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https://mathoverflow.net/questions/13793 | 3 | Can I find an analytical solution to the the length of an 2-dimensional cubic B-spline? All I can find are [chorded approximations](http://www.tinaja.com/glib/nubzlen1.pdf) and the opinion that the analytic solution is "unbearably gruesome". However, I believe if I had the solution to this elliptic integral: $\int\_0^u \sqrt{\left(b\_{1} + 2 b\_{2} t + 3 b\_{3} t^{2}\right)^{2} + \left(a\_{1} + 2 a\_{2} t + 3 a\_{3} t^{2}\right)^{2}} dt$, I'd have my answer (at least on a per-segment basis). My goal is implement it in some software, so a little up-front grue is ok. Is it really not worth pursuing? I can't believe I'm better off with a numerical solution. I'm after 64-bit accuracy if that helps with the discussion.
Sympy cannot integrate it. I have not tried Math-CAD, Maxima, or other CASs. I do not have any significant math background (if that's not already obvious).
---
Edit (background info for Greg):
This spline will be used as a control-surface for conserved quantities in a physics analysis. Flux of these quantities must also be known. The spline must be second-derivative continuous and must pass through specified and an arbitrary number of knot points. A cubic B-spline seems right for the job. It is also required that the length of the spline or any arbitrary interval along the spline also be determinable to a near-machine-precision value. Unfortunately, this needs to be done quickly because the knots of the spline and the parameterization intervals are themselves being converged-upon to meet criteria which is dependent on these lengths. And to make matters worse, this is a time-transient solution (which means another level of iteration).
I am using Python to prototype this on a 64-bit multi-core PC. (I wince as I type 'Python' because it has a reputation for being slow.) However, I am leveraging numpy (for basic array math), scipy (for FITPACK, QUADPACK, etc.), and possibly sympy (for elliptical functions) for the heavy lifting. I plan to (if I can ever find an exact closed-form "gruesome" solution) to write it up in C or FORTRAN and compile it as it's own module.
It has to run on a PC as it is not a one-time-use algorithm. It will be used by others many times over in a design environment. (I guess that's another level of iteration again.)
| https://mathoverflow.net/users/3716 | Finding the length of a cubic B-spline | You should first accept the fact that it's an elliptic integral, and therefore doesn't have an elementary expression without elliptic functions. If you had a numerical library with elliptic functions, then great. Otherwise, you need to either implement elliptic functions yourself, or implement numerical integration of your integral.
I recommend numerical integration, just because in context it is conceptually simple and reliable. Your integrand has a fairly tame form: It can't blow up, the integrand is continuous, and the integrand is also real analytic unless it touches zero. In this situation, [Gaussian integration](http://en.wikipedia.org/wiki/Gaussian_quadrature) has excellent properties. I don't feel like doing a precise calculation, but I would expect that for any choice of the coefficients, Gaussian quadrature with just 5 evaluation points already has to be fairly close to the exact answer for any choices of the coefficients.
---
The above is part of an answer, but not a complete answer you really want 64 bits of accuracy. Assuming that the integrand is real analytic, Gaussian quadrature or Clenshaw-Curtis will converge exponentially. It seems reasonable enough to use Clenshaw-Curtis, which lets you recycle evaluation points and has a predictable formula for the numerical integration weights, with more and more points until the answer looks accurate.
The only problem is in the annoying case in which the integrand touches zero, or comes close to touching zero, which can be interpreted geometrically as a point on the spline with nearly zero velocity. (Typically it looks like a cusp.) Then the integrand is NOT real analytic and these numerical methods do not converge exponentially. Or, in the near-zero case, the integrand is analytic but the exponential rate of convergence is slow. I'm sure that there are tricks available that will handle this case properly: You could cut out an interval near the bad point and do something different, or you could subtract off a known integral to tame the bad point. But at the moment I do not have specific advice for an algorithm that is both reasonable fast and reasonably convenient. Clenshaw-Curtis is convenient and usually very fast for this problem, but not all that fast in bad cases if you push it to 64 bits of precision.
Also, these methods can be thought of as a more sophisticated version of chordal approximation. Chordal approximation faces the same issue, except worse: It never converges at an exponential rate for a non-trivial cubic spline. If you want 64 bits, you might need a million chords.
---
Meanwhile, the GNU Scientific Library does have elliptic function routines. If you have elliptic functions, then again, your integral is not all that simple, but it is elementary. I don't know whether GSL or equivalent is available for your software problem. If it is, then an elliptic function formula is probably by far the fastest (for the computer, not necessarily for you).
---
In a recent comment, bpowah says "All I wanted to know is whether or not it was faster to compute the given integral numerically or exactly." Here is a discussion. Computing an integral, or any transcendental quantity, "exactly" is an illusion. Transcendental functions are themselves computed by approximate numerical procedures of various kinds: Newton's method, power series, arithmetic-geometric means, etc. There is an art to coding these functions properly. A competitive implementation of a function even as simple as sin(x) is already non-trivial.
Even so, I'm sure that it's faster in principle to evaluate the integral in question in closed form using elliptic functions. It could be hard work to do this right, because the first step is to factor the quartic polynomial under the square root. That already requires either the quartic formula (unfortunately not listed in the GNU Scientific Library even though it has the cubic) or a general polynomial solver (which **is** in GSL but has unclear performance and reliability). The solution also requires elliptic functions with complex arguments, even though the answer is real. It could require careful handling of branch cuts of the elliptic functions, which are multivalued. With all of these caveats, it doesn't seem worth it to work out an explicit formula. The main fact is that there is one, if you have elliptic functions available but not otherwise.
The merit of a numerical integration algorithm such as Gaussian quadrature (or Clenshaw-Curtis, Gauss-Kronrod, etc.) is that it is vastly simpler to code. It won't be as fast, but it should be quite fast if it is coded properly. The only problem is that the integrand becomes singular if it reaches 0, and nearly singular if it is near 0. This makes convergence much slower, although still not as slow as approximation with chords. With special handling of the near-singular points, it should still be fine for high-performance numerical computation. For instance, a polished strategy for numerical integration might well be faster than a clumsy evaluation of the relevant elliptic functions.
| 13 | https://mathoverflow.net/users/1450 | 13796 | 9,307 |
https://mathoverflow.net/questions/11675 | 9 | Given $X$, $Y$ two real Banach spaces, let's say that $(X,\ Y)$
is a *Borsuk pair* if for any continuous mapping $T$ : {$x$ $\in$
$X$ ; $||x||\leq1$} $\rightarrow$ $Y$ s.t. $T$ is **odd** on {$x$
$\in$ $X$ ; $||x||=1$}, it follows that the set $T^{-1}$( { 0 } )
is nonempty.
My conjecture is : $(X,\ Y)$ Borsuk pair $\Longleftrightarrow$
dim $Y$ < $\infty$ and dim $Y$ $\leq$ dim $X$. [Here "dim"
stands for the algebraic dimension; it may be a natural number or $\infty$.]
In other words, there is *no* Borsuk pair with $X$, $Y$ both infinite-dimensional.
Any thoughts ?
[There exist several infinite-dimensional versions of the Classical Borsuk Theorem, e.g.,
when $X$ is reflexive, $Y$ is its dual, and $T$ is a demicontinuous mapping of monotone type.]
[I think the "big" problem (if it truly exists, of course ;-)) would be when both $X$, $Y$ are infinite dimensional,
and the density character dens($X$) $\gg$ dens($Y$), i.e., when $X$ is "huge".]
| https://mathoverflow.net/users/2508 | Borsuk pairs of Banach spaces | Without loss of generality you can think that $Y=\ell^1$ (just pick an infinite sequence of linearly independent unit vectors in $Y$ and multiply them by the coefficients decaying so fast that no non-trivial $\ell^1$ combination will have any chance to be $0$). Now, construct a locally finite cover of $X$ with opposite points identified that has small diameters of the covering sets (every metric space is paracompact). Construct the corresponding continuous partition of unity and lift it back to $X$ to get symmetric continuous locally finite partition of unity. Decouple symmetric open sets consisting of 2 symmetric components. Now, for each open set in the new cover, choose a number so that the opposite open sets intersecting the sphere correspond to opposite numbers and all numbers are non-zero, but otherwise arbitrarily. Multiply the numbers by the partition of unity. Fix also any linear ordering of the symmetric pairs of neighborhoods. All we need now is a continuous mapping $F$ from the metric space of ordered finite sets of numbers with the metric that allows both finitely many small insertions and small perturbations (to find the distance between two sets, you can place their elements in the sequence in the given order adding any number of zeroes anywhere and take the minimum of the $\ell^1$ norm of the difference that can be attained in this way) to $\ell^1$ that is symmetric ($F(-x)=-F(x)$) and never $0$ on for $x\ne 0$. This is a great relief because the space $X$ we want to map is not huge anymore, it is just a fancy metric space of finite sets of numbers. The downside is that we need much more symmetry than before but still not too much ($x$ and $-x$ are still clearly distinguishable though can come arbitrarily close). Now we can raise the level of abstraction again and consider any separable metric space $X$ with an isometric involution $U$ without fixed points and try to map it to $\ell^1$ so that $F(Ux)=-F(x)$. This is trivial. Just create a countable symmetric (with respect to $U$) dense set, for each pair of symmetric points $x,Ux$ take the balls around $x$ and $Ux$ of radius $1/3$ times the distance between them, and construct an antisymmetric function that vanishes outside these 2 balls but not inside them (the plus-minus distance to the boundary, say). Multiply these functions by small numbers and put them in a sequence. You are done.
Of course, the main part of this proof is the paracompactness result. So, if you strongly dislike AC, then the question remains open (but how can you do abstract Banach spaces and dislike AC?)
---
OK, if you so strongly prefer formal language, here is a sequence of claims. For each claim, tell me if you agree with it, have a counterexample, would like to see the proof/reference, or just do not understand what I mean. I'll respond to your reaction.
Claim 1. We can take $Y=\ell^1$
Claim 2. If $Z$ is a separable metric space with an isometric involution $I$ without fixed points, then we can find a continuous maping $F$ from $Z$ to $\ell^1$ such that $F(Iz)=-F(z)$ for all $z\in Z$ and $F(z)\ne 0$ for all $z\in Z$.
Claim 3. Define the equivalence relation on the set of finite or infinite sequences of reals with finitely many non-zero terms as follows. Two sequences are equivalent if two finite sequences obtained by deleting all zero elements in the original sequences (so 0,1,0,0,2,0,3,0,0,0,...) becomes just (1,2,3)) coincide. The space $Z$ of equivalence classes of non-zero sequences with the distance defined as the infimum (actually, minimum) of the $\ell^1$ distances between class representatives is a separable metric space.
Claim 4. The mapping $I$ that maps the class of a sequence $z$ to the class of the sequence $-z$ is well defined and is an isometric involution on $Z$ without fixed points.
Claim 5. Every set can be linearly ordered.
Claim 6. There is a symmetric locally finite covering $\mathcal U$ of the unit ball in $X$ by open sets $U$ of diameter less than $1/10$ each (symmetric means that if $U$ is in the cover, then so is $-U$).
Claim 7. There exists a mapping $G:\mathcal U\to\{\pm 1\}$ such that $G(-U)=-G(U)$ when $U$ intersects the unit sphere.
Define the functions $f\_U(x)=dist(x,X\setminus U)$.
Claim 8. For each point x at least one $f\_U(x)\ne 0$ and one can find a neighborhood $V$ of $x$ and a finite subset $\mathcal V\_x\subset\mathcal U$ such that $f\_U$ is identically zero on $V$ for all $U\notin\mathcal V\_x$.
Claim 9. There exists a linear order $L$ on $\mathcal U$ such that $U\_1<U\_2$ implies $(-U\_1)<(-U\_2)$ when both $U\_1,U\_2\in \mathcal U$ intersect the unit sphere in $X$ and have non-empty intersection.
Claim 10: The mapping $H$ from the unit ball in $X$ to $Z$ that maps every point $x$ to the equivalence class of the sequence $\{G(U)f\_U(x)\}\_{U\in V\_x}$ arranged according to $L$ is well-defined, continuous, and satisfies $H(-x)=I(H(x))$ on the unit sphere.
Claim 11: The composition mapping $F\circ H$ is a continuous mapping from $X$ to $\ell\_1$ that is antisymmetric on the unit sphere and the pre-image of zero under this mapping is empty.
---
Response round 1.
Proof of claim 2. Consider a countable dense set $S$ in $Z$. For each $y\in S$, define $r=r\_y$ to be one third of the distance between $y$ and $Iy$.
Claim 2.1: The open ball $B(Iy,r)$ equals $IB(y,r)$ and these 2 balls are disjoint.
Claim 2.2: The function $f\_y(z)$ defined as $r-d(z,y)$ on $B(y,r)$, as $d(z,Iy)-r)$ on $B(Iy,r)$ and 0 everywhere else is continuous and satisfies $f\_y(Iz)=-f\_y(z)$.
Claim 2.3: For each $z\in Z$ there exists $y\in S$ such that $f\_y(s)\ne 0$.
Claim 2.4. There exists a bijection $\psi:\mathbb N\to S$ and a mapping $\eta:\mathbb N\to(0,+\infty)$ such that the sequence $\eta(n)r\_{\psi(n)}$ belongs to $\ell^1$.
Claim 2.4. The mapping $F$ that maps $z$ to the sequence $\eta(n)f\_{\psi(n)}(z)$ has the properties proclaimed in Claim 2.
Metric space of equivalence classes of sequences:
Definition 3.1. A sequence is a mapping from any linearly ordered set S to reals.
Definition 3.2. A sequence $\psi$ is non-zero and has finitely many non-zero elements if the set of $s\in S$ such that $\psi(s)\ne 0$ is finite and non-empty.
Definition 3.3. Non-zero sequences $\psi\_1:S\_1\to\mathbb R$ and $\psi\_2:S\_2\to\mathbb R$ with finitely many non-zero elements are equivalent if there is an order preserving bijection $\varphi$ between the sets $S'\_j=\{s\in S\_j:\psi\_j(s)\ne 0\}$ such that $\psi\_2\circ \varphi=\psi\_1$.
Claim 3.4. Thus introduced equivalence relation is reflexive, symmetric, and transitive.
Definition 3.4. The distance between equivalence classes $\Psi\_1$ and $\Psi\_2$ is defined as the infimum over all pairs $\psi\_1\in \Psi\_1$ and $\psi\_2\in \Psi\_2$ such that $\psi\_1$ and $\psi\_2$ are defined on the same finite set $S$ of the (finite) sums $\sum\_S|\psi\_1(s)-\psi\_2(s)|$
Claim 3.5. This distance is well-defined and satisfies three standard distance axioms.
Claim 3.6. The metric space thus constructed is separable.
| 14 | https://mathoverflow.net/users/1131 | 13798 | 9,308 |
https://mathoverflow.net/questions/13825 | 1 | I was reading [this paper](http://dmle.cindoc.csic.es/pdf/PUBLICACIONSMATEMATIQUES_1989_33_01_10.pdf) a while ago, and I couldn't figure out how to prove a lemma that was left as an exercise by only using universal properties and the definition of an abelian category.
I'll reproduce the diagram:
$$\ \ \matrix{
&&0&&0&&0
\cr&&\downarrow&&\downarrow&&\downarrow
\cr &&A\_1 & & B\_1& &C\_1
\cr &&\downarrow & &\downarrow&&\downarrow
\cr &&A\_2 & \to & B\_2 & \to & C\_2 &
\cr &&\downarrow &&\downarrow&&\downarrow
\cr 0&\to&A\_3 & \to & B\_3 & \to & C\_3
}\ \ $$
With all rows and columns exact. (This diagram lives in an abelian category).
Show that there exists an exact sequence $A\_1\to B\_1\to C\_1$ making the diagram commute.
Sure, it's not too hard with elements, I mean, it's just part of the snake lemma. However, proving it with universal properties is another story. By the universal property of the kernel, there are natural maps $A\_1 \to B\_1$, and $B\_1 \to C\_1$. Proving that this is exact is another story entirely. I believe I was able to show (I tried this a few months ago) that the top left corner (not counting zeros) is cartesian (a pullback square), but I still couldn't prove exactness.
I repeat, this is for a proof without elements. It should rely only on the definition of an abelian category and universal properties. If you happened not to click the link to the paper, the whole point is a proof *without elements*. I'd really like to see at least one proof in homological algebra actually done from the definition, just because it would be extremely instructive.
| https://mathoverflow.net/users/1353 | The sharp 3x3 lemma: a proof by universal properties? | This should follow from the Salamander lemma, which, as you found out in your [previous MO question,](https://mathoverflow.net/questions/6749/a-proof-of-the-salamander-lemma-without-mitchells-embedding-theorem) can be proved without the use of elements. It is a nice exercise, so I won't spell it out for you.
| 3 | https://mathoverflow.net/users/184 | 13830 | 9,326 |
https://mathoverflow.net/questions/13839 | 14 | On a smooth complex projective variety of $\dim X=n$, we have $n$ complex tori associated to it via $J^k(X)=F^kH^{2k-1}(X,\mathbb{C})/H\_k(X,\mathbb{Z})$ (assuming I've got all the indices right) called the $k$th intermediate Jacobian.
If $k=1$, we have $J^1(X)=H^{1,0}/H\_1$, and so $J^1(X)\cong Jac(X)$ is an abelian variety (the bilinear form is a polarization because it has to be definite on each piece of the Hodge decomposition (I think) ) and is in fact isomorphic as PPAV's to the Jacobian of the variety.
If $k=n$, we have $H^{2n-1,1}/H\_{2n-1}$, which is also a PPAV, and is in fact the Albanese of $X$.
The ones in the middle, however, the "true" intermediate Jacobians, are generally only complex tori. One example of an application is that Clemens and Griffiths proved that cubic threefolds are unirational but not rational using $J^2(X)$ for $X$ a cubic threefold.
So, what information do the intermediate Jacobians contain? I've been told that we don't really know much about that, but what is known, beyond Clemens/Griffiths?
| https://mathoverflow.net/users/622 | What do intermediate Jacobians do? | Probably you know this, but just to be sure: they receive cycle class maps from
codimension $k$ cycles. More precisely, if $Z$ is a cycle on $X$ of codimension $k$
that is cohomologically trivial, then it gives an element in $J^k(X)$. This map,
from cohomologically trivial cycles to $J^k(X)$, is a (generalized) Abel-Jacobia map.
If we let $CH^k(X)$ be the Chow group of codimension $k$ cycles on $X$, then we get
a filtration on $CH^k(X)$, namely $CH^k(X) \supset \text{ cohomologically trivial cycles}
\supset \text{ Albanese trivial cycles}.$ (Albanese trivial means "has trivial image
in $J^k(X)$.) It is conjectured that this filtration can be further extended; this is
part of the Bloch--Beilinson series of conjectures on Chow groups and motives.
On the other other hand, if $X$ is defined over a number field $K$, and we look just
at cycles defined over $K$, then it is conjectured that Albanese trivial implies rationally
trivial (i.e. any cohomologically trivial cycle that is also is in the kernel of themap to $J^k(X)$ (and *is defined over a number field* ) is in fact rationally equivalent to zero). I believe this conjecture is extremely wide open at the moment.
| 14 | https://mathoverflow.net/users/2874 | 13841 | 9,333 |
https://mathoverflow.net/questions/13714 | 24 | Assume $\Gamma$ be a Bieberbach group which acts on $\mathbb R^n$
(i.e. a discrete subgroup of isometries of $n$-dimensional Euclidean
space with a compact fundamental domain).
Denote by $M(\Gamma)$ the number of maximal finite subgroups (up to
conjugation) in $\Gamma$.
Is it true that $M(\Gamma)\le 2^n$?
Things I can do:
There is a simple geometric observation (due to Perelman) which shows
that if $N(\Gamma)$ is the number of orbits of isolated fixed point of
some subgroups of
$\Gamma$ then
$N(\Gamma)\le2^n$.
Clearly, each such point corresponds to a maximal finite subgroup.
Thus, $N(\Gamma)\le M(\Gamma)$, but in all examples I know I still
have $M(\Gamma)\le 2^n$ (and I believe it is allways true).
The formulation is completely algebraic so maybe it has a completely
algebraic solution...
| https://mathoverflow.net/users/3700 | Number of subgroups in a Bieberbach group. | Dima, I can not write a comment (yet) so I will start an answer to my
own question.
You may assume that $\Gamma$ acts by isometries, so $A=\mathbb
R^n/\Gamma$ is an Alexandrov space.
For each maximal subgroup $F$ one can take its fixed point set $S\_F$
in $\mathbb R^n$.
$S\_F$ is an (affine) subspace and image (say $E\_F$) in $\mathbb
R^n/\Gamma$ is a singular set (so called extremal subset of Alexandrov
space). The maximality of $F$ implies that $E\_F$ contains no proper
extremal sets (a smaller subset is fixed by bigger group). (In fact
$E\_F$ is a flat manifold and its has a neighborhood which isometric
for a product $E\_F\times Cone$.)
So the question boils down to finding maximal number of such extremal
sets in $A$.
A particular case of such sets are isolated singular points.
Perelman's theorem states that number of "one-point extremal sets" in
an Alexandrov space with curvature $\ge 0$ is at most $2^n$.
The
proof repeats a proof of Erdős problem: if you have $m$ points in
$\mathbb R^n$ such that all angles in all triangles $\le \pi/2$ then
$m\le 2^n$. We take homothety with coefficient 1/2
for each point, then images of convex hull don't have common internal
points (otherwise it would occur obtuse angle), then
compairing volume of convex hull and its images gives
estimate for number of points.
| 11 | https://mathoverflow.net/users/3700 | 13846 | 9,336 |
https://mathoverflow.net/questions/13843 | 17 | We have a natural number $n>1$. We want to determine whether there exist
natural numbers $a, k>1$ such that $n = a^k$.
Please suggest a polynomial-time algorithm.
| https://mathoverflow.net/users/1735 | How to quickly determine whether a given natural number is a power of another natural number? | This can be done in "essentially linear time." Check out Daniel Bernstein's website: <http://cr.yp.to/arith.html>
Especially note his papers labeled [powers] and [powers2].
| 25 | https://mathoverflow.net/users/3199 | 13852 | 9,339 |
https://mathoverflow.net/questions/13844 | 5 | Suppose we have an arbitrary function $f : \mathbb{R}^2 \to \mathbb{R}$. For any subset $s \subseteq \mathbb{R}^2$, we can define $g\_f(s)$ as the integral\* of $f$ over the region $s$. Suppose further that we have access to an oracle that will tell us the value of $g\_f(s)$ for any $s$.
Now, restrict our attention to subsets of $s$ that are the convex hull of a given subset of points $\bar x\_c \subseteq \{x\_1, \ldots, x\_N \}$ with $x\_i \in \mathbb{R}^2$. Assuming calls to the oracle are O(1), what is the complexity (in terms of $N$) of finding $\bar x\_c^\* = \arg \max\_{\bar x\_c} g\_f(conv(\bar x\_c))$? Is there a known algorithm or reduction to a known problem?
EDIT: \*Previous statement that Scott answered said "average value" here.
| https://mathoverflow.net/users/2785 | Heaviest Convex Polygon | It should be polynomial (probably O(N^3)) in the number of input points using the dynamic programming technique in my paper with Overmars et al, "Finding minimum area k-gons", Disc. Comput. Geom. 7:45-58, 1992, [doi:10.1007/BF02187823](http://dx.doi.org/10.1007/BF02187823).
The idea is: for each three points p,q,r, let W[p,q,r] be the optimal convex polygon that has p as its bottommost point (smallest y-coordinate) and qr and rp as edges. We can calculate W[p,q,r] by looking at all choices of s for which psqr is convex and combining the (previously computed) value W[p,s,q] with the weight of triangle pqr.
As described above this takes time O(N^4) but I think that, for each pair of p and q one can examine the points s and r in the order of the slopes of the lines sq and sr, keeping track of the best s seen so far and using that choice of s for each r in this slope ordering, to reduce the time to O(N^3)
| 5 | https://mathoverflow.net/users/440 | 13855 | 9,341 |
https://mathoverflow.net/questions/13850 | 1 | I was reading this question [link text](https://mathoverflow.net/questions/5772/principal-bundles-representations-and-vector-bundles)
and can't seem to see why, if $\pi: P \to B$ is a principle $G$-bundle and $$\rho:G \to GL\_n(\mathbb{C})$$ is a representation of $G$, then the total space $P \times\_{\rho} \mathbb{C}^n$ is locally trivial.
| https://mathoverflow.net/users/2612 | Local Triviality of an Associated Bundle | A principal G-bundle $\pi: P \to B$ is locally equivalent to a product. Depending on who you ask, this is either part of the definition, or a short lemma. It means that there is a cover of B by open sets U, together with bundle isomorphisms $\alpha\_U: \pi^{-1}U \to G \times U$ that are both G-equivariant, and induce transition sections in G: see [Wikipedia](http://en.wikipedia.org/wiki/Principal_bundle#Trivializations_and_cross_sections).
The associated bundle construction is a quotient of $P \times \mathbb{C}^n$ by an equivalence relation given by the actions of G on the left and right factors. Since $\alpha\_U$ is G-equivariant, the map $\alpha\_u \times id: \pi^{-1}U \times \mathbb{C}^n \to G \times U \times \mathbb{C}^n$ is also a G-equivariant bundle map, so you get bundle isomorphisms on the quotient bundles: $\alpha\_u \times\_\rho id: \pi^{-1}U \times\_\rho \mathbb{C}^n \to U \times \mathbb{C}^n$.
| 3 | https://mathoverflow.net/users/121 | 13858 | 9,342 |
https://mathoverflow.net/questions/13851 | 71 | I have a slight interest in both the inverse Galois problem and in the Monster group. I learned some time ago that all of the sporadic simple groups, with the exception of the Mathieu group $M\_{23}$, have been proven to be Galois groups over $\mathbb{Q}$.
In particular, the Monster group has been proven to be a Galois group over $\mathbb{Q}$. What techniques are used to prove such an assertion?
Is proving that $M\_{23}$ is also Galois over $\mathbb{Q}$ within reach? I assume that the same techniques do not apply, for it is a much more manageable group than the Monster.
| https://mathoverflow.net/users/1079 | The inverse Galois problem and the Monster | the monster is a nice example of how the so-called **rigidity method** for the inverse Galois problem works. There is a lot of beautiful mathematics behind this, I will sketch the different steps.
A general remark: it is known that every profinite group, i.e. every group which could be a Galois group of some field extension, is indeed the Galois group of *some* Galois extension. This is still quite elementary (a result of Leptin, also proved by Waterhouse). To make the inverse Galois question more interesting, we should therefore consider a fixed base field $K$. We can't hope that any profinite group will still be a Galois group over $K$ - every Galois group over $K$ is a quotient of the absolute Galois group of $K$, and therefore the cardinality of a Galois group over $K$ is bounded from above, whereas it is easy to see that there are profinite groups which are "strictly bigger" in cardinality. So a very reasonable question is indeed to ask whether every finite group is a Galois group over some fixed base field $K$. The most natural case is to ask the question for $K = \mathbb{Q}$, but also other base fields can be considered - for example, for $K = \mathbb{C}(t)$ the inverse Galois conjecture is true. In fact, the inverse Galois problems for different base fields $K$ are sometimes closely linked; the method which I will sketch below is a perfect illustration for this, since we will have the consider four different base fields: $\mathbb{C}(t)$, $\overline{\mathbb{Q}}(t)$, $\mathbb{Q}(t)$, and of course $\mathbb{Q}$.
(1) Start with the fact that each finite group $G$ can be realized as a Galois group over $\mathbb{C}(t)$. This follows from the theory of coverings of Riemann-surfaces; if $G$ can be generated by $n - 1$ elements, then we can realize $G$ as a quotient of the fundamental group of the punctured Riemann sphere $\pi\_1^{\text{top}}(\mathbb{P}^1(\mathbb{C}) \setminus \{P\_1,P\_2,\,\cdots,P\_n\})$, where we choose the points $P\_1,P\_2,\,\cdots,P\_n$ to be rational.
(2) We use the theory of the étale fundamental group to get an isomorphism $\pi\_1(\mathbb{P}^1(\mathbb{C}) \setminus \{P\_1,P\_2,\,\cdots,P\_n\}) = \pi\_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P\_1,P\_2,\,\cdots,P\_n\})$, which allows us to realize $G$ as a Galois group over $\overline{\mathbb{Q}}(t)$. [$\pi\_1$ is the étale fundamental group - here the profinite completion of the topological version.]
[Of course, this is already advanced material; see the [Wikipedia article](http://en.wikipedia.org/wiki/%C3%89tale_fundamental_group) for background. For a proper introduction to the theory, there is SGA 1 by Grothendieck; and the recent book "Galois groups and fundamental groups" by Tamas Szamuely is a very gentle introduction (and does all this in detail).]
(3) There exists an exact sequence
$1 \to \pi\_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P\_1,P\_2,\,\cdots,P\_n\}) \to \pi\_1(\mathbb{P}^1(\mathbb{Q}) \setminus \{P\_1,P\_2,\,\cdots,P\_n\}) \to \text{Gal}(\overline{\mathbb{Q}}|\mathbb{Q}) \to 1$
(this is a very fundamental result; see again the books I mentioned) and basically we now want to extend a surjective homomorphism from $\pi\_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P\_1,P\_2,\,\cdots,P\_n\})$ to $G$ to a surjective homomorphism from $\pi\_1(\mathbb{P}^1(\mathbb{Q}) \setminus \{P\_1,P\_2,\,\cdots,P\_n\})$ to $G$. Of course this depends heavily on the structure of the group $G$. This works for many finite simple groups; the construction is quite general, but for particular groups there is always some technical work to do to show that the method applies. In particular *it works for finite groups with a trivial centre, and a rigid system of rational conjugacy classes* - these are quite technical conditions, of course, and I will just state the definitions. An $n$-tuple of conjugacy classes $C\_1,C\_2,\,\cdots,C\_n$ of $G$ is rigid if there exists $(g\_1,g\_2,\,\cdots,g\_n) \in G^n$ such that the $g\_i$ generate $G$, $g\_1g\_2\cdots g\_n = 1$ and $g\_i \in C\_i$, and if moreover $G$ acts transitively on the set of all such $n$-tuples $(g\_1,g\_2,\,\cdots,g\_n)$. A conjugacy class $C$ of $G$ is rational if $g \in C$ implies $g^m \in C$ for all $m$ coprime to the order of $G$. I won't explain why precisely these conditions give you what you want, since it is really technical. Szamuely explains this very clearly. The conditions can be generalized, but that doesn't make it more readable...
[References: section 4.8 in Szamuely's book I mentioned above, and also Serre's wonderful book "Topics in Galois theory", which should maybe be called "Topics in inverse Galois theory" :)]
(4) The previous step allows us to descend from $\overline{\mathbb{Q}}(t)$ to $\mathbb{Q}(t)$, i.e. to realize $G$ as a Galois group of a *regular* extension - another technical notion which I won't explain, but it is not unimportant - of $\mathbb{Q}(t)$. To descend from $\mathbb{Q}(t)$ to $\mathbb{Q}$, there is [Hilbert's irreducibility theorem](http://en.wikipedia.org/wiki/Hilbert's_irreducibility_theorem), or some slight generalization (I don't remember exactly).
According to Thompson, the Monster has a rigid system of three rational conjugacy classes of orders 2, 3 and 29. So the method will apply; of course, I guess that it will be very hard to construct these conjugacy classes, and it is clear that the classification of finite simple groups has played a very big role in these developments. (But I am not a group theorist, so anyone who knows how this works is welcome to give additional information about this construction :))
So I hope this gives you an idea; I wrote this up in a hurry, so suggestions to make this clearer or more coherent (or of course corrections of details which I got wrong) are always welcome.
| 86 | https://mathoverflow.net/users/1107 | 13861 | 9,343 |
https://mathoverflow.net/questions/13577 | 3 | Let $\xi$ be a Poisson Random Measure of intensity $\mu$ (informally $\mathbb E\xi = \mu$).
(For $f \ge 0$, say) when does $\xi f = \infty?$ Kallenberg (Foundations of Modern Probabilility) claims the following:
* $\xi f$ exists iff $\mu(|f| \wedge 1) < \infty$.
It's certainly a basic result on Poisson Random Measure that $\mu(|f| \wedge 1) < \infty$ implies $\xi f < \infty$ a.s.. However, he doesn't actually seem to prove the converse, and I'm not sure what he means by the converse either. It could be either of
* $\mathbb P(\xi f < \infty) = 1$ implies $\mu(|f| \wedge 1) < \infty$
* $\mathbb P(\xi f < \infty) > 0$ implies $\mu(|f| \wedge 1) < \infty$
although perhaps these two are equivalent by Kolmogorov's 0-1 law.
Can anyone shed any light on how to prove the converse, or point me to a better reference for Poisson Random Measure?
| https://mathoverflow.net/users/3676 | Non-existence of integral with respect to Poisson Random Measure | As I mentioned in my comment, you can prove the statement and its converse by looking at the moment generating function. Supposing that f ≥ 0 and λ > 0 is a real number, the following is true for a Poisson point measure ξ with Eξ = μ,
$$\mathbb{E}\left[e^{-\lambda\xi f}\right]=\exp\left(-\mu\left(1-e^{-\lambda f}\right)\right).$$
You can calculate the probability that ξf is finite from this using monotone convergence,
$$\mathbb{P}\left(\xi f \lt \infty\right)=\lim\_{\lambda\downarrow 0}\exp\left(-\mu\left(1-e^{-\lambda f}\right)\right).$$
If μ(f∧1) <∞ then (1-e-λf) ≤ f∧1 for all λ ≤ 1, so dominated convergence gives μ(1-e-λf) → 0 as λ →0. So, E[e-λξf] → 1. This gives ξf < ∞ almost surely.
Now, for the converse statement: If μ(f∧1) = ∞ then, using (1-e-λf) ≥ ½ λ(f∧1) for λ ≤ 1 shows that μ(1-e-λf) = ∞. So, E[e-λξf]=0, giving ξf = ∞ almost surely.
And, yes, the Kolmogorov-zero one law does indeed imply that ξf < ∞ with probability 0 or 1. Splitting the space up into a countable sequence of measurable sets Sn on which both μ and f are bounded, then we want to know if the sum ∑nξ(1Snf) of independent random variables is finite, which is a tail event.
Looking at my copy of Kallenberg I see that the statement you give, and I have just proven above, is Lemma 12.13. Precisely quoting his proof:
>
> If ξ|f| < ∞ a.s. then μ(|f|∧1) < ∞ by Lemma 12.2. The converse implication was established in the proof of the same lemma.
>
>
>
Looking at his Lemma 12.2, it is the statement of the moment generating function which I just used. Like you say, it doesn't seem like he does establish the converse implication at all. However, it is a relatively simple step using my argument above.
| 5 | https://mathoverflow.net/users/1004 | 13878 | 9,349 |
https://mathoverflow.net/questions/13477 | 4 | An affine monoid is a finitely generated commutative submonoid of $\mathbb Z^k$ for some positive integer k. Let S be an affine monoid and let G(S) be the group generated by S. We say the monoid S is normal if and only if for all $g \in G(S)$ and $n \in \mathbb N \setminus \{0\}$, $ng \in S$ implies $g \in S$.
Let $S$ be the submonoid generated by the finite set $T= \{(p\_0,p\_1, \cdots ,p\_{n-1}) \in (\mathbb Z\_{\geq 0})^n: \sum\_{i=0}^{n-1}p\_i=n \,\, and \,\, \sum\_{i=0}^{n-1}i.p\_i \cong 0 \pmod n\}$. Now my question is how to prove that $S$ is normal ?
| https://mathoverflow.net/users/3649 | Normality of an affine semigroup | I worked on this for a bit with Ricky Liu, who came up with this very quick solution:
Take your set T. Suppose $\sum p\_i = kn$, where $k \geq 2$. Create the following set T': let $i$ appear $p\_i$ times. This creates a set $T'$ with at least $2n$ elements (by your first constraint and $k \geq 2$), whose sum is divisible by $n$ by your second constraint.
