How many primes stay inert in a finite (non-cyclic) extension of number fields?

In the following suppose L/K is a finite Galois extension of number fields, (maybe it works for other cases also, I don't know) By the Chebotorev density theorem when Gal(L/K) is cyclic, there are infinitely many primes in K that stay inert during this extension (cf Janus p136, Algerbaic Number Fields.) When L/K is non cyclic, an exercise from Neukirch (somewhere in Chap I) says there are at most finitely many primes that stay inert. I want to say that there are none. The reason is by a cycle description from Janus, p101, Prop 2.8,

In short, that proposition says when $delta:=Frob(frac{L/K}{beta})$, $beta|p$ is a prime in L, consider $delta$ act on the cosets of H in G, H=Gal(L/E), $Ksubset Esubset L$, then every cycle of length i corresponds to a prime factor in E with residue degree i. In particular, for inert guys we want there is only one cycle in the action. When we take H to be trivial, E=L is Galois over K, and the cosets are just the elements of G themselves.
So we want that there exist an element (the Frobenius element above p) act transitively on G, thus G is cyclic.

I wonder if this is true, then more people should have been aware of it. If it is not, is there a counter example?

gravity - Why isn't the star that created the black hole a black hole?

A supernova may actually be necessary in the creation of a stellar black hole.

At the ends of their lives the cores of massive stars are made mostly of iron-peak nuclei from which you cannot extract more fusion energy. To support their weight, these stars rely on electron degeneracy pressure - the pressure caused by the Pauli exclusion principle allowing no more than one electron to share the same quantum state.

In principle a star might be supported by degeneracy pressure forever as it gradually cools - this is the fate of most white dwarfs.

However, the core of a massive star is just too big for that to work. The density increases until all the electron are moving at close-to the speed of light and that's as high as the degeneracy pressure can get. If the core exceeds the Chandrasekhar mass, it will collapse and as it does so, the rest of the star collapses with it (a little more slowly).

The collapse is triggered by the removal of electrons by electron capture into nuclei to form neutrons. At some point enough neutrons are produced for neutron degeneracy pressure to halt or at least slow the collapse. This and the release of a lot of gravitational potential energy are ultimately what power a supernova explosion. But if the collapse is not halted then even neutron degeneracy pressure will not support the star and collapse to a black hole becomes inevitable. A black hole status is reached once a proportion of its mass is compressed inside its Schwarzschild radius $r_s = 2GM/c^2$. i.e. once its density achieves
$$ rho > frac{3M}{4pi r_s^{3}}$$
i.e. when a central mass $M$ has a density that exceeds
$$ rho > frac{3}{32pi} frac{c^6}{G^3 M^2} = 1.8times10^{19} left(frac{M}{M_{odot}}right)^{-2} {rm kg/m}^3$$
This is a ball park figure and assumes spherical symmetry and neglects any detailed GR treatment, but is more or less correct - a few times higher than typical neutron star densities.

In other words it is the density of the material that largely determines whether something becomes a black hole. The mass is only an indirect parameter.

complex geometry - A group action of the Heisenberg group with special symmetries

Suppose we look at the Heisenberg group $H_{d}$ as a matrix group of upper triangular matrices over the ring $mathbb{Z}/dmathbb{Z}$. You can even choose $d$ to be prime if you want. A natural irrep of $H_{d}$ acting on $mathbb{C}^{d}$ maps the group elements into the "shift" and "phase" operators, plus roots of unity. More specifically, the two natural generators map the orthonormal basis vectors from $j to j+1mod d$, and the Fourier transform of that operation, plus overall phases by roots of unity. The question is this:

Can you find a unit vector $v$ such that $|(v,U_g v)| = c$ for all g not in the center of $H_{d} ?$ One can solve for the constant: $c=frac{1}{sqrt{d+1}}$.

Numerics suggests that these vectors exist in all the dimensions $< 67$, hence they may exist in every dimension, but the form of the vectors contains no (obvious) hint as to how to prove this.

This problem seems extremely truculent and any help is greatly appreciated!

co.combinatorics - An Operation on Multisets

But I don't have a clear proof that the sequence always terminates in a loop. – Martin Erickson

Here is a proof that the sequence always terminates in a loop.

Let $A, B$ be consecutive arrays in the sequence, and c(X) denote the number of columns in array X.

Claim 1. $c(A) leq c(B)$.

Proof. Trivial.

Claim 2. $$sum_{i=1}^{c(B)} B[2,i] = 2 c(A).$$

Proof. By the sequence definition, the second row of each subsequent array contains the multiplicities of elements of the preceding array, and thus the sum of multiplicities equals the total number of elements in the preceding array. QED

Now, let $A', A, B$ be three consecutive arrays in the sequence.

Claim 3. Let $mgeq 5$ be an odd integer such that no element $B$ exceeds $m$. Then no element larger than $m$ can appear in subsequent arrays (while their size may eventually grow up to $2times m$).

Proof. Assume that a larger element $m'geq m+1$ appears. Without loss of generality suppose that this happens in (the bottom row of) array $C$ that immediately follows $B$. By Claim 1, we have $c(A)leq c(B)leq c(C)leq m$.

By Claim 2 and since $C$ contains $m'geq m+1$ (while the other elements are at least 1),
$$m + c(C) leq m' + (c(C)-1) leq sum_{i=1}^{c(C)} C[2,i] = 2c(B) leq 2c(C)leq 2m$$
implying that $c(B)=c(C)=m$ and $m'=m+1$.

Therefore, the top row of both $B$ and $C$ contains all integers from 1 to $m$.

The bottom row of $C$ consists of one number $m'=m+1$ and $m-1$ ones. If $m'$ appear under the number $k$, then bottom row of $B$ contains at least $m$ numbers $k$, whose sum must not exceed $2c(A)leq 2m$, implying that $kleq 2$. Consider two cases:

If $k=1$ then the bottom row of $B$ consists of all ones, implying that the elements of $A$ are the integers from $1$ to $m$ without repetitions, and hence $m$ is even, a contradiction proving that no element larger than $m$ may appear.

If $k=2$ then the bottom row of $B$ consists of all twos, implying that $c(A)=m$ and $A$ contains in each row all integers from 1 to $m$ so that the sum of its bottom row equal $1+2+dots+m = m(m+1)/2.$
The inequality $m(m+1)/2 leq 2c(A') leq 2c(A) = 2m$ then implies that $mleq 3$, that is not the case.

QED

Claim 4. The sequence always terminates in a loop.

Proof. By Claim 3, there exists an integer $m$ such that elements of the arrays in the sequence do not exceed $m$. By Claim 1 and since $c(X)leq m$ for all $X$ in the sequence, the size (and hence the top row) of arrays stabilizes to a certain $c(X)=n$. Then by Claim 2, the sum of the bottom row stabilizes to $2n$. Since, there are only a finite number of compositions of $2n$ into the sum of $n$ positive integers (namely, $binom{2n-1}{n}$), there exists only a finite number of distinct arrays that may appear after the size stabilization, implying that the sequence loops. QED

Claim 5. The length of the terminal loop is bounded by $binom{2m-1}{m}$, where $m$ is defined as in Claim 3.

Proof. See proof of Claim 4.

mg.metric geometry - How can I embed an N-points metric space to a hypercube with low distortion?

Y. Bartal has studied a related problem of embedding metric spaces to hierarchically separated trees. With $1 < mu$ being a fixed real number, a $mu$-HST is equivalent to the set of corners of a rectangle whose edges are of length $c, cmu^{-1}, cmu^{-2}, dots, cmu^{1-D}$ with the $l_infty$-metric. That is, if you think of the space as the set of bit sequences of length $D$, the distance of two sequences is $cmu^{-j}$ if they first differ in bit $j$.

Now in your question you didn't ask for $infty$-metric, but for this set of points, it doesn't really matter which metric you take because the distortion between this and either the $l_1$ or $l_2$ metric is bounded by a constant (if you fix $mu$ but $D$ can vary).

(This metric can be considered a graph metric on a special tree, that is, one where the points are some (but not necessarily all) vertexes of a tree graph with weighted edges, and the distance is the shortest path. This is where "tree" in the name comes from.)

Now Bartal's result in [1] basically says that you can embed any metric space randomly to a $mu$-HST with distortion at most $mu(2ln n+2)(1+log_mu n)$ where $n$ is the number of points. (Also, this embedding can be computed with a randomized polynomial algorithm.)

For this, you need to know what a distortion $alpha$ random embedding $f$ means. It means that for any two points $d(x,y) < d(f(x),f(y))$ is always true and that the expected value of $d(f(x),f(y))$ is at most $alpha d(x,y)$. For many applications, this is just as good as a deterministic embedding with low distortion. In fact, you can make a deterministic embedding with low distortion from it by imagining the metric $d^* $ on the original space where $d^*(x,y) = E(d(f(x), f(y))$, but this notion isn't too useful because the resulting metric does not have nice properties anymore (it's not HST). Indeed, I believe the randomness is essential here as I seem to remember reading somewhere that you can't embed a cycle graph (with equal edge weights) to a tree graph with low distortion.

Anyway, this might not really answer your question. Firstly, $D$ (the number of dimensions of the rectangle) is not determined in advance, but that's not a real problem because if you have $D$ significantly different distances in the input metric then you need at least that large a $D$ for any embedding; and with this embedding you don't need a $D$ larger than $log_mu (Delta/delta)$ where $Delta$ and $delta$ are the largest and smallest distances in the input. The real problem is that you seem to want to know a deterministic embedding, and the highest possible distortion necessary in that case, which this really doesn't tell. For example, a cycle graph with an even number $n$ of vertexes can of course be embedded isometrically to a cube of dimension $n/2$.

Survey [2] has some more references.

[1]: Yair Bartal, On Approximating Arbitrary Metrics by Tree Metrics. Annual ACM Symposium on Foundations of Computer Science, 37 (1996), 184–193.

[2]: Piotr Indyk, Jiří Matoušek, Low-distortion embeddings of finite metric spaces. Chapter 8 in Handbook of Discrete and Computational Geometry, ed. Jacob E. Goodman and Joseph O'Rourke, CRC Press, 2004.

co.combinatorics - What bijection on permutations corresponds under RS to transpose?

When you conjugate diagrams and apply RSK or other Young tableau bijections, the answer is typically bad, with some rare exceptions. The right way to think of RSK is to think of row length being continuous while columns still integer (see e.g. my "Geometric proof of the hook-length formula" paper and refs therein).

