co.combinatorics - Local-Global approach to graph theory

Does planarity qualify as a "local condition?" I'd think it should, but I don't see how to put it into the "data on small/local subgraphs" framework.

Anyway, if it does, you have of course the four-color theorem, and even better the five-color theorem, whose proofs essentially take advantage of the fact that we sort of understand how to move between "local" and "global" in topological spaces.

ETA: More generally, of course, there's the whole subfield of "structural graph theory" and its methods. I don't know that the graph minor theorem is "local-to-global" -- it's really more "local-to-a-different-kind-of-local" -- but it's probably the most important structural result.

Structural graph theory is something that I wish I knew about, but looks so horrifically technical and difficult that I'm sort of afraid to study it. There are clearly some deep patterns hidden there, though -- witness how Robertson, Seymour, and Thomas all worked on the proof of the Strong Perfect Graph Conjecture, which used a decomposition argument and had a hugely structural flavor despite being (as far as I can tell) mostly unrelated to the more topological work they'd previously done.

Tangentially, this recent preprint of Dvorak, Kral and Thomas caught my eye for exactly the "local properties" reason. Unfortunately the proof of the main theorem doesn't seem to be available yet...

reference request - Inverse limit in metric geometry

Question. Did you ever see inverse limits to be used (or even seriousely considered) anywhere in metric geometry (but NOT in topology)?

The definition of inverse limit for metric spaces is given below. (It is usual inverse limit in the category with class of objects formed by metric spaces and class of morphisms formed by short maps.)

Definition.
Consider an inverse system of metric spaces $X_n$ and short maps $phi_{m,n}:X_mto X_n$ for $mge n$;
i.e.,(1) $phi_{m,n}circ phi_{k,m}=phi_{k,n}$ for any triple $kge mge n$ and (2) for any $n$, the map $phi_{n,n}$ is identity map of $X_n$.

A metric space $X$ is called inverse limit of the system $(phi_{m,n}, X_n)$ if its underlying space consists of all sequences $x_nin X_n$ such that $phi_{m,n}(x_m)=x_n$ for all $mge n$ and for any two such sequences $(x_n)$ and $(y_n)$ the distance is defined by

$$ | (x_n) (y_n)| = lim_{ntoinfty} | x_n y_n | .$$

Why: I have a theorem, with little cheating you can stated it this way: The class of metric spaces which admit path-isometries to Euclidean $d$-spaces coincides with class of inverse limits of $d$-polyhedral spaces.
In the paper I write: it seems to be the first case when inverse limits help to solve a natural problem in metric geometry. But I can not be 100% sure, and if I'm wrong I still have time to change this sentence.

nt.number theory - What is the distribution of the $L^infty$ norm of minimal polynomials of numbers in a number field?

Given a number field $K$, I am interested in the distribution of the $L^infty$ norm of minimal polynomials (over $mathbb{Z}$) of numbers in $K$. Also, it is interesting to restrict to numbers $alpha$ with $K=mathbb{Q} (alpha)$.

All such polynomials' discriminants are a square factor away from the discriminant of $K$. So that must say something. And it makes it very interesting - the discriminant is a high-degree multivariate polynomial, so the fact that any discriminant can be achieved many times (even if we allow the distance of square factors) amazes me.

To be precise, so no one is annoyed, the question is exactly: what is the asymptotic behavior and error term for $A(x) = |{ alpha in K | parallel m(alpha)parallel_infty < x}|$, and $B(x) = |{ alpha in K | K=mathbb{Q} (alpha), parallel m(alpha)parallel_infty < x}|$, as $x rightarrow infty$?

nt.number theory - Conductor of monomial forms with trivial nebentypus

Just to augment Kevin's series of comments: I think that the conductor of the induction of some
character $chi$ over a quadratic field to $mathbb Q$ would normally equal $D N(C)^2$, where $D$ is the discriminant of the quadratic field, $C$ is the condutor of the character (an ideal in
the quadratic field) and $N$ is the norm from the quadratic field to $mathbb Q$. E.g.
in Kevin's $23$ example, one inducing a character of conductor 1 from $mathbb Q(sqrt{-23})$,
so the conductor is $23$. [Added in response to an edit in the question: This form has nebentypus equal to the Legendre symbol mod 23.]

In the Maass case one should be able to do something similar, by e.g. choosing
a prime $p equiv 1 mod 4$ such that $mathbb Q(sqrt{p})$ has non-trivial class group,
and then inducing a non-trivial character of conductor 1. [Added in response to an edit in the question: Such examples will have nebentypus equal to the Legendre symbol mod p, I think.]

[Added in response to an edit in the question:] Based on the formula above for the conductor, I think that to have square free conductor one will need to induce a character with trivial conductor, i.e. coming from the (strict) class group. I think that such an induction will always have non-trivial nebentypus, though. (The key point being that
if $H/{mathbb Q}$ is the stict Hilbert class field of the real quadratic field,
then this is a generalized dihedral extension.)

Another argument, pointed out me by a colleague, is that if the conductor is square fee
and the nebentypus is trivial, then all the local factors of the automorphic representation at primes in the conductor are Steinberg, which is not possible for the induction of a character.

[One more remark:] It seems to me that if we replace $mathbb Q$ by some well-chosen
number field $F$, then it will be possible to find an unramified quadratic extension $E$ of $F$ such that $E$ in turn admits a degree $4$ extension $K$, everywhere unramified,
so that $K$ over $F$ (a degree 8 extension) is Galois with the quaternion group
as Galois group (as opposed to a dihedral group). I think if we then take the
corresponding order 4 ideal class character of $E$ and induce it to $F$,
we get a monomial representation of $F$ with trivial determinant whose conductor
is equal to one (and in particular, is square-free).

In other words, one is a little bit "lucky" in the $mathbb Q$-case that Hilbert class
fields of real quadratic fields are dihedral over $mathbb Q$.

linear algebra - Number of independent distances between n points in d-dimensional Euclidean space?

There are $binom{n}{2}$ distances between $n$ points in $mathbb{R}^d$. Not all of them can be chosen freely if $n$ exceeds the number $n_d = d + 1$. If $n = n_d$ we obviously have $binom{d+1}{2}$ distances which can be chosen (more or less) independently (restricted only by the triangle inequality).

I see two ways to count $N_n^d$, the number of independent distances between $ngeq d$ points in $mathbb{R}^d$, which is given by $nd - binom{d+1}{2}$

The first one: $nd$ coordinates minus one translation ($d$) minus one rotation ($binom{d}{2}$):

$N_n^d = nd - d - binom{d}{2} = nd - (binom{d}{2} + d) = nd - binom{d+1}{2}$

The second one: $binom{d+1}{2}$ distances between $d + 1$ base points plus $d ( n - (d + 1) )$ distances between the remaining points and $d$ of the base points (the remaining one serving to say in which half-space with respect to the $d$ base points the point is located):

$N_n^d = binom{d+1}{2} + d ( n - (d + 1) ) = nd - d (d + 1) + binom{d+1}{2} = nd - binom{d+1}{2}$. (Is this sound?)

Observation: The binomials seem to come from two very different directions (with two seemingly different interpretations), also the term $nd$. Does this tell something deep about (Euclidean) geometry? And what?

Are there further "independent" ways to compute $N_n^d$?

at.algebraic topology - Complex vector bundles with trivial Chern classes on k-tori

As the cohomology of $(S^1)^n$ is torsion free every stable bundle on $(S^1)^n$ is
determined by Chern classes (this also follows from the $K$-theory Künneth
formula) so just as for the spheres it is an unstable problem. As for the
unstable problem unless I have miscalculated, if $(S^1)^5rightarrow S^5$ is a
degree $1$ map, then the pullback of the non-trivial $U(2)$ bundle on $S^5$ with
trivial Chern class is non-trivial. (The proof uses that the $5$'th step in the
Postnikov tower of $mathrm{BU}(2)$ is a principal fibration $K(mathbb
Z/2,5)rightarrow Urightarrow K(mathbb Z,4)times K(mathbb Z,2)$.)

Some more details of the calculation: The first and second Chern class gives a
map
$$mathrm{BU}(2)rightarrow K((mathbb Z,4)times K(mathbb Z,2)$$
which induces an isomorphism on homotopy groups in degrees up to $4$. As
$pi_i(mathrm{BU}(2))=pi_{i-1}(mathrm{SU}(2))$ for $i>2$ we get that
$pi_5(mathrm{BU}(2))=pi_4(S^3)=mathbb Z/2$. Hence, the next step $U$ in the
Postnikov tower of $mathrm{BU}(2)$ is the pullback of the path space fibration
of a morphism $K(mathbb Z,4)times K(mathbb Z,2)rightarrow K(mathbb Z/2,6)$.
In particular we have a principal fibration
$$K(mathbb Z/2,5)rightarrow
Urightarrow K(mathbb Z,4)times K(mathbb Z,2).$$
This means that for any space
$X$, the image of $[X,K(mathbb Z/2,5)]$ in $[X,U]$ is in bijection with the
cokernel of $[X,K(mathbb Z,3)times K(mathbb Z,1)]rightarrow[X,K(mathbb
Z/2,5)]$ obtained by applying $[X,-]$ to the looping of the structure map
$K(mathbb Z,4)times K(mathbb Z,2)rightarrow K(mathbb Z/2,6)$. As
$H^4(K(mathbb Z,3),mathbb Z/2)=0$ the Künneth formula shows that any map
$K(mathbb Z,3)times K(mathbb Z,1)rightarrow K(mathbb Z/2,5)$ factors
through the projection $K(mathbb Z,3)times K(mathbb Z,1)rightarrow K(mathbb
Z,3)$ and $H^5(K(mathbb Z,3),mathbb Z/2)=mathbb Z/2mathrm{Sq}^2rhoiota$ (where
$iota$ is the canonical class, $iotain H^3(K(mathbb Z,3),mathbb Z)$ and
$rho$ is induced by the reduction $mathbb Zrightarrowmathbb Z/2$). Hence,
the map $[X,K(mathbb Z,3)times K(mathbb Z,1)]rightarrow[X,K(mathbb
Z/2,5)]$ is either the zero map or given by the composite of the projection to
$H^3(X,mathbb Z)$, the reduction to $mathbb Z/2$ coefficients and
$mathrm{Sq}^2$ (I actually think it is non-zero as otherwise the cohomology of
$H^ast(mathrm{BU}(2),mathbb Z)$ would have $2$-torsion). If we apply it to
$X=(S^1)^5$ we get that $[X,K(mathbb Z,3)times K(mathbb Z,1)]rightarrow[X,K(mathbb
Z/2,5)]$ is zero provided that
$$mathrm{Sq}^2colon H^3((S^1)^5,mathbb
Z/2)rightarrow H^5((S^1)^5,mathbb
Z/2)$$
is zero. However, all Steenrod squares are zero on all of
$H^*((S^1)^n,mathbb Z/2)$. Indeed, the Künneth and Cartan formulas reduce this to
$n=1$ where it is obvious.