However, by [Erdos-Ginsberg-Ziv](http://en.wikipedia.org/wiki/Erd%C5%91s-Ginzburg-Ziv_theorem), there's a subset of $n$ elements which add to $n$, which exactly corresponds to your generator, so we're done.
| 4 | https://mathoverflow.net/users/81883 | 13879 | 9,350 |
https://mathoverflow.net/questions/13882 | 20 | Background: When Ueno builds the fully faithful functor from Var/k to Sch/k he mentions that the variety $V$ can be identified with the rational points of $t(V)$ over $k$. I know how to prove this on affine everything and will work out the general case at some future time.
The question that this got me thinking about was if $X$ is a $k$-scheme where $k$ is algebraically closed, then are the $k$-rational points of $X$ just the closed points? This is probably extremely well known, but I can't find it explicitly stated nor can I find a counterexample.
For $k$ not algebraically closed, I can come up with examples where this is not true. So in general is there some relation between the closed points and rational points on schemes (everything over $k$)?
This would give a bit more insight into what this functor does. It takes the variety and makes all the points into closed points of a scheme, then adds the generic points necessary to actually make it a legitimate scheme. General tangential thoughts on this are welcome as well.
| https://mathoverflow.net/users/14672 | Closed vs Rational Points on Schemes | If $k$ is algebraically closed and $X$ is a $k$-scheme locally of finite type, then the $k$-rational points are precisely the closed points. (See EGA 1971, Ch. I, Corollaire 6.5.3).
More generally: if $k$ is a field and $X$ is a $k$-scheme locally of finite type, then $X$ is a Jacobson scheme (i.e. it is quasi-isomorphic to its underlying ultrascheme) and the closed points are precisely the points $x \in X$ such that $\kappa(x)|k$ is a finite extension.
You should also confer the appendix of EGA 1971. There it is shown that for any field $k$ the category of $k$-schemes locally of finite type with morphisms locally of finite type is equivalent to the category of $k$-ultraschemes (a $k$-ultrascheme is locally the maximal spectrum of a $k$-algebra).
| 19 | https://mathoverflow.net/users/717 | 13886 | 9,355 |
https://mathoverflow.net/questions/13897 | 31 | In every group theory textbook I've read, the holomorph has been defined, and maybe a few problems done with it. I've also seen papers focusing on computing Hol($G$) for a specific class of $G$.
One thing I have never seen is any actual use for it. Are there major results using the holomorph of a group? Does it occur in the proof of any useful theorems?
It seems intrisically interesting to me since it allows you to treat automorphisms of a group and elements of a group uniformly, and I would definitely like to learn more about it.
| https://mathoverflow.net/users/2616 | Uses of the holomorph, Hol($G$) = $G \rtimes $ Aut($G$) | If G is abelian, then the holomorph of G is a reasonably nice group. If G is a finite elementary abelian p-group of order pn, then you can consider it to be a vector space over Z/pZ. The automorphism group is the group GL(n,p) of invertible n×n matrices over Z/pZ. The holomorph is called the affine general linear group, AGL(n,p), which can be thought of as (n+1)×(n+1) matrices [ A, v ; 0, 1 ] where A in Aut(G) ≅ GL(n,p) and v in G ≅ (Z/pZ)^n. If you restrict the automorphism group to only include GF(p^k) automorphisms, where k divides n, then you get a subgroup AGL(n/k,p^k) that is also important.
These sorts of groups are (one of) the standard examples of primitive permutation groups. Every soluble primitive permutation group has some minimal normal subgroup G that is elementary abelian, and a maximal subgroup M contained in Aut(G) = GL(n,p) that acts irreducibly on G, and the group itself is then the obvious subgroup { [ A, v ; 0, 1 ] : A in M } of AGL(n,p). Insoluble primitive groups can replace G with a non-abelian simple group or two, but a fair amount of the theory still applies.
These are all examples of the original motivation of the holomorph as the normalizer in the symmetric group of the regular representation of the group. For instance, a Sylow p-subgroup of the symmetric group on p points is regular of order p, and the Sylow normalizer is the holomorph, AGL(1,p).
Regular normal subgroups occur frequently in permutation groups and computational group theory (usually as something to be avoided due to behaving so differently), and their normalizers (aka, the whole group, since the subgroup is normal), are contained in the holomorph.
Primitive soluble groups, and in general, "irreducible" subgroups of AGL(n,p), tend to be important "boundary" examples (as in the boundary of a Schunck class) which do not have a property, but such that every quotient does. "M" is chosen to have the property, and then "G" is taken to be an irreducible M-module such that M⋉G does not have the property. It bugs me to call M the group and G the module, so in the next part M will be the module, and R the ring:
Something similar to a holomorph can be constructed from any module over a ring. You take the matrices [ r, m ; 0, 1 ] where r in R, m in M, and you get another ring where M the module becomes M the ideal; a so-called trivial extension. If instead of all of R, you just take the units of R, GL(1,R), then you get a nice group. For instance, taking R to be the p-adic integers extended by a p'th root of unity z (not already in there), and M to be R, then you get a very important pro-p-group of coclass 1, G = { [ z^i, r ; 0, 1 ] : 0 ≤ i < p, r in R }. For p=2, this is a pro-2 version of the dihedral group, and for all p its finite quotients are "mainline" p-groups of maximal class.
When G is not abelian, many of these comments still apply, but the matrix formulations are usually less enlightening. In general, the holomorph is a very nice setting in which to work with a group G and its automorphisms, with a regular normal subgroup G, with a primitive soluble group, or with various other nice examples.
| 9 | https://mathoverflow.net/users/3710 | 13914 | 9,372 |
https://mathoverflow.net/questions/2757 | 24 | For positive integers $n$ and $L$, denote by $SL\_n(Z,L)$ the level $L$ congruence subgroup of $SL\_n(Z)$, i.e. the kernel of the homomorphism $SL\_n(Z)\rightarrow SL\_n(Z/LZ)$.
For $n$ at least $3$, it is known that $SL\_n(Z,L)$ is normally generated (as a subgroup of $SL\_n(Z)$) by Lth powers of elementary matrices. Indeed, this is essentially equivalent to the congruence subgroup problem for $SL\_n(Z)$.
However, this fails for $SL\_2(Z,L)$ since $SL\_2(Z)$ does not have the congruence subgroup property.
Question : Is there a nice generating set for $SL\_2(Z,L)\ ?$ I'm sure this is in the literature somewhere, but I have not been able to find it.
| https://mathoverflow.net/users/317 | Generators for congruence subgroups of SL_2 | Hi Andy,
I don't know if you are still interested in this, but I just found the reference:
MR0049937 (14,250d)
Grosswald, Emil
On the parabolic generators of the principal congruence subgroups of the modular group.
Amer. J. Math. 74, (1952). 435--443.
It is based on the previous work of H.Frasch (1933) who gave an explicit set of free generators for principal congruence subgroups Gamma(p) in PSL(2,Z), for prime p's.
-Ignat
| 13 | https://mathoverflow.net/users/3738 | 13915 | 9,373 |
https://mathoverflow.net/questions/13909 | 25 | I'm a bit confused by the double role which sheaves play in the theory of stacks.
On the one hand, sheaves on a site are the obvious generalization of a sheaf on a topological space. On the other hand a sheaf on a site is (or better its associated category fibered in sets is) a very particular stack itself, so a generalization of a space. This is not completely confusing: more or less it amounts (I believe) to identifying a space X with the sheaf of continuos functions with values in X.
But now my question is the following. An equivalent condition for a fibered category to be a prestack is that for any two objects (over the same base object), the associated functor of arrows should be a sheaf. In particular this is true for a stack, so for any stack and any two objects in it we have a sheaf, and so a stack (over a comma category).
What is the meaning of this geometrically?
For instance take the stack $\mathcal{M}\_{g,n}$.
Giving two objects in the stack (over the same base object) means giving two families $X$ and $Y$ of stable pointed curves over the same scheme $S$, and the associated functor of arrows maps every other scheme $f \colon T \rightarrow S$ to the set of morphism between $f^\* X$ and $f^\* Y$. How should I think of the associated stack as a space?
To avoid misunderstandings I give the defition of the functor of arrows. Let $\mathcal{F}$ be a fibered category over $\mathcal{C}$. Take $U \in \mathcal{C}$ and $\xi, \eta \in \mathcal{F}(U)$. Then there is a functor $F \colon \mathcal{C}/U \rightarrow Set$ defined as follows. For a map $f \colon T \rightarrow U$ we put $F(f) = Hom(f^\* \xi, f^\* \eta)$. The action on arrows requires some diagrams to be described, but it's really the only possible one.
| https://mathoverflow.net/users/828 | Stacks and sheaves | Let me see if I understand your example correctly: you are fixing $X$ and $Y$, families
of curves over $S$, and now you are considering the functor which maps an $S$-scheme $T$
to the set of $T$-isomorphisms $f^\*X \to f^\*Y$ (where $f$ is the map from $T$ to $S$).
If I have things straight, then this functor shouldn't be so bad to think about, because it is actually representable, by an Isom scheme. In other words, there is an $S$-scheme
$Isom\_S(X,Y)$ whose $T$-valued points, for any $f:T \to S$, are precisely the $T$-isomorphisms
from $f^\*X$ to $f^\*Y$. (One can construct the Isom scheme by looking inside a
certain well-chosen Hilbert scheme.)
One way to think about this geometrically is as follows: one can imagine that two
curves over $k$ (a field) are isomorphic precisely when certain invariants coincide
(e.g. for elliptic curves, the $j$-invariant). (Of course this is a simplification,
and the whole point of the theory of moduli spaces/schemes/stacks is to make it precise,
but it is a helpful intuition.) Now if we have a family $X$ over $S$, these invariants
vary over $S$ to give a collections of functions on $S$ (e.g. a function $j$ in the
genus $1$ case), and similarly with $Y$. Now $X$ and $Y$ will have isomorphic
fibres precisely at those points where the invariants coincide, so if we look
at the subscheme $Z$ of $S$ defined by the coincidence of the invariants,
we expect that $f^\*X$ and $f^\*Y$ will be isomorphic precisely if the map $f$
factors through $Z$. Thus $Z$ is a rough approximation to the Isom scheme.
It is not precisely the Isom scheme, because curves sometimes have non-trivial
automorphisms, and so even if we know that $X\_s$ and $Y\_s$ are isomorphic for
some $s \in S$, they may be isomorphic in more than one way. So actually the
Isom scheme will be some kind of (possibly ramified) finite cover of $Z$.
Of course, if one pursues this line of intuition much more seriously, one will
recover the notions of moduli stack, coarse moduli space, and so on.
Added: The following additional remark might help:
The families $X$ and $Y$ over $S$ correspond to a map $\phi:S \to {\mathcal M}\_g
\times {\mathcal M}\_g$. The stack which maps a $T$-scheme to $Isom\_T(f^\*X,
f^\*Y)$ can then seen to be the fibre product of the map $\phi$ and the diagonal
$\Delta:{\mathcal M}\_g \to {\mathcal M}\_g \times {\mathcal M}\_g$.
In the particular case of ${\mathcal M}\_g$ the fact that this fibre product is representable is part of the condition that ${\mathcal M}\_g$ be an algebraic stack.
But in general, the construction you describe is the construction of a fibre product
with the diagonal. This might help with the geometric picture, and make the relationship to Mike's answer clearer. (For the latter:note that the path space into $X$ has a natural
projection to $X\times X$ (take the two endpoints), and the loop space is the fibre product
of the path space with the diagonal $X\to X\times X$.)
| 21 | https://mathoverflow.net/users/2874 | 13917 | 9,375 |
https://mathoverflow.net/questions/10581 | 9 | There exist a large family of noncommutative spaces that arise from the quantum matrices. These algebraic objects $q$-deform the coordinate rings of certain varieties. For example, take quantum $SU(2)$, this is the algebra $< a,b,c,d >$ quotiented by the ideal generated by
$$
ab−qba, ~~ ac−qca, ~~ bc−cb, ~~ bd−qdb, ~~ cd−qdc, ~~ ad−da−(q−q^{−1})bc,
$$
and the "q-det" relation
$$
ad−qbc−1
$$
where $q$ is some complex number. Clearly, when $q=1$ we get back the coordinate ring of $SU(2)$. In the classical case $S^2 = SU(2)/U(1)$ (the famous Hopf fibration). This generalises to the q-case: the $U(1)$-action generalises to a $U(1)$-coaction with an invariant subalgebra that q-deforms the coordinate algebra of $S^2$ - the famous Podles sphere. There exist such q-matrix deformations of all flag manifolds.
Since all such manifolds are Kahler, we can also apply Kontsevich deformation to them to obtain a q-defomation. My question is: What is the relationship between these two approaches?
Alternatively, we can apply Kostant-Souriau geometric quantization to a flag manifold. How does alegbra relate to its q-matrix deformation?
| https://mathoverflow.net/users/1095 | Kontsevich, and Geometric, Quantization and the Podles sphere | As far as I understand, the flag manifolds with Kahler structures mentioned in the question are simply coadjoint orbits of compact Lie groups with the Kirillov-Kostant-Souriau bracket, so their quantizations will yield quotients of the usual enveloping algebra $U(g)$ and will not have to do with quantum groups. I suppose that the q-spaces discussed in the question are meant to be q-deformations of these.
Here are some papers about this: arXiv:math/0206049 and arXiv:math/9807159. This is a rather subtle business: e.g., it is explained that the 2-parameter deformations (similar to the 2-parameter family of Podles spheres) do not always exist, although they do exist in type A
and in many other cases, e.g. if orbits are symmetric spaces.
| 6 | https://mathoverflow.net/users/3696 | 13920 | 9,377 |
https://mathoverflow.net/questions/9011 | 8 | Defining the translated Gaussians by $f\_t(x)=\exp(-(x-t)^2)$ for $t,x\in\Bbb{R}$, we showed that the linear span of $\{f\_t \mid 0 \le t < \epsilon\}$ is dense in $L^2(\Bbb{R})$, for any $\epsilon>0$. As a consequence, low-frequency trigonometric functions are dense in $L^2([a,b])$. The proof of the first result uses Hermite functions, and the second one follows by taking the Fourier transform of the first. ([Arxiv link](http://arxiv.org/abs/0805.3795).)
Two reviewers told us that these results "must be known" but didn't provide references.
Two questions:
* Are these results folklore?
* What's a good place to look for similar/related results?
| https://mathoverflow.net/users/674 | Approximating with translated Gaussians and low-frequency trig functions | Wiener's approximation theorem says that
Given a function $h: \mathbb{R} \to \mathbb{R}$, the set $\{\sum a\_i h(\cdot - x\_i): a\_i, x\_i \in \mathbb{R}\}$ is dense in $L^2(\mathbb{R})$ if and only if zeros of the Fourier transform of $h$ has zero Lebesgue measure.
See Wiener's book "The Fourier Integral and Certain of Its Applications" or Chandrasekharan's "Classical Fourier Transforms".
Further question when the translating parameters are restricted to a smaller set are considered by a series of authors. See [this paper](https://doi.org/10.1007/BF02384485 "Faxén, B. On approximation by translates and related problems in function theory. Ark. Mat. 19, 271–289 (1981)") for instance and the reference therein.
| 2 | https://mathoverflow.net/users/3736 | 13931 | 9,382 |
https://mathoverflow.net/questions/13934 | 5 | Suppose one wants to color the points in the plane so any two points at distance one apart are different colors. How many colors are needed?
I heard this problem when I was a kid. Back then the most I knew was that 3 is impossible and 7 is possible (tile hexagons of diameter 1-ε). I haven't heard about this problem since then and I don't know how to search for it. Is more known? Is this problem well known in certain circles?
| https://mathoverflow.net/users/27 | Coloring Points in the Plane | This is the [Hadwiger–Nelson problem](http://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem).
| 9 | https://mathoverflow.net/users/1409 | 13936 | 9,385 |
https://mathoverflow.net/questions/13868 | 5 | This question arose after I thought about Ben Webster's comments to [this question](https://mathoverflow.net/questions/10532/).
There he asked me what was my definition of a moduli problem. When I came to think of it, I never saw a precise definition like that. My understanding is along the following lines.
Roughly, say, we want to describe the moduli problem classifying objects of a certain type. In the functorial formulation, we would have a functor
$$Schemes \rightarrow Sets $$
$$ X \mapsto \{ Iso.\ Classes\ of\ some\ objects\ of\ a\ certain\ type\ defined\ over\ X.\}$$
And if this functor is representable, we say that a fine moduli space exists for this moduli problem, and even if it is not, if a certain one-one correspondence between points and objects is true over algebraically closed fields, and if a certain universal property for this is satisfied, then a coarse moduli space exists.
I suppose the above is the agreed standard terminology. Please correct me if I am wrong.
Now, the problem is that in the above definition of a moduli problem, the notion of a "functor classifying isomorphism classes of a certain type of object" is vague. We could have curves with marked points, other types of varieties with extra conditions, bundles, and so on. If we on the other hand relax the criteria and allow just any functor, then the definition becomes too broad,and any scheme will be a fine moduli space for its functor of points.
So is there a better definition, or is this all one can say? Pardon me if this was a stupid question.
| https://mathoverflow.net/users/2938 | Proper definition of a moduli problem | Since not many people have had anything to say, I thought I might make a few remarks. But beware that this is all what I've passively picked up over the years---it's not the result of an actual study of things.
I think the right definition of a moduli problem is a fibered category $p:E\to B$. This should be thought of as a family of categories parametrized by $B$, just like you think of a map $X \to S$ of spaces as being a family of spaces parametrized by $S$. Here is an example: $E$ is the category whose objects are maps of schemes $X\to S$ making $X$ a family of elliptic curves over $S$ (i.e. an abelian scheme of relative dimension 1), and whose maps are the (hopefully) evident cartesian squares; $B$ is the category of schemes, and the functor $E\to B$ sends an object $X\to S$ to $S$. The fibered structure is given by pull back: given $X\to S$ in $E$ and a map $S'\to S$ in $B$, we get the object $X\times\_S S'\to S'$ (which maps under $p$ to $S'$). If we let $E\_S$ denote the fiber of this fibered category over an object $S$, then in this example, $E\_S$ is the category of families of elliptic curves parametrized by $S$. Thus the fibered category encodes the data of all possible families of elliptic curves and how they behave under base change. So I hope my point that the fibered category *is* the moduli problem seems reasonable.
You then say the fibered category is representable if there is an object $U$ of $E$ such that for any $S$ in $B$, the pull back functor from the discrete category consisting of the set $\mathrm{Hom}(S,p(U))$ to the category $E\_S$ is an equivalence. This is pretty unlikely. For instance, it implies that each category $E\_S$ is discrete---all maps are isomorphisms and all automorphisms are the identity. This is certainly not the case with the elliptic curve example. Every elliptic curve has a nontrivial automorphism given by the inverse map with respect to the group structure.
Another version of representability of a fibered category is the following. Let $F:B\to\mathrm{Sets}$ denote the functor which sends an object $S$ to the set of isomorphism classes of objects of $E\_S$. Then the moduli problem is (weakly?) representable if the functor $F$ is representable. This definition is surely weaker in general than the one above, but it is often equivalent in the examples that people look at in algebraic geometry. For example, the two are probably equivalent if each $E\_S$ is a discrete category.
If the moduli problem is not representable, then you get into other issues, such as whether you have effective descent with respect to some topology on $B$ (i.e. $E$ is a "sheaf of categories" over $B$), and if so, stack-theoretic issues, such as whether $E$ can be represented by a category object in $E$, and if so, whether it's a groupoid object.
| 7 | https://mathoverflow.net/users/1114 | 13960 | 9,393 |
https://mathoverflow.net/questions/13251 | 11 | **Background**
Yet another homework inspired question: A scheme is reduced if no section of the structure sheaf is nilpotent. To every scheme $X$ there is a scheme $X\_{red}$ and a morphism $i: X\_{red} \rightarrow X$ such that every morphism from a reduced scheme into $X$ factors through $X\_{red}$. Hartshorne Ex. 2.2.6 guides you through the construction of this scheme. Basically you leave the topological space alone, but you mod out the nilpotents in the structure sheaf. This gives you a presheaf, which you then sheafify to get the structure sheaf of $X\_{red}$.
I have been trying to run through all of the constructions in Hartshorne from a functor of points perspective in addition to the "standard" approach, so naturally I was interested in seeing this construction as well.
From this perspective $X$ is a functor $CRing \rightarrow Sets$, namely $Hom(Spec(-),X)$ if you were using the standard definition of schemes. The functor from $F: CRing \rightarrow CRing$ taking $A$ to $A/nil(A)$ seems relevant here, so it seems natural to ask if $X \circ F: CRing \rightarrow Sets$ is the reduced scheme associated to $X$. I won't spell out the details, but this actually turns out to be true (I think at least!). This brings me to my questions.
**Questions**
Did I mess up, or does the construction above pan out?
Is there a nice characterization of the functors $CRing \rightarrow Cring$ which will give a scheme when composed with any scheme $CRing \rightarrow Sets$?
Even if there is no simple characterization of all such functors, is there a large class of such functors which is nice?
Do you have any other examples of standard constructions in algebraic geometry which are of this form?
| https://mathoverflow.net/users/1106 | Endofunctors of CRing which give schemes when composed with schemes? | Here is a nontrivial example I like. Let $W:\mathrm{Rings}\to\mathrm{Rings}$ denote the Witt vector functor of some fixed finite length. (You can consider the $p$-typical Witt vectors, for some prime $p$, but everything works with the other standard flavors.) Then the functor $W\_\*(-)=-\circ W$ is an endofunctor of the category of functors $\mathrm{Rings}\to\mathrm{Sets}$, and it takes schemes to schemes. The scheme $W\_\*(X)=X\circ W$ is the so-called arithmetic jet space of $X$, extensively studied by Buium in the case of $p$-adic formal schemes. The fiber over $p$ is the Greenberg transform.
There is a standard method for proving $W\_\*$ takes schemes to schemes (or rather for proving almost that), though it probably doesn't work for every functor $F:\mathrm{Rings}\to\mathrm{Rings}$ such that $F\_\*$ takes schemes to schemes. First you show that the category of sheaves of sets on the category of affine schemes w.r.t. the etale topology is stable under $W\_\*$. This is true since $W$ takes etale covers of rings to the same and also takes cocartesian squares of etale rings maps to the same. (These properties of $W$ are not obvious.) Then you show $W\_\*$ takes sheaf epimorphisms to the same, and etale maps of sheaves to the same. (These properties are much easier.) Therefore any etale equivalence relation on an affine scheme is sent to an etale equivalence relation on an affine scheme, and the quotient of the first is sent to the quotient of the second. Therefore the category of quasi-compact quasi-separated algebraic spaces is stable under $W\_\*$. I have no doubt you could find reasonable abstract properties on the endofunctor $W$ of Rings that allow this argument to go through.
Showing $W\_\*$ takes schemes to schemes is a bit subtler. You have to deal with quasi-compactness issues (as a right adjoint, $W\_\*$ doesn't necessarily behave well w.r.t. disjoint unions) and also the fact that it's harder to tell whether a functor is represented by a scheme than by an algebraic space. But as I said, in the Witt vector example above, it is true.
Presumably the same argument works, and is much easier, for the functor $F$ defined by $F(R)=R[t]/(t^{n+1})$. Then $F\_\*(X)$ should be the usual jet space functor of length $n$. The case $n=1$ should give the total space of the tangent bundle, at least when $X$ is smooth.
Edit: I'm reminded below that this example is just a particular case of the representability of the Weil restriction of scalars for a finite flat map $A\to B$. There you consider the endofunctor of the category of $A$-algebras given by $F(R)=B\otimes\_A R$. In particular, it's reasonable to view $W\_\*$ as a generalized Weil resitrction of scalars.
| 8 | https://mathoverflow.net/users/1114 | 13962 | 9,395 |
https://mathoverflow.net/questions/13942 | 11 | Cherry Kearton, Bayer-Fluckiger and others have results that say the monoid of isotopy classes of smooth oriented embeddings of $S^n$ in $S^{n+2}$ is not a free commutative monoid provided $n \geq 3$. The monoid structure I'm referring to is the connect sum of knots.
Bayer-Fluckiger has a result in particular that says you can satisfy these equations $$a+b=a+c, \ \ \ \ b \neq c$$
where $a,b,c$ are isotopy classes of knots and $+$ is connect sum.
When $n=1$ it's an old result of Horst Schubert's that the monoid of knots is free commutative on countably-infinite many generators.
What I'm wondering is, does anyone have an idea of how difficult it might be to compute the structure of the group completion of the monoid of knots, say, for $n \geq 3$? That's not really my question for the forum, though.
It's this: Do people have good examples where it's "easy" to compute the group-completion of a commutative monoid, but for which the monoid itself is still rather mysterious? Meaning, one where rather minimal amounts of information are required to compute the group completion? Presumably there are examples where it's painfully difficult to say anything about the group completion? For example, can it be hard to say if there's torsion in the group completion?
| https://mathoverflow.net/users/1465 | Computing the structure of the group completion of an abelian monoid, how hard can it be? |
>
> Do people have good examples where it's "easy" to compute the group-completion of a commutative monoid, but for which the monoid itself is still rather mysterious?
>
>
>
This happens all the time in K-theory $K^0(X)$, both algebraic and topological. Perhaps it is even the reason that K-theory is a useful tool.
For a striking algebraic example, take $X = \mathbb{A}^n\_k$ where $k$ is a field. Then $K^0(X)$ is the group completion of the commutative monoid $M$ of isomorphism classes of finitely generated projective modules over $R = k[x\_1, \ldots, x\_n]$. In 1955 Serre asked whether every such module was free, i.e., whether $M = \mathbb{N}$. This question became known as [Serre's conjecture](http://en.wikipedia.org/wiki/Quillen-Suslin_theorem). Serre proved in 1957 that every finitely generated projective $R$-module is *stably* free, i.e., $K^0(X) = \mathbb{Z}$. However, it was not until 1976 that Quillen and Suslin independently proved Serre's original conjecture. So between 1957 and 1976, $M$ was an example of a commutative monoid whose group completion was known but which itself was not known. This is only a historical example, because $M = \mathbb{N}$ turns out to be very simple; however, it illustrates the difficulty of the question in general.
A topological example where the commutative monoid is not so simple is given by $KO^0(S^n)$. Let us take $n$ congruent to 3, 5, 6, or 7 modulo 8, so that $KO^0(S^n) = \mathbb{Z}$ by Bott periodicity (the generator being given by the trivial one-dimensional real vector bundle). Let $T$ be the tangent bundle to $S^n$. In $KO^0(S^n)$, of course, the class of $T$ is equal to its dimension $n$. But if we let $M$ be the commutative monoid of isomorphism classes of finite-dimensional real vector bundles on $S^n$ (so that $KO^0(S^n)$ is the group completion of $M$) then the class of $T$ is not equal to the class of the trivial $n$-dimensional vector bundle unless $S^n$ is parallelizable, which only happens when $n$ is equal to (0 or 1 or) 3 or 7. So for all other values of $n$, $M$ is not simply $\mathbb{N}$; there are extra vector bundles which get killed by the group completion process. Understanding these monoids $M$ for all $n$ amounts to understanding the homotopy groups of all the groups $O(m)$, which I expect is not much easier than understanding unstable homotopy groups of spheres.
Finally, Pete's example of the monoid of cardinalities of at most countable sets and its absorbing element also makes an appearance in K-theory; here it is called the [Eilenberg swindle](http://en.wikipedia.org/wiki/Eilenberg_swindle) and it explains why we restrict ourselves to *finitely-generated* projective modules.
| 28 | https://mathoverflow.net/users/126667 | 13979 | 9,401 |
https://mathoverflow.net/questions/13989 | 9 | Suppose $E\_1$ and $E\_2$ are elliptic curves defined over $\mathbb{Q}$.
Now we know that both curves are isomorphic over $\mathbb{C}$ iff
they have the same $j$-invariant.
But $E\_1$ and $E\_2$ could also be isomorphic over a subfield of $\mathbb{C}$.
As is the case for $E$ and its quadratic twist $E\_d$. Now the question general is.
>
> $E\_1$ and $E\_2$ defined over $\mathbb{Q}$ and isomorphic over $\mathbb{C}$. Let $K$
> the smallest subfield of $\mathbb{C}$ such that $E\_1$ and $E\_2$ become isomorphic over $K$.
> What can be said about $K$. Is it always a finite extension of $\mathbb{Q}$. If so, what can be
> said about the extension $K|\mathbb{Q}$.
>
>
>
My second question is something goes something like in the opposite direction. I start again with
quadratic twists. Let $E$ be an elliptic curve over $\mathbb{Q}$ and consider the quadratic extension
$\mathbb{Q}|\mathbb{Q}(\sqrt{d})$. Describe the curves over $\mathbb{Q}$(or isomorphism classes over $\mathbb{Q}$)
which become isomorphic to $E$ over $\mathbb{Q}(\sqrt{d})$. I think the answer is $E$ and $E\_d$.
Again I would like to know what happens if we take a larger extension.
>
> Let $E$ be an elliptic curve over $\mathbb{Q}$ and $K|\mathbb{Q}$ a finite extension.
> Describe the isomorphism classes of elliptic curves over $\mathbb{Q}$ which become isomorphic
> to $E$ over K.
>
>
>
I have no idea what is the right context to answer such questions.
| https://mathoverflow.net/users/3757 | Two questions on isomorphic elliptic curves | Question 1: Putting both curves in say, Legendre Normal Form (or else appealing the lefschetz principle) shows that if the two curves are isomorphic over $\mathbf{C}$ then they are isomorphic over $\overline{\mathbf{Q}}$. Now we could say that for instance $E\_2$ is an element of $H^1(G\_{\overline{Q}}, Isom(E\_1))$ where we let $Isom(E\_1)$ be the group of isomorphisms of $E\_1$ as a curve over $\mathbf{Q}$ (as in Silverman, to distinguish from $Aut(E\_1)$, the automorphisms of $E\_1$ as an *Elliptic Curve* over $\mathbf{Q}$, that is, automorphisms fixing the identity point). However, $E\_2$ is also a principal homogeneous space for a unique curve over $\mathbf{Q}$ with a rational point, which of course has to be $E\_2$, so the cocycle $E\_2$ represents could be taken to have values in $Aut(E\_1)$. Now $Aut(E\_1)$ is well known to be of order 6,4 or 2 depending on whether the $j$-invariant of $E\_1$ is 0, 1728 or anything else, respectively. Moreover the order of the cocycle representing $E\_2$ (which we now see must divide 2, 4 or 6) must be the order of the minimal field extension $K$ over which $E\_1$ is isomorphic to $E\_2$. So $K$ must be degree 2,3,4 or 6 unless I've made an error somewhere.
Question 2: If you restrict your focus to just elliptic curves, yes your idea is right. If it's a quadratic extension, you have exactly 1 non-isomorphic companion. If you have a higher degree number field, you have nothing but composites of the quadratic case unless your elliptic curve has j invariant 0 or 1728.
Notice I am very explicitly using your choice of the word elliptic curve for both of these answers.
| 7 | https://mathoverflow.net/users/3384 | 13997 | 9,413 |
https://mathoverflow.net/questions/13995 | 12 | First of all, I know the concepts of isomorphism and equivalence between categories, and that the latter one is the more interesting one, whereas the first is rather rare and uninteresting.
Are there isomorphisms of categories, which are not trivial and not pathological? I regard the examples on [wikipedia](http://en.wikipedia.org/wiki/Isomorphism_of_categories) as trivial, because these are only reformulations of the definitions of the objects in consideration. Thus perhaps the question is: Are there nontrivial reformulations?
There are lots of nontrivial equivalences of categories (affine schemes <-> rings (dual), compact hausdorff spaces <-> unital commutative C\*-algebras (dual), finite abelian groups <-> finite abelian groups (dual), skeletons such as the algebraic extensions of function fields over fixed prime fields in the category of fields), but I wonder if these categories are actually isomorphic. Of course, in the examples of interest, you can't take the known equivalence as an isomorphism, but perhaps there is another one?
| https://mathoverflow.net/users/2841 | nontrivial isomorphisms of categories | Whether this counts as trivial is a subjective matter, but here goes.
Any adjunction
$$
F: C \to D,\ \ \ G: D \to C
$$
(with $F$ left adjoint to $G$) gives rise canonically to a monad $T = GF$ on $C$ and a "comparison" functor $K: D \to C^T$. Here $C^T$ is the category of algebras for the monad $T$. The adjunction is said to be **monadic** if $K$ is an equivalence of categories.
Now in fact, for most of the obvious examples of monadic adjunctions, the comparison is actually an *isomorphism*. For example, if $G$ is the forgetful functor from groups to sets then it's an isomorphism. The same is true if you replace groups by any other algebraic theory (rings, Lie algebras, etc).
Indeed, if you look in *Categories for the Working Mathematician*, you'll see that Mac Lane calls a functor monadic if $K$ is an *isomorphism*. He does the whole basic theory of monads with this definition. I suspect this is because $K$ really is an isomorphism in the standard examples. *CWM* was published in 1971, and since then it's become clear that Mac Lane's definition was too narrow. Whether the pioneers of monad theory (such as Beck) also used this narrow definition, I don't know.
| 16 | https://mathoverflow.net/users/586 | 14004 | 9,417 |
https://mathoverflow.net/questions/13990 | 19 | Hartshorne's famous conjecture on vector bundles say that any rank $2$ vector bundle over a projective space $\mathbb{P}^n$ with $n\geq 7$ splits into the direct sum of two line bundles.
So my questions are the following:
1) what is an evidence for this conjecture?
2)why is the condition on $n\geq 7$, but not other numbers?
3)any recent survey or reference on this conjecture?
| https://mathoverflow.net/users/2348 | Evidences on Hartshorne's conjecture? References? | This [answer](https://mathoverflow.net/questions/12688/nonsingular-normal-schemes/12689#12689) of mine briefly discusses Hartshorne conjecture and some related questions about smooth subvarieties of $\mathbb P^n$ of small codimensions. It links to Hartshorne's original paper, which I think is still the best source to answer your questions 1) and 2).
As for 3), you can also look at Zolbani's [thesis](https://edocs.uis.edu/mmaji2/www/Research/Dissertation.pdf), which has a lot more details then his research statements mentioned by Steven.
(That's all I know, I would be very interested in what's new about Hartshorne's conjecture as well).
**EDIT**: Today while answering another question I was reminded of a line of research which can be viewed as evidence for Hartshorne's conjecture: smooth subvarieties of small codimension behave cohomologically like complete intersections (this was discussed in Section 2 of Hartshorne original [paper](http://www.ams.org/bull/1974-80-06/S0002-9904-1974-13612-8/)). A paper by [Lyubeznik](http://www.jstor.org/pss/2946619), especially Section 11, has many such results, even for positive characteristic cases. It also includes many relevant references.
| 4 | https://mathoverflow.net/users/2083 | 14005 | 9,418 |
https://mathoverflow.net/questions/14003 | 4 | Consider a scaled sine function, $\sin(2\pi x/2^n)$, for some positive integer $n$. For this, I have the following linear combination.
$$ \sum\_{x=1}^{2^{n-2}} c\_x \sin(2\pi x/2^n).$$
(The upper limit to the sum is $2^{n-2}$.)
The question is whether there exist $c\_x \in \{0, \pm 1, \pm 2\}$, not all $0$, that make the above expression $0$, for infinitely many $n$?
