ra.rings and algebras - Characterizing nilpotents in a ring by a universal property

Let $e in A$ be a non-zero idempotent (and hence not nilpotent).
Then if $f(e)$ is a unit, we find that $f(e) = 1,$
and so $f(e - 1) = 0.$ Thus if $e - 1$ generates (as a two-sided ideal) the entire
ring, we find that $f$ is identically zero, and hence that $B = 0$.

Thus, if we can find a non-zero idempotent $e in A$ such that $A(1-e)A = A$,
we have a counterexample.

Note by the way that $f: = 1 - e$ is again idempotent, and so it suffices instead
to find a non-unital idempotent $f$ such that $A f A = A$.

E.g. If $A$ is simple (so that any non-zero two-sided ideal equals $A$), any non-unital and
non-zero idempotent gives a counterexample.

E.g. if $A = M_2(k)$ for some field $k$, and $f = (1 0 , 0 0)$, we are done. (I think
this is what Kevin intended to write down in his comment.)

observatory - Why do astronomers like green laser pointers?

Mostly general purpose laser pointers are used for pointing things at smaller distances eg. Diagram or Equations in Powerpoint presentations, so the power of such laser pointers is quite limited/restricted to 5mw ( Class 3A or IIIa) or 10mW in some regions.
Because of the low power and small aperture of laser pointers if you point them through empty space, you can only see them impact spot where it hits a surface.

So to point stars/planets as in astronomy club during observation sessions these low power laser pointers are not useful as there is no surface where the beam can hit & you can see the spot.

Some higher-powered laser pointers project a visible beam via scattering from dust particles or water droplets along the beam path.
Higher-power ( Class 3B or IIIb lasers: greater than 5mW) and higher-frequency (Green or Blue) lasers may produce a beam visible even in clean air because of "Rayleigh Scattering" from air molecules, especially when viewed in moderately-to-dimly lit conditions. The strong wavelength dependence of the scattering (~λ^ −4 ) means that shorter (green & blue) wavelengths are scattered more strongly than longer (red) wavelengths.

[ Rayleigh scattering is the mainly the elastic scattering (in lay term collision) of light or other electromagnetic radiation by particles much smaller than the wavelength of the light.
Rayleigh scattering results from the electric polarizability of the particles (locally seperating positive & negstive charges by small distance apart inside the atom. i.e. temporarily separating centre of positive charge and centre of negative charge). The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency (hence the green colour). The particle therefore becomes a small radiating dipole (or small sources of light) whose radiation we see as scattered light (the beam passing from pointer to star pointed). ]

The intensity of such scattering increases when these beams are viewed from angles near the beam axis. (Thats the reason, if you standing close to the person pointing stars in astronomy club, you can see the beam bright n clearly whereas the ppl standing away can barely see it. So always stand near the presenter But avoid contact with your eye. :-) )

The apparent brightness of a spot from a laser beam depends on the optical power of the laser, the reflectivity of the surface, and the chromatic response of the human eye. For the same optical power, Green Laser light will seem brighter than other colors because the human eye is most sensitive at low light levels in the green region of the spectrum (wavelength 520–570 nm). The most sensitive pigment, rhodopsin, has a peak response at 500 nm. Sensitivity decreases for redder or bluer wavelengths.

Further, Green laser pointers are of moderate power (sufficient for Rayleigh Scattering in clean Air), they are compact and relatively cheap than blue one (even though human eye is less sensitive to blue colour, blue laser are little bit more expensive than green one.)

Avoid their contact with your eyes and dont point them at aircrafts.
People have been given up to five years in jail for aiming a green laser at an aircraft.

fa.functional analysis - Saito-Wright definition of Rickart C*-algebras

A C*-algebra is Rickart if for each $xin A$ there is a projection $pin A$ so that
$R(x)=pA$.
Here the right-annihilator $R(S)$ of $Ssubset A$ is defined
as $$R(S)={ain Amid xa=0, forall xin S}$$ and $R(x)equiv R({x})$.

In:

Kazuyuki Saito and J. D. Maitland Wright. $C^∗$-algebras which are Grothendieck
spaces. Rend. Circ. Mat. Palermo (2), 52(1):141–144, 2003.

an alternative definition is studied:
define a C*-algebra to be Rickart if each maximal Abelian *-subalgebra of $A$ is Rickart (or, equivalently, monotone $sigma$-complete).
Equivalently, one may require that every Abelian *-subalgebra is contained in an
Abelian Rickart C*-algebra.

This definition is more general and seems to be sufficient for many applications.

Is this definition in fact equivalent to the original one?

This question recently came up in our investigations in the foundations of quantum theory:

Bohrification of operator algebras and quantum logic

applications - Algebraic geometry used "externally" (in problems without obvious algebraic structure).

There exists a unique map from permutations of 1,2,3... that move finitely many numbers, to their "Schubert polynomials" in ${mathbb Z}[x_1,x_2,...]$, satisfying the following recursion: $S_{id} = 1$, and if $w(i) > w(i+1)$, then $S_{w r_i} = (S_w - r_i cdot S_w) / (x_i - x_{i+1})$. (Here $r_i$ switches $i$ and $i+1$, or $x_i$ and $x_{i+1}$.)

It's not too hard to prove that these are polynomials, form a basis of the polynomial ring, have positive coefficients, and much else. The nonobvious theorem is that the structure constants (expanding a product of two basis elements in the basis) are positive. The only proofs known of this are geometric.

(This is perhaps a lame example, in that the motivation for Schubert polynomials was geometric -- they represent the classes of Schubert varieties in the cohomology rings of flag manifolds.)

star - Why are orbits elliptical instead of circular?

Assume the planet has a negligible mass compared to the star, that both are spherically symmetric (so Newton's law of gravitation holds, but this normally happens to a very good approximation anyway), and that there aren't any forces besides the gravity between them. If the first condition does not hold, then the acceleration of each is going to be towards the barycenter of the system, as if barycenter was attracting them a gravitational force with a certain reduced mass, so the problem is mathematically equivalent.

Take the star to be at the origin. By Newton's law of gravitation, the force is $mathbf{F} = -frac{mmu}{r^3}mathbf{r}$, where $mathbf{r}$ is the vector to the planet, $m$ is its mass, and $mu = GM$ is the standard gravitational parameter of the star.

Conservation Laws

Because the force is purely radial $(mathbf{F}parallelmathbf{r})$, angular momentum $mathbf{L} = mathbf{r}timesmathbf{p}$ is conserved:
$$dot{mathbf{L}} = frac{mathrm{d}}{mathrm{d}t}left(mathbf{r}timesmathbf{p}right) = m(dot{mathbf{r}}times dot{mathbf{r}}) + mathbf{r}timesmathbf{F} = mathbf{0}text{.}$$
If the initial velocity is nonzero and the star is at the origin, then in terms of the initial position and velocity, the orbit must be confined to the plane of all points with vectors $mathbf{x}$ from the origin that satisify $mathbf{L}cdotmathbf{x} = 0$. If the initial velocity is zero, then the motion is purely radial, and we can take any one of infinitely many planes that contain the barycenter and initial position.

The total orbital energy is given by
$$mathcal{E} = frac{p^2}{2m} - frac{mmu}{r}text{,}$$
where the first term part is the kinetic energy and the second term is the gravitational potential energy of the planet. Its conservation, as well as the fact that it invokes the correct potential energy, can be proven by the fundamental theorem of calculus for line integrals.

Define the Laplace-Runge-Lenz vector to be
$$mathbf{A} = mathbf{p}timesmathbf{L} - frac{m^2mu}{r}mathbf{r}text{.}$$
It is also conserved:
$$begin{eqnarray*}
dot{mathbf{A}} &=& mathbf{F}timesmathbf{L} + mathbf{p}timesdot{mathbf{L}} - frac{mmu}{r}mathbf{p} + frac{mmu}{r^3}(mathbf{p}cdotmathbf{r})mathbf{r}\
&=& -frac{mmu}{r^3}underbrace{left(mathbf{r}times(mathbf{r}timesmathbf{p})right)}_{(mathbf{r}cdotmathbf{p})mathbf{r} - r^2mathbf{p}} - frac{mmu}{r}mathbf{p} + frac{mmu}{r^3}(mathbf{p}cdotmathbf{r})mathbf{r}\
&=& mathbf{0}text{.}
end{eqnarray*}$$

Finally, let's also take $mathbf{f} = mathbf{A}/(mmathcal{E})$, which has the same units as $mathbf{r}$, and since $mathbf{L}cdotmathbf{f} = 0$, it lies along the orbital plane. As it's a conserved vector scaled by a conserved scalar, it's easy to show that $mathbf{f}$ is conserved as well, as long as $mathcal{E}neq 0$.