The Existence of Natural Satellites in Geostationary Orbits

While browsing through Physics SE, I noticed a question about satellites in geostationary orbit (unrelated to the one I'm asking here), and for a moment I interpreted it as referring to natural satellites (e.g. a moon). So I wondered: Could a natural satellite exist in geostationary orbit?

Then I stopped and thought. For large gas giants, such as Jupiter, having moons too close to the planet can be fatal (for the moon). If it ventures inside the planet's Roche limit, it's toast. But there is good news: the Roche limit depends on both the masses and densities of the primary body and the satellite. So perhaps this reason is non-applicable, as a high-mass natural satellite might be able to survive. So the question changes:

Could a sufficiently high-mass, high-density natural satellite occupy geostationary orbit over its primary body?

mathematics education - Why is a topology made up of 'open' sets?

I'm ashamed to admit it, but I don't think I've ever been able to genuinely motivate the definition of a topological space in an undergraduate course. Clearly, the definition distills the essence of many examples, but it's never been obvious to me how it came about, compared, for example, to the rather intuitive definition of a metric space. In some ways, the sparseness of the definition is startling as it tries to capture, apparently successfully, the barest notion of 'space' imaginable.

I can try to make this question more precise if necessary, but I'd prefer to leave it slightly vague, and hope that someone who has discussed this successfully in a first course, perhaps using a better understanding of history, might be able to help me out.

Added 24 March:

I'm grateful to everyone for their thoughtful answers so far. I'll have to think over them a bit before I can get a sense of the 'right' answer for myself. In the meanwhile, I thought I'd emphasize again the obvious fact that the standard concise definition has been tremendously successful. For example, when you classify two-manifolds with it, you get equivalence classes that agree exactly with intuition. Then in as divergent a direction as the study of equations over finite fields, there is the etale topology*, which explains very clearly surprising and intricate patterns in the behaviour of solution sets.

*If someone objects that the etale topology goes beyond the usual definition, I would argue that the logical essence is the same. It is notable that the standard definition admits such a generalization so naturally, whereas some of the others do not. (At least not in any obvious way.)

For those who haven't encountered one before, a Grothendieck topology just replaces subsets of a set $X$ by maps $$Yrightarrow X.$$ The collection of maps that defines the topology on $X$ is required to satisfy some obvious axioms generalizing the usual ones.

Added 25 March:

I hope people aren't too annoyed if I admit I don't quite see a satisfactory answer yet. But thank you for all your efforts. Even though Sigfpe's answer is undoubtedly interesting, invoking the notion of measurment, even a fuzzy one, just doesn't seem to be the best approach. As Qiaochu has pointed out, a topological space is genuinely supposed to be more general than a metric space. If we leave aside the pedagogical issue for a moment and speak as working mathematicians, a general concept is most naturally justified in terms of its consequences. As pointed out earlier, topologies that have no trace of a metric interpretation have been consequential indeed.

When topologies were naturally generalized by Grothendieck, a good deal of emphasis was put on the notion of an open covering, and not just the open sets themselves. I wonder if this was true for Hausdorff as well. (Thanks for the historical information, Donu!) We can see the reason as we visualize a two-manifold. Any sufficiently fine open covering captures a combinatorial skeleton of the space by way of the intersections. Note that this is not true for a closed covering. In fact, I'm not sure what a sensible condition might be on a closed covering of a reasonable space that would allow us to compute homology with it. (Other than just saying they have to be the simplices of a triangulation. Which also reminds me to point out that homology can be computed for ordinary objects without any notion of topology.)

To summarize, a topology relates to analysis with its emphasis on functions and their continuity, and to metric geometry, with its measurements and distances. However, it also interpolates between these and something like combinatorial geometry, where continuous functions and measurements play very minor roles indeed.

For myself, I'm still confused.

Another afterthought: I see what I was trying to say above is that open sets in topology provide an abstract framework for describing local properties of functions. However, an open cover is also able to encode global properties of spaces. It seems the finite intersection property is important for this, but I'm not able to say for sure. And then, when I try to return to the pedagogical question with all this, I'm totally at a loss. There are very few basic concepts that trouble me as much in the classroom.

mars - What is the next planned mission that can discover life on another planet?

The Exomars rover of ESA and Roscosmos would be the obvious answer. To be launched in 2018. It will drill 2 meters deep and as far I know is the only mission since the Vikings in the 1970's, to explicitly be equipped to find biosignatures, signs of life. But the Russians, who will land the rover, have had a very poor Mars mission success rate, ESA cooperates with them since NASA abandoned the project a few years ago, and ESA's only own landing attempt on Mars failed too, Beagle 2. And some biologists think it is more challenging to detect sparse exotic microbial life than what that rover is capable of. It weights about 1/3 of MSL Curiosity, and underground life can maybe be very local.

I want to recommend the blogger Robert Walker who writes at great length and well informed, still interestingly speculative, about possibilities for life on Mars, what one maybe should be looking for.

I should add that since 1960 SETI uses telescopes to pick up evidence of interstellar life which is powerful enough to somehow change its environment to make itself astronomically detectable. That's the other main potential except for probes in the Solar system. But not even SETI people sit up waiting for it any more.

orbit - Orbiting around a black hole

For the purposes of comparison, here's flat, Minkowski spacetime in spherical coordinates:
$$mathrm{d}s^2 = -mathrm{d}t^2 + underbrace{mathrm{d}r^2 + r^2(mathrm{d}theta^2+sin^2theta,mathrm{d}phi^2)}_text{Euclidean 3-space}text{.}$$

The soure of misunderstanding was that I wasn't clear enough, how black holes work. I always imagined them as "suckholes" like whirlpools in the water.

That is not entirely incorrect. The Schwarzschild spacetime of an uncharged, nonrotating black hole in the Gullstrand-Painlevé coordinates is
$$mathrm{d}s^2 = -mathrm{d}t^2 + underbrace{left(mathrm{d}r + sqrt{frac{2M}{r}},mathrm{d}tright)^2}_text{suckhole} + r^2(mathrm{d}theta^2+sin^2theta,mathrm{d}phi^2)text{.}$$
Where it deviates from ordinary, flat Minkowski spacetime is entirely in the middle square term. Here, the time coordinate $t$ is not the Schwarzschild time, but rather the time measured by an observer free-falling from rest at infinity. The last bit, if adjoined with the $mathrm{d}r^2$ term one would get by multiplying out the middle part, is ordinary Euclidean $3$-space written in spherical coordinates.

If you recognize from Newtonian gravity the quantity $sqrt{2M/r}$, or $sqrt{2GM/r}$ in ordinary units, as the escape velocity, then the picture is very peculiar indeed: according to an observer free-falling from rest at infinity, Euclidean space is sucked into the singularity at the local escape velocity. The event horizon is the surface at which the speed at which space is "falling" at the speed of light.

This is an additional reason why sonic black holes are good analogues to their gravitational counterparts. In a sonic black hole, there can be an actual "suckhole" that drains a low-viscosity fluid at an increasing velocity, up to and faster than the speed of sound in that fluid. This forms an acoustic event horizon that is one-way to sound and is expected to have an analogue of Hawking radiation.

The corresponding structure for charged black holes is similar, and for a rotating one more complicated, although can still be described as "sucking" with a certain additional twist that rotates the free-falling observers.

normalization - Normality of an affine semigroup

Sorry, I did not get the above answer. Can somebody please explain me.

Suppose, $q.(p_0,p_1, cdots ,p_{n-1}) in S$, where $ q in mathbb N setminus 0$
and $(p_0,p_1, cdots ,p_{n-1}) in G(S)$, why is it clear from the above answer that
$(p_0,p_1, cdots ,p_{n-1})$ is already in $S$. Some of the $p_i$'s can be negative
as well. It is not at all clear to me. Thanks in advance.

analytic number theory - PNT for general zeta functions, Applications of.

Hi Anweshi,

Since Emerton answered your third grey-boxed question very nicely, let me try at the first two. Suppose $L(s,f)$ is one of the L-functions that you listed (including the first two, which we might as well call L-functions too). (For simplicity we always normalize so the functional equation is induced by $sto 1-s$.) This guy has an expansion $L(s,f)=sum_{n}a_f(n)n^{-s}$ as a Dirichlet series, and the most general prime number theorem reads

$sum_{pleq X}a_f(p)=r_f mathrm{Li}(x)+O(x exp(-(log{x})^{frac{1}{2}-varepsilon})$.

Here $mathrm{Li}(x)$ is the logarithmic integral, $r_f$ is the order of the pole of $L(s,f)$ at the point $s=1$, and the implied constant depends on $f$ and $varepsilon$.

Let's unwind this for your examples.

1) The Riemann zeta function has a simple pole at $s=1$ and $a_f(p)=1$ for all $p$, so this is the classical prime number theorem.

2) The Dedekind zeta function (say of a degree d extension $K/mathbb{Q}$) is a little different. It also has a simple pole at $s=1$, but the coefficients are determined by the rule: $a(p)=d$ if $p$ splits completely in $mathcal{O}_K$, and $a(p)=0$ otherwise. Hence the prime number theorem in this case reads

$|pleq X ; mathrm{with};p;mathrm{totally;split;in};mathcal{O}_K|=d^{-1}mathrm{Li}(x)+O(x exp(-(log{x})^{frac{1}{2}-varepsilon})$.

This already has very interesting applications: the fact that the proportion of primes splitting totally is $1/d$ was very important in the first proofs of the main general results of class field theory.

3) If $rho:mathcal{G}_{mathbb{Q}}to mathrm{GL}_n(mathbb{C})$ is an Artin representation then $a(p)=mathrm{tr}rho(mathrm{Fr}_p)$. If $rho$ does not contain the trivial representation, then $L(s,rho)$ has no pole in neighborhood of the line $mathrm{Re}(s)geq 1$, so we get

$sum_{pleq X}mathrm{tr}rho(mathrm{Fr}_p)=O(x exp(-(log{x})^{frac{1}{2}-varepsilon})$.

The absence of a pole is not a problem: it just means there's no main term! In this particular case, you could interpret the above equation as saying that "$mathrm{tr}rho(mathrm{Fr}_p)$ has mean value zero.

4) For an elliptic curve, the same phenomenon occurs. Here again there is no pole, and $a(p)=frac{p+1-|E(mathbb{F}_p)|}{sqrt{p}}$. By a theorem of Hasse these numbers satisfy $|a(p)|leq 2$, so you could think of them as the (scaled) deviation of $|E(mathbb{F}_p)|$ from its
"expected value" of $p+1$. In this case the prime number theorem reads

$sum_{pleq X}a(p)=O(x exp(-(log{x})^{frac{1}{2}-varepsilon})$

so you could say that "the average deviation of $|E(mathbb{F}_p)|$ from $p+1$ is zero."

