How did the Mars satellites "take cover" from the recent comet flyby?

I've written an answer or two dealing with this within the last week, and I can't believe I never thought about this. Great question.

These spacecraft are in orbit around Mars, constantly in motion, so they can't very well just pack up and move to the other side of Mars to avoid getting hit with the dust.

Believe it or not, they can just pack up and move. All they have to do is alter their orbital speed - i.e. either slow down, speed up, or change their orbit so they will automatically go a different speed, and so be in a different place. They execute a series of orbital maneuvers to ensure that they are in the part of their orbit that temporarily carries them behind Mars. Orbital maneuvers aren't hard; practically any spacecraft can modify its orbit slightly, if need be. According to this,

The orbiters will be repositioned to a “safe zone,” where Mars will actually help shield the satellites from the cloud of dust particles.

In other words, the spacecraft will be moved so they are on the opposite side of Mars as the comet. They can't be there for long - after all, they orbit Mars extremely quickly. But, fortunately, they don't have to be there for long. As this says,

The satellites will remain in this region for a period of up to 40 minutes, until the threat has subsided.

Why do short a period? Well, the comet will only be in the area for a short time: NASA estimates the worst part will last roughly 20 minutes. After that, the danger will have subsided. Any particles that could damage the spacecraft will be long gone.

Did these orbital modifications affect the quality of research these satellites will be able to perform in relation to Mars?

I can't find a source for the answer to this, but I would expect there wouldn't be any problems. You should be able to undo any maneuvers that would change the orbit of the craft.

sg.symplectic geometry - Clarification of classical field theory lecture notes by P. Deligne and D. Freed

Configuration space is, by definition, the position space of your particles. Phase space, on the other hand, is the space of pairs (position, momentum). The latter has a symplectic structure; the former has a Riemannian structure.

Regarding the relationship between kinetic energy and the Riemannian structure: You will recall from your high school physics class that kinetic energy is $frac{1}{2} mv^2$. Of course the $v^2$ is really the dot product $v cdot v$, in other words it's $g(v,v)$, where $g$ is the Riemannian metric and $v$ is a tangent vector. The $frac{1}{2}$ explains the "twice the kinetic energy" part.

size - What was the greatest spatial extent of the Asteroid Belt prior to the Late Heavy Bombardment?

There have been some studies based on observations on the best preserved impact craters of that time in the inner solar system, namely those on Mercury and the Moon. In the paper, Impact-induced compositional variations on Mercury (Edgard et al. 2014), state that based on the nature of Mercury's craters, they have developed a

New dynamical simulations of the early evolution of the asteroid belt hint at the presence of additional
asteroids in a region interior to the present-day belt, known as the “E-belt”.

Through their modelling and simulations, they have determined that the 'E-belt'

The recent reevaluation of the early history of the inner Solar System suggests the primordial asteroid belt could have originally extended well into the Mars-crossing zone

The 'Hungarias' are thought to be a remnant of this belt - as shown in the diagram below

Image attribution:
By Agmartin (Own work) [CC-BY-3.0], via Wikimedia Commons

temperature - What would happen if an ice cube is left in space?

It depends on where in outer space you are.

If you simply stick it in orbit around the Earth, it'll sublimate: the mean surface temperature of something at Earth's distance from the Sun is about 220K, which is solidly in the vapor phase for water in a vacuum, and the solid-vapor transition at that temperature doesn't pass through the liquid phase. On the other hand, if you stick your ice cube out in the Oort Cloud, it'll grow: the mean surface temperature is 40K or below, well into the solid phase, so it'll pick up (or be picked up by) gas and other objects in space.

A comet is a rough approximation to an ice cube. If you think of what happens to a comet at various places, that's about what would happen to your ice cube.

ag.algebraic geometry - Spectrum of the Grothendieck ring of varieties

Here's a problem that may ultimately require just simple algebraic-geometry skills to be solved, or perhaps it's very deep and will never be solved at all. From the comments, some literature and my memory it appears this was posed by Grothendieck as part of the big program of motives.

Consider classes of complex algebraic varieties X modulo relations

    [X] - [Y] = [XY], 
    [X x Y] = [X] x [Y], 

Also, if you're familiar with taking inverse of an affine line, let's do that too:
$$ exists mathbb A^{-1}quad text{such that}quad [mathbb A] cdot [mathbb A^{-1}] = [mathbb A^0].$$

(+ if you want, you can also take idempotent completion and formal completion by A^-1).

It's not hard to see that you can add (formally) and multiply (geometric product as above) those things, so they form a ring. Let's denote this ring  Mot (It's actually very close to what Grothendieck called baby motives.)

And for things that form a ring you can study their Spec. For example, you can talk about points of the ring — each point is by definition a homomorphism to complex numbers.

Question: what are the properties of Spec Mot? How to describe its points?

For example, one point is Euler characteristics $chi in text{Spec},mathbf{Mot}$, since it's additive and multiplicative (it's even integral!) Any other homomorphism to complex numbers is thus sometimes called generalized Euler characteristics.

There's also a plane there given by mixed Hodge polynomials (that is, polynomials whose coefficients are weighted Hodge numbers $h^{p,q}_k$), since Hodge polynomial at a given point satisfies those relations too (see the references below).

As Ben says below, things would become even more interesting if we considered this ring for schemes over $mathbb Z$, because then each $q$ would give a generalized Euler characteristic $chi_q$ that counts points of $X(mathbb F_q).$

Are there any other points? Any more information?

algorithms - NP Complete for range sum constraints?

Is the following problem NP Complete?

We have $n$ variables $x_1$,$x_2$,....,$x_n$ and a set of constraints:

$sum_{i=a_1}^{b_1}x_i = h_1$

$sum_{i=a_2}^{b_2}x_i = h_2$

$sum_{i=a_3}^{b_3}x_i = h_3$

......

where $h_1$,$h_2$,...,$h_n$ are integers. We ask for an integer assignment of $x_1$,$x_2$,...,$x_n$.

The constraint matrix does not seem to be totally unimodular. Is the problem NP Complete?

amateur observing - Please Guide me to buy my first Telescope

It's a broad question, but I'll take a stab trying to provide the essentials for a good start. Also see this post which contains important additional information:

Best telescope for the viewing of Nebulae, Stars and Planets

There are many, many factors involved in choosing a telescope. You seem settled on a dobsonian. That's not the only possible choice, but a dob is a good scope to start with.

You are correct that aperture (diameter of the primary lens or mirror) is the primary driver of telescope performance. Bigger aperture = better performance. However, performance is not the only thing you should worry about. Think of a car - do you only care about speed when buying a new car?

Any size scope will show you something. I've a 50 mm (2") achromat refractor that does double duty as finderscope. I can push it up to 100x magnification, and it will show lots of craters and mountains on the Moon, will show Jupiter as a disk with two equatorial belts, will show the rings of Saturn, will show the M13 globular cluster (barely) and the M31 Andromeda galaxy (again, barely). Not bad for a little scope.

A small 150 mm ... 250 mm (6" ... 10") dob will show you more. When seeing is good, you see several belts on Jupiter, and even individual curls on the big belts. Saturn's rings start to show the divisions, such as Cassini. These days Mars is at opposition - such a scope will show you the polar ice cap and the major land features. M13 fills the eyepiece with a dust of stars. The Orion Nebula looks big and clear even in the city. The Ring Nebula is visible, etc. Under a very dark sky, you can see (barely) the 3C 273 quasar, at 2.4 billion light years.

A mid-size dob (12" ... 18") will show you even more. More details on the planets, when seeing permits. More faint fuzzies (galaxies, nebulae, star clusters) especially out and far away from the city.

A large dob (20" and over) is an amazing instrument. On those very rare occasions when seeing cooperates, Jupiter at 1000x in such a scope is unforgettable. Also, under a very dark sky out in the desert, point it anywhere and the sky is full of faint fuzzies.

So, which size should you choose? It depends on price, maintenance, and ease of use (size, etc). Let's talk about maintenance.

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As a dob owner, there are two things you must never forget: collimation, and cooling.

Collimation

It means "keeping all the optics aligned and centered". The mirrors will move around a bit, and in time will lose collimation. As a dob owner, you have to collimate the scope once in a while (and every time before you observe high-resolution targets such as planets). Once you learn how to do it, it takes about 2 minutes, it's super-quick and easy.

This is why collimation is important:

http://legault.perso.sfr.fr/collim.html

This is an introduction to dobsonian collimation:

http://www.cloudynights.com/documents/primer.pdf

Here are some simpler collimation techniques, which you can use until you acquire all the tools:

http://www.etnaastros.com/documents/No_Tools_Collimation.pdf

http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/

Also see the manual that comes with the scope.

Cooling

If you keep the scope in the house or garage, and then take it out for observation, the primary mirror will be much hotter than the night air. As a result, air convection will appear on the mirror and distort the image - just like the shimmering you see in the summer along the hot walls of a house under sunlight. The mirror is too "hot" for the ambient.

This is what happens:

http://www.garyseronik.com/?q=node/55

http://www.garyseronik.com/?q=node/69

To combat convection, take the scope outside at least 1 hour prior to observation, and let it cool off. Very small dobs (up to 6" or so) will probably work fine just with passive cooling. Small-medium and above (10" or above) basically require a fan on the primary mirror (forced cooling).

Always collimate and cool off the scope before you do an observation requiring high resolution (when observing planets, the Moon, double stars). If you don't need a lot of high resolution (like when observing galaxies, nebulae, star clusters) then collimation and cooling are not so important.

If you keep your dob collimated and cooled, it will provide great views of the planets. There is this myth out there that dobs "are not great for planets", and refractors are somehow magically "so much better for planets". This is not true. The myth exists only because of the large number of people out there who keep their dobs in terrible shape; those poor scopes underperform by a very large margin. Don't be one of those folks.

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Let me just draw a line in the sand for you: An 8" f/6 dob is a good start overall. Why is that?

At 8", it is big enough that it will be seeing-limited most of the time. In other words, it is so big, atmospheric turbulence will limit the resolution more often than the size of the scope.

It is big enough to show you some faint fuzzies even from the city. All the Messier objects should be visible no matter where you are.

It is not too big to fit on the back seat of a sedan car, if you want to drive to a place with a darker sky. It is not too big to move around in the backyard easily.

The mirror is not so big and thick as to mandate active cooling (fan on the mirror); you can pretty much get by with passive cooling only - take it out at least 1 hour before observation. However, a fan may improve things a bit anyway (you need to experiment with that, but it's not a priority at this size).