An exception: there is a "hidden symmetry" for LR-coefficients when you conjugate all three diagrams - see Hanlon-Sundaram paper (1992). Once you know this bijection, there are natural connections to RSK as described in the long Pak-Vallejo paper and a followup by Azenhas-Conflitti-Mamede (search the web for a paper and a ppt presentation). Together, these do give a complete description of your involution, but making sense of it might require quite a bit of work.

linear algebra - Can a vector space over an infinite field be a finite union of proper subspaces?

Anton Geraschenko's comment prompted me to write a new version of this short answer. I'm leaving the old version to make Anton's comment clearer (and also to increase the probability of having at least one correct answer).

NEW VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each (affine) line. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.

Indeed, as pointed out by Anton, the $K$-valued functions on $A$ which are polynomial on each line form obviously a ring $R$. This ring is a domain, because if $f$ and $g$ are nonzero elements of $R$, then there is a line on which none of them is zero, and their product is nonzero on this line.

OLD VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each finite dimensional affine subspace. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.

Indeed, we can assume that $A$ is finite dimensional, in which case the result is easy and well known.

cmb - Did cosmological inflation occur at speeds greater than $c$?

The simple answer to your question is "yes" - the universe expanded at much greater speeds than $c$ during the inflationary epoch. This period of time was very quick but very dramatic, lasting from about $10^{-36}$ to $10^{-32}$ seconds. The universe expanded, in this very short period, by a factor of $10^{26}$. That's pretty incredible, when you think about it.

Inflation was originally proposed to solve, among other things, the horizon problem - that is, why the universe is isotropic and homogenous (on a large scale). This would mean that all the parts of the universe were in "causal contact" at one point in time. Inflation is the explanation for this.

Now, what does this all have to do with the CMB? Well, the temperature of the CMB is the same throughout the universe - a toasty 2.7 Kelvin. For the temperature to be uniform, all the regions of the universe would have had to be in causal contact in some point in time; hence, inflation explains the uniform temperature of the CMB.

However, the CMB was not around during the inflationary epoch. Far from it. It formed a lot later, when the universe was at the ripe old age of $379,000$ years. But the reason that it was formed (photon decoupling) equally throughout the universe is because the conditions were roughly equal, because of inflation. The CMB was, and still is, everywhere. It was never around during inflation, and as such was not otherwise affected by it.

I hope this helps.

---

My sources for the times:

Inflationary epoch

NASA

gr.group theory - Explicit expression of an alternating polynomial in characteristic $2$?

Although the question is easy to pose, I think some background will help to motivate it, so I'll start with it.

Consider variables $X=(X_1, ldots, X_n)$ over a field $K$ and the elementary symmetric functions $T=(T_1, ldots, T_n)$ in $X$. In other words $X$ are the roots of the polynomial $Y^n + T_1 Y^{n-1} + cdots + T_n$.

A polynomial $f$ in $X$ is symmetric is $f(s X) = f(X)$ for any permutation $s$. Here $s X := (X_{s(1)}, ldots, X_{s(n)})$. Then a basic fact is that if $f(X)$ is symmetric, then $f(X) = g(T)$, for some polynomial $g$.

It is reasonable to define an alternating polynomial to be $f$ that satisfy $f(s X) = sign(s) f(X)$, where $sign(s) = pm 1$ is the signature. The "elementary" alternating polynomial is the Vandermonde polynomial $V(X) = prod_{i<j} (X_j-X_i)$, and any other alternating polynomial can be expressed as a polynomial in $T$ and $V$.

Note that $V$ is a square root of the discriminant $Delta$ of $Y^n + T_1 Y^{n-1} + cdots + T_n$ and the discriminant has an explicit formula in terms of $T$ using the Sylvester matrix.

That definition for alternating polynomials gives nothing interesting in characteristic $2$ (because then $1=-1$). The only definition that makes sense to me in characteristic $2$ is: $f$ is alternating if $f(s X) = f(X) + add.sign(s)$. Here $add.sign(s) = 0,1$ is the additive signature, i.e., equals $1$ if $s$ is odd and $0$ if $s$ is even.

I already figured out what is the "elementary" alternating polynomial $u/V$ and what is the Artin-Schreir equation it satisfies:
$u(X) = sum_{s {rm is even}} X^{n-1}_{s(1)} cdots X^0_{s(n)}$ and it satisfies the Artin-Schreier equation $X^2 + X = frac{u(X) u(s_0 X)}{Delta}$, where $s_0$ is any odd permutation (e.g., transposition), and $Delta$ is again the discriminant. (Note that $u(X) + u(s_0 X) = V$.)

My question is: Does there exist a nice formula for $frac{u(X) u(s_0 X)}{Delta}$ in terms of $T$?

linear algebra - Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?

Here's what's true instead of the claim that domains of positivity
are self-dual cones.

$mathbf{Proposition:}$
$Y$ is a domain of positivity for a nondegenerate
symmetric bilinear form $B$ if and only if it is an open cone whose dual,
according to the Euclidean inner product $E$ associated with a basis
orthonormalizing the form, is its image under reflection of $X_-$
through $X_+$, the ``negative and positive eigenspaces'' associated
with the form in this basis.

$mathbf{Proof:}$
We'll write $v,v'$ for vectors in $X$. We'll use an orthonormal basis
as described above, in which the form is diagonal with diagonal
elements $pm 1$, writing $v = (x,t)$ for a decomposition with $x$ in
the span (call it $X_+$) of the basis vectors with $B(e_i, e_i) = 1$,
and $t$ in the span (call it $X_-$) of the basis vectors with $B(e_i,
e_i) = -1$. Let $S$ be the linear map $(x,t) mapsto (x, -t)$,
i.e. reflection of the subspace $X_-$ through the subspace $X_+$.
Note that $E(x,y) := B(x,Sy)$ is a positive semidefinite
symmetric nondegenerate bilinear form.
Also, note that for all $v,v'$, $B(Sv, Sv') = B(v,v')$, i.e. the form $B$
is reflection-symmetric.

For "if": the definition of $Y^ast$ says it is
maximal such that $E(Y^ast,Y) > 0$. But since
$Y=SY$, it is also maximal such that $E(SY,Y) equiv B(Y,Y) > 0$,
i.e., it is a domain of positivity of $B$.

For ``only if'': let $Y$ be a domain of positivity for $B$. For
every $y$ in the boundary $partial Y$ of $Y$,
the hyperplane $H_y := {x: B(x,y) = 0}$ is
a supporting hyperplane for the cone $Y$, and these are all the
supporting hyperplanes. But it's standard convex geometry that the
supporting hyperplanes of a proper convex cone $Y$ are the precisely
the zero-sets of the linear functionals that constitute the boundary
of $Y$'s dual cone. We have $H_y = {x: B(x,y) equiv E(x,Sy) = 0}$;
that is, this hyperplane is just the plane normal to $Sy$ according to
the Euclidean inner product. That is to say, the vectors $Sy$, for $y
in partial Y$ generate the closure of the cone $Y^ast$ dual to $Y$
according to the Euclidean inner product $E$. I.e., $Y^ast = SY$.
$diamond$

Offline (or rather, off-math-overflow) correspondence with Will Jagy helped
stimulate this solution. He gave
another example---which I'd come up with a few weeks ago, but forgotten about---of a DOP for $xx' + yy' - zz'$---namely, the positive orthant generated by $(0, 1, 0)$, $(1, 0, 1)$
and $(1, 0, -1)$ (or in his dual description, defined by inequalities $x > z$, $x > -z$, $y > 0$), which is of course not isomorphic to an ice-cream cone, but is symmetric under reflection through the xy plane. The hypothesis that the DOPs were precisely the self-dual cones symmetric under reflection suggested itself to me, and attempts to prove the hypothesis ended up providing the proof of the proposition above.

at.algebraic topology - Sheaves over simplicial sets

Clearly looking at sheaves on the geometric realisation gives something too far
removed from the simplicial picture. This is essentially because there are too
many sheaves on a simplex have (most of which are unrelated to simplicial
ideas). What one could do is to consider such sheaves which are constructible
with respect to the skeleton filtration, i.e., are constant on each open
simplex. This can be described inductively using Artin gluing. I think it
amounts to the following for a simplicial set $F$.

For each simplex $cin F_n$ we have a set $T_c$, the constant value of the sheaf
$T$ on the interior of the simplex corresponding to $c$.

For each surjective map $fcolon [n] to [m]$ in $Delta$ the corresponding (degeneracy) map
on geometric simplices maps the interior of $Delta_n$ into (onto in fact) the interior of $Delta_m$ and
hence we have a bijection $T_{f(c)} to T_c$. These bijections are transitive with
respect to compositions of $f$'s.

For each injective map $fcolon [m] to [n]$ in $Delta$ the corresponding map on geometric
simplices maps $Delta_m$ onto a closed subset of $Delta_n$. If $jcolon Delta^o_n hookrightarrow Delta_n$ is the
inclusion of the interior we get an adjunction map $T to j_ast j^ast T$ and
$j_ast j^ast T=T_c$ where $T_c$ also denotes the constanct sheaf with value $T_c$. If
$f'colon Delta^o_m hookrightarrow Delta_n$ is the inclusion of the interior composed with $f$ we can
restrict the adjunction map to get a map $T_{f(c)}=f'^ast T to f'T_c$ and taking
global sections we get an actual map $T_{f(c)} to T_c$. These maps are transitive
with respect compositions of $f$'s.

We have a compatibility between maps coming from surjections and
injections. Unless something very funny is going on this compatibility should be
that we wind up with a function on the comma category $Delta/F$ which takes
surjections $[n] to [m]$ to isomorphisms.

There is the stronger condition on the sheaf $F$, namely that it is constant on
each star of each simplex. This means on the one hand that it is locally
constant on the geometric realisation, on the other hand that $T_{f(c)} to T_c$
is always an isomorpism.

