nt.number theory - A prime sequence can be partitioned into two sets of equal or consecutive sum

Scott Carnahan had an interesting idea; let's formalize it into an actual solution. We will show that, given $n ge 2$ a positive integer, $p_1, cdots, p_n$ the first n primes, we have some $e_1, cdots, e_n$ with $e_i = pm 1$ such that $|e_1p_1 + e_2p_2 + cdots + e_np_n| le 1$. (Note that we may further stipulate that $e_n = 1$.) A simple parity argument from here suffices to prove the conjecture.

We will prove this by induction on $n$. The cases $2 le n le 6$ are trivial to verify, and were provided already by a-boy. We now fix $n ge 7$.

We first need some asymptotics in the form of the Bertrand-Chebyshev theorem; we use the formulation that for $m > 1$ there is a prime between $m$ and $2m$.

Write $S_k = e_np_n + e_{n-1}p_{n-1} + cdots + e_{n-k+1}p_{n-k+1}$, and let $M(k)$ be the minimum of $|S_k|$ over all tuples $(e_n, e_{n-1}, cdots, e_{n-k+1})$. We stipulated earlier that $e_n = 1$, so we have $M(1) = p_n$. Two facts that will be useful to us in the future are that (1) $M(k+1) le |M(k)-p_{n-k}|$ and (2) if $|a| le |b|$, then $min{{|a+b|,|a-b|}} le |b|$.

We claim that $M(k) le p_{n-k+1}$ for $k = 1, 2, cdots, n-2$. We prove this by induction on $k$. The claim for $k = 1$ is trivial. Now if $M(k) le p_{n-k}$, then we are done, as $M(k+1) le min{{|M(k)+p_{n-k}|,|M(k)-p_{n-k}|}} le p_{n-k}$ by fact (2).

Now suppose $p_{n-k} < M(k) le p_{n-k+1}$. Write $2m+1 = p_{n-k+1} ge p_3 = 5$, so that $m > 1$. In this case we know that $m < p_{n-k} < M(k) le 2m$. But then $M(k+1) le M(k) - p_{n-k} le 2m-(m+1) = (m-1) < p_{n-k}$ as desired.

The fact that $M(k) le p_{n-k+1}$ is eminently useful.

Indeed, we may use it to dispatch of the even case immediately. Set $k = n-6$. Then we have $M(n-6) le 17$. As all sums considered in $M(n-6)$ are sums of an even number of odd terms, we in fact have $M(n-6) le 16$ and even. Now we simply note that all odd numbers between -15 and 15 are realizable as sums and differences of the first 6 primes, which is left as an easy computational exercise.

In the odd case, we consider $k = n-5$. Then $M(n-5) le 13$. For the same parity reasons as above, we have in fact $M(n-5) le 12$. And again, we note that all even numbers between -12 and 12 are realizable as sums and differences of the first 5 primes - another easy computational exercise.

The limits of $n-6$ and $n-5$ are the best possible for our small-case analysis.

If we were to establish an algorithm for this, we could just do the greedy algorithm on choosing $e_n$, then $e_{n-1}$, and so on, each time choosing $e_k$ so as to minimize $S_{k+1}$ (or randomly if $S_k = 0$). Our claim that $M(k) le p_{n-k}$ will continue to be satisfied by the greedy algorithm, as the proof of the claim does not involve changing prior $e_i$. Thus our greedy-algorithm mimium modulus must satisfy the same inequality, and we continue until we are at $n-6$ or $n-5$, then finish as in our nonconstructive proof.

fa.functional analysis - On the convolution of generalized functions

If I understand correctly what you are asking then the answer is: "No".

Here's where I may be misunderstanding: I assume that $Delta t$ is fixed. If this is correct, we can argue as follows.

Let me write $r = Delta t$ since it is fixed and I want to disassociate it from $t$. We consider the operator $A_r colon C^infty_c(mathbb{R}) to C^infty_c(mathbb{R})$ defined by

$$A_r(phi)(t) = int_{t - r}^{t + r} phi(tau) d tau$$

We want to extend this function to the space of distributions, $mathcal{D} = C^infty_c(mathbb{R})$. To do this, we look for an adjoint as per the nlab page on distributions (particularly the section operations on distributions; note that my notation is chosen to agree with that page so it's hopefully easy to compare). So for two test functions, $phi, psi in C^infty_c(mathbb{R})$ we calculate as follows:

$$begin{array}{rl} langle psi, A_r(phi)rangle &= int_{mathbb{R}} psi(t) A_r(phi)(t) d t \ &= int_{mathbb{R}} psi(t) int_{t - r}^{t + r} phi(tau) d tau d t \ &= int_{mathbb{R}} int_{t - r}^{t + r} psi(t) phi(tau) d tau d t \ &= int_{mathbb{R}} int_{tau - r}^{tau + r} psi(t) phi(tau) d t d tau \ &= int_{mathbb{R}} A_r(psi)(tau) phi(tau) d tau \ &= langle A_r(psi), phi rangle end{array}$$

When we do the switch in order of integration, it's useful to draw the region of integration in the plane (I used a table "cloth" in a restaurant here in Copacabana!). It's a diagonal swathe and looks the same when flipped about the line $x = y$ (don't do this with a table cloth unless you've taken the plates off it first.).

So $A_r$ is self-adjoint and we extend it to distributions by the formula $langle A_r(T), phi rangle = langle T, A_r(phi)rangle$.

Now we come to your question. You have $(T_lambda)$ such that $langle T_lambda, A_r(phi)rangle to langle T, A_r(phi) rangle$ for each test function $phi$. Using the definition of the extension of $A_r$, this is the same as saying that $(A_r(T_lambda)) to A_r(T)$ weakly. You want to know if this implies that $(T_lambda) to T$.

To answer this, we consider $A_r$. Unfortunately for you, it has a non-trivial kernel. Note that we are now working with distributions, so can allow things with non-compact support (otherwise it wouldn't have a non-trivial kernel). For $A_r(S) = 0$ we simply require that $S$ integrate to $0$ over every interval of length $2 r$. If we pick $S$ an integrable function with period $2 r$ then this is quite easy to arrange: almost all such functions integrate to $0$ over such intervals. In particular, $S = sin(pi t/r)$ will do.

But now we can add random amounts of $S$ to each $T_lambda$ thus ensuring that $(T_lambda)$ does not converge (okay, the amounts are not quite random: we choose them such that the resulting sequence does not converge, but this is always possible). However, since $A_r$ does not "see" $S$, $(A_r(T_lambda + alpha_lambda S))$ still converges.

As I said, I'm not convinced that I understood the question correctly so please do comment if this doesn't look right. But if it doesn't look right, please edit your question so that it's clearer that this isn't what you wanted.

galaxy - Why are stars more metallic closer as you move closer to the galactic bulge?

It has to do with the formation of the Milky Way.

At the beginning, the Milky Way was much more spherical than it is now - perhaps closer to what an elliptical galaxy is like than a spiral galaxy. Population III stars would have formed first, then quickly died out. Next came Population II stars. They formed when the galaxy was still somewhat spherical, and so they tend to inhabit the galactic spheroid/halo.

Eventually, the rotation of the Milky Way flattened out much of the remaining gas and dust, and some of the stars. When younger stars formed, they formed in the flatter disk, nearer to the center. The disk itself became smaller that the spheroid/halo. Thus, the younger Population I stars are found in the galactic disk, and are closer in.

No more stars will form in globular clusters; they are relatively dust-free and contain old, Population II stars. The same is true for the galactic halo.

_{Source: Populations & Components of the Milky Way}

The galactic bulge itself contains several populations of stars. Some may have come from the halo and thick disk (thus being metal-poor) while others may have formed together more recently from the thin disk itself (thus being metal-rich).

gr.group theory - Is an invertible biset necessarily a bitorsor?

Question

Let $G$ be a group, and let $X$ be a $G$-biset that is (weakly) invertible with respect to the contracted product. Is $X$ necessarily a bitorsor?

Background

By $G$-biset, I mean a set equipped with commuting left and right $G$-actions. There is a standard tensor product on the category of $G$-bisets called the contracted product; it is defined by $X times_G Y = X times Y / (x cdot g, y) sim (x, g cdot y)$, where $G$ acts on the left by its left action on $X$, and on the right by its right action on $Y$. The unit object is the group $G$, where $G$ acts by left and right multiplication on itself.

A left $G$-torsor is a left $G$-set $X$ such that the map $G times X to X times X$, $(g, x) mapsto (g cdot x, x)$ is a bijection. A right $G$-torsor is defined analogously. A $G$-bitorsor is a $G$-biset that is both a left and a right $G$-torsor. A $G$-bitorsor $X$ is necessarily invertible with respect to the contracted product; its inverse is the opposite $G$-bitorsor $X^{operatorname{op}}$. This bitorsor has the same objects as $X$, but $g in G$ acts on the left (resp. the right) by the right (resp. left) action of $g^{-1}$.

It follows from a simple counting argument that when $G$ is a finite group, any invertible $G$-biset is a $G$-bitorsor. Is this true for arbitrary groups (and more generally, in an arbitrary topos)? What about if we replace "invertible" by "right- (or left-) invertible"?

I can show, at least in the punctual topos (and I think it's true in general), that if $X^{operatorname{op}}$ is an inverse to $X$, then $X$ must be a $G$-bitorsor. So the question is whether a $G$-biset can have an inverse not of this form.

The reason I'm interested in this question is that I want to understand how to generalize bitorsors to higher categorical settings. A possible generalization would be an invertible profunctor, but this is only a good definition if the answer to my question is affirmative.

ag.algebraic geometry - Measure on real Grassmannians

First of all, a general fact. Any transitive homogeneous space $X$ of a
compact group $K$ has a unique $K$-invariant probability measure. Existence:
take the image $nu$ of the normalized Haar measure $m$ on $K$ under the map
$gmapsto gx_0$ (that's the construction you refer to). The measure $nu$ is
well-defined, since the mass of $m$ is finite, it does not depend on $x_0in X$ by right invariance of $m$, and is $K$-invariant by left invariance of $m$.
Uniqueness: take an arbitrary $K$-invariant measure $nu'$ on $X$, and
consider its convolution $mastnu'$ with the measure $m$ (i.e., the image of
the product of $m$ and $nu'$ under the map $(g,x)mapsto gx$). Then, on one
hand $mastnu'=nu'$ by $K$-invariance of $nu'$, on the other hand
$mastnu'=nu$ by the above construction.

Thus, since the Grassmannian in question has a transitive compact group of
automorphisms, it carries a "natural" invariant measure. So that
"platonically" it is always there - like, for instance, the Riemannian volume
on a Riemannian manifold (by the way, as mentioned before, any invariant
Riemannian metric on the Grassmannian produces the measure in question in this
way).