At f/6 it's not too hard on the eyepieces, and it's not too hard to collimate. A shorter scope (f/5 and below) produces a steeper light cone; you need higher quality (and price) eyepieces to use it, and collimation requires much better precision.

Finally, the price is right - a few hundred dollars, which is affordable for quite a few people. Take care of it and it will last you a lifetime.

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Now, if you can afford it and think you can deal with the steeper learning curve, could you buy a bigger instrument? Sure. 16", 18" - the wallet is the limit. Such big dobs are made differently - they are called "truss dobs". They don't have a big tube where everything is contained. Instead, the tube is replaced by a truss made of a bunch of poles - looks like the arm of a crane machine.

When you're done observing, you disassemble it. It all fits in a sedan car. It's not too heavy for transport, because you handle each part separately.

However, such an instrument is usually f/5 or less. Collimation is much more difficult. You need a coma corrector (like a Paracorr) to fix coma at the edge, and that's additional money. You need better eyepieces. Cooling doesn't work unless you use active cooling (fan). And it's just more expensive overall.

Can you start on a large dob like that? It's not impossible, but it requires more money and a much more significant commitment. This is not a toy that you can discard after two week-ends.

fa.functional analysis - infinitely many linear equations in infinitely many variables

The systems of this kind are fairly common in applications. For example, they naturally appear when solving boundary value problems for linear partial differential equations using the method of separation of variables.

Predictably, the problem is not meaningful for any sequences {$a_{nm}$}, {$b_m$}, but only for sufficiently well-behaved ones. If, for example, you were to consider systems of the form
$$ x_n+sum_{m=1}^{infty}a_{nm}x_m=b_n,quadmbox{such that}quad sum_nsum_m a_{nm}^2<infty quadmbox{ and }quad sum_nb_n^2<infty, $$
then this system possesses a unique solution in the Hilbert space $l_2$ such that $sum_n x_n^2<infty$ (assuming that the problem is not singular, i.e. that $det(I+A)ne0$). These requirements are too restrictive for some applications, hence there is a body of literature concerned with various kinds of regularity conditions involving {$a_{nm}$} and {$b_m$}, weaker than above, which ensure the well-posedness of the problem and enable numerical solution of such systems (which is usually done by truncation; see the appropriate accuracy estimates in F. Ursell (1996) "Infinite systems of equations: the effect of truncation", Quarterly Journal of Mechanics and Applied Mathematics, 49(2), 217--233).

One good old book that discusses these systems in some detail was written by By L. V. Kantorovich and V. I. Krylov and is called "Approximate methods of higher analysis" (New York: Interscience Publishers, 1958).

orbit - Is Earth's orbital eccentricity enough to cause even minor seasons, without axial tilt?

Very cool question. I want to get into a little bit of detail here because otherwise there would be a one-paragraph answer, and I don't think that would cut it. So here goes.

The planets in the solar systems have orbits with pretty low eccentricities (see this for more eccentricity values). At the upper end is Mercury, with an eccentricity of 0.2056. At the lower end is Venus, at 0.00677. Earth is in between but moderately low, at 0.0167. The distance between perihelion and apelion is 5 million kilometers - in an orbit with an average radius of about 150 million kilometers. Note, though, that eccentricities are always changing.

We would certainly have "seasons" if there was no axial tilt, but they most likely would not be dramatic. As Wikipedia says

Because of the increased distance at aphelion, only 93.55% of the solar radiation from the Sun falls on a given square area of land then at perihelion.

But on this page, it says

Orbital eccentricity can influence temperatures, but on Earth, this effect is small and is more than counteracted by other factors; research shows that the Earth as a whole is actually slightly warmer when farther from the sun. This is because the northern hemisphere has more land than the southern, and land warms more readily than sea.

So more land means that the planet can absorb heat better, which counteracts the change in distance.

But I think that's pretty boring. Don't you? So let's calculate how eccentric Earth's orbit would have to be for it to have seasons without axial tilt. From the chart here we can see that the highest average hemispherical temperature is 22 degrees Celsius, while the lowest is 8 degrees Celsius. What about other planets?

There's a steady downward trend as we get further away from the Sun (Venus is an exception because of its runaway greenhouse effect). For Earth, the average surface temperature is 287 (15 degrees C). So our maximum would be 295K and our minimum would be about 281 K (Kelvin = Celsius + 273). Compare that to the other planets.

I made a graph, which unfortunately I cannot paste here, that shows that to have a surface temperature of 295 K, the planet would have to be - well, not much closer than we are. Relatively, that is. Roughly 0.975 AU from the star. To have a surface temperature of 281 K, it would have to be at 1.05 AU from the star. This neglects, by the way, the different levels of land/sea absorption.

So, actually, if the Earth's orbit had an eccentricity of about 0.35, it could have some moderate seasons.

I hope this helps.

Note: I used this eccentricity calculator for the sake of accuracy and time. I would still be obliged if anyone could check my overall calculations for this answer.

planet - Is it Possible for Planetary Bodies To Exist In Close Proximity Without Adverse Effects?

Have a look at this question (Help in determining the features of an unusual, fictional star system) for one possible hypothetical scenario in which such a planet (or a satellite) could exist.

For a satellite of a giant planet, there will be tidal forces, sure, but after a few hundred million years of formation of the system, it is likely that there will be tidal locking due to the dissipative forces within the satellite, which will ensure that the effects of tides are minimal after that (or, in other words, the effects are 'static'). On a related note, I do find it interesting that Io isn't tidally locked to Jupiter yet, but this is a different question altogether.

On the other hand, I believe it actually helps to have a Jupiter-sized body around as it drives meteoroids away from the moon you live on, making life a little safer (this might be wrong, but I think this is how it will work out; also, this does not help if 'panspermia hypothesis' turns out to be correct).

Having multiple planets nearby, as shown in the picture, is a little more difficult due to many body systems not being very stable. You could possibly have a binary planetary systems and life in the moons at Lagrangian points, but it is very likely that such systems will lose some of their bodies due to perturbations from other planets or similar second order effects due to them being unstable. You can have a moon revolving around the binary planetary system some distance away as an alternative, which would be fairly stable, I guess. However, with the current planet formation hypotheses, it seems very unlikely that you will have large planets near each other, since they won't be formed in the protoplanetary discs in a way that they don't coalesce to form a single larger planet.

Also try this: http://www.stefanom.org/spc/ . You'll soon realize that having many planets close by, in a stable system is difficult.

analytic number theory - Primes as the first coefficient of a reduced indefinite quadratic form

I recommend a book by Duncan A. Buell called "Binary Quadratic Forms."

First, we discard the case where $d$ is a square. In such a case the forms represent entire arithmetic progressions. For example, with $x^2 - y^2$ and $d = 4$ we get
$ (n+1)^2 - n^2 = 2 n + 1.$ Or, with $x y$ and $d=1,$ we have $n cdot 1 = n.$

Note that we always require $$ d equiv 0,1 pmod 4.$$

For what primes $p > 0$ is there any form, reduced or not, having $p$ as a "diagonal" coefficient? With odd $p$ that does not divide $d,$ this answer is essentially quadratic reciprocity. We are demanding
$$ beta^2 equiv d pmod p $$ If we can solve this, that is $(d | p) = 1,$ we can choose either $ b = beta$ or
$ b = beta + p$ to arrange
$$ b^2 equiv d pmod {4p}. $$ But this is the condition $$ b^2 = d + 4 p c, $$
or $ b^2 - 4 p c = d .$

Note that we also have the form $$ (-p)x^2 + b x y + (-c) y^2 $$ with the same discriminant. So as long as you do not ask whether the two forms are equivalent we are in good shape.

Finally, you asked about "reduced" forms, which is to say coefficients
$$ langle a,b,c rangle $$ and discriminant $d$ with
$$ 0 < b < sqrt{d}, ; ; mbox{and} ; ; sqrt{d} - b < 2 | a | < sqrt{d} + b. $$
Here we have Lagrange's theorem that any represented number $n$ occurs as a coefficient of $x^2$ in a reduced form if $$ | n | < ; frac{1}{2} ; sqrt{d}, $$ so you have a simple answer for small primes. It is a bit of a toss-up if you have
$$ frac{1}{2} ; sqrt{d} < p < ; sqrt{d} .$$ Here I suggest creating a form and then checking the entire cycle of reduced forms in its equivalence class. The recipe for doing exactly that is on pages 21-23 of Buell.

EDIT: I'm afraid I was not sufficiently cautious as relates to primes that divide the discriminant. It is true that 2 is represented when $ d equiv 1 pmod 8$ and not when
$ d equiv 5 pmod 8.$ But as soon as we have even discriminant there is need for care. The trouble is the existence of imprimitive forms, $$ langle a,b,c rangle $$ with $$ gcd(a,b,c) neq 1 .$$ What follows is from page 75 in Buell. If $ d equiv 0 pmod {16}$ then 2 is not represented by a primitive form, but if $ d equiv 8 pmod {16}$ it is, by the class of the primitive (but not reduced) form
$$ langle 2,0,frac{-d}{8} rangle .$$ If $ d equiv 4 pmod {16}$ then 2 is not represented by a primitive form, but if $ d equiv 12 pmod {16}$ it is, by the class of the primitive (but probably not reduced) form
$$ langle 2,2,frac{4-d}{8} rangle. $$ Alright, now that I see Buell's Theorem 4.24, the case of odd primes $p$ dividing the discriminant is comparatively clean. If $$ p^2 | d$$ then only imprimitive forms represent $p.$ If $$ p parallel d$$ then $p$ is represented by a primitive form, either $$ langle p,0,frac{-d}{4 p} rangle $$ if $d$ is even, or $$ langle p,p,frac{p^2-d}{4 p} rangle $$ if $d$ is odd. My comments about the size of $p$ and reduced forms still apply.

galaxy - Can we see individual stars in other galaxies?

Yes, Edwin Hubble did that for the first time in 1919. Before that time, it was thought that the galaxies we can observe were just nearby gas nebulae located inside our Milky Way. But Hubble was able to resolve the nearby galaxies like the Andromeda nebula into individual stars. By measuring the brightness of so-called Cepheid variable stars, he was able to calculate the distance to the Andromeda galaxy. Here one uses the fact that the total power radiated by the star is related to the period of the brightness oscillations, so by observing such stars you can deduce the distance to these stars and hence the distance to the galaxy. But later it was found that there were two different types of Cepheid stars and the wrong relation had been used; the distances were actually about twice as large.

data analysis - Any freely available large stellar spectra catalog?