[Added] Some comments intended to give some kind of relation with the answer
provided by fpqc. My suggested answer is not homotopy invariant in the sense
that a weak (or even homotopy) equivalence of simplicial sets does not induce an
equivalence on the category of sheaves. This is so however if one, as per above,
adds the condition that all the maps $T_{f(c)} to T_c$ are isomorphisms. However,
that condition is not so good as many maps that are not weak equivalences
induces category equivalences (it is enough that the map induce isomorphisms on
$pi_0$ and $pi_1$). This is a well-known phenomenon and has to do with the fact
the $T_c$ are just sets. One could go further and assume that the $T_c$ are
topological spaces and the maps $T_{f(c)} to T_c$ continuous. Of course adding
the condition that these maps be homeomorphisms shouldn't be right thing to do,
instead one should demand that they be homotopy (or weak) equivalences. Again,
this shouldn't be quite it because of the transitivity conditions. We should not
have that the composite $T_{g(f(c))} to T_{f(c)} to T_c$ should be equal to
$T_{g(f(c))} to T_c$ but rather homotopic to it. Once we have opened that can of
worms we should impose higher homotopies between repeated composites. This can
no doubt be (has been) done but there seems to be an easier way out. In the
first step away from set-valued $T_c$ we have the possibility of they being
instead categories. In that case the higher homotopy conditions is that we
should have a pseudofunctor $Delta/F to mathcal{C}mathrm{at}$. Even they are
somewhat unpleasant and it is much better to pass to the associated fibred
category $mathcal{T} to Delta/F$. In the general case, and admitting that $Delta/F$ is
essentially the same things as $F$ itself, we should therefore look at
(Serre) fibrations $X to |F|$ or if we want to stay completely simplicial, Kan
fibrations $X to F$. This gives another notion of (very flabby) sheaf which now
should be homotopy invariant (though that should probably be in the sense of
homotopy equivalence of $Delta$-enriched categories).

orbit - earth is spherical , does it mean the ground on earth is like a ball?

As noted by Conrad Turner, the approximate shape of the Earth is an oblate spheroid, though it is so close to spherical that you would be pretty hard-pressed to see the difference without precise scientific equipment. This web page has some pretty good information on this topic that you might be interested in.

When we talk about this shape, we are generally speaking of the land and sea. Gas in general has the property of not only having fluid shape, like a liquid, but also having indefinite volume. Gas expands to fill the container it is in. The atmosphere is not contained per se. What keeps it from simply expanding all the way out to the moon is Earth's gravity. But this means that faster moving air particles get further away from the surface than slower ones before being pulled back by gravity. So where exactly is the edge of the atmosphere? The outer-most layer of the atmosphere, the ionosphere, is very thin, and the further out toward space you go, the thinner it is. Where exactly its edge is, and what shape it has is... complicated. Anyone who wants to make any authoritative statement on the shape or height of the edge of our atmosphere will have a few other authoritative statements to argue against.

While the exact shape of the atmosphere is difficult to define, it's pretty safe to say that it is not exactly spherical due mostly to the planet's rotation, and solar winds. The Earth's surface is also not exactly spherical due mostly to the planet's rotation, and tidal forces. Even if the entire surface was covered in water, rotation would flatten the planet's shape ever so slightly. And even if our rotation became tidally locked with the sun, and the moon... um... went away somehow, the tidal forces of the sun, would distort the spherical perfection ever so slightly. But laying all of that aside, the Earth is still probably closer to a sphere than most ball-shaped objects you have ever handled, whether you measure the "edge" of the atmosphere, the land and sea, or whatever.

pr.probability - probability puzzle - selecting a person

I have written an article about this game, and solved it numerically for the case of 10 person. I computed that the probability is the same for each of the 9 person.

In short, I have defined the following:

Let P(n,i,j,k) be the probability of at the n-th round, the k person is having the coin, while the people from i counting clockwise to k have all received the coin before.

And 3 recurrence equations are formulated:

$P (n+1,i,j,k)= frac{1}{2} P (n,i,j,k-1)+ frac{1}{2} P (n,i,j,k+1)$ ........(1)

$P (n+1,i,j,j)=frac{1}{2} P (n,i,j-1,j-1)+ frac{1}{2} P(n,i,j,j-1)$ ........(2)

....(3)

For details, please refer to:

Solving a probability game using recurrence equations and python

In this article, a(n,i) which denotes the probability of the i-th person being the head in the n-th round is found and plotted out as well.

units - What are some applicable problems with the correct usage of G?

Usually, using different units is done for convienience, as HDE 226868's answer explains. If we're talking about stars, then measuring masses relative to the Sun is sensible. The Sun is the closest star to us, and astronomers understand it much better than any other, making it the de facto standard of comparison. Similarly, on the interstellar scales the parsec is more convenient than the meter. The conversion is trivial and doesn't change anything physically. Recall $F = ma$ to see how the newton corresponds to base SI units:
$$mathrm{N}cdotleft(frac{mathrm{m}}{mathrm{kg}}right)^2 = left(frac{mathrm{kg}cdotmathrm{m}}{mathrm{s}^2}right)frac{mathrm{m}^2}{mathrm{kg}^2} = frac{mathrm{m}^3}{mathrm{kg}cdotmathrm{s}^2}text{,}$$
so we can just multiply the SI value of $G$ by
$$left(frac{1.989times 10^{30},mathrm{kg}}{1,mathrm{M}_odot}right)
left(frac{1,mathrm{pc}}{3.086times 10^{16},mathrm{m}}right)
left(frac{1,mathrm{km}}{1000,mathrm{m}}right)^2text{.}$$
But there could be another reason to use solar masses in particular. The value of Newton's gravitational constant is:
$$begin{eqnarray*}
G &=& 6.674times 10^{-11},frac{mathrm{m}^3}{mathrm{kg}cdotmathrm{s}^2}\
&=& 2.959122083times 10^{-4},frac{mathrm{AU}^3}{mathrm{M}_odotcdotmathrm{day}^2}\
% &=& 3.964015993times 10^{-14},frac{mathrm{AU}^3}{mathrm{M}_odotcdotmathrm{s}^2}\
&=& 4.300917271times 10^{-3},frac{mathrm{pc}}{mathrm{M}_odot}frac{mathrm{km}^2}{mathrm{s}^2}text{.}
end{eqnarray*}$$
The difference in precision is real. The 2010 CODATA value of $G$ is $
(6.67384pm0.00080)times 10^{-11}$ in the SI units, which has a relative uncertainty of $10^{-4}$, so we can't expect more than about $4$ significant figures. On the other hand, the standard gravitational parameter $mu = GM$ of the Sun is known to a much greater precision than either $G$ or their masses individually, so expressing $G$ in solar masses gives a more precise value. The basic reason for that is simple: it's much easier to compare the orbits of planets to each other than, say, to compare the Sun to the International Prototype of the Kilogram kept in a French vault.

Of course, masses of other stars and interstellar distances tend to have uncertainties larger than that, so this is irrelevant on that scale. Still, within the solar system, using units of solar masses (or just $GM_odot$ directly) allows one to make much more precise measurements, and similarly Earth masses for satellite orbits around the Earth.

Measuring $G$ in terms of parsecs (or light-years) and $(mathrm{km}/mathrm{s})^2$ would be more convenient any time one is dealing with galactic-scale distances, such as in the galactic rotation curve:

(Image from Wikipedia).

big picture - Examples of eventual counterexamples

Nate Eldredge has mentioned the Skewes number,and in fact it is not the only place where we can speak of counterexamples, within number theory:
The Riemann hypothesis is a fairly good case where counterexamples have been sought for through huge amounts of computations. In the paper "The $10^{13}$ first zeros of the Riemann Zeta function, and zeros computation at very large height" by Xavier Gourdon and Patrick Demichel (using an algorithm by Andrew Odlyzko), the authors have checked out the truth of the Riemann hypothesis (that all non-trivial zeroes of $zeta(s)$ are encountered whenever $s = 1/2 + mathcal{i} T, T in mathbb{R}$), from the first, up to precisely the $10^{13}$th zero. In the same paper Riemann hypothesis has been tested numerically, checking out some $10^{9}$ zeroes from heights of $T$ as large as $10^{24}$. Now then, in spite of the fact that we have available such large amount of numerical evidence, this does not constitute a proof of the Riemann hypothesis, simply because the amount of zeroes is infinite, and there is no telling (yet) on whether we might encounter some day an instance of $zeta(s) = 0, s = a + mathcal{i} T, aneq 1/2$, and we might as well wait long time for a numerical counterexample, much in the same philosophy mentioned in Nate Eldredge's answer, and all this would constitute the answer to the second part of your first question: "do different eventual counterexamples share any common features?". In the cases discussed by Eldredge and in here, the common feature of the (possible) counterexamples is that in both cases a gigantic amount of numerical evidence was (has been, in the case of the Riemann hypothesis) amassed, and still there was (there is) the possibility of finding a counterexample.

Numerical calculations are still useful, tough, because there has been instances where the calculation does not have to be carried out that far. For example, the so-called Fermat's Little Theorem states that all numbers of the form $2^{2^{n}}+1$ are primes, and Fermat carried up calculations up to $n=4$. However, when $n=5$, Euler proved that such was no longer the case, since this "Fermat Number" can be factored into 641 and 6700417.

As for your second question:
"Could we build an 'early warning system' set of heuristics for seemingly plausible theorems?"
I am going to answer with something that must be taken "with a pinch of salt", but it is the closest I can think of an answer. I want you to refer to the p vs np problem (Polynomial time computer solving as opposite to Non-polynomial time algorithms). Informally, it asks whether every problem whose solution can be quickly verified by a computer can also be quickly solved by a computer. I imagine, if someone solves this computer unsolved problem (finding a polynomial-time running algorithm for a known non-polynomial time problem), then perhaps we could answer your second question in the positive sense, at least for those theorems which involve computable problems. For other types of theorems (say, one non-expressible algorithmically) I cannot imagine at the moment how can one could set up anything that would resemble somehow an 'early warning system' (but it would be interesting, though, if someone could furnish something like that, at least for a restricted problem :-) ).

soft question - Lecture notes on representations of finite groups

Next term I am supposed to teach a course on representation of finite groups. This is a third year course for undegrads. I was thinking to use the book of Grodon James and Martin Liebeck "Representations and characters of groups", but also looking for other references.

The question is: could you advise some other books (or lecture notes)? Maybe you had a nice experience of teaching or listening to a course with a similar title? It would be really nice if this book (notes) has also exercises.

ADDED. I would like to thank everybody who answered the question, very helpful answers!!! The answer of John Mangual below contains a "universal" reference.
For the moment my favourites are Serre (very clear and short introduction of main ideas), some bits from notes of Teleman and Martin, and Etingof for beautiful exposition. My last problem is to have enough of exercises, in particular to write down a good exam. So I would like to ask if there are some additional references for exercises (with or without solutions)?

ct.category theory - Equivalence of definitions of cartesian morphisms

Let $p: Cto D$ be a functor, and let $f:yto x$ be a morphism of $C$. We say that $f$ is cartesian if the canonical map $Q:(Cdownarrow f) to P:=(Cdownarrow x)times_{(Ddownarrow p(x)} (Ddownarrow p(f))$ is a surjective (on objects) equivalence of categories. However, if we write out what the (strict 2-) pullback means, the objects are precisely the pairs of morphisms $g: zto x$ and $h:p(z)to p(y)$ such that $p(g)=p(f) circ h$. If we look at the fibres of $Q$ over objects of $A$, we see that that they are contractible groupoids.