If it helps, the above came up in a computation concerning the discrete Fourier Transform.
| https://mathoverflow.net/users/3760 | A sum involving sines | No, such $c\_x$ don't exist. Even if you replace $\lbrace 0,\pm 1,\pm 2\rbrace$ by $\mathbb Q$, this won't change. In fact, if they would exist, then, using the relation $\displaystyle \sin\frac{2\pi x}{2^n} = \frac{\zeta^x-\zeta^{-x}}{2i}$ (where $\zeta$ is a primitive $2^n$-th root of unity), the equation
$\displaystyle \sum\_{x=1}^{2^{n-2}}c\_x\sin\frac{2\pi x}{2^n}=0$
would rewrite (after multiplication by $i\zeta^{2^{n-2}}$) as a polynomial equation (with rational coefficients!) for $\zeta$ of degree $\leq 2^{n-1}$. But the only (up to a scalar coefficient) polynomial equation of degree $\leq 2^{n-1}$ that $\zeta$ satisfies is $\zeta^{2^{n-1}}+1=0$ (since $X^{2^{n-1}}+1$ is the $2^n$-th cyclotomic polynomial, and the cyclotomic polynomials are known to be irreducible over $\mathbb Q$), and it is easy to see that this is not our equation (in fact, in our equation, $\zeta^{2^{n-1}}$ and $\zeta^0$ must occur with different signs, while in $\zeta^{2^{n-1}}+1=0$ they occur with the same sign).
| 15 | https://mathoverflow.net/users/2530 | 14006 | 9,419 |
https://mathoverflow.net/questions/14001 | 2 | Let $z\_1,\ldots,z\_n$ be complex numbers of modulus one. Does it exist an increasing sequence $k\_j\in\mathbb{N}$ such that $\lim\_{j\to\infty}z\_i^{k\_j}=1$ for all i?
| https://mathoverflow.net/users/1626 | Simultaneous convergence of powers of unit complex numbers | Yes. Let $z=(z\_1,...,z\_n)$, a point on the torus $(S^1)^n$. Since the torus is compact, the sequence $z^m$ has a convergent subsequence $z^{m\_1}$, $z^{m\_2}$,..., and we may choose it so that $k\_j=m\_{j+1}-m\_j$ are nonnegative and increasing. Then ${\rm lim} z\_i^{k\_j}=1$ for all $i$.
| 7 | https://mathoverflow.net/users/3696 | 14008 | 9,420 |
https://mathoverflow.net/questions/14010 | 3 | I've been trying to understand principal bundles, and to that end have been looking at the bundle
$$
\pi: SU(2) \to \mathbb{CP}^1,~~~ (a\_{ij}) \mapsto [a\_{11},a\_{21}],
$$
with fibre $U(1)$. I assumed that the bundle would be trivial over the standard nbds $U\_1,U\_2 \subset \mathbb{C}$, but can't seem to identify the local trivializations. Now
$$
\pi^{-1}(U\_1) = \{\left( \array{a & - \overline{b}\\\
b & \overline{a}} \right)|~ a \neq 0\}, ~~~ \pi^{-1}(U\_2) = \{\left( \array{a & - \overline{b}\\\
b & \overline{a}} \right)|~ b \neq 0\},
$$
and any trivialization $\alpha\_1:\pi^{-1}(U\_i) \to U\_i \times U(1)$, will map
$$
\alpha\_1:\left( \array{a & - \overline{b}\\\
b & \overline{a}} \right) \mapsto ([a,b],h\_{a,b}^1),
$$
for some $h\_{a,b}^1$. Defining $h^1\_{a,b} = arg(a) = \frac{a}{|a|}$, and similarly $h^2$, works, but then the transition functions are not in $U(1)$.
| https://mathoverflow.net/users/2612 | Transition Functions of the Principal Bundle $SU(2) \to \mathbb{CP}^1$ | You have two charts $U\_0=\Bbb C$ and $U\_\infty=(\Bbb C\setminus 0)\cup \infty$.
The transition function on the intersection $\Bbb C\setminus 0$ is $g(z)=z/|z|$.
| 4 | https://mathoverflow.net/users/3696 | 14013 | 9,423 |
https://mathoverflow.net/questions/13965 | -1 | (This is a follow-up question from over there: [Natural models of graphs](https://mathoverflow.net/questions/11647/natural-models-of-graphs).)
(And it *has* a follow-up question over there: [Naturally definable sets of natural numbers (2): Can the circle be broken?](https://mathoverflow.net/questions/14211/naturally-definable-sets-of-natural-numbers-2-can-the-circle-be-broken))
### Motivation
I am looking for a way to syntactically characterize formulas in the first order language of Peano arithmetics that specify a natural or "truely shared" property of a set of natural numbers.
The term *natural* (formula or set) is chosen in contrast to *contingent*, alternatively *ad hoc, mereological* or even *random* (→ Kolmogorov complexity).
The canonical counter-examples are formulas of the form $x = n\_0 \vee x = n\_1 \vee ... \vee x = n\_k$ which prima facie merely list some arbitrary numbers. But of course such formulas are often enough equivalent to "natural" ones, e.g. $x = 2 \vee x = 4 $ is equivalent to $x \geq 1 \wedge x \leq 4 \wedge (\exists y)\ x = 2 \cdot y$ which specifies the natural property of "being even and greater than 0 and less than 5".
Which syntactic conditions on a formula $\Phi(x)$ would do it? Let's try to prohibit literals of the form $x = n\_0$ for some fixed natural number $n\_0$ and everything equivalent to such literals: conjunctions like $x > n\_0-1 \wedge x < n\_0+1$, literals with general terms $t(x) = t\_0$ (with $n\_0$ the unique solution) and so on.
>
> Definition: A formula $\Phi(x)$ is *natural* if there is
> a formula equivalent to it that does not contain subformulas $\phi(x)$ (with $x$ the only free variable) - or (maybe spread) conjunctions of such subformulas - that are equivalent to $x = n\_0$ for some fixed $n\_0$.
>
>
>
A consequence of this definition is that singleton sets can never be defined naturally because every formula that defines a singleton set $\lbrace n\_0 \rbrace$ is equivalent to the formula $x = n\_0$. But that's OK since I want to capture the notion of a "shared" property.
>
> Question #1a: Can anyone show by a simple (!) argument, that every formula with at least two "truth-makers" is natural as defined above (if this is the case)?
>
>
>
This would make my definition useless, since it would not capture what I tried to capture. A tricky argument would be OK (see Question #4).
>
> Question #1b: How could one try to demonstrate that a given formula is **not** natural?
>
>
>
---
>
> Question #2: Is there already
> research or relevant statements about natural
> formulas as defined above? How can the set of natural formulas be characterized otherwise?
>
>
>
Anyway, the following questions arise:
>
> Question #3: Can every finite set of natural
> numbers be defined by a natural formula?
>
>
>
---
>
> Question #4: Can every definable set
> of natural numbers be defined by a
> natural formula?
>
>
>
I.e.: Is every formula natural?
>
> Question #5: What's the classification
> of a definable set in the arithmetical
> hierarchy if only natural formulas are
> allowed?
>
>
>
I guess it might be better to "form(ul)alize" a **degree** of naturality (resp. randomness) like this: "How many literals of the form $m = m\_0$ are minimally needed to define set $M$?" But that's another question...
| https://mathoverflow.net/users/2672 | Naturally definable sets of natural numbers |
>
> Question 3: Can every finite set of natural numbers be defined by a natural formula?
>
>
>
If I understand correctly, the answer is yes: We can always rewrite $x = n\_0 \vee x = n\_1 \vee ... \vee x = n\_k$ in the form $p(x) = q(x)$ for some polynomials $p$ and $q$ with natural number coefficients, which seems to meet your criteria. Is this really the question you want to ask?
| 4 | https://mathoverflow.net/users/126667 | 14020 | 9,429 |
https://mathoverflow.net/questions/14024 | 8 | Algebraic geometry predicts a degree 2 branched cover from an elliptic curve to the projective line. What does this map look like topologically?
| https://mathoverflow.net/users/3238 | Degree 2 branched map from the torus to the sphere | One example: lay your $g$-holed torus $T$ out flat and draw a line the long way through each hole. It hits the torus in $2g + 2$ points. Consider the 180 degree rotation $w$ through that line. Now consider the space $T/w$ formed by identifying two points $P$ and $Q$ if $P = wQ$ (since $w^2 = 1$ we also have $Q = wP$). I claim that it's not too hard to see that $T/w$ is isomorphic to the projective line, and the $2g+2$ points which hit the line are the ramification points.
edit: This is of course more general than you were asking, but the picture is completely general when you're talking about topology.
edit 2: A picture of this (albeit approached from the perspective of starting on the projective line and cutting slits) can be found in section 20e of Fulton's Algebraic Topology book.
| 7 | https://mathoverflow.net/users/3384 | 14027 | 9,433 |
https://mathoverflow.net/questions/14012 | 3 | Let $k$ be a perfect field (so reduced = geometrically reduced) and $f:X\rightarrow \mathrm{Spec}(k)$ a Cohen-Macaulay morphism. Denote by $i:X\_{red}\rightarrow X$ the underlying reduced subscheme and by $\omega\_{X}$ and $\omega\_{X\_{red}}$ the relative dualizing sheaves of $X$ and $X\_{red}$ over $k$. What can one say about the relationship between these two sheaves?
One might hope that there is an "adjunction formula" relating them, but I only know the adjunction formula in the context of a pair of maps $g:Y\rightarrow X$ and $f:X\rightarrow Z$
that are flat, of finite type, and CM, so this doesn't apply to the closed immersion $i:X\_{red}\rightarrow X$ unless $X$ is already reduced (failure of flatness).
Certainly one has a trace morphism $i\_\*: i\_\*\omega\_{X\_{red}}\rightarrow \omega\_X$. Can one describe the image and kernel of $i\_\*$, say in terms of the ideal sheaf defining $i$?
| https://mathoverflow.net/users/2215 | Adjunction for underlying reduced subschemes | Dear Bryden,
Hopefully I have things straight, and there is a general formula $i^!\omega\\_X = \omega\\_{X\\_{red}}$. One then has the functorial isomorphism (of sheaves on $X$)
$RHom\_{\mathcal O\_{X\\_{red}}}({\mathcal F},\omega\_{X\\_{red}}) = Rhom\_{\mathcal O\_X}(i\_\\*{\mathcal F}, \omega\\_X),$
for a coherent sheaf $\mathcal F$ on $X\_{red}$. (Normally we would have to apply
an $Ri\_\*$ to the source of this isomorphism, to put the RHom sheaves on the
same space, and would have to have an $Ri\_\*$ in the formula on the RHS. But
$i\_\*$ is exact, and in fact just identifies sheaves on $X\_{red}$ with sheaves
on $X$ via the identification of their underlying topological spaces.
Now $RHom\_{\mathcal O\_X}(i\_\*{\mathcal F},\omega\_X) = Hom\_{\mathcal O\_X}(i\_\*{\mathcal F}, {\mathcal J}^{\bullet})$,
where ${\mathcal J}^{\bullet}$ is an injective resolution of $\omega\_X$, which
in turn equals $Hom\_{\mathcal O\_{X\_{red}}}(\mathcal F,{\mathcal J}^{\bullet}[\mathcal I]),$
where $\mathcal I$ is the ideal sheaf of $X\_{red}$ in $X$.
Finally, this last complex can be identified with
$RHom\_{\mathcal O\_{X\_{red}}}(\mathcal F, RHom\_{\mathcal O\_X}(\mathcal O\_{X\_{red}},
\omega\_X)).$
So we get the formula
$\omega\_{X\_{red}} = RHom\_{\mathcal O\_X}(O\_{X\_{red}}, \omega\_X).$
(And the derivation shows that this should be valid for any closed immersion,
provided one is in a context where the dualizing complex formalism is satisfied,
except that probably there should be some shifts in dimension in general,
because the dualizing complex probably coincides with the dualizing sheaf place
not in degree 0, but in degree $-dim X$. However, in our case the dimensions
of $X$ and $X\_{red}$ coincide, so this shift can be ignored.)
Note that, as this formula shows, $\omega\_{X\_{red}}$ could be a complex, not
just a sheaf. This is reasonable, I guess; in general, even if $X$ is CM,
this needn't imply that $X\_{red}$ is (I imagine).
If in fact $X\_{red}$ is CM, then I guess we find just one non-zero term in the formula
for $\omega\_{X\_{red}},$ and so have $\omega\_{X\_{red}} = \omega\_X[\mathcal I].$
With a bit of luck, the above is not bogus, and answers your question.
| 3 | https://mathoverflow.net/users/2874 | 14038 | 9,438 |
https://mathoverflow.net/questions/14022 | 4 | This is again a question asked to me by [this user](https://mathoverflow.net/users/3582/). He apparently quit using MO due to a busy time in personal and professional life and resulting difficulties in spending time here with patience. I am taking the liberty to ask it myself(with permission) as I consider him and his questions to be of value.
Let $A$ be unitary ring(ie ring with identity), $a \in A$ be such that for all ring homomorphisms $f : A \rightarrow B$, $B$ a unitary non-zero ring, $f (a)$ is not a unit in $B$.
[a unit in a unitary ring is an element both right and left invertible].
Does it follow that $a$ is nilpotent?
[in particular, $f(a)$ is neither left, nor right invertible for all $f : A \rightarrow B \neq 0$.]
A weaker version may be, if it $a \in A$ is such that $f (a)$ does neither left nor right invertible for all $f : A \rightarrow B \neq 0$ imply that $a$ is a nilpotent element?
| https://mathoverflow.net/users/2938 | Characterizing nilpotents in a ring by a universal property | Let $e \in A$ be a non-zero idempotent (and hence not nilpotent).
Then if $f(e)$ is a unit, we find that $f(e) = 1,$
and so $f(e - 1) = 0.$ Thus if $e - 1$ generates (as a two-sided ideal) the entire
ring, we find that $f$ is identically zero, and hence that $B = 0$.
Thus, if we can find a non-zero idempotent $e \in A$ such that $A(1-e)A = A$,
we have a counterexample.
Note by the way that $f: = 1 - e$ is again idempotent, and so it suffices instead
to find a non-unital idempotent $f$ such that $A f A = A$.
E.g. If $A$ is simple (so that any non-zero two-sided ideal equals $A$), any non-unital and
non-zero idempotent gives a counterexample.
E.g. if $A = M\_2(k)$ for some field $k$, and $f = (1 0 , 0 0)$, we are done. (I think
this is what Kevin intended to write down in his comment.)
| 8 | https://mathoverflow.net/users/2874 | 14041 | 9,440 |
https://mathoverflow.net/questions/14051 | 6 | This question was prompted by [the post here](https://mathoverflow.net/questions/14024/), and I asked this earlier, deleted it, and due to [pressure exerted by Ilya Nikokoshev](http://mathoverflow.tqft.net/discussion/194/), I am asking it again. Apologies to Pavel Etingof.
Q1. Let $\Lambda$ be a lattice in $\mathbb{C}$. We look at the behavior of the zero set of the Weierstrass $\wp$-function for this lattice. By integration around a unit cell for the lattice, we see that the number of poles and zeros are the same. So there has to be two zeros. We position the fundamental domain to be symmetric about $0$, and from the expression for $\wp$, we see that the zero should be $z$ and $-z$ in case they are distinct. Otherwise, it is a double zero, which is one of the $2$-torsion points.
>
> Now, in the case that the zero is not a double zero, can anything be said about its location from the knowledge of $\Lambda$?
>
>
>
Q2. This is stupid. . But, what is the degree of the branched covering $\wp: \mathbb{C}/\Lambda \rightarrow \mathbb{P}^1(\mathbb{C})$? I must confess that I am not good in this stuff.
| https://mathoverflow.net/users/2938 | Zeros of the Weierstrass $\wp$-function | There is an explicit formula for the zeroes: [Eichler, M.; Zagier, D. On the zeros of the Weierstrass $\wp$-function. Math. Ann. 258 (1981/82), no. 4, 399--407. [MR0650945 (83e:10031)](http://www.ams.org/mathscinet-getitem?mr=MR0650945)]
| 6 | https://mathoverflow.net/users/1409 | 14056 | 9,448 |
https://mathoverflow.net/questions/14058 | 11 | I'm interested in the structures of categories like $Rep(GL\_n), Rep(SL\_n)$, etc. of algebraic representations of an algebraic group. I understand that there should be some relation between these and the categories of representations of the corresponding Lie algebras.
However, it's not as intuitive to me what's going on here as with the case of, say, a Lie group, perhaps because the notion of a "tangent vector" is somewhat different.
So, how does one switch between the categories $Rep(G)$ and $Rep(\mathfrak{g})$ for $G$ an algebraic group and $\mathfrak{g}$ its Lie algebra---are there functors in each direction? Can this be used to prove that $Rep(G)$ is semisimple when $G$ is reductive?
In another direction, can the structure of $Rep(\mathfrak{g})$ as known from the representation theory of, say, semisimple Lie algebras give the structure of $Rep(G)$?
| https://mathoverflow.net/users/344 | How do you switch between representations of an algebraic group and its Lie algebra? | If $G$ is semisimple simply connected in characteristic zero, the differential at $1$ gives an equivalence of (tensor) categories $Rep(G)\to Rep({\mathfrak g})$. If $G$ is not semisimple, this is not the case, but this functor is always fully faithful (i.e. an equivalence onto a full subcategory) if $G$ is connected. The essential image of this functor can be described explicitly. Namely, consider the Levi decomposition ${\mathfrak g}={\mathfrak l}\ltimes {\mathfrak u}$, where ${\mathfrak l}$ is reductive and ${\mathfrak u}=Lie(U)$, where $U$ is the unipotent radical of $G$. Then the image is those finite dimensional representations of ${\mathfrak g}$ for which ${\mathfrak u}$ acts nilpotently, and the weights for ${\mathfrak l}$ are integral (i.e. descend to chartacters of the maximal torus).
| 22 | https://mathoverflow.net/users/3696 | 14064 | 9,452 |
https://mathoverflow.net/questions/14054 | 25 | Does there exist anywhere a comprehensive list of small genus modular curves $X\_G$, for G a subgroup of GL(2,Z/(n))$? (say genus <= 2), together with equations? I'm particularly interested in genus one cases, and moreso in split/non-split cartan, with or without normalizers.
Ken Mcmurdy has a list [here](http://phobos.ramapo.edu/~kmcmurdy/research/Models/index.html) for $X\_0(N)$, and Burcu Baran writes down equations for all $X\_{ns}^+(p)$ of genus <=2 in [this](http://www.mat.uniroma2.it/~baran/classlast.pdf) preprint.
| https://mathoverflow.net/users/2 | Where can I find a comprehensive list of equations for small genus modular curves? | No, there does not exist a comprehensive list of equations: the known equations are spread out over several papers, and some people (e.g. Noam Elkies, John Voight; and even me) know equations which have not been published anywhere.
When I have more time, I will give bibliographic data for some of the papers which give lists of some of these equations. Some names of the relevant authors: Ogg, Elkies, Gonzalez, Reichert.
In my opinion, it would be a very worthy service to the number theory community to create an electronic source for information on modular curves (including Shimura curves) of low genus, including genus formulas, gonality, automorphism groups, explicit defining equations...In my absolutely expert opinion (that is, I make and use such computations in my own work, but am not an especially good computational number theorist: i.e., even I can do these calculations, so I know they're not so hard), this is a doable and even rather modest project compared to some related things that are already out there, e.g. William Stein's modular forms databases and John Voight's quaternion algebra packages.
It is possible that it is a little *too* easy for our own good, i.e., there is the sense that you should just do it yourself. But I think that by current standards of what should be communal mathematical knowledge, this is a big waste of a lot of people's time. E.g., by coincidence I just spoke to one of my students, J. Stankewicz, who has spent some time implementing software to enumerate all full Atkin-Lehner quotients of semistable Shimura curves (over Q) with bounded genus. I assigned him this little project on the grounds that it would be nice to have such information, and I think he's learned something from it, but the truth is that there are people who probably already have code to do exactly this and I sort of regret that he's spent so much time reinventing this particular wheel. (Yes, he reads MO, and yes, this is sort of an apology on my behalf.)
Maybe this is a good topic for the coming SAGE days at MSRI?
**Addendum**: Some references:
>
> Kurihara, Akira
> On some examples of equations defining Shimura curves and the Mumford uniformization.
> J. Fac. Sci. Univ. Tokyo Sect. IA Math. 25 (1979), no. 3, 277--300.
>
>
>
$ \ $
>
> Reichert, Markus A. Explicit determination of nontrivial torsion structures of elliptic curves over quadratic number fields. Math. Comp. 46 (1986), no. 174, 637--658.
>
>
>
>
> [http://alpha.math.uga.edu/~pete/Reichert86.pdf](http://alpha.math.uga.edu/%7Epete/Reichert86.pdf)
>
>
>
$ \ $
>
> Gonzàlez Rovira, Josep Equations of hyperelliptic modular curves. Ann. Inst. Fourier (Grenoble) 41 (1991), no. 4, 779--795.
>
>
>
>
> [http://alpha.math.uga.edu/~pete/Gonzalez.pdf](http://alpha.math.uga.edu/%7Epete/Gonzalez.pdf)
>
>
>
$ \ $
>
> Noam Elkies, equations for some hyperelliptic modular curves, early 1990's. [So far as I know, these have never been made publicly available, but if you want to know an equation of a modular curve, try emailing Noam Elkies!]
>
>
>
$ \ $
>
> Elkies, Noam D. Shimura curve computations. Algorithmic number theory (Portland, OR, 1998), 1--47, Lecture Notes in Comput. Sci., 1423, Springer, Berlin, 1998.
>
>
>
>
> <http://arxiv.org/abs/math/0005160>
>
>
>
$ \ $
>
> An algorithm which was used to find explicit defining equations for $X\_1(N)$, $N$ prime, can be found in
>
>
>
>
> Pete L. Clark, Patrick K. Corn and the UGA VIGRE Number Theory Group, *Computation On Elliptic Curves With Complex Multiplication*, preprint.
>
>
>
>
> http://alpha.math.uga.edu/~pete/TorsCompv6.pdf
>
>
>
This is just a first pass. I probably have encountered something like 10 more papers on this subject, and I wasn't familiar with some of the papers that others have mentioned.
| 22 | https://mathoverflow.net/users/1149 | 14075 | 9,461 |
https://mathoverflow.net/questions/14076 | 77 | Let $k$ be a field. I am interested in sufficient criteria for $f \in k[x,y]$ to be irreducible. An example is Theorem A of this [paper](http://matwbn.icm.edu.pl/ksiazki/aa/aa82/aa8237.pdf) (Brindza and Pintér, *On the irreducibility of some polynomials in two variables*, Acta Arith. 1997).
Does anyone know of similar results in the same vein? How about criteria over fields other than the complex numbers?
| https://mathoverflow.net/users/2083 | Irreducibility of polynomials in two variables | A trick which works surprisingly often in my experience: If the Newton polytope of $f$ can not be written as a Minkowski sum of two smaller polytopes, then $f$ is irreducible. I think of this as a generalization of Eisenstein's criterion.
It is surprisingly easy to test whether a lattice polygon in $\mathbb{R}^2$ can be written as a Minkowski sum of smaller lattice polygon. Let $P$ be a lattice polytope. Travel around $\partial P$ and write down the vectors pointing from each lattice point to the next lattice point; call this sequence of vectors $v(P)$. For example, if our polynomial is $a y^2 + b y + c xy + d + e x + f x^2 + g x^3$, with $adg \neq 0$, then the lattice points on the boundary are $(0,2)$, $(0,1)$, $(0,0)$, $(1,0)$, $(2,0)$, $(3,0)$ so $v(P) =(\ (0,-1),\ (0,-1),\ (1,0),\ (1,0),\ (1,0),\ (-3,2)\ )$. (Note that $(1,1)$ is not on the boundary of the triangle.)
It turns out that $v(A + B)$ is simply the sequences $v(A)$ and $v(B)$, interleaved by sorting their slopes . So, if $P$ can be written as the Minkowski sum $A+B$, we must be able to partition $v(P)$ into two disjoint sub-sequences, each of which sums to zero. In the above example, this can't be done, so any polynomial of the form $a y^2 + b y + c xy + d + e x + f x^2 + g x^3$, with $adg \neq 0$ is irreducible.
As an example of a polynomial which could factor, look at $a y^2 + by + c xy + dx + e x^2$. So the boundary is $(0,2)$, $(0,1)$, $(1,0)$, $(2,0)$, $(1,1)$ with $v(P) = (\ (0,-1),\ (1,-1),\ (1,0),\ (-1,1),\ (-1, 1)\ )$. This is the interleaving of $(\ (0,-1),\ (1,0),\ (-1, 1)\ )$ and $(\ (1,-1),\ (-1,1)\ )$, so a polynomial with this Newton polytope could factor.
| 100 | https://mathoverflow.net/users/297 | 14080 | 9,464 |
https://mathoverflow.net/questions/14093 | 13 | Let ${\cal M}\_g$ be the moduli space of smooth complex genus $g$ curves, let ${\cal H}\_g\subset {\cal M}\_g$ be the hyperelliptic locus and set ${{\cal H}}'\_g$ to be the preimage of ${\cal H}\_g$ in the Teichmueller space.
While working on a problem I arrive at two results that can't be reconciled unless ${\cal H}'\_3$ is disconnected.
While it seems a bit strange to me that ${\cal H}'\_3$ should be disconnnected, I don't see why it should't be. So I'd like to ask whether this is known or known to be false.
[sorry, had to cut this into small paragraphs, otherwise the tex part wouldn't show properly.]
| https://mathoverflow.net/users/2349 | Hyperelliptic loci in Teichmueller spaces | That $\mathcal H'\_g$ can't be connected for $g \geq 3$, isn't this just the "lift to Teichmuller space" of the result that the mapping class group isn't the hyperelliptic group?
In particular, the path components of $\mathcal H'\_g$ are indexed by the cosets of the hyperelliptic group in the mapping class group. No?
edit: Anweshi, I'm not sure how you're thinking about Teichmuller space but the answer to your question can be seen in many ways, you don't *have* to use the language of orbifolds, it's just a convienient container. In my mind I suppose I think of a path in Teichmuller space as a motion of the surface -- make this concrete using Fenchel-Nielsen coordinates, for example. So if you have a path that connects one point to another there is an associated diffeomorphism of the surfaces that stretches/twists the metric appropriately and matches up the markings of the surfaces. So if you go between two points in your $\mathcal H\_g'$ covering the same point in $\mathcal H\_g$ the relating diffeomorphism is in the hyperelliptic group (since the hyperelliptic group is a subgroup of the mapping class group). This is how you `see' the cosets of the hyperelliptic group in the mapping class group as indexing $\pi\_0 \mathcal H'\_g$.
| 7 | https://mathoverflow.net/users/1465 | 14095 | 9,474 |
https://mathoverflow.net/questions/13817 | 37 | This is a followup to a previous question
[What is the right definition of the Picard group of a commutative ring?](https://mathoverflow.net/questions/13768/what-is-the-right-definition-of-the-picard-group-of-a-commutative-ring)
where I was worried about the distinction between invertible modules and rank one projective modules over an arbitrary commutative ring. I was worrying too much, because of the following theorem [Bourbaki, Commutative Algebra, Section II.5.2, Theorem 1]:
Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. The following are equivalent:
(i) $M$ is projective.
(ii) $M$ is finitely presented and locally free in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M\_{\mathfrak{p}} \cong R\_{\mathfrak{p}}^{r(\mathfrak{p})}$.
(iii) $M$ is locally free in the weaker sense and its rank function $\mathfrak{p} \mapsto r(\mathfrak{p})$ is locally constant on $\operatorname{Spec}(R)$.
(iv) $M$ is locally free in the stronger sense: there exist $f\_1,\ldots,f\_n \in R$, generating the unit ideal, such that for each $i$, $M\_{f\_i}$ is a free $R\_{f\_i}$-module.
(v) For every maximal ideal $\mathfrak{m}$ of $R$, there exists $f \in R \setminus \mathfrak{m}$ such that $M\_f$ is a free $R\_f$-module.
This answers my previous question, because the rank function of an invertible module is identically one.
In order to really feel like I understand what's going on here, I would like to see an example of a finitely generated locally free [in the weaker sense of (ii) above] module which is *not* projective. Thus $R$ must be non-Noetherian. The wikipedia article on [projective modules](http://en.wikipedia.org/wiki/Projective_module) contains some nice information, in particular sketching an example of such a module over a Boolean ring. For a Boolean ring though the localization at every prime ideal is simply $\mathbb{Z}/2\mathbb{Z}$, so it is not too surprising that there are more locally free modules than projectives.
I would like to see an example with $R$ an integral domain, if possible. It would be especially nice if you can give a reference to one of the standard texts on commutative algebra which contains such an example or at least a citation of such an example.
| https://mathoverflow.net/users/1149 | A finitely generated, locally free module over a domain which is not projective? | It is impossible to produce an example of a finitely generated flat $R$-module that is not projective when $R$ is an integral domain. See: Cartier, "Questions de rationalité des diviseurs en géométrie algébrique," [here](http://www.numdam.org/numdam-bin/item?id=BSMF_1958__86__177_0), Appendice, Lemme 5, p. 249. Also see Bourbaki Algèbre Chapitre X (Algèbre Homologique, "AH") X.169 Exercise Sect. 1, No. 13. I also sketch an alternate proof that there are no such examples for $R$ an integral domain below.
Observe that, for finitely generated $R$-modules $M$, being locally free in the weaker sense is equivalent to being flat [Bourbaki, AC II.3.4 Pr. 15, combined with AH X.169 Exercise Sect. 1, No. 14(c).]. ($R$ doesn't have to be noetherian for this, though many books seem to assume it.)
There's a concrete way to interpret projectivity for finitely generated flat modules. We begin by translating Bourbaki's criterion into the language of invariant factors. For any finitely generated flat $R$-module $M$ and any nonnegative integer $n$, the $n$-th *invariant factor* $I\_n(M)$ is the annihilator of the $n$-th exterior power of $M$.
**Lemma.** (Bourbaki's criterion) A finitely generated flat $R$-module $M$ is projective if and only if, for any nonnegative integer $n$, the set $V(I\_n(M))$ is open in $\mathrm{Spec}(R)$.
This openness translates to finite generation.
**Proposition.** If $M$ is a finitely generated flat $R$-module, then $M$ is projective iff its invariant factors are finitely generated.
**Corollary.** The following conditions are equivalent for a ring $R$: (1) Every flat cyclic $R$-module is projective. (2) Every finitely generated flat $R$-module is projective.
**Corollary.** Over an integral domain $R$, every finitely generated flat $R$-module is projective.
**Corollary.** A flat ideal $I$ of $R$ is projective iff its annihilator is finitely generated.
**Example.** Let me try to give an example of a principal ideal of a ring $R$ that is locally free in the weak sense but not projective. Of course my point is not the nature of this counterexample itself, but rather the way in which one uses the criteria above to produce it.
Let $S:=\bigoplus\_{n=1}^{\infty}\mathbf{F}\_2$, and let $R=\mathbf{Z}[S]$. (The elements of $R$ are thus expressions $\ell+s$, where $\ell\in\mathbf{Z}$ and $s=(s\_1,s\_2,\dots)$ of elements of $\mathbf{F}\_2$ that eventually stabilize at $0$.) Consider the ideal $I=(2+0)$.
I first claim that for any prime ideal $\mathfrak{p}\in\mathrm{Spec}(R)$, the $R\_{\mathfrak{p}}$-module $I\_{\mathfrak{p}}$ is free of rank $0$ or $1$. There are three cases: (1) If $x\notin\mathfrak{p}$, then $I\_{\mathfrak{p}}=R\_{\mathfrak{p}}$. (2) If $x\in\mathfrak{p}$ and $\mathfrak{p}$ does not contain $S$, then $I\_{\mathfrak{p}}=0$. (3) Finally, if both $x\in\mathfrak{p}$ and $S\subset\mathfrak{p}$, then $I\_{\mathfrak{p}}$ is a principal ideal of $R\_{\mathfrak{p}}$ with trivial annihilator.
It remains to show that $I$ is not projective as an $R$-module. But its annihilator is $S$, which is not finitely generated over $R$.
[This answer was reorganized on the recommendation of Pete Clark.]
| 44 | https://mathoverflow.net/users/3049 | 14111 | 9,479 |
https://mathoverflow.net/questions/13784 | 5 | Let $X\_t$ be an ergodic (time-homogeneous) Markov process (in discrete or continuous time) on a finite state space $\{1,\dots,n\}$. Let $T(X\_0)$ be the stopping time given by the infimum of times such that $X$ has covered the space (i.e., for all $j$ with $1 \le j \le n$ there exists some $t\_j \le T(X\_0)$ s.t. $X\_{t\_j} = j$) and $X\_{T(X\_0)} = X\_0$. Clearly $T(X\_0)$ dominates the cover time of $X$. I would expect it to be dominated in turn by the sum of the cover time and the expected [hitting time](http://en.wikipedia.org/wiki/Hitting_time) of $X\_0$ starting from a state chosen w/r/t the invariant distribution $p$.
Define the *recurrence time* as $\sum\_{X\_0} p(X\_0) \cdot \mathbb{E}\_{X\_0} T(X\_0)$, where again the first term is the invariant distribution of $X$.
Now it has been quite a while (early 2000s) since I looked at cover and hitting times, but I recall that while the [fundamental matrix](http://books.google.com/books?id=KF0LgxRCgQsC&pg=PA226) (in discrete time) or the ["deviation matrix"](http://doc.utwente.nl/62339/) (in continuous time) give lots of information about hitting times, computing the cover time is hard. I am aware of the [Matthews bound](http://arxiv.org/abs/math.PR/0005121), but I do not know of a simpler way to compute the cover time than by simulating the chain. In particular, I don't know of an analytical approach to this quantity.
I am in the same situation w/r/t the recurrence time, and it is this quantity that interests me much more than the cover time *per se*. But both are of some interest/utility to me.
So my questions are:
>
> 1. Has the recurrence time (or a similar quantity besides the cover time or first return time) been treated anywhere?
> 2. Are there known analytical results on computing or at least (besides the Matthews
> bound) bounding cover times or
> recurrence times?
>
>
>
| https://mathoverflow.net/users/1847 | Is there a way to analytically compute the recurrence time of a finite Markov process? | This is a response to a comment.
The [coupon collector's problem](http://en.wikipedia.org/wiki/Coupon_collector%27s_problem) is elementary. I don't have a particular scholarly reference in mind, but rather the technique of the proofs. There are a few proofs of the $n H\_n$ expected time to collect all coupons. One possibility is that you can compute the expected time to collect the $k$th new coupon,
$n/(n-k+1)$. That uses a lot of symmetry you don't have for a general Markov process. Here, you have transition probabilities and times on (current location, subset visited so far).
Analogous to what I did [here](https://mathoverflow.net/questions/13171/how-many-trial-picks-expectedly-sufficient-to-cover-a-sample-space/13217#13217), you can use inclusion-exclusion. The expected time to cover everything (with discrete time) is the sum of the probability that you haven't covered everything by time $t-1$, which you can express as
$$\sum\_t \sum\_{S\subset V} -1^{|S|+1}Prob(\{X\_i\}\_{i\lt t}\cap S = \emptyset) $$
where $V$ = $\{1,...,n\}$. You can switch the order of summation to get about $2^n$ analytically solvable problems about avoiding particular subsets.
$$\sum\_{S\subset V} -1^{|S|+1} A(S)$$
where $A(S)$ = expected time before you first enter $S$.
The same holds for continuous time.
| 2 | https://mathoverflow.net/users/2954 | 14122 | 9,488 |
https://mathoverflow.net/questions/13972 | 4 | This question is about a puzzle from the book of Raymond Smullyan: The Lady or the Tiger? The description of the puzzle starts in Chapter 8, p. 103, and here is the important part (copy-pasted from Bruce Ediger's webpage):
Property Q: For any combination x, the combination QxQ is specially related to x.
Property L: If x is specially related to y, then Lx is specially related to Qy.
Property V (the reversal property: If x is specially related to y, then Vx is specially related to the reverse of y.
Property R (the repetition property: If x is specially related to y, then Rx is specially related to yy.
Property Sp: If x is specially related to y, then if x jams the lock, y is neutral, and if x is neutral, then y jams the lock.
Smullyan provides a key of length 10 on p. 163, however, this is not the shortest, I know a key of length 8. Is this the shortest?
Spoiler alert: You can find this key in my comment to the accepted answer.)
| https://mathoverflow.net/users/955 | Shortest Key for the Monte Carlo Lock of Smullyan | Edited in recognition of closed-mindedness.
~~My brute force search shows no keys shorter than 10. Here are the only keys of length 10 and 11 respectively:~~
RVLVQRVLVQ
VRLVQVRLVQ
VLRVQVLRVQQ
VLVRQVLVRQQ
Curiously, there are no keys of length 12.