Simplifying

By employing the vector triple product, we can write
$$begin{eqnarray*}
frac{1}{m}mathbf{A} &=& frac{1}{m}left[p^2mathbf{r}-(mathbf{p}cdotmathbf{r})mathbf{p}right] -frac{mmu}{r}mathbf{r}\
&=& left(mathcal{E}+frac{p^2}{2m}right)mathbf{r} - frac{1}{m}left(mathbf{p}cdotmathbf{r}right)mathbf{p}\
mathcal{E}(mathbf{f}-mathbf{r}) &=& left(frac{p^2}{2m}right)mathbf{r} - frac{1}{m}left(mathbf{p}cdotmathbf{r}right)mathbf{p}text{,}
end{eqnarray*}$$
the norm-squared of which is easy to crank out:
$$mathcal{E}^2|mathbf{f}-mathbf{r}|^2 = left(mathcal{E} + frac{mmu}{r}right)^2r^2text{,}$$
where $mathcal{E}$ was used throughout to switch between kinetic and potential terms.

Why Ellipses?

Since $mathcal{E}$ is energy relative to infinity, to have a bound orbit we need $mathcal{E}<0$. Thus, from the previous section, $|mathbf{f}-mathbf{r}| = -mathcal{E}^{-1}left(mathcal{E}r + mmuright)$ and therefore
$$|mathbf{f}-mathbf{r}| + |mathbf{r}| = -frac{mmu}{mathcal{E}}text{,}$$
which defines an ellipse with foci $mathbf{0},,mathbf{f}$ and major axis $2a=-mmu/mathcal{E}$.

Why Not Circles?

The circle is a special case where the foci are the same point, $mathbf{f} = mathbf{0}$, which can be restated as
$$mathcal{E} = -frac{1}{2}frac{mmu}{r} = -frac{p^2}{2m}text{.}$$
In other words, circular orbits require the orbital energy to be the negative of the kinetic energy. This is possible, but almost certain not to hold exactly. Since any values of $mathcal{E}<0$ are allowed for bound orbits, there are many more ways to have elliptic orbits. (Although some of them would actually crash because the star and planet have positive size.)

Note that hyperbolic orbits have $mathcal{E}>0$, and we can still find the foci using the above method, though being careful with the signs. For $mathcal{E}=0$, the second focus $mathbf{f}$ is undefined because this is a parabolic orbit, and parabolas only have one focus within a finite distance from the center.

Additionally, the eccentricity vector $mathbf{e} = mathbf{A}/(m^2mu)$ is an alternative choice for the LRL vector; as the name suggests, its magnitude is the orbital eccentricity.

gravity - How much would you weigh if Earth stopped orbiting the Sun?

In a sense, this is one of those "If I break the laws of physics, what do the laws of physics say will happen" kinds of questions.

Ignoring that, a 150 pound person would still weigh about 150 pounds if the Earth miraculously stopped orbiting the Sun. Almost all of a person's weight results from resistance to the Earth's gravity field. The tidal gravitational forces from the Moon and Sun are on the order of 10^-4 ounces of force.

Note: I'm answering the question from the perspective of "what would a bathroom scale say". A doctor's scale would say a 150 pound person weighs 150 pounds regardless of whether the person is standing on the Moon, the Earth, or a very massive planet. A doctor's scale measures mass. A bathroom scale measures apparent weight.

So what would a bathroom scale say? If the Earth is at one astronomical unit from the Sun, whether the Earth is orbiting at 30 km/s or falling straight toward the Sun doesn't matter one iota. All that matters is that the Earth is one AU from the Sun. In both cases, that the Sun is 1 AU distant makes a 150 pound person weigh about $1.8 times 10^-4$ ounces less at noon or at midnight than at sunrise or sunset.

The difference between the orbiting Earth and the in-falling Earth is that these tidal forces are more or less constant in the case of the orbiting Earth. They grow ever larger in the case of the in-falling Earth as it gets closer and closer to the Sun.

These tidal forces become quite large by the time the surface of the Earth just touches the surface of the Sun. Now our 150 pound person would weigh only 73.5 pounds per a bathroom scale if he was standing at that point where the surface of the Earth just touches the Sun. He would also weigh about the same at the point on the Earth furthest from the Sun. He would however weigh considerably more, about 188 pounds, if he happens to be at a point where the Sun is on the horizon.

If, on the other hand, some mysterious, magical force makes the Earth stop orbiting the Sun and makes this change permanent (i.e., the Earth stops moving with respect to the Sun), a 150 pound person closest to the Sun would weigh about 1.4 ounces less, while a 150 pound person furthest from the Sun would weigh about 1.4 ounces more. A 150 pound person who sees the Sun at the horizon would still weigh about 150 pounds in this scenario.

Formulas to determine the illuminated phase and orientation of the Moon

The phase of the Moon (the bright limb) looks different depending on where on Earth the observer is located and the time of the observation. I've located this video and the accompanying spreadsheet that supposedly handles this sort of calculation, however I am somewhat confused on what formulas are actually used. I would greatly appreciate if someone with a bit more experience could decipher the spreadsheet a bit and perhaps write down or link me to the formulas being used.

I'm most interested in determining the illuminated phase (i.e the size of the bright limb) and the apparent orientation of the illuminated portion relative to the horizon. My goal is to understand how this type of calculation is done, in order to re-implement it in code for an application to use - displaying the typical crescent moon icon for night-time, but at the correct angle for a specific location and time.

PS! First post here, so I hope I have used the correct terminology for things.

EDIT: Eventhough the author of the spreadsheet left an answer to the topic, it doesn't feel like it really answers the question. Not to sound clueless and needy, but the question talks about formulas, which I know can be derived from the spreadsheet, however as I have no idea how to approach this, then the answer I was hoping for would help a bit more.

ct.category theory - Profunctors corresponding to "partial functors"

One thing along these lines that you can say is that if D has, and G preserves, finite limits (or more generally is flat), then so does $hat{G}$ considered as a cocontinuous functor $[D^{op},Set] to [E^{op},Set]$. Since $hat{F}^dagger : [C^{op},Set] to [D^{op},Set]$ is just precomposition with F, it preserves all limits and colimits; thus $hat{G} hat{F}^dagger$ preserves finite limits as soon as G does, without any hypothesis on F. I don't know whether this can be extended to other kinds of limits.

Regarding the more general question of whether $hat{G} hat{F}^dagger$ could be considered a "partial functor," another way to describe it is as the left Kan extension of the composite $D overset{G}{to} E hookrightarrow [E^{op},Set]$ along F. If F is fully faithful, then such an extension is an honest extension, i.e. it restricts back along F to the original functor. So one could think of it as obtained by extending G to objects not in D in the most universal way possible: it maps an object $cin C$ to the formal colimit (viewing $[E^{op},Set]$ as the free cocompletion of E) over all approximations to c by objects of D.

On the other hand, every profunctor can be obtained as a composite $hat{G} hat{F}^dagger$ for some functors G and F (not necessarily an embedding): let the intermediate category D be the two-sided discrete fibration corresponding to that profunctor. An arbitrary profunctor can be thought of as a "generalized functor," but usually not specifically a "partial functor." However, perhaps faithfulness, or full-and-faithfulness, of F implies some properties of the resulting profunctor which makes it seem more like a "partial functor."

galaxy - What is the speed of light relative to (in space)?

The light will eventually reach galaxy B even though the space between them is expanding. It is important to remember that it is the space that is expanding not galaxy B moving away from galaxy A. Imagine putting two dots on a balloon and blowing it up. It is the rubber between the dots that is stretching, not one dot moving about the balloon's surface away from the other dot.

Imagine a photon of light leaving galaxy A. After a period of time it will have travelled some distance towards galaxy B. Let's say it has travelled 1% of the distance so it has 99% to go. Space is expanding equally in front of and behind the photon, so even if it were to stop moving it would stay at 1% of the total distance. But it doesn't stop, it moves nearer. It will eventually get half way and again the space is expanding both behind and in front of the photon. Now there is the same amount of space in between the photon and galaxy A as there is between the photon and galaxy B. No matter how much expansion there is the photon is still in the middle. Eventually the photon will get all the way to galaxy B.

No matter how fast the universe is expanding and how slow the photon is moving it will eventually reach galaxy B although it may take a very very long time.

As the space the photon is in is expanding the wavelength of the photon will slowly change. This is red shift.

Now for the problem with this answer. I believe this answer is only valid if the speed of the expansion and the speed of the photon is constant, but the expansion of the universe appears to be accelerating.

geometry - How to generate random points in ell_p balls?

---

I'll assume that you're looking for a uniformly chosen random point in the ball, since you didn't state otherwise. For p=1, you're asking for a uniform random point in the cross polytope in n dimensions. That is the set

$ C_n = { x_1, x_2, ldots, x_n in mathbb{R} : |x_1| + cdots + |x_n| le 1 }. $

By symmetry, it suffices to pick a random point $(X_1, ldots, X_n)$ from the simplex

$ S_n = { x_1, x_2, ldots, x_n in mathbb{R}^+ : x_1 + cdots + x_n le 1 }$

and then flip $n$ independent coins to attach signs to the $x_i$.