However, there is one subtlety here which has so far remained unnoticed. In
order to define the Grassmannian one needs a linear structure, whereas the
orthogonal group $O(n)$ is defined in terms of the Euclidean structure.
Therefore, the "canonical" measure we are talking about is only canonical with
respect to the given Euclidean structure on the linear space $V=R^n$. So, if
we look at the problem from this point of view, we obtain a map which assigns
to any Euclidean structure on $V$ a probability measure on the Grassmannian
$Gr_k(V)$. In fact, this measure depends only on the projective class of the
Euclidean structure (i.e., on the corresponding similarity structure), which
are parameterized by the Riemannian symmetric space $S=SL(n,R)/SO(n)$
(equivalently, one can say that we consider only the Euclidean structures on
$V$ with the same volume form, whence $SL$ instead of $GL$).

Thus, we have a map $xmapstonu_x$ from $S$ to the space $P(Gr_k)$ of
probability measures on $Gr_k$. One can show that this map is an injection, so
that it can be used in order to compactify the symmetric space $S$ by taking
its closure in the weak$^*$ topology of $P(Gr_k)$. This is an example of a
so-called Satake-Furstenberg compactification, which can be defined for an
arbitrary non-compact Riemannian symmetric space. In the case of the space
$S=SL(n,R)/SO(n)$ all such compactifications are obtained by considering
rotation invariant measures on the flag space of $V$ and its equivariant
quotients (in particular, Grassmannians). In the general case the role of the
flag space is played by the so-called Furstenberg boundary, which is the
quotient of the semi-simple Lie group by its minimal parabolic subgroup. The
most recent reference for all this is the book by Borel and Ji.

The simplest non-compact symmetric space is the hyperbolic plane. In this case
the Furstenberg boundary (the associated "flag space") is just the boundary
circle in the disk model. Each point of the hyperbolic plane determines a
unique probability measure on the boundary circle invariant with respect to
the rotations around this point. These measures appear in the classical
Poisson formula for bounded harmonic functions in the unit disk (usually it is
written in terms of just a single measure corresponding to the Euclidean
center of the disk; the other measures appear in the guise of their
Radon-Nikodym derivatives with respect to this one, which is precisely the
Poisson kernel).

Can light projection on a wall travel faster than the speed of light?

Yes, projection can move faster than light. Assuming a circular wall, and projector at the center, the maximum limit will be 2 times c.

However, there is nothing moving faster than c. Different photons are moving from projector to the wall at c. Just because you rotate the projector quickly, the projection seem to move from one side to other, but actually photons are moving from projector to the wall at different angles, at different times.

Assuming the projector can be rotated 180 degrees in no time, the projection can move from one side to the other in half the time as compared to the time taken by actual light to move from one side to the other. This is assuming initial and final projections are equidistant from the projector.

You can say, if the projector is rotated in no time, then the projection is not actually moving. No time means non-zero, very small time as compared to the time taken by light from projector to the wall.

If you make the projection move faster than 2c (by making the wall not circular, or by making final point much closer to the projector as compared to the initial point, then it will not be possible to see the projection moving continuously. In that case, you will see the final projection before the moving projection completes its journey along the wall.

differential equations - Does there exist a potential which realizes this strange quantum mechanical system?

I have done some courses on quantum mechanics and statistical mechanics in the past. Since I also do math, I wonder about converge issues which are usually not such a problem in physics. One of those questions is the following. I will describe the background, but in the end it boils down to a question about ordinary differential equations.

In quantum mechanics on the real line, we start with a potential $V: mathbb{R} to mathbb{R}$ and try to solve the Schrödinger question $ihbar frac{partial}{partial t}Psi(x,t) = - frac{hbar^2}{2m}frac{partial^2}{partial x^2}Psi(x,t)+V(x)Psi(x,t)$. In many cases this can be accomplished by seperating variables, in which case we obtain the equation $EPsi(x,t) = - frac{hbar^2}{2m}frac{partial^2}{partial x^2}Psi(x,t)+V(x)Psi(x,t)$ which we try to solve for $E$ and $Psi$ to obtain a basis for our space of states together with an associated energy spectrum. For example, if we have a harmonic oscillator, $V(x) = frac{1}{2}momega^2x^2$ and we get $E_n = hbar omega (n+frac{1}{2})$ and $Psi_n$ a certain product of exponentials and Hermite polynomials. We assume that the energy in normalized such that the lowest energy state has energy $0$.

If the states of our system are non-degenerate, i.e. there is only one state for each energy level in the spectrum, then the partition function in statistical mechanics for this system is given by the sum $Z(beta) = sum_n exp(-beta E_n)$, where $beta$ is the inverse temperature $frac{1}{k_B T}$. It is clear that this sum can be divergent; in fact for a free particle ($V = 0$), it is not even well defined since spectrum is a continuum.

However, I was wondering about the following question: Is there a system such that $Z(beta)$ diverges for $beta < alpha$ and converges for $beta > alpha$ for some $alpha in mathbb{R}_{> 0}$? Am I correct in thinking that such a system is most likely an approximation of another system, which undergoes a phase transition at $beta = alpha$?

Anyway, an obvious candidate would be a potential $V$ such that the spectrum is given $E_n = C log(n+1)$ for $n geq 0$ and $C > 0$. This gets me to my main mathematical question: Does such a potential (or one with spectrum asymptotically similar) exist? If so, can you give it explicitly?

One the circle, the theory of Sturm-Liouville equations tells us that the eigenvalues must go asymptotically as $C n^2$, so in this case such problems can't occur. I don't know much about spectral theory for Sturm-Liouville equations on the real line though. The second question is therefore: What is known about the asymptotics of the spectrum of a Sturm-Liouville operator on the real line?

ag.algebraic geometry - Divisors on Proj(UFD)

Well, if you read on to Chapter 2, exercise 6.3, then it is stated that:

$$Cl(A) cong Cl(X)/mathbb Z[H]$$
here $[H]$ represents the hyperplane section. So the answer is yes.

There is a less well-known but very nice generalization. Suppose that $X$ is smooth. Let $R=A_m$ be the local ring of A at the irrelevant ideal. Then one has a (graded) isomorphism of $mathbb Q$- vector spaces:

$$CH(X)_{mathbb Q}/[H]CH(X)_{mathbb Q} cong A_*(R)_{mathbb Q}$$

Here $CH(X)$ is the Chow ring of $X$ and $A_*(R)$ is the total Chow group of $R$. Details can be found in this paper by Kurano.

reference request - Algorithms for modeling asynchronicity in Asynchronous Cellular Automata

Most cellular automata are defined as being updated synchronously. I am interested in asynchronous automata, where they do not all have to update simultaneously. I am restricting myself to cellular automata on a graph (e.g. lattice) where the cellular automata is a FSM and all of the automata on the graph are identical.

I have seen some representations of asynchronously updated networks as synchronous updating of the automata with probabilistic updating of the cells, i.e. at each time step, each cell has probability $p$ of possibly updating its state.

I have seen asynchronous models where there is a single ordered list of the individual cells with each cell firing one after the other, with the same ordering maintained over multiple cycles, or with a different ordering being generated each time after all of the cells have fired. In this scheme, a cell is guaranteed to fire at most $3$ times in $2*n$ time-steps if there are $n$ total cells or at most $x+1$ times in $x*n$ time-steps ($x$ cycles of $n$ timesteps). (Example, it fires at time $n$ in the first cycle, at any time $n+s, (0 le s le n)$ in the second cycle, and at time $2n+1$ in the third cycle, meaning it fires 3 times during the $n+2$ steps from $t=n$ to $t=2n+1$.

I have also seen sequential firing, where at each time step a single cell is chosen to be updated, with no restrictions on firing all of the cells before starting over. This schema also averages a cell firing once every $n$ time-steps, but does not restrict it from firing more frequently.

Are there other better ways to mathematically model asynchronous cellular automata? What are the pitfalls and benefits of these particular schemes? I also agree at the outset that the type of firing scheme to be used for an asynchronous system depends on the particular system being modeled; I am asking for general answers or references for algorithms to model asynchronicity.

ag.algebraic geometry - fpqc covers of stacks

It is false. I'm not sure what the comment about algebraic spaces has to do with the question, since algebraic spaces do admit an fpqc (even 'etale) cover by a scheme. This is analogous to the fact that the failure of smoothness for automorphism schemes of geometric points is not an obstruction to being an Artin stack. For example, $Bmu_n$ is an Artin stack over $mathbf{Z}$ even though $mu_n$ is not smooth over $mathbf{Z}$ when $n > 1$. Undeterred by this, we'll make a counterexample using $BG$ for an affine group scheme that is fpqc but not fppf.

First, we set up the framework for the counterexample in some generality before we make a specific counterexample. Let $S$ be a scheme and $G rightarrow S$ a $S$-group whose structural morphism is affine. (For example, if $S = {rm{Spec}}(k)$ for a field $k$ then $G$ is just an affine $k$-group scheme.) If $X$ is any $G$-torsor for the fpqc topology over an $S$-scheme $T$ then the structural map $X rightarrow T$ is affine (since it becomes so over a cover of $T$ that splits the torsor). Hence, the fibered category $BG$ of $G$-torsors for the fpqc topology (on the category of schemes over $S$) satisfies effective descent for the fpqc topology, due to the affineness requirement.

The diagonal $BG rightarrow BG times_S BG$ is represented by affine morphisms since for any pair of $G$-torsors $X$ and $Y$ (for the fpqc topology) over an $S$-scheme $T$, the functor ${rm{Isom}}(X,Y)$ on $T$-schemes is represented by a scheme affine over $T$. Indeed, this functor is visibly an fpqc sheaf, so to check the claim we can work locally and thereby reduced to the case $X = Y = G_T$ which is clear.

Now impose the assumption (not yet used above) that $G rightarrow S$ is fpqc. In this case I claim that the map $S rightarrow BG$ corresponding to the trivial torsor is an fpqc cover. For any $S$-scheme $T$ and $G$-torsor $X$ over $T$ for the fpqc topology, the functor

$$S times_{BG} T = {rm{Isom}}(G_T,X)$$
on $T$-schemes is not only represented by a scheme affine over $T$ (namely, $X$) but actually one that is an fpqc cover of $T$. Indeed, to check this we can work locally over $T$, so passing to a cover that splits the torsor reduces us to the case of the trivial $G$-torsor over the base (still denoted $T$), for which the representing object is $G_T$.

So far so good: such examples satisfy all of the hypotheses, and we just have to prove in some example that it violates the conclusion, which is to say that it does not admit a smooth cover by a scheme. Take $S = {rm{Spec}}(k)$ for a field $k$, and let $k_s/k$ be a separable closure and $Gamma = {rm{Gal}}(k_s/k)^{rm{opp}}$. (The "opposite" is due to my implicit convention to use left torsors on the geometric side.) Let $G$ be the affine $k$-group that "corresponds" to the profinite group $Gamma$ (i.e., it is the inverse limit of the finite constant $k$-groups $Gamma/N$ for open normal $N$ in $Gamma$). To get a handle on $G$-torsors, the key point is to give a more concrete description of the ``points'' of $BG$.