The only word of length 7 that does not crash under iteration is RRQRRQQ, and it evolves unboundedly. There are 74 words of length 8 that grow to over 30 letters, I think none of them cycles, and there are two eventually cycling words, one you gave and the other its pair RQVRLVQQ. The first time an odd period greater than 1 appears is at length 12, these are the originating words (all end up with period 3):
RQQVLLRLVQQQ
RLQVLLRLVQQQ
RQLVLLLRVQQQ
RQVLLLVLRQQQ
RQLVLLLVRQQQ
RQLVLLLRVLQQ
RQVLLLVLRLQQ
RQLVLLLVRLQQ
RQLLVLLRLVQQ
RQVLLLVQQRQQ
RQVLLLVLQRQQ
RQLVLLLVQRQQ
| 3 | https://mathoverflow.net/users/2368 | 14127 | 9,492 |
https://mathoverflow.net/questions/14133 | 12 | Is there a category (in the category theory sense) of non-well-founded sets (something analogous to Set, the category of sets), and has it been (well-)studied? Any references are appreciated.
| https://mathoverflow.net/users/1015 | Is there a category of non-well-founded sets? | Yes, there is. See Peter Aczel's 1988 notes on models of non-well-founded set theory, [here](http://standish.stanford.edu/pdf/00000056.pdf). The basic idea is that you can model sets as graphs (ie, as a collection of element sets together with a binary relation intuitively suggesting membership), and then the foundation axiom corresponds exactly to the assertion that the membership relation is well-founded. Allowing non-well-founded sets corresponds to allowing arbitrary relations.
These graphs are perfectly ordinary mathematical objects, and so we can collect them into a category exactly the way we do with "normal" sets. The nLab has some notes on this on [their page on pure sets](http://ncatlab.org/nlab/show/pure+set), and also on their page on [material set theory](http://ncatlab.org/nlab/show/set+theory#material_set_theory_4).
| 14 | https://mathoverflow.net/users/1610 | 14135 | 9,497 |
https://mathoverflow.net/questions/13977 | 13 | Suppose that I have a nice variety *X* over ℚp, with good reduction if you like, and a nice sheaf on *X*, say coming from a smooth group scheme *G*. I can cover *X* by some p-adic open sets *Uα*, for example the mod-p neighbourhoods coming from some model $\mathcal{X}$ of *X*. Clearly I can't expect to use Čech cohomology in a naïve way to compute the *Hi(X,G)* in terms of the cohomology of the *Uα*, because they don't overlap. But the information about how they fit together to make *X* is instead contained in the geometry of the special fibre of $\mathcal{X}$.
>
> Is there a spectral sequence which calculates *Hi(X,G)* in terms of some sort
> of cohomology of the *Uα*, and some information about how they
> fit together?
>
>
>
Motivation
----------
In the situation described above, the Leray spectral sequence gives
$$ 0 \to H^1(\mathcal{X},j\_\*G) \to H^1(X,G) \to H^0(\mathcal{X}, R^1 j\_\* G)$$
where $j\colon X \to \mathcal{X}$ is the inclusion of the generic fibre.
So in this situation I can compute *H1(X,G)* in terms of: $R^1 j\_\* G$, which I want to think of as somehow being the cohomology of the p-adic discs covering *X*; and the cohomology of $\mathcal{X}$, which I want to think of as saying how those discs fit together.
I would like to see how to generalise this to smaller p-adic neighbourhoods.
Supplementary question:
>
> Should I just go away and read a book on rigid cohomology?
>
>
>
| https://mathoverflow.net/users/3753 | Is there something like Čech cohomology for p-adic varieties? | The first comment to make is that Cech theory is really extremely general, and can be set up to compute the cohomology of any complex of abelian sheaves on any site (provided you have coverings that are cohomologically trivial). This is explained at least somewhat in SGA4, Expose 5 and EGA III, Chap 0, section 12.
I think you should be working with the rigid analytic space attached to $X$, and not with the $\mathbf{Q}\_p$-points of $X$, and the latter really has no good topology on it besides the totally disconnected one induced from the topology on $\mathbf{Q}\_p$.
Let's assume that $X$ has a model
$\mathcal{X}$ over $\mathbf{Z}\_p$ that is smooth and proper and write $\widehat{\mathcal{X}}$
for the formal completion of $\mathcal{X}$ along its closed fiber. Then the (Berthelot) generic fiber $\widehat{\mathcal{X}}^{rig}$ of $\widehat{\mathcal{X}}$ is a rigid analytic space that is canonically identified with the rigid analytification of $X$ (using properness here). Moreover, one has a "specialization morphism" of ringed sites
$$sp:X^{an}\simeq \widehat{\mathcal{X}}^{rig}\rightarrow \widehat{\mathcal{X}}$$
with the property that for any (Zariski) locally closed subset $W$ of the target, the inverse image $sp^{-1}(W)$ is an admissible open of the rigid space $X^{an}$ (called the open tube over W). In this way, coverings of the special fiber by locally closed subsets give coverings of the rigid generic fiber by admissible opens, and you can use Cech theory with these coverings and or your favorite spectral sequence to compute sheaf cohomology in the rigid analytic world. Again using properness, by rigid GAGA this cohomology agrees with usual (Zariski) cohomology on the scheme $X$ (provided your sheaf is a coherent sheaf of $\mathcal{O}\_X$-modules, say).
This idea of computing cohomology using admissible coverings of the associated rigid space is a really important one as it allows you to use the geometry of the special fiber. It occurs (allowing $\mathcal{X}$ to have semistable reduction) in the work of Gross on companion forms, of Coleman on $\mathcal{L}$-invariants and most prominently in Iovita-Coleman (see their article on "Frobenius and Monodromy operators"). This latter article might be a good place to start.
I would also highly recommend the articles of Berthelot:
<http://perso.univ-rennes1.fr/pierre.berthelot/publis/Cohomologie_Rigide_I.pdf>
<http://perso.univ-rennes1.fr/pierre.berthelot/publis/Finitude.pdf>
I'd also suggest the AWS 2007 notes by Brian Conrad for learning about rigid geometry, which seems generally quite pertinent to your situation.
For etale cohomology of rigid spaces, you might want to look at the article of Berkovich, though this would require learning about his analytic spaces.
In any case, I hope this is a good start.
| 13 | https://mathoverflow.net/users/2215 | 14146 | 9,502 |
https://mathoverflow.net/questions/14151 | 7 | I would like to learn about derived functors.
Which reference do you advise ?
| https://mathoverflow.net/users/2330 | A good place where to learn about derived functors | I have to agree strongly with Ame's answer, in part. Weibel is a great place to go for the formalism. Once you have a little bit of the formalism, though, where to go depends on interests. To really see a lot of the power of derived functors, Hartshorne chapter 3 has some good theorems and exercises using them (though he does bounce back and forth with Cech cohomology...) and Huybrecht's "Fourier Mukai Transforms in Algebraic Geometry" book is very clear (at least to me) in the first few chapters, where he discusses derived categories (Note: I do disagree with Harry, this is a SECOND step, not a first for most) and makes extensive use of them, as he's interested in talking about derived categories, and they're the most natural thing in the universe there.
Also, the following papers might help:
[Fourier Mukai Transforms and Applications to String Theory](http://arxiv.org/abs/math/0412328) talks about how FM transforms (which are compositions of derived functors) help in string theory
[Derived categories for the working mathematician](http://arxiv.org/abs/math/0001045) This is a wonderful paper, very clear about what the derived category is, and might help in conjunction with the books above.
| 7 | https://mathoverflow.net/users/622 | 14155 | 9,509 |
https://mathoverflow.net/questions/14062 | 2 | How do I factorize a polynomial $X^n - 1$ over $\mathbb{F}\_p$? In particular I need to find factors of the polynomial $X^{3^3 - 1} - 1 = X^{26} - 1$ over $\mathbb{F}\_3$.
| https://mathoverflow.net/users/507 | How to factorize X^n - 1 in Z/pZ? | If you just need a quick answer (to decide if something else is going to work how you need), then you can do this with Wolfram|Alpha. Go there:
<http://www.wolframalpha.com/>
and input "factor x^26-1" and press the "equal" button. It'll show some info about the polynomial, including the factors mod 2. In many boxes, there's a link for "Show More". Press the one attached to the factors over GF(2), and it'll show you the factors over GF(3). In this case, you get
$$(x+1) (x+2) (x^3+2 x+1) (x^3+2 x+2) (x^3+x^2+2) (x^3+x^2+x+2) (x^3+x^2+2 x+1) (x^3+2 x^2+1) (x^3+2 x^2+x+1) (x^3+2 x^2+2 x+2).$$
Annoying to have "2" instead of "-1" in GF(3), but that's the price of having a machine do your work for you.
| 10 | https://mathoverflow.net/users/935 | 14156 | 9,510 |
https://mathoverflow.net/questions/14129 | 1 | For $A \in SU(2,C)$, it is clear that $A$ is completly determined by its first row (well any row or column, but let's say first column). In the general $SU(n,C)$-case this is no longer true. In fact, it seems that for every complex $n$-vector of unit norm, there exists a family of matrices for which the vector is the first column. By working with messy simultaneous equations I seem to have shown that every two elements of this family differ by a multiple of $SU(n-1)$ embedded in the bottom right hand corner with 1 in the first entry of the first column and zeros everywhere else. I suspect that there is a bijective correspondence between elements of this family and $SU(n-1)$ but can't prove it.
Can anyone give a simple slick reworking of all of this?
| https://mathoverflow.net/users/3787 | Describing $SU(n,C)$ | Use the fact that matrices act on vectors. $SU(n)$ acts transitively on the space of unit-length vectors; the stabilizer of a point is $SU(n-1)$ by Thorny's argument. For example, for the vector $(1,0,...,0)$ the stabilizer is the subgroup $\left(\begin{matrix}1&0\\\\ 0&A\end{matrix}\right)\approx SU(n-1)$. Now by the [orbit-stabilizer theorem](http://myyn.org/m/article/orbit-stabilizer-theorem/), the space of unit-length vectors is identified with $SU(n)/SU(n-1)$. Fixing one vector $(1,0,...,0)$ fixes this identification, and then each other vector corresponds to a coset $gSU(n-1)$ which is the family you describe.
This isn't really using much about matrices or geometry; I referred to this as the "orbit-stabilizer theorem" above, but it is really just the basic structural feature of group actions. It's certainly something you can understand by yourself; if it's not immediately obvious, you can think about some simpler examples. In the group of permutations $S\_n$, consider the family of permutations that map $1\mapsto 3$ -- how do elements of this family differ from each other? You could also try extending your argument to understand the first column of matrices in $GL(n,\mathbb{R})$; you will of course find a similar answer, but the details are interestingly different. Another fun example is to consider linear functions from $\mathbb{R}\to\mathbb{R}$, and look at the family of linear functions taking $2\mapsto 7$.
| 3 | https://mathoverflow.net/users/250 | 14169 | 9,520 |
https://mathoverflow.net/questions/14177 | 29 | We know that any smooth projective curve can be embedded (closed immersion) in $\mathbb{P}^3$. By definition a projective scheme over $k$ admits an embedding into some $\mathbb{P}^n$. Can we create an upper bound for the $n$ required (perhaps by strengthening the hypotheses) necessary to create an embedding of a smooth projective dimension $k$ scheme into $\mathbb{P}^n$ much like Whitney's theorem tells us we can embed an $n$ dimensional manifold in $\mathbb{R}^{2n}$?
| https://mathoverflow.net/users/3779 | Is there a Whitney theorem type theorem for projective schemes? | Over an algebraically closed field, any projective smooth variety of dimension $n$ can be embedded in $\mathbb P^{2n+1}$. This is elementary and can be found in Shafarevich's Basic Algebraic Geometry, Chapter II, §5.4 .
Of course specific varieties might be embedded in projective spaces of lower dimension.
For an abelian variety however we have a very satisfying complete description of the situation:
For $n=1$, we can embed any abelian (=elliptic) curve in $\mathbb P^{2}$.
For $n=2$, some abelian surfaces but not all of them can be embedded in $\mathbb P^{4}$.
The others can only be embedded in $\mathbb P^{5}$. This is due to Horrocks-Mumford.
For $n\geq 3$, no abelian variety of dimension $n$ can be embedded in $\mathbb P^{2n}$. They can only be embedded in $\mathbb P^{2n+1}$. (This theorem was proved by Van de Ven.)
**Summary and references** (added later) $\quad$ Over an algebraically closed field every projective smooth variety of dimension $n$ can be embedded in $\mathbb P^{2n+1}$. The embedding dimension $2n+1$ is sharp in the sense that for every $n$ there is a projective smooth variety of dimension $n$ not embeddable in $\mathbb P^{2n}$.
[For $n=1$ the sharpness is due to the fact that smooth curves do not embed in $\mathbb P^{2}$ unless their genus is of the form $(d-1)(d-2)/2$. For $n\geq 2$ the sharpness is due to the discussion of abelian varieties above]
G. Horrocks and D. Mumford. A rank 2 vector bundle on P4 with
15,000 symmetries. Topology 12 (1973), 63-81
A. Van de Ven. On the embedding of abelian varieties in projective spaces. Ann. Mat. Pura Appl. (4), 103:127–129, 1975.
| 29 | https://mathoverflow.net/users/450 | 14183 | 9,529 |
https://mathoverflow.net/questions/14186 | 7 | Lemma 1 from [Anderson & Trapp's Shorted Operators, II](http://www.jstor.org/stable/2100462) is
> Let $A$ and $B$ be bounded operators on the Hilbert space $\mathcal H$. The following statements are equivalent:
(1) ran($A$) $\subset$ ran($B$).
(2) $AA^\* \le \lambda^2 BB^\*$ for some $\lambda \ge 0$.
(3) There exists a bounded operator $C$ such that $A = BC$.
Moreover, if (1), (2) and (3) are satisfied, there exists a unique operator $C$ so that ker($A$) = ker($C)$ and ran($C$) $\subset$ closure(ran($B^\*$)).They follow this with the statement, "the lemmas and the original proofs remain valid for operators between two Hilbert spaces."
**Question:**I would like to know if there is a similar statement for more general Banach spaces, and if so, where I might find it.
**My context:** I am considering the Banach space $\Omega = C(U\_1) \times C(U\_2)$ of continuous functions over two domains. I have a covariance operator $$K : \Omega^\* \to \Omega$$
which is decomposed as $$K = \binom{K\_{11} ~ K\_{12}}{K\_{21} ~ K\_{22}}.$$ I want to apply the above lemma to $A = K\_{21}$ and $B = K\_{22}^{1/2}$.
**Edit:** If we have a probability measure $\mathbb P$ on $\Omega$, then continuous linear functionals $\Omega^\*$ are random variables. Thus the expectation $\mathbb Efg$ for $f, g \in \Omega^\*$ is well-defined. The covariance operator is the bilinear form defined by $f(Kg) = \mathbb Efg$.
| https://mathoverflow.net/users/238 | On operator ranges in Hilbert & Banach spaces | (1) does not generally imply (3) for bounded operators between Banach spaces. The first example I have a reference for was due to Douglas and was included in ["Factorization of operators on Banach space"](http://www.jstor.org/stable/2038955) by Embry in 1973. That paper has much more that might interest you, such as the fact that a factorization holds when you have the reverse range inclusion of the adjoints.
See also: <http://www.jstor.org/stable/2043114>
| 5 | https://mathoverflow.net/users/1119 | 14191 | 9,533 |
https://mathoverflow.net/questions/14195 | 0 | (All rings assumed to be commutative and unital)
Given a ring $R$, a prime ideal $\mathfrak{p}$ of $R$, and an extension ring $S$ (the algebra map $R\to S$ is injective), are there any nontrivial sufficient conditions for $rad(\mathfrak{p}S)$ to be prime? How about when $S$ is integral over $R$? How about when it is finitely presented as an $R$-module?
EDIT: Qing Liu's comment means that I should be even more specific.
Assume the third question above ($S$ is finitely presented as an $R$-module.).
Suppose further that there exists an element $f \in R$ such that $R\_f$=$S\_f$. Then even further, suppose $f \in \mathfrak{p}$.
EDIT 2: It is now open season for adding any extra conditions on $S$ that you want. Please don't add stupid conditions like S=R. Note that the condition about $f$ is actually a condition on $S$.
| https://mathoverflow.net/users/1353 | When is the radical of the extension of a prime ideal prime? | Probably Harry has already thought of it like this, but rephrased in terms of algebraic geometry, this question is basically "under what conditions is the preimage of an irreducible variety under a map be irreducible."
When you phrase it like that, it sounds like you might be able to get a statement like "when all maximal prime ideals have prime (up to taking radical) extension" (I think this really does follow over an algebraically closed field by the Nullstellensatz, and some easy topology) but I don't think you'll really do any better than that. This at least gives you some geometric intuition (lots of maps, like smooth one with connected fibers satisfy this condition).
I have no idea how the condition that $f$ exists with $R\_f=S\_f$ is supposed to help. This says that the map is an isomorphism away from the 0 set of f, but $V(\mathfrak{p})$ is in that zero set, so you haven't gained any control.
| 4 | https://mathoverflow.net/users/66 | 14196 | 9,536 |
https://mathoverflow.net/questions/14206 | 4 | Sorry for not knowing the answers to these elementary questions:
Is the property of formulas of the first-order language of Peano arithmetic of "defining a finite set of natural numbers" goedelizable?
If so: Is the set of formulas with this property decidable, semidecidable or non-decidable?
| https://mathoverflow.net/users/2672 | Goedelizability and decidability of a property of Peano formulas | The set of (Gödel codes for) PA provably bounded formulas $\phi(x)$ is computably enumerable (c.e.). By provably bounded, I mean PA $\vdash \exists b\forall x(\phi(x)\to x \leq b)$. Indeed, you can enumerate all consequences of PA and when you find one of the shape $\exists b\forall x(\phi(x)\to x \leq b)$ then enumerate $\phi(x)$.
You can't do better than that since you can easily reduce the halting problem to the decision problem for the set of PA provably bounded formulas. Consider the formula $\phi\_T(x)$ which says "the Turing machine $T$ (with blank input) has not halted after $x$ steps," with the usual arithmetic coding of Turing machines. Then PA proves that $\phi\_T(x)$ is bounded if and only if $T$ truly halts in finite time.
Thus, the set of PA provably bounded formulas is a complete c.e. set.
| 10 | https://mathoverflow.net/users/2000 | 14207 | 9,545 |
https://mathoverflow.net/questions/14214 | 9 | background/motivation
---------------------
let *Ek* denote the modular form of level one and weight *k* with *q*-expansion given by $E\_k(q)=1- \frac{2k}{b\_k}\sum\_n \sigma\_{k-1}(n)q^n$ where σ*i* is the [divisor sum](http://en.wikipedia.org/wiki/Divisor_function) and *bk* is the *k*-th [bernoulli number](http://en.wikipedia.org/wiki/Bernoulli_number). for *k = p – 1* the the denominators of this series don't contain any powers of *p* and hence by the *q*-expansion principle *Ep–1* defines a modular form over ℤ*p*. that is, *Ep–1* ∈ *H*0(*X*0(1), ℤ*p* ⊗ ω⊗k) where *X*0(1) is the compactified modular curve for the full modular group, and ω is the pushforward of the canonical bundle on the universal elliptic curve over ℤ.
now i can use *p*-adic uniformisation to evaluate *Ep–1* via its *q*-expansion at elliptic curves with *j*-invariant satisfying | *j* | > 1; in concrete terms this means formally inverting the power series for 1/j(*q*), and then evaluating \*E*p*–1(*q*) on the result.
i don't suppose one should expect to get a well-defined value in ℂ*p* for *Ep–1* evaluated at elliptic curves for which | *j* | < 1 (as you would for an modular form over ℂ) since this would mean choosing a non-vanishing differential on the universal elliptic curve over ℤ*p*. the *norm* |*Ep–1* |, however, *is* well defined, since any two choices of basis for the canonial bundle on a given elliptic curve can only differ by a unit.
of course the context i am interested in is when |*Ep–1*| is used to define the "overconvergent" region of the modular curve. this is (roughly) defined as the region of the $j$-line satisfying |*Ep–1*|>*r* for |*r*|<1. this region will (always) include part of the |*j*|<1 region. in order to understand what |*Ep–1*|>*r* "means" in concrete terms, i was hoping to numerically compare |*E4*| and |j| for some specific ℂ*p*-values of *j* near 0 (for primes *p* = 5, 7, 11 etc. for which *j* = 0 is supersingular).
anyway, my question is:
>
>
> >
> > how can i *explicitly* evaluate |*Ep–1*| at elliptic curves with | *j* | < 1?
> >
> >
> >
>
>
>
| https://mathoverflow.net/users/2615 | how do you evaluate the p-adic modular form E_p-1 in the region |j|<1 | One has $j = E\_4^3/\Delta$. In the region $|j|\leq 1$, one is parameterizing elliptic curves with good reduction, and so $\Delta$ is a unit. Thus $|j| = |E\_4|^3$. This will help you
when $p = 5$.
When $p = 7,$ one can write $j = 1278 + E\_6^2/\Delta,$ hence $|E\_6|^2 = | j - 1728|$ on
the region $|j| \leq 1$.
For $p = 11$, these sort of explicit computations are harder (but maybe not much; see
the added material below), because there are two supersingular $j$-invariants. But the $p = 5$ and 7 cases will already be illustrative.
In the case when $p = 2$, I wrote something about this once, which appeared in an
appendix an article by Fernando Gouvea in a Park City proceedings volume. A slightly
butchered version (missing figures, among other things) can be found on my
[web-page](http://www.math.northwestern.edu/~emerton/preprints.html) (near
the bottom). You might also look at the papers of Buzzard--Calegari
for related computations, as well as my thesis (available on my web-page) and later
work by Kilford and Buzzard--Kilford. (There is, or at least once was, a cottage industry
based on combining these sorts of explicit computations with some more theoretical
estimates, to compute information about slopes of overconvergent $p$-adic modular forms
for various small primes $p$.)
Added in response to the comment below: For $p = 11$, one has
$E\_{10} = E\_4 E\_6,$
so $E\_{10}^6 = j^2(j-1728)^3 \Delta^5,$ and so when $|j| \leq 1,$ one has
$|E\_{10}|^6 = |j|^2|j-1728|^3.$ Perhaps this will help?
| 5 | https://mathoverflow.net/users/2874 | 14220 | 9,552 |
https://mathoverflow.net/questions/14211 | -1 | (follow-up to: [Naturally definable sets of natural numbers](https://mathoverflow.net/questions/13965/naturally-definable-sets-of-natural-numbers))
Every formula $\Psi(x)$ in the first-order language of Peano arithmetic defines a set of natural numbers. Some of these sets are finite, others are infinite. Every finite set $\lbrace n\_0, n\_1, ..., n\_k \rbrace$ can be defined by an equation $p(x) = q(x)$ with $p(x), q(x)$ finite polynomials in $x$ with natural coefficients. Let in the following $\phi(x)$ be such an equation *[read "phi" for "finite"]*. Infinite sets cannot be described by any $\phi(x)$.
Given a formula $\Omega(x)$ which defines an infinite set *[read "omega" for "infinite"]*. Then every formula of the form $\Omega(x) \vee \phi(x)$ or $\Omega(x)\wedge \neg\phi(x)$ defines an infinite set, too.
The motivation of the following definition is this: A formula defining an infinite set shall be called *arbitrary* if it is derived from a natural (= non-arbitrary) formula by adding or removing finitely many arbitrary elements.
**Definition** (wannabe): A formula $\Omega(x)$ is **arbitrary** iff it defines an infinite set and is equivalent
1. to a formula $\omega(x) \vee \phi(x)$ with $\phi(x) \not\rightarrow \omega(x)\ \ \ \ \ \ \ \ \ \ $ **or**
2. to a formula $\omega(x) \wedge \neg \phi(x)$ with $\omega(x) \not\rightarrow \neg\phi(x)$
where $\omega(x)$ is **not arbitrary**. (Of course, $\omega(x)$ defines an infinite set.)
On first sight, this definition seems circular:
Let $\Omega(x) \equiv \omega(x) \vee \phi(x)$ with $\phi(x) \not\rightarrow \omega(x)$.
Then $\omega(x) \equiv \Omega(x) \wedge \neg\phi'(x)$ with $\Omega(x) \not\rightarrow \neg\phi'(x)$.
Then $\Omega(x)$ is *arbitrary* iff $\omega(x)$ is **not** *arbitrary*.
Might this seemingly vicious circle not be in fact a (hidden) recursive definition (by something like "(abstract) length of formulas")?
Cannot this circle be broken? What about the intuition, that $(\exists y) x = 2 \cdot y$ is a non-arbitrary formula, but that $(\exists y) x = 2 \cdot y \vee x = 17$ is an arbitrary one?
| https://mathoverflow.net/users/2672 | Naturally definable sets of natural numbers (2): Can the circle be broken? | From what I gather from your question, your "natural formulas" would form a complete set of distinct representatives for definable subsets of N under the equivalence relation $X \mathrel{E}\_0 Y$ defined by $|(X \setminus Y) \cup (Y \setminus X)| < \aleph\_0$. Unfortunately, there is no simple way to do this; such a system of distinct representatives necessarily has very high complexity. In fact, finding a system of distinct representative for $E\_0$ over all subsets of N is precisely the same complexity as Vitali's construction of a non-measurable set!
In your question, you talk about formulas rather than the sets they define, but this is not much different (the difference is known as "lightface vs boldface" in the literature). Problems of this type are extensively studied in Descriptive Set Theory, under the heading of *Borel Equivalence Relations*. The relation $E\_0$ is in a very precise sense the simplest Borel equivalence for which has no simple system of distinct representatives, it thus plays a very important role in this theory.
A good place to get started with Descriptive Set Theory is Kechris's *Classical Descriptive Set Theory*. For your particular problem, you want to look at the "lightface theory," for which the standard reference is Moschovakis's *Descriptive Set Theory*. There are a few good surveys and books on Borel equivalence relations, but most require a fair amount of familiarity with the subject.
| 9 | https://mathoverflow.net/users/2000 | 14221 | 9,553 |
https://mathoverflow.net/questions/14210 | 2 | Given a ring $R$, a prime ideal $\mathfrak{p}$ of $R$, and an extension ring $S$ (the algebra map $R\to S$ is injective), are there any nontrivial sufficient conditions for the induced map $Spec(S) \to Spec(R)$ to be injective as a map of topological spaces (completely disregarding any of our sheaf structure)? That is, every prime that lifts lifts to a unique prime? If this problem is intractable as-is, I can add more conditions, so please don't add answers unless they are actual answers.
| https://mathoverflow.net/users/1353 | When do primes lift uniquely (provided they lift at all)? | Rather than listing various non-trivial sufficient conditions, let me give you a reference. Search EGA for the word "radiciel" (or start reading at Definition 3.5.4 of EGA I). A morphism of schemes $X\to Y$ is said to be *radiciel* (or *universally injective*) if for every field $K$, the induced map on $K$-points $X(K)\to Y(K)$ is injective. This generalizes the notion of a purely inseparable field extension.
| 2 | https://mathoverflow.net/users/1 | 14222 | 9,554 |
https://mathoverflow.net/questions/14007 | 5 | Hi,
Given a $\sigma$-unital $C^\*$-algebra $A$ and a full Hilbert $A$-module $E$, is it possible to find an approximate unit $ \{\epsilon\_i\}, i\in I$ in $A$ such that each $\epsilon\_i$ is of the form $< e\_i,e\_i>\_A$, where $e\_i \in E, \forall i\in I$? If not, what are the conditions on $E$ and $A$ for which this might be possible?
Thanks in advance.
| https://mathoverflow.net/users/3494 | Hilbert $C^*$-modules and approximate units | No, there is not always such an approximate unit. (This will be easier to formulate in terms of left modules, and with inner products linear in the first entry. The warning seems necessary due to the common convention in C\*-module theory to do the opposite.)
### Example
Let $A=B(\mathbb{C}^2)$ (linear operators), $E=\mathbb{C}^2$, with the module action given by operators acting on vectors (on the left) and the $A$-valued inner product of $x$ and $y$ in $E$ given by $<x,y>\_A(z)=<z,y>\_\mathbb{C}x$. Then $<x,y>\_A$ has rank at most one for all $x$ and $y$, so no such approximate identity exists. $\square$
---
Any finite dimensional Hilbert space with dimension at least 2 would give a slight modification of this example. Or, let $E=H$ be a separable, infinite dimensional Hilbert space, and let $A=\mathcal{K}(H)$ be the algebra of compact operators on $H$. (**Added**: Note that fullness follows from the fact that the span of the range of the inner product is the set of finite rank operators.) Or, given a C\*-algebra $B$, one could form $H\_B=B\oplus B\oplus\ldots$ with its usual right $B$-module structure and consider the analogous construction with $A=\mathcal{K}(H\_B)$ and $E=H\_B$. If $B$ is $\sigma$-unital, then so is $A$, but there will be no approximate identity of the desired form.
I do not have anything useful to say about formulating sufficient conditions for such an approximate identity to exist, but this simple example shows that lack of existence is common.
| 2 | https://mathoverflow.net/users/1119 | 14227 | 9,559 |
https://mathoverflow.net/questions/14219 | 41 | I'm looking for a "conceptual" explanation to the question in the title. The standard proofs that I've seen go as follows: use the Schubert cell decomposition to get a basis for cohomology and show that the special Schubert classes satisfy Pieri's formula. Then use the fact that basic homogeneous symmetric functions $h\_n$ are algebraically independent generators of the ring of symmetric functions, to get a surjective homomorphism from the ring of symmetric functions to the cohomology ring. Pieri's rule can be shown with some calculations, but is there any reason a priori to believe that tensor product multiplicities for the general linear group should have anything at all to do with the Grassmannian?
Maybe a more specific question: is it possible to prove that these coefficients are the same without calculating them beforehand?
One motivation for asking is that the cohomology ring of the type B and type C Grassmannians ${\bf OGr}(n,2n+1)$ and ${\bf IGr}(n,2n)$ are described by (modified) Schur P- and Q-functions which seem to have nothing(?) to do with the representation theory of the orthogonal and symplectic groups ${\bf SO}(2n+1)$ and ${\bf Sp}(2n)$. So as far as I can tell the answer isn't just because general linear groups and Grassmannians are "type A" objects.
| https://mathoverflow.net/users/321 | Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? | There are several rings-with-bases to get straight here. I'll explain that, then describe three serious connections (not just ~~Ehresmann's~~ Lesieur's proof as recounted in the OP).
The wrong one is $Rep(GL\_d)$, whose basis is indexed by decreasing sequences in ${\mathbb Z}^d$.
That has a subring $Rep(M\_d)$, representations of the Lie monoid of *all* $d\times d$ matrices, whose basis is indexed by decreasing sequences in ${\mathbb N}^d$, or partitions with at most $d$ rows.
That is a quotient of $Rep({\bf Vec})$, the Grothendieck ring of algebraic endofunctors of ${\bf Vec}$, whose basis (coming from Schur functors) is indexed by all partitions. Obviously any such functor will restrict to a rep of $M\_d$ (not just $GL\_d$); what's amazing is that the irreps either restrict to $0$ (if they have too many rows) or again to irreps!
1. Harry Tamvakis' proof is to define a natural ring homomorphism $Rep({\bf Vec}) \to H^\*(Gr(d,\infty))$, applying a functor to the tautological vector bundle, then doing a Chern-Weil trick to obtain a cohomology class. (It's not just the Euler class of the resulting huge vector bundle.)
The Chern-Weil theorem is essentially the statement that Harry's map takes alternating powers to special Schubert classes. So then it must do the right thing, but to know that he essentially repeats the Ehresmann proof.
2. Kostant studied $H^\* (G/P)$ in general, in "Lie algebra cohomology and
generalized Schubert cells", by passing to the compact
picture $H^\* (K/L)$, then to de Rham cohomology, then taking $K$-invariant
forms, which means $L$-invariant forms on the tangent space $Lie(K)/Lie(L)$.
Then he complexifies that space to $Lie(G)/Lie(L\_C)$, and identifies that
with $n\_+ \oplus n\_-$, where $n\_+$ is the nilpotent radical of $Lie(P)$.
Therefore forms on that space is $Alt^\* (n\_+) \otimes Alt^\* (n\_-)$.
Now, there are two things left to do to relate this space to $H^\* (G/P)$. One is to take cohomology of this complex (which is hard, but he describes the differential), and the other
is to take $L$-invariants as I said. Luckily those commute. Kostant degenerates the differential so as to make sense on each factor separately (at the cost of not quite getting $H^\* (G/P)$).
Theorem: (1) Once you take cohomology, $Alt^\* (n\_+)$ is a multiplicity-free $L$-representation. So when you tensor it with its dual and take $L$-invariants, you get a canonical basis by Schur's lemma. (2) This basis is the degeneration of the Schubert basis.
Theorem: (1) If $P$ is (co?)minuscule, the differential is zero, so you can skip the take-cohomology step. That is, $Alt^\* (n\_+)$ is already a multiplicity-free $L$-rep. The Schur's lemma basis has structure constants coming from representation theory. (2) In the Grassmannian case, the degeneration doesn't actually affect the answer, so the product of Schubert classes does indeed come from representation theory.
I believe the degenerate product on $H^\*(G/P)$ is exactly the one described by [Belkale-Kumar].
It's fun to see what's going on in the Grassmannian case -- $L = U(d) \times U(n-d)$, $n\_+ = M\_{d,n-d}$, and $Alt^\* (n\_+)$ contains each partition (or rather, the $U(d)$-irrep corresponding) fitting inside that rectangle tensor its transpose (or rather, the $U(n-d)$-irrep).
I think this is going to be the closest to what you want, for other groups' Grassmannians.
3. (No, 3. Silly site software!)
Belkale has [the best (least decategorified) proof](https://arxiv.org/abs/math/0208107v2) I've seen (which appeared separately as \_Invariant theory of $GL(n)$ and intersection theory of Grassmannians \_ IMRN **2004** Issue 69 (2004) 3709–3721, <https://doi.org/10.1155/S107379280414155X>). He takes three Schubert cycles meeting transversely, and for each point of intersection, constructs an actual invariant vector inside the corresponding triple product of representations. The set of such vectors is then a basis.
| 31 | https://mathoverflow.net/users/391 | 14231 | 9,563 |
https://mathoverflow.net/questions/13593 | 5 | Is it true that any monomorphism of commutative Hopf algebras over a field is injective? Moreover, is it true that any epimorphism of commutative Hopf algebras over a field is surjective?
| https://mathoverflow.net/users/3824 | Is every monomorphism of commutative Hopf algebras (over a field) injective? | It seems that Ben is nevertheless right that the answer to the first question is "NO". Let $G=SL(2,\Bbb C)$, and $B$ be the subgroup of lower triangular matrices. Then the inclusion $B\to G$ is an epi, since every algebraic representation of $B$ that extends to $G$ does so uniquely (on the nose, not just up to an isomorphism!). This follows from the fact that in any finite dimensional representation $V$ of the Lie algebra $sl(2)$, the operator $e$ is determined by $f$ and $h$. Indeed, the kernel $K$ of $e$ is spanned by vectors $v$ satisfying $hv=mv$ and $f^{m+1}v=0$ for some integer $m\ge 0$, and since $V=\Bbb C[f]K$, the operator $e$ on $V$ is uniquely determined.
So one might guess that a morphism of complex affine algebraic groups $\phi: H\to G$ is an epi if and only if $G/\phi(H)$ is connected and proper (but I did not check this).
| 2 | https://mathoverflow.net/users/3696 | 14234 | 9,566 |
https://mathoverflow.net/questions/14235 | 0 | By Heisenberg group I mean the group with presentation $H$ generated by $x$ and $y$ such that $x$ and $y$ commute with $xyx^{-1}y^{-1}$. Is there an infinite chain of subgroups
$H > H\_1 > H\_2 > \dots$ such that the index $[H\_i: H\_{i+1}]< n$ for some $n\ ?$ Thanks
| https://mathoverflow.net/users/3804 | decreasing chain of subgroups in the Heisenberg group | Assuming this *is* the discrete Heisenberg group $H=H\_3({\mathbb Z})$, as in my comment above, then here is another way of looking at Mariano's answer (I think). Take any sequence of positive integers
$n\_1 < n\_2 < \dots $ where $n\_i \vert n\_{i+1}$ for all $i$, and put
$$ H\_i = H\_3(n\_i{\mathbb Z}) $$
(Mariano's answer corresponds to taking $n\_i = 2^i$.)
| 4 | https://mathoverflow.net/users/763 | 14240 | 9,569 |
https://mathoverflow.net/questions/14257 | 5 | Forgive me for this naive question.