Now, how do you prove generalizations of the prime number theorem? There are two main steps in this, one of which is easily lifted from the case of the Riemann zeta function.

1. Prove that the prime number theorem for $L(s,f)$ is a consequence of the nonvanishing of $L(s,f)$ in a region of the form $s=sigma+it,;sigma geq 1-psi(t)$ with $psi(t)$ positive and tending to zero as $tto infty$. So this is some region which is a very slight widening of $mathrm{Re}(s)>1$. The proof of this step is essentially contour integration and goes exactly as in the case of the $zeta$-function.




2. Actually produce a zero-free region of the type I just described. The key to this is the existence of an auxiliary L-function (or product thereof) which has positive coefficients in its Dirichlet series. In the case of the Riemann zeta function, Hadamard worked with the auxiliary function $ A(s)=zeta(s)^3zeta(s+it)^2 zeta(s-it)^2 zeta(s+2it) zeta(s-2it)$. Note the pole of order $3$ at $s=1$; on the other hand, if $zeta(sigma+it)$ vanished then $A(s)$ would vanish at $s=sigma$ to order $4$. The inequality $3<4$ of order-of-polarity/nearby-order-of-vanishing leads via some analysis to the absence of any zero in the range $s=sigma+it,;sigma geq 1-frac{c}{log(|t|+3)}.$ In the general case the construction of the relevant auxiliary functions is more complicated. For the case of an Artin representation, for example, you can take $B(s)=zeta(s)^3 L(s+it,rho)^2 L(s-it,widetilde{rho})^2 L(s,rho otimes widetilde {rho})^2 L(s+2it,rho times rho) L(s-2it,widetilde{rho} times widetilde{rho})$. The general key is the Rankin-Selberg L-functions, or more complicated L-functions whose analytic properties can be controlled by known instances of Langlands functoriality.

If you'd like to see everything I just said carried out elegantly and in crystalline detail, I can do no better than to recommend Chapter 5 of Iwaniec and Kowalski's book "Analytic Number Theory."

telescope - How to compare a Catadioptrics light asorbtion with a reflector?

Comparing optical instruments is very difficult. There are numerous variables involved, the performance is not a single number, and the theory is complex. Perfect scenario to baffle amateurs. Also, it sounds like you may have been listening to the wrong rumors.

The architecture of a telescope (refractor, reflector, catadioptric) is one of the least important factors in the performance analysis, regardless of how you define "performance".

Don't forget that how you maintain and where you use the scope are very important factors. Collimation will strongly impact resolving power - so make sure it's always perfect. Light pollution will impact you much more than light absorption within the instrument. Etc.

In terms of amount of light gathered, aperture is the dominant factor. 2x the aperture is 4x the area, so it's obviously very important. The central obstruction may factor in a little, but the effect is small. Going by diameter, a typical newtonian will have a 20% central obstruction; a typical catadioptric scope will have a 40% c.o. But if you translate that into areas, the newtonian is at 4%, whereas the catadioptric is at 16% - both are very small.

In terms of resolving power, again aperture is king. Bigger aperture = more resolving power, no matter what the architecture. But poor collimation will quickly destroy your gains, so keep that scope collimated.

In terms of internal losses, the newtonian will do a bit better - it's two reflections and that's it. The catadioptric will have two or more reflections (depending on design), plus absorption / reflection on the diopters. Again, with modern instruments, the difference is tiny.

Overall, at the same aperture, in theory, the newtonian pure catoptric scope delivers a bit more light to the sensor, compared to a catadioptric scope. In practice, the difference is far outweighed by other factors and you should completely ignore it.

You should make the purchase based on more important things. What's the size of the usable field? What kind of adapters can you use? Does it provide enough back-focus? What's the aberration in different points within the field? Is the telescope too heavy for the driven mount? Is it long focal ratio, or short? (each kind has different applications) Finally but not least important, price?

It's more common to see catadioptrics used as astrographs. That's not to say that a newtonian reflector cannot do the job - but if you're still new to the field, just do what most people do and get a catadioptric, with a good mount.

---

Final note:

When doing prime focus photography, the brightness of the image is given by both aperture and focal ratio - but not all objects behave the same.

For stars and other point-like objects (smaller than the resolving power), brightness is driven by aperture. Bigger aperture = brighter stars.

For extended objects (bigger than the resolving power), such as nebulae, many galaxies, the Moon, the planets, etc., the brightness of the object is given by the focal ratio. An f/4 scope will make brighter nebulae in prime focus, compared to an f/8 scope. Aperture doesn't matter. It's just that, at the same focal ratio, the bigger scope will also make a bigger image in prime focus.

This is true only for prime focus photography. If you're doing eyepiece projection, or something else, different rules apply.

exoplanet - Do most planetary systems have fewer planets than the solar system?

Perhaps the way to answer this is ask - could we detect the planets in our solar system if we were looking at the Sun, using current technology, from distances of many light years?

The short answer is that we could detect Jupiter using the Doppler radial velocity technique, if we observed for more than 10 years (at least one orbit is required). If we were lucky, and the orientation is right, we might then also be able to detect a transit of Venus or the Earth, using a satellite observatory like Kepler. Kepler could detect Earthlike planets by the transit technique, but the solar system is not "flat" enough that you would observe multiple transiting planets.

So the answer is that we would currently have seen Jupiter and maybe one other planet. Therefore we cannot at the moment conclude that 8 planets is an unusually high number; it may be quite typical. Although we do know that solar systems can be much more densely populated with planets (in their "terrestrial planet zones") than our own.

mars - Can the martian dichotomy (SIH) and the moon's formation (GIH) be related?

Early in the solar system's history, the Sun was surrounded by a protoplanetary disk, full of gas, dust and rock. Eventually, protoplanets began to form. These were small, rocky bodies - smaller than the rocky planets today, but bigger than asteroids. Think of them as dwarf-planet-sized objects. All the terrestrial planets formed from them.

Anyway, as you might know, the object that slammed into the Earth (according to the GIH) was one such protoplanet, lovingly named Theia. The resulting debris coalesced into the Moon. The SIH tells of a similar story: Once again, a protoplanet headed on a collision course, this time towards Mars. It slammed into Mars, creating a large basin. Simple but effective.

There's one thing that can relate these two disaster tales together, and that is the protoplanetary disk. Remember, the protoplanets formed from the same disk. The differences in them stemmed from the materials they formed from. So we could relate the two suicidal protoplanets to each other via that connection. Also, most collisions in the inner solar system were due to the erratic orbits of these protoplanets, so it is likely that the fault for the collisions lies solely on the protoplanets.

I think you want something a bit different from random collisions, though; the previous reason is a bit of a cop-out. Well, there is something more interesting: the rocky planets weren't alone in the early solar system. There were still their big siblings: the gas giants. The gas giants hadn't yet grown to their full sizes, but their cores and early atmospheres were still pretty massive, and commanded a strong presence in the outer solar system. It is possible that the two objects that hit the Earth and Mars formed together farther away from the inner planets, and were gravitationally perturbed by the gas giants, swinging in from an outer orbit towards the inner solar system.

You might have heard of the Nice Model. Basically, this says that the gas giants formed closed to the inner solar system than they are now. They encountered some of these early protoplanets and flung them into the inner solar system, moving outwards in the process. thus, collisions increased in the inner solar system.

It's a "nice" explanation. Trouble is, the majority of this "flinging" happened later on in the solar system's history; it has been suggested that the gas giants did this and thus caused the Late Heavy Bombardment. Still this idea shows that there could have been interaction between the protoplanets and the gas giants.

So how does this relate the two suicidal protoplanets? Well, it means that they could have formed together, further out in the solar system, and have been brought in by a gas giant. They could even have been part of a larger object, which broke apart after various collisions. But if the two objects had formed together (even somewhere in the inner solar system), or been part of a larger object, they should be made of the same materials, right? So if we were to travel to Mars we should be able to find similar materials as we did on the Moon. So far there hasn't been much luck along that avenue, but the search will continue.

To summarize: The early solar system was chaotic. It is likely that the impacts were unrelated, and that the two suicidal protoplanets formed apart in the inner solar system, but they could have formed together further our and been perturbed by one of the gas giants. They may or may not have been formed neary each other, or even been fragments of the same object. Detailed analysis in the future could give us some clues as to their formation.

the sun - Which stars did the Sun form with?

Most stars form in clusters, so it is very likely that the Sun was part of a star cluster when it formed.

But in On the Dynamics of is Open Clusters, the relaxation time of a cluster is calculated to be in the order of $tau=4times10^7 textrm{yr}$. During that time about one hundredth of the stars will escape from the cluster (i.e. reach escape velocity). The dissipation time of a star cluster is. therefore, in the order of a few billion years. These figures were calculated for an idealised star cluster containing stars of equal mass. For more realistic clusters the dissipation time may be much shorter. Especially dwarf stars seem to dissipate faster.

It is, therefore, likely that after 4.5 billion years, the Sun has escaped from its star cluster. It would be very difficult to identify the stars with which it was formed, unless you can trace back the position of the Sun and other stars before dissipation, which is impracticable.

As far as I am aware, no sibling stars have been positively identified. The sibling stars should be similar in composition (element abundances), although they do not have to be solar twins as the masses of these stars may vary widely.

EDIT: @adrianmcmenamin mentions an article in Sky & Telescope, where a sibling has been proposed. This refers to an article on arXiv by Ramirex et al. in which the element abundances have been analysed of stars that were earlier identified as possible siblings using galactic dynamics (i.e. tracing back the movement of these stars). Of the 30 stars identified as possible solar siblings using dynamical methods, only 2 stars had similar element abundances, which in itself says something about the accuracy of these dynamical methods...

mg.metric geometry - Break Polyhedron into Tetrahedron

If I understand your question correctly, you're saying that the given information is the face structure of a 3-dimensional convex polytope, and you would like a subdivision of the polytope into tetrahedra.

Here is one way to proceed. First, subdivide all the faces into triangles. Then pick your favourite vertex $v_0$. Connect $v_0$ to each triangle belonging to a face not containing $v_0$. This subdivides your polytope into tetrahedra.

element - what is the metal distribution in our universe

From the big bang nucleosynthesis (BBN) page is a handy chart of the sources of elements:

Since basically all metals come from some stellar process, the question of elemental distribution can vary greatly between different environments depending on when you're observing that particular galaxy, star/galaxy cluster, etc. Galaxy mergers, supernova, and stellar winds can be particularly good at moving metals around a galaxy or galaxy cluster.

Interestingly Wolfram has element data with Universe abundances but I'm not sure where they derive those.

ac.commutative algebra - Formally étale at all primes does not imply formally étale.

