2, then you can set $A\_i$ to be the set where $B\_i=1$.
| 10 | https://mathoverflow.net/users/1459 | 19436 | 12,926 |
https://mathoverflow.net/questions/19435 | 7 | Hello
Suppose given a polynomial $P=Q\_1\cdots Q\_k$ of degree $n$, where each $Q\_i$ is irreducible. Suppose also that I know the Galois group $G\_i$ (over the rationals) of each irreducible factor $Q\_i$.
Is there an easy correlation between the Galois group of $P$, and the $G\_i$?
| https://mathoverflow.net/users/416 | Galois group of a product of irreducible polynomials | The Galois group of $P$ will be a subdirect product of
the $G\_i$, that is a subgroup of $G\_1\times\cdots\times G\_k$
projecting surjectively onto each of the $G\_i$.
| 19 | https://mathoverflow.net/users/4213 | 19437 | 12,927 |
https://mathoverflow.net/questions/19413 | 3 | Searching for maths tutors online finds people willing to teach up to A-level. I'm looking for help at a more advanced level.
At the moment I'm trying to teach myself category theory from downloaded lecture notes, but I have my eye on other mathematical fields including having another go at algebraic geometry once my category theory is better. However, because I'm teaching myself, if I get stuck I have nowhere to turn. By the same token, I'm doing the exercises but it's frustrating when there's no-one to tell me if I'm getting the answers right or approaching it at the right level of rigor; I find myself missing being able to submit work and get it marked.
How might one go about hiring someone who might be able to give occasional help, either online or in person (I'm in London) at this level? I'm sure university maths departments have plenty of people doing postgrad maths who might like occasional work like this, but how might I go about reaching them?
| https://mathoverflow.net/users/4793 | How should I find a tutor for math-overflow level mathematics? | Some agencies do offer undergrad/postgrad-level tuition - in principle. (I know because I used to be a tutor [for one](http://bluetutors.co.uk/)). Your problem will be finding somebody with the specific knowledge you want. So other peoples' ideas about advertising directly to maths departments will probably be more helpful.
(By the way that agency charges £34/hr for university-level tuition - to give you a rough idea of rates. A large chunk of this goes as commission so you will probably be able to offer a lower direct price.)
| 0 | https://mathoverflow.net/users/1256 | 19456 | 12,938 |
https://mathoverflow.net/questions/19453 | 4 | Suppose you have a scheme $X$ that is acted on by a torus $T$. Then the action induces a grading on the functions on $X$ by the character lattice of $T$. So for a fixed character $\lambda$, we can consider $\mathcal{O}\_{X,\lambda}$, the $\lambda$ graded part. Assuming the quotient $X/T$ exists, these graded parts should descend to quasicoherent sheaves on the quotient.
My question is, when are these sheaves line bundles?
In the basic examples I know, they are always line bundles. For example, if you take $X= \mathbb{A}^{n+1} - 0$ and $T = \mathbb{C}^\*$, then on $X$ you get the ordinary grading by homogeneous degree. When you descend to the quotient, you get the line bundles $\mathcal{O}(k)$ on $\mathbb{P}^n$. You can also take $G$ a complex semi-simple group, $B$ a Borel subgroup, $U$ the maximal unipotent. Then $G/U \rightarrow G/B$ is a torus quotient, and the graded pieces descend to line bundles. I think, a similar story is true for all homogeneous spaces, but I'm having a little trouble phrasing it in terms of torus quotients.
In fact, in these situations, these are all the line bundles.
So more generally, my question is, what properties can you require of the general $X$ so it behaves like the two examples above?
| https://mathoverflow.net/users/788 | Line Bundles on Torus Quotient | (**EDIT**: thought I had added this, guess I was wrong. As Brian points out, you definitely want all your tori to be split, or you won't have enough 1-d representations (for example, $S^1$ has no 1-d real representations); over an algebraically closed field, this is automatic.)
If $X/T$ is actually a nice scheme, and $T$ acts freely, then you can think of this as follows: the pushforward of the structure sheaf on $X$ is etale locally isomorphic to the pushforward of the structure sheaf on $X/T \times T$, so proving that your sheaf is line bundle can be reduced to the case of a trivial bundle, in which case it's obvious (you always get the structure sheaf). This actually proves it's a line bundle in the etale topology, but that's the same as being a line bundle in the Zariski topology.
This actually all works in the land of Artin stacks. In more abstract language, if you have any character of the torus, there's a line bundle on the stack $pt/T$ corresponding to this character (since a quasi-coherent sheaf on $pt/T$ is a vector bundle with $T$ action). The line bundles you're talking about are the pullback by the map $X/T\to pt/T$. This is essentially tautological; the stack $X/T$ is specifically designed so that a quasi-coherent sheaf on $X/T$ is a quasi-coherent sheaf on $X$ (of the same rank) which is $T$-equivariant. You're seeing that the structure sheaf can be $T$-equivariant in a bunch of different ways, since you can always tensor the action with a character of $T$.
| 2 | https://mathoverflow.net/users/66 | 19457 | 12,939 |
https://mathoverflow.net/questions/19458 | 10 | Question
--------
Let $G$ be a group, and let $X$ be a $G$-biset that is (weakly) invertible with respect to the contracted product. Is $X$ necessarily a bitorsor?
Background
----------
By $G$-biset, I mean a set equipped with commuting left and right $G$-actions. There is a standard tensor product on the category of $G$-bisets called the contracted product; it is defined by $X \times\_G Y = X \times Y / (x \cdot g, y) \sim (x, g \cdot y)$, where $G$ acts on the left by its left action on $X$, and on the right by its right action on $Y$. The unit object is the group $G$, where $G$ acts by left and right multiplication on itself.
A left $G$-torsor is a left $G$-set $X$ such that the map $G \times X \to X \times X$, $(g, x) \mapsto (g \cdot x, x)$ is a bijection. A right $G$-torsor is defined analogously. A $G$-bitorsor is a $G$-biset that is both a left and a right $G$-torsor. A $G$-bitorsor $X$ is necessarily invertible with respect to the contracted product; its inverse is the opposite $G$-bitorsor $X^{\operatorname{op}}$. This bitorsor has the same objects as $X$, but $g \in G$ acts on the left (resp. the right) by the right (resp. left) action of $g^{-1}$.
It follows from a simple counting argument that when $G$ is a finite group, any invertible $G$-biset is a $G$-bitorsor. Is this true for arbitrary groups (and more generally, in an arbitrary topos)? What about if we replace "invertible" by "right- (or left-) invertible"?
I can show, at least in the punctual topos (and I think it's true in general), that if $X^{\operatorname{op}}$ is an inverse to $X$, then $X$ must be a $G$-bitorsor. So the question is whether a $G$-biset can have an inverse not of this form.
The reason I'm interested in this question is that I want to understand how to generalize bitorsors to higher categorical settings. A possible generalization would be an invertible profunctor, but this is only a good definition if the answer to my question is affirmative.
| https://mathoverflow.net/users/396 | Is an invertible biset necessarily a bitorsor? | A torsor is a faithful transitive $G$-set. If the left $G$-action on $X$ is not faithful, the left $G$-action on $X\times\_G Y$ will not be faithful. If the left $G$-action on $Y$ is not transitive, the left $G$-action on $X\times\_G Y$ will not be transitive. By symmetry, it follows that a $G$-biset with a left and right inverse is a $G$-bitorsor.
| 10 | https://mathoverflow.net/users/250 | 19462 | 12,941 |
https://mathoverflow.net/questions/19392 | 4 | I ran into an expression calculating the expected value of $\exp(i t \sigma)$ where $\sigma$ is the total number of cycles in a uniformly chosen $S\_n$ element. The expression is
$$E\_n (\exp(i t \sigma)) = \Gamma(n + \exp(it)) / (\Gamma(\exp(it)) n!)$$
where $E\_n$ denotes the expectation under the uniform distribution on $S\_n$. The paper then claims that using Binet's form of Stirling approximation one can get
$$E\_n (\exp(it \sigma)) = n^{\exp(it) -1}/\Gamma(\exp(it)) (1 + o(1))$$
Then here comes the derivation I cannot understand:
using the last expression, they claim one gets the following central limit theorem
$$\lim\_{n \to \infty} E\_n(\exp(it (\sigma - \log n)/\sqrt{\log n})) = \exp( -1/2 t^2)$$
for any real $t$.
I would highly appreciate anyone who can tell me why this is true. It appears to be related to some property of the Gamma function over the complex number.
The relevant paper is Shepp and Lloyd: Ordered cycle lengths in a random permutation.
| https://mathoverflow.net/users/4923 | Generalized binomial coefficients and Gaussian density | Note that $exp(it) = 1 + it - t^2/2 + O(t^3)$ uniformly in $t \in \mathbb{R}$. Thus $n^{exp(it)-1} = exp(it \cdot \log n - \log n \cdot t^{2}/2 + O(t^3 \cdot \log n))$ and also by Taylor's theorem $1/\Gamma(exp(it)) = 1 + O(t)$ when $t$ is small (but in fact also for all real $t \in \mathbb{R}$ by periodicity). Thus $$n^{exp(it)-1}/\Gamma(exp(it)) = exp(it \cdot \log n - \log n \cdot t^2/2 +O(t^3 \cdot \log n))$$
Multiplying both sides by $exp(- it \cdot \log n)$ and substituting $t := t \cdot (\log n)^{-1/2}$ we obtain as $n \rightarrow \infty$ the desired limit, $exp(-t^2/2)$.
| 6 | https://mathoverflow.net/users/3882 | 19464 | 12,942 |
https://mathoverflow.net/questions/19466 | 6 | Let $A$ be your favorite finite dimensional algebra, and $P\_i$ be a sets of representatives for the indecomposible projectives (or [PIMs](http://en.wikipedia.org/wiki/Principal_indecomposable_module), if you like). Then we have the Cartan matrix $C$ of the algebra, whose entries are $\dim Hom(P\_i, P\_j)$. You can think of this as the matrix of the Euler form on the Grothendieck group $K^0(A-pmod)$ of projective $A$-modules.
Now, if the algebra $A$ has finite global dimension, then we can define the classes of the simple heads of these $L\_i$ in $K^0(A-pmod)$ as integer linear combinations of the $P\_i$'s, and $[P\_i]$ and $[L\_i]$ are dual bases in the Euler form. That is, the matrix $C$ is integer valued and has integer-valued inverse, i.e. it has determinant 1.
>
> To what degree is the converse of this true? Is there a weaker hypothesis than finite global dimension itself such that $det(C)=1$ and that hypothesis will imply finite global dimension?
>
>
>
The application I have in mind for this is a little more complex. I'd like to consider a graded version of this question. So, let $A$ be a graded algebra such that each degree is finite dimensional (and let say the appearing gradings are bounded below). The the graded version of $C$ is well-defined in $\mathbb{Z}((q))$, and similarly, if every simple has a resolution by projectives where only finitely many projectives generated in a given degree appear, this implies that this matrix has an inverse in $\mathbb{Z}((q))$, that is determinant with leading coefficient 1.
>
> The same question as above: can I use a hypothesis like the graded Cartan matrix having determinant with integral leading coefficient to conclude the existence of such a projective resolution?
>
>
>
| https://mathoverflow.net/users/66 | Does a finite dimensional algebra having a Cartan matrix with determinant 1 imply finite global dimension (possibly with more hypotheses)? | In general, no. See [Burgess, W. D.; Fuller, K. R.; Voss, E. R.; Zimmermann-Huisgen, B. The Cartan matrix as an indicator of finite global dimension for Artinian rings. Proc. Amer. Math. Soc. 95 (1985), no. 2, 157--165. [MR0801315](http://www.ams.org/mathscinet-getitem?mr=MR0801315)]
It does work for artin algebras of Loewy length at most two, though, and various other families, like quasi-stratified algebras or left serial (that's the main result of the paper mentioned above)
| 7 | https://mathoverflow.net/users/1409 | 19467 | 12,944 |
https://mathoverflow.net/questions/19459 | 32 | This is motivated by pure curiosity (triggered by [this question](https://mathoverflow.net/questions/19402)). A map $f:\mathbb R^n\to\mathbb R^m$ is said to be *Lebesgue-Lebesgue measurable* if the pre-image of any Lebesgue-measurable subset of $\mathbb R^m$ is Lebesgue-measurable in $\mathbb R^n$. This class of maps is terribly inconvenient to deal with but it might be useful sometimes. And maybe it is not that bad in the case $m=1$, especially if the answer to the following question is affirmative.
**Question**: Is every $C^1$ function $f:\mathbb R^n\to\mathbb R$ Lebesgue-Lebesgue measurable? If not, what about $C^\infty$ functions?
I could not figure out the answer even for $n=1$. However, there are some immediate observations (please correct me if I am wrong):
* Since the map is already Borel measurable, the desired condition is equivalent to the following: if $A\subset\mathbb R$ has zero measure, then $f^{-1}(A)$ is measurable.
* If $df\ne 0$ almost everywhere, then $f$ is Lebesgue-Lebesgue measurable (because locally it is a coordinate projection, up to a $C^1$ diffeomorphism). So the question is essentially about how weird $f$ can be on the set where $df=0$.
* If the answer is affirmative for $C^1$, it is also affirmative for Lipschitz functions (by an approximation theorem).
* The answer is negative for $C^0$, already for $n=1$. An example is a continuous bijection $\mathbb R\to\mathbb R$ that sends a Cantor-like set $K$ of positive measure to the standard (zero-measure) Cantor set. There is a non-measurable subset of $K$ but its image is measurable since it is a subset of a zero-measure set.
| https://mathoverflow.net/users/4354 | Is every smooth function Lebesgue-Lebesgue measurable? | It seems that your example of bijection that sends one Cantor set with positive measure to another Cantor set with zero measure can be made $C^\infty$.
Am I missing something?
| 20 | https://mathoverflow.net/users/1441 | 19468 | 12,945 |
https://mathoverflow.net/questions/19471 | 22 | Is the sum of two measurable set measurable? I think it is not...
| https://mathoverflow.net/users/4928 | Is the sum of 2 Lebesgue measurable sets measurable? | Evidently, there are [measure zero sets with a non measurable sum](http://www.math.wvu.edu/~kcies/prepF/89A+A/89A+A.pdf). The article begins as follows:
>
>
> >
> > Krzysztof Ciesielski,
> > Hajrudin Fejzi´c, Chris Freiling,
> >
> >
> > **Measure zero sets with non-measurable sum**
> >
> >
> >
> > >
> > >
> > > >
> > > > Abstract
> > > >
> > > >
> > > > For any C ⊆ R there is a subset A ⊆ C such that A + A has inner
> > > > measure zero and outer measure the same as C + C. Also, there is a
> > > > subset A of the Cantor middle third set such that A+A is Bernstein in
> > > > [0, 2]. On the other hand there is a perfect set C such that C + C is an
> > > > interval I and there is no subset A ⊆ C with A + A Bernstein in I.
> > > >
> > > >
> > > >
> > >
> > >
> > >
> >
> >
> > 1 Introduction.
> >
> >
> > It is not at all surprising that there should be measure zero sets, A, whose sum
> > A+A = {x+y : x ∈ A, y ∈ A} is non-measurable. Ask a typical mathematician
> > why this should be so and you are likely to get the following response:
> >
> >
> >
> > >
> > >
> > > >
> > > > The Cantor middle-third set, when added to itself gives an entire
> > > > interval, [0, 2]. So certainly there exists a measure zero set that
> > > > when added to itself gives a non-measurable set.
> > > >
> > > >
> > > >
> > >
> > >
> > >
> >
> >
> > The intuition being that an interval has much more content than is needed for
> > a non-measurable set.
> > Indeed such sets do exist (in ZFC). Sierpi´nski (1920) seems to be the first
> > to address this issue. Actually, he shows the existence of measure zero sets
> > X, Y such that X+Y is non-measurable (see [7]). The paper by Rubel (see [6])
> > in 1963 contains the first proof that we could find for the case X = Y (see also
> > [5]). Ciesielski [3] extends these results to much greater generality, showing
> > that A can be a measure zero Hamel basis, or it can be a (non-measurable)
> > Bernstein set and that A+A can also be Bernstein. He also establishes similar
> > results for multiple sums, A + A + A etc.
> >
> >
> > This paper is mainly about the statement above and the intuition behind
> > it. Below we list four conjectures, each of which seems justified by extending
> > this line of reasoning.
> >
> >
> > 1. Not only does such a set exist, but it can be taken to be a subset of the
> > Cantor middle-third set, C. (This does not seem to immediately follow
> > from any of the above proofs. Thomson [9, p. 136] claims this to be
> > true, but without proof.)
> > 2. The intuition really has nothing to do with the precise structure of the
> > Cantor set, which might lead one to conjecture the following. Suppose
> > C is any set with the property that C + C contains a set of positive
> > measure. Then there must exist a subset A ⊆ C such that A + A is
> > non-measurable.
> > 3. The intuition relies on the fact that non-measurable sets can have far
> > less content than an entire interval. Therefore, the claim should also
> > hold when non-measurable is replaced by other similar qualities. Recall
> > that if I is a set then a set S is called Bernstein in I if and only if
> > both S and its complement intersect every non-empty perfect subset
> > of I. Constructing a set that is Bernstein in an interval is one of the
> > standard ways of establishing non-measurability. Certainly, any set that
> > is Bernstein in an interval has far less content than the interval itself.
> > Therefore, we might conjecture that there is a subset A ⊆ C
> > with A+A
> > Bernstein in [0,2].
> > 4. Combining the reasoning behind the Conjectures 2 and 3, let C be any
> > set with the property that C + C contains an interval, I. We might
> > conjecture that there must exist a subset A ⊆ C such that A + A is
> > Bernstein in I.
> >
> >
> > We will settle these four conjectures in the next four sections.
> >
> >
> >
>
>
>
The paper goes on to show that conjectures 1, 2 and 3 are true, but 4 is false.
| 28 | https://mathoverflow.net/users/1946 | 19472 | 12,948 |
https://mathoverflow.net/questions/19116 | 20 | In the category of smooth real manifolds, do all small colimits exist? In other words, is this category small-cocomplete? I can see that computing push-outs in the category of topological spaces of smooth manifolds need not be manifolds, but this is not a proof.
| https://mathoverflow.net/users/4517 | Colimits in the category of smooth manifolds | I'd like to recast Reid's (excellent) answer slightly. The essence of it is the following principle:
>
> To show that a limit or colimit doesn't exist in some category, embed your category in one where limits or colimits do exist and find some diagram in the original category whose colimit in the larger category does not lie in the image of the embedding.
>
>
>
The point is, it's usually much easier to show that an object $X$ of $\mathcal{D}$ is not an object of $\mathcal{C}$ than it is to show that $\mathcal{C}$ has nothing that looks like $X$. For a simpler analogy, think of the difference between proving that $(0,1)$ is not complete versus proving that $(0,1) \subseteq \mathbb{R}$ is not closed. The essence is the same, but the latter always seems to me to be a lot easier to grasp.
Back to the principle. As stated, it's not quite strong enough. You need a condition on the embedding:
>
> Make sure that your embedding preserves those limits or colimits that already exist.
>
>
>
Again, by analogy: to prove that a metric space $X$ is not complete, we need a **continuous** map from $X$ to a complete space with non-closed image. An arbitrary map won't do.
Back to the case in hand. As the functor $M \mapsto C^\infty(M,\mathbb{R})$ is a (contravariantly) representable embedding, it preserves colimits and so is suitable for the argument to go through.
However, it does not preserve limits so if you asked the corresponding question about limits, you'd need a different embedding. It turns out, though, that there is a complete and cocomplete category in which the category of manifolds embeds preserving all limits and colimits. That is the category of [Hausdorff Froelicher spaces](http://ncatlab.org/nlab/show/Froelicher+space). Froelicher spaces may feel a little more topological than algebras so for those who, like myself, prefer topology to algebra, here's a recasting of Reid's answer using (Hausdorff) Froelicher spaces.
The key thing is that a Froelicher space is completely determined by either the smooth functions from it to $\mathbb{R}$ or the smooth curves in it (i.e. smooth functions from $\mathbb{R}$).
We take the same colimit: the pushout of
$$
\begin{matrix}
\{0\} &\to& \mathbb{R}\\
\downarrow \\
\mathbb{R}
\end{matrix}
$$
We shall show that it is the union of the $x$ and $y$ axes in $\mathbb{R}^2$, which is clearly not a manifold.
Let us write the colimit as $X$. First, we define a smooth function $F \colon X \to \mathbb{R}^2$. It is the obvious one: it sends the first copy of $\mathbb{R}$ to the $x$-axis and the second copy to the $y$-axis. As these two functions agree on $\{0\}$, this is a well-defined smooth function.
We want to show that this is an initial map. One sufficient (but not necessary) condition for this is that every smooth function $f \colon X \to \mathbb{R}$ factors through $F$.
As Reid says, a smooth function $f \colon X \to \mathbb{R}$ consists of two smooth functions $f\_1, f\_2 \colon \mathbb{R} \to \mathbb{R}$ satisfying $f\_1(0) = f\_2(0)$. Let $g \colon \mathbb{R}^2 \to \mathbb{R}$ be the function $g(x,y) = f\_1(x) + f\_2(y) - f\_1(0)$. This is smooth and we have $g(x,0) = f\_1(x) + f\_2(0) - f\_1(0) = f\_1(x)$ and, similarly, $g(0,y) = f\_2(y)$. Thus $g \circ F = f$ and so every function $X \to \mathbb{R}$ factors through the inclusion $X \to \mathbb{R}^2$. Hence the inclusion $X \to \mathbb{R}^2$ is initial. Thus we can identify $X$ with its image, that being the union of the two axes.