We consider the following lemma and its proof in Lang's algebra, Third Ed., published 1999, Chap. 20, section 4, [page 784](https://books.google.com/books?id=Fge-BwqhqIYC&pg=PA784).
>
> Every module is a submodule of an injective module.
>
>
>
For a module $M$, Lang defines its dual to be $M$^ = $Hom(M, \mathbb{Q}/\mathbb{Z})$.
This "dual", if anything, is an algebraic version of the Pontrjagin dual of a topological group $G$, which is $Hom(G, \bf{S}^1)$, where we consider the continuous homomorphisms. $\mathbb{Q}/\mathbb{Z}$ is the torsion part of $\bf{S}^1$, and it seems this restriction is done for preventing too big a dual.
The proof goes on and shows that any module is a submodule of an injective module.
However, is there any conceptual explanation of why this works, and why this mysterious appearance of a dual which looks so much like the Pontrjagin dual?
| https://mathoverflow.net/users/2938 | Injective modules and Pontrjagin duals | Why this works? Because $\mathbb{Q}/\mathbb{Z}$ is an [injective cogenerator](https://en.wikipedia.org/wiki/Injective_cogenerator) in the category of abelian groups.
The mysterious appearance can be explained if you look at it from a more general perspective. This would be Morita duality, which was motivated by classical Pontryagin duality (locally compact Hausdorff topological groups), and it replaces the role of the circle group with an injective cogenerator.
In the category Set of sets and mappings, for example, the two point set $\{0,1\}$ is an injective cogenerator. The category Gr of groups and group homomorphisms on the other hand doesn't have any cogenerator.
| 16 | https://mathoverflow.net/users/2384 | 14260 | 9,580 |
https://mathoverflow.net/questions/14264 | 0 | The integral I am interested in is:
$$t(x)=\int\_{-K}^{K}\frac{\exp(ixy)}{1+y^{2q}}dy$$
$K<\infty$, q natural number
For q=1 one can use contour integration.
So for K>1 we have :
$$\pi/2-\int\_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy $$
Where Arc has radius $K$
Is it correct that for K<1 this integral is:
$$-\int\_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy ?$$
What about K=1?
| https://mathoverflow.net/users/3589 | An integral arising in statistics(2) | For $K=1$ your arc passes through the pole of the function $\frac{exp(ixy)}{1+y^2} = \frac{exp(ixy)}{2i}\left(\frac{1}{y-i}-\frac{1}{y+i}\right)$, so you don't get a sensible value (the discontinuity of the integrand is asymptotically $\frac{c}{t}$ for $t$ around $0$).
| 1 | https://mathoverflow.net/users/2368 | 14265 | 9,582 |
https://mathoverflow.net/questions/14237 | 8 | Any finitely presented group naturally gives rise to an edge-labeled graph (the Cayley graph) and I am considering paths through this graph. Paths correspond to infinite sequences of generators, so we can ask questions about the recursiveness (or not) of these sequences\*. For example, is there a group where any recursive sequence of generators necessarily corresponds to a self-intersecting path?
Has anyone seen any questions like this answered anywhere? Any good resources for the recursive properties of finitely presented groups? Maybe just an example or two of a really badly behaved (in a recursion-theoretic sense) group?
\*To alleviate any potential confusion, an infinite sequence, {s\_i} is recursive if there is a Turing Machine that, on input n, produces s\_n.
| https://mathoverflow.net/users/3806 | How bad can the recursive properties of finitely presented groups be? | Your question is very interesting. I don't have a complete
answer.
First, let me note that in a finitely presented group, the
Cayley graph itself may not be a decidable graph, since to
know whether or not a node in the graph, which is a word in
the presentation, is trivial or not amounts exactly to the
word problem for that presentation. And since there are
finitely presented groups having an undecidable word
problem, there are finitely presented groups having an
undecidable Cayley graph.
This suggests an interesting sub-case of your problem, the
case when the Cayley graph is decidable. And in this case,
one can at least find a non-intersecting infinite path that
is *low*.
Let me explain. For any group presentation, one may form
the tree T of all finite non-intersecting paths. This is
the tree of attempts to build an infinite non-intersecting
path. Your question is equivalent to asking, when the group
is infinite, whether or not this tree has a computable
infinite branch. If the Cayley graph is decidable, then
this tree will be decidable. Thus, by the [Low Basis
Theorem](http://en.wikipedia.org/wiki/Low_basis_theorem),
it follows that there is a branch b through the tree which
is *low*, meaning that the halting problem relative to b is
Turing equivalent to the ordinary halting problem. In
particular, this branch b is strictly below the halting
problem in the Turing degrees. This shows that any
computably-presented group with a decidable Cayley graph
admits a low non-intersecting path. Such a path is close to
being computable, but perhaps not quite computable. (Even in
this very special case when the Cayley graph is decidable,
I'm not sure whether there must be a computable
non-intersecting path.)
In the general case, the argument shows that for any group presentation of an infinite group, we can find an infinite non-intersecting path s, which has low degree relative to the Turing degree of the Cayley graph (low in the technical sense). I suspect that this is the best that one can say.
There is another classical theorem in computability that
seems relevant to your question, namely, the fact that
there are computable infinite binary branching trees T,
subtrees of 2ω, having no computable
infinite branches. These trees therefore constitute
violations of the computable analogue of Konig's lemma.
This problem is very like yours, isn't it? I am intrigued
by the possibility of using that classical construction to
build an example of the kind of group you seek.
There is a natural generalization of your question beyond
the finitely presented groups, to the class of
finitely-generated but computably presented groups. That
is, consider a group presentation with finitely many
generators and a computable list of relations. It may be
easier to find a instance of the kind of group you seek
having this more general form, since one can imagine a
priority argument, where one gradually adds relations as
the construction proceeds in order to meet various
requirements that diagonalize against the computable paths.
---
Here are some resources to general decidability questions in
finite group presentations: the undecidability of the [word
problem](http://en.wikipedia.org/wiki/Word_problem_for_groups),
the [conjugacy
problem](http://en.wikipedia.org/wiki/Conjugacy_problem),
and the [group isomorphism
problem](http://en.wikipedia.org/wiki/Group_isomorphism_problem).
| 4 | https://mathoverflow.net/users/1946 | 14274 | 9,585 |
https://mathoverflow.net/questions/14251 | 6 | Introduction
------------
Let's fix $m\in \mathbb N$. For each n, the unitary group $\mathbf U(m)$ is represented in the space of tensors of rank $n$ over $\mathbb C^m$
$$V\_{n,m}=\bigotimes\_{k=1}^n \mathbb C^m$$ and the symmetric group $S\_n$ acts on $V\_{n,m}$ by permutation of factors. Now the space $V\_{n,m}$ breaks into the direct sum of subspaces $V\_{n,m}(\lambda)$ which are primary with respect to each of these actions and irreducible with respect to the joint action of $S\_n\times \mathbf U(m)$
$$V\_{n,m}=\bigoplus\_{\lambda \in \mathbb{Y}(n,m)}V\_{n,m}(\lambda)$$
Where $\lambda$ ranges over all Young diagrams of size $n$ with at most $m$ rows (So $\lambda \in \mathbb{Y}(n,m)$ means $\lambda=(\lambda\_1,\dots ,\lambda\_k)$ is a partition of $n$ with $k\le m$). The tensors from $V\_{n,m}(\lambda)$ are said to have symmetry type $\lambda$. We define the relative dimensions
$$d\_{n,m}(\lambda)=\frac{\dim V\_{n,m}(\lambda)}{\dim V\_{n,m}}$$
which tell us how tensors are distributed into symmetry types.
Motivation and Question
-----------------------
I was reading Kerov's "Asymptotic representation theory of the symmetric group and it's aplications in analysis" and was trying to provide proofs for some of the results stated there. (He does give references, which I can't reach at the moment.)
The following two theorems are due to Kerov
>
> **Theorem 1** If for each $\lambda$ we associate $x=(x\_1,\dots,x\_m)$ with $x\_k=\frac{\lambda ^{(n)}\_k-n/m}{\sqrt{n}}$. The joint distribution of $x\_k$'s as $n\to \infty$ with respect to the measure $d\_{n,m}$ on $\mathbb{Y}(n,m)$ weakly converges to an absolutely continuous measure on the cone $C\_m=\{x: x\_1\geq x\_2\geq\cdots \geq x\_m \;;\;\sum x\_k=0\}$
> with density
> $$\phi \_m(x)=c \prod \_{i < j}(x\_i-x\_j)^2 e^{-m/2 \sum x\_k^2}$$ where
> $$c=\frac{m^{(m-1)m/2}}{1!2!\cdots (m-1)!}\left(\frac{m}{2\pi}\right)^{(m-1)/2}$$
>
>
>
.
>
> **Theorem 2** Let $\lambda ^{(n)}\in \mathbb{Y}(n,m)$ be the Young diagram for which the tensors of type $\lambda ^{(n)}$ are most probable. Then $$\lim\_{n\to \infty} \frac{\lambda ^{(n)}\_k-n/m}{\sqrt{n/m}}=z\_k$$
> for each $k=1,2,\dots,m$ where $z\_1,z\_2,\dots z\_m$ are the roots of the [Hermite polynomial $H\_m(z)$](http://en.wikipedia.org/wiki/Hermite_polynomials)
>
>
>
I can prove Theorem 2 assuming Theorem 1, but I don't see a nice argument for proving the first theorem itself. Can anyone provide a sketch of the proof, or some hint how to approach Theorem 1?
| https://mathoverflow.net/users/2384 | Asymptotics of symmetry types of tensors | This is not a complete answer, but perhaps will help. The probability distribution on tuples that shows up in your Theorem 1 is well known: It is the joint probability density function for the eigenvalues of a random unitary matrix, in the standard ``Gaussian unitary ensemble". See equation (9) in the paper of Terry Tao and Van Vue
<http://arxiv.org/abs/0906.0510v9>
I have unfortunately not read that closely, but the title, "random matrices: universality of local eigenvalue statistics" suggests it might have something to say about why this distribution would appear in other places. I've looked more closely at the following paper of Okounkov.
<http://arxiv.org/abs/math-ph/0309015>
There he explains how a similar distribution does show up in a system of random partitions. See especially Section 1.4.2. There Okounkov uses the distribution coming from the ``Plancherel measure", which is slightly different then the distribution you describe: the probability of observing $\lambda$ is propositional to $\operatorname{dim} S(\lambda)^2$, where $S(\lambda)$ is the representation of the symmetric group corresponding to $\lambda$. You seem to have chosen the distribution where the probability of observing $\lambda$ is proportional to $\operatorname{dim} S(\lambda) \operatorname{dim} V(\lambda)$, where $V(\lambda)$ is an irreducible representation of $U(n)$. Also, he has $-1/2$ where you have $-m/2$ in the exponential part. But perhaps it is still related.
Anyway, if you are interested in this type of question about partitions, I highly recommend looking at these papers.
| 4 | https://mathoverflow.net/users/1799 | 14295 | 9,596 |
https://mathoverflow.net/questions/14283 | 16 | As [noted earlier](https://mathoverflow.net/questions/14278/), I found reading Weil's book "Basic Number Theory" to be a harrowing experience, and I find his writing to be intrinsically hard to understand, though it is perfectly rigorous and clean.
In any case, the second part of the book is "thoroughly unmodern", in the author's own terms. He dispenses with cohomology and builds the whole theory based on central simple algebras. And he advises the reader to make it an exercise to himself for working out the "hidden cohomology" in it.
I personally felt this as some kind of perversion(just as in the fixation for Haar measure in the first part) to show the reader that he is a powerful mathematician *could* do things precisely the way he wants, and my feeling was strengthened by the title "Basic Number Theory".
However since he is a great mathematician, and I am nobody, I feel unqualified dismiss him like that. There must be some merits/uses of doing it this way, rather than using cohomology. If somebody can enlighten me, I would be very grateful.
Again, is it the case that class field theory can be done with just $H^1$? I wanted to study the subject and attempted, but was put off for a long time by the wrong book choice, reading Weil which felt like drinking stone soup, and it takes me time to take up the subject again.
| https://mathoverflow.net/users/2938 | Central simple algebras approach to class field theory, merits of | If you are looking for a more conceptual way of understanding the central simple algebras approach to class field theory, I think that you would be well advised to look at Roquette's book [The Brauer-Hasse-Noether Theorem in Historical Perspective](http://books.google.com/books?id=ICnySc00PA4C&printsec=frontcover&dq=the+brauer+hasse+noether+roquette&source=bl&ots=iOvsihICHi&sig=i3ZbrU0CqGjahllEwBf1FFLeXgY&hl=en&ei=U0psS8bVH5SWtge908SOBg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAkQ6AEwAA#v=onepage&q=&f=false). The Brauer-Hasse-Noether Theorem, oftentimes referred to as the Albert-Brauer-Hasse-Noether theorem (Adrian Albert, an american, discovered part of the theorem independently), states that a central simple algebra defined over a number field splits globally if and only if it splits everywhere locally and is intimately connected with class field theory.
The entire book is only 80 pages long, and takes only a few hours to read. You should especially look at Chapter 6: The Brauer group and class field theory. The last few pages of this chapter make explicit the application of work done with central simple algebras to class field theory and the connection between this approach and the more modern cohomological approach.
[**Edit 1**] As Franz has pointed out, applying the theory of central simple algebras to class field theory is not just *another* approach, but rather was the *original* approach. The connection with cohomology arises as follows:
For a field extension $K/k$, denote by $Br(K/k)$ the **Relative Brauer group** of $k$ with respect to $K$. Explicitly, it is the kernel of the homomorphism $Br(k) \rightarrow Br(K)$. Then $Br(k)=\cup Br(K/k)$. It isn't too hard to show that the elements of these relative Brauer groups (for $K/k$ Galois) are in one to one correspondence with certain factor sets relative to $K$. These factor sets naturally arise as the elements of $H^2(K/k)$, ultimately yielding an isomorphism $Br(K/k)\cong H^2(K/k)$. This isomorphism allows one to translate results like the Albert-Brauer-Hasse-Noether theorem into the perhaps more familiar cohomological language.
[**Edit 2**] Apparently Roquette's book is available [online](https://www.mathi.uni-heidelberg.de/%7Eroquette/brhano.pdf) (along with many other gems).
| 16 | https://mathoverflow.net/users/nan | 14297 | 9,597 |
https://mathoverflow.net/questions/14267 | 6 | If $U \subset \mathbb C^n$ is the complement of a closed analytic subset of codimension at least three then there is a result of Cartan which says that $H^1(U,\mathcal O^{analytic}\_U)=0$, see page 133 of "Theory of Stein Spaces" by Grauert and Remmert.
Does anyone know a reference for the analogous result in the algebraic category ?
| https://mathoverflow.net/users/605 | Reference for cohomology vanishing | To expand on Emerton's [answer](https://mathoverflow.net/a/14270): Using the excision sequence, Cartan's result in the algebraic case boils down to showing the following: Let $R$ be a regular local ring, and $I$ and ideal of height at least $3$, then $H^i\_I(R)=0$ for $i\leq 2$.This follows because: $$H^i\_I(R) = \lim Ext^i(R/I^n,R)$$
And $I^n$, being height $3$, always contains a regular sequence of length $2$, so the $Ext^i$ vanishes for $i\leq 2$ by standard result (see Bruns-Herzog Cohen Macaulay book, Proposition 1.2.10 for example). This argument extends to the case of codimension at least $n$ and vanishing of $H^{n-2}$.
Incidentally, a pretty non-trivial question is to find upper bound for the vanishing of local cohomology modules, in other words, the [cohomological dimension](https://encyclopediaofmath.org/wiki/Cohomological_dimension) of a subvariety $Z$. Many strong results have been obtained after SGA, by Hartshorne, Ogus, Faltings, Huneke-Lyubeznik, etc. All those references can be found in
[Lyubeznik's paper](https://www.jstor.org/stable/2946619 "Gennady Lyubeznik, Etale Cohomological Dimension and the Topology of Algebraic Varieties, Annals of Mathematics Second Series, Vol. 137, No. 1 (Jan., 1993), pp. 71–128 (58 pages), doi:10.2307/2946619. zbMATH review at https://zbmath.org/?q=an:0811.14014") (they were mentioned in the very first page) which primarily treated the vanishing of etale cohomology.
| 4 | https://mathoverflow.net/users/2083 | 14298 | 9,598 |
https://mathoverflow.net/questions/14300 | 1 | I have another question about $SU(n)$, again I hope it's not too basic. For $n=2$, the action of $SU(2)$ on $C^2$ is free since it's equal to the group of rotations. In general, the group of rotations is properly contained in $SU(n)$, does this mean that its action on $C^n$ is no longer free.
| https://mathoverflow.net/users/3787 | Freeness of the Canonical $SU(n)$ Action | $SU(n)$ acts freely on $\Bbb C^n\setminus 0$ (not on $\Bbb C^n$) for $n=1,2$, but not for higher $n$ (the stabilizer of a point is $SU(n-1)$). If this is what you are asking about, the answer is "yes". (This IS a basic question, though.)
| 1 | https://mathoverflow.net/users/3696 | 14302 | 9,601 |
https://mathoverflow.net/questions/13374 | 4 | Everybody knows that equality is reflexive: $\forall(x)(x=x)$. But should reflexivity of equality be taken as an axiom of logic or as a theorem of set theory?
If you choose the former then you probably need the axiom of extensionality: $\forall(x)\forall(y)(x=y\leftrightarrow\forall(z)(z\in x\leftrightarrow z\in y))$.
If you choose the latter then probably $x=y$ is just an abbreviation for $\forall(z)(z\in x\leftrightarrow z\in y)$.
What I'm trying to do is to write down proofs for basic facts about set theory, but I'm not so sure which logical axioms and rules of inference should I take for granted.
Thanks.
| https://mathoverflow.net/users/3554 | Is reflexivity of equality an axiom or a theorem? | I managed to write down a proof for the reflexivity of equality using only the definition of equality in terms of membership and the rules of natural deduction.
1. Premise: $\forall x\_0\forall x\_1\left(\left(x\_0=x\_1\right)\leftrightarrow\forall x\_2\left(\left(x\_2\in x\_0\right)\leftrightarrow\left(x\_2\in x\_1\right)\right)\right)$
2. Assumption (1): $\forall x\_0\forall x\_1\left(\left(x\_0=x\_1\right)\leftrightarrow\forall x\_2\left(\left(x\_2\in x\_0\right)\leftrightarrow\left(x\_2\in x\_1\right)\right)\right)$
3. Universal elimination (2): $\forall x\_1\left(\left(k\_0=x\_1\right)\leftrightarrow\forall x\_2\left(\left(x\_2\in k\_0\right)\leftrightarrow\left(x\_2\in x\_1\right)\right)\right)$
4. Universal elimination (3): $\left(\left(k\_0=k\_0\right)\leftrightarrow\forall x\_2\left(\left(x\_2\in k\_0\right)\leftrightarrow\left(x\_2\in k\_0\right)\right)\right)$
5. (Subproof 1) Premise: $\left(k\_2\in k\_0\right)$
6. (Subproof 1) Assumption (5): $\left(k\_2\in k\_0\right)$
7. (Subproof 2) Premise: $\left(k\_2\in k\_0\right)$
8. (Subproof 2) Assumption (7): $\left(k\_2\in k\_0\right)$
9. Biconditional introduction (5, 6, 7, 8): $\left(\left(k\_2\in k\_0\right)\leftrightarrow\left(k\_2\in k\_0\right)\right)$
10. Universal introduction (9): $\forall x\_2\left(\left(x\_2\in k\_0\right)\leftrightarrow\left(x\_2\in k\_0\right)\right)$
11. Biconditional elimination (4, 10): $\left(k\_0=k\_0\right)$
12. Universal introduction (11): $\forall x\left(x=x\right)$
In a similar way I also wrote down the proof for the symmetry and for the transitivity of equality.
| 1 | https://mathoverflow.net/users/3554 | 14304 | 9,602 |
https://mathoverflow.net/questions/14278 | 19 | I remember reading Weil's "Basic Number Theory" and giving up after a while. Now I find myself thinking of it (thanks to [some comments](https://mathoverflow.net/questions/13106/map-of-number-theory/13134#13134) by Ben Linowitz).
Right from the very beginning, Weil uses the fact that when you have a locally compact topolgocal group $G$ and a locally compact subgroup $H$, in addition to the Haar measures on $G$ and $H$, there exists a "Haar measure" on the coset space $G/H$, with some properties.
For instance, the upper half plane $\mathbb H$ is the quotient $\operatorname{SL}\_2(\mathbb R)/{\operatorname{SO}\_2(\mathbb R)}$ and the usual measure there which gives rise to the usual hyperbolic metric, is arising in this way.
I originally assumed this theorem and went ahead(but not much) with that book.
I want to have a reference for the above theorem. A reference which is not written by Weil. I find him very hard to penetrate. This should exclude Bourbaki's "Integration", as I supppose it would be heavily influenced by him, and thus is a horrible book (note to Harry: this is personal opinion; spare me the brickbats).
I had originally seen the construction of Haar measure on H. Royden's "Real Analysis", in which he is not considering any quotients.
| https://mathoverflow.net/users/2938 | Haar measure on a quotient, References for | The book I always look at for such things is Nachbin, The Haar Integral, which is short, and has a whole chapter on Integration on Locally Compact Homogeneous Spaces.
And a plus: he gives you a choice of reading the proof of the existence and uniqueness of the Haar integral according to Weil or according to Henri Cartan.
| 27 | https://mathoverflow.net/users/930 | 14305 | 9,603 |
https://mathoverflow.net/questions/14296 | 3 | Following on from this question [link text](https://mathoverflow.net/questions), how are the line bundles of a complex flag variety indexed? Are they still labeled by the integers? If so, why? A representation theory explanition in terms of the homogenous space description of the variety $U(n)/U(k\_1) \times \cdots \times U(k\_m)$ would be the most useful.
Also, is there an 'obvious' reason why the line bundles should be associated to a representation of $U(k\_1) \times \cdots \times U(k\_m)$, as opposed to a representation of another quotient description of the variety. For example, $\mathbb{CP}^{n-1}$ can be described as a $SU(n)/U(n)$ or as $S^{2n-1}/U(1)$, but the tangent bundle can only be described as associated to a representation of $U(n)$. In general, for a principal $G$-bundle $P$, what vector bundles over $P$ can be described as associated to a representation of $G$? All of them?
| https://mathoverflow.net/users/1648 | Indexing the Line Bundles of a Flag Manifold | Attempting to answer your last question: given a (topological, Lie, algebraic, etc.) group G and a closed subgroup P in G, the category of G-equivariant vector bundles on X=G/P is equivalent to the category of representations of P. So to a representation of P one can assign a vector bundle over G/P, and even a G-equivariant vector bundle. In addition, any line bundle over a generalized flag manifold of a simply connected semisimple group G has a unique, up to an isomorphism, G-equivariant structure. So line bundles over a generalized flag manifold are classified by one-dimensional representations of the parabolic subgroup.
| 5 | https://mathoverflow.net/users/2106 | 14308 | 9,606 |
https://mathoverflow.net/questions/14266 | 23 | I'm wondering if there is a Quillen model structure on the category of simplicial sets which generalizes the usual model structure, but where every simplicial set is fibrant? I want to use this to do homological algebra for commutative monoids, but first let me explain some background, my motivation, and articulate more precisely what I am after.
Background
----------
A Quillen Model structure on a category has three classes of morphisms: fibrations, weak equivalences, and cofibrations. This structure allows one to do many advanced homotopical constructions mimicking the homotopy theory of (nice) topological spaces. There is a notion of Quillen equivalence between model categories which consists of a particular adjunction between the two model categories in question. This gives you "equivalent homotopy theories" for the two model categories in question.
The usual Model structure on [simplicial sets](http://en.wikipedia.org/wiki/Simplicial_set) has fibrations the [Kan fibrations](http://en.wikipedia.org/wiki/Kan_fibration), the weak-equivalences are the maps which induces isomorphisms of homotopy groups, and the cofibrations are the (levelwise) inclusions. This is equivalent to the usual model category of topological spaces, and the Quillen equivalence is realized by the adjoint pair of functors: the geometric realization functor and the singular functor.
Quillen model categories are also useful for doing homological algebra, and particularly for working with derived categories. For reasonable abelian categories there are several nice (Quillen equivalent) model category structures on the category of (possibly bounded) chain complexes which one can use which reproduce the derived category. More precisely the homotopy category of the model category is the derived category and the "homotopical constructions" I mentioned above, in this case, correspond to the notion of (total) derived functor.
This story is further enriched by the Dold-Kan correspondence which is an equivalence between the categories of positively graded chain complexes of, say, abelian groups and the category of simplicial abelian groups, a.k.a. simplicial sets which are also abelian group objects. This in turn is Quillen equivalent to a model category of topological abelian groups.
Previous MathOverflow Question and Progress
-------------------------------------------
Previously I asked a question on MO about doing [homological algebra for commutative monoids](https://mathoverflow.net/questions/430/homological-algebra-for-commutative-monoids). I got many fascinating and exciting answers. Some were more or less "here is something you might try", some were more like "here is a bit that people have done, but the full theory hasn't been studied". After hearing those answers I'm much more excited about the field with one element. But still, in the end none of the answers really had what I was after. Then Reid Barton asked [his MO question](https://mathoverflow.net/questions/13518/is-in-simplicial-commutative-monoids-group-completion).
This got me thinking about commutative monoids and simplicial commutative monoids again. I had some nice observations which have lead me to the question at hand.
1. The first is that while a simplicial abelian group is automatically a [Kan simplicial set](http://ncatlab.org/nlab/show/Kan+complex) (i.e. if you forget the abelian group structure what have sitting in front of you is a Kan simplicial set) this is *not* the case for simplicial commutative monoids. A simplicial commutative monoid does not have to be a Kan simplicial set.
2. Next, I realized that the (normalized!) [Dold-Kan correspondence](http://ncatlab.org/nlab/show/Dold-Kan+correspondence) still seems to work. You can go from a simplicial commutative monoid to a "complex" of monoids, ( and back again It is just an adjunction. Thanks, Reid, for pointing this out!).
3. If your simplicial commutative monoid is also a Kan complex, then under the Dold-Kan correspondence you get a complex where the bottom object is a commutative monoid, but all the rest are abelian groups (this was pointed out to me by Reid Barton in a conversation we had recently). Thus the theory of *topological* commutative monoids (which was one of the suggested answers to my previous question), which has a nice model category structure ([see Clark Barwick's answer to an MO question](https://mathoverflow.net/questions/11059/model-structure-homotopy-pushouts-in-topological-monoids/11068#11068)), should be equivalent to a theory of chain complexes of this type. It shouldn't model arbitrary complexes of commutative monoids.
4. If you have a simplicial set which is *not* necessarily a Kan complex you can still define the naive simplicial homotopy "groups" $\pi\_iX$. I put "goups" in quotes because for a general simplicial set these are just pointed sets. If your complex is Kan, these are automatically groups (for $i>0$). If your simplicial set is a simplicial abelian group, these are abelian groups (for all $i$) and they are precisely the homology groups of the chain complex you get under the Dold-Kan correspondence. If your simplicial set is a commutative monoid, but not Kan, then they are commutative monoids and are again the "homology" monoids of the chain complex you get under the Dold-Kan correspondence.
The Question
------------
All this suggests that there should be a Quillen model structure on simplicial commutative monoids in which the weak equivalences are the $\pi\_{\bullet} $-isomorphisms, where here $\pi\_{\bullet} $ denotes the naive simplicial version, i.e. these are commutative monoids, not groups. I'm sure if such a thing was well known then it would have been mentioned as an answer to my previous question. I'd really like to see something that generalizes the usual theory of abelian groups. That way if we worked with simplicial abelian groups and construct derived functors we would just reproduce the old answers. As a stepping stone, there should be a companion model structure on simplicial sets which, I think, is more likely to be well known.
One of the properties that I think this hypothetical model structure on simplicial sets should have is that *every* simplicial set is [fibrant](http://en.wikipedia.org/wiki/Fibrant_object), not just the Kan simplicial sets.
>
> **Question**: Is there a model structure on simplicial sets in which every simplicial set is fibrant, and such that the weak equivalences between the Kan complexes are exactly the usual weak equivalences? Specifically can the weak equivalences be taken to be those maps which induce $\pi\_\*$-isomorphism, where these are the naive simplicial homotopy **sets**?
>
>
>
If this model structure exists, I'd like to know as much as possible about it. If you have any references to the literature, I'd appreciate those too, but the main question is as it stands.
| https://mathoverflow.net/users/184 | A Peculiar Model Structure on Simplicial Sets? | So the short answer is that there is not such a model structure. The difficulty arises in trying to show that the class of weak equivalences has all of the necessary properties; in particular, even two-of-three does not hold for the naive definition. The first difficulty arises even before that: on ordinary simplicial sets we can arrange for a model of every set that is "minimal" on the $\pi\_0$-level, meaning that every connected component has exactly one $0$-simplex. In simplicial commutative monoids we can no longer do this. However, we could assume that in order to be a weak equivalence we need to be a $\pi\_\*$-isomorphism when choosing any (coherently chosen) basepoints.
For the purposes of our discussion we are going to assume that $\pi\_\*$-equivalences use the model $S^n = \Delta^n/\partial\Delta^n$. (This is the model that most closely mimicks the boundary maps in the Dold-Kan correspondence.) Now let $X = S^2$, and let $Y$ be $S^2$ with an extra $0$-simplex connected by a $1$-simplex to the original basepoint. (So it looks like a balloon on a string.) We define a map $X\rightarrow Y$ to be the inclusion of $S^2$ in the obvious manner, and a map $Y\rightarrow X$ to be collapsing the extra $1$-simplex back down. Then the composition of these two maps is the identity on $X$, so obviously a weak equivalence. The map $X\rightarrow Y$ is also a weak equivalence, because adding the "string" can't add any new homotopy groups to $X$. However, the map $Y\rightarrow X$ is not a weak equivalence, as $\pi\_2Y$ based at the extra point is a one-point set but $\pi\_2X$ at its image is a two-point set.
The problem arose because in order to show that $\pi\_\*$ was invariant of basepoint in the usual Kan complex model we needed to be able to "pull back" simplices along paths in the simplicial set, which used the Kan condition. The new model does not have such a condition, and thus we can't necessarily pull things back.
Another observation along these lines. Take any connected simplicial set $X$, and let $Y$ be $X$ with a "string" added to it at any basepoint. Then $\*\rightarrow Y$ (including into the new point) is a weak equivalence, and $X\rightarrow Y$ (including into itself) is a weak equivalence. Thus in the homotopy category, $X$ is isomorphic to a point (and thus the homotopy category is just the category of sets) ... which is presumably not desired.
-- The Bourbon seminar
| 12 | https://mathoverflow.net/users/1378 | 14310 | 9,608 |
https://mathoverflow.net/questions/14311 | 27 | I was reading [this question](https://mathoverflow.net/questions/731/points-in-algebraic-geometry-why-shift-from-m-spec-to-spec) on why algebraic geometry looks at prime ideals instead of only maximal ideals, and I understand Anton's answer, but I'm a little confused as to how this fits with Hilbert's Nullstellensatz - affine algebraic sets are in bijection with radical ideals, not prime ideals, and it seems like we'd want the extra information we'd get by looking at "RadSpec(R)" (my own imagined notation). Also, the preimage of a radical ideal is radical, so there isn't the same objection as to maximal ideals - "RadSpec" would also be a contravariant functor.
So, why not radical ideals instead of only prime ideals, and what kinds of things could we say about RadSpec(R) even if they aren't very interesting?
| https://mathoverflow.net/users/1916 | MaxSpec, Spec, ... "RadSpec"? Or, why not look at all radical ideals? | The short answer is that every radical ideal is the intersection of the prime ideals containing it, so that the pullback map on Specs determines the pullback map on your wouldbe RadSpecs.
Note that a similar reason underlies the success of looking only at maximal ideals in classical algebraic geometry: a finitely generated algebra over a field is a Hilbert-Jacobson ring: every radical ideal is the intersection of the *maximal* ideals containing it.
I won't rule out the possibility that RadSpec could be useful for something, though...
| 25 | https://mathoverflow.net/users/1149 | 14312 | 9,609 |
https://mathoverflow.net/questions/14314 | 51 | The fact that a commutative ring has a natural topological space associated with it is still a really interesting coincidence. The entire subject of Algebraic geometry is based on this simple fact.
Question: Are there other categories of algebraic objects that have interesting natural topologies that carry algebraic data like the Zariski topology on a ring (spectrum)? If they exist, what are they and how are they used?
| https://mathoverflow.net/users/1353 | "Algebraic" topologies like the Zariski topology? | Yes, there are plenty of such things.
[In the following, "compact" implies "locally compact" implies "Hausdorff".]
1) To a Boolean algebra, one associates its Stone space, a compact totally disconnected space.
(Via the correspondence between Boolean algebras and Boolean rings, this is a special case of the Zariski topology -- but with a distinctive flavor -- that predates it.)
2) To a non-unital Boolean ring one associates its Stone space, a locally compact totally
disconnected space.
3) To a commutative C\*-algebra with unit, one associates its Gelfand spectrum, a compact space.
4) To a commutative C\*-algebra without unit, one associates its Gelfand spectrum, a locally compact space.
6) To a commutative Banach ring [or a scheme over a non-Archimedean field, or...] one
associates its Berkovich spectrum (the bounded multiplicative seminorms).
7) To a commutative ring R, one associates its real spectrum (prime ideals, plus orderings on the residue domain.)
8) To a field extension K/k, one associates its Zariski Riemann surface (equivalence classes of valuations on K which are trivial on k).
This is by no means a complete list...
**Addendum**: I hadn't addressed the second part of your question, i.e., explaining what these things are used for. Briefly, the analogy to the Zariski spectrum of a commutative ring is tight enough to give the correct impression of the usefulness of these other spectra/spaces: they are topological spaces associated (cofunctorially) to the algebraic (or algebraic-geometric, topological algebraic, etc.) objects in question. They carry enough information to be useful in the study of the algebraic objects themselves (sometimes, e.g. in the case of Stone and Gelfand spaces, they give complete information, i.e., an anti-equivalence of categories, but not always). In some further cases, one can get the anti-equivalence by adding further structure in a very familiar way: one can attach structure sheaves to these guys and thus get a class of "model spaces" for a certain species of locally ringed spaces -- e.g., Berkovich spectra glue together to give Berkovich analytic spaces.
| 50 | https://mathoverflow.net/users/1149 | 14315 | 9,611 |
https://mathoverflow.net/questions/14273 | 16 | When I read it for the first time, I found the whole slog towards proving the Prime Number Theorem and the final success to be magnificent. So I am curious about more general results.
We talk of complex $\zeta$-functions and $L$-functions. As a preliminary list, we fix the following list. But feel free to add to it.
$1$. Riemann $\zeta$-function.
$2$. Dedekind $\zeta$-function for a number field.
$3$. Artin $L$-functions for a character of the Galois group of some number field.
$4$. Zeta functions of algebraic varieties over number fields; for getting analytic continuation up to $s = 0$, we for instance fix the zeta function of an elliptic curve defined over $\mathbb{Q}$ which is modular in the sense of Eichler-Shiumura.
I am worried about $4$ here, since the requirement that the zeta function has a pole at $s=1$ is not fulfilled, and thus we fail to capture the main term from the residue there. Is something possible in this case? If so, we can consider that and also more general zeta functions of algebraic varieties over any finitely generated field, zeta functions of Galois representations, etc.. which have analytic continuation up to $s=0$ and the question could be extended to those cases as well.