From Devroye's book Non-uniform random variable generation (freely available on the web at the link above, see p. 207 near the beginning of Chapter 5), we can pick a point in the simplex uniformly at random by the following procedure:

• let $U_1, ldots, U_n$ be iid uniform(0,1) random variables


• let $V_1, ldots, V_n$ be the $U_i$ reordered so that $V_1 le V_2 le cdots le V_n$ (the "order statistics"); let $V_0 = 0, V_{n+1} = 1$


• let $X_i = V_i - V_{i-1}$
  

So do this to pick the absolute values of the coordinates of your points; attach signs chosen uniformly at random, and you're done.

This of course relies on the special structure of balls in $ell^1$; I don't know how to generalize it to arbitrary $p$.

Can I pose any bounary data for the wave equation on $[0,infty)$ for given Cauchy data?

I am just going to answer here the problem actually posed in the title.

The one dimensional wave equation can be re-written as $partial_upartial_v phi = 0$, where $u$ and $v$ are the null variables $x + t$ and $x - t$. The initial data then is prescribed on $u+v geq 0, u-v = 0$, while the boundary is $u+v = 0, u-v geq 0$.

So we see that the wave equation implies that the function $psi(x,t) = partial_u phi(x,t)$ solve a transport equation with negative velocity (cf. my second comment above). Thus if your boundary condition is given such that $partial_u phi(x,t)$ is well-determined along the boundary by just the data given there, you will reach an inconsistency. This is one of the ways of seeing why you cannot prescribe simultaneously Dirichlet and Neumann conditions at the same time.

(On the other hand, to make the IBVP well-posed, you need to specify $partial_v phi(x,t)$ along the boundary, since it satisfies a transport equation with positive velocity. That just one of Dirichlet or Neumann conditions suffice follows from the fact that $partial_v phi(0,t)$ can be solved from $partial_u phi(0,t)$ [transported from Cauchy data] and the boundary data.)

ultrapowers - Ultraproducts of finite cyclic groups

I think the answer is no. The ultraproduct $U$ is naturally a quotient of ${mathbb Z}^{infty}$, the direct product of countably many copies of ${mathbb Z}$. In the obvious quotient map, the image of the direct sum is zero. Now, it is enough to show that:

Claim: Any homomorphism $ phi: {mathbb Z}^{infty} to {mathbb Z}$ that vanishes on the direct sum is identically zero.

Proof: (I learned this from a book by T.Y.Lam):

For a prime number $p$, let $A_p$ be the set of elements in ${mathbb Z}^{infty}$ of the form $(a_0, pa_1, p^2a_2,...)$, i.e. the elements whose $i$-th coordinate is divisible by $p^i$. Any element $x in A_p$ can be decomposed as

$x= y+z= (a_0, pa_1, dots, p^{n-1}a_{n-1}, 0,0, dots ) + p^n (0,0,..,0, a_n, pa_{n+1},..)$

Now, $y$ is in the direct sum, hence $phi(y)=0$. Also $phi(z) in p^n {mathbb Z}$, which
implies that $phi (x) in cap_{n=1}^{infty} p^n {mathbb Z} = { 0 }$

Now, choose two distinct primes $p$ and $q$. Since $gcd(p^n,q^n)=1$, it is easy to see that
$A_p+A_q= {mathbb Z}^{infty}$. This implies that $phi equiv 0$.

dg.differential geometry - Geodesics on a Grassmannian

This answer is a little bit redundant with the other two answers given so far, but here goes anyway.

It is easier to describe the real Grassmannian case. We can look at the Grassmannian $text{Gr}(n,k)$, and suppose that $2k le n$; if not then you can pass to the opposite Grassmannian. If $V$ and $W$ are two $k$-planes in $mathbb{R}^n$, then there a set of $k$ 2-dimensional planes that are each perpendicular to each other and each intersect $V$ and $W$ in a line. Call the angles between these lines $theta_1,ldots,theta_k$. Then in the connecting geodesic $V_t$ with $V_0 = W$ and $V_1 = V$, the angles are instead $ttheta_1,ldots,ttheta_k$. This is an explicit description that is basically equivalent to Mariano's remark about invariant complements.

The complex version has the same system of angles, but complexified lines, 2-planes, and $k$-planes. The "angle" between two lines in a 2-plane can be defined as the geometric angle between their slopes plotted on the Riemann sphere. Actually these angles are twice as large as the angles in the real case in the previous paragraph, but that makes no difference.

---

I was vague on the positions of the planes. The orthogonal projection of $V$ onto $W$ has a singular value decomposition, and the singular values are $cos theta_1,ldots,cos theta_n$. The orthogonal projection the other way is the transpose, or Hermitian transpose in the complex case, so it has the same singular values. The corresponding singular vectors are the lines $V cap P_k$ and $W cap P_k$. So the actual explicit work of finding the geodesic comes from solving a singular value problem, or equivalently an eigenvalue problem.

Errata to Principles of Algebraic Geometry

The Principles of Algebraic Geometry is a great book with, IMHO, many typos and mistakes. Why don't we collaborate to write a full list of all of its typos, mistakes etc? My suggestions:

Page 10 top, definition of $mathcal{O}_{n,z}$ is wrong (or at least written in a confusing way)

Page 15, change of coordinates given for the projective spaces only work when $i < j$. It states that the given transitions also work in the case when $j< i$.

Page 27, need to put a bar on the second entry of the $h_ij(z)$ operator defined. Also, shouldn't the title of this section be geometry of complex manifolds, instead of calculus on complex manifolds?

Page 35, definition of what is a sheaf is wrong. The gluing condition should be for any family of open sets, not just for pairs of open sets! (I've seem PhD students presenting this definition of sheaf on pg seminars...)

Page 74, writes $D(psi wedge e)$, but $psi$ and $e$ are in two different vector spaces, and one cannot wedge vectors in different vector spaces... I guess they mean tensor product.

Page 130, definition of divisor: it says it's a linear combination of codim 1 of irreducible subvarieties. By linear it means over $mathbb{Z}$ not over the complex numbers (better should say, like Hartshorne, that $Div$ is the free abelian group generated by the irreducible subvarieties).

Page 180, equation (*) has target a direct sum of line bundles, not tensor.

Page 366, when it says "supported smooth functions over $mathbb{R}^n$, are these complex valued or real valued functions?

Page 440 top equation. Is it really correct?

Page 445 Second phrase of hypercohomology section; it says sheaves of abelian sheaves. Probably means set of abelian sheaves.

lo.logic - Reductio ad absurdum or the contrapositive?

My answer applies to standard logic only.

In terms of standard logic, proofs by contrapositive and by contradiction are "equivalent" in that they are both logically valid, and two logically valid propositions are equivalent to each other.

On the other hand, it is certainly true that every proof by contrapositive can be phrased as a proof by contradiction. Indeed, since the latter is perhaps a bit more intuitive, it is often used as a justification of the former when it is needed e.g. in calculus courses:

We wish to show $A implies B$. Suppose we know that $lnot B implies lnot A$. Suppose further that $B$ is false. Then $lnot B$ is true, so $lnot A$ is true, so $A$ is false, contrary to our assumption.

Suppose a proposition can be proved by contraposition. As above, there is then a standard recipe for modifying the proof to give a proof by contradiction. However, if you compare the two proofs you find that the one by contradiction merely has the above two line argument appended to it, so it is just slightly longer without any additional content. For this reason, when it is possible to give a direct proof of $lnot B implies lnot A$ it is preferable to do so, rather than casting it as a proof by contradiction.

However, proof by contradiction is a more powerful technique in the informal sense that some proofs are more difficult to phrase using contraposition. (I don't want to say impossible, because as above, both "techniques" are simply logically valid arguments, so may be inserted in a proof at any point.)

What makes contradiction potentially more powerful? (This is a question that you have to face when you teach introduction to proofs classes, as I have. I wouldn't have had as ready an answer before.) I think it is because we get to assume two things rather than one. Namely, instead of just assuming $lnot B$ and using that one assumption to work our way to $lnot A$, we get to assume both $A$ and $lnot B$ and play them off one another in order to derive a contradiction.

Here is an example of this. Suppose we wish to prove that $sqrt{2}$ is irrational. First, let's phrase this as an implication:

For all $x in mathbb{R}$, $x^2 = 2 implies x not in mathbb{Q}$.

Or, contrapositively:

For all $x in mathbb{R}$, $x in mathbb{Q} implies x^2 neq 2$.

Taking the contrapositive was not so helpful! What we need to do is work from both ends at once:

Suppose that $x in mathbb{Q}$ and $x^2 = 2$. Now we are in business; we can work with this. (I omit the proof since I assume that everyone knows it.)

Here is another difference between the two proofs, which I didn't notice until I thought about this answer: the contrapositive of the statement

$forall x in S, P(x) implies Q(x)$

is

$forall x in S, lnot Q(x) implies lnot P(x)$:

note that the quantifier has not changed. (Of course we might have an existential quantifier instead, and the discussion would be the same. Anyway, in practice most mathematical propositions do begin with a universal quantifier.)

However, the negation of the statement is

$exists x in S | P(x) wedge lnot Q(x)$.

Note that the quantifier has changed from $forall$ to $exists$, which is a key feature of the above proof.