Claim: If $A$ is a $k$-algebra and $B$ is an $A$-algebra, then to give a $G$-torsor structure to ${rm{Spec}}(B)$ over ${rm{Spec}}(A)$ is the same as to give a right $Gamma$-action on the $A$-algebra $B$ that is continuous for the discrete topology such that for each open normal subgroup $N subseteq Gamma$ the $A$-subalgebra $B^N$ is a right $Gamma/N$-torsor (for the fpqc topology, and then equivalently the 'etale topology).

Proof: Descent theory and a calculation for the trivial torsor. QED Claim

Example: $A = k$, $B = k_s$, and the usual (right) action by $Gamma$.

Corollary: If $A$ is a strictly henselian local ring then every $G$-torsor over $A$ for the fpqc topology is trivial.

Proof: Let ${rm{Spec}}(B)$ be such a torsor. By the Claim, for each open normal subgroup $N$ in $Gamma$, $B^N$ is a $Gamma/N$-torsor over $A$. Since $A$ is strictly henselian, this latter torsor is trivial for each $N$. That is, there is a $Gamma/N$-invariant section $B^N rightarrow A$. The non-empty set of these is finite for each $N$, so by set theory nonsense with inverse limits of finite sets (ultimately not so fancy if we take $k$ for which there are only countably many open subgroups of $Gamma$) we get a $Gamma$-invariant section $B rightarrow A$. QED Corollary

Now suppose there is a smooth cover $Y rightarrow BG$ by a scheme. In particular, $Y$ is non-empty, so we may choose an open affine $U$ in $Y$. I claim that $U rightarrow BG$ is also surjective. To see this, pick any $y in U$ and consider the resulting composite map

$${rm{Spec}} mathcal{O}_{Y,y}^{rm{sh}} rightarrow BG$$
over $k$. By the Corollary, this corresponds to a trivial $G$-torsor, so it factors through the canonical map ${rm{Spec}}(k) rightarrow BG$ corresponding to the trivial $G$-torsor. This latter map is surjective, so the assertion follows. Hence, we may replace $Y$ with $U$ to arrange that $Y$ is affine. (All we just showed is that $BG$ is a quasi-compact Artin stack, if it is an Artin stack at all, hardly a surprise in view of the (fpqc!) cover by ${rm{Spec}}(k)$.)

OK, so with a smooth cover $Y rightarrow BG$ by an affine scheme, the fiber product

$$Y' = {rm{Spec}}(k) times_{BG} Y$$
(using the canonical covering map for the first factor) is an affine scheme since we saw that $BG$ has affine diagonal. Let $A$ and $B$ be the respective coordinate rings of $Y$ and $Y'$, so by the Claim there is a natural $Gamma$-action on $B$ over $A$ such that the $A$-subalgebras $B^N$ for open normal subgroups $N subseteq Gamma$ exhaust $B$ and each $B^N$ is a $Gamma/N$-torsor over $A$. But $Y' rightarrow {rm{Spec}}(k)$ is smooth, and in particular locally of finite type, so $B$ is finitely generated as a $k$-algebra. Since the $B^N$'s are $k$-subalgebras of $B$ which exhaust it, we conclude that $B = B^N$ for sufficiently small $N$. This forces such $N$ to equal $Gamma$, which is to say that $Gamma$ is finite.

Thus, any $k$ with infinite Galois group does the job. (In other words, if $k$ is neither separably closed nor real closed, then $BG$ is a counterexample.)

gravity - Can dark matter be found in the shape of planets, galaxies etc.?

Planets and stars, no. Globular clusters and galaxies, yes.

Small scales

To condense into such relatively compact objects as planets, stars, and even the more diffuse star-forming clouds, particles need to be able to dissipate their energy. If they don't do this, their velocities prohibit them from forming anything.

"Normal" particles, i.e. atoms, do this by colliding. When atoms collide, they're excited, and when they de-excite, they emit radiation which leaves the system, carrying away energy. In this way, an ensemble of particles can relax into a less energetic system, eventually condensing into e.g. a star. Additionally, the collisions cause more energetic particles to donate energy to the less energetic ones, making the ensemble reach thermodynamic equilibrium, i.e. all particles have the same energy on average.

Dark matter is, by definition, unable to collide and radiate, and hence, on such small scales as stars and planets, particles that enters a potential well with a given energy will maintain that energy. They will thus accelerate toward the center, then decelerate after its closest approach to the center, and finally leave the system with the same energy as before (if it was unbound to begin with). This makes it impossible for collisionless matter to form such small objects.

Large scales

On the scale of galaxies, however, various relaxation mechanisms allows dark matter to form structure. This is the reason that in pure N-body simulations of the Universe, such as the Millennium Simulation, you will see galaxies. The sizes of these structures depend on the resolution, but are measured in millions of Solar masses.

The relaxation mechanisms include:

Phase mixing

This is sort of like galaxy arms winding up, but in phase space rather than real space.

Chaotic mixing

This happens when particles come so close that their trajectories diverge exponentially.

Violent relaxation

The two mechanisms listed above assume a constant gravitational potential $Phi$, but as the systems relaxes, $Phi$ changes, giving rise to an additional relaxation process. For instance, more massive particles tend to transfer more energy to their lighter neighbors and so become more tightly bound, sinking towards the center of the gravitational potential. This effect is known as mass segregation and is particularly important in the evolution of globular star clusters.

Landau damping

For a perturbation/wave with velocity $v_p$, if a particle comes with $vgg v_p$, it will overtake the wave, first gaining energy as it falls into the potential, but later losing the same amount of energy as it climb up again. The same holds for particles with $vll v_p$ which are overtaken by the wave.
However, particles with $vsim v_p$ (i.e. that are near resonance with the wave) may experience a net gain or loss in energy.
Consider a particle with $v$ slightly larger than $v_p$. Depending on its phase when interacting with the wave, it will be either accelerated and move away from resonance, or decelerated and move closer to resonance. The latter interact more effectively with the wave (i.e. be decelerated for a longer time), and on average there will thus be a net transfer of energy from particles with $v gtrsim v_p$ to the wave. The opposite is true for particles with $v$ slightly smaller than $v_p$

You can read more about these mechanisms in Mo, Bosch, & White's Galaxy Formation and Evolution.

ag.algebraic geometry - is there a push-forward of closed subschemes?

It sounds as though what you want is the closure of the image of $F$ under $f$. (That is, the minimal closed subscheme of $Y$ factoring $f$.)

If $X =$ Spec $A$, and $Y =$ Spec $B$, and $F =$ Spec $A/I$, and $f$ corresponds to the ring map $f':Bto A$, then we can consider the preimage $J$ of $I$ under $i'$. Consider the set of primes in $B$ containing $J$. Of course, any prime in the image of $F$ under $f$ must contain $J$, since its preimage under the ring map has to contain $I$. Thus, Spec $B/J$ contains the preimage. You can check that it's the biggest such ideal, noting that in order for us to have a map $B/J to A/I$, $J$ should be contained in the preimage of $I$.

Being the closure of the image of $F$ under $f$ is a universal property of sorts (in particular, it's unique), which more or less allows us to argue that this construction generalizes to non-affine schemes. (Just apply this local construction, and uniqueness tells us that the local constructions glue together.)

mg.metric geometry - Is there any algorithm for determining 3d position in such case?

What you see in the second image is a projection of the first, after a rotation. So treat it exactly like that. Meaning, you have 4 lines in $mathbb{R}^3$. You know their equations. Furthermore, you have another 4 lines in $mathbb{R}^2$ corresponding to the second image.

To solve, parameterize the family of possible 4 lines in $mathbb{R}^3$ that project onto the 4 lines in $mathbb{R}^2$. Find the matrix that takes the original 4 lines to generic quadruple in the mentioned family. Now write equations to ensure that matrix is actually a rotation. You should have enough information for there to be at most one quadruple that can actually be gotten from the original lines. If not, then your question has multiple answers.

mg.metric geometry - Can a discrete set of the plane of uniform density intersect all large triangles?

This question is infuriating. I think I've made some progress, and would like to hear other's thoughts. Therefore, I am making this post a community wiki:

---

First of all, any triangle of area A contains a rectangle of area A/2. Proof: let the triangle be ABC, with AC the longest side. Let P and Q be the midpoints of AB and BC, and let R and S be the feet of perpendiculars from P and Q to AC. Then PQSR is a rectangle of the required area. Conversely, a rectangle of area A contains a triangle of area A/2. So we may instead ask whether there are large rectangles in the complement.

This is convenient because specifying a triangle involves 6 parameters, while specifying a rectangle has only 5. So this cuts our search space down a dimension. I find the most conveninent parameters for a rectangle to be the length of the longer side, L, the area, A, the angle of the longer side, $theta$, and one of the vertices $(x,y)$. I'll call the type of the rectangle $(L, A, theta)$, forgetting the translation parameters.

---

I now have a conjectural solution, although I have no idea how to prove that it works. I call this the sunflower configuration, because it was inspired by pattern of florets at the center of a sunflower.

Let $tau$ be the Golden ratio. Consider the sequence of points $(sqrt{k} cos(2 pi tau k), sqrt{k} sin(2 pi tau k))$. A circle of radius $R$ around $0$ contains $R^2$ points, and has area $pi R^2$, so the density is right.

Why do I think this is reasonable? There is a gorgeous property of the Golden ratio: if you look at the sequence $k tau, (k+1) tau, (k+2) tau, ..., ell tau$ in $mathbb R/mathbb Z$, then the largest gap between any two consecutive angles is $dfractau{ell-k}$. So multiples of $2 pi tau$ are very well distributed around the unit circle. I learned this from Volume 2 of the Art of Computer Programming; I'll try to find a reference later if no one else does.

I don't have a strategy yet for proving that the sunflower pattern doesn't contain large triangles (or rectangles). But, where ever I try to place one, heuristics indicate that this equidistribution of angles destroys me.

---

My first strategy, aiming to prove a "no", was to overlay several lattices which were all rotations of each other. Let's fix $A$ once and for all, our goal is to exclude a rectangle of size $A$. I first set out to see when a lattice could contain a rectangle of type $(L, A, theta)$.

Translate the rectangle so that one of the short sides touches $(0,0)$. Then the rectangle contains a circular wedge of radius $L$ and angle approximately $A/L^2$ with no lattice points. In other words, there is no $(p,q)$ with $sqrt{p^2+q^2} le L$ and $|theta - tan^{-1}(p/q)| le ccdot A/L^2$, where $c$ is a constant I have not computed.

There are now a bunch of nuisances, having to do with the presence of that $tan^{-1}$ and the fact that people who do Diophatine approximation usually ask for $q$ to be small, not $sqrt{p^2+q^2}$. Passing over all of the details, the set of $theta$'s for which such a rectangle exists should look something like the union of all intervals in the $L$-th Farey sequence whose length is greater than $A/L^2$. Heuristics give me that the size of this is $c/A$ for $c$ a (different) constant that I haven't computed.