For a non-astronomy personal project I would like to have a large set of optical stellar spectra combined with absolute magnitude when available. I search for a dataset that contains 10k - 10M objects, with total compressed size below 100Gb (ideally below 10Gb), is freely available via ftp/http/rsync and machine-readable. Presence of most 'naked eye visible' stars and their individual names in the set is desired, but not required.

I know that some digital survey data are freely available via FTP, so maybe some reasonable spectral data catalog exists as well?

Note, I'm planning to download the data for local processing, so retrieving them through on-line forms (or even automated http interfaces) is undesirable.

The best I was able to find is this catalog . Dropping aside that FITS is not something I'm used to (this apparentrly can be rectified) it is too small. I'd like at least 1k stars, 10k preferable. There is apparently a lot of surveys in 0.1-10k objects range that are focused on specific type of stars (say, nearby M-class dwarfs) and stellar libraries containing spectral data on representative objects from various classes. However I'd like a 'representative set' here, which contains stars of various nature in proportion similar to their 'natural abundance' in some region. An example would be the set of stars of solar neighborhood above, but again it's too small.

ag.algebraic geometry - Proving existence of non-special divisors of a given degree d on compact Riemann surfaces

I have a simple question. Let $C$ be a compact Riemann surface of genus, say $g >= 2$, to avoid silly cases.

I think it should be true, but I want to prove the following concretely:

"there exists a divisor $D$ on $C$ of degree $g-1$, that is non-special."

(For those who do not know what special divisors are: a divisor is called special if it has $h^0 (D) >0$ and $h^1 (D) >0$.)

Notice that by the Riemann-Roch, for this degree $g-1$ case we immediately have $h^0 (D) = h^1 (D) = 0$. This is, in fact, equivalent to $D$ being non-special, when $deg D = g-1$.

Is there an interesting (or any) way to prove this? I believe it should be fairly easy, and maybe I am very dumb so that I can't immediately produce a proof.

More generally, if this is possible, if the degree is a given $d$, when do we see that there exists a non-special or special divisor of given degree $d$ on a given compact Riemann surface?

What is a singularity? What is at the center of a black hole? Specifically regarding space-time

This is more of a question for the Physics stack, but I'll give it a shot, since it's fairly basic.

You need to understand something before we begin. The theoretical framework we have to gauge and answer this sort of thing is called General Relativity, which was proposed by Einstein in 1915. It describes things such as gravity, black holes, or just about any phenomena where large densities of mass or energy are involved.

There's another chapter in Physics called Quantum Mechanics. This describes, usually, what happens at very small scales - things that are super-tiny.

Both GR and QM are fine in their own way. Both are tested against reality and work very well. But they are not compatible with each other. Meaning: you cannot describe a phenomenon from a GR and a QM perspective, both at once. Or meaning: we don't have a coherent set of equations that we could write down, and then "extract" out of them either a GR-like view of reality, or a QM-like view.

The problem is, the center of a black hole is both very high mass density and very high gravity (and therefore right in the field of GR), and very small (and therefore "quantum-like"). To properly deal with it, we'd have to reconcile GR and QM and work with both at once. This is not possible with current physics.

We pretty much have to stick to GR only for now, when talking about black holes. This basically means that anything we say about the center of a black hole is probably incomplete, and subject to further revision.

A star dies, collapses into a black hole, what is at the center? The
star's mass compacted into the size of the plank length or something
similarly small? Is there really nothing at the center of a black
hole?, surely the core collapsed into something, just really small
right?

According to General Relativity, it collapses all the way down to nothing. Not just "very small", but smaller and smaller until it's exactly zero in size. Density becomes infinite.

You can't say "Plank length" because, remember, we can't combine GR and QM, we just don't know how. All we have here is GR, and GR says it goes all the way down.

I'm using words such as "size" (which implies space) and "becomes" (which implies time). But both space and time in the context of a black hole are very seriously warped. The "becoming" of a black hole all the way down to the zero-size dot is a reality only for the unlucky observer that gets caught in it. But for a distant, external observer, this process is slowed and extended all the way to plus infinity (it's only complete after an infinitely long time). Both observers are correct, BTW.

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EDIT:

So, when we are saying "density is infinite and size is zero at the singularity", this language applies to the unfortunate observer being dragged down in the middle of the initial collapse of the star.

But from the perspective of the distant observer, a black hole is still a chunk of mass (the original star) in a non-zero volume (the event horizon of the BH). To this observer, the density of that object is finite, and its size is definitely not zero. From this perspective, anything falling into the BH never quite finishes falling, but just slows down more and more.

Both observers are correct. So, keep in mind, when I talk about "infinite density", that's the inside observer point of view.

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What is a singularity? Is it just the warping of space time that makes
it this way?

You get a singularity whenever there's a division by zero in the equations, or when the equations misbehave somehow at that point. There are many different kinds of singularities in science.

http://en.wikipedia.org/wiki/Mathematical_singularity

In the context of a black hole, the center is said to be a gravitational singularity, because density and gravity are suggested to become infinite, according to the GR equations.

GR says: when you have a lump of matter that's big enough, it starts to collapse into itself so hard, there's nothing to stop it. It keeps falling and falling into itself, with no limit whatsoever. Extrapolate this process, and it's easy to see that the size of it tends to zero, and density tends to an infinite value.

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EDIT:

Put another way - if density becomes large enough, gravity is so huge, no other force is strong enough to resist it. It just crushes all barriers that matter raises to oppose further crushing. That lump of matter simply crushes itself, its own gravity pulls it together smaller and smaller... and smaller... and so on. According to current theories, there's nothing to stop it (QM might stop it, but we cannot prove it, because we don't have the math). So it just spirals down in a vicious cycle of ever-increasing gravity that increases itself.

Space and time are really pathologic inside the event horizon. If you are already inside, there's no way out. This is not because you can't move out fast enough, but because there's really no way out. No matter which way you turn, you're looking towards the central singularity - in both space and time. There is no conceivable trajectory that you could draw, starting from the inside of the event horizon, that leads outside. All trajectories point at the singularity. All your possible futures, if you're inside the event horizon, end at the central singularity.

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So, why the center of a black hole is called a "singularity"? Because all sorts of discontinuities and divisions by zero jump out of the equations, when you push math to the limit, trying to describe the very center of a black hole, within a GR frame.

http://en.wikipedia.org/wiki/Gravitational_singularity

Speaking in general, physicists don't like singularities. In most cases, this is an indication that the mathematical apparatus has broken down, and some other calculations are necessary at that point. Or it might indicate that new physics are taking place there, superseding the old physics.

One last thing: just because we don't have a combined GR/QM theory to fully describe the center of black holes, that doesn't mean a pure GR research in this area is "wrong" or "useless". It doesn't mean one could imagine some arbitrary fantasy taking place inside a black hole.

Astronomers these days are starting to observe cosmic objects that are very much like black holes, and their observed properties are in very close accord with what GR predicts for such things. So research in this field must continue, because it's clearly on the right track, at least in the ways we can verify today in astronomy.

What's the largest non-spherical astronomical object in the universe?

Let me see, some objects are listed because of their records. First, nebulae, which are huge clouds of dust and gas. The largest in the Milky Way's neighborhood is the Tarantula Nebula (30 Doradus, NGC 2070 or Caldwell 103).:

This is the largest nebula in the Milky Way's neighborhood, at 300 light-years (almost 100 pc). But if we are in the universal record, that would be NGC 604, which is 740 light-years (240 pc) across.:

Next, galaxies, which come in different shapes. Spirals first, and what like called2voyage has said, is NGC 262. Because he can't provide a pic, I will!:

NGC 262 is a giant Seyfert spiral, and is 1.3 million light-years across, more than two times larger than the mistakenly clamed largest spiral last January 2013, NGC 6872.:

I don't know why they claimed this one as largest, at 522,000 light-years across only is less than half as that of NGC 262.

But the hands-down king of all galaxies is IC 1101, at 5.6 million light-years across, and is the record holder since 1989.:

A size comparison of this monster:

Next, structures. aneroid has the right answer. It's name is Hercules-Corona Borealis Great Wall, some 10 billion light-years across.

ag.algebraic geometry - Unusual ray tracing

Background

Ray tracing is very common in computational geometry and the problem is then to find the point of intersection between the equation of a line and the equation of a plane in 3D.

The parametric form of the line is given by

$mathbf{p}_mathrm{line}=mathbf{p}_mathrm{a} + xi (mathbf{p}_mathrm{b}-mathbf{p}_mathrm{a})$

and the plane can be defined by

$mathbf{p}_mathrm{plane} cdot mathbf{n}+mathrm{d}=0$,

where $mathbf{p}_mathrm{plane}$ is a point on the plane and $mathbf{n}$ is the normal vector to the plane.

Combining these two equations $(mathbf{p}_mathrm{line}=mathbf{p}_mathrm{plane})$ gives a convenient expression for the desired point from

$xi=frac{-mathrm{d}-mathbf{p}_mathrm{a} cdot mathbf{n}}{(mathbf{p}_mathrm{b}-mathbf{p}_mathrm{a}) cdot mathbf{n}}$.

---

Question

I now consider the problem of finding the intsersection(s) between an ellipse and a plane in 3D. Is there an effective way to perform this without an iterative scheme?

at.algebraic topology - Oriented Cobordism Rings

There is no torsion other than 2-primary torsion in the oriented bordism ring. One has that after inverting 2, the oriented bordism ring is a polynomial algebra on generators in degrees which are multiples of 4:
$$
Omega^{SO}_*[1/2] = mathbb{Z}[1/2, x_4, x_8, x_{12}, ldots]
$$
If I remember correctly, this (and the answers to many bordism-related questions) can be found in Stong's "Notes on cobordism theory".

Telescope size to detect Ceres (Newtonian)

Ceres varies between mag 9.3 and mag 6.7 which are below the nominal naked eye limiting magnitude of ~5. The gain in limiting magnitude for something like 10x50 binoculars (~4.5 magnitudes) would bring Ceres even at its faintest above the limiting magnitude of about 9.5.

ag.algebraic geometry - What does being Analytically Isomorphic imply for classification of singularities on curves?

Analytically equivalent (over C) implies, for instance, topologically equivalent. All information items Charles Siegel mentions (number of branches, tangencies, etc, and also the classification of double points) are topological invariants so in particular they are analytic invariants and can be read off the completion indeed.