Finally, you ask why we would prefer one technique over another since both are equivalent. But of course we do this all the time, according to convenience and taste: e.g. induction, strong induction and well-ordering are all logically equivalent, but we use all three. We could e.g. phrase all induction proofs as appeals to the Well-Ordering Principle, but in many cases that would amount to inserting a tiresome rigamarole "Consider the set $S = { n in mathbb{N} | P(n) text{is false} }$..." which does not add to the clarity or concision of the proof.

distances - What is a parsec and how is it measured?

A parsec (abbreviated pc) is a unit of distance used by astronomers, cosmologists, and astrophysicists.

1 parsec is equal to $3.08567758 times10^{16}$ meters, or $3.26163344$ light years (ly).

A few typical scales to keep in mind:

1) The disks of galaxies like the Milky Way are a few 10's of kpc (that's pronounced kiloparsecs, which are 1000's of parsecs) in size. The dark matter halos surrounding them extend out to almost an order of magnitude further.

2) Galaxy clusters come in at around 1 Mpc (that's pronounced megaparsecs, or millions of parsecs), and are the largest bound objects in the universe. Pictured below is the galaxy cluster Abell 2218.

3) The closest star to the planet Earth (not counting the sun, of course) is the star system Alpha Centauri, at a distance of just about 1.3 parsecs. While you may think that this is incredibly close (and it is by cosmological standards), it would nonetheless still take us 4.24 years to travel to it if we traveled at the speed of light.

st.statistics - Linear Regression Coefficients W/ X, Y swapped

Well, I think Mike McCoy's answer is "the right answer," but here's another way of thinking about it: the linear regression is looking for an approximation (up to the error $epsilon$) for $y$ as a function of $x$. That is, we're given a non-noisy $x$ value, and from it we're computing a $y$ value, possibly with some noise. This situation is not symmetric in the variables -- in particular, flipping $x$ and $y$ means that the error is now in the independent variable, while our dependent variable is measured exactly.

One could, of course, find the equation of the line that minimizes the sum of the squares of the (perpendicular) distances from the data points. My guess is that the reason that this isn't done is related to my first paragraph and "physical" interpretations in which one of the variables is treated as dependent on the other.

Incidentally, it's not hard to think up silly examples for which $B_x$ and $B_y$ don't satisfy anything remotely like $B_x cdot B_y = 1$. The first one that pops to mind is to consider the least-squares line for the points {(0, 1), (1, 0), (-1, 0), (0, -1)}. (Or fudge the positions of those points slightly to make it a shade less artificial.)

Another possible reason that the perpendicular distances method is nonstandard is that it doesn't guarantee a unique solution -- see for example the silly example in the preceding paragraph.

(N.B.: I don't actually know anything about statistics.)

nt.number theory - What are the local Langlands conjectures nowadays, for connected reductive groups over a $p$-adic field?

First, I'd like to second the reference given by JT: David Vogan, "The local Langlands conjecture", appearing in Representation Theory of Groups and Algebras (J. Adams et al., eds. Contemporary Mathematics 145. American Mathematical Society, 1993. It can be found on Vogan's webpage

Vogan's article contains a very nice exposition of the local Langlands conjectures, and Arthur's local conjectures, and Vogan's own reformulations which I enjoy. In Conjecture 1.9, Vogan gives the local Langlands conjecture, as the OP has given it. Then, in Conjecture 1.12, Vogan gives a refinement describing L-packets, in the language of perverse sheaves (which the OP may or may not like). Adams, Barbasch, and Vogan proved this refinement for real reductive groups, and Vogan's article is certainly influenced by this.

Later, in Conjecture 4.3, Vogan gives a more detailed version of Langlands original conjectures. In Conjecture 4.15, Vogan gives a refinement, which seems equivalent to some conjectures of Arthur, though I'm not sure. This applies to most cases of interest.

To be specific, regarding $SL_2$ over a $p$-adic field, one must -- in addition to a Weil-Deligne representation $phi$ into $PGL_2(C)$ -- give an irreducible representation of the component group of the centralizer of the image of $phi$.

For example, consider an irreducible constituent of an unramified principal series of $SL_2(Q_p)$, whose Weil-Deligne representation $phi$ sends (geometric, but who cares) Frobenius to the class of a diagonal matrix $diag(-1, 1)$ in $PGL_2(C)$. Note that this matrix is centralized not only by diagonal matrices in $PGL_2(C)$, but also by the Weyl element (since we work in $PGL_2(C)$ and not just $GL_2(C)$). The centralizer of the image of $phi$ will be the group $N_{hat G}(hat T)$ normalizing a maximal torus in $hat G = PGL_2(C)$, I think. Its component group has order $2$. Since a group of order $2$ has two irreps, there are in fact two irreps of $SL_2(Q_p)$ with this Langlands parameter. This fills out the whole L-packet -- the two irreps occur as constituents in the same principal series in this case.

I think the most helpful treatment of L-packets for $SL_2$ can be found in the recent paper of Lansky and Raghuram, "Conductors and newforms for $SL(2)$", published in Pac. J. of Math, 2007. It's very explicit and considers every case thoroughly, and in a way directly relevant to modular forms. There you can find proven the statements you mention about the nontrivial L-packets being related to the two hyperspecial compact subgroups -- it's also related to the fact that "generic" has two possible meanings for $SL_2$, and representations can be generic for one orbit of character, and not for the other.

Mechanisms of binary/multiple star formation

I don't have enough reputation to comment...

I think this might help you understand the formation or binary and more stars systems. This of course is not the only possible method but it might explain the systems with big mass differences.

As the initial rotation speed increases (marked in the videos as beta) you will see how the protoplanetary disk breaks up and form more than one protostar.

http://www.astro.ex.ac.uk/people/mbate/Animations/prestellar_discs.html

Star names and the HYG Database

What about BSC5p for stars brighter than magnitude 6.5? The "alt_name" "is usually the Flamsteed and/or Bayer constellation-based name for the star".

Here a list of more advanced catalogs.

Here a list of some common star names for Hipparcos numbers. Probably not much different from your query result.

On Wikipedia a probably more complete list of more or less wide-spread common star names.

Those common names often are not unique, and not all common names are widely accepted. That's probably why they aren't provided in the star catalogs.

The only way I could imagine to get a solution is to convert the Wikipedia table to a csv file, and to import it to a database table used as a relation from the various catalog names to the proper names. That will probably need quite a bit of manual work.

None of the naming systems is complete, not even necessarily unique. Stars frequently turn out to be not one, but several stars, or no star at all. At some point it's easier to use 'scientifics'.

co.combinatorics - How to describe a tree? (depth, degree, balance, ... what else?)

Mind maps, as desribed on their wikipedia page, are a way of mapping or placing a graph structure onto a collection of data.

Items can be linked together with directed edges and with a label on the edge describing the relationship. Each data item at a vertex can taken on multiple tags (coloring) to describe their type.

If you have a Linux distribution that has the KDE (Kommon Desktop Environment, as opposed to the CDE Common Desktop Environment in Solaris) environment, you can see an implementation of mind maps in a note taking and note organizing software package

BasKet Notepads http://basket.kde.org/

The wikipedia page for BasKet is rather sparse and uninformative at http://en.wikipedia.org/wiki/BasKet_Note_Pads and does not really describe the full potential of the note organizing software.

The Mind Map software seems to be made for "rapid collaborating" and "brainstorming", very fuzzy words that seem to match the warm soft fuzzyness of the software. I have played with it, but it is poorly structured and not amazingly useful for organizing my research information.

It is, however, very useful for laying out quick hierarchical diagrams or tree diagrams. It does not easily allow one to export the graph structure in a useful and easy to reuse file format.

As for describing the structure: just look at it as a graph. Do you have any one-way oriented relationships on it? (e.g. links such as PARENT-OF, REFERS-TO, DERIVED-FROM, COMES-AFTER) If so, then you have a directed graph, otherwise you have an undirected graph.

How many elements are there? That is the number of vertices.

How many relations/links are there? That is the number of edges.

How many edges are there connecting each vertex? The number of edges on a vertex is the degree of the vertex. In a directed graph, you can have out-degree for outward-linking edges and in-degree for inward linking edges. What is the fewest number of edges? What is the largest number of edges on a vertex?

Draw a histogram of how many vertices have zero edges (free disconnected vertices), how many have one, two, etc. List them in order and you have the degree sequence of the graph.

Look at all of the elements; can you reach them all from one to another by following edges? Then you have a single connected graph. If you have separate islands that are not linked, count the number of islands as the number of components in the graph. Recursively describe each of these islands as listed above.

This is a simple way to. start. Treat the diagram as a graph and describe it in good detail. Could you please provide more details about what you are doing, or perhaps an example structure?

Moon orbits crossing each other

Though it's probably very unlikely in a solar system as old as ours (~5 billion years), I would not put it out of the realm of possibilities. Most of the collisions of this scale and magnitude have already happened - the early days of the formation of the solar system was violent and full of collisions between protoplanets.