Using the module of Kähler differentials, it is easy to show that $Rto S$ is formally unramified if and only if the induced maps $Rto S_{mathfrak{p}}$ are formally unramified for all primes $mathfrak{p}subset S$.

Consider a presentation of $S$ over $R$ as $R[X]/I$ in generators and relations, where $R[X]:=R[X_m]_{min M}$ is a polynomial ring in a possibly infinite family of indeterminates indexed by $M$, and $Isubset R[X]$ is an ideal. Fix a family of generators of $I=(F_j)_{jin J}$ indexed by $J$, again not necessarily finite.

It is enough to show that $Rto S$ is formally smooth. This is equivalent to showing that there exists a morphism of $R$-algebras that is a splitting for the canonical projection $pi:R[X]/I^2 to R[X]/I=S$, which will necessarily be unique because $Rto S$ is formally unramified.

Let $overline{X}_m$ denote the image of $X_m$ in $R[X]/I^2$. We must find elements $delta_min I/I^2$ such that $(forall jin J)F_j(X_m + delta_m)=0$. We rewrite this using Taylor's formula as $$bar{F}_j+ sum_{min M}overline{frac{partial F_j}{partial X_m}}delta_m=0.$$

Rearranging, we get a system of equations indexed by $J$
$$(*)_{jin J} qquad sum_{min M}overline{frac{partial F_j}{partial X_m}}delta_m=-overline{F}_j.$$

We wish to find a unique solution for this system in the $delta_m$. Since $Omega_{S/R}=0$, each $dX_min Omega_{R[X]/R}$ is an $S$-linear combination $dX_m=s_{m,1}dF_{j_{m,1}}+cdots + s_{m,h_m}dF_{j_{m,h_m}}$. If we use the $s_{m,k}$ as coefficients to form $S$-linear combinations of the equations $(*)_{j_k}$, for each $m$, we get an equation of the form $$(**)_m qquad delta_m=-(s_{m,1}overline{F}_{j_{m,1}}+cdots + s_{m,h_m}overline{F}_{j_{m,h_m}}).$$

Showing that these define solutions for all of the equations $(*)_j$ is not immediate, but it is a local question on $S$. However, our local rings $S_{mathfrak{p}}$ are all formally étale, so the local conditions are satisfied. Then this proves the global claim.

(Note: This is not my proof. I've paraphrased the proof communicated to me by Mel Hochster.)

Edit: Fixed LaTeX using Scott's suggestion.

distances - How near to Earth do comets pass?

The answer is that they don't come very close. As Wikipedia notes,

It is rare for a comet to pass within 0.1 AU (15,000,000 km; 9,300,000 mi) of Earth.

Even better, though, is an actual list of some of the closest approaches of comets to Earth. The closest one listed, Comet Lexell in 1770, came 0.0151 AU away from Earth. The list only shows 20 comets have come with 0.1 AU of Earth. So Earth has not had a close encounter with a comet (like Mars is having) in a long time, if ever.

This site says that C/1491 came within an incredible 0.0094 AU in 1491, but I'll take the IAU site over it. And it does admit that that comet has an uncertain orbit, so that figure could be wrong.

But when comets come even a tiny bit near to Earth, there are some articles written about them. See here and here.

As for the distances to famous comets:

Closest Ever

• Halley's - 0.0334 AU


• Great Comet of 1760 - 0.0682 AU


• Tempel-Tuttle - 0.0229 AU

Also, some near-Earth asteroids used to be comets. Extinct comets have lost a lot (if not all) of their ice and so may be classified as asteroids. This blurs the lines between the two groups. Some extinct comets have become near-Earth asteroids. This explores the issue pretty well. So if you include extinct comets as "comets," you may have comets that have come even closer to Earth.

Appropriate battery for overnight viewing to drive telescope and GOTO?

I have a Plettstone 18" Dobsonian telescope with GOTO and clock drive that runs off a 12 volt cigarette lighter adapter. What is an appropriate battery to use to run this overnight while viewing all night long and then charge during the day for the following night's use?

In the past I've used deep cycle lead acid batteries, but these are heavy and can fail if not carefully maintained with a float charge. Also, I don't need cranking power of a car, just sustained energy to drive the scope. A typical Optima YellowTop deep cycle is rated 75 ampHours. I don't know what the draw in Amps of my telescope is, but imagine it is less than an amp.

I've also seen these Battery Tender Lithium Iron Phosphate types but wonder if their 10-12 ampHour rating is enough and whether they are meant to be drained over a whole night? Some say if you get below 8 volts the battery becomes broken, but I doubt the telescope would still run at such a low voltage.

What would a reasonable ampHour rating be? What kind of battery is a good fit for astronomy equipment overnight?

What would happen if a planet is removed from the solar system?

Short answer: the solar system would be less stable in the short run due to the massive gravitational influence of Jupiter vanishing.

Long answer: The current layout of the solar system is actually remarkably well balanced; without outside interference, it'll stay stable for a good long while. Anything that changes the current equilibrium will force the system to adjust until it finds a new equilibrium. Removing a huge mass will certainly affect this in some way, but it's difficult to say exactly how without running a few hundred simulations. Jupiter's gravitational influence kept the asteroid belt from coalescing into a planet, so that might happen, for one.

Homotopy Limits over Fibered Categories

Suppose I have a small category $ mathcal{C} $ which is fibered over some category $mathcal{I}$ in the categorical sense. That is, there is a functor $pi : mathcal{C} rightarrow mathcal{I}$ which is a fibration of categories. (One way to say this, I guess, is that $mathcal{C}$ has a factorization system consisting of vertical arrows, i.e. the ones that $pi$ sends to an identity arrow in $mathcal{I}$, and horizontal arrows, which are the ones it does not. But there are many other characterizations.)

Now let $F : mathcal{C} rightarrow smathcal{S}$ be a diagram of simplicial sets indexed by $mathcal{C}$. My question concerns the homotopy limit of $F$. Intuition tells me that there should be an equivalence

$$ varprojlim_{mathcal{C}} F simeq varprojlim_{mathcal{I}} left (varprojlim_{mathcal{C}_i} F_i right ) $$

where I write $mathcal{C}_i = pi^{-1}(i)$ for any $i in mathcal{I}$, $F_i$ for the restriction of $F$ to $mathcal{C}_i$ and $varprojlim$ for the homotopy limit.

Intuitively this says that when $mathcal{C}$ is fibered over $mathcal{I}$, I can find the homotopy limit of a $mathcal{C}$ diagram of spaces by first forming the homotopy limit of all the fibers, realizing that this collection has a natural $mathcal{I}$ indexing, and then taking the homotopy limit of the resulting diagram.

Does anyone know of a result like this in the model category literature?

Update: After reading the responses, I was able to find a nice set of exercises here which go through this result in its homotopy colimit version.

pr.probability - Are there nonequivalent randomnesses?

Probably you are interested in an answer coming from measure theory or probability, and I would be interested in such an answer also, but let me give you an answer from set theory, which I believe does in fact offer a precise answer, with something like the sense that I believe you intend in your question.

In set theory, the method of forcing is fundamentally about different kinds of randomness or genericity. Specifically, in order to carry out a forcing argument, we first must specify the notion of forcing or genericity that will be used. This is done by specifying a certain partial order or Boolean algebra, whose natural topology will be used to determine a notion of dense sets. Generic objects will be those objects that are in many dense sets. In the extreme case in which forcing is applied, one has a model of set theory V, and considers adjoining an object G that is V-generic, in the sense that G is a filter containing elements from every dense set in V. The generic extension V[G] is a new model of set theory containing this new generic object G. The overall construction is something like a field extension, because V sits inside V[G], which is the smallest model of ZFC containing V and G, and everything in V[G] is constructible algebraically from G and objects in V. Forcing was first used to prove the consistency of ZFC with the negation of the Continuum Hypothesis, and the notion of forcing used had the effect of adding ω₂ many new real numbers. The generic object was a list of ω₂ many new reals, forcing CH to fail.

The point is that in order to control the nature of the forcing extension V[G], one must carefully control the forcing notion, that is, the notion of randomness that will be used to build V[G].

It turns out that many of the most fruitful forcing notions involve a notion of genericity or randomness on the reals. For example, with Cohen forcing, a V-generic real is one that is a member of every co-meager set in V. A random real is a member of every Lebesgue measure one set in V. There are dozens of different forcing notions (see the list of forcing notions), and these have been proved to be fundamentally different, in the sense that a generic filter for one of these is never generic with respect to another. With forcing, we can add Cohen reals, random reals, Mathias reals, Laver reals, dominating reals and so on, and these notions have no generic instances in common.

There isn't any probability space here to speak of, and being V-generic for a particular notion of forcing is not equivalent to any probabilistic property. Rather, one is tailoring the notion of forcing to describe a certain kind of randomness or genericity that is then used to build the forcing extension. Most of the detailed care in a forcing argument is about choosing the forcing notion and making sure that it works as desired.

Thus, since each notion of forcing corresponds fundamentally to a notion of genericity or randomness, I take the proofs that the different notions of forcing are different as an answer to your question.

lattices - Models with SLE scaling limit

From Cardy's article http://arxiv.org/abs/cond-mat/0503313

---

Some important special cases are therefore:

$kappa = −2$: loop-erased random walks (proven in [24]);

$kappa = 8/3$: self-avoiding walks, as already suggested by the restriction property, Sec. 3.5.2;
unproven, but see [22] for many consequences;

$kappa = 3$: cluster boundaries in the Ising model, however as yet unproven;

$kappa = 4$: BCSOS model of roughening transition (equivalent to the 4-state Potts
model and the double dimer model), as yet unproven; also certain level lines of a
gaussian random field and the ‘harmonic explorer’ (proven in [23]); also believed to
be dual to the Kosterlitz-Thouless transition in the XY model;

$kappa = 6$: cluster boundaries in percolation (proven in [7]);

$kappa = 8$: dense phase of self-avoiding walks; boundaries of uniform spanning trees
(proven in [24]).

It should be noted that no lattice candidates for κ > 8, or for the dual values κ < 2, have
been proposed.

ca.analysis and odes - Are there space filling curves for the Hilbert cube ?

Well there is indeed a "simple" construction of such a space filling curve.

Let $gamma:[0;1]rightarrow [0;1]^2$ be a space filling curve. Then one can obtain a space filling curve for $[0;1]^3$ by postcomposing with $id_mathbb{R}times gamma$. Then one can postcompose with $id_{mathbb{R}^2}times gamma$ and so on. Note that the first coordinates didn't change in the last step.

Putting all this together we get a map

$f:[0;1]rightarrow [0;1]^mathbb{N} qquad tmapsto (pr_1circ gamma circ (pr_2circ gamma)^{n-1}(t))_{nin mathbb{N}}, $
where $pr_i$ denote the obvious projections. This map can be seen as the infinite composition of the maps above. By the definition of the product topology this map is continuous.