As I said, this is merely a recasting of Reid's answer. I post it partly to make it more topological in feel, but mainly to expose the general principle which Reid uses.
| 17 | https://mathoverflow.net/users/45 | 19473 | 12,949 |
https://mathoverflow.net/questions/19398 | 1 | It is provable that $f\_\lambda\to f\Rightarrow f\_\lambda\*g\to f\*g$ if $g$ has a compact support (shown in my textbook). In my particular case, $g=u(t+\triangle t)-u(t-\triangle t)$. Does for that particular case, $f\_\lambda\*g\to f\*g\Rightarrow f\_\lambda\to f$?
In other words: It is easy to prove that the existence of $\lim\_{\lambda\to\lambda\_0}\langle f\_\lambda(t),\phi(t)\rangle$ implies existence of $\lim\_{\lambda\to\lambda\_0}\langle f\_\lambda(t),\int\_{t-\triangle t}^{t+\triangle t}\phi(\tau)d\tau\rangle$ where $\phi(t)$ denotes test function, and $f$ can be a distribution (since $\int\_{t-\triangle t}^{t+\triangle t}\phi(\tau)d\tau$ is a test function itself). I am wondering does the converse hold, i.e. does the existence of latter limit imply existence of the former (note that not all test functions are expressible as $\int\_{t-\triangle t}^{t+\triangle t}\phi(\tau)d\tau$)?
It may be a silly question, but it would be of much use to know the answer: proof if it holds, or a counterexample if doesn't.
| https://mathoverflow.net/users/4925 | On the convolution of generalized functions | If I understand correctly what you are asking then the answer is: "No".
Here's where I may be misunderstanding: I assume that $\Delta t$ is fixed. If this is correct, we can argue as follows.
Let me write $r = \Delta t$ since it is fixed and I want to disassociate it from $t$. We consider the operator $A\_r \colon C^\infty\_c(\mathbb{R}) \to C^\infty\_c(\mathbb{R})$ defined by
$$
A\_r(\phi)(t) = \int\_{t - r}^{t + r} \phi(\tau) d \tau
$$
We want to extend this function to the space of distributions, $\mathcal{D} = C^\infty\_c(\mathbb{R})$. To do this, we look for an adjoint as per the nlab page on distributions (particularly the section [operations on distributions](http://ncatlab.org/nlab/show/distribution#operations_on_distributions_4); note that my notation is chosen to agree with that page so it's hopefully easy to compare). So for two test functions, $\phi, \psi \in C^\infty\_c(\mathbb{R})$ we calculate as follows:
$$
\begin{array}{rl}
\langle \psi, A\_r(\phi)\rangle &= \int\_{\mathbb{R}} \psi(t) A\_r(\phi)(t) d t \\
&= \int\_{\mathbb{R}} \psi(t) \int\_{t - r}^{t + r} \phi(\tau) d \tau d t \\
&= \int\_{\mathbb{R}} \int\_{t - r}^{t + r} \psi(t) \phi(\tau) d \tau d t \\
&= \int\_{\mathbb{R}} \int\_{\tau - r}^{\tau + r} \psi(t) \phi(\tau) d t d \tau \\
&= \int\_{\mathbb{R}} A\_r(\psi)(\tau) \phi(\tau) d \tau \\
&= \langle A\_r(\psi), \phi \rangle
\end{array}
$$
When we do the switch in order of integration, it's useful to draw the region of integration in the plane (I used a table "cloth" in a restaurant here in Copacabana!). It's a diagonal swathe and looks the same when flipped about the line $x = y$ (don't do this with a table cloth unless you've taken the plates off it first.).
So $A\_r$ is self-adjoint and we extend it to distributions by the formula $\langle A\_r(T), \phi \rangle = \langle T, A\_r(\phi)\rangle$.
Now we come to your question. You have $(T\_\lambda)$ such that $\langle T\_\lambda, A\_r(\phi)\rangle \to \langle T, A\_r(\phi) \rangle$ for each test function $\phi$. Using the definition of the extension of $A\_r$, this is the same as saying that $(A\_r(T\_\lambda)) \to A\_r(T)$ weakly. You want to know if this implies that $(T\_\lambda) \to T$.
To answer this, we consider $A\_r$. Unfortunately for you, it has a non-trivial kernel. Note that we are now working with distributions, so can allow things with non-compact support (otherwise it wouldn't have a non-trivial kernel). For $A\_r(S) = 0$ we simply require that $S$ integrate to $0$ over every interval of length $2 r$. If we pick $S$ an integrable function with period $2 r$ then this is quite easy to arrange: almost all such functions integrate to $0$ over such intervals. In particular, $S = \sin(\pi t/r)$ will do.
But now we can add random amounts of $S$ to each $T\_\lambda$ thus ensuring that $(T\_\lambda)$ does not converge (okay, the amounts are not quite random: we choose them such that the resulting sequence does not converge, but this is always possible). However, since $A\_r$ does not "see" $S$, $(A\_r(T\_\lambda + \alpha\_\lambda S))$ still converges.
As I said, I'm not convinced that I understood the question correctly so please do comment if this doesn't look right. But if it doesn't look right, please edit your question so that it's clearer that this isn't what you wanted.
| 3 | https://mathoverflow.net/users/45 | 19484 | 12,956 |
https://mathoverflow.net/questions/19475 | 6 | I'm going to postpone the motivation for this question because the question itself involves no complicated maths and may well have a very simple solution so I don't want to put anyone off with high falutin' symbols.
Here's the question. I have a smooth curve $c \colon (0,1) \to \mathbb{R}^2$ which does not intersect the $x$-axis. As $t \to 0$, this curve approaches the $x$-axis. I want to find out if it has a point of impact. At my disposal, I have any smooth function $f \colon \mathbb{R}^2 \to \mathbb{R}$ which is constant along the $x$-axis. I know that for any such $f$, $f \circ c \colon (0,1) \to \mathbb{R}$ extends to a smooth function $[0,1) \to \mathbb{R}$ (that is, all one-sided derivatives exist at the origin and are the limits of the corresponding derivatives as we approach the origin) with value $f(0,0)$ at $0$. So:
>
> Is there some $f$ (satisfying the condition) such that the composition $f \circ c$ tells me that $c$ approaches a particular point on the $x$-axis?
>
>
>
If the answer to that is "yes", then my follow-up question is about the derivatives of $c$ at the point of approach.
Here are some comments and partial results:
1. I'm allowed to use **any** information about the compositions $f \circ c$: their values, their derivatives, and so forth.
2. The $y$-value of $c$ easily extends to a smooth function $[0,1) \to \mathbb{R}$ since the second projection $\mathbb{R}^2 \to \mathbb{R}$ is one of our detectors. Moreover, it extends taking the value $0$ at $t = 0$.
3. I can show that the $x$-value of $c$ is bounded as $t$ approaches $0$. If it weren't, I could stick bump functions along the image of $c$ with disjoint support and that were $0$ on the $x$-axis. As they have disjoint support, their sum, call it $f$, is a smooth function and is at the disposal of my detection agency. So there is some sequence $(t\_n) \to 0$ such that $f \circ c(t\_n) = 1$, but $f(0,0) = 0$ so this violates my condition.
4. I can show that if $c$ approaches the $x$-axis with any sort of speed then I can detect its point of impact (and all derivatives). To do this, I use the function $g \colon (x,y) \mapsto x y$. So long as *some* derivative of the (extended) $y$-value of $c$ is non-zero at $0$, I can use this to find out the $x$-value by differentiating $g \circ c$ that many times.
5. If $c$ approaches the $x$-axis infinitely slowly, has a point of impact, and the $x$-value extends to a smooth function $[0,1) \to \mathbb{R}$ (so then $c$ extends to a smooth function $[0,1) \to \mathbb{R}^2$) then I cannot detect the actual point of impact. This is because I can use the chain rule to find the value of any derivative of $f \circ c$ and each term vanishes because either it involves a derivative of the $y$-value (assumed zero) or it involves a pure $x$-derivative of $f$ (zero by assumption on $f$ as we're then on the $x$-axis).
So it seems to me that it's a reasonable conjecture that I can't show that $c$ has a point of impact, if it approaches infinitely slowly. However, the above is not a proof of that fact.
---
**Motivation** I'm trying to finish off the details of an example of a Froelicher space (<http://ncatlab.org/nlab/show/Froelicher+space>). The space in question, let's call it $X$, is the quotient of the plane by the $x$-axis. By the rules for taking quotients, the smooth functions on this space are simply the smooth functions on $\mathbb{R}^2$ which are constant along the $x$-axis. I want to work out the smooth curves. Let $c \colon \mathbb{R} \to X$ be a smooth curve. It's straightforward to show that $c$ lifts to a smooth curve $U\_c \to \mathbb{R}^2$, where $U\_c$ is the complement of the preimage of the collapsed point. As $U\_c$ is (easily shown to be) open, it decomposes as a union of intervals. On each interval, $c$ is a smooth curve in $\mathbb{R}^2$ which approaches the $x$-axis at the end-points. So the question is as to what can be said as $c$ gets near one of these end-points. That's the source of this question.
---
**Edit** Added in response to Bjorn's answer. The underlying question is:
>
> What are the smooth functions $c \colon \mathbb{R} \to \mathbb{R}^2/\mathbb{R}$?
>
>
>
(Blah, blah, [Froelicher space structure](http://ncatlab.org/nlab/show/Froelicher+space), blah, blah)
Thus the point of the question is not "I have a curve, what is it?" but "Which curves can I get?". However, I figured that the question "What are the smooth curves in $\mathbb{R}^2/\mathbb{R}$?" wouldn't get much interest, but something about extending smooth curves in $\mathbb{R}^2$ might!
**Also** If, as I suspect, I can get curves with no definite "point of impact", can I limit how bad these curves must be in some way? Can I put some bound on their ($x$-)derivatives, or at least limit how fast they go to infinity?
| https://mathoverflow.net/users/45 | Can I detect the point of impact without looking at it? | Andrew's comments showed me that in my first answer I was misunderstanding several aspects of his question. Since I am still not entirely sure that I am capturing the spirit of the problem, let be begin this answer by stating in my own words (in very dry mathematical terms) what I interpret the question(s) to be, so that he or someone else can correct me if necessary.
---
Let $\mathcal{F}$ be the set of smooth functions $f \colon \mathbf{R}^2 \to \mathbf{R}$ whose restriction to the $x$-axis is constant. Let $\mathcal{C}$ be the set of smooth functions $c \colon (0,1) \to \mathbf{R}$ such that for every $f \in \mathcal{F}$,
* the function $f \circ c \colon (0,1) \to \mathbf{R}$ extends to a smooth function $[0,1) \to \mathbf{R}$ (in the sense that all one-sided derivatives exist at the origin and equal the one-sided limits of the corresponding derivatives), and
* $\lim\_{t \to 0^+} f(c(t)) = f(0,0)$.
**Question 1:** If $c \in \mathcal{C}$, must $\lim\_{t \to 0^+} c(t)$ exist? (Actually, Andrew is also asking more generally what one can say about $c$ if the limit does not exist.)
**Question 2:** Is there a rule (function) $\mathcal{C} \to \mathcal{F}^r$ for some positive integer $r$, say taking $c$ to $(f\_{1,c},\ldots,f\_{r,c})$, such that $f\_{1,c}$ is independent of $c$, and $f\_{2,c}$ depends only on $f\_{1,c} \circ c$, and $f\_{3,c}$ depends only on $f\_{1,c} \circ c$ and $f\_{2,c} \circ c$, and so on, together with a rule (function) $R$ that takes as input a sequence of smooth functions $(g\_1,\ldots,g\_r)$ from $(0,1)$ to $\mathbf{R}$ and outputs a point in $\mathbf{R}^2$ such that for every $c \in \mathcal{C}$, we have $R(f\_{1,c}\circ c,\ldots,f\_{r,c} \circ c) = \lim\_{t \to 0^+} c(t)$ whenever the latter exists?
**Question 3:** Same as Question 2, but with $\lim\_{t \to 0^+} c'(t)$ in place of $\lim\_{t \to 0^+} c(t)$.
---
**Answer to Question 1:** No. A negative answer was essentially given already by Andrew himself. Namely, let $c(t) := (\sin 1/t,e^{-1/t})$. This $c(t)$ has the following properties: the $x$-coordinate is bounded, the derivatives of the $x$-coordinate grow at most polynomially in $1/t$ as $t \to 0^+$, and the $y$-coordinate and derivatives of the $y$-coordinate decay to $0$ exponentially as $t \to 0^+$. As explained by Andrew, for any $f \in \mathcal{F}$, the chain rule shows that $f(c(t))$ extends to a smooth function $[0,1) \to \mathbf{R}$ whose value at $0$ is $f(0,0)$ and whose higher derivatives at $0$ are all $0$. $\square$
**Answer to Questions 2 and 3:** Yes. In fact, we can do it with $r=2$, and with both $f\_{1,c}$ and $f\_{2,c}$ independent of $c$. Namely, use $f\_1(x,y)=y$ and $f\_2(x,y)=e^x y$. From $f\_1 \circ c$ and $f\_2 \circ c$, we may recover not only the $y$-coordinate of $c$ as $f\_1 \circ c$, but also the $x$-coordinate of $c$ as $\log (f\_2(c(t))/f\_1(c(t)))$. So the whole function $c(t)$, and hence any property of $c(t)$, can be detected. $\square$
| 4 | https://mathoverflow.net/users/2757 | 19492 | 12,958 |
https://mathoverflow.net/questions/19478 | 23 | Let $K$ and $L$ be two subfields of some field. If a variety is defined over both $K$ and $L$, does it follow that the variety can be defined over their intersection?
| https://mathoverflow.net/users/4948 | Fields of definition of a variety | Yes, if varieties are interpreted as subvarieties closed subschemes of base extensions of a fixed ambient variety scheme (e.g., affine space or projective space).
More precisely, suppose that $k \subseteq F$ are fields and the variety $X$ is an $F$-subvariety a closed subscheme of $\mathbf{P}^n\_F$. Say for a field $K$ with $k \subseteq K \subseteq F$ that "**$X$ is defined over $K$**" if $X$ is the base extension of some subvariety of $\mathbf{P}^n\_K$. Then $X$ has a minimal field of definition $E$ with $k \subseteq E \subseteq F$, characterized by the property that for any field $K$ with $k \subseteq K \subseteq F$, we have that $X$ is defined over $K$ if and only if $K$ contains $E$.
The same statement holds if $\mathbf{P}^n$ is replaced by any fixed $k$-variety $k$-scheme $Y$.
(Note: this answer does not contradict Pete's. This is just a different interpretation of the question.)
**EDIT:** As Brian points out, I was indeed assuming that my varieties were closed in the ambient space. The statement about minimal field of definition is not even true for open subschemes in characteristic $p$. For example, if $k=\mathbb{F}\_p$ and $F=k(t)$ and $Y=\operatorname{Spec} k[x]$ and $X=\operatorname{Spec} F[x,1/(x-t)]$, then $X$ is the base extension of $\operatorname{Spec} F^{p^n}[x,1/(x^{p^n}-t^{p^n})]$, and hence is definable over $F^{p^n}$ for all $n$, but not over the intersection of all these fields, which is just $k$.
On the other hand, the intersection of any *finite* number of fields of definition is still a field of definition.
I have generalized to schemes as suggested by Brian.
| 20 | https://mathoverflow.net/users/2757 | 19494 | 12,960 |
https://mathoverflow.net/questions/19490 | 37 | I have often heard in the folk-lore that Feynman Path Integral can be used to compute geometric invariants of a space.
Coming from a background of studying Quantum Field Theory from the books like that of Weinberg, I have myself used Feynman Path Integrals to compute scattering of particles.
Earlier I had done courses in Riemannian Geometry and these days I am also doing courses in Algebraic Topology and hence I think it would be very educative if I can see how exactly the calculation of topological invariants that one does here are related to Feynman's ideas.
It would be helpful if someone can give me references which explain (hopefully starting with simple examples!) how one can use path integrals in geometry.
| https://mathoverflow.net/users/2678 | Doing geometry using Feynman Path Integral? | Try:
Witten, Quantum field theory and the Jones polynomial
Witten, The index of the Dirac operator in loop space
I have found both of these papers quite difficult to understand. I don't know any easier references, and would greatly appreciate it if anybody could suggest some.
Anyway, I guess the basic idea is very simple: Take a manifold, consider some space of "fields" on the manifold (for example a space of sections of a vector bundle), do "integrals" over this space of fields. The results should be invariants of your manifold --- this is not always true, but this is the idea or the hope, anyway.
Edit: I want to also add that (T)QFT has applications not just to geometry/topology but also representation theory. For example check out these [nice](http://www.math.utexas.edu/users/benzvi/GRASP/lectures/NWTFT.html) [notes](http://ncatlab.org/nlab/show/Northwestern+TFT+Conference+2009) of David Ben-Zvi.
| 18 | https://mathoverflow.net/users/83 | 19498 | 12,962 |
https://mathoverflow.net/questions/19496 | 3 | Hello all, it is well-known by transcendence results or Galois theory that famous geometric problems such as trisecting an angle or "squaring the circle" (i.e. given a disk of radius 1 construct a square of equal area) are not solvable by compasses-and-ruler only
constructions.
On the other side, it is equally well-known, by approximation results such as Weierstrass', that given any $\varepsilon >0$
there is a definite construction process that yields an approximate solution which is correct up to $\varepsilon$.
Of course, the obvious solution is to compute the coordinates of the points you need up to
the precision you need, and then place the points.
This solution however relies on some classical function tables (cosine, arc cosine or the decimal expansion of $\sqrt{\pi}$)
and I am wondering if there is a more "purely geometric solution" needing no calculator or tables.
Specifically, for angle trisection, one could ask the following :
Define explicitly a compasses-and-ruler only algorithm with the following properties :
Initial data : a circle with center O and radius 1 cm, two points I and J on that circle such that IOJ is a straight angle, and a point
M on the arc between I and J. Let us call N the point on that arc such that the angle ION is one third of IOM. The algorithm
must return a point N' which is undistinguishable from N to the naked eye, and must not rely on any calculator or tables.
Either that question is interesting or it isn't. If it isn't, the "shortest number of steps" solution has a large number of steps and is only a complicated reformulation of the compute-coordinates-with-calculator method.
| https://mathoverflow.net/users/2389 | Approximate solutions for trisecting the angle or squaring the circle | Well, for trisection it's very simple. You could divide angle into $2^n$ parts, then just take $\lfloor\frac{2^n}{3}\rfloor$ parts. Of course it could be made as close to one third as you want, but might be hard to do.
For circling the square - draw the $2^n$-gon, then a rectangle with sides $a\_n \cdot 2^n$ and $R/2$ where $a\_n$ - is the side of the $2^n$-gon$, and R - is the radius of inscribed circle. then it's easy to transform rectangle into square.
I think it's not harder then constructing the 65537-gon
| 8 | https://mathoverflow.net/users/1888 | 19500 | 12,964 |
https://mathoverflow.net/questions/19505 | 39 | I have studied differential geometry, and am looking for basic introductory texts on Riemannian geometry. My target is eventually Kähler geometry, but certain topics like geodesics, curvature, connections and transport belong more firmly in Riemmanian geometry.
I am aware of earlier questions that ask for basic texts on differential geometry (or topology). However, these questions address mainly differential geometry. I'm more interested in Riemannian geometry here.
| https://mathoverflow.net/users/nan | Introductory text on Riemannian geometry | Personally, for the basics, I can't recommend John M. Lee's "Riemannian Manifolds: An Introduction to Curvature" highly enough. If you already know a lot though, then it might be too basic, because it is a genuine 'introduction' (as opposed to some textbooks which just seem to almost randomly put the word on the cover).
However, right from the first line: "If you've just completed an introductory course on differential geometry, you might be wondering where the geometry went", I was hooked. It introduces geodesics and curvature beautifully and is very readable.
I think the first chapter might be available on the author's website.
| 33 | https://mathoverflow.net/users/4281 | 19508 | 12,970 |
https://mathoverflow.net/questions/19454 | 6 | For $i, j \in \{ 1, \ldots, n \}$, let $X\_{i,j}$ be a real-valued random variable uniformly distributed on the interval $[0,1]$. The $X\_{i,j}$ are independent.
Let $A\_{i,j}$ be the indicator random variable of the event that $X\_{i,j}$ is a local maximum, i. e. it is the largest of the five random variables $X\_{i,j}, X\_{i,j+1}, X\_{i,j-1}, X\_{i-1,j}, X\_{i+1, j}$. For the sake of not having to think about boundary conditions, interpret all coordinates modulo $n$.
Then it's clear that $E A\_{i,j} = 1/5$, by symmetry. It's also not hard to see that:
* $E (A\_{i,j} A\_{i,j+1}) = 0$, since we can't have local maxima both at $(i,j)$ and at $(i,j+1)$. (The case where $X\_{i,j} = X\_{i,j+1}$ can be ignored since it occurs with probability zero.
* I believe $E (A\_{i,j} A\_{i,j+2}) = 2/45$ and $E(A\_{i,j} A\_{i+1,j+1}) = 1/20$. (In any case, these are clearly constants, and their exact values don't matter.)