In fact, one hypothesis for the formation of the moon was that it formed from the debris of just such a collision.

Additionally, the axial tilt of Uranus could also be explained by the collision of the planet with a protoplanet billions of years ago.

Is the likelihood zero? No. But I would bet money that collisions between planets and asteroids/comets are much more probable.

rt.representation theory - Is "semisimple" a dense condition among Lie algebras?

The "Motivation" section is a cute story, and may be skipped; the "Definitions" section establishes notation and background results; my question is in "My Question", and in brief in the title. Some of my statements go wrong in non-zero characteristic, but I don't know that story well enough, so you are welcome to point them out, but this is my characteristic-zero disclaimer.

Motivation

In his 1972 talk "Missed Opportunities" (MR0522147), F. Dyson tells the following story of how mathematicians could have invented special and much of general relativity long before the physicists did. Following the physicists, I will talk about Lie groups, but I really mean Lie algebras, or maybe connected Lie groups, ...

The physics of Galileo and Newton is invariant under the action of the Galileo Group (indeed, this is the largest group leaving classical physics invariant and fixing a point), which is the group $G_infty = text{SO}(3) ltimes mathbb R^3 subseteq text{GL}(4)$, where $text{SO}(3)$ acts as rotations of space, fixing the time axis, and $mathbb R^3$ are the nonrelativistic boosts $vec x mapsto tvec u$, $tmapsto t$. This group is not semisimple. But it is the limit as $cto infty$ of the Lorentz Group $G_c = text{SO}(3,1)$, generated by the same $text{SO}(3)$ part but the boosts are now
$$ t mapsto frac{t + c^{-2}vec u cdot vec x}{sqrt{1 - c^{-2}u^2}} quad quad vec x mapsto frac{vec x + tvec u}{sqrt{1 - c^{-2}u^2}} $$
Since semisimple groups are easier to deal with than nonsemisimple ones, by Occam's Razor we should prefer the Lorentz group. Fortunately, Maxwell's equations are not invariant under $G_infty$, but rather under $G_c$, where $c^{-2}$ is the product of the electric permititivity and magnetic permeability of free space (each of which is directly measurable, giving the first accurate measurement of the speed of light). Actually, from this perspective, $c^{-2}$ is the fundamental number, and we should really think of the Galileo Group as a limit to $0$, not $infty$, of something.

But of course from this perspective we should go further. Actual physics is invariant under more than just the Lorentz group, which is the group that physics physics and a point. So Special Relativity is invariant under the Poincare Group $P = G_c ltimes mathbb R^{3+1}$. Again this is not semisimple. It is the limit as $r to infty$ of the DeSitter Group $D_r$, which in modern language "is the invariance group of an empty expanding universe whose radius of curvature $R$ is a linear function of time" (Dyson), so that $R = rt$ in absolute time units. Hubble measured the expansion of the universe in the first half of the twentieth century.

Anyway, I'm curious to know if it's true that every Lie group is a limit of a semisimple one: how typical are these physics examples? To make this more precise, I'll switch to Lie algebras.

Definitions

Let $V$ be an $n$-dimensional vector space. If you like, pick a basis $e_1,dots,e_n$ of $V$, and adopt Einstein's repeated index notation, so that given $n$-tuples $a^1,dots,a^n$ and $b_1,dots,b_n$, then $a^ib_i = sum_{i=1}^n a^ib_i$, and if $v in V$, we define the numbers $v^i$ by $v = v^ie_i$; better, work in Penrose's index notation. Anyway, a Lie algebra structure on $V$ is a map $Gamma: V otimes V to V$ satisfying two conditions, one homogeneous linear in (the matrix coefficients) of $Gamma$ and the other homogeneous quadratic:
$$ Gamma^k_{ij} + Gamma^k_{ji} = 0 quadtext{and}quad Gamma^l_{im}Gamma^m_{jk} + Gamma^l_{jm}Gamma^m_{ki} + Gamma^l_{km}Gamma^m_{ij} = 0 $$
Thus the space of Lie algebra structures on $V$ is an algebraic variety in $V otimes V^* otimes V^*$, where $V^*$ is the dual space to $V$.

If $Gamma$ is a Lie algebra structure on $V$, the corresponding Killing form $beta$ is the symmetric bilinear pairing $beta_{ij} = Gamma^k_{im}Gamma^m_{jk}$. Then $Gamma$ is semisimple if and only if $beta$ is nondegenerate. Nondegeneracy is a Zariski-open condition on bilinear forms, since $beta$ is degenerate if and only if a certain homogeneous-of-degree-$n$ expression in $beta$ vanishes (namely $sum_{sigma in S_n} (-1)^sigma prod_{k=1}^n beta_{i_k,j_{sigma(k)}} = 0$ as a map out of $V^{otimes n} otimes V^{otimes n}$, where $S_n$ is the symmetric group on $n$ objects and $(-1)^sigma$ is the "sign" character of $S_n$). Since $beta$ is expressed algebraically in terms of $Gamma$, semisimplicity is a Zariski-open condition on the variety of Lie algebra structures on a given vector space.

Incidentally, Cartan classified all semisimple Lie algebras (at least over $mathbb C$ and $mathbb R$) up to isomorphism, and the classification is discrete. So any two semisimple Lie algebras in the same connected component of the space of semisimple structures are isomorphic.

My Question

Is the space of semisimple Lie algebra structures on $V$ dense in the space of all Lie algebra structures on $V$? (I.e. if $Gamma$ is a Lie algebra structure on $V$ and $U ni Gamma$ is an open set of Lie algebra structures, does it necessarily contain a semisimple one?) This is really two questions. One is whether it is Zariski-dense. But we can also work over other fields, e.g. $mathbb R$ or $mathbb C$, which have topologies of their own. Is the space of semisimple Lie algebra structures on a real vector space $V$ dense with respect to the usual real topology?

(The answer is no when $dim V = 1$, as then the only Lie algebra structure is the abelian one $Gamma = 0$, which is not semisimple, and it is no when $dim V = 2$, as there are nontrivial two-dimensional Lie algebras but no semisimple ones. So I should ask my question for higher-dimensional things.)

Edit: I have posted the rest of these as this follow-up question.

If the answer is no in general, is it possible to (nicely) characterize the Lie algebra structures that are in the closure of the semisimple part?

A related question is whether given a nonsemisimple Lie algebra structure, are all its nearby semisimple neighbors isomorphic? Of course the answer is no: the abelian Lie algebra structure $Gamma = 0$ is near every Lie algebra but in general there are nonisomoprhic semisimple Lie algebras of the same dimension, and more generally we could always split $V = V_1 oplus V_2$, and put a semisimple Lie algebra structure on $G_1$ and a trivial one on $V_2$. So the converse question: are there any nonsemisimple Lie algebras so that all their semisimple deformations are isomorphic? Yes, e.g. anyone on the three-dimensional vector space over $mathbb C$. So: is there a (computationally useful) characterization of those that are?

If the answer to all my questions are "yes", then it's probably been done somewhere, so a complete response could consist of a good link. The further-further question is to what extent one can deform representations, but that's probably pushing it.

universe - Why does a planet rotate and revolve?

As the planets evolve during their protoplanetary stage and accrete materials from the protoplanetary disks, which are gravitationally collapsing interstellar dust and gases, these accreted particles retain some of the angular momentum from the materials they form from and being in constant motion.

    

      ^{Generated image (virtual fly-by) from a simulation of the accretion period of the protoplanetary disk, showing preservation of
      angular momentum in the orbit around a Jupiter-size planet, as it clears its neighborhood. (Source: Frédéric Masset)}

One nice description for this angular momentum preservation, and why the planets appear to rotate faster than their surrounding protoplanetary disk goes like this:

Conservation of angular momentum explains why an ice skater spins more
rapidly as she pulls her arms in. As her arms come closer to her axis
of rotation, her [rotation] speed increases and her angular momentum remains the
same. Similarly, her rotation slows when she extends her arms at the
conclusion of the spin.

^{Source: Scientific American article on Why and how do planets rotate? (George Spagna)}

So it could be described as this axial rotation of planets resulting in conservation of the angular momentum of the materials in the protoplanetary disk, forming during the accretion period of the planetary system as the protoplanets gain in weight, and preserve this angular momentum due to inertia of their radial velocity.

planet - What day/night cycles, climate and seasons would experience Alpha Centauri Bb inhabitants?

This is just a partial answer (so far), but it will address at least part of your question.

1.To what temperatures the dark surface of such planet could be heated? Is there possibility of liquid water?

Wikipedia gives a good estimate for this: 1,200 degrees Celsius. For a planet, that's pretty hot! Any water there would evaporate very quickly. In fact, it is expected that a lot of the planet's surface is molten. Now, does this apply to the back side of the planet? I would think so. While only the front side of the planet would bask in the warmth of Alpha Centauri B, the heat should dissipate throughout the entire planet.

2.Will the radiation of the second star be enough to provide normal day-like illumination and heating?