So, one lattice can't save me. The above suggests that, for every $L$, there will be a positive length set of $theta$'s for which rectangles of type $(L, A, theta)$ exist. Of course, we already knew that a single lattice couldn't work.

What about two lattices, rotated by some phi? I think I still lose. As $L$ grows, the set of theta's which work stays of size $1/A$, but becomes more and more spread out. Eventually, I would expect that it would contain two points that differ by phi. This part is very nonrigorous, though. In particular, I'm not sure whether we might be able to save things for some very special phi.

That's about how far I've gotten. I tried some other ideas, but didn't get anywhere. I'm not even sure which answer I think is right.

dg.differential geometry - Is there a good (co)homology theory for manifolds with corners?

Recall that a (smooth) manifold with corners is a Hausdroff space that can be covered by open sets homeomorphic to $mathbb R^{n-m} times mathbb R_{geq 0}^m$ for some (fixed) $n$ (but $m$ can vary), and such that all transition maps extend to smooth maps on open neighborhoods of $mathbb R^n$.

I feel like I know what a "differential form" on a manifold with corners should be. Namely, near a corner $mathbb R^{n-m} times mathbb R_{geq 0}^m$, a differential form should extend to some open neighborhood $mathbb R^{n-m} times mathbb R_{> -epsilon}^m$. So we can set up the usual words like "closed" and "exact", but then Stokes' theorem is a little weird: for example, the integral of an exact $n$-form over the whole manifold need not vanish.

In any case, I read in D. Thurston, "Integral Expressions for the Vassiliev Knot Invariants", 1999, that "there is not yet any sensible homology theory with general manifolds with corners". So, what are all the ways naive attempts go wrong, and have they been fixed in the last decade?

As always, please retag as you see fit.

Edit: It's been pointed out in the comments that (1) I'm not really asking about general (co)homology, as much as about the theory of De Rham differential forms on manifolds with corners, and (2) there is already a question about that. Really I was just reading the D. Thurston paper, and was surprised by his comment, and thought I'd ask about it. But, anyway, since there is another question, I'm closing this one as exact duplicate. I'll re-open if you feel like you have a good answer, though. -Theo Edit 2: Or rather, apparently OP can't just unilaterally close their own question?

How do you prove that a field is isomorphic to C(x)?

I think that your question is formulated to broad to give a precise answer.

I assume that K is given in the form F(x₁,...,x_k)/(f₁,..,f_m). If your equations f_1...f_m are nice enough (low degre etc.) then several computer algebra packages can tell you transcendence degree of K/F using e.g. Groebner Bases. If this tr. degree is different from one then you are done, otherwise K is the function field of a smooth projective curve C/F.

Now the function field of C/F is isomorphic to F(x) if and only if C is isomorphic to ℙ¹ .

If F is algebraically closed then it suffices to show that C has genus 0. If char(K)=0 this is can be done relatively easy: following the proof of the lemma of the primitive element you can write K=F(x,y)/f. One can consider y as a function on C and this yields a morphism g:C→ ℙ¹. The Riemann-Hurwitz formula (RH) gives you an easy recipe to calculate the genus of C.
(Most of the details are explained in Fulton's book on algebraic curves, if only care about RH you might also read the first two chapters of Silverman's book on Arithmetic of elliptic curves.)
If F is alg. closed, but char(K)>0 this recipe also works, except that the calculation of the entries in the RH formula is harder.
So if F is alg closed then a computer can do the job.

If F is not alg. closed you have a harder problem. You need to use the following criterion
a smooth projective curve C is isomorphic to ℙ¹
if and only if
the genus of C is zero and C has a point with coordinates in F.

So besides the genus calculation you need to show that the curve C has at least one point with coordinates in F. This is in general a hard problem, if F is finite this is known and if F is a number field there is a good criterion to check whether C has a point with coordinates or not, for arbitrary F this seems hard.

Examples ℂ[x,y]/(x^2+y^2+1) is isomorphic to ℂ[z], but ℝ[x,y]/(x^2+y^2+1) is not isomorphic to ℝ[z], since the conic x^2+y^2+1 has non ℝ-points.

ct.category theory - Graph properties, categorically defined

There are some subtle point concerning the definition of this category. It has different properties if you define "graph" in a different way.

If you allow your graphs to have loops, then the category of all graph becomes a topological category over SET. In particular all (small) limits and colimits exist in this category. Sometimes it is best to consider only the graphs with loops at all vertices (if you just don't draw them this subcategory can be identified with the category of all simple graphs). The product in this category is the tensor product of graphs. A connected object (in a topological category an object is connected if all morphisms into discrete objects are constant) in this category is exactly a connected graph.

The subcategory of graphs without loops is not that well behaved from the point of view of category theory, But graphs without loops have their merits too: A graph morphism from $(V,E)$ to the complete graph without loops on $n$ vertices is the same thing as a $n$-vertex-coloring of $(V,E)$.

observation - Could Venus or Mercury have a moon that we haven't detected?

It's unlikely that either Mercury or Venus could have moons to begin with. Both of these planets are pretty close to the Sun — and in general, this prevents moons from finding stable orbits.

If a moon were too close to the planets, it would fall within the Roche limit and be torn apart by tidal forces. If a moon were too far from the planets, it would fall outside the Hill sphere and be pulled into the Sun.

The zones where moons around these planets could be stable over billions of years is probably so narrow that no body was ever captured into orbit when the planets were first being accreted.

Now, it gets even more complicated in Mercury's case. Its Hill sphere extends about 3000 miles (~4828 km) — that is, any satellite more than 3000 miles from the planet will be pulled away by the Sun's gravity.

However, its Roche limit extends to 3600 miles (~5794 km), so any satellite within this would be ripped apart. Venus lacks this excuse, but the region between where the Sun would pull away a satellite and the Roche limit is probably too small for a satellite to happen to have formed there.

Next, you have to realize that inner planets generally don't have moons. Solar wind pressure usually sends dust and debris flying away from the planets, so they don't have any spare material for moons.

Remember that Earth's Moon is thought to have formed from a collision with the Mars-sized planet Theia. This makes Earth an oddity in itself. Meanwhile, Mars is thought to have captured Phobos and Deimos from the asteroid belt — so Mars and its moons didn't condense from the same dust mass.

With all this, it'd be extremely surprising if we found moons on either Mercury or Venus.

ag.algebraic geometry - Real solutions to underdetermined system of polynomial equations

If you're dealing with finding solutions to a specific system, the traditional method is cylindrical algebraic decomposition, described for instance in Algorithms in Real Algebraic Geometry
by Saugata Basu, Richard Pollack, Marie-Françoise Roy downloadable HERE. This method is widely implemented in many computer algebra packages. An alternative method is the critical point method, which was implemented in Maple by Mohab Safey El Din (LINK). (Both methods eventually use Groebner bases b the way.)

This might let you experiment, but I doubt that these methods alone can give you all you need to know about a family of systems. Solving over the reals is harder than over the complexes (because the interplay between algebra and geometry is less straightforward). There is active research on the algorithmic aspects of the real Nullstellensatz, (e.g. M. Coste, H. Lombardi, M.-F. Roy. Dynamical method in algebra: efective Nullstellensatze, Annals of Pure and Applied Logic, 111, 203-256 (2001)), but the complexity is even worse than Groebner bases.

nt.number theory - Asymptotic density of k-almost primes

As not necessarily proven results were asked for, I have found the following quite accurate:

$$N_k(x):= mid{nleq x : Omega(n)=k}mid sim Rebigg(frac{2^{1-k}alpha e^{1+e}xlog(1+e+log(2^{1-k}alpha x))^{beta}}{beta!(1+e+log(2^{1+e}alpha x)}bigg)$$
for $1 leq kleq lfloor log_2 (x) rfloor$, where $log_2$ is $log$ base $2$, $gamma$ is Euler's constant,
$beta=1+e+ log alpha +(1+e+ log alpha) ^{1/pi}$, and $$alpha=frac{1}{2} rm{erfc}bigg(-frac{k-(2e^{gamma}+frac{1}{4})}{(2e^{gamma}+frac{1}{4})sqrt{2} }bigg)-2rm{T}bigg(bigg(frac{k}{(2e^{gamma}+frac{1}{4})}-1bigg),e^{gamma}-frac{1}{4}bigg)\ $$
where $rm{erfc}$ is the complementary error function and $rm{T}$ is the Owen T-function.

In integral form,

$$alpha= frac{1}{pi}int_{(-3+8e^gamma)/(sqrt{2}(1+8e^gamma))}^infty e^{-t^2}rm{d} t +int_0^{1/4 - e^gamma}frac{e^{-(3 - 8e^gamma)^2(1+t^2)/(2(1+8e^gamma)^2)}}{1+t^2}rm{d} t.$$

As $krightarrow infty$, $alpharightarrow 1$, so

$$lim_{k rightarrow infty}N_{k}(x cdot 2^{k-1})simfrac { {e^{e+1}} xloglog( {e^{e+1}} x)^{beta}}{log( {e^{e+1}} x)beta!},$$
where $beta=log(e^{e+1})+log(e^{e+1})^{1/pi}.$

For $kleqslant 3$, improvements to the above can certainly be made, but as $krightarrow infty$ (or more correctly, as $krightarrow lfloor log_2 (x) rfloor$), the formulae above, as far as have been tested, seem to be fairly accurate.

For convenience, I include the following Mathematica code:

cdf[k_, x_] :=
Re[N[
(2^-k E^(1 + E) x Log[1 + E + Log[2^-k x (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])]]^(1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] +4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])] + (1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])])^(1/[Pi])) (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma]))/((1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])] + (1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])])^(1/[Pi]))!
(1 + E + Log[2^-k x (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2] (1 + 8 E^EulerGamma))] +
4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma), 1/4 - E^EulerGamma])]))]];

landau[k_, x_] := N[(x Log[Log[x]]^(-1 + k))/((-1 + k)! Log[x])];

actual[k_, x_] := N[Sum[1, ##] & @@ Transpose[{#, Prepend[Most[#], 1], PrimePi@
Prepend[ Prime[First[#]]^(1 - k) Rest@FoldList[Times, x, Prime@First[#]/Prime@Most[#]],
x^(1/k)]}] &@Table[Unique[], {k}]];

I warmly welcome any criticism or comments on the above, and apologise in advance if I have made any serious errors.

Note: Table code included as requested:

a = 7;
x = 10^a;
kk = 20;
TableForm[Transpose[{Table[x, {x, 1, kk}], Table[Round[landau[k, x]], {k, 1, kk}], 
Table[Round[cdf[k, x]], {k, 1, kk}], Table[actual[k, x], {k, 1, kk}]}], 
TableHeadings -> {None, {"k  ", "Landau", "CDF   ", "Actual"}}, 
TableSpacing -> {2, 3, 0}]

physics - Angle Maximizing the Distance of a Projectile

Added 11/23/12: ChrisJB has given a really wonderful answer that avoids both derivatives and trigonemetric identities by considering geometry not in the (rotated) $xy$ plane but in the (rotated) $v_xv_y$ plane. It took me a while to understand why, in that answer, the distance traveled is proportional to the area of a triangle rather than a trapezoid, so I'm appending below my own original answer a slightly longwinded version of ChrisJB's, for the benefit of others who are as slow as I am. Credit (and upvotes) for it, however, should go entirely to ChrisJB. (You can skip now straight to the bottom.)