* $E(A\_{i,j} A\_{k,l}) = 1/25$ for all choices of $i, j, k, l$ other than the ones I already listed and those that clearly are related to them by symmetry. That is, $A\_{i,j}$ and $A\_{k,l}$ are independent unless the rook-neighborhoods of $(i,j)$ and $(k,l)$ overlap.
From this we can compute the mean and variance of $M\_n = \sum\_{i=1}^n \sum\_{j=1}^n A\_{i,j}$, the total number of maxima. Of course $EM = n^2/5$. The variance is a bit harder to compute and I haven't actually written out the computation, but it ought to be asymptotic to $c^2 n^2$ for some positive $c$. (It is the sum of $\Theta(n^2)$ covariances each of order 1.)
Let $\tilde{M}\_n = (M\_n-n^2/5)/(cn)$. Then $\tilde{M}\_n$ has mean $0$ and variance $1$. Is it true that the sequence $\{ \tilde{M\_n} \}\_{n=1}^\infty$ converges in distribution to the standard normal? Intuitively this seems like it should be true -- we're adding up a bunch of small, almost-independent contributions. If it's true, how can this be proven?
This problem came from last year's qualifying exam in probability at Penn; it's been making the rounds around here over the past few days but nobody seems to remember how to do it.
| https://mathoverflow.net/users/143 | Limit law for the number of local maxima in a square lattice of IID random variables | There are quite a few extensions of the Central Limit Theorem to dependent random variables whose dependence is controlled. This includes the case of a sequence of sums of identically distributed random variables whose dependency graphs have uniformly bounded degrees. ["On Normal Approximations of Distributions in Terms of Dependency Graphs"](https://projecteuclid.org/journals/annals-of-probability/volume-17/issue-4/On-Normal-Approximations-of-Distributions-in-Terms-of-Dependency-Graphs/10.1214/aop/1176991178.full) is overkill, but it includes a sort of Berry-Esseen result bounding the error of the normal approximation.
My guess is that the expected answer on the qualifying exam was not a proof of that extension of the CLT, but was instead, "Recall this locally dependent version of the Central Limit Theorem from class. See that the indicator variables are only locally dependent." I'd be happy to be wrong about that, though.
| 5 | https://mathoverflow.net/users/2954 | 19516 | 12,977 |
https://mathoverflow.net/questions/18797 | 13 | I'm sure this is standard but I don't know where to look. Let $M$ be a contractible compact smooth $n$-manifold with boundary. Does it have to be homeomorphic to $D^n$? What about diffeomorphic?
[UPDATE: the answer is well-known to be negative as many people kindly pointed out. But actually I assume more about the manifold, namely the following:]
There is a Riemannian metric on $M$ such that every two points are connected by a unique shortest path. So $M$ can be contracted to a point $p\in M$ by sending every point along a shortest path to $p$. These paths can bend along the boundary and can merge because of this. But they are relatively nice (namely $C^{1,1}$) curves and their first derivatives depend continuously on their endpoints. Given all this, can one conclude that $M$ is a disc?
ADDED: These curves are of course gradient curves of a function (the distance to $p$) which is $C^1$ and has no critical points in the interior of $M$, except at $p$.
| https://mathoverflow.net/users/4354 | Contractible manifold with boundary - is it a disc? | Given a function $\psi:\mathbb R\to \mathbb R$,
set
$$\Psi=\psi\circ\mathrm{dist}\_ {\partial M},\ \ \ \ \ f=\Psi\cdot(R-\mathrm{dist}\_ p)$$
for some fixed $R>\mathrm{diam}\ M$.
Further,
$$d\,f =
(R-\mathrm{dist}\_ p)\cdot d\,\Psi-\Psi\cdot d\,\mathrm{dist}\_ p$$
Thus, we may choose smooth increasing $\psi$,
such that $\psi(0)=0$
and it is constant outside of little nbhd of $0$ so that
$\Psi$ is smooth.
(It is possible since the function $\mathrm{dist}\_ {\partial M}$ is smooth and has no critical points in a small neighborhood of $\partial M$.)
Note that $d\,\Psi$ is positive muliple of $d\,\mathrm{dist}\_ {\partial M}$.
Thus $d\_x\,f=0$ means that geodesic from $x$ to $p$ goes directly in the direction of minimizing geodesic from $x$ to $\partial M$, which can not happen.
Now we can apply Morse theory for $f$...
| 3 | https://mathoverflow.net/users/1441 | 19522 | 12,981 |
https://mathoverflow.net/questions/19521 | 11 | This question was inspired by
[How to prove that the subrings of the rational numbers are noetherian?](https://mathoverflow.net/questions/19480/how-to-prove-that-the-subrings-of-the-rational-numbers-are-noetherian/19481#19481)
which some people found too routine to be of interest. So I have decided to liven things up a bit with the following questions. In the interest of full disclosure, I have not thought seriously about these questions, and I think that I probably could answer at least some of them myself, but I do think they are interesting and, if I may say so, educational.
Find all (commutative!) fields $K$ such that every (unital!) subring $R$ of $K$ is:
a) a principal ideal domain.
b) a Dedekind domain.
c) a Noetherian domain.
I mean here to be asking three different questions, one for each condition. Evidently the classes of such fields are nondecreasing from a) to b) and from b) to c).
If you would like to answer the question with a), b) or c) replaced by some other standard property of commutative rings — especially if it yields a different class of fields than in the first three questions — please feel free.
**Addendum**: How about
d) a Dedekind domain if it is integrally closed?
e) a PID if it is integrally closed?
| https://mathoverflow.net/users/1149 | For which fields K is every subring of K…? | Regarding question (c), I can tell you exactly which integral domains have only Noetherian subrings by quoting the aptly titled [*Integral domains with Noetherian subrings*](https://doi.org/10.1007/BF02567320 "Commentarii Mathematici Helvetici 45, 129–134 (1970)") by Robert Gilmer:
If $K$ is the field of fractions and $\operatorname{char}(K)=0$, we just need $[K:\mathbb{Q}]<\infty$.
If $\operatorname{char}(K)=p$ with prime subfield $k$, we need $K$ to be either finite or a finite algebraic extension of a $k[X]$ for some transcendental $X$.
I guess this pretty much restricts the answers to questions (a) and (b)….
| 14 | https://mathoverflow.net/users/3143 | 19523 | 12,982 |
https://mathoverflow.net/questions/19527 | 3 | Let *m* the n-dimensional Lebesgue measure on $R^n$.By definition of product measure, on each borelian set E
$m(E)=\inf \left(\sum\_{j=1}^\infty m(R\_j),\:\: E\subseteq \bigcup R\_j , \:\:R\_j \text{ rectangles}\right)$
It is also true that lebesgue measures are regular, so
$m(E)=\inf \left(m(U), E\subseteq U, \: U \text{ open set} \right)$.
Can I say that also holds
$m(E)=\inf \left(\sum\_{j=1}^\infty m(B\_j),\:\: E\subseteq \bigcup B\_j , \:\:B\_j \text{ balls}\right)$ or not?
| https://mathoverflow.net/users/4928 | Lebesgue measure of a set | It follows from Vitali's covering theorem but not in an entirely trivial
fashion. We can reduce to the case where $E$ is open of finite measure.
The set of all open balls contained in $E$ is then a Vitali cover. By Vitali's
covering theorem there is a sequence of disjoint balls $(B\_n)$
whose union is a subset $U$ of $E$ with the same measure as $E$.
Thus $F=E-U$ is a set of Lebesgue measure zero.
Let $\epsilon>0$. There is a sequence of open cubes covering $F$
with total measure $<\epsilon$. Circumscribe these with balls
and we get a sequence of open balls covering $F$ with total measure $< c\_n\epsilon$
where $c\_n>0$. Interweaving this sequence with the $(B\_n)$ we get
a sequence of balls covering $E$ of total measure $< m(E)+c\_n\epsilon$.
| 3 | https://mathoverflow.net/users/4213 | 19532 | 12,985 |
https://mathoverflow.net/questions/17306 | 4 | To say that I am a novice at deformation theory is to grossly overestimate my abilities in this area. I've come across the following theorem in a paper, and I'd like to know how far one is able to generalize it:
### Theorem
Let $R$ be a complete DVR, $X$ a proper smooth curve over $R$, and $D$ a simple divisor on $X$. Let $\bar f:\bar Y\rightarrow \bar X$ be a tame covering of $\bar X$ with branch locus contained in Supp($\bar D$). Then there exists a unique lifting of $\bar f$ to a D-tame covering $f:Y\rightarrow X$. In addition, $f$ is a tame covering of $X$, and the branch locus of $f$ is contained in Supp($D$).
### Notation
I'm not really sure how much of it is standard, so here's what I mean by the terms in the theorem:
A simple divisor on $X$ - a divisor that has no multiple components when base-changed to any geometric point of the base scheme.
Tame covering of integral varieties over a field ($f:\bar Y\rightarrow \bar X$ in our theorem) - what you think: the ramification indices in codim 1 are coprime to the characteristic.
D-tame covering of schemes over a DVR ($f:Y\rightarrow X$ in the theorem) - if for every (natural number) $k$, base change to $R/m^k$ is $D\times\_{R}R/m^k$ -tame in the following sense:
D-tame covering of schemes over an Artin local ring, $\Lambda$ ($R/m^k$ in the previous notation) - a finite, flat morphism which is etale away from Supp($D$)., and such that if $x$ is in Supp($D$), and t is a local equation for D at $x$, then $f\_\ast(O\_Y)\_x$ is a t-tame extension of $O\_x(X)$. What's that? $A'$ in the following, plays the role of $O\_x(X)$ here.
t-tame extensions of a ring $A'$ s.t. $A'/Nil(A')$ is a DVR with parameter $\check t$ (and assume (0) is primary in $Spec(A')$, meaning $Spec(A')$ has no imbedded components)- tame when base changed to $A'/tA'$ (where $t$ is some element going to $\check t$). $A'/tA'$ is an Artin ring. So what do I mean by tameness of extensions over Artin rings? Well:
A tame extension of an Artin ring $\Lambda$ - it is a (finite) product of $\Gamma\_i$'s, such that for each $\Gamma\_i$ it is free over $\Lambda$ of rank $e\_if\_i$; the extension of residue fields is $f\_i$; $e\_i$ is prime to the characterstic of the residue field of $\Lambda$; and $\Gamma\_i$ contains a principal ideal $J$ such that $J^{e\_i}$=0, and such that $\Gamma\_i/J$ is free over $\Lambda$ of rank $f$.
### Question
Phew, that was a lot of work... So - other than finding the whole thing very confusing, here is a concrete set of questions:
1. Can this be generalized to $X$ a higher dimensional scheme?
2. What would be a good reference to read for this type of deformation theory?
3. Well - just about any insight and context you can give would be great (whether deformation theoretic or about the various generalizations of tameness here.)
This is a bit long winded, but if nothing else - it was good exercise for me to write this.
| https://mathoverflow.net/users/3238 | Deformations of Tame Coverings | A paper I'm reading now is a PERFECT reference for this: "Deformation of tame admissible covers of curves" by Stefan Wewers is written in an expository style. (corollary 3.1.3 is exactly the theorem stated in the question.)
| 3 | https://mathoverflow.net/users/2665 | 19537 | 12,989 |
https://mathoverflow.net/questions/19530 | 30 | There seems to be some confusion over what the tangent space to a singular point of an orbifold is.
On the one hand there is the obvious notion that smooth structures on orbifolds lift to smooth $G$-invariant structures on $\mathbb R^n$ ($G$ being the finite group so that the orbifold is locally (about some specific point $x$) the quotient of $\mathbb R^n$ by the action of $G$). One might be tempted to consider cone points as differentiable spaces (that is, subsets of some $\mathbb R^k$, inheriting their differential structure by restriction), however, we are told, for example, that $\mathbb R^2/\mathbb Z\_3$ and $\mathbb R^2/\mathbb Z\_4$ are distinct as orbifolds, so it is not the case that cone points can be modeled merely with cone-like subsets of some $\mathbb R^k$. The definition in which 'smooth' means 'lifts to $G$-invariant smooth' distinguishes these two cones, as the set of functions with 3-fold symmetry and the set of functions with 4-fold symmetry, in $\mathbb R^2$, are distinct. The third item in Satake's seminal paper [On a Generalization of the Notion of Manifold] corroborates this, giving $C^\infty$ forms of degree $p$ at a singularity $x$ as those $C^\infty$ $p$-forms in $\mathbb R^n$ which are invariant under $G\_x$. If we require the same property of vectors, that is, that they lift to $G$-invariant vectors in $\mathbb R^n$, then we have that the dimension of the tangent space of an orbifold is the dimension of the invariant subspace upstairs. In particular the dimension tends to drop at the singular points. For example, the dimension of the tangent space at the singularity in $\mathbb R^2/\mathbb Z\_3$ is 0. This notion of vector agrees with the notion of vector as derivation on the germ of smooth functions. In this case smooth functions in $\mathbb R^2$ which have 3-fold symmetry necessarily have vanishing derivatives at the origin.
On the other hand, one finds descriptions of smooth orbifolds as objects which have tangent bundle-like structures, which are locally $\mathbb R^n/G$. It is not clear what this means as far as smooth structures go, but the explanation above of $\mathbb R^2/\mathbb Z\_3$ having a 0 dimensional tangent space at the cone point seems to contradict the notion that the tangent-like space at the singularity in $\mathbb R^2/\mathbb Z\_3$ *is* $\mathbb R^2/\mathbb Z\_3$, whatever that means.
It is also said that manifolds with boundary can be viewed as orbifolds, which have isotropy group reflection by $\mathbb Z\_2$ along their boundaries. It would be nice to include the note that the differentiable structures are different. Specifically, smooth manifolds with boundaries have tangent spaces along their boundaries which are the same dimension as the manifold. In contrast, the same topological space as an orbifold with $\mathbb Z\_2$ structure group along the boundary should have a tangent space which is one dimension less than the dimension at a generic point, if the definition of tangent space follows Satake's guideline. Indeed, smooth functions in $\mathbb R^n$ which locally have symmetry by reflection through a codimension 1 hyperplane, have vanishing partial derivatives in the normal direction.
I am asking for concurrence or correction and clarification, since I am still not certain I have the correct notion of tangent space to an orbifold, although I'm fairly confident in the first given here.
| https://mathoverflow.net/users/2031 | What is meant by smooth orbifold? | Disclaimer: I don't talk to people about orbifolds. This answer may not represent the opinions of orbifolders.
As I understand it, the orbifold $\mathbb R^n/G$ is characterized by how manifolds map to it,† not by how it maps to manifolds. In particular, the orbifold *is not* determined by the ring of smooth functions on it (i.e. the ring of $G$-invariant smooth functions on $\mathbb R^n$). In particular, the ring of 3-fold symmetric smooth functions on $\mathbb R^2$ *is isomorphic* to the ring of 4-fold symmetric functions on $\mathbb R^2$.
Given that the smooth functions on an orbifold don't "remember" everything about it, it seems unreasonable to define the tangent space at a point in terms of derivations of smooth functions. However, we have another construction of the tangent space at a point: equivalence classes of smooth curves through that point. Since this definition has to do with maps *into* the space rather than maps *out of* it, we expect it to play well with orbifolds. Any smooth curve through the cone point of $\mathbb R^n/G$ lifts to a curve in $\mathbb R^n$, and any curve in $\mathbb R^n$ induces a curve in $\mathbb R^n/G$, so the tangent space to $\mathbb R^n/G$ should be the same as the tangent space to $\mathbb R^n$. Well, not exactly, since a curve in $\mathbb R^n/G$ can lift to a curve in $\mathbb R^n$ in $G$ different ways. So the tangent space to $\mathbb R^n/G$ at the cone point should really be the quotient of the tangent space of $\mathbb R^n$ by the action of $G$.
So far, it seems like I'm arguing that the tangent space to $\mathbb R^n/G$ at the cone point should be *the orbifold* $\mathbb R^n/G$. But I actually want to say that the tangent space should be the tangent space of $\mathbb R^n$, together with the action of $G$. The reasoning is that the vector space together with the action of the residual group is independent of how you express $\mathbb R^n/G$ as a quotient by a finite group. In other words, even though an isomorphism of orbifolds $\mathbb R^n/G\cong M/G'$ does not induce isomorphisms $\mathbb R^n\cong M$ or $G\cong G'$, it does induce isomorphisms of tangent spaces and residual groups in such a way that respects the actions of the residual groups on the tangent spaces. Once your definition of tangent space is canonically related to the orbifold, you should be welcome to think of it however you like.
† Roughly, a map from a manifold $M$ to $\mathbb R^n/G$ should be the same thing as a map from $M$ to $\mathbb R^n$, except that maps that differ by the action of $G$ should be regarded as the same map. More precisely, I think a map from $M$ to $\mathbb R^n/G$ should consist of a $G$-fold covering space of $M$ with a $G$-equivariant map to $\mathbb R^n$.
| 27 | https://mathoverflow.net/users/1 | 19542 | 12,994 |
https://mathoverflow.net/questions/19397 | 6 | I often hear mention of two theorems, [Mostow's rigidity theorem](http://en.wikipedia.org/wiki/Mostow_rigidity) and [Liouville's theorem on conformal mappings](http://en.wikipedia.org/wiki/Liouville%2527s_theorem_%2528conformal_mappings%2529), which superficially sound similar: they say that a set of geometric structures is, if nonempty, big in dimension 2, but small in dimension greater than 2.
(For Mostow's theorem, the set of structures in question is the set of hyperbolic metrics on a manifold; for Liouville's, it's the set of germs of flat metrics in a conformal equivalence class.)
I know that hyperbolic and conformal geometry are closely connected, at least in dimension 2. I'm curious as to whether this analogy is hinting at one such connection. Is there a "good reason" for this analogy?
| https://mathoverflow.net/users/2819 | Analogy of Liouville conformal mapping theorem with Mostow rigidity? | There is a connection in some way: if I remember right, you usually prove Mostow rigidity by looking at the hyperbolic space, which is the universal cover of your hyperbolic manifold, then you consider its boundary, which is the flat conformal sphere. Any isometry of the hyperbolic space induces a conformal transformation of its boundary, and vice-versa. But I don't think that you can derive Mostow rigidity from Liouville's conformal theorem.
| 1 | https://mathoverflow.net/users/4961 | 19544 | 12,996 |
https://mathoverflow.net/questions/19526 | 3 | I ran into a "well-known identity" on page 345 of Shepp and Lloyd's [On ordered cycle lengths in a random permutation](http://www.jstor.org/pss/1994483):
$$\int\_x^{\infty} \frac{\exp(-y)}y dy = \int\_0^x \frac{1-\exp(-y)}y dy - \log x - \gamma, $$
where $\gamma$ is the Euler constant. I am clueless as to how it is derived. Any reference to the derivation of such formulae would suffice, but an explicit solution will also be appreciated.
| https://mathoverflow.net/users/4923 | Reference request for a "well-known identity" in a paper of Shepp and Lloyd | You can apply WZ theory to such identities. In particular, both sides satisfy
$$x\*z''(x) + (x+1)z'(x)$$
Picking $x=1$ as the initial condition (since the DE is regular there, that helps), we see that both sides evaluate to $Ei(1,1)$ and their derivatives both evaluate to $-1/e$, so they are equal.
I got that differential equation using Maple's *PDEtools[dpolyform]* function, which uses Groebner bases over differential polynomials to 'solve' this problem. All the rest is classical analysis (as in [A course of modern analysis](http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521588072) by Whittaker and Watson, 1926 - which is unfortunately not material that is taught very much anymore, I certainly had to learn a lot of that 'on my own').
[Edit: fixed an error in the evaluation of the derivative, I pasted in the wrong line]
| 6 | https://mathoverflow.net/users/3993 | 19546 | 12,998 |
https://mathoverflow.net/questions/19547 | 7 | Say we have three infinite sequences $\{a\_i\},\{b\_i\},\{c\_i\}$ of natural numbers, satisfying the equations $$a\_1+b\_1=c\_1,\dots, a\_n+b\_n=c\_n,\dots $$
Assume further that $gcd(a\_i,b\_i,c\_i)=1$ for each $i$ and that $(a\_i,b\_i,c\_i)\neq (a\_j,b\_j,c\_j)$ for all $i,j$. Now let's define $S$ as the set of primes $p$ which divide $a\_ib\_ic\_i$ for at least one $i$. From the S-unit theorem we know that $S$ has to be infinite.
Now the question is: Can $S$ be sparse? This can be taken to mean Dirichlet density zero, for example.
I haven't thought much about this but there are reasons to believe the answer is yes, indeed if there are infinitely many Mersenne primes $q=2^p-1$ then the equations $1+q=2^p$ give such a sparse $S$. However, I am looking for an unconditional result.
| https://mathoverflow.net/users/2384 | An S-unit equation, with S an infinite and sparse set of primes. | Yes, in fact, you can make $S$ grow as slowly as you like. This follows, for example, from the fact that there exist 3-term arithmetic progressions of primes $(p,q,r)$ with $\min(p,q,r)$ arbitrarily large. For each such arithmetic progression, you can take $(a\_i,b\_i,c\_i)=(p,r,2q)$. Now just choose these arithmetic progressions one at a time, making sure that the primes involved each time are much larger than any primes appearing earlier.
| 10 | https://mathoverflow.net/users/2757 | 19554 | 13,001 |
https://mathoverflow.net/questions/19485 | 2 | The famous hopf theorem says that a smooth map from a oriented closed dimension $p$ manifold to $S^{p}$ is homotopic if and only if $f$ and $g$ have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':
* Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .
* Theorem B: If two Pontryagin manifold induced by $f$ and $g$ are frame cobordant, the $f$ and $g$ are homotopic (smooth).
* Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping $f$.
First, it is well-know that if $f$ and $g$ is smooth homotopic, then they have the same brower degree.
Second, we need to prove that if $f$ and $g$ have the same degree, then they are homotopic. From above three theorems, we only have to prove that $f$ and g have the frame cobordant Pontryagin manifold. Since $\operatorname{dim} M=p=$ dimension of $p$-sphere, so the corresponding frame are of $0$ dim, i.e. discrete points in $M$, so if we define $\operatorname{sgn}(x)=1$ or $-1$ for $x$ in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?
Note : we have the notations and definitions as in Milnor's book.
| https://mathoverflow.net/users/4437 | a small questions about hopf theorem | I realize that this doesn't answer your question, but there is also an approach using the methods of homotopy theory and CW complexes. If $M$ is a closed smooth orientable $p$-manifold, then $M$ is homeomorphic to a finite CW complex with cells of dimension $\leq p$, and $H^p(M)=\mathbb{Z}$.
We may construct a $K(\mathbb{Z},p)$ as a CW complex by taking $S^p$ and attaching cells of dimension $\geq p+2$ to kill the higher homotopy groups (for example, $K(\mathbb{Z},2)= \mathbb{CP}^{\infty}$ has a cell decomposition with one cell in every even dimension, and $2$-skeleton $S^2$). Let $i:S^p\hookrightarrow K(\mathbb{Z},p)$ be the induced inclusion, representing the fundamental class $[S^p]\in H^p(S^p)$.
If we have maps $f\_i: M \to S^p$, $i=0,1$, such that the induced maps $H^p(f\_i): H^p(S^p)\to H^p(M)$ are equal, then this implies that $H^p(f\_0)([S^p])=H^p(f\_1)([S^p])$, and therefore that the maps $i\circ f\_i: M\to K(\mathbb{Z},p)$ are homotopic (by Brown Representability, Thm. 4E.1 Hatcher), and thus are realized by a map $F: M\times [0,1] \to K(\mathbb{Z},p)$, $F\_{| M\times i}=f\_i$. By the Cellular Approximation Theorem (Theorem 4.8 Hatcher), the map $F$ may be homotoped rel $M\times \{ 0, 1 \}$ to a map $F': M\times [0,1] \to K(\mathbb{Z},p)^{(p+1)}=S^p$, the $p+1$ skeleton of $K(\mathbb{Z},p)$, and thus $f\_0\simeq f\_1$.
| 3 | https://mathoverflow.net/users/1345 | 19562 | 13,008 |
https://mathoverflow.net/questions/16128 | 4 | Bayesian probabilities are usually justified by the Cox theorems, that can be written this way:
*Under some technical assumptions (continuity, etc, etc...), given a set $P$ of objects $A, B, C, \ldots$, with a boolean algebra defined over it with operations $A \wedge B$ (and) and $A | B$ (or) such that*:
1) $A \wedge B = B \wedge A$
2) $A \wedge (B \wedge C) = (A \wedge B) \wedge C$
3) $A | (B \wedge C) = (A|B) \wedge (A|C)$
*and a "valuation":*
$f : P \rightarrow \mathcal{R}$
*there is a strictly monotonic "regraduation function"* $R : \mathcal{R} \rightarrow \mathcal{R}$ *such that, for*:
$R(f(A\wedge B)) = R(f(A)) + R(f(B))$ (sum rule)
and
$R(f(A|B)) = R(f(A) ) R(f(B))$ (product rule)
This theorem allows one to show that any system designed to "evaluate" boolean expressions consistently with a single real number redunds in the laws of classical probability (this can be seen shortly here: arxiv:physics/0403089 and more thoroughly here: arxiv:abs/0808.0012)
Recently this has been extended for valuations of the type $f : P \rightarrow \mathcal{R}^2$ in <http://arxiv.org/abs/0907.0909> and they proved that there are just 5 canonical valuations compatible with the underlying Boolean algebra (one of them giving a complex field structure to the "valuation" field).
My question is: is it possible/interesting/feasible to classify at least a class of valuations of the type:
$f : P \rightarrow W$
where W is a continuous manifold? If we retrict our attention to $\mathcal{R}^n$ for example, is there, for each n, a set of canonical valuations to which all others can be reduced after a regraduation?
If this can be done, are those nice rules for inference in some sense? Are they useful as inference tools in specific situations?
| https://mathoverflow.net/users/757 | Generalized Cox Theorems, valuations on boolean sets, bayesian probabilities and posets | Thanks for noting our work in this area.
This has been worked out in even more detail here:
Lattice duality: The origin of probability and entropy. Neurocomputing. 67C: 245-274. DOI: 10.1016/j.neucom.2004.11.039
<http://knuthlab.rit.albany.edu/papers/knuth-neurocomp-05-published.pdf>
Its fundamental application to a wide array of problems is first noted and described here
arXiv:physics/0403031v1
which was, in part, the inspiration for the quantum mechanics paper you kindly note above:
arXiv:0907.0909
Last, I have a more recent derivation of Bayesian probability theory from lattices with simple graphical proofs:
arXiv:0909.3684
This work is literally a derivation of measure theory from much the more basic symmetry of associativity. It is immediately applicable to Boolean lattices of statements, which results in Bayesian probability theory. And, as we have shown, results in information theory and the complex Feynman rules of Quantum Mechanics.
Cheers
Kevin
| 5 | https://mathoverflow.net/users/4963 | 19563 | 13,009 |
https://mathoverflow.net/questions/19568 | 7 | If you get your PhD in math , and then work for 1 or 2 years in a non-academic institution and then turn to apply for postdoc or tenure-track position in math like usual, is there any disadvantage (I mean for your application for postdoc or tenure-track position)?
An appendix: I just want to make sure whether or not I can't or it is difficult to get reference letter, take conference or give talks (in the future) because you are not in academic institution. This is the most important for someone (like me) who will returen to academic job.(but for some reasons he can't now)
| https://mathoverflow.net/users/2391 | Is there any disadvantage from non-academic job turn to academic job in math | My personal opinion is that such a career path can contribute a lot to mathematics, because such candidates can often be informed by a more practical or utilitarian focus in their mathematical research, providing an important and invigorating perspective. For example, for someone to arrive at hard-core mathematical research in financial mathematics after having actually worked in investment banking would be extremely useful. And surely there is a similar situation with many other areas of applied mathematics. In areas tending towards pure mathematics, however, probably it becomes more difficult to make the case that the time spent away from research mathematics was beneficial.
In any case, a person applying for a position from such a situation would naturally have a disadvantage in terms of publications and research accomplishments in comparison with the competing candidates. If the idea were to take a non-academic position with the goal of eventually returning to academic research, therefore, then it would seem advisable to keep up one's research as best as one can. Even a publication or two would be fairly convincing evidence of one's true mathematical nature.
I do know several examples of people who spent a long time away from academic mathematics and then returned to a successful academic career in pure mathematics (even in set theory!). I know of several mathematians who had entire careers in business or computer software before returning to academic research and becoming tenured professors. So indeed it is possible. But surely this is far from usual, and certainly the more typical pattern is that once someone leaves mathematical research, unfortunately, they do not return.
| 17 | https://mathoverflow.net/users/1946 | 19569 | 13,013 |
https://mathoverflow.net/questions/19548 | 1 | It is common that you have some interesting object (set, group, algebra or something, whatever) which has certain properties, structure etc. You may try describe it in pure algebraic way. Sometimes You cannot find representation different from this you are working on.
How to distinguish its algebraic properties form properties arising from particular representation? Is there any systematic procedure?
Are there only trivial answers such as: "find another model/representation", or "every object may be regarded as example of different algebraic structures"?
Or maybe it is enough to describe something without notions from representation ( finite dimensionality, matrix indices etc)?
---
**Examples**: suppose you have an algebraic structure given by a set $M$ with some operations on it. It has a representation in a matrix algebra. Suppose that this representation has following properties:
1. for every matrix $M$ from the representation $\det(M) = 1$
2. for every matrix from the representation
$\mathrm{Tr}(M) = 0$
3. the whole representation is a vector space of dimension $n$ with basis given by set of $n$ independent matrices
4. there is a matrix $S$ in the representation for which $S^2=id$
5. for certain matrices $A$, $B$ and $C$,
there is the relation $AB^2 - C = C^2$
etc.
If we regard its as a group only property 4 will be representation independent. But if we talking about vector space obviously property 3 is crucial.
We have some freedom to choose what is important: so if we are talking about abstract group property 3 may be called particular property of representation, whilst when we are in vector spaces, property 4 may be particular. Property 2 may be called fundamental if we are talking about "matrix groups" etc. What about property 5? Is it important?
Suppose we are able to wrote relations in a way which is purely algebraic ( for example for matrices we may write relations which not must be narrowed to matrix operations, we may interpret it as a general algebraic property of more general object, as relation $S^2 =\mathrm{id}$). And it lead to some interesting conclusions. How to be sure, the property we are research is not only property of chosen representation? What with property 5?
Of course we may regard the same object as a group and as a vector space simultaneously. In this way however certain properties may be considered as detached one from each other. Is this the only way?
| https://mathoverflow.net/users/3811 | How to distinguish property of particular representation from property of algebraic structure? | I'm not sure I completely understand what you're asking, but here is some information that appears to be relevant.
In the context you're describing, you have two languages: the pure language L0 of groups and the augmented language L1 of groups together with a linear representation over some field (see note). You seem to be asking the following:
* When does a sentence in L1 (i.e. a property of groups with linear representations) equivalent to a sentence in L0 (i.e. a pure property of groups)?
The [Robinson Consistency Theorem](http://en.wikipedia.org/wiki/Robinson%27s_joint_consistency_theorem) answers this, at least in part.
Let $G$ be a group and let L2 be the language of groups augmented with a constant for every element of $G$. Let T2 be the complete theory of $G$ in L2 and let T0 = T2 ∩ L0 be the complete theory of $G$ in L0. The difference between T0 and T2 is that statements in T2 are allowed to mention particular elements of $G$. So if $x$ is an element of order $2$ in $G$, then that fact is recorded in T2 but not in T0. However, the fact that $G$ has an element of order 2 is recorded in T0 since that fact does not explicitly mention any element of $G$.
Now let φ be any statement in the language L1 of groups with linear representations. The Robinson Consistency Theorem says that if T0 ∪ T1 ∪ {φ} is consistent then so is T2 ∪ T1 ∪ {φ}, where T1 ⊆ L1 is the theory of linear representations (see note). Stated differently, T2 ∪ T1 ⊦ φ if and only if T0 ∪ T1 ⊦ φ. (Recall that T ⊦ φ iff T ∪ {¬φ} is inconsistent.) The models of T2 are precisely the [elementary extensions](http://en.wikipedia.org/wiki/Elementary_equivalence#Elementary_substructures_and_elementary_extensions) of $G$. Thus the following are equivalent:
* φ is a consequence of some purely group theoretic property of $G$
* φ is true for every linear representation of an elementary extension of $G$
Sometimes we can say more. If φ is an existential statement in L1 (i.e. φ is equivalent to a sentence of the form ∃x,y,z,...φ0(x,y,z,...) where φ0 is quantifier free) then we don't have to check all elementary extensions of $G$. Thus, for such φ, the following are equivalent:
* φ is true in all linear representations of $G$
* φ is a consequence of some purely group theoretic property of $G$
Of course, the Robinson Consistency Theorem is not particular to groups and linear representations of groups; the same reasoning applies in all sorts of contexts.
---
Note: I'm assuming that all languages are first-order (possibly with multiple sorts). There are various ways to formulate linear representations in first-order logic, but none are completely satisfactory. For fixed dimension $n$, one can add a sort $F$ for the field elements together with functions $a\_{i,j}:G\to F$ for the entries of the matrices of the representations. Then T1 consists of all field axioms together with all required identities between the $a\_{i,j}$.
| 4 | https://mathoverflow.net/users/2000 | 19578 | 13,018 |
https://mathoverflow.net/questions/19574 | 3 | Most people define a function, f(n) on N recursively. I think I can calculate f(n) without dealing with f(n-r) for any 0 < r < n. How do I know that my method isn't still going through the same calculations needed to find f(n-1) (or whatever previous terms are required to find f(n) recursively) -- ?
1. If my method takes many fewer calculations than the recursive way of calculating it does that show that I am not relying on f(n-r) for any 0 < r < n? What would "many fewer" have to mean for this to be significant?
2. The number of calculations my method takes still depends on n, just like the recursive way of calculating f(n), does that alone mean that the methods are pretty much the same?
3. If my method takes the same number (or more) calculations than recursive way of calculating f(n) is there any other way of telling if my method is not, in some way, duplicating the recursive way of calculating f(n)?
Examples:
f(n) is recursively defined to be f(n) = f(n-1) + 1 and f(1) = 1
Then f(n) = n. Clearly, f(n) = n is a much faster way to find f(2876) rather than counting up from 1.
f(n) is recursively defined to be f(n) = f(n-1) + f(n-2) This is a linear recurrence and has a closed-form solution. $F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}={{\varphi^n-(-1/\varphi)^{n}} \over {\sqrt 5}}\,$
(from wikipedia)
$S(n,k)=kS(n−1,k)+S(n−1,k−1)$ with $S(n,n)=S(n,1)=1$ (Stirling numbers of the second kind) Almost seems like it's a linear recurrence... but we need to know about k-1. These numbers are defined as "the number of ways to partition a set of n objects into k groups" so, if I have written a few programs to find S(n,k) from that definition I want to know if I "must" find the values in the linear recurrence along the way...
But, I was trying to keep it more general to make it interesting?
One more example:
$C(n,k)=C(n−1,k)+C(n−1,k−1)$ with $C(n,n)=C(n,1)=1$ but $C(n,k) = \frac{n!}{k!(n-k!)}$, most people like the 2nd one better of you want to look at large values of n and k>1...
| https://mathoverflow.net/users/2907 | Can you tell if you have escaped from a recursive definition? | You inquire about comparing your algorithm to a given recursive algorithm, but the more fundamental question would seem to be how good is your algorithm just by itself?
There are numerous ways to measure the efficacy of a computational algorithm using the ideas of [computational complexity](http://en.wikipedia.org/wiki/Computational_complexity_theory). That is, you should measure the complexity of your algorithm by the intensely studied classes of P, NP, PSPACE, EXP, and so on. That is, if you have an algorithm to calculate a function f, the important thing to look at is where does your algorithm sit with respect to these complexity measures: is it polynomial time? exponential time? Is there a nondeterministic polynomial time algorithm? Is there a PSPACE algorithm?
You inquire about comparing your algorithm to a given recursive algorithm. For such a comparison, one should use the measures of complexity theory. If these two algorithms have the same computational complexity, then they are equivalent by these measures, even if your algorithm does not exactly amount to performing the same computation, and the two methods would be equivalent in terms of their computational cost. But if one algorithm finds itself in a lower complexity classification, then it will inevitably be superior in the general case.
| 8 | https://mathoverflow.net/users/1946 | 19580 | 13,020 |
https://mathoverflow.net/questions/19529 | 7 | Has anybody done computations of such a theory? Is there a place I could look up and see what the answers are for low crossing knots?
| https://mathoverflow.net/users/4304 | Computations of the Link homology categorifying the second colored Jones polynomial | Slava Krushkal and I have an alternative approach set inside of Dror Bar-Natan's universal construction. It should agree with results obtained by Webster and Frenkel, Stroppel Sussan. Computations are reasonable in our setting. We hope to place the paper on the arxiv shortly.
| 3 | https://mathoverflow.net/users/4960 | 19582 | 13,022 |
https://mathoverflow.net/questions/19552 | 6 | Let $A$ be a Noetherian commutative ring and $I$ an ideal in $A$. It is pretty much trivial to see that every free $A/I$-Module is obtained from a free $A$-module by tensoring over $A$ with $A/I$: Just choose preimages for every generator and take the free module generated by them. I'm wondering whether the same remains true if I replace ‘free’ by ‘projective and finitely generated’.
One way to prove this would be the following: Let $X=\operatorname{Spec} A$, $Z=\operatorname{Spec}(A/I)$, and $U=Z^c$. If there were something like an exact sequence
$$K\_0(U) \rightarrow K\_0(X) \rightarrow K\_0(Z) \rightarrow 0$$
the claim would follow, right? Does such a sequence exist? If I'm not mistaken, such a sequence exists if $A$ is a Dedekind ring. Just to make it clear, here is the precise question:
Question: Given a Noetherian commutative ring $A$, an ideal $I$ and a projective, finitely generated $A/I$-module $M$, does there exist a projective, finitely generated $A$-module $N$ such that $N/IN=M$?
| https://mathoverflow.net/users/473 | Does every projective A/I-module come from A? | Thomas already [gave](https://mathoverflow.net/a/19564) an example, but let me make a general point I wish he had said: lots of rings with complicated sets of projectives are quotients of rings with simple sets of projectives. For example, any finitely generated projective module over a polynomial ring in any field is free (this is actually a hard theorem; it was proved by Suslin and Quillen in 1976 ([Projective modules over polynomial rings](https://doi.org/10.1007/BF01390008))).
So, if the answer to your question were affirmative, every projective module over a finitely generated commutative $k$-algebra for any field $k$ would be free, which is extremely false. In particular, you would have proven that every vector bundle on an affine variety is trivial, which as Tom points out, is not the case.
| 17 | https://mathoverflow.net/users/66 | 19583 | 13,023 |
https://mathoverflow.net/questions/19571 | 3 | Let $M$ and $N$ be two topological manifolds. Denote by $[M,N]^{\text{diff}}$ and $[M,N]^{\text{cont}}$ the set of homotopy classes of differential and continuous maps respectively. Is it true that $[M,N]^{\text{diff}}=[M,N]^{\text{cont}}$ ? Any reference? Thank you.
**Edit:**
Thanks to the comments below, I should ask whether it is true when $M$ and $N$ are differentiable manifolds.
| https://mathoverflow.net/users/2348 | Homotopy classes of differential maps VS those of continuous maps | There's no way this can be literally true:
$$[M,N]^{diff} = [M,N]^{cont}$$
Most of the continuous functions from $M$ to $N$ are not differentiable. So there's no way the above equality can be an equality of sets. I think what you want to ask is if the inclusion:
$$[M,N]^{diff} \to [M,N]^{cont}$$
a bijection? This is answered affimatively in Hirsch's "Differential Topology" textbook. It boils down to a smoothing argument, that every continuous function can be uniformly approximated by a $C^\infty$-smooth function and the smoothing is unique up to a small homotopy. The argument goes further, to state the the space of continuous functions has the same homotopy-type as the space of $C^\infty$ functions. The smoothing argument can be done with bump functions and partitions of unity, and also via a standard convolution with a bump function argument ("smoothing operators").
| 10 | https://mathoverflow.net/users/1465 | 19592 | 13,027 |
https://mathoverflow.net/questions/19589 | 12 | Background
----------
Inside the Temperley-Lieb algebra $TL\_n$ (with loop value $\delta=-[2]$ and standard generators $e\_1,\ldots,e\_{n-1}$), the Jones-Wenzl idempotent is the unique non-zero element $f^{(n)}$ satisfying
$$ f^{(n)}f^{(n)} = f^{(n)} \quad \textrm{and} \quad e\_i\;f^{(n)} = 0 = f^{(n)}e\_i \quad \textrm{for each } i.$$
Consider the Iwahori-Hecke algebra $\mathcal{H}\_n$, $n\ge3$, normalized so that $(T\_i-q)(T\_i+q^{-1})=0$, where $q$ is generic. Let $\mathcal{I}$ be the two-sided cellular ideal generated by canonical basis element
$$C\_{121} = T\_1T\_2T\_1-qT\_1T\_2-qT\_2T\_1+q^2T\_1+q^2T\_2-q^3.$$
The assignment $\mathcal{H}\_n \rightarrow TL\_n$ given by $T\_i \mapsto e\_i + q$ is a surjective $\mathbb{C}(q)$-algebra homomorphism with kernel $\mathcal{I}$.
We can lift the generators $e\_i$ in $TL\_n$ to the Kazhdan-Lusztig elements $C\_i=T\_i-q \in \mathcal{H}\_n$. In fact, we have $C\_{121} = C\_1C\_2C\_1 - C\_1$, hence the relation down below. Rescaling a bit, $E=-\frac{1}{[3]!}C\_{121}$ is an idempotent, corresponding to the partition $(1,1,1)$. Actually, all of the primitive idempotents in $\mathcal{H}\_n$ that correspond to Young diagrams with more than two rows live in the ideal $\mathcal{I}$.
Now, any preimage of $f^{(n)}$ in the Hecke algebra (call it $F^{(n)}$) satisfies
$$F^{(n)}F^{(n)} \equiv F^{(n)} \quad \textrm{and} \quad C\_iF^{(n)} \equiv 0 \equiv F^{(n)}C\_i \quad (\operatorname{mod} \mathcal{I})$$
Question
--------
>
> Can we choose $F^{(n)}$ to be an idempotent in $\mathcal{H}\_n$?