I would say that the first step to classify singularities is equisingularity, which over C is the same as topological classification. The analytic classification of is extremely complicated, even for plane curves. For each topological class there is an analytic moduli space, but it need not be irreducible, equidimensional, or even separated.

For plane curve singularities, there are a few cases where analytically equivalent and topologically equivalent are the same thing. These are the so-called simple, or Du Val, singularities, namely "A-singularities" or double points, whose equations have the form $y²-x^n=0$, "D-singularities" or $x(y^2-x^n)=0$ and $E_6:y^3-x^4=0$, $E_7:y(y^2-x^3)=0$, $E_8:y^3-x^5=0$.

A good old reference for the analytic classification of plane curves is Zariski's booklet ''Le problème des modules pour les branches planes.'' As introduction to the theory of plane curve singularities you have Casas-Alvero ''Singularities of plane curves'', and Laudal-Pfister's LNM ''Local moduli and singularities'' is also recommendable.

Is it a coincidence that both the sun and moon look of same size from earth?

The coincidence isn't so much that they appear very similar sizes from Earth, but that we are alive to see them at the point in time in which they appear very similar sizes. The moon is slowly moving away from the Earth, and at some point in the future the moon will be unable to totally eclipse the sun and conversely, if you could step far into prehistory, you would be able to see the moon with a much greater angular diameter than you see it now.

Most research I've found on the topic seem to be unavailable through my institute, however I did find one paper, "Outcomes of tidal evolution", which references results from Goldreich's research on the subject.

This qualitative description of the eventual disruption of the Earth-Moon system is confirmed by the results of Goldreich's numerical integration, which showed that the moon will recede to 75 Earth radii, when spin-orbit synchronism will be reached; then the Moon's orbit will decay steadily inward because of the influence of the Sun.

For reference, the Moon is currently at a distance of approximately 60.3 Earth radii. As such, the moon will steadily move away until synchronism would be reached, and from that point begin to recede towards the Earth due to the tidal affects of the Sun on the Earth disturbing the synchronization. It would seem that at some eventual point in the far distant future, it will return to this coincidental position once again.

Counselman III, Charles C. "Outcomes of tidal evolution." The Astrophysical Journal 180 (1973): 307-316.

What is the direction of a comet's dust tail before and after perihelion?

Comet tail always aims away from the sun, as it's torn from the comet's gas cloud by solar wind. It trails slightly behind, as the comet moves along its orbit while the gas travels directly away, but that's relatively minor - the speed at which the gas and particles are pulled away forming the tail is much higher than the comet's orbital velocity.

Your visualisation shows it trailing behind the comet as if it was moving in some gaseous/liquid medium that stops the lighter particles while letting the heavy comet head travel ahead. This is not the case - as the comet is far away from the sun, the gas cloud just forms its atmosphere and travels with the comet. As it approaches though, the intensity of solar wind increases and it pulls away the atmosphere which can't be protected against it by magnetic field or strong gravity, as the comet has neither.

spectroscopy - Can spectroscopes identify minerals?

Yes, minerals can be observed using spectroscopy on a telescope, typically from their thermal signature. We can determine what type of mineral based on the elements that make up the mineral as well as the type of crystal structure, which will give a different spectral signature

ca.analysis and odes - a^b = b^a when a is not equal to b.

Given your original function, $f(x) = x^{x+1} - (x+1)^x = 0$, you can define the Taylor Series expansion around some value v to give an expansion such that f( x) = a for some a, which in itself isn't helpful. However, there exists an inversion, using the Lagrange Inversion Theorem, which allows you to specify the a which your f( x) will return and gives you the corresponding x.
The first few terms of the series are:
$a+frac{x-f[a]}{f'[a]}-frac{f''[a] (x-f[a])^2}{2 f'[a]^3}+frac{left(3 f''[a]^2-f'[a] f^{(3)}[a]right) (x-f[a])^3}{6 f'[a]^5}+frac{left(-15 f''[a]^3+10 f'[a] f''[a] f^{(3)}[a]-f'[a]^2 f^{(4)}[a]right) (x-f[a])^4}{24 f'[a]^7}$
Remember, though: this is HIGHLY volatile when you're not near the actual value. Using 3 makes the value explode upwards, and using 2 makes it explode in the negative direction. However, a value of 2.2 will converge to the value you're looking for. You can use the Inversion Theorem to calculate more terms in the inverse series and use it to calculate your constant to arbitrary precision, given enough computational time.

Of course, this isn't what you're looking for (i.e. closed form or 'nice' series), but it's the best that I've found for this particular problem. Unfortunately, it converges so slowly with respect to computational time that it probably becomes more convenient for you to use the secant method to get any sort of good approximation.

--Gabriel Benamy

ca.analysis and odes - inverse Laplace transform of $delta_1(cdot)$

Let's try to find a function $psi(x)$ such that for Laplace transform $tilde{f}(p)=int_0^{infty} f(y) e^{-py} dy$ one has $f(x)=int_0^{infty} tilde{f}(p)psi(px)dp$ (here we do not specify classes of functions, for which this should hold).

In other words, $tilde{psi}(p)=delta_1(p)$ in the sense of distributions.

Or, Stiltjes transform $int_0^{infty} frac{psi(t)}{t+y} dt=e^{-y}, y>0$.

(just represent $frac1{t+y}=int_0^{infty} e^{-q(t+y)}dq$ and change order of integration).

So the question(s) is(are):

Does such function exist, if it exists, what are its properties, where is it written about all this stuff and so on.

How to prove that a set of facets are all the facets of a convex polytope.

The unhelpful suggestion is to perform the complete dual cone calculation (using available software) and then see if the answer matches your initial guess. Unfortunately, my pessimistic intuition is that in the general case, you cannot expect to do much better than this, even in the case where the initial guess is correct. You might get lucky optimizing in random directions in the dual space and find a hyperplane you missed, but in high dimensions there are too many directions and things are just subtle and elusive.

Three things that make the problem easier:

1. If the polytope P is known to be simple,


2. if the intersection of half-spaces Q is known to be simplicial, or


3. if the dimension is small.

The first case was covered by the answer given by Hugh Thomas: just verify that each vertex is a vertex of Q. In the second case the same verification suffices by duality, where the roles of facet-defining hyperplanes and extreme points (genuine vertices) are exchanged. In the third case you can construct the entire face lattice inductively. The hard case is when the dimension is high, which means that even if there are relatively few vertices and facets, the number of intermediate faces can be unmanageable.

Unfortunately it is not even easy in general to verify whether one of the first two conditions holds. An instructive example is the case where Q is a cube and P is obtained by deleting a pair of opposite vertices from Q (giving a flattened octahedron). In this case every vertex of P is a vertex of Q, every facet of Q restricts to a facet of P, P looks simple when checked by Q, and Q looks simplicial when checked by P, but they are not equal, P is not simple, and Q is not simplicial. (Fortunately, the dimension is low!)

Having expressed pessimism about there being any good solution, let me at least offer a bad one—likely to run much too slowly on any interesting example—that essentially does construct the face lattice (inefficiently) as suggested for low dimensions. We assume that it has already been verified (not difficult) that every vertex defining P does lie within Q (and therefore that every hyperplane defining Q lies outside the interior of P) but for the purposes of induction we will not insist that every every "vertex" is an extreme point of P or that every hyperplane gives a facet of Q; when we encounter such redundancies we will silently discard them for the purposes of that stage of the algorithm. (On the other hand, we do require that vertices are distinct and hyperplanes are distinct, and we delete repetitions before doing anything else.) What we will verify instead is that every vertex of Q is in the list for P, and that every facet of P comes from the list of hyperplanes for Q. If that ever fails, we report it and quit.

Firstly, every polygon is both simple and simplicial, which makes the case of two dimensions easy: Recursively eliminate any vertex of P that does not lie on two lines of Q, or any line of Q that does not contain two vertices of P. If anything is left when you are done (again, assuming P was contained in Q), they were always equal.

Now, suppose you have a solution in dimension d with which you are happy. In dimension $d+1$, you do as follows: For each hyperplane H in turn, identify the set of vertices incident to it, and verify that they span it affinely. (Otherwise H is redundant, so just continue on to the next hyperplane.) The convex hull of these vertices defines a polytope P' of dimension d within H, and the (largely redundant) intersection of all other halfspaces with H defines a polytope Q' which contains P'. The polytopes P and Q are equal if and only if P' and Q' are equal for every non-redundant hyperplane H.

Lather, rinse, repeat.

A learning roadmap for Representation Theory

I second the suggestion of Fulton and Harris. It's a funny book, and definitely you want to keep going after you finish it, but it's a good introduction to the basic ideas.

You specifically might be happier reading a book on algebraic groups.

While I third the suggestion of Ginzburg and Chriss, I wouldn't call it a "second course." Maybe if what you really wanted to do was serious, Russian-style geometric representation theory, but otherwise you might want to try something a little less focused, like Knapp's "Lie Groups Beyond an Introduction."

If you want Langlandsy stuff, then Ginzburg and Chriss is actually a bit of a tangent; good enrichment, but not directly what you want, since it skips over all the good stuff with D-modules. Look in the background reading for the graduate student seminar we're having in Boston this year: http://www.math.harvard.edu/~gaitsgde/grad_2009/

nt.number theory - Quadratic Diophantine equations solver

This function in Mathematica find them all:

f[n_] := Reduce[ Total[Table[x[i]^2, {i, 1, n}]] + 1 == x[0]^2 && 
             (Table[x[i], {i, 1, 3}] /. List -> LessEqual), 
              Table[x[i], {i, 0, n}], Integers]   

Invoke with:

f[1], f[3] ... etc

ag.algebraic geometry - Is there an analogue Beilinson-Bernstein localization for quantized enveloping algebra

[Edit: removed an attribution after Shizhuo's correction]

Kremnizer gave a nice course where he worked through the examples of G=SL_2, G/B=CP^1 in complete detail. I have some incomplete notes if you want to email me (I'd rather not post them online since they are still being revised).

To answer your question briefly about what the notion of quantum differential operators are for the flag variety, here is a rough outline:

1) First, define quantum differential operators on G. This is done by constructing the so-called Heisenberg double (just another name in this specific situation for the semi-direct product, also called smash prodcut) D(U_q,O_q) of the quantum group U_q with its dual Hopf algebra O_q, where U_q acts on O_q by the left-regular action (X.f)(y):=f(S(X)y), here X is the antipode.