The short answer here is no. In fact, when compared to our solar system, nothing about the radiation Alpha Centauri Bb gets is normal. It is situated extremely close to Alpha Centauri B - 0.04 AU. Given that both Alpha Centauri A and Alpha Centauri B are similar to the Sun, I think it's clear that this exoplanet will not receive a "normal" (normal here meaning similar to Earth) amount of radiation from Alpha Centauri B. Conversely, Alpha Centauri Bb is 11 AU away from Alpha Centauri A - much greater than the distance between it and Alpha Centauri B (a bit more than the distance between Saturn and the Sun). This means that the planet will receive only a tiny fraction of light from Alpha Centauri A relative to Alpha Centauri B.

4.Will such planet experience seasons and how they would be arranged?

Seasons depend on the tilt of a planet's axis. Unfortunately, it is hard (if not impossible) to measure the tilt of the axis of Alpha Centauri Bb (I haven't been able to find any measurements). I would assume that it would experience some seasons, but I don't know how much they would extend.

As for question 3 - I'm not quite sure what you mean by "calendar." If you mean an artificial calendar, I would think that any beings on the planet (life there is extremely unlikely) would only use Alpha Centauri B to determine their years (just over three days and five hours).

Why can I see the whole moon during various non-full-moon phases?

Here's an image that describes the phenomenon I'm asking about. The very thin sliver to the left is of course the surface of the moon as lighted by the sun. But the rest of the moon is also faintly lighted -- by what?

I'm guessing that some amount of "earthshine" -- sunlight reflected from the earth -- is the culprit here, but I've found no authoritative confirmation online. It would be great to have a link to a source that explains what's really going on.

orbital elements - How do I calculate the positions of objects in orbit?

I'm not sure how much this will help, but I can give you this passage from Wikipedia:

Under ideal conditions of a perfectly spherical central body, and zero perturbations, all orbital elements, with the exception of the Mean anomaly are constants, and Mean anomaly changes linearly with time[dubious – discuss], scaled by the Mean motion, $n=sqrt{frac{mu } {a^3}}$. Hence if at any instant $t_0$ the orbital parameters are $[e_0,a_0,i_0,Omega_0,omega_0,M_0]$, then the elements at time $t_0+delta t$ is given by $[e_0,a_0,i_0,Omega_0,omega_0,M_0+ndelta t]$.

In other words, most of the orbital elements (save $M_0$) are constant under very short intervals.

planet - What astronomical observations would give conclusive proof of alien life?

There are several ways we are coming closer to answering the question "is there life elsewhere in the universe?". One is by first understanding very well the origin life on our own planet. Another is by trying to look for radio signals from extraterrestrial civilizations. Another is to try to search for habitable exoplanets, and eventually examine the habitable ones in detail via direct imaging. Another is too look for evidence of life or life that existed in the past right in our own solar system (for exmaple on Mars, Titan, Europa, etc).

1) Have I missed any other astronomy-related approaches to answering this question?

2) Short of radio contact from an ET civilization, or a visit by aliens, or the discovery of life/past-life in our solar system, what would give conclusive evidence of alien life in the galaxy?

For example:

In 20 or so years when we develop the technology to directly image exoplanets, we will be able to determine their chemical composition and characterize their structure. If astronmers who study exoplanets found what they are looking for -evidence of oxygen and other biomarker gases - this might be highly suggestive of life, but not conclusive evidence. What kinds of observations of an exoplanets could enable us to say with ceartainty "there is life on this planet"? Short of telescopic observations of truly unbelievable detail, I cannot think of a way that we can look at a planet and say "yes there is life on it, or no there is not" other than perhaps if seeing lights on the dark side of the planet.

Edit: Actually, I forgot the obvious scenario where there are trees/grass on the exoplanet. But I don't think the plant life is required to be a different color from the rest of the terrain, and a planet with aliens does not have to have plant life. So we might not be able to detect life visually.

mathematics education - How seriously do professors take teaching evaluations?

I'm not sure whether I should really post these few thoughts of mine because they do not directly address the question as it was asked but every time somebody starts talking about students evaluations I feel a strong urge to say something like that, so why not now and here? As usual, I believe I'm guilty of violating the "no discussions on MO" rule, so feel free to downvote.

I find the current system of term end student evaluations extremely counterproductive. On one hand, most students are just plainly incompetent to answer many of the typical questions (like "rate your professor's knowledge of the subject"). On the other hand, it creates an unhealthy situation when instead of getting a constructive criticism during the course when the adjustments can easily be made, we just get emotional comments on our performance when it is too late to do anything and more often than not in the format that is impossible to interpret in any reasonable way. Out of each 50 evaluations I got only 3 or 4 contained something that I could use to adjust my teaching. The rest either contained pure rating numbers that didn't tell me anything I hadn't known myself already (yes, I did know if the topic was hard or easy, I did know whether I prepared my lectures carefully or just improvised, etc.) or the comments like "This is the best teacher ever!" or "He should be fired immediately!" (sometimes I got both in the same class), which also carried no real content whatsoever.

I agree that it is a great idea to get the feedback from the students and I admit that some students may be afraid to express themselves openly, but it would be much better just to have a small forum type website for each course where every enrolled student would be able to publicly post any concerns or comments about the course in a free style format as soon as they arise with or without signing it (both options should be available) and the teacher would be able to read those and to respond and decide whether any action/adjustment is due. Such records would also tell much more about everybody's teaching to any promotion committee or whoever else who might have a real interest in what is going on.

Technically it is not so hard but I've never seen anything like that done. Instead, we have that system of meaningless average numbers of subjective ratings of rather poorly defined things. Indeed, what is my "being organized", or "preparedness", or "willingness to help"? I believe I can read and speak basic English but I'm just unable to assign any meanings to such general words, much less to assign a numerical value to the qualities they are assumed to represent. I understand the sentences like "You never returned homework #2" or "Your lecture on continuity was very clear", but not the sentence "Your overall preparedness was 5 on the scale from 1 to 10".

The most unpleasant thing is that the administration tends to take those numbers rather seriously sometimes. I don't care: I've got my tenure and my salary is high enough but it is a pity to see young people who are afraid to teach $varepsilon-delta$ in calculus courses because it may irritate the students and result in bad evaluations.

As to the requested background information, I'm currently a full professor of mathematics at University of Wisconsin-Madison but I spent more than 10 years at Michigan State University before that. In Michigan we were required to do evaluations for every course taught; Wisconsin requires just to do them "now and then" (not less than one in 3 semesters or something like that; I still have to figure the exact rule out). I wouldn't say they were taken very seriously by the faculty in either place but I cannot speak for everyone. I have never tried to figure out who wrote what for the lack of time and curiosity for such things but I believe Leonid that it may be possible to do in a small size class.

nt.number theory - Alternative axiom to induction

A crucial difference between non-Euclidean geometries and "non-inductive" models of PA- is that any model of PA- contains a canonical copy of the true natural numbers, and in this copy of N, the induction schema is true. In other words, PA is part of the complete theory of a very canonical model of PA-, and as such it seems much more natural (so to speak) to study fragments of PA rather than extensions of PA- which contradict induction.

To make this a little more precise (and sketch a proof), the axioms of PA- say that any model M has a unique member 0_M which is not the successor of anything, and that the successor function S_M: M to M is injective; so by letting k_M (for any k in N) be the k-th successor of 0_M, the set {k_M : k in N} forms a submodel of M which is isomorphic to the usual natural numbers, N. Any extra elements of M not lying in this submodel lie in various "Z-chains," that is, infinite orbits of the model M's successor function S_M.

(In the language of categories: the "usual natural numbers" are an initial object in the category of all models of PA-, where morphisms are injective homomorphisms in the sense of model theory.)

So, while PA seems natural, I'm not sure why there would be any more motivation to study PA- plus "non-induction" than there is to study any of the other countless consistent theories you could cook up, unless you find that one of these non-inductive extensions of PA- has a particularly nice class of models.

nt.number theory - Intuition for the last step in Serre's proof of the three-squares theorem

The Lemma in the aswer by Bjorn Poonen can be sharpened to give an exact relationship
between $mathrm{den}(x)$ and $mathrm{den}(x')$.

Lemma 1.
$~$Let $f=f_2+f_1+f_0inmathbb{Z}[X]=mathbb{Z}[X_1,ldots,X_n]$ ($ngeq 1$),
with each $f_i$
homogeneous of degree $i$.
Let $y,vinmathbb{Z}^n$, where $v$ is primitive and $f_2(v)neq0$,
and define $F=AT^2+BT+C:=f(y+Tv)inmathbb{Z}[T]$.
Suppose that the two zeros $t$ and $t'$ of $F$ are rational,
so that the two rational points $x=y+tv$ and $x'=y+t'v$ are zeros of $f$.
If $x=a/b$ and $x'=a'/b'$ are the reduced representations (with $b,b'>0$),
and $xneq y$ (which is certainly true if $xnotinmathbb{Z}^n$),
then
begin{equation*}
b' ,=, mathrm{sgn}(A)cdot
frac{f_2(x-y)}
{mathrm{gcd}(A,B,C),mathrm{gcd}(a-by)^2}cdot b~; tag{1}
end{equation*}
exchanging $x$ and $x'$ gives the analogous identity (provided $yneq x'$).