Using the more common definition of a cartesian morphism, we must show that any pair of morphisms $(g, h)$ as above uniquely determines an arrow $ell:zto y$ such that $fcirc ell= g$ and $p(ell)=h$.

I see how the first definition implies the existence of such a map, but how does it determine the map's uniqueness (up to more than a contractible space of choices)?

gr.group theory - Infinite groups which contain all finite groups as subgroups

My favorite is Thompson's group $V$.

My favorite picture of $V$ is to take a set $X$ which is a disjoint union of subsets $L$ and $R$ having fixed bijections $l:Xrightarrow L$ and $r:Xrightarrow R$. Finite words in $l$ and $r$ map $X$ to "fragments" of $X$. Two "fragments" $W$ and $U$ that are the images, respectively, of words $w$ and $u$ in $l$ and $r$ are connected by the bijection $uw^{-1}:Wrightarrow U$. The two "fragments" will be disjoint iff neither of $w$ nor $u$ is a prefix of the other. If this condition holds, let $(w,u)$ represent the permutation on $X$ that is $uw^{-1}$ on $W$, is $wu^{-1}$ on $U$, and is the identity elsewhere. The group $V$ is generated by all such $(w,u)$. It is finitely presented and contains all finite groups.

Other f.p. groups containing all finite groups are known as Houghton groups. See the section on Houghton groups K. S. Brown "Finiteness properties of groups" in Journal of Pure and Applied Algebra, 44 (1987), 45-75. I forget how they are indexed, but from about n=3 on up, they are all f.p.

set theory - Minimum cover of partitions of a set

If $k=2$, the answer is $2[(n+1)/2]-1$.

If $k=2$, then there are $nchoose2$ pairs, and each partition gets at most $[n/2]$ of them, so you can't do better than ${nchoose2}/[n/2]$, which is $2[(n+1)/2]-1$. So we have to show that we can achieve $2[(n+1)/2]-1$.

First let $n=2m-1$ be odd. Let the first partition be 1-with-$n$, 2-with-$(n-1)$, ...,
$(m-1)$-with-$(m+1)$,$m$-by-itself. Get the other partitions by repeatedly adding 1 to each number in the previous partition, working modulo $n$.

E.g., for $n=7$, the first partition is 1-7, 2-6, 3-5, 4, and the others are
2-1, 3-7, 4-6, 5; 3-2, 4-1, 5-7, 6; 4-3, 5-2, 6-1, 7; 5-4, 6-3, 7-2, 1;
6-5, 7-4, 1-3, 2; and 7-6, 1-5, 2-4, 3.

Now if $n=2m$ is even, just take the solution for $n=2m-1$ and in each partition pair $n$ up with the singleton. E.g., when $n=8$, the solution starts 1-7, 2-6, 3-5, 4-8;
2-1, 3-7, 4-6, 5-8; etc.

As Douglas notes, this is a problem of factoring the symmetric graph. For $k=2$ we're factoring it into 1-factors, and undoubtedly what I've written above is well-known.

ag.algebraic geometry - Gromov-Witten theory and compactifications of the moduli of curves

I can give individual answers to a lot of your questions, but I can't answer any of them completely, nor can I fit all these answers together into a coherent whole.

For string theory, there does seem to be something special about the Deligne-Mumford compactification. Morally, what's going on is this: string theorists are allowing cylinder-shaped submanifolds of their Riemann surfaces to become infinitely long. The only finite energy fields on such infinitely long submanifolds are constant, so you can replace the long cylinder with a node. (Likewise, morally, if you allow vertex operators at two marked points to come together, you should take their operator product. This is what bubbling when marked points collide does for you.)

Somewhat more technically: The first step in (bosonic) string theory is to compute the partition function of the nonlinear sigma model as a function on the space of metrics on your worldsheet. This function on metrics descends to a section of some line bundle on the moduli stack of complex structures on the worldsheet. When you can compute it at all, you can show that this section has exactly the right pole structure it needs to be a section of the 13th power of the canonical bundle tensored with the 2nd power of the dual of the line bundle corresponding to the boundary divisor of the Deligne-Mumford compactification. (There's an old Physics Report by Phil Nelson that explains this pretty well, although not with anything you'd call a proof. Should also credit Belavin & Knizhik, who did the initial calculations.)

There's a somewhat more modern perspective on this (Zwiebach, Sullivan, Costello,...) that says that the generating function of string theory correlation functions for smooth Riemann surfaces satisfies a certain equation (a "quantum master equation"), which gives instructions for how to extend the theory to nodal Riemann surfaces. Different master equations give different recipes for extending to the boundary, if I understand your advisor correctly.

People have played around with other compactifications. There are a lot of different compactifications of the stack of smooth marked curves. People have already mentioned a few of them. David Smyth has some cool results which classify the "stable modular" compactifications of the stack of curves (http://arxiv.org/abs/0902.3690 ). For compactifications of the moduli of maps, the only one that comes immediatley to mind is Losev, Nekrasov, and Shatashvili's "freckled instanton compactification", in which IIRC, you allow zeros and poles to collide and cancel each other out.

grants - "How not to write an NSF proposal" poster: does any one know where to find it online?

When I was recently at MSRI, they had a big poster of humorous advice on how to get an NSF proposal rejected (for example: "Definitely don't worry about sending it to an appropriate program officer. Surely it will find its way to the right person eventually."). I thought it was a very good summation of how to write a proposal. But I just tried to find it on Google, and failed (probably because I'm not remembering what the real title was).

Does anyone know where this list is available online?

terrestrial planets - Why Did Mars lose its Magnetic Field?

Our own magnetic field is generated by convection currents in Earth's liquid outer core.

A useful summary from Physics.org:

Differences in temperature, pressure and composition within the outer core cause convection currents in the molten metal as cool, dense matter sinks whilst warm, less dense matter rises.

This flow of liquid iron generates electric currents, which in turn produce magnetic fields.

The spiralling caused by the Coriolis force means that separate magnetic fields created are roughly aligned in the same direction.

Mars used to have a liquid iron core, but it was never as extensive as Earth's and has long since solidified.

Once our own core cools enough to solidify, we too will lose our magnetic field.

cosmology - To what extent are structure formations unexplained?

The info you reported are not opposite.
They refers to different scales on structure formation, and are both true.
As you said, ΛCDM model is very useful to explain large scale structures, as galaxies, clusters, and larger.
This means that the model fits really well to the observables, first of all the CMB map and its anisotropy.

What the model can NOT explain is the original formation of the structures, at the observed rate (take a look at this).
We know the age of universe (within this model), and we know the age of the observed galaxies. Then it results that galaxies have formed much faster then in a "simple" ΛCDM scenario. We can not explain how the primordial structures started to collapse and to grow in what now we see as galaxies.
To account for this observation we need to invoke dark matter primordial fluctuations. But this is an assumption, and does not give us the original spectrum of the fluctuations, neither it explain how it is formed. And this is a parameter that the ΛCDM model can not account for, at least at this stage.

reference request - Proofs without words

$$arctan frac{1}{3} + arctan frac{1}{2} = arctan 1$$

It's easy to generalize this to

$$ arctan frac{1}{n} + arctan frac{n-1}{n+1} = arctan 1, text{ for } n in mathbb{N}$$

which can further be generalized to

$$ arctan frac{a}{b} + arctan frac{b-a}{b+a} = arctan 1, text{ for } a,b in mathbb{N}, a leq b $$

Edit: A similar result relating Fibonacci numbers to arctangents can be found here and here.

Is the influence of gravity instantaneous?

The first question as stated has a rather trivial answer:

"If the sun magically disappeared, instantly, along with all its influences, how long would it take its gravity to stop having an effect on us?"

Since the Sun's gravity is among its influences, it would instantly stop having an effect on us. That's just part of the magical situation, and doesn't even involve any physics. A bit more interesting is the question without the bolded part.

In general relativity, changes in the gravitational field propagate at the speed of light. Thus, one might expect that the magical and instant disappearance of the Sun would not affect earth for about eight minutes, since that's how long light from the Sun takes to reach Earth.

However, this is mistaken because the instant disappearance of the Sun itself violates general relativity, as the Einstein field equation enforces a kind of local conservation law on the stress-energy tensor analogous to the non-divergence of the magnetic field in the electromagnetism: in any small neighborhood of spacetime, there are no local sources or sinks of stress-energy; it must come from somewhere and go somewhere. Since the magical instant disappearance of the Sun violates general relativity, it does not make sense to use that theory to predict what happens in such a situation.

Thus, the Sun's gravity instantly ceasing any effect on the Earth is just as consistent with general relativity as having any sort of time-delay. Or to be precise, it's no more inconsistent.

My big question, now, is: "How do we know it's instant?"

It's not instant, but it can appear that way.

We can't possibly move an object large enough to have a noticeable gravitational influence fast enough to measure if it creates (or doesn't create) a doppler-like phenomenon.

We don't have to: solar system dynamics are quite fast enough. An simple calculation due to Laplace in the early nineteenth century concluded that if gravity aberrated, Earth's orbit would crash into the Sun on the time-scale of about four centuries. Thus gravity does not aberrate appreciably--more careful analyses concluded that in the Newtonian framework, the speed of gravity must be more than $2times10^{10}$ the speed of light to be consistent with the observed lack of aberration.

This may seem quite a bit puzzling with how it fits with general relativity's claim that changes in the gravitational field propagate at the speed of light, but it's actually not that peculiar. As an analogy, the electric field of a uniformly moving electric charge is directed toward the instantaneous position of the charge--not where the charge used to be, as one might expect from a speed of light delay. This doesn't mean that electromagnetism propagates instantaneously--if you wiggle the charge, that information will be limited by $c$, as the electromagnetic field changes in response to your action. Instead, it's just something that's true for uniformly moving charges: the electric field "anticipates" where the change will be if no influence acts on it. If the charge velocity changes slowly enough, it will look like electromagnetism is instantaneous, even though it really isn't.