2) O_q(G) has a sub-algebra called O_q(B) which is a quantization of the functions on the Borel.

3) In sufficiently nice cases in algebraic geometry, one can identify D(X/G)-modules where X is some variety with G-action as D(X)-modules M, together with an O(G) co-action, and a compatibility condition. The case G/B is such a situation classically, so one defines D_q(G/B)-modules to be D_q(G)-modules with a O_q(B) co-action plus compatibility.

By the way, there are some papers of Varagnolo and Vasserot, notably http://arxiv.org/abs/math/0603744, which discuss D_q(G) and might introduce you to some tricks people use in the area.

exposition - Introductory text for the non-arithmetic moduli of elliptic curves

I'm looking for an introduction to the non-arithmetic aspects of the moduli of elliptic curves. I'd particularly like one that discusses the $H^1$ local system on the moduli space (whether it's $Y(1)$ or $Y(2)$ or whatever doesn't matter) from the Betti point of view ($SL_2(Z)$, representations of the fundamental group, etc) and the de Rham point of view (Picard-Fuchs equation, hypergeometric functions, etc). This is for a student who has taken classes in algebraic topology and complex analysis and who is just learning algebraic geometry, so I would prefer something that's as down to earth as possible, not a full-blown sheaf-theoretic treatment (i.e. with $R^1f_*$, D-modules, etc). This is a beautiful classical subject, so I can't believe there aren't really great expositions out there, but I can't think of even one!

[Edit 2010/01/21: Thanks to every who suggested references below. I'll probably suggest Clemens's book and Hain's notes. But after looking at them, I realize that I'd really like something even more basic, with no algebraic geometry (cubic curves) and no holomorphic geometry (Riemann surfaces). I just want the moduli spaces of homothety classes of lattices in C (maybe plus some level structure), viewed first as a topological space and later as a differentiable manifold, together with the H^1 local systems, viewed first as a representation of the fundamental group and later as a vector bundle with connection. Probably this is so easy that no one ever bothered to write it down, but on the chance that that's not the case, consider this a renewed request in more precise form.]

at.algebraic topology - Whitehead products on manifolds

I guess that by the Whitehead Lie algebra, you mean the homotopy group Lie algebra $pi_*(Omega X)simeq pi_{*-1}(X)$ maybe tensored by the reals $R$. In that case there is a theorem of Felix-Halperin-THomas, called the dichotomy theorem which tells you that either this Lie algebra is finite-dimensional (and the space is said to be "elliptic"), or it is very big in the sense that the ranks of $pi_k(X)$ grows exponentially with k (and the space is then called "hyperbolic"). If the Euler characteristic of the manifold is negative then the space is always hyperbolic/ Moreover when the space is hyperbolic the Whitehead Lie algebra is very far from being abelian: actually its radical is finite dimensional. Therefore any manifold with negative euler characteristic has an non abelian infinite dimensional homotopy Lie algebra.

To generalize what Ryan says, actually any connected sum of two simply connected manifolds $M$ and $N$ is hyperbolic unless the cohomology of both $M$ and $N$ are truncatated polynomial algebras on a single genrator (like the sphere or $CP(n)$). In particular the connected sum of 3 or more closed manifolds not having the rational homotopy type of a sphere is hyperbolic.

Another example of a non abelian Whitehead Lie algebra but finite dimensional, is the one associated to a manifold $M$ obtained as an $S^5$-bundle with base $S^3times S^3$ and where the euler class of the bundle is the fundamental class of the base (or any non zero multiple of it). In that case the Whitehead rational Lie algebra $pi_*(M)otimes Q$ is of dimension $3$ with basis $x,y,[x,y]$ where $x$ and $y$ are in degree $3$ and $[x,y]$ is in degree $5$. Thus this manifold M is elliptic. Interestingly enough, the cohomology algebra of M is isomorphic to that of the connected sum $W$ of two copies of $S^3times S^8$, but $W$ is hyperbolic.

co.combinatorics - Existence of a zero-sum subset

For every finite set of real numbers, $S ={a_1,a_2,...,a_n}$ which satisfies the condition of the problem, there corresponds a directed graph $G_S$ with the following construction:

Each element of $S$ is associated with a vertex in $G_S$, according to the rule that if $a_i in S $ then $ V_i in G_S$. A vertex $V_i in G_S$ is connected to another vertex $V_k in G_S$ by an edge $E_j$ if and only if $a_i + a_j = a_k$. Clearly, the existence of a closed cycle in $G_s$ with no repeat edges corresponds to a zero sum subset of $S$.

A useful fact, which I refer to as the "vertex replacement trick", is that if $V_i rightarrow(E_j)rightarrow V_k$, then by the defining property of $S$, we also have $V_j rightarrow(E_i)rightarrow V_k$.

I will say that a path $P$ is "well behaved" if no index of $S$ appears more than once in $P$, unless it occurs in an adjacent vertex/edge pair of the form $ cdots rightarrow V_i rightarrow (E_i) rightarrow cdots$

Choose an arbitrary vertex $V_a in G_S$, and choose an arbitrary incoming edge to $V_a$, say $E_b$. Then the initial path is:

$$P=V_c rightarrow (E_b) rightarrow V_a $$

Unless $0 in S$, then every possible initial path is well behaved.

Add one vertex/edge pair to $P$, if $P$ is still well behaved then repeat until it is not. Due to the defining property of $S$, it's always possible to find another vertex to add to any given path, and since $S$ is finite, this process must eventually terminate.

$Claim:$ If $P$ is well behaved, but $P'=V_y rightarrow (E_x) rightarrow P$ is not, then $P'$ either contains a closed cycle with no repeat edges, or it can be transformed into one that does by applying the vertex replacement trick.

$Proof:$ If the index $x$ does not appear in $P$ but $y$ does, then we can use the vertex replacement trick to ensure that $V_y in P$, and thus obtain a closed cycle, which by virtue of $P$ being well behaved, is guaranteed to have no repeat edges:

$$V_y rightarrow E_x rightarrow cdots rightarrow V_y$$

If the index $y$ does not appear in $P$ but $x$ does, then similarly, we use the vertex replacement trick to obtain:

$$V_x rightarrow E_y rightarrow cdots rightarrow V_x$$

In the case that $x=y$, we obtain:

$$V_x rightarrow E_x rightarrow cdots rightarrow V_x$$

Otherwise, in the case that the indices $x$ and $y$ both appear in $P$, choose the element with index of $x$ or $y$ which occurs first in $P$, and if that element is an edge, turn it into a vertex using the trick. Then remove all the elements of $P$ which occur after that vertex, and return to one of the previous cases where only one of the indices appears in $P$.

EDIT: This is a massive update/rewrite of an unfinished answer of mine from years ago. My apologies if this makes the old comments no longer relevant.

solar system - What are the current observational constraints on the existence of Nemesis?

Nemesis is a hypothetical companion to the Sun on a very eccentric, long-period orbit. The star supposedly returns every few tens of millions years, driving comets into the inner solar system and causing extinction events. Given our very stringent observational limits from infrared surveys (such as WISE), is its existence definitively ruled out?

Precise relation between prime number theorem and zero-free region

I was wondering about the following, and I was hoping that some expert here could answer, rather than me indulging in a search for a needle in the haystack of formulas in books like Titchmarsch.

Notation:

• $zeta(s)$ is the Riemann zeta function.


• $f : mathbb R^+ rightarrow (0,1/2)$ is such that $zeta(s)$ does not vanish between $s = 1+it$ and $s=1 - f(t) + it$.


• $pi(x)$, $Li(x)$ as in wikipedia.

Assuming the above data, suppose the version of the prime number theorem that can be proven is:

$$ pi(x) = Li(x) + Oleft(G(x)right) $$

Question:

Can G(x) be given a closed form expression showing its precise(if and only if) dependence on $f(t)$?

Heuristics: When $f = 0$, $G(x) = x mathrm{e}^{-asqrt{ln x}}$ and when $f = 1/2$, $G(x) = sqrt x ln x$. So possibly there would be a term like $x^{1-f(x)}$ in a putative expression for $G(x)$.

dg.differential geometry - Twisting an L-infinity-morphism with "non-associated" Maurer-Cartan elements

Background

Suppose we are given $L_infty$-algebras $(g,Q)$ and $(g',Q')$ and an $L_infty$-morphism $F$ from $(g,Q)$ to $(g',Q')$. Furthermore, we have a Maurer-Cartan element $pi$ of $(g,Q)$.
One can twist $(g,Q)$ with the Maurer-Cartan element $pi$ and obtains a new $L_infty$-algebra that we call $(g,Q_pi)$. Furthermore, we can construct a Maurer-Cartan element $pi'$ of $(g',Q')$ by the formula

$pi' = sum_{n=1}^infty frac{1}{n!} F_n(pi, ldots , pi)$,

where $F_n$ is the $bigwedge^n g rightarrow g'$-part of $F$. I don't know whether there is a (better) term, so I call $pi$ and $pi'$ associated Maurer-Cartan elements

One can twist the morphism $F$ with the Maurer-Cartan elements $pi$ and $pi'$ and obtain an $L_infty$-morphism $F_pi$ from $(g,Q_pi)$ to $(g',Q'_{pi'})$. The references I found are
Dolgushev: A Proof of Tsygan's Formality Conjecture for an Arbitrary Smooth Manifold (section 2.4) and Yekutieli: Continuous and Twisted L_infinity Morphisms
(section 3).

Question

Given $L_infty$-algebras $(g,Q)$ and $(g',Q')$, an $L_infty$-morphism $F$ from $(g,Q)$ to $(g',Q')$, a Maurer-Cartan elements $pi$ of $(g,Q)$ and a Maurer-Cartan element $omega$, $omeganeq pi'$, of $(g',Q')$. Can one construct an $L_infty$-morphisms between $(g,Q)$ twisted with the Maurer-Cartan element $pi$ and $(g',Q')$ twisted with the Maurer-Cartan element $omega$, where the Maurer-Cartan elements are not "associated"? I.e. can one construct an $L_infty$-morphism between $(g,Q_pi)$ to $(g',Q'_omega)$?

ag.algebraic geometry - Verdier duality via Brown representability?