>
>
>
When $n=2$, the map is an isomorphism and we have no choice.
$$F^{(2)} = \frac{1}{[2]}(T\_1+q^{-1}),$$
which projects onto the $q$-eigenspace for $T\_1$. In other words, it is the idempotent corresponding to the partition $(2)$.
| https://mathoverflow.net/users/813 | Do Jones-Wenzl idempotents lift to anything interesting in the Hecke algebra? | Yes. No problem. This is the $q$-analogue of the symmetriser. In terms of $T\_i$ we have
$$ \frac{1}{[n]!}\sum\_{\pi\in S\_n} q^{\ell(\pi)} T\_\pi$$
where for $T\_\pi$ we take a reduced word for $\pi$ and $\ell(\pi)$ is the length of a reduced word.
There is another presentation for the Hecke algebra which I am used to writing as
generators $u\_i$ and defining relations
$$u\_i^2=[2]u\_i$$
$$u\_iu\_j=u\_ju\_i\qquad\text{for $|i-j|>1$}$$
$$u\_iu\_{i+1}u\_i-u\_i=u\_{i+1}u\_iu\_{i+1}-u\_{i+1}$$
Strictly speaking this is the subring of the Hecke ring which is invariant under the bar involution. This algebra is defined over $\mathbb{Z}[\delta]$. The Hecke algebra is the algebra over $\mathrm{Z}[q,q^{-1}]$ obtained by the specialisation $\delta\mapsto q+q^{-1}$.
Then we have $u\_i=-C\_i$.
The Temperley-Lieb algebra is the quotient by $u\_iu\_{i\pm 1}u\_i=u\_i$.
This has an involution given by $u\_i\leftrightarrow \delta-u\_i$.
To define the idempotents we need to divide by $[n]!$.
Define $R\_i(k)=1-\frac{[k]}{[k+1]}u\_i$. Then these satisfy the Yang-Baxter equation
$$R\_i(r)R\_{i+1}(r+s)R\_i(s)=R\_{i+1}(s)R\_i(r+s)R\_{i+1}(r)$$
Using this you can write everything (well a lot, at least) explicitly.
For example the idempotents can be written
$$1\qquad R\_1(1)\qquad R\_1(1)R\_2(2)R\_1(1)\qquad R\_1(1)R\_2(2)R\_1(1)R\_3(3)R\_2(2)R\_1(1)$$
and so on. You get another sequence of idempotents by applying the involution. Under the involution we have $R\_i(k)\leftrightarrow R\_i(-k)$. The three string idempotent is
$$\left(\frac{u\_1}{\delta}\right)\left(1-\delta u\_2\right)\left(\frac{u\_1}{\delta}\right)$$
which gives the relation for the Temperley-Lieb algebra.
| 8 | https://mathoverflow.net/users/3992 | 19594 | 13,029 |
https://mathoverflow.net/questions/7089 | 8 | There's an old problem in 4-manifold theory that, as far as I know, doesn't have a name associated with it and really deserves a name.
Let $M$ be a smooth 4-manifold with boundary. Let $S$ be a smoothly embedded 2-dimensional sphere in $\partial M$. Assume $S$ does not bound a ball in $\partial M$, but $S$ is null-homotopic in $M$. Does $S$ bound a smooth 3-ball in $M$? Perhaps you need to replace $S$ by another non-trivial $S'$ in $\partial M$ before you can find a 3-ball in $M$ bounding it?
You could think of this as the co-dimension one analogue to Dehn's lemma for 4-manifolds. Usually when people talk about a Dehn lemma for 4-manifolds they're interested in the co-dimension 2 analogue.
Does this problem / conjecture have a name? If not, do you have a good name for it? Do you know of anywhere in the literature where this issue is investigated?
Off the top of my head the only vaguely related things I know about in the literature is a 1975 paper of Swarup's.
| https://mathoverflow.net/users/1465 | A problem/conjecture related to 4-manifolds that deserves a name. What name does it deserve? | I think that the conjecture is wrong. The following leads to counterexamples in the topological category and probably also smoothly: Take a closed oriented 4-manifold N with infinite cyclic fundamental group and remove an open neighborhood of a generating circle. Then you get a 4-manifold M with boundary $S^1 \times S^2$ where only the generators can be represented by embedded 2-spheres. If S denotes $pt \times S^2$ then the following hold:
1. S is null homotopic in M via Poincare duality and the exact sequence of the pair.
2. S bounds an embedded 3-ball in M if and only if N can be decomposed into $S^1 \times S^3$, connected sum with a simply connected manifold.
Hambleton and I found topological 4-manifolds N for which the intersection form is not extended from the integers, in particular 2 doesn't hold. With Friedl and Melvin we showed later that our examples don't have a smooth structure. But now I remember a discussion with Fintushel and Stern who mentioned that they constructed a smooth counterexample to 2 (and hence to the conjecture).
| 10 | https://mathoverflow.net/users/4625 | 19596 | 13,031 |
https://mathoverflow.net/questions/19584 | 70 | One of the more misleadingly difficult theorems in mathematics is that all finitely generated projective modules over a polynomial ring are free. It involves some of the most basic notions in commutative algebra, and really sounds as though it should be easy (the graded case, for example, is easy), but it's not. The question at least goes as far back as Serre's FAC, but it wasn't proved until 1976, by Quillen in [Projective modules over polynomial rings](https://doi.org/10.1007/BF01390008) **EDIT:** and also independently by Suslin.
I decided that this is the sort of fact that I should know a rough outline of how to prove, but the paper was not very helpful. Usually when someone kills off a famous conjecture in 5 pages, it's because they've developed some fantastic new piece of machinery people didn't have before. And, indeed, Quillen is famous for inventing some fancy and wonderful machinery, and the paper is only 5 pages long, but as far as I can tell, none of that fancy machinery actually appears in the proof.
>
> So, what was it that Quillen saw, that Serre missed?
>
>
>
| https://mathoverflow.net/users/66 | What is the insight of Quillen's proof that all projective modules over a polynomial ring are free? | Here is a summary of what I learned from a nice expository account by Eisenbud (written in French), can be found as number 27 [here](https://web.archive.org/web/20120710030443/http://www.msri.org/%7Ede/papers/index.html).1
First, one studies a more general problem: Let $A$ be a Noetherian ring, $M$ a finite presented projective $A[T]$-module. When is $M$ *extended* from $A$, meaning there is $A$-module $N$ such that $M = A[T]\otimes\_AN$?
The proof can be broken down to 2 punches:
**Theorem 1** (Horrocks) If $A$ is local and there is a monic $f \in A[T]$ such that $M\_f$ is free over $A\_f$, then $M$ is $A$-free
(this statement is much more elementary than what was stated in Quillen's paper).
**Theorem 2** (Quillen) If for each maximal ideal $m \subset A$, $M\_m$ is extended from $A\_m[T]$, then $M$ is extended from $A$
(on $A$, locally extended implies globally extended).
So the proof of Serre's conjecture goes as follows: Let $A=k[x\_1,\dotsc,x\_{n-1}]$, $T=x\_n$, $M$ projective over $A[T]$. Induction (invert all monic polynomials in $k[T]$ to reduce the dimension) + further localizing at maximal ideals of $A$ + Theorem 1 show that $M$ is locally extended. Theorem 2 shows that $M$ is actually extended from $A$, so by induction must be free.
Eisenbud's note also provides a very elementary proof of Horrocks's Theorem, basically using linear algebra, due to Swan and Lindel (Horrocks's original proof was quite a bit more geometric).
As Lieven [wrote](https://mathoverflow.net/a/19598), the key contribution by Quillen was Theorem 2: patching. Actually the proof is fairly natural, there is only one candidate for $N$, namely $N=M/TM$, so let $M'=A[T]\otimes\_AN$ and build an isomorphism $M \to M'$ from the known isomorphism locally.
It is hard to answer your question: what did Serre miss (-:? I don't know what he tried. Anyone knows?
1*Eisenbud, David*, Solution du problème de Serre par Quillen–Suslin, Semin. d’Algebre, Paul Dubreil, Paris 1975-76 (29eme Annee), Lect. Notes Math. 586, 9-19 (1977). [DOI: 10.1007/BFb008711](https://doi.org/10.1007/BFb008711), [PDF on the MSRI website](https://web.archive.org/web/20120610175858/http://www.msri.org/%7Ede/papers/pdfs/1977-006.pdf), [ZBL0352.13005](https://zbmath.org/?q=an:0352.13005), [MR568878](https://mathscinet.ams.org/mathscinet-getitem?mr=0568878).
| 60 | https://mathoverflow.net/users/2083 | 19603 | 13,037 |
https://mathoverflow.net/questions/19599 | 7 | Regarding the sphere as complex projective line (take $(0,0,1)$ as the infinite point), the Gauss map of a smooth surface in the 3 dimensional space pulls a complex line bundle back on the surface.
My question is, what the bundle is?
(In the trivial case, if the surface is sphere itself, the bundle is just the tautological line bundle.)
Does the chern class (Of course the first one) of this bundle depend on the embedding of the surface?
(The Jacobian determinant of Gauss map is just the Gauss curvature, hence is intrinsic. Also its degree is the Euler $\chi$, so I ask for more...)
If yes, how much does the chern class/bundle reflect the geometry of embedding?
There may be something to make the question meaningless, such as there is no cannonical way to identify a sephere with the projective line... But as a beginner in learning geometry, I am still curious to it...
*edit* Thank Sergei for your answer, thank you
| https://mathoverflow.net/users/2913 | About the Gauss map of a surface in euclidean 3 space | I assume that your surface is closed. Suppose you have a fixed vector bundle $\xi$ over $S^2$ (no matter which one). You have an oriented surface $M$ embedded in $\mathbb R^3$, which defines the Gauss map $\nu:M\to S^2$, which defines the vector bundle $\nu^\*\xi$ on $M$. You want to know whether $\nu^\*\xi$ depends on the embedding.
No it does not. Indeed, $\deg\nu=\chi(M)/2$ regardless of the embedding. Two maps $f\_1,f\_2:M\to S^2$ having the same degree are homotopic. And homotopic maps induce the same bundle.
Concerning the Chern class, we have $c\_1(\nu^\*\xi)=\nu^\*(c\_1(\xi))$ by definition. So, if you identify the top cohomology with integers, then $c\_1(\nu^\*\xi)$ is (the same number as) $\deg(\nu) c\_1(\xi)=\frac12\chi(M)c\_1(\xi)$.
| 10 | https://mathoverflow.net/users/4354 | 19606 | 13,039 |
https://mathoverflow.net/questions/19607 | 12 | What I said is Lusztig's conjecture about representation of quantum group at root of unity and representation of Lie algebra at positive characters.
It seems that Andersen-Jantzen-Soergel ever wrote a book on this conjecture.
Is it solved? Any recent development? I am looking for reference talking about it.
Thank you
| https://mathoverflow.net/users/1851 | Is Lusztig's conjecture solved? | The result of that book is that the conjecture is true for sufficiently large, but unspecified characteristic. (First fix a Dynkin type.)
More recently Peter Fiebig has given actual bounds. See
An upper bound on the exceptional characteristics for Lusztig's character formula
by Peter Fiebig arXiv:0811.1674v2 at <http://arxiv.org/pdf/0811.1674v2>
| 14 | https://mathoverflow.net/users/4794 | 19609 | 13,041 |
https://mathoverflow.net/questions/19635 | 7 | Given a smooth projective complex variety $X$, instead of using Mumford's GIT to construct the moduli of rank $n$ topologically trivial vector bundles, we can also take the gauge theory approach.
To classify all topologically trivial vector bundles is same as to classify all possible complex structures on the topologically trivial vector bundle $X\times \mathbb{C}^n$ module the ambiguity from the choice of basis. And a complex structure is determined by $\bar\partial=\bar\partial\_0+\eta$, where $\bar\partial\_0$ corresponds to the holomorphic structure as a trivial vector bundle, and $\eta$ is an anti-holomorphic $End(\mathbb{C}^n)$ valued one form, such that $d\eta+\eta\wedge\eta=0$. Let $M$ denote the set of all such $\eta$, and $G$ denote the gauge group with fiber $GL(\mathbb{C}^n)$, then $G$ acts on $M$. Now, set-theoretically the orbits will correspond to the set of isomorphism classes of rank $n$ topologically trivial holomorphic vector bundles, but the quotient topology seems rather bad (non-Hausdorff).
My question is using this approach
(1)Whether stability condition is reflected in this construction?
(2)Can we obtain the same moduli space as using Mumford's GIT?
(3)I guess this is a natural approach. So is there any article or book about this?
The assumption of topologically trivial should not be necessary, just to simplify some notation. We only need to fix the underlying topological vector bundle.
| https://mathoverflow.net/users/4975 | Gauge theory construction of moduli of vector bundles | In the case of a nonsingular algebraic curve, I guess this is the point of Atiyah and Bott's Yang-Mills on Riemann surfaces, or the work of Narasimhan and Seshadri. Choose a Hermitian pairing on your complex bundle and then search for a unitary connection with central curvature. There is a really great book by Kobayashi "The Differential Geometry of Complex Vector Bundles" that explores this approach in more generality.
Once you have reduced to unitary connections with central curvature you can understand them as representations of a central extension of the fundamental group of the manifold, which then puts you in a finite dimensional situation with a compact group acting. The quotient corresponds to the geometric invariant theory quotient.
| 8 | https://mathoverflow.net/users/4304 | 19637 | 13,053 |
https://mathoverflow.net/questions/19652 | 9 | I am currently writing up some notes on the max-plus algebra $\mathbb{R}\_{\max}$ (for those that have never seen the term "max-plus algebra", it is just $\mathbb{R}$ with addition replaced by $\max$ and multiplication replaced by addition. For some reason, authors whose main interest is control-theoretic applications never seem to use the term "tropical", and I have been reading from such authors). There is a nice result which says the following:
$\textbf{Theorem.}$ Let $G$ be a directed graph on $n$ vertices such that each arc $(i,j)$ in $G$ has a real weight $w(i,j)$. Define the $n \times n$ matrix $A$ by $(A)\_{ij} = w(i,j)$ if $(i,j)$ is an arc, and $(A)\_{ij} = -\infty$ otherwise. Then for each $k > 0$, the maximum weight of a path of length $k$ from vertex $i$ to vertex $j$ is given by $(A^{\otimes k})\_{ij}$ (here, $A^{\otimes k}$ is just the $k$th power of $A$, computed using the $\mathbb{R}\_{\max}$ operations).
This result is certainly analagous to the standard result that the $ij$-entry of the $k$th power of the adjacency matrix gives the number of walks of length $k$ from vertex $i$ to vertex $j$. When writing up my notes I found myself claiming that the above theorem provides some evidence that $\mathbb{R}\_{\max}$ is in fact a natural setting in which to study weighted digraphs, since there is no natural definition of an ``adjacency matrix'' of a weighted digraph (in the usual setting of $\mathbb{R}^{n \times n}$) that gives useful information about the weights. This seemed like too strong of a claim, especially since I am no expert in networks or combinatorial optimization. This leads to the question:
$\textbf{Question.}$ Is there a standard matrix (in $\mathbb{R}^{n \times n}$) associated with a weighted digraph that is analogous to the adjacency matrix and captures in a useful way the weights of the arcs?
$\textbf{Clarification:}$ By "analogous to the adjacency matrix" I mean a matrix that is defined simply in terms of the graph (vertices, arcs, and weights). I imagine there are all sorts of matrices associated to weighted digraphs so that computers can be used to analyze networks. But I am not interested in, say, a matrix that requires a complicated algorithm to compute its entries.
| https://mathoverflow.net/users/4977 | Is there an "adjacency matrix" for weighted directed graphs that captures the weights? | It looks like in your definition the weight of a path is the **sum** of the weights of its edges. In many combinatorial applications a natural definition of the weight of a path is the **product** of the weights of the edges, and there one uses precisely the weighted adjacency matrix $A\_{ij} = w(i, j)$ (as an element of the usual $\mathbb{R}$). This is the definition relevant to, for example, the theory of Markov chains, where $w(i, j)$ is a transition probability.
One way to get information about sums of weights is to use $B\_{ij} = e^{w(i, j)}$, but what you'll get in the end is a sum of exponentials of weights instead of (direct) information about the maximum or minimum weight. I think one can instead consider $B\_{ij}(t) = e^{t w(i, j)}$ and in the "low-temperature" limit as $t \to \infty$ this approaches the tropical result; the largest term will dominate. (I think physicists call these things partition functions.)
| 10 | https://mathoverflow.net/users/290 | 19653 | 13,063 |
https://mathoverflow.net/questions/19649 | 18 | Miranda's book on Riemann surfaces ignores the analytical details of proving that compact Riemann surfaces admit nonconstant meromorphic functions, preferring instead to work out the algebraic consequences of (a stronger version of) that assumption. Shafarevich's book on algebraic geometry has this to say:
> A harmonic function on a Riemann surface can be conceived as a description of a stationary state of some physical system: a distribution of temperatures, for instance, in case the Riemann surface is a homogeneous heat conductor. Klein (following Riemann) had a very concrete picture in his mind:
>
>
>
> *"This is easily done by covering the Riemann surface with tin foil... Suppose the poles of a galvanic battery of a given voltage are placed at the points $A\_1$ and $A\_2$. A current arises whose potential $u$ is single-valued, continuous, and satisfies the equation $\Delta u = 0$ across the entire surface, except for the points $A\_1$ and $A\_2$, which are discontinuity points of the function."*
Does anyone know of a good reference on Riemann surfaces where a complete proof along these physical lines (Shafarevich mentions the theory of elliptic PDEs) is written down? How hard is it to make this appealing physical picture rigorous? (The proof given in Weyl seems too computational and a little old-fashioned. Presumably there are now slick conceptual approaches.)
| https://mathoverflow.net/users/290 | "Physical" construction of nonconstant meromorphic functions on compact Riemann surfaces? | For this and most other things about Riemann surfaces, I recommend Donaldson's [Notes on Riemann surfaces](http://www2.imperial.ac.uk/~skdona/RSPREF.PDF), which are based on a graduate course I was once lucky enough to see, and which may eventually make it into book format.
In his account, the "main theorem for compact Riemann surfaces" says that one can solve $\Delta f = \rho$ for any 2-form $\rho$ with integral zero. He describes this as the equation for a steady-state temperature distribution. A full proof is given, but I wouldn't describe it as slick: this is still a substantial result in analysis.
| 14 | https://mathoverflow.net/users/2356 | 19658 | 13,067 |
https://mathoverflow.net/questions/19661 | 4 | Let P(X,Y,Z) be a homogeneous polynomial in ℂ[X,Y,Z] whose locus M in ℂℙ2 is a nonsingular curve of genus ≥ 2.
Define M to be *maximally symmetric* if the following is **not** true:
---
There exists a continuous family { Pt | t ∊ [0,1] } of homogeneous polynomials in ℂ[X,Y,Z] such that 1), 2), and 3) hold:
1) P0 = P.
2) The locus Mt in ℂℙ2 of each Pt is nonsingular.
3) There is a group G such that the ambient isometry groups
Gt := IsomA(Mt) are all isomorphic to G for 0 ≤ t < 1, but G1 contains G as a proper subgroup.
Here the "ambient isometry group" IsomA(Mt) of a projective curve M in ℂℙ2 means the subgroup of Isom(ℂℙ2) = PSU(3) that carries M to itself.
---
Question: I'd like pointers to the literature regarding what may be known about a classification of such "maximally symmetric" projective curves up to ambient isometry, their defining polynomials, and *especially* their ambient isometry groups.
---
| https://mathoverflow.net/users/5484 | Maximally symmetric smooth projective varieties in CP^2 | By the same averaging trick that shows that finite-dimensional complex representations of a finite group are unitary with respect to some inner product, your question is equivalent to the one obtained by replacing ambient isotropy groups with linear automorphism groups in the sense of algebraic geometry. Here *linear* means induced by a linear automorphism of $\mathbf{P}^2$. Actually, for a smooth plane curve of degree $d>3$, all automorphisms are linear, by
H. C. Chang, On plane algebraic curves, *Chinese J. Math.* **6** (1978), 185-189.
Fix $d>3$. Let $\mathcal{H}\_d$ be the moduli space of smooth degree-$d$ curves in $\mathbf{P}^2$, so $\mathcal{H}\_d$ is an open subscheme of some projective space. Then there is a stratification of $\mathcal{H}\_d$ into finitely many locally closed subschemes such that the automorphism group is constant on each piece. (This could also be stated in terms of the automorphism group scheme of the universal curve over $\mathcal{H}\_d$.) In these terms, you are asking for the $0$-dimensional strata, or equivalently the smooth plane curves such that in a punctured neighborhood of the corresponding point of $\mathcal{H}\_d$ the automorphism group is strictly smaller.
The analogous question with $\mathcal{H}\_d$ replaced by the full moduli space $\mathcal{M}\_g$ of curves of genus $g>1$ has been much studied. The direct analogue of your maximally symmetric curves in this setting are the curves said to have "many automorphisms" in Section 3 of the article
Jürgen Wolfart, The obvious part of Belyi's theorem and curves with many automorphisms, pp. 97-112 in: *Geometric Galois actions 1*, edited by L. Schneps and P. Lochak, LMS Lecture Notes Series **342**, Cambridge Univ. Press, 1997.