Especially if we postcompose $f$ with the projection on the first $n$ coordinates, we just get a space filling curve (see above). Let us show, that a arbitrary element $x=(x_i)_{iin mathbb{N}}in [0;1]^mathbb{N}$ lies in the Image of $f$. We already know, that for each $n$ there is a element $y^n$ in Im$(f)$ agreeing with $x$ in the first $n$ coordinates.

As $[0;1]$ is compact, Im$(f)$ is compact and hence closed ($[0;1]^mathbb{N}$ is Hausdorff).

And $lim_{ntoinfty}y^n=x$. Hence $xin$ Im$(f)$. So $f$ is a continuous surjective map $[0;1]rightarrow [0;1]^mathbb{N}$.

co.combinatorics - What is the minimum N for which there exist N points in the plane that cannot be covered by any number of non-overlapping closed unit discs?

I was told this puzzle last friday by Peter Winkler (who had mentioned that it was told to him by a japanese fellow who is perhaps the one you are referring to).

The solution in the $n leq 10 $ case is to consider the tiling of the plane by unit height hexagons. Inscribe within each of these hexagons a unit circle. This grid of circles has density > 0.90 on the plane, and so if you randomly place this grid on the plane you accordingly have expected number of points covered > 9 (out of the 10), and this implies exists an arrangement that covers 10. (theres a few details missing from this probabilistic method argument, but you get the basic idea).

I believe for the $n>10$ case we have some way of computing an upper bound on the density of a sphere packing on the plane that rules it out in general. (or something to that extent)

distances - Why is the Eagle Nebula so "static"?

To add to Florin Andrei's answer, with an image height of 7,000 pixels for 14 light years, that's 17.5 light hours per pixel. That's 20 billion kilometres per pixel. To make a change in a single pixel over that time, something of that size must have either changed composition dramatically (to give a different colour or opacity) or it must have moved by a comparable distance.

Given the timeframe, that's a billion kilometres per year, or 123,000 kilometres per hour. (77,000 miles per hour, if you prefer) Few things that large are moving that fast relative to their neighbours.

pr.probability - linear recurrence relations with random coefficients

Are there such things as recurrence equations with random variable coefficients. For example, $$W_n=W_{n-1}+Fcdot W_{n-1}$$ where $F$ is a random variable. I tried to see if I could make sense of it using the simplest possible case of $F$ being a uniform discrete random variable on 2 points but I didn't get far because even if the initial data is not random the succeeding terms in the sequence are and each term seems to live on a different space. I couldn't figure out what space $W_n$ and $W_{n-1}$ should live on. A google search turned up nothing for the obvious keywords "random recurrence equation".

Edit in response to Alekk's answer: More specifically suppose I wanted to find the probability $P(W_{200}>3000)$. Is there a way to compute the distribution of $W_{200}$ explicitly given the distribution of $F$ and some non-random initial data $W_0$?

Edit: $F$ does not depend on $n$ and to make things even more explicit lets say $F$ has the distribution $P(F=2f)=dfrac{1}{2}, P(F=-f)=dfrac{1}{2}$.

co.combinatorics - Strong induction without a base case

My example is the classical proof that sqrt(2) is irrational.

More generally, many proofs that proceed by showing that there are no minimal counterexamples exemplify your phenomenon. The method of no-minimal-counterexamples is exactly the same as strong induction, but where one proves the required implication by contradiction. In many applications of this method, it is often clear that the smallest numbers are not counterexamples, and this would not ordinarily regarded as a separate base "case".

In the classical proof that sqrt(2) is irrational, for example, we suppose sqrt(2) = p/q, where p is minimal. Now, square both sides and proceed with the usual argument, to arrive at a smaller counterexample. Contradiction! This amounts to a proof by strong induction that no rational number squares to 2, and there seems to be no separate base case here.

People often carry out the classical argument by assuming p/q is in lowest terms, but the argument I just described does not need this extra complication. Also, in any case, the proof that every rational number can be put into lowest terms is itself another instance of the phenomenon. Namely, if p/q is a counterexample with p minimal, then divide by any common factor and apply induction. There seems to be no separate base case here where it is already in lowest terms, since we were considering a minimal counterexample. Perhaps someone objects that there is no induction here at all, since one can just divide by the gcd(p,q). But the usual proof that any two numbers have a gcd is, of course, also inductive: considering the least linear combination xq+yp amounts to strong induction, again with no separate base case.

real analysis - Cardinality of Equivalence Classes of Cauchy Sequences

It's the same as the size of the real numbers. Here's a rough sketch of the proof.

For each element of (0,1) (which has the same cardinality as the reals), I'm going to construct a distinct sequence of rationals that converges to 0.

Think of an element of (0,1) in binary, so as an infinite sequence of 0s and 1s. For definiteness, we assume the sequence is never eventually constant 1 (this just gives a well defined bijection between (0,1) and the sequences, since, e.g., .1000000... = .0111111... ). Given such a sequence of 0s 1s $a_1, a_2, ...$, create the rational sequence $(-1^{a_1}(1), -1^{a_2} (1/2), -1^{a_3} (1/3), ..., -1^{a_n} (1/n),...)$.

Convince yourself all these sequences of rational numbers are distinct.

Thus, the size of the reals is less than or equal to the number of rational sequences converging to 0. For a bound in the other direction, note that the collection of ALL sequences of rationals = $prod_{mathbb{N}} mathbb{Q}$ and $|prod_{mathbb{N}} mathbb{Q}| = |mathbb{Q}|^{|mathbb{N}|} = |mathbb{N}|^{|mathbb{N}|} leq |2^{mathbb{N}}|^{|mathbb{N}|} = |2^{mathbb{N}}| =$ the size of the reals.

By the Cantor-Schroeder-Bernstein theorem, the set off all rational sequences converging to 0 has the same cardinality as the reals.

What is visible light colour output of different stars?

Compare the effective temperature of the stars with Planck's law, and with a light bulb (2000 to 3300 K): The color of a light bulb corresponds to spectral class L (red brown star) or class M (red star, including dwarfs and giants); violet, and some blue are missing.

Light of Brown Dwarfs (spectral class T) would look red from their black body radiation, but due to their chemical composition the actual color may vary a bit over different tones of red. They would mainly feel hot, and look dull red at the same time, similar to glowing iron. Colors can hardly be distinguished under these lighting conditions.

Some of the very hottest stars (spectral class O) may ressemble "black" light, looking bluish, and causing fluorescence due to to the shift of the spectrum into the ultraviolet, including x-rays.
Without protective ozone/atmosphere layer, this wouldn't be healthy.

This may apply to some degree also to spectral class B (blue white stars); plasma cores of lightnings are of a similar temperature (up to 30,000 K or more) as the effectve temperature of class B or class O stars.

Any spectral class in-between (A, F, G, K) would look almost white, a little bluish for hot stars, yellowish for cooler stars, after some adaption of the eyes.

set theory - A question about ordinal definable real numbers

The original problem solves in the positive: there is a model of ZFC in which there exists a countable OD (well, even lightface $Pi^1_2$, which is the best possible) set of reals $X$ containing no OD elements. The model (by the way, as conjectured by Ali Enayat at http://cs.nyu.edu/pipermail/fom/2010-July/014944.html) is a $mathbf P^{<omega}$-generic extension of $L$, where $mathbf P$ is Jensen's minimal $Pi^1_2$ real singleton forcing and $mathbf P^{<omega}$ is the finite-support product of $omega$ copies of $mathbf P$.

A few details. Jensen's forcing is defined in $L$ so that $mathbf P =bigcup_{xi<omega_1} mathbf P_xi$, where each $mathbf P_xi$ is a ctble set of perfect trees in $2^{<omega}$, generic over the outcome $mathbf P_{<xi}=bigcup_{eta<xi}mathbf P_eta$ of all earlier steps in such a way that any $mathbf P_{<xi}$-name $c$ for a real ($c$ belongs to a minimal countable transitive model of a fragment of ZFC, containing $mathbf P_{<xi}$), which $mathbf P_{<xi}$ forces to be different from the generic real itself, is pushed by $mathbf P_{xi}$ (the next layer) not to belong to any $[T]$ where $T$ is a tree in $mathbf P_{xi}$. The effect is that the generic real itself is the only $mathbf P$-generic real in the extension, while examination of the complexity shows that it is a $Pi^1_2$ singleton.

Now let $mathbf P^{<omega}$ be the finite-support product of $omega$ copies of $mathbf P$. It adds a ctble sequence of $mathbf P$-generic reals $x_n$. A version of the argument above shows that still the reals $x_n$ are the only $mathbf P$-generic reals in the extension and the set ${x_n:n<omega}$ is $Pi^1_2$. Finally the routine technique of finite-support-product extensions ensures that $x_n$ are not OD in the extension.

Addendum. For detailed proofs of the above claims, see this manuscript.

Jindra Zapletal informed me that he got a model where a $mathsf E_0$-equivalence class $X=[x]_{E_0}$ of a certain Silver generic real is OD and contains no OD elements, but in that model $X$ does not seem to be analytically definable, let alone $Pi^1_2$. The model involves a combination of several forcing notions and some modern ideas in descriptive set theory recently presented in Canonical Ramsey Theory on Polish Spaces. Thus whether a $mathsf E_0$-class of a non-OD real can be $Pi^1_2$ is probably still open.

Further Kanovei's addendum of Aug 23.
It looks like a clone of Jensen's forcing on the base of Silver's (or $mathsf E_0$-large Sacks) forcing instead of the simple Sacks one leads to a lightface $Pi^1_2$ generic $mathsf E_0$-class with no OD elements. The advantage of Silver's forcing here is that it seems to produce a Jensen-type forcing closed under the 0-1 flip at any digit, so that the corresponding extension contains a $mathsf E_0$-class of generic reals instead of a generic singleton. I am working on details, hopefully it pans out.

Further Kanovei's addendum of Aug 25.
Yes it works, so there is a generic extension $L[x]$ of $L$ by a real in which the
$mathsf E_0$-class $[x]_{mathsf E_0}$ is a lightface $Pi^1_2$ (countable) set with no OD elements. I'll send it to Axriv in a few days.

Further Kanovei's addendum of Aug 29. arXiv:1408.6642

mg.metric geometry - Can Lipschitz maps increase the Lebesgue dimension ?

Let me add to the examples by Benoit and Victor another Cantor example, this time a straightforward naive one rather than ingenious.