Hello,

I wonder if the techniques introduced in Neemans paper:
"The Grothendieck duality theorem via Bousfield's techniques and Brown representability "
can be used to establish Verdier duality. More precisely:

Consider the unbounded, derived category $D(M)$ of $mathbb{Q}$ vector spaces on a compact complex manifold $M$ . I would like to show that $Rf_!$ has a right adjoint.
In order to use Brown representability one has to show that $D(M)$ is compactly generated.
i.e. there exists a set of objects $c_i$ that commutes with direct sums:
$$Hom(c_i,bigoplus x_j)=bigoplus Hom(c_i,x_j)$$
and generates $D(M)$:
$$forall c_i Hom(c_i,x)=0 Rightarrow x=0$$

My problem is that i can't find such a set of generators. I first tried shifts of

$$i_*mathbb{Q}$$ where $i$ is the inclusion of an open subset. However these do neither commute with coproducts nor are they generators (they can not see sheaves without global sections).
My second try was shifts of
$$i_!mathbb{Q}$$
these are generators, but again they do not seem to respect coproducts.

Can someone give a set of compact generators? Or is this approach to Verdier duality doomed anyway?

nt.number theory - Variants of Waring's problem

Waring's problem (previously asked about here) asks, for each integer $k ge 2$, what is the smallest integer $g(k)$ such that any positive integer can be written as a sum of $g(k)$ kth powers. We have $g(2) = 4$, $g(3) = 9$, etc. A (harder) variant asks what the smallest integer $G(k)$ is such that all sufficiently large integers can be written as a sum of $G(k)$ kth powers.

I have two related questions:

1. What is known if we relax the condition ``any positive integer'' and only require a positive-density subset? More precisely, we look for the smallest $g'(k)$ for which there is some $S subset mathbb{Z}_{>0}$ of positive density such that any $x in S$ can be written as $g'(k)$ $k$th powers. Then we have $g'(2) = 3$, while $G(2) = 4$; and $g'(3) = 4$, while it is only known that $4 le G(3) le 7$. Is anything known about $g'(k)$ for k = 4,5, or larger?




2. For fixed k, is there an efficient algorithm that, given n, writes n as a sum of $g(k)$ kth powers? What about decomposing n into the minimal number of kth powers for that n? (Here `efficient' means polynomial in log(n).)

Edit: Wikipedia says that ``In the absence of congruence restrictions, a density argument suggests that G(k) should equal k + 1.'' So perhaps this is the answer to (1)?

reference request - nth-order generalizations of the arithmetic-geometric mean

You want Maclaurin's inequality. Given $n$ positive numbers $a_1, a_2,dots,a_n$,
write
$$
(x+a_1)(x+a_2)cdots(x+a_n) = x^n + S_1x^{n-1} + cdots + S_{n-1}x + S_n,
$$
so $S_i$ is the $i$-th elementary symmetric function of $a_1,dots,a_n$.
For $i = 1,dots,n$, set $A_i = S_i/binom{n}{i}$. When
$n = 2$, $A_1 = (a_1+a_2)/2$ and $A_2 = a_1a_2$. Maclaurin's inequality is that
$$
A_1 geq sqrt{A_2} geq sqrt[3]{A_3} geq cdots geq sqrt[n]{A_n},
$$
where the inequality signs are all strict unless $a_1,dots,a_n$ are all equal.
The inequality of the outer terms, $A_1 geq sqrt[n]{A_n}$, is the arithmetic-geometric mean inequality for $n$ positive numbers.

From a list of $n$ positive numbers $a_1,dots,a_n$ we have produced another list of $n$ positive numbers $A_1,sqrt{A_2},dots,sqrt[n]{A_n}$. The construction can be repeated.

Theorem: All the terms in the list tend to the same limit.

Off the top of my head I can't recall a reference where this is proved.
It was studied by Meissel in 1875 for $n = 3$.

For example, if we start with the three numbers 1, 2, 3 then
after 4 iterations the three numbers we get all look like 1.9099262335 to 10 digits after the decimal point.

[Edit: Here is a proof of the common limit, based on Will Sawin's first comment below to my answer. Order the numbers $a_1,dots,a_n$ so that $a_1 geq cdots geq a_n > 0$. By Maclaurin's inequality (or really just the arithmetic-geometric mean inequality)
$A_1 geq sqrt[n]{A_n}$ and we will bound
$A_1 - sqrt[n]{A_n}$ from above in terms of $a_1 - a_n$ by bounding $A_1$ from above and $sqrt[n]{A_n}$ from below using just $a_1$ and $a_n$. To bound $A_1$ from above,
$$
A_1 = frac{a_1 + cdots + a_n}{n} leq frac{(n-1)a_1 + a_n}{n} = a_1 - frac{a_1 - a_n}{n}
$$
and to bound $A_n$ from below we write $A_n = a_1cdots a_n geq a_n^n$, so
$$
sqrt[n]{A_n} geq a_n.
$$
Therefore
$$
0 leq A_1 - sqrt[n]{A_n} leq left(a_1 - frac{a_1 - a_n}{n}right) - a_n = left(1 - frac{1}{n}right)(a_1 - a_n).
$$
Start from an $n$-tuple $(a_1,a_2,dots,a_n)$ which is ordered so that $a_1 geq cdots geq a_n > 0$ and construct the $n$-tuple $(A_1,sqrt{A_2},dots,sqrt[n]{A_n})$ and keep repeating this,
which produces a sequence of $n$-tuples $(a_1^{(k)},a_2^{(k)},dots,a_n^{(k)})$ for $k = 0,1,2,dots$, where $a_i^{(0)} = a_i$. Let's look at the sequence of first coordinates $a_1^{(k)}$. An arithmetic mean of positive numbers is bounded above by the largest number, so $a_1 = a_1^{(0)} geq a_1^{(1)} geq a_1^{(2)} geq cdots > 0$. Therefore
the sequence $a_1^{(k)}$ converges as $k rightarrow infty$. (The limit is positive because the sequence of last coordinates $a_n^{(k)}$ is non-decreasing and $a_1^{(k)} geq a_n^{(k)} geq a_n^{(0)} = a_n$ for all $k$.) The above calculation shows
$$
0 leq a_1^{(k)} - a_n^{(k)} leq left(1 - frac{1}{n}right)(a_1^{(k-1)} - a_n^{(k-1)}),
$$
so $0 leq a_1^{(k)} - a_n^{(k)} leq (1 - 1/n)^k(a_1 - a_n)$. Letting $k rightarrow infty$ we see the sequence of last coordinates $a_n^{(0)},a_n^{(1)},a_n^{(2)},dots$ converges to the limit of the sequence of first coordinates $a_1^{(0)}, a_1^{(1)}, a_1^{(2)},dots$. Since $a_1^{(k)} geq a_i^{(k)} geq a_n^{(k)}$, each intermediate sequence $a_i^{(0)},a_i^{(1)},a_i^{(2)},dots$ converges to the same limit.
]

meteor - Do meteorites streak across the sky in the same direction?

Once while camping with my family as a child we all watched a meteor shower in the summer (in the month of August).

All the meteorites streaked across the sky in the same direction.

Obviously, this was the result of the Earth passing thru a group of meteors but I've always wondered.

Do meteorites always streak in the same direction? While I know it's not likely to be exactly the same direction, but generally speaking. Would one say meteorites always travel for example: east to west?

I thought this was a plausible fact since the Earth is travelling around the Sun in a constant direction, and additionally it's always rotating in the same direction.

While meteorites can travel in all directions. Does the Earth's gravity pull them straight down towards the surface resulting in the streak to be always in the same direction.

ag.algebraic geometry - Rationality of GIT quotients

A useful general result is the 'no-name lemma' stating that when a reductive group G acts linearly on two vectorspaces V and W 'almost freely' (that is, the stabilizer subgroup of a general point is trivial), then the GIT-quotients V/G and W/G are stably rational (that is, V/G x C^m and W/G x C^m are birational for some m and n).

Btw. Katsylo used it in the rationality of genus 3 curves you mentioned.

C;early, the following implications hold

rational ==> stably rational ==> unirational

and counterexamples to the other implications exist (Artin-Mumford for a unirational non stably rational variety and Colliot-Thelene, Sansuc and Swinnerton-Dyer for a non-rational stably rational one).

As to PGL_n : here the 'canonical' example of a vectorspace having an almost free PGL_n-action is couples of nxn matrices under simultaneous conjugation. Hence, by the NNL any other almost free GIT-quotient is stably rational to it.

Here the best result known is that when n divides 420=2^2x3x5x7 then such quotients are stably rational. For couples of matrices under simultaneous conjugation rationality is known for n<= 4 but even for the cases n=5 and n=7 only stably rationality is known. 'Retract rationality' (a lot weaker than stable rationality) is known for all squarefree n by a result of David Saltman.

ag.algebraic geometry - Degrees of etale covers of stacks

This is probably pretty basic, but as I said before I'm just beginning my way in the language of stacks.

Say you have an etale cover X->Y of stacks (in the etale site). Is there a standard way to define the degree of this cover? Here's my intuition: if X and Y are schemes, we can look etale locally and then this cover is Yoneda-trivial in the sense of http://front.math.ucdavis.edu/0902.3464 , meaning that etale locally it is just a disjoint unions of (d many) pancakes. Can we do this generally? Is there some "connectivity" conditions on Y for this to work? Is there a different valid definition for degree of an etale cover of stacks?

Yoneda-Triviality

I figured since nobody answered so far, maybe I should write down what a possible Yoneda-triviality condition could mean for stacks:

Def: Call f:X->Y (stacks) Yoneda-trivial if there exists a set of sections of f, S, such that the natural map Y(Z)xS->X(Z) is an isomorphism (or maybe a bijection?) for any connected scheme Z.

But I'm still clinging to the hope that there's a completely different definition out there that I'm just not aware of.

ag.algebraic geometry - Questions Suggested by the Parabolic Subgroup Definition

i)-ii) If the subgroup isn't closed, there's no reason in general for the quotient space to even be an algebraic variety. So for the definition to even make sense, there has to be some guarantee that $G/P$ is a variety. A standard result says that if $H subset G$ is a closed subgroup of a linear algebraic group, then the quotient $G/H$ is a quasi-projective variety. You can find these results, for example, in Borel's text on linear algebraic groups. I don't know of an example of a non-closed subgroups whose quotient both exists and is not complete.

(iii) Pretty much by definition, $G/P$ is what's known as a partial flag variety. For the classical groups, you can directly verify that you get flag-like objects in this way, and for more general groups this is taken as the definition. However, if by flag variety you mean the full flag variety, then just pick any parabolic subgroup which is not a Borel subgroup. The simplest example occurs for $G = GL_3$ where you have parabolic subgroups corresponding the variety of lines in $mathbb{C}^3$ (i.e., $mathbb{P}^2$) and its dual, the variety of planes in $mathbb{C}^3.$

(iv) I'm not sure exactly what you're asking for here. Just examples of varieties that aren't quotients of linear algebraic groups?

the sun - Why there are three equations governing blackbody emission?