We have the original prime number theorem, in the following form,
$\pi (x) = \frac{x}{log x} + O(error\ term)$.
which is proved by integration of the Riemann zeta function along a rectangular contour including $s = 1$, and letting the vertical edges get longer and longer, and estimating the integrals. The reference I have in mind is Ram Murty, Problems in Analytic Number Theory, or, J. Ayoub's book.
Now the questions:
>
> What is the known about a similar prime number theorem for more general zeta and $L$-functions?
>
>
>
I could imagine that it will be pretty straightforward for the Dedekind zeta function. But then I am curious about the error term.
For the rest of the cases, does a similar proof of the PNT carry over? More importantly,
>
> What is the "meaning" of the prime number theorem in these general cases?
>
>
>
Finally,
>
> What are some applications(to other problems) of such a prime number theorem proved with a good error term?
>
>
>
The applications of the original PNT are of course well-known, as for instance given in the books of Titchmarsch and Heath-Brown, or Ivic, or in the more modern book of Iwaniec and Kowalski.
| https://mathoverflow.net/users/2938 | PNT for general zeta functions, Applications of. | Hi Anweshi,
Since Emerton answered your third grey-boxed question very nicely, let me try at the first two. Suppose $L(s,f)$ is one of the L-functions that you listed (including the first two, which we might as well call L-functions too). (For simplicity we always normalize so the functional equation is induced by $s\to 1-s$.) This guy has an expansion $L(s,f)=\sum\_{n}a\_f(n)n^{-s}$ as a Dirichlet series, and the most general prime number theorem reads
$\sum\_{p\leq X}a\_f(p)=r\_f \mathrm{Li}(x)+O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
Here $\mathrm{Li}(x)$ is the logarithmic integral, $r\_f$ is the order of the pole of $L(s,f)$ at the point $s=1$, and the implied constant depends on $f$ and $\varepsilon$.
Let's unwind this for your examples.
1) The Riemann zeta function has a simple pole at $s=1$ and $a\_f(p)=1$ for all $p$, so this is the classical prime number theorem.
2) The Dedekind zeta function (say of a degree d extension $K/\mathbb{Q}$) is a little different. It also has a simple pole at $s=1$, but the coefficients are determined by the rule: $a(p)=d$ if $p$ splits completely in $\mathcal{O}\_K$, and $a(p)=0$ otherwise. Hence the prime number theorem in this case reads
$|p\leq X \; \mathrm{with}\;p\;\mathrm{totally\;split\;in}\;\mathcal{O}\_K|=d^{-1}\mathrm{Li}(x)+O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
This already has very interesting applications: the fact that the proportion of primes splitting totally is $1/d$ was very important in the first proofs of the main general results of class field theory.
3) If $\rho:\mathcal{G}\_{\mathbb{Q}}\to \mathrm{GL}\_n(\mathbb{C})$ is an Artin representation then $a(p)=\mathrm{tr}\rho(\mathrm{Fr}\_p)$. If $\rho$ does not contain the trivial representation, then $L(s,\rho)$ has no pole in neighborhood of the line $\mathrm{Re}(s)\geq 1$, so we get
$\sum\_{p\leq X}\mathrm{tr}\rho(\mathrm{Fr}\_p)=O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
The absence of a pole is not a problem: it just means there's no main term! In this particular case, you could interpret the above equation as saying that "$\mathrm{tr}\rho(\mathrm{Fr}\_p)$ has mean value zero.
4) For an elliptic curve, the same phenomenon occurs. Here again there is no pole, and $a(p)=\frac{p+1-|E(\mathbb{F}\_p)|}{\sqrt{p}}$. By a theorem of Hasse these numbers satisfy $|a(p)|\leq 2$, so you could think of them as the (scaled) deviation of $|E(\mathbb{F}\_p)|$ from its
"expected value" of $p+1$. In this case the prime number theorem reads
$\sum\_{p\leq X}a(p)=O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$
so you could say that "the average deviation of $|E(\mathbb{F}\_p)|$ from $p+1$ is zero."
Now, how do you prove generalizations of the prime number theorem? There are two main steps in this, one of which is easily lifted from the case of the Riemann zeta function.
1. Prove that the prime number theorem for $L(s,f)$ is a consequence of the nonvanishing of $L(s,f)$ in a region of the form $s=\sigma+it,\;\sigma \geq 1-\psi(t)$ with $\psi(t)$ positive and tending to zero as $t\to \infty$. So this is some region which is a very slight widening of $\mathrm{Re}(s)>1$. The proof of this step is essentially contour integration and goes exactly as in the case of the $\zeta$-function.
2. Actually produce a zero-free region of the type I just described. The key to this is the existence of an auxiliary L-function (or product thereof) which has positive coefficients in its Dirichlet series. In the case of the Riemann zeta function, Hadamard worked with the auxiliary function $ A(s)=\zeta(s)^3\zeta(s+it)^2 \zeta(s-it)^2 \zeta(s+2it) \zeta(s-2it)$. Note the pole of order $3$ at $s=1$; on the other hand, if $\zeta(\sigma+it)$ vanished then $A(s)$ would vanish at $s=\sigma$ to order $4$. The inequality $3<4$ of order-of-polarity/nearby-order-of-vanishing leads via some analysis to the absence of any zero in the range $s=\sigma+it,\;\sigma \geq 1-\frac{c}{\log(|t|+3)}.$ In the general case the construction of the relevant auxiliary functions is more complicated. For the case of an Artin representation, for example, you can take $B(s)=\zeta(s)^3 L(s+it,\rho)^2 L(s-it,\widetilde{\rho})^2 L(s,\rho \otimes \widetilde {\rho})^2 L(s+2it,\rho \times \rho) L(s-2it,\widetilde{\rho} \times \widetilde{\rho})$. The general key is the Rankin-Selberg L-functions, or more complicated L-functions whose analytic properties can be controlled by known instances of Langlands functoriality.
If you'd like to see everything I just said carried out elegantly and in crystalline detail, I can do no better than to recommend Chapter 5 of Iwaniec and Kowalski's book "Analytic Number Theory."
| 9 | https://mathoverflow.net/users/1464 | 14318 | 9,614 |
https://mathoverflow.net/questions/14341 | 50 | [A recent answer](https://mathoverflow.net/questions/14311/maxspec-spec-radspec-or-why-not-look-at-all-radical-ideals/14319#14319) motivated me to post about this. I've always had a vague, unpleasant feeling that somehow lattice theory has been completely robbed of the important place it deserves in mathematics - lattices seem to show up everywhere, the author or teacher says "observe that these \_\_\_\_ form a complete lattice" or something similar, and then moves on, never to speak of what that might imply. But, not currently knowing anything about them, I can't be sure. What would be a good place to learn about lattice theory, especially its implications for "naturally occurring" lattices (subgroups, ideals, etc.)?
| https://mathoverflow.net/users/1916 | Good lattice theory books? | A good, user-friendly, modern, introductory textbook is Davey and Priestley's *Introduction to Lattices and Order*.
Incidentally, Gian-Carlo Rota used to say much the same thing as you, Zev: that lattice theory had been robbed of its rightful place in mathematics.
| 34 | https://mathoverflow.net/users/586 | 14342 | 9,628 |
https://mathoverflow.net/questions/14324 | 9 | Warning: I'll be using the "pre-$\lambda$-ring" and "$\lambda$-ring" nomenclature, as opposed to the "$\lambda$-ring" and "special $\lambda$-ring" one (although I just used the latter a few days ago on MO). It's mainly because both sources use it, and I am (by reading them) slowly getting used to it.
Let $G$ be a finite group. The [Burnside ring](http://en.wikipedia.org/wiki/Burnside_ring) $B\left(G\right)$ is defined as the Grothendieck ring of the category of finite $G$-sets, with multiplication defined by cartesian product (with diagonal structure, or at least I have difficulties imagining any other $G$-set structure on it; please correct me if I am wrong).
For every $n\in\mathbb{N}$, we can define a map $\sigma^n:B\left(G\right)\to B\left(G\right)$ as follows: Whenever $U$ is a $G$-set, we let $\sigma^n U$ be the set of all multisets of size $n$ consisting of elements from $U$. The $G$-set structure on $\sigma^n U$ is what programmers call "map": an element $g\in G$ is applied by applying it to each element of the multiset. This way we have defined $\sigma^n U$ for every $G$-set $U$; we extend the map $\sigma^n$ to all of $B\left(G\right)$ (including "virtual" $G$-sets) by forcing the rule
$\displaystyle \sigma^i\left(u+v\right)=\sum\_{k=0}^i\sigma^k\left(u\right)\sigma^{i-k}\left(v\right)$ for all $u,v\in B\left(G\right)$.
Ah, and $\sigma^0$ should be identically $1$, and $\sigma^1=\mathrm{id}$. Anyway, this works, and gives a "pre-$\sigma$-ring structure", which is basically the same as a pre-$\lambda$-ring structure, with $\lambda^i$ denoted by $\sigma^i$. Now, we turn this pre-$\sigma$-ring into a pre-$\lambda$-ring by defining maps $\lambda^i:B\left(G\right)\to B\left(G\right)$ by
$\displaystyle \sum\_{n=0}^{\infty}\sigma^n\left(u\right)T^n\cdot\sum\_{n=0}^{\infty}\left(-1\right)^n\lambda^n\left(u\right)T^n=1$ in $B\left(G\right)\left[\left[T\right]\right]$ for every $u\in B\left(G\right)$.
Now, let me quote two sources:
Donald Knutson, *$\lambda$-Rings and the Representation Theory of the Symmetric Group*, 1973, p. 107: "The fact that $B\left(G\right)$ is a $\lambda$-ring and not just a pre-$\lambda$-ring - i. e., the truth of all the identities - follows from [...]"
Michiel Hazewinkel, *[Witt vectors, part 1](http://arxiv.org/abs/0804.3888)*, 19.46: "It seems clear from [370] that there is no good way to define a $\lambda$-ring structure on Burnside rings, see also [158]. There are (at least) two different choices giving pre-$\lambda$-rings but neither is guaranteed to yield a $\lambda$-ring. Of the two the symmetric power construction seems to work best."
(No, I don't have access to any of these references.)
For a long time I found Knutson's assertion self-evident (even without having read that far in Knutson). Now I tend to believe Hazewinkel's position more, particularly as I am unable to verify one of the relations required for a pre-$\lambda$-ring to be a $\lambda$-ring:
$\lambda^2\left(uv\right)=\left(\lambda^1\left(u\right)\right)^2\lambda^2\left(v\right)+\left(\lambda^1\left(v\right)\right)^2\lambda^2\left(u\right)-2\lambda^2\left(u\right)\lambda^2\left(v\right)$ for $B\left(G\right)$.
What also bothers me is Knutson's "conjecture" on p. 113, which states that the canonical (Burnside) map $B\left(G\right)\to SCF\left(G\right)$ is a $\lambda$-homomorphism, where $SCF\left(G\right)$ denotes the $\lambda$-ring of super characters on $G$, with the $\lambda$-structure defined via the Adams operations $\Psi^n\left(\varphi\left(H\right)\right)=\varphi\left(H^n\right)$ (I think he wanted to say $\left(\Psi^n\left(\varphi\right)\right)\left(H\right)=\varphi\left(H^n\right)$ instead) for every subgroup $H$ of $G$, where $H^n$ means the subgroup of $G$ generated by the $n$-th powers of elements of $H$. This seems wrong to me for $n=2$ and $H=\left(\mathbb Z / 2\mathbb Z\right)^2$ already. And if the ring $B\left(G\right)$ is not a $\lambda$-ring, then this conjecture is wrong anyway (since the map $B\left(G\right)\to SCF\left(G\right)$ is injective).
Can anyone clear up this mess? I am really confused...
Thanks a lot.
| https://mathoverflow.net/users/2530 | Is the Burnside ring a lambda-ring? + conjecture in Knutson p. 113 | I've just gone and looked up [158] (Gay, C. D.; Morris, G. C.; Morris, I.
Computing Adams operations on the Burnside ring of a finite group.
J. Reine Angew. Math. 341 (1983), 87--97.
On p. 90, at the end of section 2, they say: "Knutson conjectured that the Adams operations on SCF(G) inherited from A(G) [=Burnside ring of G] are given by [the formula you mentioned, involving the subgroup generated by nth powers of a subgroup $K$]. We will show that this is correct if $K$ is cyclic, but not true in general."
I haven't looked at it carefully, but they appear to give some more complicated looking formulas for the action of Adams operations on super-characters, valid in some cases.
They don't seem to mention Knutson's claim that the Burnside ring is a lambda-ring (not merely pre-lambda).
| 6 | https://mathoverflow.net/users/437 | 14351 | 9,636 |
https://mathoverflow.net/questions/14338 | 38 | I'm community wikiing this, since although I don't want it to be a discussion thread, I don't think that there is really a right answer to this.
From what I've seen, model theorists and logicians are mostly opposed to GCH, while on the other end of the spectrum, some functional analysis depends on GCH, so it is much better tolerated among functional analysts. In fact, I considered myself very much +GCH for a while, but Joel and Francois noted some interesting stuff about forcing axioms, (the more powerful ones directly contradict CH).
What is the general opinion on GCH in the mathematical community (replace GCH with CH where necessary)? Does it happen to be that CH/GCH doesn't often come up in algebra?
Please don't post just post "I agree with +-CH". I'd like your assessment of the mathematical community's opinion. Maybe your experiences with mathematicians you know, etc. Even your own experiences or opinion can work. I am just not interested in having 30 or 40 one line answers. Essentially, I'm not looking for a poll.
Edit: GCH=Generalized Continuum Hypothesis
CH= Continuum Hypothesis
CH says that $\aleph\_1=\mathfrak{c}$. That is, the successor cardinal of $\aleph\_0$ is the continuum. The generalized form (GCH) says that for any infinite cardinal $\kappa$, we have $\kappa^+=2^\kappa$, that is, there are no cardinals strictly between $\kappa$ and $2^\kappa$.
Edit 2 (Harry): Changed the wording about FA. If it still isn't true, and you can improve it, feel free to edit the post yourself and change it.
| https://mathoverflow.net/users/1353 | What is the general opinion on the Generalized Continuum Hypothesis? | There is definitely a not-CH tendency among set theorists with a strong Platonist bent, and my impression is that this is the most common view. Many of these set theorists believe that the large cardinal hierarchy and the accompanying uniformization consequences are pointing us towards the final, true set theory, and that the various forcing axioms, such as PFA, MM etc. are a part of it.
Another large group of set theorists working in the area of inner model theory have GCH in all the most important models that they study, and regard GCH as one of the attractive regularity features of those inner models.
There is a far smaller group of set theorists (among whom I count myself) with a multiverse perspective, who take the view that set theory is really about studying all the possible universes that we might live in, and studying their inter-relations. For this group, the CH question is largely settled by the fact that we understand in a very deep way how to move fom the CH universes to the not-CH universes and vice versa, by the method of forcing. They are each dense in a sense in the collection of all set-theoretic universes.
| 46 | https://mathoverflow.net/users/1946 | 14353 | 9,638 |
https://mathoverflow.net/questions/14358 | 6 | If one views a group as a one object category with the elements of the group as morphisms then a natural transformation between functors of such categories is an inner automorphism, i.e. if we have two group homomorphisms $f,g: A\to B$ then a natural transformation $\eta :f\to g$ is just an element $b\in B$ such that $f(a)\cdot b = b \cdot g(a)$ which can be rewritten as $f(a)=b \cdot g(a)\cdot b^{-1}$. This isn't the only way to turn groups into categories. Another way is to take the elements of the group as objects and to have a morphism $h\_a:a\to b$ if $h\cdot a=b$. If we view groups in this way then are the natural transformations again something nice like inner automorphisms?
| https://mathoverflow.net/users/nan | groups as categories and their natural transformations | The comments thread is getting a bit long, so here's an answer. The category $C(G)$ that David associates to a group $G$ (by his second recipe) has the elements of $G$ as its objects, and exactly one morphism between any given pair of objects. It's what category theorists call an indiscrete or codiscrete category, and graph theorists call a complete graph or clique. You can form the indiscrete category on any set: it doesn't need a group structure.
A functor from one indiscrete category to another is simply a function between their sets of underlying objects. In particular, given groups $G$ and $H$, a functor from $C(G)$ to $C(H)$ is simply a function from $G$ to $H$. That's any function (map of sets) whatsoever -- it completely ignores the group structure.
Given indiscrete categories $C$ and $D$ and functors $P, Q: C \to D$, there is always exactly one natural transformation from $P$ to $Q$. In particular, given groups $G$ and $H$ and functors $P, Q: C(G) \to C(H)$, there is always exactly one natural transformation from $P$ to $Q$.
| 11 | https://mathoverflow.net/users/586 | 14364 | 9,644 |
https://mathoverflow.net/questions/14361 | 11 | In "Quivers, perverse sheaves and quantized enveloping algebras," Lusztig defines a category of perverse sheaves on the moduli stack of representations of a quiver. These perverse sheaves are defined as summands of the pushforwards of the constant sheaves on stacks of quiver representations along with a choice of invariant flag (and thus, by definition are supported on the nilpotent locus in the moduli stack). They're mostly of interest since they categorify the canonical basis.
**My question is:** Is there a stratum in this stack where the pull-back of one of these sheaves is not the trivial local system?
Now, in finite type, this is not a concern, since each stratum is the classifying space of a connected algebraic group, and thus simply connected. But I believe in affine or wild type this is no longer true; this was at least my takeaway from the latter sections of "Affine quivers and canonical bases." However, I got a little confused about the relationship between the results of the two papers mentioned above, since they use quite different formalisms, so I hold out some hope that the local systems associated to symmetric group representations aren't relevant to the perverse sheaves for the canonical basis. Am I just hoping in vain?
| https://mathoverflow.net/users/66 | What do the local systems in Lusztig's perverse sheaves on quiver varieties look like? | For affine quivers, except cyclic ones, there are always perverse sheaves attached to nontrivial local systems.
If you just need an example, I recommend you to read McGerty's paper [math/0403279](http://arxiv.org/abs/math/0403279), before Lusztig's paper, where the Kronecker quiver case is studied in detail.
I also wrote a survey paper
Crystal, canonical and PBW bases of quantum affine algebras, in *Algebraic Groups and Homogeneous Spaces*, Ed. V.B.Mehta, Narosa Publ House. 2007, 389–421.
| 10 | https://mathoverflow.net/users/3837 | 14366 | 9,646 |
https://mathoverflow.net/questions/14118 | 7 | Let p be a prime and let K be a field containing the p'th roots of unity. Let E be an elliptic curve over K. We consider the the moduli problem $Y\_E(p)$, which sends L to set of elliptic curves F/L, and symplectic isomorphisms $\phi:E[p] \rightarrow F[p]$. We know that this moduli problem is representable by a curve over $K$, and we let the compactification of this curve be $X\_E(p)$. We know $X\_E(p)$ is a twist of $X(p)$. Similarly, we can construct $X\_E(p^2)$, and we see that $X\_E(p^2)$ is a normal cover of $X\_E(p)$. I think the Galois group of $X\_E(p^2)/X\_E(p)$ is $(Z/pZ)^3$. If that is the case, then given any K point of $X\_E(p)$, we can look at fiber over this point. This fiber is defined over K, hence it will define a field extension of K, with Galois group a subset of $(Z/pZ)^3$.
This means, if we have E and F defined over K, with $E[p] \equiv F[p]$, then we should be able to construct a cyclic extension of K of order p. What is that extension?
| https://mathoverflow.net/users/92 | Cyclic extensions coming from E[p] \equiv F[p], | The answer is that in fact this construction does not produce cyclic extensions! The problem is that $X\_E(p^2) \to X\_E(p)$ is not generically Galois; it is so only after extension of the ground field.
Here is a more detailed explanation. Assume that $p \ge 3$ and that $p \nmid \operatorname{char} K$. Then $Y\_E(p^2) \to Y\_E(p)$ is a torsor not under $(\mathbb{Z}/p\mathbb{Z})^3$, but under the étale group scheme $G$ corresponding to the Galois module of $\mathbb{F}\_p$-linear endomorphisms $g \colon E[p] \to E[p]$ of trace zero. Explicitly, $g \in G(\overline{K})$ maps $(F,\Phi) \in Y\_E(p^2)(\overline{K})$ to $(F,\Phi')$ where $\Phi'(x):=\Phi(x+g(px))$. (Any $\Phi'\colon E[p^2] \to F[p^2]$ with the same restriction to $E[p]$ as $\Phi$ arises from some $g \in \operatorname{End} E[p]$ in this way, and the trace-zero condition is what guarantees that $\Phi'$ is symplectic.) Finally, this Galois module is typically irreducible, in which case it does not have $\mathbb{Z}/p\mathbb{Z}$ as a quotient.
| 9 | https://mathoverflow.net/users/2757 | 14367 | 9,647 |
https://mathoverflow.net/questions/14238 | 16 | In question #7656, Peter Arndt asked [why the Gamma function completes the Riemann zeta function](https://mathoverflow.net/questions/7656/why-does-the-gamma-function-complete-the-riemann-zeta-function) in the sense that it makes the functional equation easy to write down. Several of the answers were from the perspective of Tate's thesis, which I don't really have the background to appreciate yet, so I'm asking for another perspective.
The perspective I want goes something like this: the Riemann zeta function is a product of the local zeta functions of a point over every finite prime $p$, and the Gamma function should therefore be the "local zeta function of a point at the infinite prime."
**Question 1:** Can this intuition be made precise without the machinery of Tate's thesis? (It's okay if you think the answer is "no" as long as you convince me why I should try to understand Tate's thesis!)
Multiplying the local zeta functions for the finite and infinite primes together, we get the Xi function, which has the nice functional equation. Now, as I learned from Andreas Holmstrom's [excellent answer to my question about functional equations](https://mathoverflow.net/questions/2040/why-are-functional-equations-important), for the local zeta functions at finite primes the functional equation
$$\zeta(X,n-s) = \pm q^{ \frac{nE}{2} - Es} \zeta(X, s)$$
(notation explained at [the Wikipedia article](http://en.wikipedia.org/wiki/Weil_conjectures#Statement_of_the_Weil_conjectures)), which for a point is just the statement $\frac{1}{1 - p^s} = -p^{-s} \frac{1}{1 - p^{-s}}$, reflects Poincare duality in etale cohomology, and the hope is that the functional equation for the Xi function reflects Poincare duality in some conjectural "arithmetic cohomology theory" for schemes over $\mathbb{Z}$ (or do I mean $\mathbb{F}\_1$?).
**Question 2:** Can the reflection formula for the Gamma function be interpreted as "Poincare duality" for some cohomology theory of a point "at the infinite prime"? (Is this question as difficult to answer as the more general one about arithmetic cohomology?)
| https://mathoverflow.net/users/290 | Why does the Gamma function satisfy a functional equation? | Your questions are a part of what Deninger has been writing about for 20 years. He's proposed a point of view that sort of explains a lot of things about zeta functions. It's important to say that this explanation is more in a theoretical physics way than in a mathematical way, in that, as I understand it, he's predicted lots of new things which he and other people have then gone on to prove using actual mathematics. I guess it's kind of like the yoga surrounding the Weil conjectures before Dwork and Grothendieck made actual cohomology theories that had a chance to do the job (and eventually did). It's pretty clear to me that he's put his finger on something, but we just don't know what yet.
Let me try to say a few things. But I should also say that I never worried too much about the details, because the details he has are about a made up picture, not the real thing. (If he had the real thing, everyone would be out of a job.) So my understanding of the actual mathematics in his papers is pretty limited.
Question 1: He gives some evidence that Euler factors at both finite and infinite places should be seen as zeta-regularized characteristic polynomials. For the usual Gamma function, see (2.1) in [1]. For the Gamma factors of general motives, see (4.1) in [1]. For the Euler factors at the finite places, see (2.3)-(2.7) in [2]. He gives a description that works simultaneously at the finite and infinite places in (0.1) of [2]. Beware that some of this is based on an artificial cohomology theory that is designed to make things uniform over the finite and infinite places. (Indeed, at the risk of speaking for him, probably the whole point was to see what such a uniform cohomology theory would look like, so maybe one day we'll be able to find the real thing.)
Question 2: He expects his cohomology theory to have a Poincare duality which is "compatible with respect to the functional equation". See the remarks and references in [3] between propositions 3.1 and 3.2.
I'd recommend having a look at [3]. It's mainly expository. Also, I remember [4] being a good exposition, but I don't have it in front of me now, so I can't say much. He also reviews things in section 2 of his recent Archive paper [5].
[1] "On the Gamma-factors attached to motives", Invent. Math. 104, pp 245-261
[2] "Local L-factors of motives and regularized determinants", Invent. Math. 107, pp 135-150
[3] "Some analogies between number theory and dynamical systems on foliated spaces", Proceedings of the ICM, Vol. I (Berlin, 1998), pp 163-186
[4] "Evidence for a cohomological approach to analytic number theory", First ECM, Vol. I (Paris, 1992), pp 491-510
[5] "The Hilbert-Polya strategy and height pairings", arxiv.org
| 13 | https://mathoverflow.net/users/1114 | 14373 | 9,652 |
https://mathoverflow.net/questions/14374 | 7 | By computing the sum of all Bernoulli numbers via Borel summation (I learned this technique from Varadarajan's excellent book *Euler through time. A new look at old themes*, 2006) I found that $$\sum B\_n = \int\_0^\infty \frac{t}{e^{2t}-e^t} dt$$ and discovered numerically that this expression equals $\zeta(2)-1$. The web is not very good for finding out where this can be found in print. Where should I look, and how can equations such as $$\zeta(2) = 1 + \int\_0^\infty \frac{t}{e^{2t}-e^t}\ dt$$ be proved?
| https://mathoverflow.net/users/3503 | Integral expression for zeta(2) | The starting point is the integral
$$
\Gamma(s) = \int\_{0}^{\infty}e^{-x}x^{s-1}dx
$$
for the gamma function. Make the change of variable $x = nu$ with $n$ an arbitrary positive integer. Then
$$
\Gamma(s)n^{-s} = \int\_{0}^{\infty}e^{-nu}u^{s-1}du
$$
and summing over $n$ from $n = 1$ yields
$$
\Gamma(s)\zeta(s) = \int\_0^{\infty}\frac{1}{e^u - 1}u^{s-1}du.
$$
This formula was the starting point of one of Riemann's two proofs of the functional equation. I am not certain who discovered it first, but it may have been Abel.
Substituting $s = 2$ gives
$$
\zeta(2) = \int\_{0}^{\infty}\frac{u}{e^u - 1}du
$$
and so
$$
\zeta(2) =
\int\_{0}^{\infty}\frac{ue^u}{e^{2u} - e^u}du =
\int\_{0}^{\infty}\frac{u(e^u - 1) + u}{e^{2u} - e^u}du =
\int\_{0}^{\infty}\left(ue^{-u} + \frac{u}{e^{2u} - e^u}\right)du =
1 + \int\_{0}^{\infty}\frac{u}{e^{2u} - e^u}du.
$$
| 13 | https://mathoverflow.net/users/3304 | 14381 | 9,657 |
https://mathoverflow.net/questions/14371 | 19 | I am now supposed to organize a tiny lecture course on algebraic geometry for undergraduate students who have an interest in this subject.
I wonder whether there are some basic algebraic geometry texts considering the level of undergraduate students who have not learnt commutative algebra or homological algebra; they just know linear algebra and basic abstract algebra.
I am looking for some textbooks which provide a lot of examples (more computations using linear algebra and calculus). Actually, I am also looking for some textbooks based on very basic mathematics but which talk a little bit about a modern view point.
Thanks in advance!
| https://mathoverflow.net/users/1851 | Looking for an introductory textbook on algebraic geometry for an undergraduate lecture course | [An invitation to algebraic geometry](http://www.amazon.fr/Invitation-Algebraic-Geometry-Karen-Smith/dp/0387989803) by Karen Smith is excellent; it is very intuitive, and does everything over the complex numbers. For absolute newcomers, this is probably the best introduction.
[Algebraic curves](http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf) by William Fulton is a classic, quite easily readable for beginners, and free available online in pdf! (He recently published the third edition on his site.)
The books by Reid, Miranda and Hulek are also good. Reid does many explicit examples.
| 28 | https://mathoverflow.net/users/1107 | 14382 | 9,658 |
https://mathoverflow.net/questions/14350 | 11 | Let $R$ be normal, local ring of dimension at least $2$. Let $M$ be a [reflexive](https://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7588#7588) $R$-module and let $A=Hom\_R(M,M)$. Suppose $A$ has [finite global dimension](https://en.wikipedia.org/wiki/Global_dimension). Then one can view $A$ as a *weak* non-commutative desingularization of $R$ (note that, a) there is a natural map $R\to A$ and b) in the commutative case, finite global dimension implies regularity, hence the name).
This concept imitates Van den Bergh's definition of non-commutative crepant resolution (NCCR). His definition arises from a proof of dimension $3$ case of [Bondal-Orlov conjecture](https://doi.org/10.1215/S0012-7094-04-12231-6). This is a long story, but an excellent account of the reasons behind the definition can be found in Section 4 of this [paper](https://arxiv.org/abs/math/0211064). For existence of NCCR in some high dimensions case, check out [this](https://arxiv.org/abs/0911.2659).
Now, non-commutative crepant resolution does not always exist (the above papers proves the equivalence, in some cases, with existence of projective crepant resolutions, which exists rarely in high dimensions). What we do know from Hironaka, in characteristic $0$ at least, is that resolution of singularity exist. So:
**Question**: Does weak non-commutative desingularizations of $R$ (as specified in the first paragraph) always exist?
Some discussions (I am a beginner in this, so feel free to correct me):
1. Why weak? By Morita equivalence, to ensures the desingularization is an isomorphism on the regular locus one needs $M$ to be free on that locus of $R$.
2. If one requires extra conditions (like $M$ being an generator-cogenerator) then there are examples when such desingularization does not exist (in some paper by Iyama which I forgot the name). I have not seen an example in the generality above.
3. There are positive results in diemension $0,1$, but I care about normal rings, so we start in dimension $2$.
4. Some people like Van den Bergh or Lieven le Bruyn probably know. May be they are even on MO!
I would appreciate even heuristic reasons for one way or another.
**EDIT**: The question is now resolved, by Lieven's answer below (as expected (:). I will provide a little bit more details in case someone is interested: Van den Bergh and Stafford [proved](https://arxiv.org/abs/math/0612032) that, in characteristic $0$, if $A$ is a non-commutative crepant resolution, then $R$ has rational singularity. The definition of NCCR is stronger, but if, for example, $R$ is Gorenstein of dimension $2$, it coincides with my version. So a counterexample is something like $R=k[x,y,z]/(x^3+y^3+z^3)$, which is a non-rational hypersurface.
| https://mathoverflow.net/users/2083 | Existence of non-commutative desingularizations | There are already counter-examples in dimension 2. If you take a 2-dml non-rational singularity, then there cannot exist a non-commutative resolution in your sense. In fact, any 2-dml nc-resolution your sense is also an nc-resolution in Michel's sense and so must have rational singularities by a [result of Toby Stafford and Michel](https://arxiv.org/abs/math/0612032).
In dimension two you can resort to ancient stuff such as the book 'Graded orders' by Fred, Michel and me, an online scanned version can be found [here](http://matrix.uantwerpen.be/lieven.lebruyn/LeBruyn1988f.pdf) ([Wayback Machine](https://web.archive.org/web/20201112163640/http://matrix.uantwerpen.be/lieven.lebruyn/LeBruyn1988f.pdf)). Lemma IV.2.3 says that a tame order (such as your End(M)) of global dim 2 is 'moderated regular' (defn IV.1.4) and hence a tame order of finite representation type which are classified further in the book to yield rational central singularities. That lemma (and dfn IV.1.1 and thm IV.1.2) also shows that your and Michel's dfns coincide in dim 2.
Further, I didn't understand your remark about an Iyama counterexample in the case M is projective? In that case End(M) is Azumaya, so if it has finite gldim, then the center has to be regular too.
| 8 | https://mathoverflow.net/users/2275 | 14389 | 9,663 |
https://mathoverflow.net/questions/14388 | 10 | Jack symmetric polynomials are known to be generalizations of Schur functions $\chi\_\lambda$, for which powerful Weyl determinant formulas are known.
Are there any generalizations of two determinant formulas for general Jack symmetric $P^\alpha\_\lambda(x)$ functions?
The first determinant (Jacobi-Trudi) formula represents the character of the irrep GL(N) given by the partition $\lambda$
$$
\chi\_\lambda(x)=\det\_{i,j} s\_{\lambda\_i-i+j}
$$
where $s\_k$ are elementary Schur function
and the second one gives the same function as determinant
$$
\chi\_\lambda(x)=\frac{\det\_{i,j} x\_i^{\lambda\_j+N-j}}{\det\_{i,j} x\_i^{N-j}}
$$
Jack symmetric polynomials are natural generalizations of Schur polynomials, and probably, to operate with them it would be useful to have as simple as possible analogs of Weyl formulas.
| https://mathoverflow.net/users/3840 | Jack polynomials as determinants | A natural analog would be a representation of the Jack polynomial as an alternating sum of eigenfunctions of the Calogero-Sutherland operator over an orbit of the symmetric group. This can be done even in the q-deformed case (i.e. for Macdonald polynomials and Macdonald operators). Such formula was conjectured by Felder and Varchenko and proved in the paper
arXiv:q-alg/9603022, see formula (5-5).
| 18 | https://mathoverflow.net/users/3696 | 14393 | 9,666 |
https://mathoverflow.net/questions/14404 | 53 | Has somebody translated J.-P. Serre's "Faisceaux algébriques cohérents" into English? At least part of it?
In a fit of enthusiasm, I started translating it and started TeXing. But after section 8, I got tired and stopped.
However if somebody else already took the trouble, I would be most grateful. I do not know a word of French(except maybe faisceau), and forgot whatever I learned in the process of translation very quickly.
This is made community wiki, as I do not want to get into rep issues. Please feel free to close this if you think this qn is inappropriate for MO(I have added my own vote for closing, in case this helps). I would be happy to receive answers in comments.
| https://mathoverflow.net/users/2938 | Serre's FAC in English | Together with some help from my friend, I translated FAC into English. I didn't have so much time to proofread it, so probably there are some mistakes.
It can be found here: [FAC](http://achinger.impan.pl/fac/fac.pdf), [Source](http://achinger.impan.pl/fac/fac.tar.gz).
| 89 | https://mathoverflow.net/users/3847 | 14417 | 9,683 |
https://mathoverflow.net/questions/14423 | 18 | I'm trying to learn about moment maps in symplectic topology (suppose our Lie group is $G$ with Lie algebra $\mathfrak g$, acting on the symplectic manifold $(M,\omega)$ by symplectomorphisms). I'm having a hard time, and I've realized this is because I don't have a good conceptual understanding of the Lie bracket, either on the Lie algebra $\mathfrak g$, or on the group of symplectomorphisms of $(M,\omega)$, or on the space of functions $\mathcal C^\infty(M,\mathbb R)$. Therefore I can't "visualize" the Hamiltonian condition, which requires that the linear map $\mathfrak g \rightarrow \mathcal C^\infty(M,\mathbb R)$, which exists when the action by $G$ is "exact," be a Lie algebra homomorphism.
Please tell me how you personally understand/intuit/conceptualize this situation, both the Lie bracket stuff and moment maps more generally! Any help is greatly appreciated.
**EDIT**: I didn't realize how non-standard some of this terminology is, so my question might be confusing. I call the action $\rho: G \rightarrow {\rm Symp}(M,\omega)$ "exact" if the image of the induced map $\rho: {\rm Lie}(G) \rightarrow {\rm Lie}({\rm Symp}(M,\omega))$ is contained in the sub-lie-algebra of Hamiltonian vector fields. The condition that was confusing me, I now realize, is just a technical point: that we choose a set of representative Hamiltonian functions for the image $\rho({\rm Lie}(G))$ which is a sub-Lie-algebra of $\mathcal C^\infty(M,\mathbb R)$ with its Poisson bracket. Thanks to all the helpful answers I think I understand this much better now.