Here's something that at least avoids taking derivatives.

Let's start with a warm-up on flat ground. If you fire a projectile with vertical velocity $v_y$ and horizontal velocity $v_x$, the amount of time it spends in the air is $T=2v_y/g$ and the distance it travels is $D=v_xT$. As a function of firing angle $theta$, we have $v_y=v_0sintheta$ and $v_x=v_0costheta$. Setting $v_0=g=1$ to clean out the clutter, we have

$$D=2sinthetacostheta = sin(2theta),$$

which is clearly maximized, taking the value $1$, when $theta = pi/4$.

Now suppose the ground slopes down at angle $phi$. Let's rotate it up flat, and imagine firing at angle $theta' = theta + phi$. (We're not taking derivatives, so there should be no confusion in using the notation $theta'$.) The obvious problem is, gravity no longer points straight down. Instead it has a vertical component $g_y = gcosphi$, pointing down, and a horizontal component $g_x = gsinphi$, pointing to the right. In terms of their effect, the vertical component $g_y$ is a new (and reduced) gravity, while the horizontal component $g_x$ acts as a kind of additional magnetic force on the projectile, accelerating it in the $x$ direction. Thus the amount of time a projectile fired with vertical velocity $v_y$ spends in the air is $T=2v_y/g_y$, much as before, while the horizontal (actually downhill) distance it travels is now

$$D = v_xT + {1over2}g_xT^2 = 2v_y(v_xg_y + v_yg_x)/g_y^2.$$

We have $v_y = v_0costheta'$ and $v_x = v_0sintheta'$. In this case it's convenient to adopt the clutter-cleaning convention $v_0 = cosphi$, which leaves us with

$$D=2sintheta'(costheta'cosphi + sintheta'sinphi)=2sintheta'cos(theta'-phi),$$

using the angle addition formula $cos(x-y) = cos x cos y + sin x sin y$. The formula $2sin xcos y = sin(x+y)+sin(x-y)$ turns this into

$$D = sin(2theta'-phi) + sinphi = sin(2theta + phi) + sinphi,$$

which this time is maximized, taking the value $1+sinphi$, when $2theta+phi = pi/2$, which is to say, when $theta = pi/4 - phi/2$.

Added 11/15/12: Oops, I just fixed a minor mistake: the correct clutter-cleaning convention is $v_0 = cosphi$, not $sinphi$. I should have realized this right away from the fact that it needs to agree with the flat-ground convention, $v_0=1$, when $phi=0$. (I had elsewhere kept my sines and cosines straight using, for example, the fact that $g_x$ should be negligible for $phiapprox0$.) The full factor in $D$ that begs to be set equal to $1$ is $v_0^2/gcos^2phi$. If you stick with the convention $v_0=g=1$, you find that the maximum downhill distance, as a function of $phi$ is

$$D_max = sec^2phi+secphitanphi,$$

whose horizontal component is

$$H_max = D_max cosphi = secphi + tanphi.$$

Added 11/23/12: This is my longwinded version of ChrisJB's answer.

If you rotate the system so the ground is flat, you'll be firing at angle $theta' = theta + phi$ into a medium where gravity points down and to the right at angle $phi$. In the velocity plane, the trajectory starts at $P=(v_0costheta', v_0sintheta')$ and follows a straight line at angle $pi/2 - phi$ to the $v_x$ axis, through a point $Q$ on the $v_x$ axis, down to a point $P'$ with $v_y$ coordinate $-v_0sintheta'$. (It's easy enough to work out the $v_x$ coordinates of $Q$ and $P'$, but it's unnecessary to do so.) Denoting points $A=(0,v_0sintheta')$ and $A'=(0,-v_0sintheta')$ on the $v_y$ axis, we find that the total (downhill) distance traveled by the projectile is proportional to the area of the trapezoid $APP'A'$. (This is because, for a given $phi$, changes in velocity are proportional to changes in time.) If you draw the trapezoid, it's easy to see that its area is 4 times the area of the triangle $triangle OPQ$, $O=(0,0)$ being the origin. (This is the "easy to see" point that took me a while to see. If someone with the wherewithal to do so could insert an actual picture here, I would very much appreciate it.) The angle at $Q$ is fixed at $pi/2 - phi$ and the length of the side opposite $Q$ is fixed at $OP=v_0$. It doesn't require calculus to conclude that the triangle's area is maximized when $Q$ is the apex of an isosceles triangle, i.e., when $theta' = pi/4 + phi/2$, which translates back to $theta = pi/4 - phi/2$.

lo.logic - Where are we working when we prove metamathematical theorems?

There are many flavors of "meta" in logic. Most make very minimal use of the metatheory. For example, the Montague Reflection Principle in Set Theory says the following:

Metatheorem. For every formula $phi(x)$ of the language of Set Theory, the following is provable: If $phi(a)$ is true then there is an ordinal $alpha$ such that $a in V_alpha$ and $V_alpha vDash phi(a)$.

This is in fact an infinite collection of theorems, one for each formula $phi(x)$. Gödel's theorem prevents ZFC from proving all of these instances at the same time. The proof of this metatheorem is by induction on the length of $phi(x)$. The requirements on the metatheory are very minimal, all you need is enough combinatorics to talk about formulas, proofs, and enough induction to make sense of it all. This is by far the most common flavor of "meta" in logic. There really is not much to it. Most logicians don't worry about the metatheory in this context. Indeed, it is often irrelevant and, like most mathematicians, logicians usually believe in the natural numbers with full induction.

However, sometimes there is need for a much stronger metatheory. For example, the metatheory of Model Theory is usually taken to be ZFC. Gödel's Completeness Theorem is a nice contrasting example.

Completeness Theorem. Every consistent theory is satisfiable.

Recall that a theory is consistent if one cannot derive a contradiction form it, while a theory is satisfiable if it has a model. The consistency of a theory is a purely syntactic notion. When the theory is effective, consistency can be expressed in simple arithmetical terms in a manner similar to the one discussed above. However, satisfiability is not at all of this form since the models of a theory are typically infinite structures. This is why this theorem is usually stated in ZFC. For the Completeness Theorem, one can often get by with substantially less. For example, the system WKL₀ of second-order arithmetic suffices to prove the Completeness Theorem for countable theories (that exist in the metaworld in question), but such a weak metatheory would be nothing more than a major inconvenience for model theorists.

Sometimes there are multiple choices of metatheory and no consensus on which is most appropriate. This happens in the case of Forcing in Set Theory, which has three common points of view:

• Forcing is a way to extend a countable transitive model of ZFC' to a larger model of ZFC' with different properties. Here ZFC' is a finite approximation to ZFC, which makes this statement non-vacuous in ZFC because of the Reflection Principle above.




• Forcing is a way to talk about truth in an alternate universe of sets. Here, the alternate universe is often taken to be a Boolean-valued model, formalized within the original universe.




• Forcing is an effective way to transform a contradiction some type of extensions of ZFC into contradictions within ZFC itself. Here, ZFC could be replaced by an extension of ZFC.

The last point of view, which is the least common, is essentially arithmetical since it only talks about proofs. The second point of view is well suited for hard-core platonists who believe in one true universe of sets. The first point of view, which is probably most popular, essentially takes ZFC as a metatheory much like model theorists do.

Since there is no consensus, set theorists will often talk as if both $V$ and its forcing extension $V[G]$ are absolutely real universes of sets. Although this point of view is hard to justify formally, it is possible to makes sense of it using any one of the three viewpoints above and it has the advantage that it makes it easier to express ideas that go into forcing constructions, which is what set theorists really want to talk about.

Finally, the idea of analyzing which systems prove the consistency of other systems is very common in logic. In Set Theory, these systems often take the form of large cardinal axioms. In Second-Order Arithmetic these systems often take the form of comprehension principles. In First-Order Arithmetic these systems often take the form of induction principles. Together these form an incredibly long consistency strength hierarchy stretching from extremely weak basic arithmetical facts to incredibly deep large cardinal axioms. A very significant part of logic deals with studying this hierarchy and, as Andrej Bauer commented, logicians are usually very aware of where they are sitting in this hierarchy when proving metatheorems.

tag removed - Generalizations of "standard" calculus

I realize this question was posted a while ago, but I would like to make a note on your edit stating,

After looking around some more I found time scales, which are pretty much what I was thinking of in the second part of my question (though many of the answers people have provided are along the same general lines). I'm surprised I don't hear more about this in analysis - unifying discrete and continuous should make it a pretty fundamental concept!

Time scale calculus is fairly new development. It came about in 1989 in Hilger's Ph.D. thesis (he initially called it a measure chain). At the time of writing his thesis, mathematicians did not embrace his idea. Then, in the early 90s, the ideas were brought to the U.S. and now there are papers being written about time scale all over the world. However, I suppose there are still some mathematicians who are not studying/teaching time scale.

If you are looking for resources on this, you might consider taking a look at [1].

Furthermore, I see above that you are trying to define the exponential function for "discrete(difference) calculus" (if I am understanding your posting correctly).

Note that for the continuous case, we have that $(e^{at})^{'}=ae^{at}$

Now, consider $Delta (1+ alpha)^{t} = (1+ alpha)^{t+1} - (1 + alpha)^{t}= (1+alpha)(1+alpha)^{t}- (1+alpha)^{t} = ((1+alpha)-1)(1+alpha)^{t}=alpha(1+alpha)^{t}$.

Thus, the discrete exponential should actually be $(1+ alpha)^{t}$.

[2] is a good resource for discrete calculus.

Also, [1] discusses exponentials in time scale. There is a delta and nabla exponential form. You may consider checking that out.

Now, to your fractional calculus question, I believe I recall my professor mentioning some problem with fractional calculus and time scale. I cannot remember what it was exactly, but it was an open problem regarding fractional calculus. (Note: There are many open problems in time scale.)

I hope this helps!

[1] Bohner, M. & Peterson, A. (2001). Dynamic Equations on Time Scales: An Introduction with Applicaitons. Boston, MA: Birkhäuser.

[2] Kelley, W. & Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.

set theory - Axiom of Infinity needed in Cantor-Bernstein?

Actually, the usual proof of the Cantor-Schröder-Bernstein Theorem does not use the Axiom of Infinity (nor the Axiom of Powersets).