Historically, two people (or groups of people) independently came up with different equations to model the blackbody equations in different parts of the spectrum. Rayleigh-Jeans law (classically derived) is valid for longer wavelengths and Wien's law (not Wien's displacement law) is valid for shorter wavelengths.

The Planck Distribution approaches the two laws in its limits, as in, for shorter wavelengths it is approximately equivalent to Wien's law and for longer wavelengths, it is approximately equivalent to Rayleigh-Jeans law. However, the quantum mechanically formulated Planck's law is accurate at all wavelengths, so it is the one that should always be used. One can use the other laws, for example, for brightness temperature definitions in radio astronomy, they use the RJ law for convenience, but that's because the wavelengths are long enough, and the approximation of the Planck's law gives the same result. (https://en.wikipedia.org/wiki/Planck%27s_law#Approximations)

Wien's displacement law only relates the peak wavelength to the temperature, which is a different law completely, though Planck's law (differentiating the function to find the maximum) also gives the same result. Though in this case, the displacement law is not approximate. It is accurate and can be used whenever you want.

nt.number theory - Density of monogenic number fields?

A recent article of Bhargava and Shankar, "Binary quartic forms having bounded invariants, and the boundedness of the average rank of elliptic curves" (http://arxiv.org/abs/1006.1002), addresses, among many other related questions, the density of monogenic cubic orders, counted by the height (slightly modified) of their invariants $I$, $J$.

This height is almost the same as the height of the discriminant of the polynomial to which they are associated with. So this answers the first case above for cubic fields.

Theorem 4.1 (Bhargava, Shankar):

Let $N_3^{(0)}(X,delta)$ (resp. $N_3^{(1)}(X,delta)$) denote
the number of cubic submonogenized rings $(C,x)$ of index $n$ with
positive (resp. negative) discriminant such that $H(C,x)lt X$ and
$nlt X^delta$, where $delta leq 1/4$. Then we have
$$N_3^{(0)}(X,delta)=displaystylefrac{4}{45}X^{5/6,+,2delta/3}+O(X^{5/6});$$
$$N_3^{(1)}(X,delta)=displaystylefrac{16}{45}X^{5/6,+,2delta/3}+O(X^{5/6}).$$

If I understand correctly, this means that the density for case (1) above (restricted to cubic fields) is in fact 0.

The article goes on to study quartic rings, but the result, Theorem4.8, is a bit more difficult to understand. But if I do, it says (a bit more than) that the density of monogenic quartic fields counted by discriminant is, once more, 0. Note that in the theorem the count is by the invariants of resolvent, which is like the discriminant of the resolvent, which is a constant factor away from the discriminant of the quartic polynomial to which it is associated with.

So, an educated guess is that the answer to (1) above is 0 for all degrees, and then in general (as stated above without degree). On the other hand, the discriminant of an $n$ degree polynomial is a multivariate polynomial, and I suspect that sieve theory can prove that for every constant degree there is a positive density (counted as in (2)) of polynomials that have discriminant with bounded squareful part. This in turn would prove that there is a positive density of polynomials giving orders of bounded index in their respective maximal orders. Taking this educated guess a little further, I suspect the answer to (2) is greater than 0, contrary to (1).

co.combinatorics - Is there a known formula for the number of SSYT of given shape with partition type?

Let $s_{lambda}$ and $m_{lambda}$ be the Schur and monomial symmetric functions indexed by an integer partition $lambda$ ($ell(lambda)$ is the number of parts of $lambda$ and $m_i(lambda)$ is the multiplicity of part $i$). By the hook-content formula we have:
$$
s_{lambda}(1^n) = prod_{uin lambda} frac{n+c(u)}{h(u)},
$$
where $c(u)$ and $h(u)$ are the content and hook length of the cell $uin lambda$.

Using $s_{lambda} = sum_{mu} K_{lambda mu} m_{mu}$ where $K_{lambda mu}$ is the Kostka number, the number of semistandard Young tableaux of shape $lambda$ and type $mu$. Then we get $sum_{mu} K_{lambda mu} m_{mu}(1^n)=prod_{uin lambda} frac{n+c(u)}{h(u)}$. This counts semistandard Young tableaux of shape $lambda$ and any type.

Does the sum $sum_{mu}K_{lambda,mu}$ have a known formula for $ell(lambda)geq 2$? This would be the number of semistandard Young tableaux of shape $lambda$ with partition type.

reference request - Expository treatment of Schubert Cells Paper

I was wondering about the paper by Bernstein, Gel'fand, and Gel'fand on Schubert Cells. This paper is fairly old(and often cited) so I figured someone must have represented this material. In particular, I was wondering if this was treated in an expository paper. More generally, I was wondering if there was a paper that explained the usefulness of the Schubert Calculus for representation theory, and even better one that talked about how Schubert Calculus came into the picture for BBD, again hopefully in an expository way.

Thanks in advance!

gr.group theory - What can be said about a group from its presentation?

Almost nothing can reliably be said about a group just from a presentation in finite time. (In fact, the abelianisation is just about the only thing one can reliably compute.) Most strikingly, there is no algorithm to recognise whether a given presentation represents the trivial group. More generally, one cannot in general solve 'the word problem' - ie, there is no algorithm to determine whether a given element is non-trivial. See Chuck Miller's survey article for details.

(Update. I inserted the word 'reliably' above in deference to Joel David Hamkins' fair comment. (Update 2. I then inserted the phrase 'in finite time' to be strictly correct, in an effort to head off further argument.) It is true that, in many special cases, there is information that can be read off from a specific presentation. This is more or less the topic of combinatorial group theory! But I want to emphasise that you can do nothing with an arbitrary presentation.)

On the other hand, there is a growing realisation that, surprisingly, if one is given a solution to the word problem (by an oracle, say) then one can compute quite a lot of information. Daniel Groves and I proved that, in these circumstances, one can determine whether the group in question is free. Nicholas Touikan generalised this to show that one cam compute the Grushko decomposition.

photography - What are the last images from the Galileo orbiter before impacting Jupiter?

NASA's Galileo orbiter at Jupiter ended its mission in 2003 by intentionally dive into the gas planet. Did it take and transmit any close up images before it ceased to function? If not, why? Are there any images from Galileo, or its separate impactor probe, which show the Jovian clouds from close enough range that their topology is discernible?

On all images I've seen, Jupiter looks like a perfect sphere because of its huge size and far distances from which it has been imaged. But if two spacecrafts have dived into it, there should be close up images of its clouds, right?

homological algebra - Hilbert Syzygy Theorem - Induction step

A proof using Gröbner bases is in Using algebraic geometry by David A. Cox, John B. Little, Donal O'Shea, Theorem 2.1.

However, I was always sure that there should be (at least in the graded case) an inductive proof along the lines of Atiyah-Macdonald's proof of Hilbert--Serre theorem, namely by induction considering the 4-term exact sequence
$$0to K_ito M_ito M_{i+1}to L_ito0$$
where $K_n$ and $L_n$ are the kernel and the cokernel for the operator of multiplication by $x_n$ (these are modules over the polynomial ring in $n-1$ variables), but something escapes me at the moment, so I just leave it here as a wish....

gr.group theory - Conjugating a subgroup of a group into a proper subgroup of itself

The following question came up in the class I'm teaching right now. There definitely exist groups $G$ with subgroups $H$ such that there exists some $g in G$ such that $g H g^{-1}$ is a proper subgroup of $H$. For instance, let $G$ be the (big) permutation group of $mathbb{Z}$ (by the big permutation group, I mean that elements of $G$ can move infinitely many elements of $mathbb{Z}$). Let $H subset G$ be the big permutation group of $mathbb{N}$ and let $g in G$ be the permutation that takes $n in mathbb{Z}$ to $n+1 in mathbb{Z}$. Then $g H g^{-1}$ is a proper subgroup of $H$.

The same sort of trick produces many examples like this. However, a feature of all of them is that $G$ is "big" in some way -- for instance, $G$ is not finitely presentable. By Higman's embedding theorem, you can embed such a $G$ into a finitely presentable group, so there exist examples where $G$ is finitely presentable. However, in all the examples I can come up with, the subgroup $H$ is not finitely presentable. I'm pretty sure that there exist examples in which $H$ is finitely presentable. Does anyone know of one? Even better, are there examples in which both $G$ and $H$ are of finite type (ie have compact $K(pi,1)$'s)?

posets - Does ⋄ generate all De Morgan algebras?

(Asked by Nathaniel Hellerstein on the Q&A board at JMM)

This question is about De Morgan algebras (see also Wikipedia), which are something like Boolean algebras, but with a different weaker sense of the complement ∼. Namely, a De Morgan algebra is a bounded distributive lattice with an involution ∼ satisfying de Morgans laws.

Let ⋄ be the four element De Morgan algebra that is not a Boolean algebra, pictured below.

    1

i      j

    0

where ∼0 = 1, ∼1 = 0, but ∼i = i and ∼j = j, so i and j and self-dual with respect to ∼. This algebra seems to express one of the fundamental differences between De Morgan algebras and Boolean algebras.

Question. Does the algebra ⋄ generate all De Morgan algebras, in the sense that every De Morgan algebra is a subalgebra of a homomorphic image of a product of ⋄?

Please see the related Birkhoff's HSP Theorem in universal algebra, concerning varieties of algebras closed under H, S, and P (homomorphic image, subalgebra and product).

(Edit: I edited the question to express the question as I understood it. I'm not sure whether the OP intended SHP as stated or HSP, which would conform with Birkhoff's theorem. Probably it was intended to take the variety generated by ⋄, that is, close {⋄} under H, S and P. The question then is whether this is equal to the class of all De Morgan algebras. Please revert if my edits are off-base.-JDH]

The ⋄ algebra can also be defined in terms of the usual 2 element Boolean algebra { f, τ } by using pairs denoted a/b, with the ∧ and ∨ operations defined coordinate-wise, but where, as mentioned by Dorais, the operation ∼ exchanges coordinates in addition to negating them, making for a "twisted square".