Remark.
$~$For any $u=(u_1,ldots,u_n)inmathbb{Z}^n$ we write $mathrm{gcd}(u):=mathrm{gcd}(u_1,ldots,u_n)$.

Proof. $~$Since $tv=x-y=(a-by)/b=(c/b)v$,
where $c=pm,mathrm{gcd}(a-by)$ and $mathrm{gcd}(c,b)=1$,
we have $t=c/b$, and similarly $t'=c'/b'$ with $c'=pm,mathrm{gcd}(a'-b'y)$
and $mathrm{gcd}(c',b')=1$. The leading coefficient of $F$ is $A=f_2(v)=(b/c)^2f_2(x-y)$
($cneq 0$ because $a-by=b(x-y)neq 0$).
By Gauss lemma $F=d(bT-c)(b'T-c')$ with $d=mathrm{sgn}(A)cdotmathrm{gcd}(A,B,C)$.
Then $dbb'=A=(b/c)^2f_2(x-y)$ gives us $b'$ expressed
as in the lemma.$~$
Done.

The Lemma in Bjorn Poonen's post is an immediate consequence.

Lemma 1 is about the geometric background of Davenport-Cassels lemma:
it relates the reduced representations of two rational zeros of $f$
that lie on an integral line $L=y+mathbb{Q}v$
whose direction vector $v$ is an anisotropic vector of the quadratic form $f_2$.
(An integral line is an affine line in $mathbb{Q}^n$
that contains an integral point and hence infinitely many integral points.)
It is not required that one or the other of the two zeros is non-integral,
the lemma says something interesting even when both zeros are integral.
Also, the two zeros may coincide, in which case the line $L$
is tangent to the quadric ${f=0}$.

If we actually write out the other identity mentioned in the lemma
and then compare the two identities,
we obtain the identity (supposing $yneq x,x'$)
begin{equation*}
f_2(x-y)f_2(x'-y)
,=, bigl(mathrm{gcd}(A,B,C),mathrm{gcd}(a-by),mathrm{gcd}(a'-b'y)bigr)^2~.tag{2}
end{equation*}
It is not in the least surprising that $f_2(x-y)f_2(x'-y)$ is a square:
if $q$ is any quadratic form on a vector space $V$ over some field $K$,
and $uin V$ and $lambda,muin K$,
then it is trivial that $q(lambda u)q(mu u)=bigl(lambdamu q(u)bigr)^2$.
It may seem slightly surprising that $f_2(x-y)f_2(x'-y)$ is a square of an integer,
but this is
a consequence of $A=dbb'$: in the situation of the lemma
the trivial identity satisfied by a general quadratic form reads
begin{equation*}
f_2(x-y)f_2(x'-y) ,=, Bigl(frac{c}{b},frac{c'}{b'}f_2(v)Bigr)^2~,
end{equation*}
and substituting $f_2(v)=A=dbb'$ yields
begin{equation*}
f_2(x-y)f_2(x'-y) ,=, (dcc')^2 ,=, C^2 ,=, f(y)^2
end{equation*}
(this time without the restriction $yneq x,x'$).
Let $L_{mathbb{Z}}$ denote the set of all integral points on the line $L$:
$L_{mathbb{Z}}:=Lcapmathbb{Z}^n=y+mathbb{Z}v$.
For any $zin L_{mathbb{Z}}$
the identity (2) still holds when $y$ is replaced by $z$,
but we must be careful and write it as
begin{equation*}
f_2(x-z)f_2(x'-z)
,=, bigl(mathrm{gcd}(A,B_z,C_z),mathrm{gcd}(a-bz),mathrm{gcd}(a'-b'z)bigr)^2~,
end{equation*}
because the coefficients $B_z$ and $C_z$ of $F_z=f(z+Tv)$ depend on $z$.
However,
note that the coefficient $A_z=A=f_2(v)$,
as well as the greatest common divisor of the coefficients of $F_z$
(the content of $F_z$),
$mathrm{gcd}(A,B_z,C_z)=left|Aright|/bb'=left|dright|$, do not depend on $z$.
For $zin L_{mathbb{Z}}$ we define $c(z),c'(z)inmathbb{Z}$
by $x-z=bigl(c(z)/bbigr)v$ and $x'-z=bigl(c'(z)/b'bigr)v$.
Since $mathrm{gcd}(a-bz)=left|c(z)right|$ and $mathrm{gcd}(a'-b'z)=left|c'(z)right|$,
we have
begin{equation*}
f_2(x-z)f_2(x'-z) ,=, bigl(d,c(z),c'(z)bigr)^2 ,=, C_z^2 ,=, f(z)^2~,
qquadquad zin L_{mathbb{Z}},.
end{equation*}
Remark. $~$Idiot me!
This is just a very special case of the general power-of-a-point theorem,
which does not rely on specific factorization properties of integers
and is almost trivial to prove:

Let $K$ be a field,
let $fin K[X] = K[X_1,ldots,X_n]$ be of degree $m$ (where $m, ngeq 1$),
and denote by $f_m$ the homogeneous component of $f$ of degree $m$.
Let $L$ be an affine line in $K^n$ with a direction vector $v$, where $f_m(v)neq 0$.
Let $yin L$, and define $F_y := f(y+Tv)in K[T]$, a polynomial of degree $m$.
Suppose that $F_y$ has $m$ zeros (counting multiplicities) $t_1$, $ldots$, $t_m$ in $K$.
Then the points $x_i=y+t_ivin L$, $1leq ileq m$, are zeros of $f$,
the multiset of the $x_i$'s does not depend on the choice of $yin L$,
and
begin{equation*}
f_m(y-x_1)f_m(y-x_2)cdots f_m(y-x_m) ,=, f(y)^m~.
end{equation*}

The independence is easy: if $z=y+sv$,
then $F_z$ has the zeros $t_i-s$, whence $z+(t_i-s)v = y+t_iv = x_i$.
The leading coefficient of $F_y$ is $f_m(v)$ and its constant term is $f(y)$.
From $F_y=f_m(v)(T-t_1)cdots(T-t_m)$ we get $f(y)=(-1)^m t_1cdots t_m f_m(v)$,
whence $f_m(y-x_1)cdots f_m(y-x_m) = bigl((-t_1)cdots(-t_m)f_m(v)bigr)^m = f(y)^m$.

We digress.
Let's return to the situation in Lemma 1.

We regard the point $y$ as fixed, serving as an origin of $L_{mathbb{Z}}$.
Let us determine $c(z)$ for a general point $z=y+kvin L_{mathbb{Z}}$, $kinmathbb{Z}$:
from
begin{equation*}
frac{c(y+kv)}{b},v ,=, x-(y+kv) ,=, (x-y)-kv ,=, frac{c(y)-kb}{b},v
end{equation*}
we see that
begin{equation*}
c(y+kv) ,=, c(y) - kb~. tag{3}
end{equation*}
For any $zin L_{mathbb{Z}}$ we have
begin{equation*}
f_2(x-z) ,=, frac{c(z)^2}{b^2}f_2(v)
,=, frac{c(z)^2}{b^2},dbb'
,=, frac{db'}{b},c(z)^2
,=, frac{e_0}{b_0},c(z)^2~, tag{4}
end{equation*}
where $b_0=b/mathrm{gcd}(b,db')$ and $e_0=db'/mathrm{gcd}(b,db')$.
Since $mathrm{gcd}bigl(b,c(z)bigr) = 1$, and hence $mathrm{gcd}bigl(b_0,c(z)bigr) = 1$, it follows that
begin{equation*}
mathrm{den}bigl(f_2(x-z)bigr) ,=, b_0 qquadquad text{for every $zin L_{mathbb{Z}}$},.
end{equation*}
Combining (3) and (4) we obtain
begin{equation*}
f_2bigl(x-(y+k)vbigr) ,=, frac{e_0}{b_0}bigl(c(y)-kbbigr)^2~, qquadquad kinmathbb{Z},.
end{equation*}
In the special case $x=x'$, when the line $L$ is a tangent of the quadric ${f=0}$,
we have $b=b'$, whence
begin{equation*}
f_2(x-z) ,=, d,c(z)^2~, qquadquad zin L_{mathbb{Z}},,
end{equation*}
thus $f_2(x-z)$ is an integer for every integral point $z$ in $L$.

The discussion above has demonstrated that the identity (1) and its brethren have uses
unrelated to Davenport-Cassels lemma.
Now we return to applications of (1) to Davenport-Cassels lemma and (a little way) beyond it.
The assumptions $xinmathbb{Q}^nsetminusmathbb{Z}^n$ and $0<left|f_2(x-y)right|<1$
imply the premises $f_2(v)neq 0$ and $xneq y$ of Lemma 1
and then yield the instantaneous result $b'<b$.
But besides the $f_2(x-y)$ in the numerator on the right hand side of (1)
there are also the factors $mathrm{gcd}(A,B,C)$ and $mathrm{gcd}(a-by)^2$ in the denominator
whose product can be greater than $1$ and can help make $b'$ smaller than $b$
even when $left|f_2(x-y)right|geq 1$.
This leads to the idea of walking with Aubry 'on the far side':
when there is no integral point $y$ satisfying $left|f_2(x-y)right|geq 1$
we may still find an integral point $y$
so that we can make a step along the line $L$
from a zero $x=a/b$ of $f$ to a zero $x'=a'/b'$ of $f$ with $b'<b$.
The folowing two examples attest that this idea actually works.