By the usual proof, I mean the one found on Wikipedia, for example. Using the notation from that proof, the main point of contention is whether we can form the sets $C_n$ and $C = bigcup_{n=0}^infty C_n$. These sets exist by comprehension:

$C_0 = {x in A: forall y in B,(x neq g(y))}$

$C_n = {x in A: exists s,(s:{0,dots,n}to A land s(0) in C_0 land (forall i < n)(s(i+1) = g(f(s(i)))) land s(n) = x}$,

where abbreviations such as $s:{0,dots,n}to A$ should be replaced by the equivalent (bounded) formulas in the language of set theory. The definition of $C_n$ is uniform in $n$, and so

$C = {x in A: exists n,(mathrm{FinOrd}(n) land x in C_n)}$,

where $x in C_n$ should be replaced by the above definition and $mathrm{FinOrd}(n)$ is an abbreviation for "$n$ is zero or a successor ordinal and every element of $n$ is zero or a successor ordinal." An alternate definition of $C$ is

$C = {x in A: forall D,(C_0 subseteq D subseteq A land g[f[D]] subseteq D to x in D)}$,

which shows that $C$ is $Delta_1$-definable. The rest of the proof uses induction on finite ordinals, but since the definition of the sets $C_n$ is uniform these are special cases of transfinite induction.

In conclusion, it looks like the proof could work in Kripke-Platek Set Theory — which has neither the Axiom of Infinity nor the Axiom of Powersets — provided that the two definitions of $C$ given above are provably equivalent. I haven't tried to check whether the two definitions are provably equivalent but, in any case, the proof can be carried out in Kripke-Platek Set Theory with $Sigma_1$-Comprehension.

soft question - A single paper everyone should read?

I am surprised to see that so many people suggest meta-mathematical articles, which try to explain how one should do good mathematics in one or the other form. Personally, I usually find it a waste of time to read these, and there a few statements to which I agree so wholeheartedly as the one of Borel:

"I feel that what mathematics needs least are pundits who issue prescriptions or guidelines for presumably less enlightened mortals."

The mere idea that you can learn how to do mathematics (or in fact anything useful) from reading a HowTo seems extremely weird to me. I would rather read any classical math article, and there are plenty of them. The subject does not really matter, you can learn good mathematical thinking from each of them, and in my opinion much easier than from any of the above guideline articles. Just to be constructive, take for example (in alphabetical order)

• Atiyah&Bott, The Yang-Mills equations over Riemann surfaces.


• Borel, Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de groupes de Lie compacts.


• Furstenberg, A Poisson formula for semi-simple Lie groups.


• Gromov,Groups of polynomial growth and expanding maps.


• Tate, Fourier analysis in number fields and Hecke's zeta-functions.

I am not suggesting that any mathematician should read all of them, but any one of them will do. In fact, the actual content of these papers does not matter so much. It is rather, that they give an insight how a new idea is born. So, if you want to give birth to new ideas yourself, look at them, not at some guideline.

nt.number theory - A bound for the Manin constant

A useful reference on this topic (which maybe you know?) is "The Manin Constant",
by Agashe, Ribet, and Stein, available here. On p. 3 they write

B. Edixhoven also has unpublished results (see [Edi89]) which assert that the
only primes that can divide $c_E$ are 2, 3, 5, and 7; he also gives bounds that are
independent of $E$ on the valuations of $c_E$ at 2, 3, 5, and 7. His arguments rely
on the construction of certain stable integral models for $X_0(p^2 )$.

The reference [Edi89] is to Edixhoven's thesis, which is unpublished, but
which can be found here
at Edixhoven's website. At the end of the introduction to his thesis, Edixhoven writes

Finally,
in the last section, we derive some results concerning the constant
"c" attached to a strong Weil curve $E$. Manin conjectured that $c=1$. It
is known that $c$ is a positive integer and Mazur proved that only 2 and
primes where $E$ has additive reduction can divide $c$. Our methods show
that primes $p>7$ where $E$ has additive reduction divide $c$ at most once,
and in fact, for most of the possible reduction types (=Kodaira symbols), not (Theorem 4.6.3 and the remarks following this theorem). It might well
be that a computation involving the period lattices of normalized newforms
can solve the problem in the case of potentially good, ordinary reduction of type $II, III$ or $IV$. It should also be tried to get bounds on the exponents
of 2, 3, 5 and 7 in $c$.

This suggests that the statement of Agashe, Ribet, and Stein is perhaps a little strong,
in that Edixhoven does not give uniform bounds at all primes, but only at primes $p > 7$,
with the suggestion that one could also hope to obtain such bounds at 2, 3, 5, and 7.
You might try writing to Edixhoven for clarification, or perhaps also to one of Agashe, Ribet, or Stein to find out more precisely what they had in mind.

big list - Nontrivial question about Fibonacci numbers?

Here you have some of the coolest ones I have heard of:

1) Let $a$ be a positive integer. Then $a$ is a Fibonacci number if and only if at least one member of the set {$5a^{2}-4, 5a^{2}+4$} is a perfect square.

I think the result is original with Prof. Ira Gessel.

2) Let $phi$ denote the Euler totient function. Prove that $phi(F_{n}) equiv 0 pmod{4}$ if $n geq 5$.

The proof consists of an unexpected application of Lagrange's theorem in Group Theory. Guess there are some other ways to prove it, but that approach will always remain my cup of tea. The problem was posed and solved in the Monthly in the 70's (if my memory serves me right). Look for all entries by Clark Kimberling in that magazine and you'll surely find it.

3) Can you find $(a,b,c) in mathbb{N}^{3}$ such that $2 < a < b < c$ and $F_{a} cdot F_{b} = F_{c}$?

This problem would be trivial if instead of the $cdot$ we had placed a plus sign there. In any case, there is no need to panic with this proposal. All you need to recall is the corresponding primitive divisor theorem.

4) Ben Linowitz mentioned above a beautiful result by Professor Florian Luca, namely:

There aren't any perfect numbers in the Fibonacci sequence.

I read the paper in my junior year and I didn't find it that hard to follow. The easy part of this cute note resides in the proof of the fact that there are no even perfect numbers in the Fibonacci sequence. Guess this result is interesting enough to deserve consideration in those lectures that you intend to give. If this proposal is not exactly your idea of excitement, you can take a look at some of the other papers by Professor Florian. He writes a lot about recurrence sequences. Another theorem of his, closely related with the subject matter of this discussion, ascertains that

There is no non-abelian finite simple group whose order is a Fibonacci number.

5) Last but not least... Prove that the sequence {$F_{n+1}/F_{n}$}$_{n in mathbb{N}}$ converges and use this fact to derive the continued fraction development for the golden ratio.

This one should be well-known, yet it would be nice to see what your students come up with...

Added (Nov 20/2010) I've just noticed that the Fibonacci Assn. has made available the articles published in The Fibonacci Quarterly between 1963 and 2003. I'm sure you will find plenty of additional material among those files that they have so generously released for our enjoyment. For instance, the seminal paper by J. H. E. Cohn that K. Buzzard mentions below can be found here.

gn.general topology - The Closure-Complement-Intersection Problem

Background

Let $A$ be a subset of a topological space $X$. An old problem asks, by applying various combinations of closure and complement operations, how many distinct subsets of $X$ can you describe?

The answer is 14, which follows from the observations that $Cl(Cl(A))=Cl(A)$, $neg(neg (A)=A$ and the slightly harder fact that

$$Cl(neg (Cl(neg (Cl(neg (Cl(A))))=Cl(neg (Cl(A))$$
where $Cl(A)$ is the closure of $A$ and $neg (A)$ is the complement. This makes every expression in $Cl$ and $neg$ equivalent to one of 14 possible expressions, and all that remains is to produce a specific choice of $A$ which makes all 14 possiblities distinct.
This problem goes by the name Kuratowski's closure-complement problem, since it was first stated and solved by Kuratowski in 1922.

The Problem

A very similar problem recently came up in a discussion that was based on a topological model for modal logic (though the logical connection is unrelated to the basic question). The idea was to take a subset $A$ of $mathbb{R}$, and to consider all possible expressions on $A$ consisting of closure, complement, and intersection. To be clear, we are allowed to take the complement or closure of any subset we have already constructed, and we are allowed to intersect any two subsets we have already constructed.

The question: Is this collection of subsets always finite?

A potentially harder question: If there are multiple starting subsets $A_1$, $A_2$, $A_3$... (finite in number), is this collection of subsets always finite?

The first question is essentially the Kuratowski question, with the added operation of intersection. There is also the closely related (but slightly stronger) question of whether there are a finite number of formally distinct expressions on an indeterminant subset $A$ (or a collection of subsets $A_1$, $A_2$...).

Some Thoughts

My guess to both questions is yes, but the trick is showing it. I can take an example of a set $A$ which realizes all 14 possibilities from Kuratowski's problem, and show that the collection of distinct subsets I can construct from it is finite. However, just because such a set captures all the interesting phenomena which can happen when closing and complementing, doesn't mean this example is missing a property that is only important when intersecting.

It also seems difficult to approach this problem formally. The problem is that there are many non-trivial intersections of the 14 expressions coming from Kuratowski's theorem. Then, each of these intersections could potentially have its own new set of 14 possible expressions using closures and complements. In examples, the intersections don't contribute the full number of 14 new sets, but its hard to show this aside from case-by-case analysis.

qa.quantum algebra - Does the canonical basis of a tensor product of quantum group representations span the isotypic components of tilting modules?

It's a theorem of Lusztig that if one takes two (or more) representations of a quantum group at a non-root-of-unity choice of $q$, and look at the canonical basis of their tensor product, then each isotypic component is spanned by a subset of the canonical basis (i.e. each basis vector lies in an isotypic component) [EDIT: I mean the canonical basis of the tensor product in the sense of Lusztig's paper "Canonical bases on tensor products" not the tensor product of the canonical bases].

When one reduces this tensor product at a root of unity instead it's no longer necessarily semi-simple. But if we assume that each of the tensor factors does remain simple, it will be a tilting module, and have a canonical direct sum decomposition in "isotypic" components corresponding to tilting modules. These are quite different from the isotypic components at a generic value of $q$. Are these still spanned by canonical basis vectors?

EDIT: I would be basically equally happy if there were a split (but not canonically) filtration whose individual spaces were spanned by canonical basis vectors such that the successive quotients where tilting modules.

sg.symplectic geometry - Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity?

I just wanted to expand two points of Greg's answer. Both are rather trivial additions, but took me a short while to understand, so I'm putting them here for completeness's sake and my own future reference.

First, here's a picture of Greg's idea of fixing the problem with Moser's trick. In his notation, we have volume forms $mu_{alpha,t} = mu^t mu_alpha^{1-t}$. They are defined because the two volume form must have the same sign everywhere (since everything is orientation-preserving; note that because of this I doubt that this argument can be generalized to symplectic forms in higher dimensions).

The obvious, but wrong, solution (described in my comments to his answer) would be to apply Moser's theorem to $mu_{alpha,0}$ in this notation. This would correspond to flowing along the horizontal axes of the picture below. However, Greg's idea is to flow in a different direction: we fix $alpha$ and vary $t$. Strictly speaking, we should also show that the resulting flow $phi_{alpha,1}$ will be smooth, but this should follow from the proof of Moser's theorem quite easily.