In particular, if we present ${\rm Lie}(G)$ (assumed finite dimensional, semi-simple, etc) by Lie algebra generators (with some relations), then we can probably just choose appropriate elements in $\mathcal C^\infty(M,\mathbb R)$ for these generators, and then the rest of the map from ${\rm Lie}(G)$ to $\mathcal C^\infty(M,\mathbb R)$ is just forced on us, and this gives a Hamiltonian action? Is that right?
| https://mathoverflow.net/users/492 | Understanding moment maps and Lie brackets | This question is (at least as I read it) about the Poisson bracket; the Poisson bracket is a Lie bracket structure on the functions on a symplectic manifold.
So how should one think about Poisson bracket? Well, remember that for every function on a symplectic manifold, one has a Hamiltonian vector field $X\_f$. One way to think about this is that if your symplectic manifold is the phase space of a physical system (the space of possible positions and momenta), and $f$ is the energy function, then the resulting vector field is the derivative of the time evolution of the system.
The Poisson bracket $\{f,g\}$ is defined to be $X\_f(g)$, the derivative of $g$ along the vector field $X\_f$. That is **the Poisson bracket of $f$ and $g$ is the time derivative of $g$ if you use $f$ as the energy function**. Remarkably, this operation is anti-symmetric, and defines a Lie algebra structure.
As you mentioned, a moment map is equivalent to a Lie algebra homomorphism from $\mathfrak{g}$ to the space of functions on your manifold. I'm not sure how you're exactly supposed to visualize that, but let me explain how I think about it.
So, imagine you have your favorite G-action on a symplectic manifold, preserving the symplectic structure. Taking derivative, you get a map of Lie algebras from $\mathfrak{g}$ to vector fields on your manifold. It sounds from your question like how to think about such Lie algebra homomorphisms is actually what is confusing you. This just says that if you were to integrate the vector fields coming from $\mathfrak{g}$, you would get the group G (or maybe a finite cover).
Now, each of these vector fields corresponds under the symplectic form to a 1-form. If your manifold has no $H^1$, then you can integrate these to functions, but of course, you can't do this uniquely; it's only unique up to a constant. So taking all of these lifts, you get a vector space of functions which is $\dim \mathfrak{g}+1$ dimensional (assuming $G$ acted faithfully). This is closed under Lie bracket, so it's a finite dimensional Lie algebra $\tilde {\mathfrak{g}}$ with a map $\tilde {\mathfrak{g}}\to {\mathfrak{g}}$.
It might be that $\tilde {\mathfrak{g}}\cong {\mathfrak{g}}\times \mathbb{R}$ as a Lie algebra, in which case you can get a moment map by picking a splitting of the map above, or it might not, in which case you don't have a moment map. If $\mathfrak{g}$ is semi-simple, then the latter case is impossible, so you always have a moment map.
| 15 | https://mathoverflow.net/users/66 | 14427 | 9,690 |
https://mathoverflow.net/questions/14412 | 1 | Hello,
While reading the article Matrix Factorization Techniques for Recommender Systems I came across the following description:
"Matrix factorization models map both users and items to a joint latent factor space of dimensionality f, such that user-item interactions are modeled as inner products in that space."
Could someone explain to me --or guide me to an article -- what is meant by a "joint latent factor space of dimensionality f".
The articles is available [here](http://research.yahoo.com/files/ieeecomputer.pdf).
Thank you
| https://mathoverflow.net/users/3845 | Matrix Factorization Model | I'll give an example from politics. Let's say you have a legislative body, such as the House of Representatives in the U. S. Congress. Over a period of time, the members of the House will vote on many bills and thereby accrue a voting history. Let's encode votes as numbers: a vote for a bill is 1, a vote against a bill is -1, and an abstention (no vote) is 0. Also, let's label the representatives $R\_1,\ldots,R\_m$, and label the bills $B\_1,\ldots,B\_n$. We thus have, for each pair $i,j$ of numbers with $0 <i\leq m$, $0 < j \leq n$, a vote $V\_{ij}\in \{-1,0,1\}$, namely how congressperson $R\_i$ voted on bill $B\_j$. This gives us a big $m\times n$ "vote matrix" $V$ whose entries are the votes $V\_{ij}$.
Now, it's true that each congressperson has an opinion on every bill, or dually, that each bill appeals to different congresspeople in different amounts (possibly negative). However, that description misses an important aspect of the situation: a congressperson's voting behavior can be approximated using much fewer parameters than a list of all his votes. Also, a bill's tendency to appeal to different people can be approximated using much fewer parameters than a list of all the people who voted for and against it. Indeed, it's not really *bills* that congresspeople have opinions on; it's issues and policies. On the flip side, a given bill will implement various types of policies and address various issues, and that's really what determines who will like it and how much.
Thus, to describe voting behaviors, what you really need is (1) a list of policies $P\_1,\ldots,P\_f$ (the *latent factors*), (2) for each congressperson $R\_i$, a degree of preference $S\_{ik}$ for each policy $P\_k$, and (3) for each bill $B\_j$, a value $C\_{kj}$ describing to what extent it implements policy $P\_k$. Let's continue the convention that positive values for $S\_{ik}$ or $C\_{kj}$ indicate accordance and negative values indicate opposition. To each congressperson $R\_i$, we can assign the vector $S\_i = (S\_{i1},\ldots,S\_{if})$ (which we might call the "policy vector" for that congressperson), and to each bill $B\_j$, we can assign the vector $C\_j = (C\_{1j},\ldots,C\_{fj})$ (which we might also call the "policy vector" for that bill).
For what follows, I'm going to use a very simplistic (but somewhat plausible) mathematical model. (Your article uses a more complicated and more realistic model, taking bias into account, for example.) Also, the model makes a lot more sense if congresspeople are asked to state their degree of preference for each bill rather than simply voting "yes" or "no," so that the "votes" $V\_{ik}$ take values in $\mathbb{R}$. When deciding how to vote on a bill, a congressperson may consider how well it correlates with her opinions and make a decision based on that. With a lot of vigorous hand waving and wishful thinking, the outcome of this process can be described very simply in terms of policy vectors: the vote $V\_{ij}$ is simply the dot product $S\_i\cdot C\_j = \sum\_k S\_{ik} C\_{kj}$. (I'll leave it as an exercise to show that's not completely ridiculous, even if unlikely to be exactly true.) Another way of saying this is that if $S$ is the matrix with entries $S\_{ik}$ and $C$ is the matrix with entries $C\_{kj}$, then $V = SC$. In other words, our knowledge about legislative policies as latent factors induces a factorization of the matrix $V$.
One merit of the above approach is that although it's a bit too simple, it leads to well understood mathematics. For it to be useful, the number of policies $f$ should be much smaller than $m$ and $n$, the numbers of congresspeople and bills. In that case, the factorization $V = SC$ means that $V$ has rank $f$, which is small compared to its dimensions. In practice, admitting that the description in terms of policies as latent factors can only be a good approximation, not exact, this means that $V$ is well approximated by a low-rank matrix. Factorizations can be obtained from that observation alone using standard matrix tools like the singular-value decomposition. In particular, the policy vectors can be found even before you have any idea what the "policies" should be. (In other words, you don't have to sit down and make a list of policies you think are important and figure out what the policy vectors must be from that; you can use a standard algorithm which will determine the policies for you. Of course, it won't *name* the policies, but if you need to, you can compare policy vectors to figure out what real-world policies the algorithmically extracted policies approximate.)
| 6 | https://mathoverflow.net/users/302 | 14437 | 9,697 |
https://mathoverflow.net/questions/10512 | 45 | [I have rewritten this post in a way which I hope will remain faithful to the questioner and make it seem more acceptable to the community. I have also voted to reopen it. -- PLC]
There are many ways to approach noncommutative geometry.
What are some of the most important currently known approaches? Who are the principal creators of each of these approaches, and where are they coming from? E.g., what more established mathematical fields are they using as jumping off points? What problems are they trying to solve?
Two examples:
1) The Connes school, with an approach from C$^\*$-algebras/mathematical physics.
2) The Kontsevich school, with an approach from algebraic geometry.
| https://mathoverflow.net/users/2938 | Theories of Noncommutative Geometry | In accordance with the suggestion of Yemon Choi, I am going to suggest some further delineation of the approaches to "Non-commutative Algebraic Geometry". I know very little about "Non-commutative Differential Geometry", or what often falls under the heading "à la Connes". This will be completely underrepresented in this summary. For that I trust Yemon's summary to be satisfactory. (**edit by YC**: BB is kind to say this, but my attempted summary is woefully incomplete and may be inaccurate in details; I would encourage anyone reading to investigate further, keeping in mind that the NCG philosophy and toolkit in analysis did not originate and does not end with Connes.)
Also note that much of what I know about these approaches comes from two sources:
1. The paper by [Mahanta](https://arxiv.org/abs/math/0501166)
2. My advisor A. Rosenberg.
Additionally, much useful discussion took place at [Kevin Lin's](https://mathoverflow.net/questions/7917/non-commutative-algebraic-geometry) question (as Ilya stated in his answer).
I think a better break down for the NCAG side would be:
A. Rosenberg/Gabriel/Kontsevich approach
----------------------------------------
Following the philosophy of Grothendieck: "to do geometry, one needs only the category of quasi-coherent sheaves on the would-be space" (**edit by KL**: Where does this quote come from?)
In the famous [dissertation of Gabriel](http://www.numdam.org/item/?id=BSMF_1962__90__323_0), he introduced the injective spectrum of an abelian category, and then reconstructed the commutative noetherian scheme, which is a starting point of noncommutative algebraic geometry. Later, A. Rosenberg introduced the left spectrum of a noncommutative ring as an analogue of the prime spectrum in commutative algebraic geometry, and generalized it to any abelian category. He used one of the spectra to reconstruct any quasi-separated (not necessarily quasi-compact), commutative scheme. (Gabriel-Rosenberg reconstruction theorem.)
In addition, Rosenberg has described the NC-localization (first observed also by Gabriel) which has been used by him and Kontsevich to build NC analogs of more classical spaces (like the NC Grassmannian) and more generally, noncommutative stacks. Rosenberg has also developed the homological algebra associated to these 'spaces'. Applications of this approach include representation theory (D-module theory in particular), quantum algebra, and physics.
**References in this area** are best found through the MPIM Preprint Series, and a large collection is linked [here](https://ncatlab.org/nlab/show/Alexander%20Rosenberg). Additionally, a book is being written by Rosenberg and Kontsevich furthering the work of their previous [paper](https://arxiv.org/abs/math/9812158). Some applications of these methods are used [here](https://doi.org/10.1007/s11464-008-0027-8), [here](https://www.cambridge.org/core/journals/bulletin-of-the-australian-mathematical-society/article/on-representations-of-quantum-groups-uqfmkh/072A7D6AF15FBF14545E25A90F64E22C), [here](https://arxiv.org/abs/math/0301090), and [here](https://arxiv.org/abs/math/0403276). The first two are focusing on representation theory, the second two on non-commutative localization.
Kontsevich/Soibelman approach
-----------------------------
They might refer to their approach as "formal deformation theory", and quoting directly from their [book](https://www.math.ksu.edu/%7Esoibel/Book-vol1.ps)
>
> The subject of deformation theory can be defined as the "study of moduli spaces of structures...The subject of this book is formal deformation theory. This means $\mathcal{M}$ will be a formal space(e.g. a formal scheme), and a typical category $\mathcal{W}$ will be the category of affine schemes..."
>
>
>
Their approach is related to $A\_{\infty}$ algebras and homological mirror symmetry.
**References that might help** are the papers of [Soibelman](https://www.lanl.gov/errors/system-notification.php). Also, I think this is related to the question [here](https://mathoverflow.net/questions/13005/what-is-formal). (Note: I know hardly anything beyond that this approach exists. If you know more, feel free to *edit* this answer! Thanks for your understanding!)
(**Some comments by KL**: I am not sure whether it is appropriate to include Kontsevich-Soibelman's deformation theory here. This kind of deformation theory is a very general thing, which intersects some of the "noncommutative algebraic geometry" described here, but I think that it is neither a subset nor a superset thereof. In any case, I've asked some questions related to this on MO in the past, see [this][22] and [this][23].
However, there is the approach of noncommutative geometry via categories, as elucidated in, for instance, [Katzarkov-Kontsevich-Pantev][24]. Here the idea is to think of a category as a category of sheaves on a (hypothetical) non-commutative space. The basic "non-commutative spaces" that we should have in mind are the "Spec" of a (not necessarily commutative) associative algebra, or dg associative algebra, or A-infinity algebra. Such a "space" is an "affine non-commutative scheme". The appropriate category is then the category of modules over such an algebra. Definitively commutative spaces, for instance quasi-projective schemes, are affine non-commutative schemes in this sense: It is a theorem of van den Bergh and Bondal that the derived category of quasicoherent sheaves on a quasi-projective scheme is equivalent to a category of modules over a dg algebra. (I should note that in my world everything is over the complex field; I have no idea what happens over more general fields.) Lots of other categories are or should be affine non-commutative in this sense: [Matrix factorization categories][25] (see in particular [Dyckerhoff][26]), and probably various kinds of Fukaya categories are conjectured to be so as well.
Anyway I have no idea how this kind of "noncommutative algebraic geometry" interacts with the other kinds explained here, and would really like to hear about it if anybody knows.)
Lieven Le Bruyn's approach
--------------------------
As I know nearly nothing about this approach and the author is a visitor to this site himself, I wouldn't dare attempt to summarize this work.
As mentioned in a comment, his website contains a plethora of links related to non-commutative geometry. I recommend you check it out [yourself](https://web.archive.org/web/20181221093452/http://win.ua.ac.be:80/%7Elebruyn/index.html).
Approach of Artin, Van den Berg school
--------------------------------------
Artin and Schelter gave a regularity condition on algebras to serve as the algebras of functions on non-commutative schemes. They arise from abstract triples which are understood for commutative algebraic geometry. (Again edits are welcome!)
Here is a nice report on [Interactions between noncommutative algebra and algebraic geometry](https://www.birs.ca/workshops/2005/05w5035/report05w5035.pdf). There are several people who are very active in this field: Michel Van den Berg, James Zhang, Paul Smith, Toby Stafford, I. Gordon, A. Yekutieli. There is also a very nice page of Paul Smith: [noncommutative geometry and noncommutative algebra](https://sites.math.washington.edu//%7Esmith/Research/research.html), where you can find almost all the people who are currently working in the noncommutative world.
**References:** [This][16] paper introduced the need for the regularity condition and showed the usefulness. Again I defer to [Mahanta][17] for details.
Serre's FAC is the starting point of noncommutative projective geometry. But the real framework is built by Artin and James Zhang in their famous paper [Noncommutative Projective scheme][18].
Non-commutative Deformation Theory by Laudal
--------------------------------------------
Olav Laudal has approached NCAG using NC-deformation theory. He also applies his method to invariant theory and moduli theory. (Please edit!)
**References** are on his page [here][19] and [this][20] paper seems to be a introductory article.
Apologies
---------
Without a doubt, I have made several errors, given bias, offended the authors, and embarrassed myself in this post. Please don't hold this against me, just edit/comment on this post until it is satisfactory. As it was said before, the [nlab][21] article on noncommutative geometry is great, you should defer to it rather than this post.
Thanks!
[16]: [https://books.google.com/books?hl=en&lr=&id=\_BnSoQSKnNUC&oi=fnd&pg=PA33&dq=%252522Artin%252522+%252522Some+algebras+associated+to+automorphisms+of+elliptic+curves%252522+&ots=hRXnP7udMW&sig=t77CnWnsYPHhuonQQffrSXedyj0#v=onepage&q="Artin"](https://books.google.com/books?hl=en&lr=&id=_BnSoQSKnNUC&oi=fnd&pg=PA33&dq=%252522Artin%252522+%252522Some+algebras+associated+to+automorphisms+of+elliptic+curves%252522+&ots=hRXnP7udMW&sig=t77CnWnsYPHhuonQQffrSXedyj0#v=onepage&q=%22Artin%22) "Some algebras associated to automorphisms of elliptic curves"&f=false
[17]: <https://arxiv.org/abs/math/0501166>
[18]: <https://web.archive.org/web/20121023193142/http://www.ingentaconnect.com/content/ap/ai/1994/00000109/00000002/art01087>
[19]: <https://web.archive.org/web/20181103123848/http://folk.uio.no:80/arnfinnl/>
[20]: <https://web.archive.org/web/20080425144650/http://folk.uio.no/arnfinnl/Noncom.alg.geom.pdf>
[21]: <https://ncatlab.org/nlab/show/noncommutative%20geometry>
[22]: [What is a deformation of a category?](https://mathoverflow.net/questions/5364/what-is-a-deformation-of-a-category)
[23]: [Deformation theory and differential graded Lie algebras](https://mathoverflow.net/questions/385/deformation-theory-and-differential-graded-lie-algebras)
[24]: <https://arxiv.org/abs/0806.0107>
[25]: [Matrix factorizations and physics](https://mathoverflow.net/questions/9733/matrix-factorizations-and-physics)
[26]: <https://arxiv.org/abs/0904.4713>
| 28 | https://mathoverflow.net/users/348 | 14443 | 9,702 |
https://mathoverflow.net/questions/14450 | 11 | Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$. Define $\mathcal Z(\mathfrak g)$ to be the center of the universal enveloping algebra $\mathcal U\mathfrak g$, and define $(\mathcal S\mathfrak g)^{\mathfrak g}$ to be the ring of invariant elements of the symmetric algebra $\mathcal S\mathfrak g$ under the induced adjoint action of $\mathfrak g$. (Clearly $\mathcal Z(\mathfrak g) = (\mathcal U\mathfrak g)^{\mathfrak g}$ via the adjoint action.) The *Duflo isomorphism* is an isomorphism of algebras $\mathcal Z(\mathfrak g) \cong (\mathcal S\mathfrak g)^{\mathfrak g}$. At the level of vector spaces, the trick is to realize that the PBW map $\mathcal U\mathfrak g \to \mathcal S \mathfrak g$ is an isomorphism of $\mathfrak g$-modules. For the isomorphism of algebras in the semisimple case, see for example [my unedited notes on the class by V. Serganova](http://math.berkeley.edu/~theojf/QuantumGroups10.pdf).
(I read [here](http://www.sfb45.de/events/the-duflo-isomorphism-and-its-relatives) that this isomorphism can be realized as a composition of the PBW vector-space isomorphism $\mathcal U\mathfrak g \to \mathcal S\mathfrak g$ with the map $\mathcal S\mathfrak g \to \mathcal S\mathfrak g$ given by $x \mapsto \sinh(x/2)/(x/2)$. But it's not at all obvious that this composition is even well-defined or linear when restricted to $\mathcal Z(\mathfrak g)$. I should mention that $\mathcal Z$ is not a functor, I think. The PBW isomorphism is non-canonical, although a canonical version can be given via the symmetrization map, and I guess on the center it is canonical.)
When $\mathfrak g$ is semisimple of rank $n$, at least, one can further show that $(\mathcal S\mathfrak g)^{\mathfrak g} \cong \mathbb C[x\_1,\dots,x\_n]$, although you have some choice about how to make this isomorphism. Thus, at least when $\mathfrak g$ is semisimple, $\mathcal Z(\mathfrak g)$ is a polynomial ring.
But any polynomial ring can be given a Hopf structure. By choosing an isomorphism with $\mathbb C[x\_1,\dots,x\_n]$, we can take the Hopf structure generated by $\Delta: x\_i \mapsto x\_i \otimes 1 + 1 \otimes x\_i$. In fact, this structure doesn't depend quite on the full choice of isomorphism. A Hopf structure on a commutative algebra $R$ is by definition the same as an algebraic group structure on $\text{Spec}(R)$. But $\text{Spec}(\mathbb C[x\_1,\dots,x\_n])$ is $n$-dimensional affine space — the algebra isomorphisms of $\mathbb C[x\_1,\dots,x\_n])$ are precisely the affine maps — so picking a commutative group structure is the same as picking an origin. (For certain values of $n$ there are also non-commutative group structures on affine $n$-space, and so non-cocommutative Hopf structures on the polynomial ring. For example, the group of upper-triangular matrices with $1$s on the diagonal is affine.)
My question is whether this Hopf structure can be picked out canonically.
>
> **Question:** If $\mathfrak g$ is a finite-dimensional Lie algebra over $\mathbb C$, can the center $\mathcal Z(\mathfrak g)$ of the universal enveloping algebra be given a canonical (cocommutative) Hopf algebra structure? If no, how much extra structure on $\mathfrak g$ is needed?
>
>
>
Here by "canonical" I of course don't mean that there is a unique one, so you may make choices once and for all. But there should be some definition/construction that does not require the user to make any choices to implement it. By "extra structure" I mean either extra structure (an invariant metric, for example) or extra properties (semisimplicity, for example).
My suspicion is that the answer is "yes" for a *metric Lie algebra*, which is a Lie algebra $\mathfrak g$ along with a choice of an invariant nondegenerate metric, i.e. a chosen isomorphism of $\mathfrak g$-modules $\mathfrak g \cong \mathfrak g^\*$. Metric Lie algebras include the semisimples and the abelians, and certain extensions of these (in fact, I believe that there is a structure theorem that any metric Lie algebra is a metric extension of semisimples and abelians, but don't quote me), but generally there are many choices of metric (e.g. any metric on an abelian Lie algebra $\mathfrak a$ is invariant, so there are $\mathfrak{gl}(\dim \mathfrak a)$ many choices).
The motivation for my question is this: by studying Vassiliev invariant and/or perturbative Chern-Simons theory, Bar Natan and others have defined a certain commutative and cocommutative Hopf algebra $A$ of "diagrams". Any choice of metric Lie algebra $\mathfrak g$ determines an algebra homomorphism $A \to \mathcal Z(\mathfrak g)$. I would like to know if this can be made into a Hopf algebra homomorphism.
| https://mathoverflow.net/users/78 | Is there a canonical Hopf structure on the center of a universal enveloping algebra? | In the semisimple case, one has the Harish-Chandra isomorphism between the center ${\mathcal Z}(\mathfrak g)$ and $(S{\mathfrak h})^W$, where ${\mathfrak h}$ is a Cartan subalgebra of ${\mathfrak g}$ and $W$ is the Weyl group. On $S{\mathfrak h}$, there is a natural inner product $(f,g)=f(\partial)g(x)|\_{x=0}$ induced by the natural inner product on ${\mathfrak h}$.
Let $I\subset (S{\mathfrak h})^W$ be the augmentation ideal. Let $E$ be the orthogonal complement of the ideal $I^2$ in $(S{\mathfrak h})^W$ (the square of $I$), and let $E\_+$ be the positive degree part of $E$. Then $(S{\mathfrak h})^W=SE\_+$, so we can define the Hopf algebra structure on $(S{\mathfrak h})^W$ by declaring that $E\_+$ consists of primitive elements.
This was explained to me by Kostant. I don't know, however, if it is sufficiently canonical, or helpful in the question related to the Bar-Natan construction.
| 8 | https://mathoverflow.net/users/3696 | 14457 | 9,708 |
https://mathoverflow.net/questions/14245 | 3 | Consider the optimization problem
$$\min\_x ||Ax||\_1 + \lambda||x-b||^2,$$
where $A \in \mathbb{R}^{n \times n}$, $x,b \in \mathbb{R}^n$ and $\lambda$ is strictly greater than 0. (This problem is closely related to the "lasso" problem in basis pursuit.) Can anything be said about the value of $\lambda$ for which $Ax^\*$ is sparsest? Clearly some values are bad: for instance, if $\lambda$ is huge and $b$ is dense then it is unlikely that $Ax^\star$ will be very sparse.
In other words: among all $\lambda > 0$ there is at least one value $\lambda^\star$ such that $||Ax^\star(\lambda)||\_0$ is minimized. Are there, say, bounds on $\lambda^\star$ in terms of $A$ and $b$? I'd also be interested in results pertaining to basis pursuit or other similar problems.
*Edit:* I'm primarily interested in problems where ideal sparsity cannot be achieved, i.e., $||Ax^\star(\lambda^\star)||\_0 > 0.$ (Assume that $A$ is square w/ full rank and $b \ne 0$.)
| https://mathoverflow.net/users/1557 | Maximizing Sparsity in l1 Minimization? | The ultimate sparseness occurs when $Ax^\*(\lambda)=0$, which is the case when the minimizer $x^\*$ is the projection of $b$ onto $\ker A$. For this to happen, $\lambda$ must be small enough so that the restriction of $A$ to the orthogonal complement of its kernel is bounded from below by a constant greater than $2\lambda \mathrm{dist}(b,\ker A)$. Here the lower bound for operator is understood in the $\ell^2\to\ell^1$ norm.
| 1 | https://mathoverflow.net/users/2912 | 14461 | 9,710 |
https://mathoverflow.net/questions/14446 | 8 | Let $C$ be a smooth curve of degree $d$ in $\mathbb{P}^2$ over $\mathbb{C}$. Say $C$ is defined by $p(x,y,z)=0$, with $p$ a homogeneous degree $d$ polynomial.
In vector calculus one learns that the gradient of $p$ is normal to $C$ at every point of the curve.
In algebraic geometry, the invertible sheaf associated to the normal bundle $N\_{C|\mathbb{P}^2}$ to $C$ in $\mathbb{P}^2$, is given by $\mathcal{O}\_{\mathbb{P}^2}(d) \_{|C}$.
Is there any relationship between the gradient and the bundle or the sheaf?
| https://mathoverflow.net/users/1724 | Normal bundle to a curve in P^2 | Yes, there is a strong relationship between the two.
First, let's work locally in affine space rather than in projective space (it makes more
sense to work locally just because we are dealing with a sheaf, which is defined locally).
So I will consider a non-homogen
Working without a metric (as one does in at least the algebraic aspects of algebraic geometry),
it is perhaps better to talk not about the gradient of $f$, but its exterior derivative
$df$, given by the same formula: $df = f\_x dx + f\_y dy.$ Since this is differential form
valued, we will compare it with the conormal bundle to the curve $C$ cut out by $f = 0$.
Now the exterior derivative can be thought of simply as taking the leading (i.e. linear) term of $f$.
On the other hand, if $\mathcal I$ is the ideal sheaf cutting out the curve $C$, then the
conormal bundle is $\mathcal I/\mathcal I^2$. (If $f$ is degree $d$, then $\mathcal I = \mathcal O(-d)$, and so this can be rewritten as $\mathcal O(-d)\\_{| C}$, dual to the normal
bundle $\mathcal O(d)\\_{| C}$.) Now $f$ is a section of $\mathcal I/\mathcal I^2$ (over the affine patch on which we are working), so we may certainly regard it as a section of $\mathcal I/\mathcal I^2$; this section
*is* the (image in the conormal bundle to $C$ of) the exterior derivative of $f$.
The formula $\mathcal I/\mathcal I^2$ for the conormal bundle is thus simply a structural
interpretation of the idea that we compute the normal to the curve by taking the leading term
of an equation for the curve.
| 9 | https://mathoverflow.net/users/2874 | 14470 | 9,716 |
https://mathoverflow.net/questions/14456 | 18 | Lagrange proved that every positive integer is a sum of 4 squares.
Are there general results like this for rings of integers of number fields? Is this class field theory?
Explicitly, suppose a number field is formally real. Denote its ring of integers by $Z$. Is it true that for every algebraic integer $x$ in $Z$ either $x$ or $-x$ is a sum of squares?
| https://mathoverflow.net/users/nan | sum of squares in ring of integers | To address the particularities of this question for number fields, the basic theorem is attributed to Hilbert, Landau and Siegel. First of all, any nonzero sum of squares in a number field has to be totally positive (that is, it is positive in all real embeddings). Hilbert (1902) conjectured that in any number field, a totally positive element is a sum of 4 squares in the number field. This was proved by Landau (1919) for quadratic fields and by Siegel (1921) for all number fields.
This sounds superficially like a direct extension of Lagrange's theorem, but there is a catch: it is about field elements, not algebraic integers as sums of squares of algebraic integers. A totally positive algebraic integer in a number field $K$ need not be a sum of 4 squares of *algebraic integers* in $K$. The Hilbert-Landau-Siegel theorem only says it is a sum of 4 squares of algebraic numbers in $K$.
For instance, in $\mathbf{Q}(i)$ all elements are totally positive in a vacuous sense (no real embeddings), so every element is a sum of four squares. As an example,
$$
i = \left(\frac{1+i}{2}\right)^2 + \left(\frac{1+i}{2}\right)^2.
$$
This shows $i$ is a sum of two squares in $\mathbf{Q}(i)$.
It is impossible to write $i$ as a finite sum of squares in ${\mathbf Z}[i]$ since
$$
(a+bi)^2 = a^2 - b^2 + 2abi
$$
has even imaginary part when $a$ and $b$ are in $\mathbf{Z}$. Thus any finite sum of squares in $\mathbf{Z}[i]$ has even imaginary part, so such a sum can't equal $i$.
Therefore it is false that every totally positive algebraic integer in a number field is a sum of 4 squares (or even any number of squares) of *algebraic integers*.
Here are some further examples:
1. In $\mathbf{Q}(\sqrt{2})$, $5 + 3\sqrt{2}$ is totally positive since
$5+3\sqrt{2}$ and $5-3\sqrt{2}$ are both positive. So it must be a sum of at most four squares in this field by Hilbert's theorem, and with a little fiddling around you find
$$
5 + 3\sqrt{2} = (1+\sqrt{2})^2 + \left(1 + \frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2.
$$
It is impossible to write $5 + 3\sqrt{2}$ as a sum of squares in the ring of integers $\mathbf{Z}[\sqrt{2}]$ because of the parity obstruction we saw for $i$ as a sum of squares in $\mathbf{Z}[i]$: the coefficient of $\sqrt{2}$ in $5 + 3\sqrt{2}$ is odd.
2. In $\mathbf{Q}(\sqrt{2})$, $\sqrt{2}$ is not totally positive (it becomes negative when we replace $\sqrt{2}$ with $-\sqrt{2}$), so it can't be a sum of squares in this field. But in the larger field $\mathbf{Q}(\sqrt{2},i)$, everything is totally positive in a vacuous sense so everything is a sum of at most four squares in this field by the Hilbert-Landau-Siegel theorem. And looking at $\sqrt{2}$ in $\mathbf{Q}(\sqrt{2},i)$, we find
$$
\sqrt{2} = \left(1 + \frac{1}{\sqrt{2}}\right)^2 + i^2 + \left(\frac{i}{\sqrt{2}}\right)^2.
$$
Hilbert made his conjecture on totally positive numbers being sums of four squares as a theorem, in his *Foundations of Geometry*. It is Theorem 42. He says the proof is quite hard, and no proof is included. A copy of the book (in English) is available at the time I write this as <http://math.berkeley.edu/~wodzicki/160/Hilbert.pdf>. See page 83 of the file (= page 78 of the book).
Siegel's work on this theorem/conjecture was done just before the Hasse-Minkowski theorem was established in all number fields (by Hasse), and the former can be regarded as a special instance of the latter.
Indeed, for nonzero $\alpha$ in a number field $K$, consider the quadratic form $$Q(x\_1,x\_2,x\_3,x\_4,x\_5) = x\_1^2+x\_2^2+x\_3^2+x\_4^2-\alpha{x}\_5^2.$$ To say $\alpha$ is a sum of four squares in $K$ is equivalent to saying $Q$ has a nontrivial zero over $K$. (In one direction, if $\alpha$ is a sum of four squares over $K$ then $Q$ has a nontrivial zero over $K$ where $x\_5 = 1$. In the other direction, if $Q$ has a nontrivial zero over $K$ where $x\_5 \not= 0$ then we can scale and make $x\_5 = 1$, thus exhibiting $\alpha$ as a sum of four squares in $K$. If $Q$ has a nontrivial zero over $K$ where $x\_5 = 0$ then the sum of four squares quadratic form represents 0 nontrivially over $K$ and thus it is universal over $K$, so it represents $\alpha$ over $K$.) By Hasse-Minkowski, $Q$ represents 0 nontrivially over $K$ if and only if it represents 0 nontrivially over every completion of $K$.
Since any nondegenerate quadratic form in five or more variables over a local field or the complex numbers represents 0 nontrivially, $Q$ represents 0 nontrivially over $K$ if and only it represents 0 nontrivially in every completion of $K$ that is isomorphic to ${\mathbf R}$. The real completions of $K$ arise precisely from embeddings $K \rightarrow {\mathbf R}$. For $t \in {\mathbf R}^\times$, the equation
$x\_1^2+x\_2^2+x\_3^2+x\_4^2-t{x}\_5^2 =0$ has a nontrivial real solution if and only if $t > 0$, so $Q$ has a nontrivial representation of 0 in every real completion of $K$ if and only if $\alpha$ is positive in every embedding of $K$ into ${\mathbf R}$, which is what it means for $\alpha$ to be totally positive. (Strictly speaking, to be totally positive in a field means being positive in every ordering on the field. The orderings on a *number field* all arise from embeddings of the number field into $\mathbf R$, so being totally positive in a number field is the same as being positive in every real completion.)
Siegel's paper is "Darstellung total positiver Zahlen durch Quadrate, Math. Zeit. 11 (1921), 246--275, and can be found online at <http://gdz.sub.uni-goettingen.de/en/dms/loader/img/?PPN=PPN266833020_0011&DMDID=DMDLOG_0022>.
| 51 | https://mathoverflow.net/users/3272 | 14473 | 9,719 |
https://mathoverflow.net/questions/14481 | -2 | Hi,
Could someone explain to me or point out some documentation on how to compute a given percentile from a histogram ?
| https://mathoverflow.net/users/3867 | calculate percentiles from a histogram | A histogram gives you the number $n\_i$ of observations between some $x\_i$ and $x\_{i+1}$. I'm assuming a total of $n$ observation. So, to get an approximation for the upper $p$-percentile, you want to find the maximal $j$ such that $\sum\_{i=j}^\infty n\_i\geq p\*n$. Then, the empirical upper $p$-percentile is between $x\_j$ and $x\_{j+1}$.
| -1 | https://mathoverflow.net/users/2933 | 14482 | 9,724 |
https://mathoverflow.net/questions/13873 | 6 | The question I will ask makes sense in much more generality, but I will leave the translation to the experts, since I'm only looking for a special case (and it would not surprise me if the answer does not generalize). I will give some background, and then ask my question as a conjecture, set apart from the main text.
Let $\mathbb R^n$ have its usual metric, and pick a differential one-form (= vector field) $B$ (the "magnetic potential") and a differential zero-form (= function) $C$ (the "electric potential"). Then consider the following second-order ODE for parameterized paths $\gamma: [0,T] \to \mathbb R^n$:
$$ 0 = \ddot \gamma + dB \cdot \dot\gamma + dC \quad\quad \text{(EOM)} $$
I'll let you pick the signs for how the two-form $dB$ eats the vector $\dot\gamma$; just be consistent.
Then (EOM) is nondegenerate, and so a solution is determined by its initial conditions $(\dot\gamma(0),\gamma(0))$. For each $T \in \mathbb R$, let $\phi\_T: \mathbb R^{2n} \to \mathbb R^n$ be the "flow by time $T$" (actually, it is defined only on an open subset of $\mathbb R^{2n}$, given by $\phi\_T(v,q) = \gamma(T)$ for the solution $\gamma\\,$ to (EOM) with initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$. Then $\phi\_T$ is smooth; in fact, it is smooth in the $T$ variable as well. This follows from some standard fundamental result in ODEs, for which I don't have a good reference.
A path $\gamma: [0,T] \to \mathbb R^n$ is *classical* if it satisfies (EOM); its *duration* is the number $T$. We can also consider paths with negative duration by flowing backwards, although we will not need to do so.
**Definition:** A point $(v,q) \in \mathbb R^{2n}$ is *focal for duration $T$* iff ($\phi\_T(v,q)$ is defined and) $\det(\partial \phi\_T(v,q)/\partial v) = 0$; i.e. fix the $q$, think of $\phi\_T(-,q)$ as a function of $v$ only, and ask that its differential is degenerate. By identifying $(v,q)$ with its classical path, we will talk about "focal (classical) paths" for given durations.
It is a standard results (see e.g. Milnor's *Morse Theory*) that for a given point $(v,q) \in \mathbb R^{2n}$, the durations $T\in \mathbb R$ for which it is focal are discretely separated. Note that every $(v,q)$ is focal for duration $T=0$.