Consider at and just after stage $0$ a closed interval $I$ with the standard (Euclidean) metrics but of length $frac{11}{10}$. At stage $k>0$ remove the center open interval of length $frac 1{2^{k-1}times 11^k}$ of each interval left after the previous stage $k-1$. After all stages $0 1 ldots$, in the remaining set $C$ in addition to the Euclidean metrics consider also the following pseudo-metrics:

$$d(x y) = |x-y| - s_{x y}$$

where $s_{x y}$ is the sum of the lengths of all removed intervals which are between points $x y$. The identity map from Euclidean $C$ to $C$ with the pseudo-metric $d$ is Lipschitz with constant 1. Let $C'$ be the metric space induced by $C$. Then $C'$ is homeomorphic to a nondegenerated closed interval, and the map induced by the identity on $C$ is Lipschitz with constant 1.

Actually, C' is isometric with the unit Euclidean interval $[0;1]$.

fibre bundles - What manifold has $mathbb{H}P^{odd}$ as a boundary?

This question is motivated by What manifolds are bounded by RP^odd? (as well as a question a fellow grad student asked me) but I can't seem to generalize any of the provided answers to this setting.

Allow me to give some background. Take all (co)homology groups with $mathbb{Z}_2$ coefficients.

Given a smooth compact manifold $M^n$, let $w_i = w_i(M)in H^i(M)$ denote the Stiefel-Whitney classes of (the tangent bundle of) M. Let $[M]in H_n(M)$ denote the fundamental class (mod 2). Consider the Stiefel-Whitney numbers of $M$, defined as the set of all outputs of $ langle w_{i_1}...w_{i_k} , [M] rangle$. Of course this is only interesting when $sum i_{j} = n$.

Pontrjagin proved that if $M$ is the boundary of some compact n+1 manifold, then all the Steifel-Whitney numbers are 0. Thom proved the converse - that if all Stiefel-Whitney numbers are 0, then $M$ can be realized as a boundary of some compact n+1 manifold.

As a quick aside, the Euler characteristic $chi(M)$ mod 2 is equal to $w_n$. Hence, we see immediately that if $chi(M)$ is odd, then $M$ is NOT the boundary of a compact manifold.

As an immediate corollary to this, none of $mathbb{R}P^{even}$, $mathbb{C}P^{even}$, nor $mathbb{H}P^{even}$ are boundaries of compact manifolds.

Conversely, one can show that all Stiefel-Whitney numbers of $mathbb{R}P^{odd}$, $mathbb{C}P^{odd}$ and $mathbb{H}P^{odd}$ are 0, so these manifolds can all be realized as boundaries.

What is an example of a manifold $M$ with $partial M = mathbb{H}P^{2n+1}$ (and please assume $n>0$ as $mathbb{H}P^1 = S^4$ is obviously a boundary)?

The question for $mathbb{R}P^{odd}$ is answered in the link at the top. The question for $mathbb{C}P^{odd}$ is similar, but slightly harder:

Consider the (standard) inclusions $Sp(n)times S^1rightarrow Sp(n)times Sp(1)rightarrow Sp(n+1)$. The associated homogeneous fibration is given as

$$Sp(n)times Sp(3)/ Sp(n)times S^1rightarrow Sp(n+1)/Sp(n)times S^1rightarrow Sp(n+1)/Sp(n)times Sp(1),$$ which is probably better recognized as

$$S^2rightarrow mathbb{C}P^{2n+1}rightarrow mathbb{H}P^{n}.$$

One can "fill in the fibers" - fill the $S^2$ to $D^3$ to get a compact manifold $M$ with boundary equal to $mathbb{C}P^{2n+1}$.

I'd love to see $mathbb{H}P^{odd}$ described in a similar fashion, but I don't know if this is possible.

Assuming it's impossible to describe $mathbb{H}^{odd}$ as above, I'd still love an answer along the lines of "if you just do this simple process to this often used class of spaces, you get the manifolds you're looking for".

Thanks in advance!

Do Roche limits apply to black holes?

The Roche limit applies when a smaller body that would be held together by its own self-gravity is in the gravitational field of another, such that the tidal forces of the latter are stronger than the self-gravity of the latter, thus destroying the smaller body.

However, the gravitational tidal forces of a black hole are always finite, except at the internal singularity. This is a problem because the self-gravity of a black hole, in the sense of the acceleration a mass would need to remain stationary on its surface, is infinite¹. Thus, we shouldn't expect for a large black hole to destroy another through gravitational tidal forces.

Put another way, the Roche limit occurs when particles from the smaller body can escape them... but they can't escape the event horizon of a black hole. Thus, the black holes will either orbit or merge, which is what happens in numerical simulations.

^{1There is a separate concept of surface gravity of a black hole that's essentially this re-scaled by the gravitational time dilation factor, and thus kept finite.}

ag.algebraic geometry - Are there Néron models over higher dimensional base schemes?

This is really a comment in response to JBorger and Qing Liu's questions about existence of Néron models after blowing up or altering the base, but is too long for the comment box.

In general, Néron models do not exist over bases of dimension greater than 1, even allowing alterations of the base. This non-existence seems quite robust - it does not help if you allow Néron lft models, or allow your Néron model to be an algebraic space, or…

The simplest example is probably to take $S = operatorname{Spec} mathbb{C}[[u,v]]$ (complete, regular, local,...), and to let $C/S$ be the nodal curve in weighted projective space $mathbb{P}(1,1,2)$ over $S$ given by the affine equation
$$y^2 = (x-1)(x-1-u)(x+1)(x+1+v).$$
If you let $J$ be the jacobian of the generic fibre of $C/S$, then $J$ does not admit a Néron model over $S$, or even over $S’$ where $S’ rightarrow S$ is proper surjective (for example, an alteration). I do not know a very short proof of this latter fact; it can be found in
http://arxiv.org/abs/1402.0647

More generally, given a nodal curve over a regular separated base, the jacobian will usually not admit a Néron model. There are two cases where a Néron model clearly does exist: if the curve is of compact type, or if it arises as pullback along a smooth morphism from a curve over a DVR. It turns out that these two situations, together with `combinations of the two’, are in some sense the only situations where Néron models do exist. Moreover, altering the base will usually make no difference to the existence of a Néron model. More precise statements can be found in the above reference.

spectroscopy - How does one determine the effective temperature of a star from its spectrum?

Temperature ($T_{eff}$) can be quite tricky to determine accurately as it interrelates to a number of other fundamental measurements.

Firstly, remember that the spectrum we observe from stars are pin-point, they give us the entire overall result and not a specific location or part of the star. We need to dissect the various parts to arrive at the fundamental parameters. We arrive at our results by iterating the values of the fundamental parameters until a model spectrum matches the true spectrum we observe. The issue is, like you say, the existence of a whole lot of uncertainties.

The first of these (although it doesn't have a large effect) is the Uncertainty Principle itself. This creates natural line broadening due to the emitted photon having a range of frequencies. The width of the line is determined by;

$$Delta E approx frac{h}{T_{decay}}$$

where $Delta E$ is the uncertainty in the energy,
$h$ is the Planck constant, and
$T_{decay}$ is the amount of time the electron stays in a high energy state before decaying.

Fundamental parameters

The rotation of the star causes a Doppler shift effect on the line spectra making it broaden. The faster the rotation, the broader (yet smaller) the line. Like the Uncertainty Principle, this is natural broadening as it doesn't impact the abundance of any particular element in the star.

Measuring the rotational velocity ($V_{proj}$) depends on both its axis of rotation and our line of sight to the star. Therefore, we use a combination of both velocity about the equator ($v_e$) and the star’s polar inclination ($i$) to determine the projected radial velocity;

$$V_{proj} = v_e sin i$$

Temperature ($T_{eff}$)impacts wavelength in such a way that higher temperatures impart higher random motions on the atoms. When these photons collide with an atom, they can cause the atom to become ionised, i.e. lose an electron. Different energy levels (and therefore temperature) will create different abundances at the various ionisation stages of atoms.

The temperature of the stellar photosphere decreases as we move away from the core. Therefore the line profile represents a range of temperatures. The wings of the line arise from deeper, hotter gas which displays a larger range of wavelengths due to increased motion. The higher the temperature, the broader the wings of the line profile ([Robinson 2007, pg 58][1]).

Here you can see the effect of various temperature values on the synthetic spectral line of FE I 6593 A. Red: $T_{eff}$ = 4000K; Black: $T_{eff}$ = 5217K; Blue: $T_{eff}$ = 6000K;

Microturbulence ($v_{mic}$) is the non-thermal localised random motion of the stellar atmosphere. It works in a similar way to temperature - an increase in the motion of atoms creates a wider range of wavelengths observed and therefore broader line profiles.

In strong lines, saturation can occur when there aren’t any more photons to be absorbed. As microturbulence in these areas increases, it presents more opportunities for photons to be absorbed. This broadens the wings of the line profile increasing the overall strength of the line. We can use this fact to determine $v_{mic}$, by ensuring that the strength of the lines (equivalent width) have no correlation with their abundances.

Finally, surface gravity which is a function of the star’s mass and size:

$$log g = log M - 2 log R + 4.437$$

with $M, R$ being in solar units and $g$ in cgs.

A star with a higher mass but smaller radius will invariably be denser and under greater pressure. By definition, denser gas has a higher number of atoms per unit of area (abundance), leading to stronger spectral lines.

A gas under pressure provides more opportunities for free electrons to recombine with ionised atoms. For a given temperature, ionisation is expected to decrease with an increase of surface gravity, in turn increasing the abundance of atoms in the neutral or low ionisation states.

The measuring of $T_{eff}$

As we've seen, there are a number of ways in which a spectrum of a star can be altered. The one you're interested in is temperature. As temperature is interrelated to all the other fundamental parameters, we need to treat them together as a whole and tease out the value of $T_{eff}$.

We begin with a synthetic spectrum and modify its properties iteratively until it matches the shape of the star’s spectrum. Adjustments of one parameter will invariably affect the others. The spectra will match when the temperature, surface gravity, and microturbulence values (amongst others) are correct. This is obviously very time consuming although programs exist to help.

Atmospheric properties can also be determined by other less time consuming means. Photometric colours can be used as a proxy for temperature, and absolute magnitudes for surface gravity. However, these determinations can suffer from inaccuracies due to interstellar extinction and are at best a close approximation.

[1] Robinson, K. 2007, Spectroscopy: The Key to the Stars (Springer)

st.statistics - discrete stochastic process: exponentially correlated Bernoulli?

There is a question that was asked on stackoverflow that at first sounds simple but I think it's a lot harder than it sounds.

Suppose we have a stationary random process that generates a sequence of random variables x[i] where each individual random variable has a Bernoulli distribution with probability p, but the correlation between any two of the random variables x[m] and x[n] is α^|m-n|.

How is it possible to generate such a process? The textbook examples of a Bernoulli process (right distribution, but independent variables) and a discrete-time IID Gaussian process passed through a low-pass filter (right correlation, but wrong distribution) are very simple by themselves, but cannot be combined in this way... can they? Or am I missing something obvious? If you take a Bernoulli process and pass it through a low-pass filter, you no longer have a discrete-valued process.