                                         Moser flow,
                                t        defines phi_alpha,t
        (mu_{alpha,1} = mu)   1 ^          ^
                                |          .
                                |          .
                                |          .
                                |          .
   (mu_{alpha,0} = mu_alpha)  0 ----------------> alpha
 (phi_{alpha,0} = phi_alpha)    

(Note that, for all t, $mu_{0,t}=mu_{1,t}=mu$)

---

Secondly, Greg's answer implies that there is an isotopy $psi_t: D^2 to D^2$ such that each $psi_t$ is volume preserving (or, equivalently, a symplectomorphism), $psi_0$ is the identity, and $psi_1=psi$. Here's how we find the time-dependent Hamiltonian such that this is the Hamiltonian flow. Let X_t be the time-dependent vector field that is the derivative of the flow $psi_t$. The fact our flow is volume-preserving is equivalent to the fact that the 1-form $iota_{X_t} omega$ is be closed for all $t$. Since we are on a disk, this form will also be exact. So, let $H_t$ be the function such that $dH_t = iota_{X_t} omega$. The Hamiltonian flow of the function $H_t$ is precisely $psi_t$; this is immediate from the definitions.

nt.number theory - Galoisian sets of prime numbers

In order to add my bit to the already rich content on this site, here is a
nice family of galoisian extensions $K_l|{bf Q}$ with group ${rm GL}_2({bf F}_l)$ (indexed by primes $lneq5$) for which a ``reciprocity law'' can be
written down explicitly. I've come across this family recently while writing
an expository article.

Let $E$ be the elliptic curve (over $bf Q$) of conductor $11$ defined by
$y^2+y=x^3-x^2$, with associated modular form

$$eta_{1^2,11^2}=qprod_{k>0}(1-q^k)^2(1-q^{11k})^2=sum_{n>0}c_nq^n.$$
Let $K_l={bf Q}(E[l])$, which is thus galoisian over $bf Q$ and unramified at
every prime $pneq11,l$.

One can deduce from cor.1 on p.308 of Serre (Inventiones 1972) that for every
prime $lneq5$, the representation

$$rho_{E,l}:{rm Gal}(K_l|{bf Q})rightarrow{rm GL}_2({bf F}_l)$$
we get upon choosing an ${bf F}_l$-basis of $E[l]$ is an isomorphism;
cf. the online notes on Serre's conjecture by Ribet and Stein. Shimura did
this for $lin[9,97]$ (Crelle 1966).

Suppose henceforth that $l$ is a prime $neq5$ and that $p$ is a prime
$neq11,l$. The characteristic polynomial of $rho_{E,l}({rm Frob}_p)in{rm GL}_2({bf F}_l)$ is

$$T^2-bar c_pT+bar pin{bf F}_l[X].$$
The prime $p$ splits completely in $K_l$ if and only if ${rm Frob}_p=1$ in
${rm Gal}(K_l|{bf Q})$, which happens if and only if
$$rho_{E,l}({rm Frob}_p)=pmatrix{1&0cr0&1}.$$
If so, then $p,c_pequiv1,2pmod l$ but not conversely, for the matrix
$displaystylepmatrix{1&1cr0&1}$ also has the characteristic polynomial
$T^2-bar2T+bar1$. But these congruences on $p,c_p$ do rule out an awful lot
of primes as not splitting completely in $K_l$.

In summary, we have the following ``reciprocity law" for $K_l$ :

$$hbox{($p$ splits completely in $K_l$)} quadLeftrightarrowquad E_p[l]subset E_p({bf F}_p),$$
where $E_p$ is the reduction of $E$ modulo $p$. Indeed, reduction modulo $p$
identifies $E[l]$ with $E_p[l]$ and the action of ${rm Frob}_p$ on the former
space with the action of the canonical generator $varphi_pin{rm Gal}(bar{bf F}_p|{bf F}_p)$ on the latter space. To say that $varphi_p$
acts trivially on $E_p[l]$ is the same as saying that $E_p[l]$ is contained in
the ${bf F}_p$-rational points of $E_p$. The analogy with the multiplicative
group $mu$ is perfect:
$$hbox{($pneq l$ splits completely in ${bf Q}(mu[l])$)} quadLeftrightarrowquad mu_p[l]subset mu_p({bf F}_p)$$
($Leftrightarrow l|p-1Leftrightarrow pequiv1pmod l$), where $mu_p$ is not the $p$-torsion of $mu$ but the reduction of $mu$ modulo $p$.

I requested Tim Dokchitser to compute the first ten $p$ which split completely
in $K_7$, and his instantaneous response was 4831, 22051, 78583, 125441,
129641, 147617, 153287, 173573, 195581, and 199501.

It is true that all this (except the list of these ten primes) was known before Serre's conjecture was proved
(2006--9) or even formulated (1973--87), but I find this example a very good
illustration of the kind of reciprocity laws it provides.

I hope you enjoyed it as much as I did.

ag.algebraic geometry - Computing fundamental groups and singular cohomology of projective varieties

The trick I know (learned it from Ron Livne) is to project it to some space with known homotopy / homology, throw away the ramification and branch loci to get a covering map (and you better pray it's Galois - otherwise the mess is even bigger) , and then bring the ramification back as extra relations.

e.g. here is a computation of the homotopy group of an elliptic curve E:

You have a degree 2 projection to a P¹ with four ramification points. The homotopy group of P¹ minus the four branch points is freely generated by loops about 3 of these points.

Claim: the homotopy group of E minus the ramification locus the kernel of the map from the free group on three generators: F(a,b,c) to Z / 2,
given by adding the powers on all the letters and taking mod 2 (e.g. abbac^-1b maps to 4 mod 2 = 0).

Sketch of proof: think of a,b,c, as paths in E minus the ramification points which have to glue to a closed loop, and to project to the generators of the homotopy of P¹ minus the branch points (i.e. they are "half loops" / sheet interchange about the ramification points).

Finally we have "fill" the ramification points - i.e. to bring the extra relations a², b², c². After adding these relations, our group is generated by ab, ba, ac, ca, bc, cb. Hence - since e.g. (bc)(cb) = 1 - it is generated by ab, ac, bc; hence - since (ab)(bc) = (ac) - it is generated by ab, ac. We now observe that the map which sends x to axa is the map sending an element to the inverse; which shows as that

(ab)(ca)(ab)^-1 = (ab)(ca)(ba) = (bc)^-1 ba = (cb)(ba) = ca.

Note that this is the simplest example one can give - this is a painful trick.

ho.history overview - When and why did the postdoctoral position originate?

In know that there is some variation among fields for when post-docs became popular. My understanding is that the "post-doc" institution as a whole began around the beginning of the last century, but didn't become common until many decades later. In the physics (and, I think, math) world, there was an explosion of post-docs in the US after WWII. During the war, the government had created many national labratories, and most of the top physicists took time away from their other academic pursuits to help in the war effort. When the war ended, the government was still very interested in funding physics research (Cold War!) in particular, and academics in general (G.I. Bill!). Thus, in the US anyway, the middle of the 20th century saw a dramatic rise in both the demand and supply for researchers. But the number of professorships didn't match pace, hence the growing number of post-docs. Moreover, many of the leading senior physicists at the time (Oppenheimer, Bethe, etc.) strongly encouraged the creation of post-doc positions as a way for the young people (Feynman, Dyson, Schwinger, etc.) to communicate their theories. As is often said: the best way to send an idea is to wrap it in a person.

This story is very well told in D. Kaiser, Drawing theories apart: the dispersion of Feynman diagrams in postwar physics, University of Chicago Press, 2005.

One other comment is important to make. The structure of the academy (and in particular the names of ranks) can vary wildly from country to country, so that "assistant professor" in some countries means "post-doc" in others. For example, Denmark in the last decade has made a conscious shift away from the "German" model, in which there are very few "professors", each of whom over see many full-time senior researchers, and towards a more "American" model where the rank "(full) professor" is more freely given.

ag.algebraic geometry - Generating the derived category with line bundles

The following lemma is useful and well-known:

LEMMA If $L^{pm 1}$ is ample on proper scheme over a field $k$, then some number of powers $mathcal{O},L,...,L^{m}$ generate the unbounded derived category of quasi-coherent sheaves $D(X)$ (or split generate the subcategory of perfect complexes).

QUESTION: What about a converse? Suppose that I know some number of powers of $L$ generate $D(X)$. Then can I conclude that $L^{pm 1}$ is ample?

The best I can do so far is see that the restriction of $L$ to any integral curve
$C$ in $X$ has non-zero degree. (Since by adjunction $mathcal{O},L,...,L^{m}$ generates $D(C)$, but if $L$ had degree $0$ on $C$, there would be something orthogonal $mathcal{O},L,...,L^{m}$, for instance a generic line bundle of degree $g-1$ having no cohomology.)

Something I don't know yet: does the degree of $L$ must have the same sign on all curves?
This would be useful for numerical tests of ampleness.

Note: I think that one doesn't need properness in the above lemma, but I am willing to assume it to get a converse. It makes life easier when restricting to closed subschemes.

Note 2: When saying a collection of objects generates a triangulated category with all coproducts, like $D(X)$, one usually means that you take the smallest triangulated subcategory closed under all coproducts and containing the the collection. Once you have all coproducts, then idempotents automatically split, by a standard argument called, I think, the Eilenberg swindle. If you are working with a smaller triangulated category having only finite coproducts, like perfect complexes on a scheme, then the smallest triangulated subcategory containing a collection might not be 'thick', in the sense that some idempotents might not split, so in this case one usually adds in the missing summands. To emphasize this, some people speak of 'split generation'.

dg.differential geometry - Is the volume form on an oriented Riemannian manifold parallel?

These are important questions.
When one starts to define connections, one usually imposes the Leibnitz condition on it. But I am proceeding too fast. First, let's recall what a connection is.
Here we come to one very confusing point, at least, it was for me, when I started learning differential geometry: There are various ways of defining connections.
Just to name some of them:

1. Define the Levi-Civita connection on a (Riemannian) manifold.
   The Levi-Civita connection is an affine connection that preserves the metric structure, i.e.,
   $nabla g = 0$ and that is torsion-free, i.e., $nabla_{X}Y-nabla{Y}X = [Y, X]$, where $[. , .]$ denotes the Lie-bracket of two elements of $Gamma(TM)$. Then one usually starts by showing that, by this definition, the Levi-Civita connection (hereafter called LC connection) is uniquely determined.




2. However, one can even generalize this approach by at first considering linear connections, then metric ones and then affine ones. After all that one can turn to the derivation of the Levi-Civita connection. Personally, I like this approach pretty much because it shows very well why the LC connection is so important. Namely, it eases computations for the Lie-derivative which normally does not require the notion of the LC connection at all. Also there are many fascinating mathematical objects, such as Killing fields which can be calculated (or at least defined) effectively, if one introduces the notion of the LC connection.