Wolfart's article contains many references to related work, and mentions some nice theorems. For instance: a smooth projective curve of genus greater than $1$ over $\mathbf{C}$ has many automorphisms if and only if it is a Galois cover of $\mathbf{P}^1$ ramified only above $0,1,\infty$.
| 5 | https://mathoverflow.net/users/2757 | 19672 | 13,079 |
https://mathoverflow.net/questions/19633 | 10 | For each $d$, I have a matrix $M$ with values
$$
M\_{ij} = \begin{cases}
\frac{4ij}{d} - \binom{2d}{d} & i \neq j & \\\\
\frac{4i^2}{d} - \binom{2d}{d} -
\frac{\binom{2d}{d}}{\binom{d}{i}^{2}} & i = j
\end{cases}
$$
I want to show that, for every $d=2,3,\ldots$, the matrix is negative-definite.
**An elegant answer has been provided by fedja, without needing to look at the determinants**
My approach is to compute the determinant of the upper left-square matrix of size $k$, for each $k=1,2,\ldots,d$. The value for this is
$$
D\_k = (-1)^k\left[\frac{\binom{2d}{d}^k}{\prod\_{i=1}^{k}\binom{d}{i}^2}\right]
\left[\sum\_{j=1}^{k}\left(\binom{d}{j}^2 - \frac{4j^2\binom{d}{j}^2}{d\binom{2d}{d}}\right)
- \frac{4}{d\binom{2d}{d}}
\sum\_{1 \leq i < j \leq k}(j-i)^2\binom{d}{i}^2\binom{d}{j}^2\right]
$$
Hence, $D\_k$ has sign $(-1)^k$ if this expression is positive
$$
\sum\_{j=1}^{k}\left(\binom{d}{j}^2 - \frac{4j^2\binom{d}{j}^2}{d\binom{2d}{d}}\right)
- \frac{4}{d\binom{2d}{d}}
\sum\_{1 \leq i < j \leq k}(j-i)^2\binom{d}{i}^2\binom{d}{j}^2
$$
I can re-write this as
$$
\sum\_{j=0}^{k}\binom{d}{j}^2 - \frac{2}{d\binom{2d}{d}}
\sum\_{0 \leq i,j \leq k}(j-i)^2\binom{d}{i}^2\binom{d}{j}^2\,,
$$
and, then, after observing that $\sum\_{i=0}^{d}i\binom{d}{i}^2 = \frac{d}{2}\binom{2d}{d}$,
I can deduce that this will certainly be positive when the following sum is positive
$$\sum\_{0 \leq i, j \leq k}(i-(j-i)^2){\binom{d}{i}}^2{\binom{d}{j}}^2$$
Any advice/techniques would be appreciated.
| https://mathoverflow.net/users/4974 | Showing a matrix is negative definite [formerly Showing a sum is always positive] | It's easier to prove the result about the matrix without resorting to determinants. What we need is the inequality
$$
\left(\sum\_{i=0}^d 2ix\_i\right)^2\le d{2d\choose d}\left(\sum\_{i=0}^d x\_i\right)^2
+d\sum\_{i=0}^d \frac{{2d\choose d}}{{d\choose i}^2}x\_i^2
$$
Now recall that $\sum\_{i=0}^d (2i-d)^2{d\choose i}^2=\frac{d^2}{2d-1}{2d\choose d}$ (elementary computation with generating functions; should have some combinatorial proof as well) and that $(A+B)^2\le \frac{2d-1}dA^2+\frac{2d-1}{d-1}B^2$. Putting
$$
A=\sum\_{i=0}^d(2i-d)x\_i\,,\qquad B=d\sum\_{i=0}^d x\_i
$$
we see that we can use Cauchy-Schwarz to get
$$
A^2\le \frac{d^2}{2d-1}\sum\_{i=0}^d \frac{{2d\choose d}}{{d\choose i}^2}x\_i^2
$$
so we just need to check that $d^2\frac{2d-1}{d-1}\le d{2d\choose d}$ for $d\ge 2$, which is rather trivial.
| 18 | https://mathoverflow.net/users/1131 | 19677 | 13,083 |
https://mathoverflow.net/questions/19393 | 3 | This question is rather vague. Are there any natural situations which involve Laurent polynomials of the form
$$\sum q^{a\_i}\in\mathbb{Z}[q,q^{-1}]$$
where the $a\_i$'s are either Euler characteristics of some spaces (possibly all subspaces of one fixed space), or more generally, indices of some elliptic operators? I've stumbled across such a beast, but am unsure how to interpret it. I was thinking at first that it was an element of $K\_{S^1} (pt)$ or something, but in that case the exponents are telling us about which $S^1$ representations show up in the appropriate bundles, not the indices of the operators, right? Is there some relation with the index? (Please tell me if I'm talking nonsense! I don't really know this K-theory stuff).
Maybe the answer I'm looking for doesn't involve K-theory, anyway. Does anyone have any ideas? I'd love to hear about any and everything!
| https://mathoverflow.net/users/492 | Euler characteristics and operator indices as exponents for Laurent polynomials | 1. Knot polynomials like the Jones polynomial
2. Perturbative expansions of Feynman Integrals
3. Heat kernel asymptotics, and other universal polynomials in characteristic classes.
4. Generating functions associated to combinatorial problems.
5. Poincare Polynomials of Topological Spaces.
6. Hilbert Polynomials.
7. Certain families of orthogonal polynomials, generally associated to representation
theory.
8. The A-polynomial
I am sure there are more.
| 6 | https://mathoverflow.net/users/4304 | 19681 | 13,085 |
https://mathoverflow.net/questions/19692 | 15 | I once heard from a graduate student that the ABC conjecture implies the Riemann hypothesis. I can't find a reference for this, but given the department the student is from I tend to believe he might know about these things.
I looked through [Goldfeld's paper](http://www.math.columbia.edu/~goldfeld/ABC-Conjecture.pdf) which shows that certain bounds on Shafarevich-Tate groups plus the Generalized Riemann Hypothesis for L-functions associated to certain modular forms imply a form of the ABC conjecture. The article mentions nothing about the opposite implication.
>
> What is the known realationship between ABC, BSD, and GRH? Are any two known to imply the third?
>
>
>
| https://mathoverflow.net/users/4872 | Is the ABC conjecture known to imply the Riemann hypothesis? | I am pretty sure that the answer to the question is no: no two of those big conjectures are known to imply the third. But I feel somewhat sheepish giving this as an answer: what evidence can I bring forth to support this, and if nothing, why should you believe me?
The only thing I can think of is that in the function field case, ABC and GRH are fully established, but only parts of BSD are known.
(Maybe I should also admit that I didn't know anything about the connection between ABC and bounds on Shafarevich-Tate groups of elliptic curves in terms of the conductor until I glanced just now at the paper of Goldfeld the OP linked to. The fact that you can build examples of large Sha from triples of integers with large ABC exponent is amazing to me.)
**Addendum**: I feel especially confident that ABC and GRH do not imply BSD, at least not the part of BSD that asserts finiteness of Shafarevich-Tate groups. The first two conjectures are essentially analytic in nature, whereas the finiteness of Sha is deeply arithmetic. It seems extremely unlikely.
Moreover, ABC is really hard, in the sense that for all of the results of the form "X implies ABC" that I've ever seen, X includes a statement which is ABC-like in the sense that it gives a uniform bound on one arithmetic quantity in terms of another. For example, ABC is known to be of a similar flavor to the Szpiro Conjecture (and implies it), but so far as I know it is only known to be implied by a more-explicitly-ABC-like [Modified Szpiro Conjecture](http://modular.fas.harvard.edu/mcs/archive/Fall2001/notes/12-10-01/12-10-01/12-10-01.html). Admittedly bounding Sha in terms of the conductor, as in Goldfeld's work, is only vaguely ABC-like, but to an arithmetic geometer like me these bounds still feel very "analytic"; I can't see any connection at all between this and BSD. So I doubt that GRH (let me say ERH, so that I more or less know what I'm talking about -- i.e., Dedekind zeta functions) plus BSD is known to imply ABC.
| 17 | https://mathoverflow.net/users/1149 | 19694 | 13,093 |
https://mathoverflow.net/questions/19697 | 1 | If $P, Q$ are prime, and $P > Q$, then let $K$ be the set of all numbers $(P-Q)$. Is there a way to determine $\frac{|K|}{|\mathbb{Z}^+|}$? Is this even a converging value? What kind of numbers are in set $K$?
So far:
if $P-Q = d$ is odd, then $P, Q$ are of different parity and $Q = 2$, so $d = P-2$.
But, if $d$ is even, then $P, Q$ are both odd, which means finding primes that are $d$ away from each other, where $d$ is an even number. For how many values of $d$ is this possible?
Though this seems similar to the twin primes conjecture, note that here we only ask if a value of $d$ is possible, not how many such pairs there are.
Sorry if this is in fact a trivial problem, I'm not very experienced in mathematics.
| https://mathoverflow.net/users/4990 | Possible values for differences of primes | Since there are infinitely many primes, the set $K$ is certainly infinite, so in the expression $\frac{|K|}{|\mathbb{Z}^+|}$, you are attempting to divide two infinite cardinalities. This is not a meaningfully defined operation.
Not so much is known about the set $K$ unconditionally. However, an old conjecture of [Alphonse de Polignac](http://en.wikipedia.org/wiki/Alphonse_de_Polignac) states that for every positive integer $k$, there are infinitely many pairs of primes $p$ and $q$ such that $p-q =2k$. This is significantly stronger than saying that every even positive integer lies in $K$. de Polignac's conjecture is in turn a special case of a much broader conjecture which is, however, still widely believed to be true and often used by 21st century mathematicians to prove conditional results: [Schinzel's Hypothesis H](http://en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H).
On the other hand, $K$ can only contain an odd positive integer $n$ if $n+2$ is prime. The set of such numbers has density zero. So, assuming Schinzel / de Polignac, the set $K$ has asymptotic density $\frac{1}{2}$, i.e.,
$\lim\_{N \rightarrow \infty} \frac{|K \cap [1,N]|}{N} = \frac{1}{2}$.
Perhaps an actual analytic number theorist can tell us how close we are to knowing this density result unconditionally.
| 2 | https://mathoverflow.net/users/1149 | 19699 | 13,097 |
https://mathoverflow.net/questions/18884 | 18 | Not long back I asked [a question](https://mathoverflow.net/questions/17599/existence-of-multi-variable-p-adic-l-functions) about the existence of p-adic L-functions for number fields that are not totally real; and I was told that when the number field concerned has a nontrivial totally real or CM subfield, then there is a construction due to various people including Coates-Sinnott and Katz.
But my favourite number field at the moment is K = $\mathbb{Q}(\sqrt[3]{2})$, and sadly K contains no totally real or CM subfield, so for trivial reasons $L(n, \chi) = 0$ for every Groessencharacter $\chi$ of $K$ and every $n \le 0$. So in this case the above constructions just give zero. When I learnt this, I thought "that can't be the whole story, what about higher derivatives at 0"? Asking around, I was told about Stark's conjectures, which apparently predict that the leading term at $s = 0$ of the L-function of any GC of K should be the product of an explicit transcendental regulator and an algebraic number (which, if I've understood this right, should lie in the field $\mathbb{Q}$(values of $\chi$).)
My question is this: assuming Stark's conjecture, can we construct a distribution on the Galois group of the maximal unramified-outside-p abelian extension of K whose evaluation at any locally constant character of this group gives the algebraic part of the leading term at 0 of the L-series of the corresponding Groessencharacter?
| https://mathoverflow.net/users/2481 | Stark's conjecture and p-adic L-functions | Conjecturally, the answer is yes, but the amount of work required is not trivial at all. The general set-up is roughly as follows: the special values of $L$-functions (in your case, for Tate motives) are predicted by the Tamagawa Number Conjecture, and by the Equivariant Tamagawa Number Conjecture (ETNC) when one wishes to incorporate the action of some Galois group (as you do). The ETNC links the leading term at 0 of the $L$-function to the determinants of some cohomological complexes. However, in order to do this coherently for any locally constant character (again, as you want to), one needs rather strong hypotheses on the complexes involved: namely, they need to be semisimple at $\rho$ if one wishes to interpolate at $\rho$. Here, the bad news start: showing that a complex is semisimple at $\rho$ amounts to Leopoldt's conjecture for Tate motives so is in general very hard. But things should be fine in your favourite case. In this way, you can construct (leading terms of) $p$-adic $L$-functions for Tate motives.
If you are ready to admit all conjectures, all of this has been known for quite some time, see for instance B.Perrin-Riou Fonctions $L$ $p$-adiques des représentations $p$-adiques section 4.2 Conjecture CP(M). For a more recent and more flexible formulation, I recommend D.Burns and O.Venjakob On the leading terms of Zeta Isomorphism... section 3.2.2 and 3.2.3.
| 7 | https://mathoverflow.net/users/2284 | 19715 | 13,110 |
https://mathoverflow.net/questions/19716 | 1 | Hello,
where can I read about some basic properties of twisted D-Modules? I would like to know, a reference, that describes how to glue these modules together/pull them back/push them forward.
| https://mathoverflow.net/users/2837 | TDO basic facts reference request | [Twisted Differential operators](http://www.math.harvard.edu/~gaitsgde/grad_2009/BB%2520-%2520Jantzen.pdf)
| 2 | https://mathoverflow.net/users/1851 | 19718 | 13,111 |
https://mathoverflow.net/questions/19705 | 6 | So from the $\overline{\partial}$-Poincare lemma, there is a short exact sequence of sheaves on $X = \mathbb{P}^1$
$$0 \to \Omega \to A^{1,0} \to Z^{1,1} \to 0$$
where $\Omega$ is the sheaf of holomophic 1-forms, $A^{1,0}$ is the sheaf of (1,0)-forms, $Z^{1,1}$ is sheaf of closed (1,1)-forms and the surjection is apply $\overline{\partial}$. The higher cohomology of $A^{1,0}$ vanishes which gives an isomorphism:
$$H^1(X,\Omega) \cong \frac{\Gamma(Z^{1,1})}{\overline{\partial}\Gamma(A^{1,0})} =: H\_{\overline{\partial}}^{1,1}(X)$$
Both groups are isomorphic to $\mathbb{C}$. The first group can be described explicitly via Cech cohomology and the standard cover of $X$; in suitable coordinates a generator is $\dfrac{dz}{z}$.
What is the image of $\dfrac{dz}{z}$ in $H\_{\overline{\partial}}^{1,1}(X)$? More specifically, can you write down a global closed (1,1) form that represents the class of $\dfrac{dz}{z}$?
I know that being able to avoid descriptions such as these is one of the great virtues of Cech Cohomology, but I guess some stubborn part of me would like to see the Dolbeault description.
| https://mathoverflow.net/users/7 | Dolbeault Cohomology of $\mathbb{P}^1$ | I wrote a [blog post](http://sbseminar.wordpress.com/2010/01/12/residues-and-integrals/) about almost exactly this question. I'll give a summary here:
Since $H^{1,1}(X)$ is one dimensional, I could answer your question by giving anythng with the correct integral. However, I'll try to give you the kind of cocycle which actually comes out of the proof of Dolbeaut-Cech equality.
Your cocycle isn't $dz/z$ but, rather, $dz/z$ with a specific choice of open cover of $X$. Lets say your choice is $U\_1 \cup U\_2$, where $U\_1 = \{ z : z \neq \infty \}$ and $U\_2 = \{ z : z \neq 0 \}$. Refine your cover to $V\_1 \cup V\_2$, where $V\_1 = \{ z : |z| > r \}$ and $V\_2 = \{ z : |z| < r^{-1} \}$ for some $r < 1$.
Let $\theta\_1$ and $\theta\_2$ be $1$ forms on $V\_1$ and $V\_2$ such that $\theta\_1|\_{V\_1} - \theta\_2|\_{V\_2} = dz/z$. Then $\overline{\partial} \theta\_1$ and $\overline{\partial} \theta\_2$ have equal restrictions to $V\_1 \cap V\_2$. The $(1,1)$-form you are looking for is their common value, which I'll call $\omega$.
---
Let's first do a fake solution. **A real solution would look like a $C^{\infty}$ smearing out of this one.**
We'll work in the degenerate case $r=1$, so we are only gluing along a circle, not an annulus. We'll take $\theta\_1 = (1/2) \ \overline{z}\ dz$ and $\theta\_2 = -(1/2) \ dz / (\overline{z} z^2)$. Notice that both $\theta\_1$ and $\theta\_2$ restrict to $dz/z$ on the unit circle, but are constructed to extend smoothly to the appropriate discs.
So $\overline{\partial} \theta\_1 = (1/2) d \overline{z} d z$ and $\overline{\partial} \theta\_2 = (1/2) dz d \overline{z} / (\overline{z}^2 z^2)$. Our $\omega$ is formed by gluing these two differential forms together.
---
A **genuine smooth solution** would be like this, but would interpolate smoothly between these two. If you push forward in a brute force manner, you'll get something with [bump functions](http://en.wikipedia.org/wiki/Bump_function) in it.
If you are more clever, you may discover the solution
$$\theta\_1 = \frac{dz}{z} \left( 1- \frac{1}{1+z \overline{z}} \right)$$
and
$$\theta\_2 = - \frac{dz}{z} \left( \frac{1}{1+z \overline{z}} \right).$$
You should check that $\theta\_1|\_{V\_1} - \theta\_2|\_{V\_2} = dz/z$ and that $\theta\_i$ is smooth and well-defined on $U\_i$.
Then
$$\overline{\partial} \theta\_1 = \overline{\partial} \theta\_2 = \frac{dz \ d\overline{z}} {(1+z \overline{z})^2}.$$
This is, as Scott guessed, the **[Fubini-Study](http://en.wikipedia.org/wiki/Fubini-Study_metric)** form.
| 7 | https://mathoverflow.net/users/297 | 19720 | 13,112 |
https://mathoverflow.net/questions/19312 | 33 | I want to know exactly how derived functor cohomology and Cech cohomology can fail to be the same.
I started worrying about this from Dinakar Muthiah's [answer](https://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/4217#4217) to [an MO question](https://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology "equivalence of Grothendieck-style versus Cech-style sheaf cohomology"), and Brian Conrad's comments [1](https://mathoverflow.net/questions/19290/why-do-gerbes-live-in-h2#comment37725_19290) [2](https://mathoverflow.net/questions/19290/why-do-gerbes-live-in-h2#comment37731_19290) to [another MO question](https://mathoverflow.net/questions/19290/why-do-gerbes-live-in-h2 "Why do gerbes live in H^2?"). Let $\mathcal{F}$ be a sheaf of abelian groups on a space $X$. (Here I want to be a little vague about what a "space" means. I'm thinking of either a scheme or a topological space.) Then [Čech cohomology](http://en.wikipedia.org/wiki/%C4%8Cech_cohomology) of $X$ with respect to a cover $U \to X$ can be defined as cohomology of the complex
$$ \mathcal{F}(U) \to \mathcal{F}(U^{[2]}) \to \mathcal{F}(U^{[3]}) \to \dotsb $$
where $U^{[ n ]} = U \times\_X U \times\_X \dotsb \times\_X U$. The total Čech cohomology of $X$, $\smash{\check H}^{ \* }(X, \mathcal{F}) $, is then given by taking the colimit over all covers $U$ of $X$. Now if the following condition is satisfied:
>
> **Condition 1**: For sufficiently many covers $U$, the sheaf $\mathcal{F}\rvert\_{U^{[ n ]}}$ is an acyclic sheaf for each $n$
>
>
>
then this cohomology will agree with the derived functor version of sheaf cohomology. We have,
$$\smash{\check H}^{ \* }(X, \mathcal{F}) \cong H^\*(X; \mathcal{F}).$$
I am told, however, that even if $\mathcal{F}\rvert\_U$ is acyclic this doesn't imply that it is acyclic on the intersections. It is still okay if this condition fails for some covers as long as it is satisfied for enough covers. However I am also told that there are spaces for which there is *no cover* satisfying condition 1.
Instead you can replace your covers by hypercovers. Basically this is an augmented simplicial object $$V\_\bullet \to X$$ which you use instead of the simplicial object $U^{[ \bullet +1 ]} \to X$. There are some conditions which a simplicial object must satisfy in order to be a hypercover, but I don't want to get into it here. You can then define cohomology with respect to a hypercover analogously to Čech cohomology with respect to a cover, and then take a colimit. This seems to always reproduce derived functor sheaf cohomology.
So my question is **when is this really necessary?**
>
> **Question 1**: What is the easiest example of a scheme and a sheaf of abelian groups (specifically representable ones such as $\mathbb{G}\_m$) for which Čech cohomology of that sheaf and derived functor cohomology disagree?
>
>
> **Question 2**: What is the easiest example of a (Hausdorff) topological space and a reasonable sheaf for which Čech cohomology and derived functor cohomology disagree?
>
>
>
---
I also want to be a little flexible about what a "cover" is supposed to be. I definitely want to allow interesting Grothendieck topologies, and would be interested in knowing if passing to a different Grothendieck topology changes the answer. It changes both the notion of sheaf and the notion of Čech cohomology, so I don't really know what to expect.
Also, I edited question 1 slightly from the original version, which just asked about quasi-coherent sheaves. Brian Conrad kindly pointed out to me that for any quasi-coherent sheaf the Čech cohomology and the sheaf cohomology will agree (at least with reasonable assumptions on our scheme, like quasi-compact quasi-separated?) and that the really interesting case is for more general sheaves of groups.
| https://mathoverflow.net/users/184 | Example Wanted: When Does Čech Cohomology Fail to be the same as Derived Functor Cohomology? | Q1: A very simple example is given in Grothendieck's Tohoku paper "Sur quelques points d'algebre homologique", sec. 3.8. *Edit:* The space is the plane, and the sheaf is constructed by using a union of two irreducible curves intersecting at two points.