(I can't create tags, so please retag as appropriate... stochastic-process?)

teaching - Undergraduate Derivation of Fundamental Solution to Heat Equation

I think the term fundamental solution (at least sometimes) conventionally includes the integral around your $K$. I will assume this. If I recall correctly then the following argument is from "Partial Differential Equations" by Strauss.

A particularly simple solution follows from the self-similarity principle, i.e.

If $u(x,t)$ is a solution then so is $u(cx, a c^2t)$

This suggests looking for a particular solution of the form $K(x,t) = g(p)$, where $p = frac{x}{sqrt{4at}}$

Substituting $g$ into the heat equation leads to the differential equation

$$g''+frac{p}{2}g' = 0 $$

Then the fundamental solution as above follows from solving this.

mass - How massive can a star be at birth?

I found a text that suggests a Population III star can have a mass between 200 and 10^5$M_{Sun}$. There you can read:

"...This would apply for VMOs larger than $M_c$ $approx$ 200 $M_{Sun}$. Stars larger than 150$M_{Sun}$ are termed supermassive objects (SMOs)...".

It seems supermassive stars like R136a1 (with masses in excess 150$M_{Sun}$, the accepted upper limit for a non-pop III star) can be formed by the collision of massive stars.

So, It is not difficult to imagine, say, two 150$M_{Sun}$ monsters colliding and forming a 300$M_{Sun}$ object.

ag.algebraic geometry - Learning about Lie groups

For someone with algebraic geometry background, I would heartily recommend Procesi's Lie groups: An approach through invariants and representations. It is masterfully written, with a lot of explicit results, and covers a lot more ground than Fulton and Harris. If you like "theory through exercises" approach then Vinberg and Onishchik, Lie groups and algebraic groups is very good (the Russian title included the word "seminar" that disappeared in translation). However, if you want to learn about the "real" side of Lie groups, both in linear and abstract manifold setting, my favorite is Godement's "Introduction à la théorie des groupes de Lie".

Several of the books mentioned in other answers are devoted mostly or entirely to Lie algebras and their representations, rather than Lie groups. Here are more comments on the Lie group books that I am familiar with. If you aren't put off by a bit archaic notation and language, vol 1 of Chevalley's Lie groups is still good. I've taught a course using the 1st edition of Rossmann's book, and while I like his explicit approach, it was a real nightmare to use due to an unconscionable number of errors. In stark contrast with Complex semisimple Lie algebras by Serre, his Lie groups, just like Bourbaki's, is ultra dry. Knapp's Lie groups: beyond the introduction contains a wealth of material about semisimple groups, but it's definitely not a first course ("The main prerequisite is some degree of familiarity with elementary Lie theory", xvii), and unlike Procesi or Chevalley, the writing style is not crisp. An earlier and more focused book with similar goals is Goto and Grosshans, Semisimple Lie algebras (don't be fooled by the title, there are groups in there!).

data analysis - Classification of Planets?

The data from Kepler has taught us that there is a much larger number of "Neptune-class" planets out there than we previously thought. I wonder, however, if this "dominance" of Neptunes is because of the wide range of radii we use to define this class. If I'm not mistaken, we classify a planet as "Earth-like" when it is in the range 0.75 to 1.25 R(Earth), "Super-earth" from about 1.25 to 2.0 R(Earth) and "Neptune class" from 2.0 to 6.0 R(Earth) (a much larger range which would stand out even more if we knew masses). Are we maintaining the nomenclature to classify planets of different characteristics or are we just "solar system biased"?

Why is Milnor K-theory not ad hoc?

To help answer Question 1, Milnor proved a local-global theorem for Witt rings of global fields. Recall that The Grothendieck-Witt ring $widehat{W}(k)$ of a field $k$ is the ring obtained by starting with the free abelian group on isomorphism classes of quadratic modules and moding out by the ideal generated by symbols of the form $[M]+[N]-[M']-[N']$, whenever $[M]oplus[N]simeq [M']+[N']$. The multiplication comes from tensor product of quadratic modules. There is a special quadratic module $H$ given by $x^2-y^2=0$. This is the hyperbolic module. The Witt ring $W(k)$ of a field $k$ is the quotient of $widehat{W}(k)$ by the ideal generated by $[H]$.

Now, the main theorem of Milnor's paper is that there is a split exact sequence $$0rightarrow W(k)rightarrow W(k(t))rightarrow bigoplus_pi W(overline{k(t)}_pi)rightarrow 0,$$ where $pi$ runs over all irreducible monic polynomials in $k[t]$, and $overline{k(t)}_pi$ denotes the residue field of the completion of $k(t)$ at $pi$.

The morphisms $W(k(t))rightarrow W(overline{k(t)}_pi)$ come from first the map $W(k(t))rightarrow W(k(t)_pi)$. Then, there is a map $W(k(t)_pi)rightarrow W(overline{k(t)}_pi)$ that sends the quadratic module $upi x^2=0$ to $ux^2=0$, where $u$ is any unit of the local field.

Interestingly, Milnor $K$-theory is not used in the proof. However, the proof for Witt rings closely models the proof of a similar fact for Milnor $K$-theory: the sequence $$0rightarrow K_n^M(k)rightarrow K_n^M(k(t))rightarrowbigoplus_pi K_{n-1}^M(overline{k(t)}_pi)rightarrow 0.$$

The important new perspective is the formal symbolic perspective, which was already existent for lower $K$-groups, but is very fruitful for studying the Witt ring as well.

fa.functional analysis - What are some interesting sequences of functions for thinking about types of convergence?

I'm thinking about the basic types of convergence for sequences of functions: convergence in measure, almost uniform convergence, convergence in Lp and point wise almost everywhere convergence. I'm looking for examples of sequences of functions that converge in one or more of these ways, but fail for others. I keep seeing the same examples over and over and I'd like to think about some new ones. Here are the examples I've seen:

$f_n=chi_{[n, n+1]}$

$f_n=chi_{A_{n}}$ where $A_1 = [0,1]$, $A_2 = [0,1/2]$, $A_3 = [1/2,1]$, $A_4 = [0,1/4]$, $A_5 = [1/4,1/2]$, $A_6 = [1/2,3/4]$, $A_7 = [3/4,1]$, $A_8 = [0,1/8]$ ...

$f_n= nchi_{[1/n,2/n]}$

These are a great set of examples since they let you give a counterexample for the relations between the types of convergence when needed, but I would like to know of some more.

(EDIT: I added a bit to the 2nd example to make it more clear.)

amateur observing - What are some night sky objects I could see with my Celestron UpClose 20x50 Porro Binocular?

I bought this binocular because I've read countless times that one should start with a pair of binoculars before diving into telescopes. I've seen details of the moon and I can notice Venus round shape on some nights. I also saw Jupiter and it's two larger moons, as faint as the faintest stars, but this was on a beach trip so the sky was clearer than what I'm used to.

I live on a place with a 7 or 6 on the Bortle Scale, is there something else I should try to see?

nt.number theory - Global fields: What exactly is the analogy between number fields and function fields?

Sure, here's a overview.

Suppose you have a ring R over a field k, then, by the magic of algebraic geometry, you can think about it in a geometric way. You do this by defining points as epimorphisms R to k and finding out that a lot of geometric intuition plays out nicely.

Now if you start with a field, the above procedure gives you just a single point, so it's more interesting to find ring inside it — people usually take the ring of integers inside the field, which is uniquely defined.

Now the amazing thing is that you can perform this exact procedure either on number fields like Q or on function fields like F_p(t) and it gives you a very similar geometric structure.

For example, you can talk about completion of your ring by some maximal ideal and this corresponds to considering infinitesimal geometry around a single point. For number fields that would be something like Q_p while for function fields that would be F_p[[t]]. Not if you think how Q_p is basically F_p formally extended by p you notice the techniques wors the same in both cases.

E.g. the theory of ramification is basically the theory of extending either F_p[[t]] or Q_p. (There are important differences though — F_p[[t]] can be extended with F_{p^2}[[t]])

ct.category theory - Definition of Category of Locales

Well, as you say, these two categories are isomorphic, so it's going to be hard to say how they differ! They only differ in the names the maps are given.

Maybe it would help to recap the definitions. I'll take them from p.39-40 of Peter Johnstone's book Stone Spaces.

A frame is a complete lattice $A$ satisfying the infinite distributive law
$$
a wedge bigvee S = bigvee { a wedge s | s in S }
$$
($a in A, S subseteq A$). A homomorphism of frames is a function preserving finite meets and arbitrary joins. This defines the category Frm of frames.

Note (as you did) that every homomorphism of frames has a right adjoint.

The category Loc of locales is the opposite of the category of frames. Morphisms in Loc are called continuous maps.

Then Johnstone says: "We adopt the convention that if $f: A to B$ is a continuous map of locales, we shall write $f^*: B to A$ for the corresponding frame homomorphism, and $f_*: A to B$ for the right adjoint of $f^*$."

So in Johnstone's convention (which is the one I know), the elements of Loc$(A, B)$ are identified with frame homomorphisms $B to A$. In Borceux's convention, the elements of Loc$(A, B)$ are identified with order-preserving maps $A to B$ that are right adjoint to frame homomorphisms.

I guess the other thing to say is that when you're dealing with ordered sets, adjoints are genuinely unique (not just unique up to isomorphism). So taking the right adjoint of a frame homomorphism is a bijective process.

I don't know what else to say. It's really just a matter of naming.

Sum of product of Fourier series

This is a consequence of the Dominated Convergence Theorem, see This question.

In details:
$$
sum_{wneq0}sum_{s} a_s a_{s+Nw}
$$
is better written
$$
sum_{s}sum_{z} b_{s,z}^N
$$
where
$$
b_{s,z}^N = a_{s}a_{z} mbox{ when } z=s mbox{ mod } N,, Nneq0 mbox{ and } b_{s,z}^N=0 mbox{ otherwise.}
$$
Now, $|b_{s,z}^N|leq |a_{s}||a_{z}|$ for all $N$, and $$sumsum|a_{s}||a_{z}|=(sum |a_{s}|)^2<infty.$$
On the other hand, for any fixed $z, s$
$$
lim_{Nto infty} b_{s,z}^N = 0 ,(mbox{ since } a_{s+Nw} to 0 mbox{ with }N)
$$
Now apply the DCT to conclude.

ag.algebraic geometry - Learning About Schubert Varieties

There is one particular fact that greatly helps me understand Schubert varieties and Schubert cells, and the serves as kind-of an introduction: They are a generalization of row echelon form for matrices. If $V subseteq F^n$ is a $k$-subspace of the standard $n$-space over a field $F$, then it is well-known that $V$ has a unique basis of row vectors in reduced row echelon form. The shape of the form divides the set of all $V$ (the Grassmannian) into cells which are affine spaces. The closures of such a cell — itself plus lower cells — is a Schubert variety. For example, one pattern of RREF for a 2-plane in $F^4$ looks like this:
$$begin{pmatrix} 1 & * & 0 & * \\ 0 & 0 & 1 & * end{pmatrix}$$
This is clearly an affine cell. It is equally easy to show (in the case of a Grassmannian) that the cells are bijective with the $k$-subsets of an $n$-set.