3. One can even start more generally, by considering fiber bundles. This approach was taken by the French mathematician Charles Ehresmann (1905-1979). This approach has one fundamental advantage: Generality, however, it lacks the easy-to-grasp-effect (I like to assign a definition this value if it has a very intuitive meaning). Interestingly, this approach does yield the LC connection as a special case when one reduces fiber bundles to vector bundles and considers the bundles $pi : TM longrightarrow M$ as a special case and requires the base manifold $M$ to be equipped with a symmetric, positive definite 2-times covariant tensor field, i.e., the metric tensor $(g_{ij})_{i,j}$.

Why am I writing all this? Simply to sum up what we already know and how beautifully, at least, in my opinion, all these definitions harmonize.

To come to your question (I apologize for relying so much on your patience):

If one has finally obtained a manifold that is equipped with a Riemannian metric, then one goes further and asks what the term orientation means for general Riemannian manifolds. Although for submanifolds of $mathbb{R}^{n}$ one has a very intuitive picture of what orientability means. We could expand this picture by making use of the Nash Embedding theorem, but it turns that this is not necessary.

Example: Consider for instance the so-called Möbius strip. It is a typical example of a non-orientable two-dimensional submanifold of the Euclidean 3-space.

Then, one can ask further what it means for a manifold to be orientable. Consider now the tangent and the cotangent bundle. It is now possible to associate to each vector space a certain orientation if one introduces one starting point for that. Consider a basis that is called "unit basis". Define the orientation of this basis to be positive. Then, one imposes further conditions on the unit basis, namely orthonormality w.r.t the Riemannian metric. This way, one can define an orientation on the cotangent bundle $TM^{*}$ as well (Duality principle).

Now we simply ask what the Riemannian volume form is: It is certainly $SO(n)$-invariant, because we have chosen an orientation and require the 1-forms that constitute a basis of $TM^{*}$ to be positively oriented (c.f. what I said above).

OK, now to question one:
Via the Leibnitz rule and the duality principle it is possible to define an induced connection on the cotangent bundle and, by that way, further expanding this definition to the k-th exterior power of the cotangent bundle.

Namely, from the tensor product spaces can be obtained the k-th exterior cotangent space. I would like to refer you to the book Introduction to Kähler Manifolds which contains and excellnt treatment of exactly this. Via the equivalence relation one then breaks down the Leibnitz rule to the k-h exterior power cotangent bundle. The rules is exactly what Matt has written above.

Let's come to question two: We know that the volume form is $SO(n)$ invariant, where $n=dim(M)$ Since we define the volume form $omega := dx^{1} wedge dots wedge dx^{n}$, with the $dx^{i}$ for $i=1, dots, n$ to be the product of the basis of the cotangent bundle, we can make the following thought.

We know how to "differentiate" the basis w.r.t to the LC connection. When we calculate all that, we finally obtain that the volume form is parallel. This was how I learned it. Of course one can use representation theory here which is, in my eyes, very interesting and also more elegant.

If there are questions, don't hesitate to contact me.

set theory - solving $f(f(x))=g(x)$

This question is of course inspired by the question How to solve f(f(x))=cosx
and Joel David Hamkins' answer, which somehow gives a formal trick for solving equations of the form $f(f(x))=g(x)$ on a bounded interval. [EDIT: actually he can do rather better than this, solving the equation away from a bounded interval (with positive measure)].

I've always found such questions ("solve $f(f(x))=g(x)$") rather vague because I always suspect that solutions are highly non-unique, but here are two precise questions which presumably are both very well-known:

Q1) Say $g:mathbf{R}tomathbf{R}$ is an arbitrary function. Is there always a function $f:mathbf{R}tomathbf{R}$ such that $f(f(x))=g(x)$ for all $xinmathbf{R}$?

Q2) If $g$ is as above but also assumed continuous, is there always a continuous $f$ as above?

The reason I'm asking is that these questions are surely standard, and perhaps even easy, but I feel like I know essentially nothing about them. Apologies in advance if there is a well-known counterexample to everything. Of course Q1 has nothing to do with the real numbers; there is a version of Q1 for every cardinal and it's really a question in combinatorics.

EDIT: Sergei Ivanov has answered both of these questions, and Gabriel Benamy has raised another, which I shall append to this one because I only asked it under an hour ago:

Q3) if $g$ is now a continuous function $mathbf{C}tomathbf{C}$, is there always continuous $f$ with $f(f(x))=g(x)$ for all $xinmathbf{C}$?

EDIT: in the comments under his answer Sergei does this one too, and even gives an example of a continuous $g$ for which no $f$, continuous or not, can exist.

Related MO questions: f(f(x))=exp(x) and other functions just in the middle between linear and exponential, and Does the exponential function has a square root.

books - Introduction to wavelets?

A very gentle introduction is Boggess & Narcowich, A First Course in Wavelets with Fourier Analysis. I should warn you, though, they're pretty fast and loose with the hypotheses of their theorems. You'll be fine if you've studied advanced linear algebra, and especially fine if you already know some Fourier analysis.

There's also Mallat's A Wavelet Tour of Signal Processing: The Sparse Way. I haven't read very much of it, so I don't have a strong opinion on it yet. It's definitely more difficult than Boggess & Narcowich, but then it probably has about ten times the content (no exaggeration).

reference request - Proofs without words

This should really be a comment on Marco Radeschi's answer from Feb 22 involving the area formula for spherical triangles, but since I'm new here I don't have the reputation to leave comments yet.

In reply to Igor's comment (on Marco's answer) wondering about an analogous proof for the area formula of hyperbolic triangles: there is one along similar lines, and you're rescued from non-compactness by the fact that asymptotic triangles have finite area. In particular, the proof in the spherical case relies on the fact that the area of a double wedge with angle $alpha$ is proportional to $alpha$; in the hyperbolic case, you need to replace the double wedge with a doubly asymptotic triangle (one vertex in the hyperbolic plane and two vertices on the ideal boundary) and show that if the angle at the finite vertex is $alpha$, then the area is proportional to $pi - alpha$. That follows from similar arguments to those in the spherical case (show that the area function depends affinely on $alpha$ and use what you know about the cases $alpha=0,pi$).

Once you have that, then everything follows from the picture below, since you know the area of the triply asymptotic triangle and of the three (yellow, red, blue) doubly asymptotic triangles.

(That picture is slightly modified from p. 221 of this book, which has the whole proof in more detail.)

at.algebraic topology - Five lemma in HoTop* and arbitrary pointed model categories

Let $textbf{HoTop}^*$ be the homotopy category of pointed topological spaces. In the following, the word "isomorphism" shall always mean isomorphism in $textbf{HoTop}^*$, i.e. pointed homotopy equivalence. All constructions like cone or suspensions are pointed/reduced.

A triangle $Xto Yto Zto Sigma X$ is called distinguished if it is isomorphic in $textbf{HoTop}^*$ to a triangle of the form $Xstackrel{f}{to} Yhookrightarrowtext{C}ftoSigma X$, where $text{C}ftoSigma X$ is the map collapsing $Y$ to a point.

Problem:

Let $matrix{X & to & Y & to & Z & to & Sigma Xcrdownarrowalpha &&downarrowbeta&&downarrowgamma &&downarrow&Sigmaalphacr X^{prime} & to & Y^{prime} & to & Z^{prime} & to & Sigma X^{prime}}$ be a morphism of distinguished triangles such that $alpha$ and $beta$ are isomorphisms. Is it true that $gamma$ is an isomorphism, too?

Suggestions:

For a morphism of triangles as above (where $alpha$ and $beta$ are not necessarily isomorphisms), the morphism $gamma^*: [Z^{prime},-]to [Z,-]$ is equivariant with respect to $[Sigmaalpha]^*: [Sigma X^{prime},-]to [Sigma X,-]$. (edit: this is wrong -- see below) Therefore, I thought one could apply theorem 6.5.3 in Hoveys book on Model Categories. Unfortunately, there seems to be a gap at the end of the proof, as already pointed out here.

Therefore, I have the following

Questions:

(1) Am I misunderstanding something in Hovey's proof of 6.5.3(b), or is there really a gap in it? If it is a gap: Do you have any suggestions on how to fix the proof?

(2) If the proof can't be fixed in this generality: Do you have suggestions on how to prove the statement above only for $textbf{HoTop}^*$?

Edit:

(1) The usual proof of this fact for triangulated categories does not work here, because there one uses the fact that $[X,-]$ is abelian-group valued for any $X$ and uses the classical five lemma together with Yoneda to conclude that $gamma$ is an isomorphism. This doesn't seem to work here.

(2) Since partial morphisms of distinguished triangles in $textbf{HoTop}^*$ can always be completed to morphisms of triangles, we can reduce to the case where $alpha$ and $beta$ both equal the identity. Therefore, we have a commutative diagram (in $textbf{HoTop}^*$, i.e. a homotopy commutative diagram in $textbf{Top}^*$)

$matrix{X & to & Y & to & Z & to & Sigma Xcrdownarrow & text{id}_X &downarrow & text{id}_Y&downarrow&gamma&downarrow&text{id}_{Sigma X}cr X & to & Y & to & Z& to & Sigma X}$

and we have to prove that $gamma$ is a homotopy equivalence.

Hovey's proof

The way Hovey proceeds in his proof is as follows: We know the following things:

(1) $gamma^*: [Z,-]to [Z,-]$ is $[Sigma X,-]$-equivariant

(2) Two maps $c,din[Z,W]$ are equal in $[Y,W]$ if and only if they lie in the same $[Sigma X,W]$-orbit.

From (2) and the commutativity of the middle square it follows that for any $hin [Z,W]$ there is some $rhoin[Sigma X,W]$ such that $gamma^*(h)=h.rho$; in other words $gamma^*$ doesn't change the $[Sigma X,-]$-orbit.

Now, suppose there are $g,hin [Z,W]$ such that $gamma^*(h)=gamma^*(g)$. Then, again by the commutativity of the middle square, there is some $alphain [Sigma X,W]$ such that $g = h.alpha$. Thus, by (1), $gamma^*(g) = gamma^*(h).alpha = gamma^*(g).alpha$, and so $alphaintext{Stab}(gamma^*(g))$.

The point is that Hovey now wants to show that $text{Stab}(gamma^*(g))=text{Stab}(g)$; this would imply $alphaintext{Stab}(g)$, and thus $h = g.alpha^{-1} = g$ as required. The inclusion $text{Stab}(gamma^*(g))supsettext{Stab}(g)$ is obvious. For the other inclusion, I have no idea how to prove it.

Do you see how one can fix the proof?

FINAL EDIT

I made a mistake in proving that for any morphism of triangles $(alpha,beta,gamma)$ the morphism $gamma^*$ is equivariant with respect to $(Sigmaalpha)^*$. This is wrong.

So what remains is the question on how to fix the proof of theorem 6.5.3 in Hovey's book. Any suggestions?

Thank you.