Gravity does this even better: the gravitational field of a uniformly accelerating mass is toward its current position. Thus, gravity "anticipates" where the mass will be based on not just current velocity, but also acceleration. Thus, if conditions are such that the acceleration of gravitating bodies changes slowly (as is the case in the solar system), gravity will look instantaneous. But this is only approximately true if the acceleration changes slowly--it's just a very good approximation under the conditions of the solar system. After all, Newtonian gravity works well.

A detailed analysis of this can be found in Steve Carlip's Aberration and the Speed of Gravity, Phys.Lett.A 267:81-87 (2000) [arXiV:gr-qc/9909087].

If he was wrong, how do we know it's not?

We have a lot of evidence for general relativity, but the best current evidence that gravitational radiation behaves as GTR says it does is Hulse-Taylor binary. However, there is no direct observation of gravitational radiation yet. The connection between the degree of apparent cancellation of velocity-dependent effects in both electromagnetism and gravity, including its connection with the dipole nature of EM radiation and quadrupole nature of gravitational radiation, can also be found in Carlip's paper.

Where to publish computer computations

You should upload the notebook along with the sources of the actual article to the arXiv, of course.

The official (but rather hard to find) advice on this from the arXiv is to place your code in a directory called /aux/. (This is problematic for windows users.)

You can see an example of this in my recent paper on the extended Haagerup subfactor. A footnote in the text of the article explains that the code is available along with the source download. It would also be appropriate to use the comments field in an arxiv submission to explain that source code is available in the /aux/ directory.

ag.algebraic geometry - Moduli spaces of coherent sheaves on K3s

This follows from a result of Yoshioka. In Theorem 8.1 of this paper Yoshioka showed that every moduli space of coherent sheaves on a K3 surface $X$ is deformation equivalent to an appropriate Hilbert scheme of points of $X$. Since every K3 is deformation equivalent to an elliptic K3 it follows that their Hilbert schemes are deformation equivalent and so you get the statement that you wanted.

lie groups - Adjoint orbits of small subspaces in Lie algebras

I've been studying isometric quotients (by compact Lie groups) of compact simply connected homogeneous spaces $G/H$ and their inherited curvature. One of the issues that continually arises is the following problem:

Let $mathfrak{g} = mathfrak{h}oplusmathfrak{p}$ be an orthogonal decomposition with respect to a Ad G invariant inner product on $mathfrak{g}$. When is there a $gin G$ such that $Ad(g)mathfrak{p}cap mathfrak{p} = {0}$?

Of course, if the dimension of $mathfrak{p}$ is too large (equivalently, if the dimension of $mathfrak{h}$ is too small), then they must intersect nontrivially for dimension reasons. This leads us to the general form of the question:

Let $G$ be a compact Lie group and let $Vsubseteq mathfrak{g} = Lie(G)$ be a vector subspace (but not neccesarily a subalgebra). Assume dim($V)leq frac{1}{2}$ dim($mathfrak{g})$. Is there a $gin G$ such that $Ad(g)Vcap V = {0}$?

Of course, if there is one $g$, an entire neighborhood around $g$ will have this property. But

when is there an open and dense set of $g$ such that $Ad(g)Vcap V={0}$?

(If it helps, feel free to assume that $G$ is simple. I'm pretty sure (but haven't been able to prove) that one can reduce the question to simple $G$ anyway.)

Note that for $G=SU(2)$ or $G=SO(3)$, there is an open and dense set of such $g$. This follows because in this case, $Ad:Grightarrow SO(mathfrak{g})$ is surjective and, of course the usual $SO(n)$ action on $mathbb{R}^n$ acts transitively on k-subspaces. Further, the set of all elements of $SO(3)$ that fix a given line in $mathbb{R}^3$ is isomorphic to an $SO(2)$, which is codimension 2 in $SO(3)$.

However, for any other $G$, this trick doesn't work as $Ad$ will fail to be surjective.

Thank you in advance for your time and feel free to add more tags as appropriate!

dg.differential geometry - Uniformization theorem for 2-manifolds

There is an excellent expository article by Peter Scott titled "The Geometries of 3-Manifolds." The first section is spent talking about uniformization of surfaces (which is what you are interested in), and he also discusses 2-Oribfolds, which are what one gets when the group action is properly discontinuous but not free. The rest of the article is certainly good reading, but not really relevant to what you asked. Most importantly though, he sketches a proof of uniformazation for surfaces. The basic idea is that the only group which acts nicely on $S^2$ is $mathbb{Z}/2mathbb{Z}$, with quotient $P^2$. Furthermore, discrete groups of isometries of $mathbb{R}^2$ with compact quotient are isomorphic to a group with a finite index subgroup isomorphic to $mathbb{Z} oplus mathbb{Z}$, and hence the quotient is either a torus or a Klein bottle.Then one can classify the isometries of the hyperbolic plane, $mathbb{H}^2$, and show that $mathbb{Z} oplus mathbb{Z}$ cannot be a discrete, orientation preserving subgroup of $Isom(mathbb{H}^2)$, and hence the torus and Klein bottle cannot admit a hyperbolic structures. Furthermore, it is easy to show (with the classification of hyperbolic isometries) that other surfaces do admit hyperbolic structures.

Also, to give you an idea of why free and properly discontinuous are important, it helps to see a few examples. First, consider $mathbb{R}^2$ and let $G$ be the group generated by rotation around the origin through the angle $2pi/n$. This action has a fixed point at the origin. Topologically, the quotient is just $mathbb{R}^2$, but is has a cone point with angle $2pi/n$ at the origin, so geometrically it is different. This type of space is what is known as an orbifold.

Without proper discontinuity, things get even worse. Consider $S^1$ and the group $G$ generated by rotation through an irrational multiple of $pi$. The resulting quotient is not even Hausdorff (the orbit of a point under $G$ is dense in $S^1$), and we generally want to avoid that sort of thing.

In dimension 3, things get a little harder, but geometrization still works. A really good reference for this dimension is William Thurston's "3 Dimensional Geometry and Topology," which is easily one of my favorite math books ever written.

In dimension 4, the same sort of approach is a lot harder. The reason is that in dimensions 2 and 3, we have a fair deal of control as to what sorts of groups show up as fundamental groups of (closed) manifolds. It can be shown (I don't recall a reference for this off the top of my head, perhaps someone else has it) that for any finitely presented group $G$, there exists a closed 4-manifold $M$ with $pi_1(M) = G$. So, we don't have the kind of set-up like in dimension 2 where we know which fundamental groups arise, and then show that they are isometries of one of three model spaces.

Hope this helps.

lattices - Is there any literature on multivariable theta functions?

The theta function of a lattice is defined to be
$$ vartheta_Lambda = sum_{vinLambda} q^{{Vert vVert}^2}$$
which yields as a coefficient of q^k the number of vectors of norm-squared k.

On the other hand, the Jacobi theta function is given by
$$ vartheta(u,q) = sum_{n=-infty}^infty u^{2n}q^{n^2}$$
and we have the obvious fact that if $Lambda = mathbb{Z}$ with its usual intersection form, then $vartheta(1,q)$ is the theta function for that lattice.

We also have the fact that $vartheta_{Lambda_1 oplus Lambda_2} = vartheta_{Lambda_1}vartheta_{Lambda_2}$, and so we can decompose our theta functions into products of theta functions of primitive lattices.

Combining these facts, it is not entirely ridiculous to hope that there is some way to write, for a lattice of rank k, a ``theta function'' of the form
$$vartheta(u_1, ldots, u_k, q)$$
such that $vartheta(1, ldots, 1, q)$ is the ordinary theta function of the lattice. In some sense, the u-variables keep track of the basis elements of the lattice which immediately raises the question as to well-definedness of such an idea; it is worth noting that for the lattice $Lambda = bigoplus_imathbb{Z}$ that this definition does make sense.

So is there any literature on such objects? Do they make sense for lattices which are not just sums of copies of ℤ? Do they have nice relations akin to those of normal theta functions?

it.information theory - Entropy of a general prob. measure

It is not. If a probability measure on $mathbb{R}$ is absolutely continuous and has density $f$, then "entropy" usually refers to the differential entropy, defined in the Wikipedia page falagar linked to. If the probability measure has discrete support, entropy is defined by an analogous formula, given in this Wikipedia page. In the most classical treatments, these are the only situations covered at all.

However, both of these are special cases of the more general notion of relative entropy that Helge and robin girard pointed out: in the continuous case the reference measure ($nu$ in Helge's notation, $mu$ in robin's notation) is Lebesgue measure, and in the discrete case the reference measure is counting measure on the support.

pr.probability - Are there interesting problems involving arbitrarily long time series of small matrices?

Are there well-known or interesting applied problems (especially of the real-time signal processing sort) where arbitrarily long time series of small (say $d equiv dim le 30$ for a nominal bound, and preferably sparse) matrices arise naturally?

I am especially interested in problems that can be mapped onto a setup in which for each event of a reasonably nice point process on $mathbb{R}$ (the simplest two such processes would be a Poisson or discrete-time process) there is an associated pair $(j,k) in {1,dots,d}^2$. In this case time-windowed sums $N_{jk}(t)$ of the various pairs can be formed in an obvious way (although there may be plenty of subtlety or freedom in the windowing itself): these supply such a matrix time series.

Each such pair $(j,k)$ could be regarded as a transition from server $j$ to another (possibly identical) server $k$ in a closed queue with $d$ servers and infinitely many clients. It is not hard to see that in the setting of communication networks, this framework amounts to a very general form of traffic analysis. Such an application should not be considered for an answer: it's already been covered.

A slightly more restrictive but simpler example is where the pairs $(j,k)$ are inherited from a cadlag random walk on the root lattice

$A_{d-1} :=left {x in mathbb{Z}^d : sum_{j=1}^d x_j = 0right }$.

Examples of this sort would also be of considerable interest to me.

gr.group theory - Residual finiteness for graph manifold groups

As far as I'm aware, every proof of this fact is essentially the same as Hempel's original proof. I don't know whether it's "simple" enough for you! The key point is that the fundamental group G of a Seifert-fibred piece has the following property.

Property. There exists an integer K such that for any positive integer n there is a finite-index normal subgroup G_n of G such that any peripheral subgroup P intersects G_n in KnP.

It's not too hard to prove. There's a nice account in a paper by Emily Hamilton (which generalizes Hempel's result).

The other important fact is that peripheral subgroups in Seifert-fibred manifold groups are separable (ie closed in the profinite topology, for any non-experts out there).

Using these two pieces of information, you can piece together finite quotients of Seifert-fibred pieces into a virtually free quotient of π₁ of the graph manifold in which your favourite element doesn't die.