**Proposition:** Let $\gamma$ be a classical path of duration $T$. Then it is non-focal if and only if it extends to a family of classical paths smoothly parametrized by the boundary positions $(\gamma(0),\gamma(T))$.
**Sketch of Proof:** Being focal for duration $T$ is a closed condition on $\mathbb R^{2n}$, so we can vary $\gamma(0) = q$ while remaining non-focal. But for non-focal paths we can vary $\gamma(T)$ via the inverse function theorem.
Anyway, pick $q \in \mathbb R^n$, and $v = B(q)$ (or $-B(q)$ depending on your sign convention: for experts, I want the momentum to vanish). Then for some $\epsilon>0$, for all $T\in (0,\epsilon)$, $(v,q)$ is non-focal for duration $T$. Thus, for each $T \in (0,\epsilon)$, I can find an open neighborhood $q \in \mathcal O\_0 \subseteq \mathbb R^n$ and another open neighborhood $\mathcal O\_1 \subseteq \mathbb R^n$ so that for $(q\_0,q\_1) \in \mathcal O\_0 \times \mathcal O\_1$, there is a non-focal classical path $\gamma$ of duration $T$ with $\gamma(0) = q\_0$, $\gamma(T) = q\_1$, depending smoothly on the boundary conditions, and such that the classical path of duration $T$ and initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$ is contained within this family.
Note that as $T \to 0$, the classical path with initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$ ends at a point very close to $q$. I don't know if I can take $\mathcal O\_1$ to actually contain $q$.
I would like to reverse the direction of choices: I'd like to pick $\mathcal O\_0,\mathcal O\_1$ first.
>
> **Question/Conjecture:** Let $q \in \mathbb R^n$. Then there exist open neighborhood $\mathcal O\_0,\mathcal O\_1 \subseteq \mathbb R^n$, with $q \in \mathcal O\_0,\mathcal O\_1$, and $\epsilon>0$ such that:
>
>
> 1. There exists a family of classical paths $\gamma$ with boundary values varying in $\mathcal O\_0,\mathcal O\_1$ and with duration varying in $(0,\epsilon)$. I.e. let $\Delta = \{ (T,t) \in \mathbb R^2 : T \in (0,\epsilon), t\in [0,T] \}$; then there is a smooth function $\gamma: \mathcal O\_0 \times \mathcal O\_1 \times \Delta \to \mathbb R^n$ with: (a) $\gamma(q\_0,q\_1,T,-)$ is classical for each $(q\_0,q\_1,T) \in \mathcal O\_0 \times \mathcal O\_1 \times (0,\epsilon)$, and (b) $\gamma(q\_0,q\_1,T,0) = q\_0$ and $\gamma(q\_0,q\_1,T,T) = q\_1$.
> 2. For each $T \in (0,\epsilon)$, the classical path of duration $T$ with initial conditions $(B(q),q)$ appears as some $\gamma(q,q\_1,T,-)$.
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For comparison, the corresponding theorem about geodesics on a Riemannian manifold is standard: around any point you can find a small neighborhood such that any two points in the neighborhood can be connected by a unique geodesic that does not leave the neighborhood. In fact, it follows from the proposition and the observation that changing the duration of a geodesic for fixed boundary conditions amounts just to a linear reparameterization.
| https://mathoverflow.net/users/78 | Do there exist small neighborhoods in a classical mechanical system without pairs of focal points? | For a sufficiently large particle energy, the original problem can be transformed to a problem of geodesic motion as follows:
The motion of a classical particle in an external magnetic field in n-dimensions can be
seen as a symplectic reduction of a geodesic motion in n+1 dimensions (Rn \* S1) through the
Kaluza-Klein construction, given for example in section 7.6 of [Marsden's book.](http://books.google.com/books?id=I2gH9ZIs-3AC&printsec=frontcover&dq=Marsden+%2B+Introduction+to+mechanics+and+symmetry&source=bl&ots=STIWna6Ib8&sig=qrkEUDWwL1mbzkoNPA8OPiAWyEY&hl=en&ei=ps5uS8KQDpWCnQPN3ZnSBA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CBEQ6AEwAQ#v=onepage&q=&f=false) The remaining problem is a geodesic motion in an electric potential field.
Now suppose that there exists a region in the vicinity of the origin where the
electric potential is bounded from above, and the particle's energy (The value of the
Kaluza-Klein Hamiltonian which is a constant of motion) is larger than the maximum potential.
In this case, the trajectories in this region are equivalent up to a reparametrization
to a free geodesic motion in the Jacobi metric (see section 7.7).
Thus, in this case, the original problem is equivalent to a Riemannian problem.
| 3 | https://mathoverflow.net/users/1059 | 14494 | 9,730 |
https://mathoverflow.net/questions/14484 | 6 | I was looking through my notes for a homotopy theory course and found the following mysterious statement (*K* is of course the Eilenberg-Maclane space):
$$H^{n+1}(K(\mathbb Z\_p,n);\mathbb Z\_p) \cong \mathbb Z\_p.$$
(This would be obvious if *n+1* were replaced with *n*. This is supposed to imply that the natural transformations $H^n(X; \mathbb Z\_p)\to H^{n+1}(X; \mathbb Z\_p)$ are all multiples of the Bockstein homomorphism).
I'm at a loss trying to understand why. Spectral sequences haven't been covered yet, so there should be some simple reason. Also, is there a way to see the Bockstein in all this?
Thank you!
| https://mathoverflow.net/users/2467 | The (n+1)-st cohomology of K(Z/p,n). | Universal coefficient theorem + $H\_{n+1}(K(Z\_p,n);Z)=0$. An elementary way to see the latter is that the single $n+1$ cell added to $S^n$ to kill $p$ times the generator is not a cellular cycle.
| 12 | https://mathoverflow.net/users/3874 | 14497 | 9,732 |
https://mathoverflow.net/questions/14514 | 5 | Let G be a nontrivial finite group. Does there exist an irreducible representation of G of dimension greater than or equal to the cardinality of G?
[Edited for clarity. -- PLC]
| https://mathoverflow.net/users/3876 | How big can the irreps of a finite group be (over an arbitrary field)? | **EDIT:** Part 4 added. **EDIT2:** Second proof of Part 4 added.
**1.** The answer is no (as long as we are working over a field - of any characteristic, algebraically closed or not). If $k$ is a field and $G$ is a finite group, then the dimension of any irreducible representation $V$ of $G$ over $k$ is $\leq \left|G\right|$. This is actually obvious: Take any nonzero vector $v\in V$; then, $k\left[G\right]v$ is a nontrivial subrepresentation of $V$ of dimension $\leq\dim\left(k\left[G\right]\right)=\left|G\right|$. Since our representation $V$ was irreducible, this subrepresentation must be $V$, and hence $\dim V\leq\left|G\right|$.
**2.** Okay, we can do a little bit better: Any irreducible representation $V$ of $G$ has dimension $\leq\left|G\right|-1$, unless $G$ is the trivial group. Same proof applies, with one additional step:
If $\dim V=\left|G\right|$, then the map $k\left[G\right]\to V,\ g\mapsto gv$ must be bijective (in fact, it is surjective,
since $k\left[G\right]v=V$, and it therefore must be bijective since $\dim\left(k\left[G\right]\right)=\left|G\right|=\dim V$), so it is an isomorphism of representations (since it is $G$-equivariant), and thus $V\cong k\left[G\right]$. But $k\left[G\right]$ is not an irreducible representation, unless $G$ is the trivial group (in fact, it always contains the $1$-dimensional trivial representation).
**3.** Note that if the base field $k$ is algebraically closed and of characteristic $0$, then we can do much better: In this case, an irreducible representation of $G$ always has dimension $<\sqrt{\left|G\right|}$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $\left|G\right|$, and one of these representations is the trivial $1$-dimensional one). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then the bound $\dim V\leq \left|G\right|-1$ can be sharp (take cyclic groups).
**4.** There is a way to improve **2.** so that it comes a bit closer to **3.**:
**Theorem 1.** If $V\_1$, $V\_2$, ..., $V\_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $\dim V\_1+\dim V\_2+...+\dim V\_m\leq\dim A$.
(Of course, if $A$ is the group algebra of some finite group $G$, then $\dim A=\left|G\right|$, and we get **2.** as a consequence.)
*First proof of Theorem 1.* At first, for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, the (left) representation $V\_i^{\ast}$ of the algebra $A^{\mathrm{op}}$ (this representation is defined by $a\cdot f=\left(v\mapsto f\left(av\right)\right)$ for any $f\in V\_i^{\ast}$ and $a\in A$) is irreducible (since $V\_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{\mathrm{op}}$ (since we can choose some nonzero $u\in V\_i^{\ast}$, and then the map $A^{\mathrm{op}}\to V\_i^{\ast}$ given by $a\mapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V\_i^{\ast}$ and therefore equal to $V\_i^{\ast}$ due to the irreducibility of $V\_i^{\ast}$). Hence, by duality, $V\_i$ is isomorphic to a subrepresentation of the (left) representation $A^{\mathrm{op}\ast}=A^{\ast}$ of $A$. Hence, from now on, let's assume that $V\_i$ actually *is* a subrepresentation of $A^{\ast}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$.
Now, let us prove that the vector subspaces $V\_1$, $V\_2$, ..., $V\_m$ of $A^{\ast}$ are linearly disjoint, i. e., that the sum $V\_1+V\_2+...+V\_m$ is actually a direct sum. We will prove this by induction over $m$, so let's assume that the sum $V\_1+V\_2+...+V\_{m-1}$ is already a direct sum. It remains to prove that $V\_m\cap \left(V\_1+V\_2+...+V\_{m-1}\right)=0$. In fact, assume the contrary. Then, $V\_m\cap \left(V\_1+V\_2+...+V\_{m-1}\right)=V\_m$ (since $V\_m\cap \left(V\_1+V\_2+...+V\_{m-1}\right)$ is a nonzero subrepresentation of $V\_m$, and $V\_m$ is irreducible). Thus, $V\_m\subseteq V\_1+V\_2+...+V\_{m-1}$. Consequently, $V\_m$ is isomorphic to a subrepresentation of the direct sum $V\_1\oplus V\_2\oplus ...\oplus V\_{m-1}$ (because the sum $V\_1+V\_2+...+V\_{m-1}$ is a direct sum, according to our induction assumption).
Now, according to Theorem 2.2 and Remark 2.3 of [Etingof's "Introduction to representation theory"](http://math.mit.edu/~etingof/replect.pdf), any subrepresentation of the direct sum $V\_1\oplus V\_2\oplus ...\oplus V\_{m-1}$ must be a direct sum of the form $r\_1V\_1\oplus r\_2V\_2\oplus ...\oplus r\_{m-1}V\_{m-1}$ for some nonnegative integers $r\_1$, $r\_2$, ..., $r\_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V\_1\oplus V\_2\oplus ...\oplus V\_{m-1}$ must be one of the representations $V\_1$, $V\_2$, ..., $V\_{m-1}$. Since we know that $V\_m$ is isomorphic to a subrepresentation of the direct sum $V\_1\oplus V\_2\oplus ...\oplus V\_{m-1}$, we conclude that $V\_m$ is isomorphic to one of the representations $V\_1$, $V\_2$, ..., $V\_{m-1}$. This contradicts the non-isomorphy of the representations $V\_1$, $V\_2$, ..., $V\_m$. Thus, we have proven that the sum $V\_1+V\_2+...+V\_m$ is actually a direct sum. Consequently, $\dim V\_1+\dim V\_2+...+\dim V\_m=\dim\left(V\_1+V\_2+...+V\_m\right)\leq \dim A^{\ast}=\dim A$, and Theorem 1 is proven.
*Second proof of Theorem 1.* I just learnt the following simpler proof of Theorem 1 from §1 Lemma 1 in [Crawley-Boevey's "Lectures on representation theory and invariant theory"](http://www.amsta.leeds.ac.uk/~pmtwc/repinv.pdf):
Let $0=A\_0\subseteq A\_1\subseteq A\_2\subseteq ...\subseteq A\_k=A$ be a composition series of the regular representation $A$ of $A$. Then, by the definition of a composition series, for every $i\in \left\lbrace 1,2,...,k\right\rbrace$, the representation $A\_i/A\_{i-1}$ of $A$ is irreducible.
Let $T$ be an irreducible representation of $A$. We are going to prove that there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A\_I/A\_{I-1}$ (as representations of $A$).
In fact, let $I$ be the smallest element $i\in \left\lbrace 1,2,...,k\right\rbrace$ satisfying $A\_iT\neq 0$ (such elements $i$ exist, because $A\_kT=AT=T\neq 0$). Then, $A\_IT\neq 0$, but $A\_{I-1}T=0$. Now, choose some vector $t\in T$ such that $A\_It\neq 0$ (such a vector $t$ exists, because $A\_IT\neq 0$), and consider the map $f:A\_I\to T$ defined by $f\left(a\right)=at$ for every $a\in A\_I$. Then, this map $f$ is a homomorphism of representations of $A$. Since it maps the subrepresentation $A\_{I-1}$ to $0$ (because $f\left(A\_{I-1}\right)=A\_{I-1}t\subseteq A\_{I-1}T=0$), it gives rise to a map $g:A\_I/A\_{I-1}\to T$, which, of course, must also be a homomorphism of representations of $A$. Since $A\_I/A\_{I-1}$ and $T$ are irreducible representations of $A$, it follows from Schur's lemma that any homomorphism of representations from $A\_I/A\_{I-1}$ to $T$ is either an isomorphism or identically zero. Hence, $g$ is either an isomorphism or identically zero. But $g$ is not identically zero (since $g\left(A\_I/A\_{I-1}\right)=f\left(A\_I\right)=A\_It\neq 0$), so that $g$ must be an isomorphism, i. e., we have $T\cong A\_I/A\_{I-1}$.
So we have just proven that
**(1)** For every irreducible representation $T$ of $A$, there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A\_I/A\_{I-1}$ (as representations of $A$).
Denote this $I$ by $I\_T$ in order to make it clear that it depends on $T$. So we have $T\cong A\_{I\_T}/A\_{I\_T-1}$ for each irreducible representation $T$ of $A$. Applying this to $T=V\_i$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, we see that $V\_i\cong A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$. Hence, the elements $I\_{V\_1}$, $I\_{V\_2}$, ..., $I\_{V\_m}$ of the set $\left\lbrace 1,2,...,k\right\rbrace$ are pairwise distinct (because $I\_{V\_i}=I\_{V\_j}$ would yield $V\_i\cong A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}=A\_{I\_{V\_j}}/A\_{I\_{V\_j}-1}\cong V\_j$, but the representations $V\_1$, $V\_2$, ..., $V\_m$ are pairwise nonisomorphic), and thus
$\sum\limits\_{i=1}^{m}\dim\left(A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}\right)=\sum\limits\_{\substack{j\in\left\lbrace 1,2,...,k\right\rbrace ;\ \\ \text{there exists }\\ i\in\left\lbrace 1,2,...,m\right\rbrace \\ \text{ such that }j=I\_{V\_i}}}\dim\left(A\_j/A\_{j-1}\right)$
$\leq \sum\limits\_{j\in\left\lbrace 1,2,...,k\right\rbrace}\dim\left(A\_j/A\_{j-1}\right)$ (since $\dim\left(A\_j/A\_{j-1}\right)\geq 0$ for every $j$, so that adding more summands cannot decrease the sum)
$=\sum\limits\_{j=1}^{k}\dim\left(A\_j/A\_{j-1}\right)=\sum\limits\_{j=1}^{k}\left(\dim A\_j-\dim A\_{j-1}\right)$.
Since $\dim\left(A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}\right)=\dim V\_i$ for each $i$ (due to $A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}\cong V\_i$) and $\sum\limits\_{j=1}^{k}\left(\dim A\_j-\dim A\_{j-1}\right)=\dim A$ (in fact, the sum $\sum\limits\_{j=1}^{k}\left(\dim A\_j-\dim A\_{j-1}\right)$ is a telescopic sum and simplifies to $\dim A\_k-\dim A\_0=\dim A-\dim 0=\dim A-0=\dim A$), this inequality becomes $\sum\limits\_{i=1}^{m}\dim V\_i\leq\dim A$. This proves Theorem 1.
| 28 | https://mathoverflow.net/users/2530 | 14516 | 9,740 |
https://mathoverflow.net/questions/14293 | 3 | This is a follow-up question to [my previous one](https://mathoverflow.net/questions/951) where I was trying to understand the classes of Legendrian immersions of circles into [contact manifolds](http://en.wikipedia.org/wiki/Contact_geometry).
I'm interested in classifying *isotropic* immersions (of spheres, in my particular application, if that makes a difference) into contact manifolds up to regular homotopy through isotropic immersions. Let $(M,\xi)$ be a contact (2n+1)-manifold; here $\xi$ is a distribution of $2n$-planes in $TM$. An isotropic immersion of a manifold N (with dimension $k\leq n$, for $k=n$ this is a *Legendrian* immersion) is an immersion $f:N\rightarrow M$ where the image of $df\_x:T\_x N\rightarrow T\_{f(x)}M$ lies in the plane $\xi\_{f(x)}$ for all $x$.
Mike Usher's answer to my previous question directed me to [a paper by Ekholm, Etnyre and Sullivan](http://arxiv.org/abs/math/0210124), in particular section 3.3 (p.19), which defines the *rotation class* of a Legendrian $f$ as follows.
If $\alpha$ is a contact 1-form on M, that is Ker $\alpha=\xi$, then $d\alpha\_p|\xi\_p$ is a symplectic form on $\xi\_p$. If one chooses any complex structure $J$ for $\xi$ that is compatible with the symplectic structure, then complexifying $df$ to be $df\_{\mathbb C}:TL\otimes \mathbb{C}\rightarrow\xi$ is a fiberwise bundle isomorphism.
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> > The rotation class is the homotopy class of $(f,df\_{\mathbb C})$ in the space of complex fiberwise isomorphisms $TL\otimes\mathbb{C}\rightarrow \xi$ and is denoted $r(f)$. An h-principle for Legendrian immersions implies that $r(f)$ is a complete invariant for $f$ up to regular homotopy through Legendrian immersions.
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**First**, I can see how the above definition generalizes to isotropic immersions (namely, instead of bundle isomorphisms, I'll get bundle monomorphisms), but how do I figure out what the possible rotation classes in that setting are? Actually, I'm not exactly sure how to do it in the Legendrian case either. It seems one has to understand certain homotopy classes of bundle maps, which I don't know anything about. (It's OK if the answer is just a reference to a section of a book on bundle theory or something)
**Second**, what happens when $\xi$ is not coorientable, that is, there is no global $\alpha$? In this case, I don't even know how to define a complex structure compatible with $\xi$ which seems to be what the rotation class relies on.
| https://mathoverflow.net/users/353 | Homotopy classes of complex bundle maps and isotropic immersions into contact manifolds | jc, you'll have fun working out answers to examples of your first question. I'm only going to address the Legendrian case. If I didn't make mistakes, I'll conclude that Legendrian immersions of $S^n$ into $S^{2n+1}$ are all regular homotopic if $n$ is even, while if $n$ is odd they are classified by a rotation *number*. For the second question, I'll just suggest that you think about 1-forms valued in the orientation line bundle.
The h-principle you quoted (cf. Eliashberg-Mishachev's book) tells us that we need to understand homotopy classes $[f,l]$, where $f$ is a map $L \to M$ and $l \colon TL\otimes \mathbb{C}\to f^\* \xi$ an injective bundle map. Understanding the set $[L,M]$ of homotopy classes of unbased maps is an obstruction theory problem, and in general pretty intractable, but $[S^k,M]$ is the set of $\pi\_1(M)$-orbits in $\pi\_k(M)$. The fibre of the forgetful map $[f,l]\mapsto [f]$ is exactly the set of injective bundle maps $l \colon TL\otimes \mathbb{C}\to f^\ast \xi$ (this is an application of the covering homotopy property of bundles, bearing in mind that the injective bundle maps themselves form a fibre bundle).
Now fix $f$. In the Legendrian case, what we're looking is $\pi\_0$ of the space $I(f)$ of isomorphisms $TL\otimes \mathbb{C}\to f^\ast \xi$. In terms of classifying maps, $\pi\_0 I(f)$ is the set of homotopy classes of homotopies from $t\_{\mathbb{C}}$, the composite of the tangent map $t\colon L \to BO(n)$ with $BO(n)\to BU(n)$, to $\tilde{\xi}f$. Necessary conditions for $I(f)\neq \emptyset$ are that, for all $j$, $p\_j(TL) = \pm f^\ast c\_{2j}(\xi)$ (I forget what the sign should be) and $2 f^\ast c\_{2j+1}(\xi)=0$. When $L=S^n$, you have to compare two elements of $\pi\_n BU(n)=\pi\_{n-1}U(n) = \pi\_{n-1}U(\infty)$ which is $\mathbb{Z}$ or $0$ according to whether $n$ is even or odd. Since the Pontryagin classes of $S^n$ are zero by the signature theorem, I think what you actually have to check is whether $f^\* c\_{n/2}(\xi)=0$.
If one isomorphism exists, there will be more: we can compose with any complex automorphism of $T^\ast L\otimes \mathbb{C}$. Up to homotopy, automorphisms correspond to isomorphism classes of vector bundles over $S^1\times L$ extending $T^\ast L\otimes \mathbb{C}\to L$, and so one has to classify such extensions. This is another obstruction theory problem. When $L=S^n$, any two extensions agree over $S^1 \vee L$, and over the last $(n+1)$-cell they differ by a map $S^{n+1} \to BU(n)$.
So, when $L=S^n$, there's a freely transitive action of $\pi\_{n+1}BU(n)=\pi\_n U(n) = \pi\_n U(\infty)$ on $I(f)$.
Examples: say $M=S^{2n+1}$, $\xi$ any contact structure, $L=S^n$, $f$ any map (necessarily homotopically trivial). Then $f^\ast \xi$ is trivial, so $I(f)$ is non-empty. It is a singleton if $n$ is even, and is a freely transitive $\mathbb{Z}$-set if $n$ is odd. When $n=1$, the $\mathbb{Z}$ should be the classical rotation number.
| 3 | https://mathoverflow.net/users/2356 | 14519 | 9,742 |
https://mathoverflow.net/questions/14508 | 18 | Suppose that $f$ is a weight $k$ newform for $\Gamma\_1(N)$ with attached $p$-adic Galois representation $\rho\_f$. Denote by $\rho\_{f,p}$ the restriction of $\rho\_f$ to a decomposition group at $p$. When is $\rho\_{f,p}$ semistable (as a representation of
$\mathrm{Gal}(\overline{\mathbf{Q}}\_p/\mathbf{Q}\_p)$?
To make things really concrete, I'm happy to assume that $k=2$ and that the $q$-expansion of $f$ lies in $\mathbf{Z}[[q]]$.
Certainly if $N$ is prime to $p$ then $\rho\_{f,p}$ is in fact crystalline, while
if $p$ divides $N$ exactly once then $\rho\_{f,p}$ is semistable (just thinking about the Shimura construction in weight 2 here, and the corresponding reduction properties of $X\_1(N)$
over $\mathbf{Q}$ at $p$). For $N$ divisible by higher powers of $p$, we know that these representations are de Rham, hence potentially semistable. Can we say more? For example,
are there conditions on "numerical data" attached to $f$ (e.g. slope, $p$-adic valuation of $N$, etc.) which guarantee semistability or crystallinity over a specific
extension? Can we bound the degree and ramification of
the minimal extension over which $\rho\_{f,p}$ becomes semistable in terms of numerical
data attached to $f$? Can it happen that $N$ is highly divisible by $p$ and yet $\rho\_{f,p}$ is semistable over $\mathbf{Q}\_p$?
I feel like there is probably a local-Langlands way of thinking about/ rephrasing this question, which may be of use...
As a possible example of the sort of thing I have in mind: if $N$ is divisible by $p$ and $f$ is ordinary at $p$ then $\rho\_{f,p}$ becomes semistable over an abelian extension of
$\mathbf{Q}p$
and even becomes crystalline over such an extension provided that the Hecke eigenvalues
of $f$ for the action of $\mu\_{p-1}\subseteq (\mathbf{Z}/N\mathbf{Z})^{\times}$ via the diamond operators
are not all 1.
| https://mathoverflow.net/users/2215 | Galois representations attached to newforms | The right way to do this sort of question is to apply Saito's local-global theorem, which says that the (semisimplification of the) Weil-Deligne representation built from $D\_{pst}(\rho\_{f,p})$ by forgetting the filtration is precisely the one attached to $\pi\_p$, the representation of $GL\_2(\mathbf{Q}\_p)$ attached to the form via local Langlands. Your suggestions about the $p$-adic valuation of $N$ and so on are rather "coarse" invariants---$\pi\_p$ tells you everything and is the invariant you really need to study.
So now you can just list everything that's going on. If $\pi\_p$ is principal series, then $\rho$ will become crystalline after an abelian extension---the one killing the ramification of the characters involved in the principal series. If $\pi\_p$ is a twist of Steinberg by a character, $\rho\_{f,p}$ will become semistable non-crystalline after you've made an abelian extension making the character unramified. And if $\pi\_p$ is supercuspidal, $\rho\_{f,p}$ will become crystalline after a finite non-trivial extension that could be either abelian or non-abelian, and figuring out which is a question about $\pi\_p$ (it will be a base change from a quadratic extension if $p>2$ and you have to bash out the possibilities).
Seems to me then that semistable $\rho$s will show up precisely when $\pi\_p$ is either unramified principal series or Steinberg, so the answer to your question is (if I've got everything right) that $\rho\_{f,p}$ will be semistable iff either $N$ (the level of the newform) is prime to $p$, or $p$ divides $N$ exactly once and the component at $p$ of the character of $f$ is trivial. Any other observations you need should also be readable from this sort of data in the same way.
One consequence of this I guess is that $\rho\_{f,p}$ is semi-stable iff the $\ell$-adic representation attached to $f$ is semistable at $p$.
| 15 | https://mathoverflow.net/users/1384 | 14521 | 9,744 |
https://mathoverflow.net/questions/14509 | 6 | Out of idle curiosity, I'm wondering about all the various idempotent constructions we have in mathematics (they seem to be generally referred to as a "closure" or "completion"), and how some of them are related (e.g., the radical of an ideal and the closure of a subset of $k^n$ in the Zariski topology, via the Nullstellensatz - the radical and the topological closure both being idempotent).
So, one answer per post, but if you have two concepts which are related, I guess it'd be okay to put them together. For the sake of the completeness (ha ha) of this list, I'll add "radical" and "topological closure".
EDIT: My bad - I should have looked around more first. There's [this list](http://en.wikipedia.org/wiki/Closure_%28mathematics%29) at Wikipedia and [this list](http://ncatlab.org/nlab/show/completion#list_of_completions_2) at nLab. Well, I'm sure there's plenty more concepts out there, so if you think of any more, feel free to add them. But let's focus on how some of these concepts are related - e.g., does one kind of completion arise in terms of another? What are some general ways in which completions and closures arise?
| https://mathoverflow.net/users/1916 | Various concepts of "closure" or "completion" in mathematics | In general, if $A\subset B$ is a full reflective subcategory, then each object $a\in A$ is isomorphic to its image under the reflector. This seems to include many cases: $\mathbf{Ab}\subset \mathbf{Grp}$, $\mathbf{CompMet}\subset\mathbf{Met}$, $\mathbf{Top}\_{n+1}\subseteq \mathbf{Top}\_{n}$ (as in [Why is Top\_4 a reflective subcategory of Top\_3?](https://mathoverflow.net/questions/9504/why-is-top4-a-reflective-subcategory-of-top3)), etc.
EDIT: Following Pete L. Clark's comment, here is a clarification: The subcategory $A$ above is called reflective if the inclusion functor $A\subset B$ has a left adjoint, and full if this inclusion functor is full. In case $A$ is a reflective subcategory, the left adjoint to the inclusion functor is called a reflector.
| 7 | https://mathoverflow.net/users/2734 | 14533 | 9,753 |
https://mathoverflow.net/questions/14505 | 0 | An n-component slice link is a link that bounds n disjoint discs in B^4. And the 4-genus of a link is defined to be the minimal genus of orientable surfaces bounded by it in B^4.
My question is: if the link bounds a surface with zero genus in B^4, is it necessarily a slice link? If not, any counter examples?
| https://mathoverflow.net/users/1537 | If the 4-genus of a link is zero, is it a slice link? | The Hopf link bounds a cylinder $S^1 \times [0,1]$ in $B^4$, and it's not slice since the two components have a non-zero linking number.
| 5 | https://mathoverflow.net/users/1465 | 14537 | 9,757 |
https://mathoverflow.net/questions/14529 | 18 | The Riemann–Stieltjes integral $\int\_a^b f(x)\,dg(x)$ is a generalization of the Riemann integral. It is e.g. heavily used as a starting point for stochastic integration. The approximating Riemann–Stieltjes sums are analogous to the Riemann sums $\sum\_{i=0}^{n-1} f(c\_i)(g(x\_{i+1})-g(x\_i))$ where $c\_i$ is in the $i$-th subinterval $[x\_i,x\_{i+1}]$.
The Riemann-sums can be very intuitively visualized by rectangles that approximate the area under the curve.
See e.g. [Wikipedia:Riemann sum](https://en.wikipedia.org/wiki/Riemann_sum).
Unfortunately, I cannot find respective intuitive visualizations of the Riemann–Stieltjes sums.
**My question:** Could anyone provide me with some literature, pictures, links, or especially tools (e.g. Mathematica or even Excel) with which I could play around to get a similar intuition for this more general kind of integral?
| https://mathoverflow.net/users/1047 | Visualization of Riemann–Stieltjes Integrals | It's fairly easy to visualize the Riemann–Stieltjes integral $\int\_a^b f(t)\,dg(t)$ [I changed the name of the integration variable for convenience below] if $f\ge0$ and $g$ is nondecreasing. Just draw the graph of the curve $(x,y)=(g(t),f(t))$. The integral is just the area below the curve. (Whenever $g$ makes a jump at some $t\_0$, fill in the gap by setting $y=f(t\_0)$ there.) The Riemann–Stieltjes sums are now easy to visualize as sums of areas of rectangles (details left as an exercise).
| 34 | https://mathoverflow.net/users/802 | 14539 | 9,759 |
https://mathoverflow.net/questions/14518 | 35 | This is related to Anweshi's question about [theories of noncommutative geometry](https://mathoverflow.net/questions/10512/).
Let's start out by saying that I live, mostly, in a commutative universe. The only noncommutative rings I have much truck with are either supercommutative, almost commutative (filtered, with commutative associated graded), group algebras or matrix algebras, none of which really show many of the true difficulties of noncommutative things.
So, here's my (somewhat pithy) question: what's noncommutative geometry good for?
To be a bit more precise, I have a vague sense that $C^\*$ stuff is supposed to work well in quantum mechanics, but I'm somewhat more interested in more algebraic noncommutative geometry. What sorts of problems does it solve that we can't solve without leaving the commutative world? Why should, say, a complex algebraic geometer learn some noncommutative geometry?
| https://mathoverflow.net/users/622 | Applications of noncommutative geometry | Charles,
a couple of reasons why a complex algebraic geometer (certainly someone who is
interested in moduli spaces of vector bundles, as your profile tells me) might
at least keep
an open verdict on the stuff NC-algebraic geometers (NCAGers from now on) are trying to do.
in recent years ,a lot of progress has been made towards understanding moduli spaces of
semi-stable representations of 'formally smooth' algebras (think 'smooth in the NC-world).
in particular when it comes to their etale local structure and their rationality.
for example, there is [this book](http://win.ua.ac.be/~lebruyn/LeBruyn2007a.pdf), by someone.
this may not seem terribly relevant to you until you realize that some of the more
interesting moduli spaces in algebraic geometry are among those studied. for example, the
moduli space of semi-stable rank n bundles of degree 0 over a curve of genus g is the moduli
space of representations of a certain dimension vector over a specific formally smooth algebra,
as Aidan Schofield showed. he also applied this to rationality results about these spaces.
likewise, the moduli space of semi-stable rank n vectorbundles on the projective plane with Chern
classes c1=0 and c2=n is birational to that of semi-simple n-dimensional representations of the free algebra
in two variables. the corresponding rationality problem has been studied by NCAG-ers (aka 'ringtheorists'
at the time) since the early 70ties (work by S.A. Amitsur, Claudio Procesi and Ed Formanek). by their results, we NCAGers,
knew that the method of 'proof' by Maruyama of their stable rationality in the mid 80ties, couldn't possibly work.
it's rather ironic that the best rationality results on these moduli spaces (of bundles over the
projective plane) are not due to AGers but to NCAGers : Procesi for n=2, Formanek for n=3 and 4 and
Bessenrodt and some guy for n=5 and 7. together with a result by Aidan Schofield these results show
that this moduli space is stably rational for all divisors n of 420.
further, what a crepant resolution of a quotient singularity is to you, is to NCAGers the moduli space of certain
representations of a nice noncommutative algebra over the singularity.
likewise, when you AGers mumble 'Deligne-Mumford stack', we NCAGers say 'ah! a noncommutative algebra'.
| 26 | https://mathoverflow.net/users/2275 | 14541 | 9,761 |
https://mathoverflow.net/questions/14530 | 4 | How can we calculate the 4-genus of a link L? The 4-genus is defined to be the minimal genus of orientable surface bounded by L in B^4. Is there any routine method to calculate that?
Especially, any good idea how to calculate the 4-genus of a 2-bridge link? Thanks.
| https://mathoverflow.net/users/1537 | 4-genus of a 2-bridge link | You might get interesting lower bounds using [Rasmussen's s-invariant](http://arxiv.org/abs/math.GT/0402131), and his calculations of [the KR homology of 2-bridge links](http://arxiv.org/abs/math.GT/0508510).
Or you might not. As Ryan says, there are various ways of bounding genus above and below, but unless you get lucky and two of them match, it can be hard to tell. The 4-genus of torus knots was an open problem for a scandalously long time (if my history is right, it was first proven in the 90's!).
| 5 | https://mathoverflow.net/users/66 | 14544 | 9,764 |
https://mathoverflow.net/questions/14501 | 9 | The background for this question is that I know that many students starting to learn algebra focus on the standard construction of the quotient group $G/N$ instead of working with the universal property. Therefore it would be nice to give another construction, thus avoiding cosets, but so that the universal property is clear. This idea came to me years ago, but I've never found something.
For another example, there is a very nice and intuitive construction of the localization of a ring: $S^{-1} A := A[\{X\_s\}\_{s \in S}] / (s X\_s = 1)$. The idea is: Invent new elements, and force them to be inverses to your elements of $S$. The universal property follows from the universal property of quotient and free algebra.
Perhaps quotient groups are a bit too elementary so that there could be another nice construction (please don't answer when you've just got this to say) ... anyway, any ideas are welcome :). I'm pretty sure that in this case there is no *better* one, but perhaps *another* one. I play around with the Cayley representation of $G$ ...
Sorry if this is too elementary for you ;-)
| https://mathoverflow.net/users/2841 | alternative construction of the quotient group | In order to make the construction of the quotient, it seems that it would be helpful to have at hand some quotients of the group $G$, described in some way other than formally via cosets.
Here is one possible approach:
Generalizing the regular action of $G$ on itself, we have the action of $G$ on its
power set (given by translating a set).
If $S \subset G$ is any subset, we can look at the orbit of $S$ under $G$, which is
some subset $\mathcal O\_S$ of the power set of $G$, and we are given a homomorphism
$G \to Perm(\mathcal O\_S)$ (the group of permutations of $\mathcal O\_S$). Let $G'$ be
the image of $G$ under this homomorphism.
If we now take $S$ to be a normal subgroup $N$ of $G$, then $G' = G/N$. I wonder how hard this is to show directly, in terms of the universal property?
It is easy to see at least that $G \rightarrow G'$ has the property that $N$ is in its kernel. Can one show directly in this set-up that any map $G \rightarrow G''$ having
$N$ in its kernel factors through the explicitly constructed quotient $G'$?
I don't see a pithy argument straight away, but it doesn't seem too unfeasible to me.
| 3 | https://mathoverflow.net/users/2874 | 14554 | 9,772 |
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