Q2: Cech cohomology and derived functor cohomology coincide on a Hausdorff paracompact space (the proof is given in Godement's "Topologie algébrique et théorie des faisceaux"). I don't know of an example on a non paracompact space where they differ.
| 36 | https://mathoverflow.net/users/1784 | 19723 | 13,114 |
https://mathoverflow.net/questions/19651 | 4 | The relevant paper is ["An estimate of the remainder in a combinatorial central limit theorem" by Erwin Bolthausen](https://doi.org/10.1007/BF00533704 "Z. Wahrscheinlichkeitstheorie verw Gebiete 66, 379–386 (1984)"). I would like to understand the estimate on page three right before the sentence "where we used independence of $S\_{n-1}$ and $X\_n$":
$$\begin{align}E|f'(S\_n) - f'(S\_{n-1})|
&\le E \bigg(\frac{|X\_n|}{\sqrt{n}} \big(1 + 2|S\_{n-1}| + \frac{1}{\lambda} \int\_0^1 1\_{[z,z+\lambda]} (S\_{n-1} + t \frac{X\_n}{ \sqrt{n}}) dt\big)\bigg) \\
&\le \frac{C}{\sqrt{n}} \big(1 + \delta(\gamma, n-1) / \lambda\big)\end{align}$$
that is, where $\delta(\gamma, n-1)/\lambda$ shows up, which is the error term in the Berry–Esséen bound.
Here $S\_n = \sum\_{i=1}^n X\_i / \sqrt{n}$ and $X1, \ldots, X\_n$ are iid with $E X\_i =0$, $E X\_i^2 = 1$, and $E|X\_i|^3 = \gamma$. Furthermore, denote $\mathcal{L}\_n$ to be the set of all sequences of $n$ random variables satisfying the above assumptions, then
$ \delta(\lambda, \gamma,n) = \sup \{ |E(h\_{z,\lambda} (S\_n)) - \Phi(h\_{z,\lambda})|: z \in \mathbb{R}, X\_1, \ldots, X\_n \in \mathcal{L}\_n \}$
and $h\_{z, \lambda}(x) = ((1 + (z-x)/\lambda) \wedge 1) \vee 0$ and $\delta(\gamma, n)$ is a short hand for $\delta(0,\gamma, n)$, and $h\_{z,0}$ is interpreted as $1\_{(-\infty, z]}$. I am mainly interested in verifying the second inequality, so I don't need to reproduce the definition of $f$ here, but it is related to $h$.
This paper is freely available online through Springer. thanks in advance.
| https://mathoverflow.net/users/4923 | Stein's method proof of the Berry–Esséen theorem | If you take expectation first with respect to $S\_{n-1}$, then by Fubini's theorem the last term gives
$$
E \left[\frac{|X\_n|}{\sqrt{n}}\frac{1}{\lambda} \int\_0^1 P\left(z-t\frac{X\_n}{\sqrt{n}}
\le S\_{n-1} \le z-t\frac{X\_n}{\sqrt{n}} + \lambda\right) dt\right].
$$
Now if $Y$ is a standard Gaussian random variable and $a\in \mathbb{R}$, then
$$
P(a\le S\_{n-1} \le a+\lambda) \le P(a\le Y \le a+\lambda) + 2\delta(\gamma,n-1) \le \frac{\lambda}{\sqrt{2\pi}} + 2\delta(\gamma,n-1),
$$
so the expectation above is bounded by $\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{2\pi}}+2\frac{\delta(\gamma,n-1)}{\lambda}\right)$.
| 4 | https://mathoverflow.net/users/1044 | 19727 | 13,117 |
https://mathoverflow.net/questions/19729 | 1 | I would like to preface by saying that I have no significant experience working with set theory, so I'm probably making an intuitive mistake. I have figured out where the mistake probably is, but I can't figure out why it IS a mistake. I figured that this was the best outlet to ask my question.
I was reading about the Continuum Hypothesis on Wikipedia recently, particularly about the fact that it's undecidable in ZFC -- which means that it is undecidable whether or not there is an infinite set whose size is strictly between that of the natural numbers and that of the real numbers. Now, intuitively, a proof that it cannot be disproved would immediately give way to the fact that no counterexample could be constructed (for such a counterexample would disprove it, which is impossible), and thus it must therefore be true. But that's not where I'm going with this.
Cantor proved that the rational numbers are countable -- that there exists a counting method such that, given any integer, you could determine the unique rational number which corresponds to it, and given any rational number, you could determine the unique integer which corresponds to it. The proof is fairly cool, but that's not where I'm going, either. Essentially, this demonstrated that $\aleph\_0^2 = \aleph\_0$. Then, he went on to prove that all *algebraic* numbers were countable, which proved the stronger statement that for any finite *n*, $\aleph\_0^n = \aleph\_0$. But yet, the cardinality of the real numbers is still strictly greater.
He explicitly determined the cardinality of the real numbers as $2^{\aleph\_0}$, or, strictly speaking, for any $\alpha > 1, \mathfrak c = \alpha^{\aleph\_0}$, because, no matter which base you're in, the number of real numbers doesn't change. This means that the cardinality of the real numbers is strictly *exponential*, whereas the cardinality of the countable numbers is strictly *polynomial*.
This is where my confusion arises. If I construct a set whose size after an infinite number of steps is bounded by *any* polynomial, it is countable, whereas a set whose size is greater than *every* polynomial would *not* be countable.
The [Adleman–Pomerance–Rumely\_primality\_test](http://en.wikipedia.org/wiki/Adleman%25E2%2580%2593Pomerance%25E2%2580%2593Rumely_primality_test) has running time, for a given *n*, of $n^{O\log(\log(n))}$, which is of super-polynomial running time -- there exists no polynomial that is strictly greater than that function. However, it is also sub-exponential -- there exists no exponential function that is strictly *less* than it, either. Therefore, it exists between the polynomials and exponentials.
Using this, I can construct a set of numbers whose size after *n* steps is equal to $f(n) = n^{O\log(\log(n))}$ by appending approximately f'(n) unique values to the end of the set. I have now explicitly constructed an infinite set whose size is $\aleph\_0^{\log(\log(\aleph\_0))}$, have I not? And, as I said before, it is, eventually, larger than any set whose size grows polynomially. But it is also smaller than every set whose size grows exponentially.
Of course, it also turns out that this set is countable -- I can give you the numbers in the set, if you so wish. But that's only part of the problem.
Using the same method, I can also construct a set whose size grows exponentially, if I only change my function to $f(n) = 2^n$. I then append f'(n) unique values to the end of my set and, voila, as I take a countably infinite number of steps, namely $\aleph\_0$, the size of my set becomes $2^{\aleph\_0}$, the cardinality of the continuum -- but, as before, I can tell the *n*th item in my set, and if you give me any item in my set, I can tell you exactly where it lies.
**Thus is my question: What did I do wrong? Either I have shown that $2^{\aleph\_0} = \aleph\_0$, which is exceedingly unlikely, or my assumption that my new set's size is equal to $2^{\aleph\_0}$ is incorrect, and I cannot understand why.**
Any help would be appreciated, thanks!
--Gabriel Benamy
| https://mathoverflow.net/users/1982 | Explicitly constructing an infinite set with particular size | The problem lies with your interpretation of the notation $2^{\aleph\_0}$ and $\aleph\_0^n.$ These do not mean that your set is constructed from finite sets with a specified rate of growth.
Let $A$ be the cardinality of a set. e.g. $A = 2$ or $A=\aleph\_0.$
$A^n$ is the cardinality of the n-fold cartesian product of a set with cardinality $A.$
$2^{A}$ is the cardinality of the power set, i.e. the set of all subsets, of a set with cardinality $A.$
Your interpretation in terms of rates of growth does not work. The reason for this notation is that when $A$ is finite, the arithmetic interpretation works. That is, if $S$ is a set with $k$ elements, then $S^n$ has $k^n$ elements, and there are $2^k$ subsets of $S.$
| 8 | https://mathoverflow.net/users/4872 | 19731 | 13,119 |
https://mathoverflow.net/questions/19648 | 2 | Again from the Shepp and Lloyd paper "ordered cycle lengths in a random permutation", I found this puzzling equality. This one might require access to the paper itself since it's quite a mouthful:
In equation (15), they claimed it is straightforward that if there is an $F\_r$ such that
$$\int\_0^1 \exp(-y/\xi) dF\_r(\xi) = \int\_y^{\infty} \frac{E(x)^{r-1}}{(r-1)!} \frac{\exp(-E(x) -x)}{x} dx $$
then $F\_r$ will have moments $G\_{r,m}$.
Here
$$G\_{r,m} = \int\_0^{\infty} \frac{x^{m-1}}{m!} \frac{E(x)^{r-1}}{(r-1)!} \exp(-E(x)-x) dx$$
and
$$E(x) = \int\_x^{\infty} \frac{e^{-y}}{y} dy$$
which is related to the thread [Reference request for a "well-known identity" in a paper of Shepp and Lloyd](https://mathoverflow.net/questions/19526/reference-request-for-a-well-known-identity-in-a-paper-of-shepp-and-lloyd)
It looks to me like some sort of Laplace transform, but I can't manage to get the algebra to work, because of the inverse exponent $y/\xi$ with respect to $\xi$.
I will be happy enough if one can tell me why we are looking at the transform $\int\_0^1 \exp(-y/\xi) dF\_r(\xi)$ instead of the usual moment generating function $\int\_0^1 \exp(-y \xi) dF\_r(\xi)$, or maybe it's a typo?
| https://mathoverflow.net/users/4923 | method of moments and Laplace transform from Shepp and Lloyd | This is a general fact: assume that $X$ is a positive random variable and that, for a given nonnegative function $g$, $\displaystyle E(\mathrm{e}^{-y/X})=\int\_y^{\infty}g(x)\mathrm{d}x$ for every positive $y$. Then $\Gamma(s+1)E(X^s)=\displaystyle\int\_0^{\infty}x^sg(x)\mathrm{d}x$ for every positive $s$.
To prove this, integrate the equality $\displaystyle\int\_0^{\infty}\mathrm{e}^{-y/x}y^{s-1}\mathrm{d}y=\Gamma(s)x^s$ over $x>0$ with respect to the distribution of $X$ and change the order of integration in the LHS.
| 7 | https://mathoverflow.net/users/4661 | 19737 | 13,124 |
https://mathoverflow.net/questions/19732 | 12 | While doing some research, I came up with a problem of proving that
$ f(a,b,c)=\begin{cases}1 &\text{ if }A(a,b)=c\\ \\\\ 0 &\text{ otherwise }\end{cases} $
is primitively recursive ($A$ is the Ackermann's function).
Any references, ideas or proofs?
(This may not be a good MO question, but since the participants in problem-solving sites listed in MO posting FAQ failed to solve it - I posted it there before - I was hoping for a solution here.)
| https://mathoverflow.net/users/4925 | Ackermann-related function | Here's a sketch of an argument which I expect could be made into a proof.
The key fact is that the Ackermann function fails to be primitive recursive only because it grows so quickly. More formally:
**Claim**. There exists a Turing machine T and a primitive recursive function f(a, b, c) (which is an increasing function of c) such that on input (a, b), T computes A(a, b) in at most f(a, b, A(a, b)) steps.
"Proof". Starting with the expression "A(a, b)", repeatedly expand terms of the form A(x, y) with the recursive definition, but do not simplify any of the resulting additions. The length of the string increases at every step, if we agree that the symbol "+" is "longer" than "A". The resulting string is a formal sum of positive integers, so its length is bounded by (a multiple of) A(a, b); hence the number of steps is also bounded above in terms of A(a, b), and we may perform each step in time polynomial in A(a, b).
Now, we can simulate a given Turing machine for a fixed number of steps using a primitive recursive function. We may therefore compute the graph of the Ackermann function with a primitive recursive function as follows: Given a, b, c,
* Compute f(a, b, c).
* Simulate the Turing machine T on input (a, b) for f(a, b, c) steps.
* If T has halted, then return whether c is equal to the output of T. If T has not halted, then c < A(a, b) so return false.
| 8 | https://mathoverflow.net/users/126667 | 19742 | 13,128 |
https://mathoverflow.net/questions/19747 | 14 | Let $P$ be a polyhedron which satisfies the following three conditions:
1. $P$ is built out of regular hexagons and regular pentagons.
2. Three faces meet at each vertex.
3. $P$ is topologically a sphere.
An easy Euler characteristic argument tells you that $P$ has exactly twelve pentagonal faces.
An example of a polyhedron like this is a truncated icosahedron (soccer ball for those of us in the States, football for everyone else). In this case, the pentagonal faces are arranged with some nice symmetry, and the polyhedron has icosahedral symmetry.
Another (trivial) example is the regular dodecahedron, which again has icosahedral symmetry.
Here's my question: Is this symmetry forced? What, if anything, can be said in general about the symmetry of a polyhedron which satisfies the above three conditions?
Edit: Since the discussion below points out that there are precisely two polyhedra which satisfy the above conditions, a suitable evolution of the question, which has already begun to be discussed below, is this: What symmetry groups can a polyhedron have if one or more of the above conditions are relaxed?
| https://mathoverflow.net/users/5000 | The Symmetry of a Soccer Ball | Only soccer ball or dodecahedron.
---
Clearly 3 hexagons can not meet at one vertex.
Thus we have only 3 choices for one vertex:
* 3 pentagons
* 2 pentagons + 1 hexagon
* 1 pentagons + 2 hexagon
Note that if $[pq]$ is an edge then $p$ has the same type as $q$ (the type is determined by angle at $[pq]$).
Thus the polyhedron is completely determined by one vertex.
Further:
* Once you have a vertex of the first type you have a regular dodecahedron.
* If you have a vertex of the second type then you will get one hexagon surrounded by pentagons. Then it is easy to see that you can not continue.
* For the third type you will get a soccer ball or "truncated icosahedron" as some people call it :)
| 13 | https://mathoverflow.net/users/1441 | 19753 | 13,134 |
https://mathoverflow.net/questions/19684 | 21 | In the study of number theory (and in other branches of mathematics) presence of Hecke Algebra and Hecke Operator is very prominent.
One of the many ways to define the Hecke Operator $T(p)$ is in terms of double coset operator corresponding to the matrix $ \begin{bmatrix} 1 & 0 \\ 0 & p \end{bmatrix}$ .
On the other hand Hecke Algebra $\mathcal{H}(G,K)$ associated to a group $G$ of td-type ( topological group, such that every neighborhood of unity contains a compact open subgroup), where $K$ is a compact open subgroup of $G$ is defined as the space of locally constant compactly supported $K$ bi-invariant functions on $G$. Convolution product makes it an associative algebra.
I was told that the hecke algebra $\mathcal{H}(Gl(2,\mathbb{Q}\_p) , Gl(2,\mathbb(Z)\_p))$ corresponds to the classical algebra of hecke operators attached to $p$ via Satake Isomorphism Theorem. Using Satake Isomorphism theorem I can show $\mathcal{H}(Gl(2,\mathbb{Q}\_p) ,Gl(2,\mathbb(Z)\_p))$ is commutative and finitely generated over $\mathbb{C}$.
So my question is how one uses Satake Isomorphism Theorem (or otherwise) to see this?
And secondly in general what is the relation between hecke operators and hecke algebra?
| https://mathoverflow.net/users/4291 | Relation between Hecke Operator and Hecke Algebra | The fact that Hecke operators (double coset stuff coming from $SL\_2(\mathbf{Z})$ acting on modular forms) and Hecke algebras (locally constant functions on $GL\_2(\mathbf{Q}\_p)$) are related has nothing really to do with the Satake isomorphism. The crucial observation is that instead of thinking of modular forms as functions on the upper half plane, you can think of them as functions on $GL\_2(\mathbf{R})$ which transform in a certain way under a subgroup of $GL\_2(\mathbf{Z})$, and *then* as functions on $GL\_2(\mathbf{A})$ ($\mathbf{A}$ the adeles) which are left invariant under $GL\_2(\mathbf{Q})$ and right invariant under some compact open subgroup of $GL\_2(\widehat{\mathbf{Z}})$.
Now there's just some general algebra yoga which says that if $H$ is a subgroup of $G$ and $f$ is a function on $G/H$, and $g\in G$ such that the $HgH$ is a finite union of cosets $g\_iH$, then you can define a Hecke operator $T=[HgH]$ acting on the functions on $G/H$, by $Tf(g)=\sum\_i f(gg\_i)$; the lemma is that this is still $H$-invariant.
Next you do the tedious but entirely elementary check that if you consider modular forms not as functions on the upper half plane but as functions on $GL\_2(\mathbf{A})$, then the classical Hecke operators have interpretations as operators $T=[HgH]$ as above, with $T\_p$ corresponding to the function supported at $p$ and with $g=(p,0;0,1)$. Because the action is "all going on locally" you may as well compute the double coset space locally, that is, if $H=H^pH\_p$ with $H\_p$ a compact open subgroup of $GL\_2(\mathbf{Q}\_p)$, then you can do all your coset decompositions and actions locally at $p$.
Now finally you have your link, because you can think of $T$ as being the characteristic function of the double coset space $HgH$ which is precisely the sort of Hecke operator in your Hecke algebra of locally constant functions. Furthermore the sum $f(gg\_i)$ is just an explicit way of writing convolution, so everything is consistent.
I don't know a book that explains how to get from the classical to the adelic point of view in a nice low-level way, but I am sure there will be some out there by now. Oh---maybe Bump?
| 24 | https://mathoverflow.net/users/1384 | 19757 | 13,137 |
https://mathoverflow.net/questions/19745 | 2 | How to obtain an upperbound for knots up to k crossings?
I think I've found something which involves the genus but I'm not sure.
| https://mathoverflow.net/users/5001 | Counting knots with fixed number of crossings | There are some known exponential bounds on the number. For example, if kn is the number of prime knots with n crossings, then Welsh proved in "On the number of knots and links" (MR1218230) that
>
> 2.68 ≤ lim inf (kn)1/n ≤ lim sup (kn)1/n ≤ 13.5.
>
>
>
The upper bound holds if you replace kn by the much larger number ln of prime n-crossing links.
Sundberg and Thistlethwaite ("The rate of growth of the number of prime alternating links and tangles," MR1609591) also found asymptotic bounds on the number an of prime alternating n-crossing links: lim (an)1/n exists and is equal to (101+√21001)/40.
| 10 | https://mathoverflow.net/users/428 | 19759 | 13,139 |
https://mathoverflow.net/questions/19638 | 1 | Hello all, let $n$ be an integer $\geq 2$ and let $\alpha$ be an algebraic number
of degree $n$. Let $R$ be the ring of algebraic integers in ${\mathbb Q}(\alpha)$, and
let $B$ be the subset of $R$ containing the elements whose degree is exactly
$n$. Any $\beta \in B$ has a minimal polynomial
$X^n+b\_{n-1}X^{n-1}+ \ldots + b\_1X+b\_0$. Identifying this latter polynomial
with the uple $(b\_0,b\_1, \ldots ,b\_{n-1})$ allows us to view $B$ as a subset
of ${\mathbb Z}^n$. I define a combinatorial subvariety $V$ of dimension
at most $r$ of ${\mathbb Z}^n$ to be a subset of $Z^n$ such that
there is a set of indices $I \subseteq \lbrace 1,2, \ldots , n \rbrace$ with
$|I|=n-r$ and the projection $p:V \to {\mathbb Z}^{n-r},
(v\_1,v\_2, \ldots ,v\_n) \mapsto (v\_i)\_{i\in I}$ is constant.
My question is : what is the smallest $r$ such that
there is an infinite subset $B' \subset B$ corresponding to
a subvariety of dimension at most $r$ ?
In other words, we are asking for infinitely many elements in $B$,
whose minimal polynomials are ``as similar as possible".
An easy case is when $\alpha=a^{\frac{1}{n}}$ for some
$a \in {\mathbb Q}$, because the rational multiples of $\mathbb \alpha$
correspond to a subvariety of dimension 1, so that $r=1$ in this case.
| https://mathoverflow.net/users/2389 | Infinite collection of elements of a number field with very similar annihilating polynomials | For $n>4$, almost all fields of degree $n$ will have $r>1$:
Fix a field $K$ with discriminant $D\_0$. Fix the $n-1$ coefficients $b\_{n-1},...,b\_{i+1}, b\_{i-1},..., b\_0$. The discriminant of the polynomial $x^n+b\_{n-1}x^{n-1}+...$ is a polynomial $D(b\_i)$ in the single variable $b\_i$, and is of degree at least $4$.
If this polynomial is squarefree, as it will be for almost all $n-1$ fixed coefficients, then the hypersurface $D\_0y^2 = D(b\_i)$ has genus at least $1$, and hence finitely many integer points.
But, every polynomial defining the same field must have the same discriminant up to a square factor, and hence $r > 1$.
Going back on my comment above: since the degree of the discriminant (multivariate) polynomial is large (linear in the number of variables) the equation $D(b\_0,...,b\_{n-1}) = D\_0y^2$ will probably have only a finite number of solutions for most $D\_0$, if $r$ is much smaller than $n$.
Therefore, my new pessimistic conjecture is that for almost all fields you will have $r \gg n$.
Note: $r \le n-1$ - in any number field there are always an infinite number of algebraic integers with trace 0.
| 3 | https://mathoverflow.net/users/2024 | 19774 | 13,148 |
https://mathoverflow.net/questions/19766 | 17 | A homotopy (limits and) colimit of a diagram $D$ topological spaces can be explicitly described as a geometric realization of simplicial replacement for $D$.