Note on separability of peripheral subgroups. Of course, Scott proved that Seifert-fibred manifold groups are LERF. But, by a pretty argument of Long and Niblo, a subgroup is separable if and only if the double along it is residually finite. In particular, you can deduce peripheral separability from the easier fact that Seifert-fibred manifold groups are residually finite.

differential equations - maximum decay rate

Harald has already shown that the $1/t$ decay is the best "general decay rate" for, say, smooth data with compact support. I claim that a slightly stronger result can hold: if you have smooth data with compact support, and the evolution is such that there exists a constant $C$ such that your solution decays faster than $C/t^{1+epsilon}$, your initial data must be trivial.

Harald's answer already tells you that, after fixing an origin, the spherical mean of the solution cannot decay faster than $1/t$, unless it vanishes. So if the solution has our supposed decay, the spherical mean around any point is 0. This is equivalent to saying that the spherical Radon transform of your initial data is 0. So if your data is in a sufficiently nice class (smooth with compact support will do), you can invert the spherical Radon transform and get that your initial data must be 0.

Another way to see this is to take the inverse scattering point of view. Look at the fundamental solution of the wave equation. If you multiply it by $t$ and formally take the limit as $t=rto infty$, and consider the function $f(x,e) = lim t u(t,x + re)$ (where $e$ is a unit vector), you see that $f$ is the Radon transform of the data. For data with compact support, the $limsup$ and $liminf$ are well-defined (by the general decary rate of $1/t$), and they both go to 0 if the $u$ decays any faster. So taking the inverse transform you have that the data must be 0 also.

big picture - Integrable dynamical system - relation to elliptic curves

From seminar on kdV equation I know that for integrable dynamical system its trajectory in phase space lays on tori. In wikipedia article You may read (http://en.wikipedia.org/wiki/Integrable_system):

When a finite dimensional Hamiltonian
system is completely integrable in the
Liouville sense, and the energy level
sets are compact, the flows are
complete, and the leaves of the
invariant foliation are tori. There
then exist, as mentioned above,
special sets of canonical coordinates
on the phase space known as
action-angle variables, such that the
invariant tori are the joint level
sets of the action variables. These
thus provide a complete set of
invariants of the Hamiltonian flow
(constants of motion), and the angle
variables are the natural periodic
coordinates on the torus. The motion
on the invariant tori, expressed in
terms of these canonical coordinates,
is linear in the angle variables.

As I also know that elliptic curve is in fact some kind of tori, then there natural question arises: Are tori for quasi-periodic motion in action-angle variables of some dynamical systems related in any way to algebraic structure like elliptic curve? Maybe some small dynamical systems and some elliptic curves are related in some way?

The most interesting in this matter is for me the size of space of elliptic functions: its quite small, every elliptic curve is rational function of Weiestrass function, and its derivative. Has this property any analogy in integrable dynamical systems theory?

As isomorphic elliptic curves shares some invariants, it is also interesting it they have any "dynamical meaning".

Is Sahara desert a good place to build radio/optical telescopes?

No, the Sahara isn't a good place to build telescopes.
The Atacama desert is used because it is at high altitude, which means that there is less atmosphere to get in the way. Other telescopes are located on mountaintops for the same reason.
The Sahara is mostly at sea level. It's also very hot, so you get lots of turbulence due to rising air, which distorts the image.
The Sahara is also undesirable for other reasons: no infrastructure, unstable regimes, lots of erosion due to sand storms.

natural satellites - Why does Jupiter have so many moons?

Bigger is better.

Most moons, especially those of gas giants, are not "formed", they are just "captured" (unlike our Moon, which could have been captured, but probably was formed in a much more exciting way).

Jupiter is the most massive planet in the solar system. It stands to reason that it has a larger region of gravitational influence (where its influence outweighs the force due to the other planets and the sun). So, it's easy for it to capture rocky masses.

If you have a look at the contours on the following image (Ignore the Lagrange points marked on it, I only want the contours)

the circular area around the Earth is more or less the area (there's a velocity dependence here which I'm not getting into) in which a moon-like body can form a reasonably stable orbit. The size of the small "well" will increase as the planet moves farther from the sun, and also when the planet is more massive.

Jupiter is both pretty far away from the Sun, and is very massive. This leads to a huge sphere of influence.

The asteroid belt may have something to do with this too, but I doubt it (it's pretty far away). However, if we assume the "half-baked planet formation" theory for the formation of the belt, Jupiter may have leeched off much of the mass that would have otherwise become part of that planet during the formative period.

rt.representation theory - References for Lie superalgebras

For a quick, self-learning introduction you can take a look at Alberto Elduque's talks and papers in

Alberto Elduque’s Research

starting first with the talk called "Simple modular Lie superalgebras; Encuentro Matemático Hispano-Marroquí (Casablanca, 2008)."

solar system - Class presentation on the Sun, ideas?

Tomorrow I have to give a [high school] class presentation on a topic of the solar system. I have thought of the Sun, as I have always believed it to be the central part of the Solar System. So I am asking here, is there any creative ideas that I can present? I do not want to make an speech and plainly read out the Wikipedia page. So, are there any ideas, like I thought of telling how we came to measure the distance between Earth and Sun. Is that possible to explain it to high school students, if so, where I can find more information regarding it? Are there such questions, facts, which I can present, preferably if there are some very simple demonstrations or pre-calculus maths involved?

ag.algebraic geometry - Intuition about schemes over a fixed scheme

I am taking a first course on Algebraic Geometry, and I am a little confused at the intuition behind looking at schemes over a fixed scheme. Categorically, I have all the motivation in the world for looking at comma categories, but I would like to make sense of this geometrically.

Here is one piece of geometric motivation I do have: A family of deformations of schemes could be thought of as a morphism $X rightarrow Y$, where the fibers of the morphism are the schemes which are being deformed, and these are indexed by the scheme Y.

This is all well and good, but I am really interested in Schemes over $Spec(k)$ are thought of as doing geometry "over" $k$. I know that their is a nice "schemeification functor" taking varieties over $k$ to schemes over $k$, but this is somewhat unsatisfying. All that I see algebraically is that $k$ injects into the ring of global sections of the structure sheaf of $X$, but this does not seem like much of a geometric condition...

Any help would be appreciated.

EDIT: The answers I have received so far have been good, but I am looking for something more. I will try to flesh out an example that give the same style of answer that I am asking for:

The notion of a k-valued point: Every point $(a,b)$ of the real variety $x^2 + y^2 - 1 = 0$ has a corresponding evaluation homomorphism $mathbb{R}[x,y] rightarrow mathbb{R}$ given by $x mapsto a$ and $y mapsto b$. Since $a^2 + b^2 - 1 = 0$, this homomorphis factors uniquely through $mathbb{R}[x,y]/(x^2 + y^2 -1)$. So the real valued points of the unit circle are in one to one correspondence with the homomorphisms from $mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $mathbb{R}$. Similarly, the complex valued points are in one to one correspondence with the homomorphisms from $mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $mathbb{C}$. Actually, points of the unit circle valued in any field $k$ are going to be in one to one correspondence with homomorphisms from $mathbb{Z}[x,y]/(x^2 +y^2 -1)$ to $k$.

Dualizing everything in sight, we are saying that the $k$- valued points of the scheme $Spec(mathbb{Z}[x,y]/(x^2 +y^2 -1))$ are just given by homomorphisms from the one point scheme $Spec(k)$ into $Spec(mathbb{Z}[x,y]/(x^2 +y^2 -1))$.

EDIT 2: Csar's comment comes very close to answering the question for me, and I will try and spell out the ideas in that comment a little better. I wish Csar had left his comment as an answer so I could select it.

So it seems to me that the most basic reason to think about schemes over a field $k$ is this:

I already spelled out above why $k$-valued points of a scheme are important. But a lot of the time, a morphism from $Spec(k)$ to $X$ will point to a generic point of $X$, not a closed point. Different morphisms can all point to the same generic point. For instance the dual of the any injection $mathbb{Z}[x,y] rightarrow mathbb{C}$ (of which there are many), will all "point" to the generic point of $mathbb{Z}[x,y]$.

On the other hand if we are looking at $mathbb{Q}[x,y]$, this is a $mathbb{Q}$ scheme. $Spec(mathbb{Q})$ is also a $mathbb{Q}$ scheme via the identity map. A morphism of $mathbb{Q}$ schemes from $Spec(mathbb{Q})$ to $mathbb{Q}[x,y]$ will correspond to a closed $mathbb{Q}$ -valued point! So this is a nice geometric reason for looking at schemes over a fixed field $k$: $k$ morphisms of $Spec(k)$ to $X$ correspond to $k$-valued closed points of $X$.

It will take some work for me to internalize the vast generalization that S-schemes entail, but I think this is a good start. Does everything I said above make sense?

ap.analysis of pdes - Regularity of reflection coefficients (or more generally the scattering transform)

Consider the Schrodinger operator $L(q) = -partial_x^2 + q(x)$ where the potential $q$ is a real-valued function of a real variable which decays sufficiently rapidly at $pm infty$.

We define the scattering data in the usual way, as follows:

The essential spectrum of $L(q)$ is the positive real axis $[0,infty)$ and it has multiplicity two. The Jost functions $f_pm(cdot,k)$ corresponding to $L(q)$ solve $L(q)f_pm = k^2f_pm$ with $f_pm(x,k) sim e^{ikx}$ as $x to pm infty$.

The reflection coefficients $R_pm(k)$ are defined so that $f_pm = bar{f_mp} + R_pm(k)f_mp$ where here overbar denotes complex conjugate. The intuition is that $R$ measures the amount of energy which is reflected back to spatial $infty$ when a wave with spatial frequency $k$ that originates at spatial $infty$ interacts with the potential $q$.

The scattering transform (the map from $q$ to the scattering data, of which $R_+$ is a part) and its inverse are important in the theory of integrable PDE.

My question is the following:
What is the regularity of the map $q mapsto R_+$? Is it continuously differentiable?

To answer this question, we first must specify the spaces that $q$ and $R_+$ live in. I don't really care so long as they are reasonable spaces, for example take $q$ in weighted $H^1$ where the weight enforces a rapid decay at $pm infty$.

Does the category Monoid of monoids have finite coproducts?

Yes. More generally, any category of algebras (in the sense of universal algebra), such as groups, rings, vector space, Lie algebras, ..., has all (small) limits and colimits. See for instance p.210 (end of section IX.1) of Mac Lane's book Categories for the Working Mathematician.