This is not a complete explanation because a Grassmannian is just the simplest type of flag variety, but it captures the basic geometric idea.

As for references, there is an interesting mini-review for combinatorialists on page 398 of Stanley, Enumerative Combinatorics, Vol 2.

ct.category theory - Completeness and cocompleteness of the Kleisli category

In general, the Kleisli category will be neither complete nor cocomplete nor cartesian closed, even if $C$ is.

You can think of the Kleisli category as the full subcategory of algebras whose objects are the free algebras. So take for example $T$ to be the monad coming from the free-forgetful adjunction between abelian groups and sets. In order for $mathbb{Z} times -: FreeAb to FreeAb$ to have a right adjoint (as required by cartesian closure), it would have to preserve all colimits, for example coproducts. But $mathbb{Z} times (mathbb{Z} oplus mathbb{Z})$ is not isomorphic to $(mathbb{Z} times mathbb{Z}) oplus (mathbb{Z} times mathbb{Z})$.

Now let's tackle completeness. We can use the same example; the basic idea is that free abelian groups are not closed under (infinite) products. (In fact, a famous but nontrivial result, due to Kurosh I believe, is that the countably infinite power of $mathbb{Z}$ is not free abelian.) But to apply this idea, one should first check that any limit in $Kl(T)$ really is constructed just as it would be in $Alg(T)$ or in $C$. That's not hard: first note that the underlying functor $U: Kl(T) to C$ is representable (in fact, $U cong hom(F(1), -)$), so $U$ preserves any limits that exist in $Kl(T)$. Also, $U$ reflects isomorphisms (i.e., if $U(g)$ is an isomorphism in $C$, then $g$ is an isomorphism in $Kl(T)$). It follows that $U$ both preserves and reflects limits, and so any limit in $Kl(T)$ is constructed as it would be in $Alg(T)$ or $C$ (here, $Set$).

Finally, cocompleteness. This time I don't think the example above works quite as easily, but one example that does work is to consider the Kleisli category for the free-forgetful adjunction between $mathbb{Z}_6$-modules and sets. I claim that the coequalizer of the pair $id, mu_3: mathbb{Z}_6 to mathbb{Z}_6$ ($mu_3$ is multiplication by 3) does not exist in the category of free $mathbb{Z}_6$-modules. I chose this particular coequalizer because it is an example of "splitting an idempotent"; here the idempotent is $mu_3$. The virtue of splittings of idempotents is that they are preserved by any functor whatsoever, and reflected by any functor that reflects isomorphisms. Given this fact, it follows that splittings of idempotents in $Kl(T)$, if they exist, are constructed just as they would be in $Alg(T)$ or in $C$. But the idempotent splitting of this pair of maps in $mathbb{Z}_6$-Mod is $mathbb{Z}_2$, which is not free. It follows that the coequalizer doesn't exist in the Kleisli category.

intuition - Taylor expansion to show that for Stratonovich stochastic calculus the chain rule takes the form of the classical one

As nobody seems to be able to give any kind of answer to that problem over there at math.stackexchange I post this question here:

How can I show with a heuristic argument based on a Taylor expansion that for Stratonovich stochastic calculus the chain rule takes the form of the classical (Newtonian) one?

The intuition goes like this: Concerning Ito calculus the fact that dX^2 = dt results via a Taylor expansion in Ito's lemma - this fact should stay the same with Stratonovich but it should somehow cancel out in there - I just don't know how...

orbit - Explanation for the mathematics behind Venus' retrograde motion

The really simple explanation is that Venus' orbital period is 225 days. Earth's is 365 days.

That ratio is 13:8 (well, more accurately 12.9777777777..:8) so you can see that your complete cycle is 13 orbits of Venus to 8 of Earth, but within that cycle our view of Venus travelling round the sun gives the pretty pattern as shown in your image.

This YouTube video is quite a nice visualisation - should make it easy to understand.

pr.probability - When does a pointwise CLT hold?

Feller states the Berry-Esseen theorem in the following way. Let the $X_k$ be independent variables with a common distribution $F$ such that $$E[X_k]=0, E[X_k^2]=sigma^2>0, E[|X_k|^3]=rho<infty,$$ and let $F_n$ stand for the distribution of the normalized sum $$(X_1+ dots X_n)/(sigma sqrt{n}).$$ Then for all $x$ and $n$ $$|F_n(x)-N(x)| leq frac{3rho}{sigma^3 sqrt{n}}.$$

The expression you are interested in is
$$left|frac{F_n(epsilon)-F_n(-epsilon)-N(epsilon)+N(-epsilon)}{epsilon}right|,$$
which is less than
$$left| frac{F_n(epsilon)-N(epsilon)}{epsilon} right| + left| frac{F_n(-epsilon)-N(-epsilon)}{epsilon} right|,$$
which by Berry-Esseen is bounded by
$$2frac{3rho}{epsilon sigma^3 sqrt{n}}.$$
So, if $epsilonsqrt{n}$ goes to infinity, then you are good.

I realize this isn't what you asked, in that you wanted conditions on $X$, and this instead gives you conditions on $epsilon_n$. Still, perhaps it'll help.

Reference: Feller, An Introduction to Probability Theory and Its Applications, Volume II, Chapter XVI.5.

redshift - Have we detected galaxies which have red-shifted beyond the visible light range?

Yes, of course. Many, many examples. Telescopes work in the infrared, far-infrared and there are even samples of galaxies that are selected on the basis of their mm emission.

The most distant galaxies detected now have redshifts of 7 or more. This means the wavelength of their light has been stretched by a factor $1+z$ - i.e. by a factor of 8. Thus light in the visible range, say 500nm, now appears at wavelength of 4 microns, in the infrared.

Telescopes that work in this range include the Spitzer space telescope; and many ground-based telescopes. Observations of highly redshifted galaxies are routinely made at infrared wavelengths on telescopses all around the world.

Galaxies are also detected in the far infrared by the Herschel satellite or at mm (getting on for microwave) wavelengths by JCMT or the ALMA telecope.

computational complexity - What impact would P!=NP have on the characterization of BQP?

David is right about one thing. Scott had a discussion about this on his blog and I was also involved.

On the one hand, many complexity theorists simply also assume that BQP does not contain NP, just as they assume that P does not contain NP. The evidence for the former is not as dramatic as that for the latter, but there is at least an oracle separation. I.e., there is an oracle A such that BQP^A does not contain NP^A. Now, there are some famous cases where two complexity classes are equal or there is an inclusion, even though there is also a credible oracle separation. But the oracle separations for BQP vs NP seem realistic. Besides, apart from tangible evidence, I for one consider BQP to be surprisingly powerful but not incredibly powerful. It's my intuition partly because I expect BQP to be realistic and I don't expect the universe to be perverse. I think of BQP as an extension of randomized computation based on quantum probability.

On the other hand, P vs PSPACE is already an unfathomable open problem. The two main barrier results for P vs NP, Baker-Gill-Solovay and Razborov-Rudich, apply to P vs PSPACE equally well. Since PSPACE contains both NP and BQP, if you were to show that either one does not equal P, then in particular you would show that PSPACE does not equal P. Actually, I don't know a good reason to try to prove that P ≠ NP rather than to first prove that P ≠ PSPACE, since the latter is at least formally easier.

mg.metric geometry - Ordered geometries from convex subsets of the plane

Convex sets give the same geometries if and only if they are projectively equivalent. In particular, it is only conics that give $H^2$.

It is more natural to work in the projective plane $P^2(mathbb{R})$. Then we define a set to be convex if its intersection with any line is empty or connected. We are given two open convex sets $C_1$ and $C_2$ in the plane with a bijection $phi:C_1to C_2$ which is order preserving in the sense that $a$ and $b$ separates $c$ and $d$ (on some projective line) if and only if $phi(a)$ and $phi(b)$ separates $phi(c)$ and $phi(d)$ (on some projective line).

We would like to prove that $phi$ is a projective transformation.

The following attempt uses the theorem of Desargues together with the fundamental theorem of projective geometry.

Claim. $phi$ can be extended to the whole of $P^2(mathbb{R})$ such that lines are mapped to lines.

Consider any point $xnotin C_1$. We locate $phi(x)$ by using the Theorem of Desargues as follows. Choose three lines through $x$ that intersect $C_1$ in three connected sets $c_1$, $c_2$, $c_3$. Choose points $a_i$ and $b_i$ on chord $c_i$. Then the triangles $triangle a_1 a_2 a_3$ and $triangle b_1 b_2 b_3$ are in perspective, so by the theorem of Desargues, the intersection points $p_{ij}$ of the lines $a_i a_j$ and $b_i b_j$ are collinear. This can be done in such a way that the $p_{ij}$ are all in $C_1$. For instance, the $c_i$ has to be chosen sufficiently close together, and each triangle has to be chosen so that its points are "almost collinear", with the two lines of collinearity intersecting inside $C_1$.

Then this picture of the triangles in perspective, without the point $x$, can be transferred to $C_2$ using $phi$. All incidences are preserved, so by the converse of the theorem of Desargues, the three connected sets $phi(c_i)$ lie on concurrent lines. Define $phi(x)$ to be the point of concurrency. It is easy to see that the definition is independent of which chords through $x$ are used.

We are halway there. It remains to prove that the extended $phi$ preserves collinearity.

So let $x,y,z$ be collinear in $mathbb{R}^2$. We would like to show that $phi(x), phi(y), phi(z)$ are collinear.

If at least two of $x,y,z$ are in $C_1$, it is clear that their images will also be collinear. So assume without loss of generality that $y,znotin C_1$.

The case $xin C_1$ is simple: the chord of $C_1$ through $x,y,z$ maps to the chord of $C_2$ through $phi(x)$ and $phi(y)$, and also to the chord through $phi(x)$ and $phi(z)$. Thus $phi(x),phi(y),phi(z)$ are collinear.

If on the other hand, $xnotin C_1$, we again use the theorem of Desargues. Find triangles inside $C_1$ such that the points in which their corresponding sides intersect, all lie on the line through $x,y,z$. (As before, it is easy to see that this is possible.) By the converse of Desargues, the triangles are in perspective. Transfer this picture with $phi$ to the plane in which $C_2$ lives. We again get two triangles in perspective, and by Desargues, $phi(x)$, $phi(y)$, $phi(z)$ are collinear.

We have shown that lines are mapped onto lines. By (a very special case of) the fundamental theorem of projective geometry, $phi$ is a projective transformation.