     1 = τ/τ

i = τ/f    j = f/τ

     0 = f/f

~(a/b)=(~b/~a)
(a/b)∧(c/d) = (a∧c)/(b∧d)
(a/b)∨(c/d) = (a∨c)/(b∨d)

Applications of Noncommutative Geometry

Charles,

a couple of reasons why a complex algebraic geometer (certainly someone who is
interested in moduli spaces of vector bundles, as your profile tells me) might
at least keep
an open verdict on the stuff NC-algebraic geometers (NCAGers from now on) are trying to do.

in recent years ,a lot of progress has been made towards understanding moduli spaces of
semi-stable representations of 'formally smooth' algebras (think 'smooth in the NC-world).
in particular when it comes to their etale local structure and their rationality.
for example, there is this book, by someone.

this may not seem terribly relevant to you until you realize that some of the more
interesting moduli spaces in algebraic geometry are among those studied. for example, the
moduli space of semi-stable rank n bundles of degree 0 over a curve of genus g is the moduli
space of representations of a certain dimension vector over a specific formally smooth algebra,
as Aidan Schofield showed. he also applied this to rationality results about these spaces.

likewise, the moduli space of semi-stable rank n vectorbundles on the projective plane with Chern
classes c1=0 and c2=n is birational to that of semi-simple n-dimensional representations of the free algebra
in two variables. the corresponding rationality problem has been studied by NCAG-ers (aka 'ringtheorists'
at the time) since the early 70ties (work by S.A. Amitsur, Claudio Procesi and Ed Formanek). by their results, we NCAGers,
knew that the method of 'proof' by Maruyama of their stable rationality in the mid 80ties, couldn't possibly work.

it's rather ironic that the best rationality results on these moduli spaces (of bundles over the
projective plane) are not due to AGers but to NCAGers : Procesi for n=2, Formanek for n=3 and 4 and
Bessenrodt and some guy for n=5 and 7. together with a result by Aidan Schofield these results show
that this moduli space is stably rational for all divisors n of 420.

further, what a crepant resolution of a quotient singularity is to you, is to NCAGers the moduli space of certain
representations of a nice noncommutative algebra over the singularity.

likewise, when you AGers mumble 'Deligne-Mumford stack', we NCAGers say 'ah! a noncommutative algebra'.

mg.metric geometry - Intrinsic metric with no geodesics

There is a very simple example of an intrinsic, complete metric space that is not geodesic (read in Ballmann's "Lectures on Spaces of Nonpositive Curvature": it is the graph on two vertices $x,y$, linked by edges $e_n$ of length $1+1/n$.

Of course it does not answer your question, but it may be possible to improve this example to one that does. Call $X_1$ the graph described above, and define $X_{n+1}$ from $X_n$
as follows: $X_n$ has a vertex $x'$ for each vertex $x$ of $X_n$, plus a vertex $v_e$ for each edge $e$ of $X_n$. For each edge $e=(xy)$ of $X_n$ we define edges $f_e^n$ and $g_e^n$ of $X{n+1}$: $f_e^n$ connects $x'$ to $v_e$ and has length $(1+1/n)$ times the original length of $e$, and $g_e^n$ does the same
but replacing $x'$ by $y'$.

Now it should be possible to construct the desired example by a limiting process. For example, take all vertices along the construction: the distance between any two of this points is constant as long it is defined, so we get a metric space. Its completion might be what you want (but I a not so sure of that after witting these lines).

linear algebra - Cubic spline of a two-variable function

There is a nice book by Les A. Piegl and Wayne Tiller, which is intended as an introduction to NURBS, but in the first chapters also provides theoretical background for uni- and bivariate Beziers und Splines.

The NURBS Book
Les A. Piegl, Wayne Tiller
Springer
ISBN-10: 3540615458
ISBN-13: 978-3540615453

gn.general topology - Is there a "natural" characterization of when X × βN is normal?

The space $Xtimesbetamathbb{N}$ is normal if and only if $X$ is normal and $mathfrak{c}$-paracompact. This follows from results of Morita (Paracompactness and product spaces, MR132525), where he generalizes Dowker's characterization of countable paracompactness.

First note that $Xtimesbetamathbb{N}$ is normal if and only if $Xtimes K$ is normal for every separable compact Hausdorff space $K$. This is because every separable compact Hausdorff space is a perfect image of $betamathbb{N}$.

Morita's Theorem 2.2 shows that if $X$ is normal and $mathfrak{c}$-paracompact, then $X times K$ is normal for every compact Hausdorff space $K$ of weight at most $mathfrak{c}$. Hence, $Xtimes K$ is normal for every separable compact Hausdorff space $K$ since these all have weight at most $mathfrak{c}$.

Morita's Theorem 2.4 shows that a space $X$ is normal and $mathfrak{c}$-paracompact if (and only if) $Xtimes[0,1]^{mathfrak{c}}$ is normal. Since the space $[0,1]^{mathfrak{c}}$ is a separable compact Hausdorff space, this closes the implication loop.

gr.group theory - 〈x,y : x^p = y^p = (xy)^p = 1〉

More generally, let $a,b,c in mathbb{Z}^+$ and define the group

$Delta(a,b,c) = langle x,y,z | x^a = y^b = z^c = xyz = 1 rangle$.

These groups were studied by von Dyck in the late 19th century and are sometimes called the von Dyck groups. The most basic fact about them is that $Delta(a,b,c)$ is infinite iff $frac{1}{a} + frac{1}{b} + frac{1}{c} leq 1$. (The groups you ask about are when $p = a = b = c$. Thus $Delta(2,2,2)$ is finite, and for $p > 2$, $Delta(p,p,p)$ is infinite.)

Perhaps the nicest way to see this is to realize $Delta(a,b,c)$ as a discrete group of isometries of a simply connected surface of constant curvature. More precisely, consider a geodesic triangle with angles $frac{pi}{a}$, $frac{pi}{b}$, $frac{pi}{c}$. Then, according to whether $frac{1}{a} + frac{1}{b} + frac{1}{c}$ is greater than, equal to, or less than $1$, these triangles live either in the Riemann sphere, the Euclidean plane or the hyperbolic plane.

Now $Delta(a,b,c)$ has as a homomorphic image the group generated by three elements $x$,$y$,$z$, each of which is the composition of reflection through two adjacent sides of the triangle. Indeed, an easy calculation shows that $x$, $y$, $z$
satisfy the relations defining $Delta(a,b,c)$, so that it must be a homomorphic image of it. (In fact the abtract group is isomorphic to the isometry group, but that is a little more delicate to show.) Now there is a corresponding tesselation of the space obtained by repeatedly reflecting copies of one fundamental triangle across each of the sides. If you consider the overgroup $tilde{Delta}(a,b,c)$ generated by the reflections themselves and not the rotations -- so that $Delta(a,b,c)$ is the index $2$ subgroup consisting of orientation-preserving isometries -- then it is immediately clear that $tilde{Delta}(a,b,c)$ acts transitively on the triangles in the tesselation. Since the Euclidean and hyperbolic plane each have infinite volume, there are clearly infinitely many triangles in the tesselation, so $tilde{Delta}(a,b,c)$ is infinite, and therefore so is its index $2$
subgroup $Delta(a,b,c)$.

In the case when $frac{1}{a} + frac{1}{b} + frac{1}{c} > 1$, this argument can be modified to show that $Delta(a,b,c)$ is finite, but in this case a reasonable alternative is brute force, since this is a well-known family of groups: the finite isometry groups of $3$-dimensional Euclidean space (namely $C_n$, $D_n$, $S_4$, $A_4$, $A_5$).

Also either of both of the families of groups $Delta$ and $tilde{Delta}$ are often called triangle groups.

real analysis - Why is this generality in Vitali's Lemma useful?

In Vitali's Lemma it uses outer measure rather than measure. What are some results that depend on it this theorem applying to sets with only outer measure rather than measurable sets?

Vitali's Lemma:
Let $E$ be a set of finite outer measure and $G$ a collection of intervals that cover $E$ in the sense of Vitali. Then given $varepsilon> 0$ there is a finite disjoint collection of intervals in $G$ such that $m^*(E - bigcup_{n=1}^N I_n) < varepsilon$.

I'm trying to learn this theorem and I keep replacing "outer measure" with "measure" and I want a reason to stop doing that.

Space time and aging - Astronomy

Einsteins general theory of relativity explains time dilation caused by gravity-emitting objects. As one experiences more gravity, time will flow slower. That means that "standing" on jupiter, which isn't possible due to the lack of surface, will cause you to move through time faster; But you do not age slower. Suppose your Lifespan is 80 years. On Jupiter you still would live 80 years but the time that has passed on earth during your 80 years would be slightly more than that. The same counts vice versa for the moon.

quivers - Matrices into path algebras

Somewhat related to this, you have the rather new field of Leavitt Path Algebras, which takes a field $K$ and a directed graph $E$, generates its extended graph $E'$ (add to $E$ its own edges reversed), and finally computes the Leavitt path algebra of $E$, $L(E)$, as the path algebra $KE'$ modulo some relations called (CK1) and (CK2), inherited from the $C*$-algebras setting.

These associative algebras provide us simultaneously with a purely algebraic analog of $C*$-algebras of graph and a generalization of the Leavitt algebras (associative algebras which do not satisfy the IBN property).

The full matrix rings over $K$ of order $n$ then arise as the Leavitt path algebras of the graphs with $n$ (consecutive) vertices and $n-1$ arrows, one between every pair of consecutive vertices.

Another simple example of Leavitt path algebra is the ring of Laurent polynomials over $K$, $K[x,x^{-1}]$, which appears associated to the graph with one vertex and a single loop.

The theory of LPAs is a beautiful one because it allows us to identify ring-theoretic properties of associative algebras from the graph-theoretic properties of their associated graphs in a visual and straightforward way.

Some references:

G. Abrams, G. Aranda Pino. "The Leavitt path algebra of a graph", J. Algebra 293 (2), 319-334 (2005). (Available at http://agt.cie.uma.es/~gonzalo/papers/AA1_Web.pdf).

P. Ara, M.A. Moreno, E. Pardo. "Nonstable K-Theory for graph algebras", Algebra Repr. Th. DOI 10.1007/s10468-006-9044-z (electronic).
(Available at http://www.springerlink.com/content/pu701474q5300m63/).

G. Abrams, G. Aranda Pino, F. Perera, M. Siles Molina. "Chain conditions for Leavitt path algebras".
(Available at http://agt.cie.uma.es/~gonzalo/papers/AAPS1_Web.pdf).

K.R. Goodearl. "Leavitt path algebras and direct limits", Contemp. Math. 480 (2009), 165-187.