Explicitly, the initial monoid ("0-fold coproduct of monoids") is the one-element monoid. The coproduct $A * B$ of two monoids $A, B$ is constructed similarly to the coproduct of two groups (often called their "free product"). That is, it's the free monoid generated by all the elements of $A$ together with all the elements of $B$, quotiented out by all the relations that hold in $A$ and all the relations that hold in $B$.

lo.logic - Basis of l^infinity

The question is about the complexity of the simplest possible Hamel basis of l^infty, and this is a perfectly sensible thing to ask about even in a context where one wants to retain the Axiom of Choice. That is, we know by AC that there is a basis---how complex must it be?

Such a question finds a natural home in descriptive set theory, a subject concerned with numerous similar complexity issues. In particular,
descriptive set theory provides tools to make the question
precise.

My answer is that one can never prove a negative answer to the question, because it is consistent with the ZFC axioms of set theory that
Yes, one can concretely exhibit a Hamel basis of l^infty.

To explain, one natural way to measure the complexity of sets of reals
(or subsets of R^infty) is with the projective hierarchy.
This is the hierarchy that begins with the closed sets (in
R, say, or in R^omega), and then iteratively closes under
complements and projections (or equivalently, continuous
images). The Borel sets appear near the bottom of this
hierarchy, at the level called Delta¹₁, and then the
analytic sets Sigma¹₁ and co-analytic sets Pi¹₁, and so
on up the hierarchy. Sets in the projective hierarchy are
exactly those sets that can be given by explicit definition
in the structure of the reals, with quantification only
over reals and over natural numbers. If we were to find a
projective Hamel basis, then it will have been
exhibited in a way that is concrete, free of
arbitrary choices. Thus, a very natural way of making the
question precise is to ask:

Question. Does l^infty have a Hamel basis that is
projective?

If the axioms of set theory are consistent, then they are
consistent with a positive answer to this question. This is
not quite the same as proving a positive answer, to be
sure, but it does mean that no-one will ever prove a
negative answer to the question.

Theorem. If the axioms of ZFC are consistent, then
they are consistent with the existence of a projective
Hamel basis for l^infty. Indeed, there can be such a basis
with complexity Pi¹₃.

Proof. I will prove that under the set-theoretic assumption
known as the Axiom of Constructibility V=L, there is a
projective Hamel basis. In my answer to question about Well-orderings of the reals, I explained that
in Goedel's constructible universe L, there is a definable
well-ordering of the reals. This well-ordering has
complexity Delta¹₂ in the projective hierarchy. From this
well-ordering, one can easily construct a well-ordering of
l^infty, since infinite sequences of reals are coded
naturally by reals. Now, given the well-ordering of
l^infty, one defines the Hamel basis as usual by taking all
elements not in the span of elements preceding it in the
well-order. The point for this question is that if the
well-order has complexity Delta¹₂, then this definition
of the basis has complexity Pi¹₃, as desired. QED

OK, so we can write down a definition, and in some
set-theoretic universes, this definition concretely
exhibits a Hamel basis of l^infty. There is no guarantee,
however, that this definition will work in other models of
set theory. I suspect that one will be able to find other
models of ZFC, in which there is no projective Hamel basis
of l^infty. It is already known that there might be no
projective well ordering of the reals (a situation that
follows from large cardinals and other set theoretic
hypotheses), and perhaps this also implies that there is no
projective Hamel basis. In this case, it would mean that
the possibility of exhibiting a concrete Hamel basis is
itself independent of ZFC. This would be an
interesting and subtle situation. To be clear, I am not
referring here merely to the existence of a basis requiring
AC, but rather, fully assuming the Axiom of Choice, I am
proposing that the possibility of finding a
projective basis is independent of ZFC.

Conjecture. The assertion that there is a projective
Hamel basis of l^infty is independent of ZFC.

I only intend to consider the question in models of ZFC, so
that l^infty has a Hamel basis of some kind, and the only
question is whether there is a projective one or not. In this situation, the fact that AD seems to imply that there is no Hamel basis is not relevent, since that axiom contradicts AC.

Apart from this, I also conjecture that there can never be
a Hamel basis of l^infty that is Borel. This would be a lower bound on the complexity of how concretely one could exhibit the basis.

galois theory - algebraic numbers of degree 3 and 6, whose sum has degree 12

Here's an example making Gerry's suggestion explicit (I was curious what an example would look like): $a = {omega}2^{1/3}$, $b = 2^{1/6}$, where $omega$ is a nontrivial cube root of unity. By PARI, $a + b$ has minimal polynomial
$$
x^{12} - 8x^9 + 18x^8 + 12x^7 + 20x^6 - 72x^5 + 276x^4 - 232x^3 + 180x^2 + 24x + 4.
$$

Without giving it a lot of thought, you might think the minimal polynomial might be something like
$$
((x-2^{1/3})^6-2)((x - {omega}2^{1/3})^6-2)((x - {omega}^22^{1/3})^6-2)
$$
and this polynomial of degree 18 does have rational coefficients, but it factors as
$$
(x^6 - 4x^3 - 18x^2 - 12x + 2)
(x^{12} - 8x^9 + 18x^8 + 12x^7 + 20x^6 - 72x^5 + 276x^4 - 232x^3 + 180x^2 + 24x + 4).
$$
Yes, the second factor is the polynomial above. The multiplicative relations among
$a$ and $b$ account for such breaking up.

ag.algebraic geometry - Poincaré duality for smooth projective varieties over finite fields

The main case can be found in Milne's article specifically Theorems 1.13, 1.14 on page 310.
The idea, briefly, is as follows: Given a sheaf $F$ on a variety $X$ over a finite field $k$, then over an algebraic closure $bar{k}$ of $k$, the group $H^i_{et}(bar{X}, F)$ becomes a $Gal(bar{k}/k)$-module. There is a spectral sequence involving the $H^j(Gal(bar{k}/k), H^i_{et}(bar{X}, F))$ which converges to $H^n_{et}(X,F)$. This is true over any perfect field.

When you have duality over $bar{k}$ (e.g. $X$ smooth proper and $F$ nice), combine it with duality in Galois cohomology (in our case, the group is very simple: $hat{Z}$) to get duality over $k$.
The duality theorems now reflect the $k$: if Poincare duality for $X$ of dimension $d$ over $bar{k}$ pairs $H^i$ with $H^{2d-i}$, over $k$ the pairing will be between $H^i$ and $H^{2d +m -i}$ where $m$ is the cohomological dimension (assumed finite) of the Galois group ($m=1$ in the case of a finite field).

Hope this helps.

nt.number theory - What are the prime ideals in rings of cyclotomic integers?

Theorem: Let $alpha$ be an algebraic integer such that $mathbb{Z}[alpha]$ is integrally closed, and let its minimal polynomial be $f(x)$. Let $p$ be a prime, and let

$displaystyle f(x) equiv prod_{i=1}^{k} f_i(x)^{e_i} bmod p$

in $mathbb{F}_p[x]$. Then the prime ideals lying above $p$ in $mathbb{Z}[alpha]$ are precisely the maximal ideals $(p, f_i(alpha))$, and the product of these ideals (with the multipicities $e_i$) is $(p)$. (Theorem 8.1.3.)

In this particular case we have $f(x) = Phi_n(x)$. When $(p, n) = 1$, its factorization in $mathbb{F}_p[x]$ is determined by the action of the Frobenius map on the elements of order $n$ in the multiplicative group of $overline{ mathbb{F}_p }$, which is in turn determined by the minimal $f$ such that $p^f - 1 equiv 0 bmod n$ as described in Chandan's answer. (This $f$ is the size of every orbit, hence the degree of every irreducible factor.) When $p | n$ write $n = p^k m$ where $(m, p) = 1$, hence $x^n - 1 equiv (x^m - 1)^{p^k} bmod p$. Then I believe that $Phi_n(x) equiv Phi_m(x)^{p^k - p^{k-1}} bmod p$ and you can repeat the above, but you'd have to check with a real number theorist on that. (Edit: Indeed, it's true over $mathbb{Z}$ that $Phi_n(x) = frac{ Phi_m(x^{p^k}) }{ Phi_m(x^{p^{k-1}}) }$.)

at.algebraic topology - Functorial Whitehead Tower?

The nth stage of the Whitehead tower of X is the homotopy fiber of the map from X to the nth (or so) stage of its Postnikov tower, so you can use your functorial construction of the Postnikov tower plus a functorial construction of the homotopy fiber (such as the usual one using the path space of the target).

The nth stage of the Whitehead tower of X is also the cofibrant replacement for X in the right Bousfield localization of Top with respect to the object Sⁿ (or so). Since Top is right proper and cellular this localization exists by the result of chapter 5 of Hirschhorn's book on localizations of model categories. You might look there to see how the cofibrant replacement functor is constructed. With some care you should be able to define functorially the maps in the tower as well.

(BTW, the Postnikov tower can similarly be obtained functorially by a left Bousfield localization of Top.)

at.algebraic topology - Functoriality of Poincaré duality and long exact sequences

Hi all,

Today I was playing with the cohomology of manifolds and I noticed something intriguing. Although I might just have been caught out by a couple of enticing coincidences, it feels enough like there might be something going on that I thought I'd put it out here.

Let $M$ be an $n$-manifold with boundary $partial M$. We write out the long exact homology sequence for the pair $(M, partial M)$:

$$cdots to H_k(M) to H_k(M, partial M) to H_{k - 1}(partial M) to cdots$$

Let's apply Poincaré duality termwise, and keep the arrows where they were out of sheer faith. What we get is

$$cdots to H^{n - k}(M, partial M) to H^{n - k}(M) to H^{n - k}(partial M) to cdots$$

Surprisingly, this is the long exact cohomology sequence for the pair $(M, partial M)$! To my mind, two things here are weird. The first is that intuitively, any functoriality properties Poincaré duality possesses should be arrow-reversing. The second is that we have a shift - but not a shift by a multiple of 3. So the boundary map in the homology sequence 'maps' to something that doesn't change degree in the cohomology sequence.

Let's play the same game with the Mayer-Vietoris sequence. For simplicity, suppose now $M$ is without boundary. Write $M = A cup B$ where $A$ and $B$ are $n$-submanifolds-with-boundary of $M$ and $A cap B$ is a submanifold of $M$ with boundary $partial A cup partial B$. Then we have

$$cdots to H_k(A cap B) to H_k(A) oplus H_k(B) to H_k(M) cdots$$

Hitting it termwise with Poincaré duality, and cruelly and unnaturally keeping the arrows where they are once again, we get

$$cdots to H^{n - k}(A cap B, partial A cup partial B) to H^{n - k}(A, partial A) oplus H^{n - k}(B, partial B) to H^{n - k}(M) to cdots$$

This looks unfamiliar, but by looking at cochains it's not hard to see that there actually is a long exact sequence with these terms. However, this time we don't have the weird shift.