However, a homotopy colimit can also be described as a derived functor of limit. A model category structure can be placed on the category $\mathrm{Top}^I$, where $I$ is a small index category, where weak equivalences and fibrations are objectwise, so that $\mathrm{colim} : \mathrm{Top}^I \leftrightarrow \mathrm{Top} : c$ form a Quillen pair, where $c$ is the diagonal functor taking an object $A$ to the constant diagram at $A$. Then the homotopy colimit can be described as a derived functor for $\mathrm{colim}$: take a cofibrant replacement $QD$ for a diagram $D$, then compute $\mathrm{colim}(QD) = \mathrm{hocolim}(D)$. It turns out that two cofibrant replacements will give weakly equivalent homotopy colimits. As such, you would suspect that this choice is not really important.
This leaves two questions: firstly, is it necessary in most cases to construct homotopy colimits explicitly, or are its properties as a homotopical functor enough? Secondly, do any problems arise from the fact that homotopy colimit is well-defined only up to weak equivalence (through the derived functor angle)? Do cases ever arise where a more canonical definition is required?
Context: I am reading through Goodwillie's "Calculus II: Analytic Functors." There the explicit simplicial construction is used, and in particular it is needed that certain maps from holim(D) are fibrations (Definition 1.1a, for example). However, being a fibration is not invariant under weak equivalence. Does this reflect that properties of this particular choice of holim are needed, or that the paper itself is too rigid? Can these arguments be made with a non-canonical choice of holim?
I apologize ahead of time for the vague question: I've been trying to read up in this subject area for a few months now, and this has been a stumbling block.
| https://mathoverflow.net/users/2532 | Homotopy colimits/limits using model categories | In the context of Goodwillie's paper, he's got an explicit natural transformation $f:holim\_I(X)\to holim\_J(X|\_J)$, where $X:I\to Top$ is a functor to spaces, and $J\subset I$ is a subcategory of $I$. With the construction of holim he's using, this map is always a fibration.
What if you tried to use a different construction of holim? Then maybe you get a map $f'$ which is not a fibration anymore. In that case, you could still have taken the *homotopy fiber* of $f'$, and this would be a notion which is invariant under weak equivalence. That is, you could (functorially) replace $f'$ with a fibration via the path construction, and take the fiber of that.
Of course, the homotopy fiber is exactly the thing he wants here. In fact, he's manufactured the situation exactly so that the homotopy fiber he wants is just the fiber of this map.
(It's worthwhile to note that in his setting, the category $I$ (which is a cube) has an initial object $\varnothing$. This means that the evident map $holim\_I(X)\to X(\varnothing)$ is a weak equivalence. In other words, $holim\_I(X)$ is really just $X$ evaluated at $\varnothing$, but modified so that it maps to (and fibers over) $holim\_J X|\_J$.)
| 14 | https://mathoverflow.net/users/437 | 19780 | 13,152 |
https://mathoverflow.net/questions/19721 | 2 | This is sort of a follow-up to:
[Gauge theory construction of moduli of vector bundles](https://mathoverflow.net/questions/19635/gauge-theory-construction-of-moduli-of-vector-bundles)
If I have a complex compact algebraic curve with at worst nodal singularities, is there an analytic description of holomorphic structures on the trivial bundle in terms of (0,1) forms satisfying some constraint? presumably one wants the forms to pick up singular behavior of some kind at the node. Also, is the description well-behaved in families as we smooth the curve? for instance, in the sense that whatever infinite-dimensional vector space is "locally constant" as we smooth the curve.
Other mild singularity types would be interesting too!
| https://mathoverflow.net/users/3500 | gauge theory construction of vector bundles on singular varieties | There are subtleties even in the simplest case - $C$ a compact, irreducible complex curve with one node, $Pic\_0(C)$ the Picard variety of line bundles of degree $0$ - so why not start there?
Pulling back line bundles via the normalisation map $\nu\colon \tilde{C}\to C$ defines a map $Pic\_0(C)\to Pic\_0(\tilde{C})$. The latter space is a complex torus of dimension $g(\tilde{C})$. To recover a line bundle $L$ from $\nu^\*L$ we also need an isomorphism $\nu^\ast L\_p\to \nu^\ast L\_q$, where $p$ and $q$ are the two points of $\tilde{C}$ that lie over the node of $C$. In this way, one sees that $Pic\_0(C)$ is a $\mathbb{C}^{\ast}$-bundle over $Pic\_0(\tilde{C})$.
So far as I can see, it's straightforward to give a gauge-theoretic description of $Pic\_0(C)$: it consists of pairs consisting of a flat, unitary connection (or equivalently a Cauchy-Riemann operator, or equivalently a holomorphic structure) in a complex line bundle of degree zero over $\tilde{C}$, with an isomorphism $I$ of the fibres over $x$ and $y$, modulo gauge transformations respecting $I$.
One can compactify $Pic\_0(C)$ to a $\mathbb{C}P^1$ bundle over $Pic\_0(\tilde{C})$ in a natural way, which also makes good sense gauge-theoretically. By gluing the zero-section to the infinity-section, covering the map on the base given by translation by the divisor $q-p$, one constructs a variety with normal crossing singularities which is isomorphic to the complex points of the compactified Picard scheme, parametrizing torsion-free sheaves of rank 1 on $C$. You could think about what this means in gauge-theory terms. There's a large literature on such compactified Picard (and Jacobian) varieties and their behaviour in families; see e.g. L. Caporaso, *A compactification of the universal Picard variety over the moduli space of stable curves*, J. Amer. Math. Soc. 7 (1994), no. 3, 589--660; MR1254134.
For higher rank stable bundles, there's a bewildering array of papers on assorted compactifications (some of which are moduli spaces, some not). I'll refer you only to an extraordinary paper by Donaldson that might give you some clues as to what to expect: *Gluing techniques in the cohomology of moduli spaces*, Topological methods in modern mathematics (Stony Brook, NY, 1991), 137--170, Publish or Perish, Houston, TX, 1993; MR1215963.
| 4 | https://mathoverflow.net/users/2356 | 19781 | 13,153 |
https://mathoverflow.net/questions/19775 | 32 | There are a couple of ways to define an action of $\pi\_1(X)$ on $\pi\_n(X)$. When $n = 1$, there is the natural action via conjugation of loops. However, the picture seems to blur a bit when looking at the action on higher $\pi\_n$. All of them have the flavor of the conjugation map, but are more geometric than algebraic, and in some cases work is needed to show the map is well defined. Here are a couple I have seen:
There is a homotopy equivalence $f : S^n \to S^n \vee I$. taking the basepoint of $S^n$ to the endpoint of the unit interval "far away" from $S^n$. Given a path $\alpha$ from $x\_0$ to $x\_1$, one can get a basepoint changing homomorphism $\pi\_n(X,x\_0) \to \pi\_n(X,x\_1)$ by taking $g : S^n \to X$ and mapping it to $(g \vee \alpha) \circ f$. If $\alpha$ is a loop this gives an action of $\pi\_1$
Another way to proceed may be to look at elements of $\pi\_n(X,x\_0)$ as homotopy classes of maps $I^n \to X$ that send $\partial I^n$ to $x\_0$. Then a base change homomorphism could be obtained by using a path $\alpha$ to define a map $I^n \cup (\partial I^n \times I) \to X$, which can be filled in to a map $I^{n+1} \to X$. Then the action would be to take the face opposite the original $I^n \subset I^{n+1}$.
These both define the same standard action of $\pi\_1$ on $\pi\_n$, but lose the algebraic flavor of the group action and instead have this stronger geometric feel, which can make working with the action a bit cumbersome. Are there other ways of looking at this action that are more algebraic?
Perhaps, can something be done wherein $\pi\_0(Y)$ acts on $\pi\_n(Y)$, where $Y$ is some sufficiently nice space like $\Omega X$, and does this coincide with the above defined actions? Is this a useful way of viewing the action?
| https://mathoverflow.net/users/2532 | Different way to view action of fundamental group on higher homotopy groups | If G is a topological group, then the group acts on itself by conjugation, and this action is base-point-preserving. In particular, for an element $g \in \pi\_0(G)$ and a higher homotopy element $\alpha \in \pi\_{n-1} G = [S^n, G]$, one can check that the conjugate $g \alpha g^{-1}$ is well-defined and defines an action of $\pi\_0(G)$ on $\pi\_{n-1} G$. The space G is weakly equivalent to the loop space of the classifying space BG, and under this equivalence the conjugation action is taken to the action of $\pi\_1 BG$ on $\pi\_n BG$.
(Unfortunately, this doesn't work directly for the conjugation action of the loop space on itself because it is not strictly basepoint-preserving; one needs to use that there is a natural homotopy from a loop $\gamma \* e \*\gamma^{-1}$ to $e$ to produce the action.)
Any path-connected based space X is weakly equivalent to the classifying space of a simplicial group G; specifically, the Kan loop group of a weakly equivalent simplicial set. Even more, there is a Quillen equivalence between the homotopy theories of spaces and simplicial groups.
(Kan's original paper can be found here: <http://www.jstor.org/pss/1970006>)
| 20 | https://mathoverflow.net/users/360 | 19787 | 13,157 |
https://mathoverflow.net/questions/19791 | 14 | Is infinite (say complex) projective space a scheme? More generally, can schemes have infinite cardinal dimension? It seems that infinite dimensional projective space is not a manifold, since it is not locally Euclidean for any R^n.
Related question. If inifinite projective space is a scheme, then take a nonclosed point. Taking the closure of this nonclosed point, can we get infinite dimensional subschemes? Sorry for I'm quite foreign to schemes.
| https://mathoverflow.net/users/nan | Infinite projective space | Starting with the affine case, if you try to define infinite dimensional affine space as Spec of k{x1,x2,...], then you realise that this is not a vector space of countable dimension, but something much larger. If you want a vector space over k of countable dimension, then this will not be a scheme, but instead will be an ind-scheme. A similar description should hold in the projective case.
Edit: Regarding why I am saying that Spec(k[x1,x2,...]) is too big: A (k-)point of Spec(k[x1,x2,...]) is an infinite sequence a1,a2,... of elements of k. If I wanted a vector space of countable dimension, then I should be asking for sequences a1,a2,... of elements of k, only finitely many of which are non-zero. This latter space is the inductive limit of affine n-space as n tends to infinity.
| 13 | https://mathoverflow.net/users/425 | 19795 | 13,162 |
https://mathoverflow.net/questions/19740 | 11 | This is maybe a dumb question. $SL\_2(\mathbb{R})$ has a natural action on the upper half plane $\mathbb{H}$ which is transitive with stabilizer isomorphic to $SO\_2(\mathbb{R})$. For this reason, people sometimes write $\mathbb{H}$ as the coset space $SL\_2(\mathbb{R})/SO\_2(\mathbb{R})$.
Now, it's clear how this description recovers the topology of $\mathbb{H}$: it's just the quotient topology. But can you recover either the Riemann surface structure or the hyperbolic metric on $\mathbb{H}$ from this description? How much of the structure of $SL\_2(\mathbb{R})$ and $SO\_2(\mathbb{R})$ do you need to do this, if it's possible?
| https://mathoverflow.net/users/290 | How do you recover the structure of the upper half plane from its description as a coset space? | **Edit:** I should have put a short version of the answer in the beginning, so here is how the various structures are recovered. To get a smooth manifold structure on the quotient, you use the fact that $SL\_2(\mathbb{R})$ is a real Lie group and $SO\_2(\mathbb{R})$ is a closed subgroup. To get a hyperbolic structure, you use the fact that $SL\_2(\mathbb{R})$ is isomorphic to an orthogonal group of signature (n,1) for some n (giving a transitive action on hyperbolic n-space). To get a complex structure, you use the fact that $SL\_2(\mathbb{R})$ is isomorphic to an orthogonal group of signature (2,m) for some m (giving an action on a hermitian symmetric space).
As others have noted, you can get a bijection on points using the Iwasawa decomposition, and you can get a hyperbolic structure using the exceptional isomorphism $PSL\_2(\mathbb{R}) \cong SO\_{2,1}^+(\mathbb{R})$. First, I'd like to clean up the Iwasawa treatment a bit. Any element of $SL\_2(\mathbb{R})$ can be uniquely decomposed as BK, where K is a rotation and B is upper triangular with positive diagonal. Any rotation K fixes i, so we should consider what elements B do. A bit of fiddling shows that $\begin{pmatrix} \sqrt{y} & x/\sqrt{y} \\ 0 & 1/\sqrt{y} \end{pmatrix} \cdot i = x+iy$.
We can view the exceptional isomorphism in another way that makes the complex structure more apparent, by viewing the hyperbolic plane as the Grassmannian $O\_{2,1}(\mathbb{R})/(O\_2(\mathbb{R}) \times O\_1(\mathbb{R}))$. From the standpoint of special relativity, this is the space of timelike lines through the origin in $\mathbb{R}^{2,1}$. Taking a quotient of the total space of these lines (minus origin) by positive rescaling, we find that this space is isomorphic to the space of pairs of antipodal points of norm -1. In particular, we have an isomorphism of the Grassmannian with the quotient of the hyperboloid with two sheets (i.e., solutions of the equation $x^2 + y^2 - z^2 = -1$) by the antipodal automorphism.
One way to explain the origin of the complex structure is by the fact that all Grassmannians of the form $O(2,n)/(O(2) \times O(n))$ are [hermitian symmetric spaces](http://en.wikipedia.org/wiki/Symmetric_space#Hermitian_symmetric_spaces), and the hyperbolic plane is just the case $n=1$. The 2 in $O(2)$ is essential, because the orthogonal group action is what yields the ninety degree rotation in the tangent space of any point, and this is what endows the quotient with an almost complex structure. If you want to see more about hermitian symmetric spaces than the Wikipedia blurb, I recommend looking in chapter 1 of Milne's [introduction to Shimura varieties](http://www.jmilne.org/math/xnotes/index.html).
Finally, I'd like to point out Deligne's description of the upper half plane as a moduli space of structured elliptic curves. Points on H parametrize elliptic curves with an oriented basis of first homology (as mentioned a few times in our class). If you want to say it is a fine moduli space, you need a functor that it represents, and it is unfortunately a bit complicated. The functor takes as input the category of complex analytic spaces, and for any such space S, it gives the set of isomorphism classes of elliptic curves over S (i.e., diagrams $E \underset{\pi}{\leftrightarrows} S$ of complex analytic spaces, where $\pi$ is smooth and proper with one-dimensional genus one fibers and the leftward map is a section) equipped with an isomorphism $R^1\pi\_\*\underline{\mathbb{Z}} \cong \underline{\mathbb{Z} \times \mathbb{Z}}$ that induces the canonical identity $R^2\pi\_\*\underline{\mathbb{Z}} \cong \underline{\mathbb{Z}}$ on exterior squares. Here, the underscore indicates a constant sheaf. The functor also takes morphisms to "the evident diagrams". To be honest, I have never seen a complete proof that this functor is represented by the complex upper half plane, although it seems to be more a question of doing lots of writing than an honest theoretical problem. You can probably do it using the fact that H is a classifying space of polarized Hodge structures, as Kevin Buzzard mentioned in the comments.
| 4 | https://mathoverflow.net/users/121 | 19796 | 13,163 |
https://mathoverflow.net/questions/19783 | 6 | (repost from the topology Q&A board)
I have a (T\_1), Normal, countably paracompact space X. I would like to know if every locally countable open cover of X (i.e. an open cover such that every x in X has a neighbourhood which intersects only countably many members of the cover) has a locally finite refinement.
My suspicion is that the answer is a resounding no, but every time I try to construct a counterexample it starts to seem more plausible.
If the answer *does* turn out to be yes I'd love to know if it generalises from aleph\_0 to arbitrary cardinals.
| https://mathoverflow.net/users/4959 | Countable paracompactness, normality and locally countable open covers | In Caryn Navy's thesis under Mary Ellen Rudin she constructed several spaces that are normal, countably paracompact and paralindelöf (every cover has a locally countable refinement) but not paracompact.
All such spaces provide counterexamples (we can refine a cover without a locally finite refinement to a locally countable one and then we cannot continue...)
I'm not sure (as I do not have access to the PhD-thesis in question, and I only know it from references like <http://www1.elsevier.com/homepage/sac/opit/10/article.pdf>) whether these examples are all under extra set-theoretic assumptions (like MA + non-CH) or whether there are absolute ones as well.
| 4 | https://mathoverflow.net/users/2060 | 19798 | 13,165 |
https://mathoverflow.net/questions/19768 | 4 | I was reading Mehta and Seshadri's paper "Moduli of vector bundles on curves with parabolic structures".
In the second paragraph, they wrote:
"Suppose that $H$ mod $\Gamma$ has finite measure ($H$ is the complex upper half plane, and $\Gamma$ is a discrete group). Let $X$ be the smooth projective curve containing $H$mod$\Gamma$ as an open subset and $S$ the finite subset of $X$ corresponding to parabolic and elliptic fixed points under $\Gamma$."
I am not sure about what parabolic and elliptic mean here. And why does such an $X$ exist? If I take a Riemann surface and remove several small balls from it, when is it biholomorphic to another Riemann surface removing several points?
| https://mathoverflow.net/users/4975 | Upper half plane quotient by a discrete group | I like to think about this geometrically. $H/\Gamma$ is a topological metric space. At most points of $H$ (that are not fixed by any element of $\Gamma$, the quotient looks just like the hyperbolic plane $H$ itself. The singularities come from elliptic elements of $\Gamma$, i.e., (locally) rotations, where you get a cone metric (locally like the metric on a piece of paper rolled into a cone). Remove those points and consider the conformal structure coming from the resulting Riemannian metric. The ends of this surface look like removable singularities (locally like $\mathbf{C}$ minus a point in their conformal structure), and so can be filled in uniquely. $X$ is the result of filling in the points, and $S$ is the set of filled in points.
The points in $S$ come both from the removed cone points and from some points at infinity, the elliptic points as Charlie explains above.
You have to be careful to distinguish between removing a point (leaving a removable singularity) and removing a small ball (which is different conformally). Removing a small ball always gives a surface with infinite hyperbolic area when uniformized, and is never equivalent to a compact surface minus points.
| 7 | https://mathoverflow.net/users/5010 | 19799 | 13,166 |
https://mathoverflow.net/questions/19760 | 12 | Suppose $E/\mathbb{Q}$ is an elliptic curve with rank zero. According to the conjecture of Birch and Swinnerton-Dyer, the special value $L(1,E\_{/\mathbb{Q}})$ should be equal (up to some harmless factors) to the order of the Tate-Shafarevich group Sha$(E/\mathbb{Q})$. Now, suppose $\chi$ is a Dirichlet character, and $K/\mathbb{Q}$ is the cyclic extension cut out by $\chi$. My question is:
What is the precise (conjectural) relation between the *individual* special values $L(1,E\times\chi^i)$ and Sha$(E/K)$?
More precisely, we have $L(s,E\_{/K})=\prod\_{i=0}^{\mathrm{ord}(\chi)} L(s,E\times \chi^i)$. If $E/K$ has rank zero, is there a decomposition of Sha$(E/K)$ with respect to an action of $\mathrm{Gal}(K/\mathbb{Q})$ such that the individual pieces of this decomposition have orders given by the individual values $L(1,E \times \chi^i)$?
| https://mathoverflow.net/users/1464 | Decomposition of Tate-Shafarevich groups in field extensions | Fix a prime p which doesn't divide the degree of K over ${\mathbb Q}$, and let ${\mathcal O}$ denote the ring of integers of ${\mathbb Q}\_p(\chi)$ i.e. an extension of ${\mathbb Q}\_p$ containing the values of $\chi$. Then the group algebra ${\mathcal O}[G]$ decomposes into a direct sum of 1-dimensional pieces over ${\mathcal O}$, one for each power of $\chi$.
Then $Sha(E/K)[p^\infty] \otimes {\mathcal O}$ being an ${\mathcal O}[G]$-module inherits such a decomposition. Concretely, the $\chi^i$-component of $Sha(E/K)[p^\infty] \otimes {\mathcal O}$ is the subset where $G$ acts by $\chi^i$.
This $\chi^i$-component is then a reasonable candidate to compare to the $p$-adic valuation of the algebraic part of $L(E,\chi^i,1)$.
Some further comments:
-Note that extending scalars to ${\mathcal O}$ increases the size of the modules so this has to be taken into account.
-The component corresponding to the trivial character is the invariants under $G$, and when $G$ has size prime-to-p this is simply $Sha(E/{\mathbb Q})[p^\infty] \otimes {\mathcal O}$ (which is good).
-To make a precise relationship between the $L$-value and Sha, you need to take into account the other terms in BSD. Namely:
\*The torsion-term should work out exactly as above (decomposing into $\chi$-components).
\*The periods have to be considered (which was ignored above in my vague phrase "the algebraic part of").
\*The Tamagawa numbers give me pause -- possibly there is an analogous $\chi$-decomposition, but I don't see it now.
\*Lastly, if K is ramified over ${\mathbb Q}$ then the discriminant of K appears in the BSD quotient (in the denominator which increases the size of Sha). To handle this, I imagine what should be done is that rather then considering the L-value alone, consider the L-value times the Gauss sum of the character. (By the conductor-discriminant formula this should give exactly the extra powers of p needed.)
| 8 | https://mathoverflow.net/users/86179 | 19821 | 13,180 |
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