Now is there anything going on here, or just happenstance? Is there really a sense in which Poincaré duality is functorial with respect to long exact sequences? If so, what's the 'Poincaré dual' of the long exact sequence of the pair $(M, N)$ where $N$ is a tamely embedded submanifold of $M$?

Edit: Realised that in the final l.e.s. the arrows should actually go the other way, which is slightly less impressive. Even so...

gn.general topology - What does the property that path-connectedness implies arc-connectedness imply?

A space X is path-connected if any two points are the endpoints of a path, that is, the image of a map [0,1] to X. A space is arc-connected if any two points are the endpoints of a path, that, the image of a map [0,1] to X which is a homeomorphism on its image. If X is Hausdorff, then path-connected implies arc-connected.

I was wondering about the converse: What properties must X have if path-connected implies arc-connected? In particular, what are equivalent properties?

ag.algebraic geometry - Hodge Standard Conjecture in Positive Characteristic

Probably too late, but...

(1) Not quite. The Hodge index theorem only works for $mathbb{C}$. To extend it to all characteristic $0$ ground fields, you need the Lefschetz principle and the comparison theorem.

(2) As J. S. Milne mentioned, the wikipedia article was mistaken (it has since been fixed). It isn't known how to extend Segre-Grothendieck to higher dimension varieties.

(3) It seems that only surfaces are known so far. The Lefschetz conjecture for projective spaces and Grassmannians follows from intersection theory, but I'm not aware of such results for the Hodge standard conjecture.

bad reduction for elliptic curves

I would like to stress something that, I think, has not been point out very explicitly in any of the nice answers above.

The question that you are asking, if I understand it correctly, is a local question. So let's start with, say, a $p$-adic field $K$, and with the equation $f(x,y)=0$ of (the affine piece of) an elliptic curve $E$ with coefficients in the ring of integers $R$ of $K$.

To define the reduction of $E$ modulo the maximal ideal $mathfrak{m}$ of $R$, one cannot follow the naive approach of simply considering the reduction mod $mathfrak{m}$ of the equation $f(x,y)=0$.

Instead one should look at what Tate called a minimal Weierstrass equation $f_{rm min}(x,y)=0$ for $E$ (LNM 476 pag. 39), and define the reduction $bar E$ of $E$ as the object obtained by considering the reduction mod $mathfrak{m}$ of $f_{rm min}(x,y)=0$.

The point now is that the minimal equation is NOT stable by taking field extensions: if $L/K$ is a finite extension then $f_{rm min}(x,y)$ need not still be a minimal equation for the base change of $E$ to $L$ (but it is if $L/K$ is unramified). Therefore it might very well happens that while $f_{rm K, min}(x,y)=0$ has a singular reduction mod $mathfrak{m}$ the corresponding thing does not hold for $f_{rm L, min}(x,y)=0$ (the meaning of the subscripts is obvious I hope).

I tried to come up with an explicit example illustrating this phenomena. I failed to find one in a reasonable amount of time. But I am sure they can be found (easily?) in the literature.

ag.algebraic geometry - About the intersection of two vector bundles

I´m looking for information about the intersection of two vector bundles (principally trivial bundles, but no necessarily). I´m trying to make a picture (literally) of reflexive finite generated modules.

Another related topic is sub-budles of a vector bundles.

All suggestions are wellcome!

---

Edit: I try to be more specific.

Suppose two sub-bundles of a bundle over a topological space X. We can do the intersection of their vector fibers in each point of base space and collect this intersection vector fibers along the base space. Then we can give it a topology, restriction of the bigger vector bundle.

What´s about of this "submodule of sections"?

Is it another vector bundle?

Has another interesting structure?

Black Holes emitting Hawking radiation

The escape velocity is c at the event horizon of a black hole. Above the event horizon the escape velocity is below c. Escaping particles of the Hawking radiation form above the event horizon; that's why they can escape, if they are pointing towards a sufficiently narrow angle to vertical upward, and if they are sufficiently energetic.

Escaping particles form as virtual particle-antiparticle pairs in the "infalling" coordinate system: One of the two particles forms outside the event horizon; the counterpart forms below the event horizon. Thus the originally virtual particles cannot annihilate, and therefore become real particles; one particle can escape; the counterpart falls toward the singularity.

The energy needed to form the escaping particle, and its remaining kinetic energy after escape, is subtracted from the mass of the black hole.

The described mechanism probably works also for both particles forming very close above the event horizon if the tidal forces are high enough to separate the virtual particle pair, before it can annihilate.

Formation of virtual particles is due to the Heisenberg uncertainty applied to time and energy: Very short time intervals require energy uncertainty, leading to short-lived particle-antiparticle pairs.

Non-commutative geometry from von Neumann algebras?

The Gelfand transform gives an equivalence of categories from the category of unital, commutative $C^*$-algebras with unital $*$-homomorphisms to the category of compact Hausdorff spaces with continuous maps. Hence the study of $C^*$-algebras is sometimes referred to as non-commutative topology.

All diffuse commutative von Neumann algebras acting on separable Hilbert space are isomorphic to $L^infty[0,1]$. Hence the study of von Neumann algebras is sometimes referred to as non-commutative measure theory.

Connes proposed that the definition of a non-commutative manifold is a spectral triple $(A,H,D)$. From a $C^*$-algebra, we can recover the "differentiable elements" as those elements of the $C^*$-algebra $A$ that have bounded commutator with the Dirac operator $D$.

What happens if we start with a von Neumann algebra? Does the same definition give a "differentiable" structure? Is there a way of recovering a $C^*$-algebra from a von Neumann algebra that contains the "differentiable" structure on our non commutative measure space? This would be akin to our von Neumann algebra being $L^infty(M)$ for $M$ a compact, orientable manifold (so we have a volume form). Or are von Neumann algebras just "too big" for this?

One of the reasons I am asking this question is Connes' spectral characterization of manifolds (arXiv:0810.2088v1) which shows we get a "Gelfand theory" for Riemannian manifolds if the spectral triples satisfy certain axioms. Connes starts with the von Neumann algebra $L^infty(M)$ instead of the $C^*$-algebra $C(M)$.

st.statistics - comparing multiple contingency tables, independent data

suppose I have $K$ different methods for forecasting a binary random variable, which I test on independent sets of data, resulting in $K$ contingency tables of values $n_{ijk}$ for $i,j=1,2$ and $k=1,2,...,K$. How can I compare these methods based on the contingency tables? The general case would be nice, but $K=2$ is also very interesting.

I can think of a few approaches:

• compare the tetrachoric coefficients of the $K$ tables. this would be especially useful if there were something like Fisher's R-to-Z transform for the tetrachoric coefficient.


• compute some statistic on each of the tables, and compare those random variables (I'm not sure if this is a standard problem or not),


• something like Goodman's improvement of Stouffer's method, but I cannot access this paper, and was hoping for something a little more recent (more likely to have the latest-greatest, plus computer simulations).

any ideas?

edit: originally had proposed computing 'chi-square' statistic on each table, which is woefully underinformed. the resultant statistic is only distributed as a Chi-square under the null hypothesis of no predictive ability, which I do not want to assume. If there is any test I would like to perform here, it is of the null that the different contingency tables represent the same level of predictive ability.

more information (at request of Kjetil): a number of problems could be posed in this way, I believe. my problem:
I have a black box method of forecasting dichotomous y from a vector of input X (if you must know, it is weather-related, but I can say no more...) The method seems to work well when looking at (X,y) pairs from the time period 1970 - 1990, but maybe doesn't work as well on data from the period 1990 to present. I get two contingency tables, one for 1970-1990, the other for 1990-2010. I want to be able to detect whether the forecasting technique has gone bad.
As an extension of this problem, you can imagine that the black box was constructed by a third party, and they possibly tuned it around 1990, using all the data available until then. the problem then becomes one of in-sample versus out-of-sample predictive ability.
As an extension of this extension, pretend, for the moment, that the third party is withholding the X, and publishes only the predictions of the y, and the actual values of the y (although the latter may be verified through other data streams), thus one cannot even use the X to diagnose the overfitting problems.

another form of the problem might be as follows: given a black box method of forecasting dichotomous y from vector of predictors X, and dichotomous X_0, does the value of X_0 affect the quality of prediction? one could construct 2 contingency tables for the values of X_0, and compare them.

etc.

oc.optimization control - Solving NP problems in (usually) Polynomial time?

Hmm, it's not an NP-complete problem, but hopefully it's still relevant to (4) and to a question I think is implicit in (2).

It's well-known that linear programming is in P, but in practice the simplex algorithm (which is exponential in the worst case) is usually the fastest method to solve LP problems, and it's virtually always competitive with the polynomial-time interior point methods. However, sampling uniformly from some space of problems is an unrealistic model, so average-case analysis isn't convincing. Spielman and Tang introduced the notion of "smoothed analysis" to remedy this, and showed that the simplex algorithm has polynomial smoothed time complexity. Glancing at Spielman's page, it looks like this has been applied to the knapsack problem, although the link to the paper is broken.

Re (1): What do you mean by "small?" :) I suspect that the heuristics would fail to help much if you took a random instance of, say, 3SAT with the right number of clauses and variables, and reduced this to an instance of TSP. But you'd get some polynomial-size blowup, so...

Re (3): It's correct that the existence of such an algorithm, on its own, would imply neither P = NP nor P != NP. But for "practical considerations" it might be hugely important, depending on what the constants were, and it would certainly spur investigation into whether there was a worst-case polynomial algorithm along the same lines.

ETA: Actually, here's a construction for an NP-complete problem and an algorithm which unconditionally runs in average-case polynomial time. The problem is the union of a language in P and an NP-complete language (solvable in exp(n) time), such that the number of instances of size n of the first problem is something like exp(n^3), while the number of instances of the second problem is exp(n).

So the interesting thing about (3) is what the existence of an average-case polynomial algorithm for every problem in NP would tell us. And there the answer is still "nothing," but it's conceivable that we could prove P = NP under this assumption.

By the way, Impagliazzo talks about some of these issues (although not all of them; the paper's 15 years old and predates Spielman-Tang, for instance) in perhaps the greatest survey paper ever written. I highly recommend it.