But first, a definition.
A quadratic form $q$ on $mathbb{Q}^n$ is said to be Euclidean
if for every $xinmathbb{Q}^nsetminusmathbb{Z}^n$
there exists $yinmathbb{Z}^n$ such that $0 < left|q(x-y)right| < 1$.

For the first example let $q(X_1,X_2,X_3)=X_1^2+X_2^2+5X_3^2$
and $f(X)=q(X) - m$, where $m$ is a positive integer.
The positive-definite quadratic form $q$ is not Euclidean on $mathbb{Q}^3$,
since for any $xinmathbb{Z}^3+bigl(frac{1}{2},frac{1}{2},frac{1}{2}bigr)$
and any $yinmathbb{Z}^3$ we have $q(x-y)geq 7/4$.
Consider the set $F$ of all points $x$ in the cube
$bigl{(x_1,x_2,x_3)inmathbb{R}^3 bigm| 0leq x_1,x_2,x_3leq frac{1}{2}bigr}$
at which $q(x)geq 1$;
the set $F$ ('the far side') is shown as the darker shaded part of the cube
in the following figure:

The function $q(1-x_1,1-x_2,1-x_3)$ of a point $(x_1,x_2,x_3)$ in the set $F$
attains its largest value $8-2sqrt{5}$ at the point $P=bigl(0,0,1/sqrt{5}bigr)$.
$quad$Suppose that $xinmathbb{Q}^3setminusmathbb{Z}^3$ is a zero of $f$,
and let $y:=mathrm{round}(x)$.
If $q(x)<1$ (we certainly have $q(x)>0$) then fine,
we make a step to the next zero of $f$ with smaller denominator
this side of the Euclidean horizon.
Otherwise $q(x-y)geq 1$, we are on the far side, and must tread more carefully.
Let $x=a/b=(a_1,a_2,a_3)/b$ be the reduced representation.
We claim that since $q(x)=m$ is an integer, the denominator $b$ is odd.
Suppose that $b$ is even;
then at least one of $a_1$, $a_2$, $a_3$ is odd,
and so in $q(x)=q(a)/b^2$ the numerator $q(a)$ is congruent to $1$, $2$, or $3$ modulo $4$,
while the denominator is divisible by $4$, contradiction.
Note that $b$ being odd implies $0 leq left|x_i-y_iright| < frac{1}{2}$, $i=1,2,3$.
Now we choose an integral point $z$, close to the integral point $y$.
If $a_1-by_1$ is even, then we set $z_1:=y_1$.
If $a_1-by_1$ is odd, we let $z_1$ be a second closest integer to $x_1$
(there are two possible choices for $z_1$ iff $x_1$ is an integer,
and we may choose either of them);
then $z_1=y_1pm1$, $a_1-bz_1$ is even, and $left|x_1-z_1right|=1-left|x_1-y_1right|$.
In either case we have $left|x_1-z_1right|leq1 - left|x_1-y_1right|$.
The coordinates $z_2$ and $z_3$ are chosen analogously.
We step to the next zero $x'=a'/b'$ of $f$ along the line laid through the points $x$ and $z$.
Since $bigl(left|x_1-y_1right|,left|x_2-y_2right|,left|x_3-y_3right|bigr)in F$,
it follows that
$$q(x-z)leq qbigl(1-left|x_1-y_1right|,1-left|x_2-y_2right|,1-left|x_3-y_3right|bigr)
leq 8-2sqrt{5} < 4~.$$
By the choice of the point $z$ all three coordinates of $a-bz$ are even,
thus $mathrm{gcd}(a-bz)geq 2$,
and Lemma 1 tells us that $b'<b$.
$quad$Done.

For the second example we consider $q(X_1,X_2,X_3)=X_1^2+X_2^2+2X_3^2$
and $f(X)=q(X)-m$ with $m$ a positive integer.
The quadratic form $q$ is barely non-Euclidean:
for every $xinmathbb{Q}^3setminusmathbb{Z}^3$ there is $y=mathrm{round}(x)inmathbb{Z}^3$
such that $q(x-y)leq 1$.
The problem is that there exist points $x$ for which $q(x-y)=1$ is the best we can do:
if $xin M := mathbb{Z}^3+bigl(frac{1}{2},frac{1}{2},frac{1}{2}bigr)$,
then $q(x-y)geq 1$ for any $yinmathbb{Z}^3$.
Note that $q(x)$ is an odd integer for every $xin M$,
thus there do exist integers $m$, all of them odd,
so that $f$ has rational zeros,
but woe, it also has a rational zero,
with all three coordinates precisely halfway between consecutive integers,
at which we get stuck, because there is no Euclidean step from it to another zero.
On the other hand, if $m$ is even we never get stuck,
there is always a Euclidean step from a non-integral zero;
that is, though $f$ is not Euclidean, it is 'conditionally' Euclidean
on the set of its zeros.
$quad$Now suppose that $m$ is odd and that $f$ has rational zeros,
and that we walked ourselves into a point $x=a/2in M$.
In this case the trick we have used in the preceding example does not work,
because $mathrm{gcd}(a-2y)^2leq f_2(x-y)$ for any $yinmathbb{Z}^3$;
we must seek help from the other factor $mathrm{gcd}(A,B,C)$
in the denominator on the right hand side of (1).
Note that we can round the point $x$ to any of the eight integral points
$x+frac{1}{2}(delta_1,delta_2,delta_3)$, where $delta_1,delta_2,delta_3in{-1,1}$;
let $y$ be one of these eight points.
We have $v=(delta_1,delta_2,delta_3)$,
and $F(T) = f(y+Tv) = q(v)T^2 + 2langle y,vrangle T + q(y) - m$,
where $langletext{-},text{-}rangle$
is the bilinear form associated with the quadratic form $q$,
$langle X,Yrangle = X_1Y_1+X_2Y_2+2X_3Y_3$.
The leading coeffient $A=q(v)=4$ is even,
the next coefficient $B=2langle y,vrangle$ is also even,
thus it remains to make $C=q(y)-m$ even.
But this is easy: choose $delta_1$ and $delta_2$ so
that one of $y_1$, $y_2$ is even and the other one is odd.
Such a choice makes $mathrm{gcd}(A,B,C)geq 2$,
and since $q(x-y)=1$, we can step to a zero $x'=a'/b'$ of $f$ with $b'<2$,
that is, to an integral zero (whence $mathrm{gcd}(A,B,C)=2$).
$quad$Done.

the sun - Can our Sun become a black hole

No, the sun won't ever become a black hole.

The choice between the three fates of stars (white dwarf, neutron star, black hole) is entirely determined by the star's mass.

A star on the main sequence (like most stars, including our sun) is constantly in a balance between the inward pressure of gravity and the outward pressure of the energy generated by the hydrogen fusion that makes it "burn".¹ This balance stays relatively stable until the star runs out of whatever its current fuel is - at that point, it stops burning, which means there's no longer outward pressure, which means it starts collapsing. Depending on how much mass there is, it might get hot enough as it collapses to start fusing helium together. (If it's really massive, it might continue on to burn carbon, neon, oxygen, silicon, and finally iron, which can't be usefully fused.)

Regardless of what its final fuel is, eventually the star will reach a point where the collapse from gravity is insufficient to start burning the next fuel in line. This is when the star "dies".

White dwarfs

If the star's remains² mass less than 1.44 solar masses (the Chandrasekhar limit³), eventually gravity will collapse the star to the point where each atom is pushed right up against the next. They can't collapse further, because the electrons can't overlap. While white dwarfs do shed light, they do so because they are extremely hot and slowly cooling off, not because they're generating new energy. Theoretically, a white dwarf will eventually dim until it becomes a black dwarf, although the universe isn't old enough for this to have happened yet.

Neutron stars

If the collapsing star is above the Chandraskhar limit, gravity is so strong that it can overcome the "electrons can't overlap" restriction. At that point, all the electrons in the star will be pushed into combining with protons to form neutrons. Eventually, the entire star will composed primarily of neutrons pushed right up next to each other. The neutrons can't be pushed into occupying the same space, so the star eventually settles into being a single ball of pure neutrons.

Black holes

Black holes are the step beyond neutron stars, although they're worth discussing in a bit more detail. Everything, in theory, has a Schwarzschild radius. That's the radius where a ball of that mass would be so dense that light can't escape. For example, the Schwarzschild radius for Earth is about 9mm. However, for all masses smaller than somewhere between 2-3 times the mass of the sun, it's impossible to squeeze the matter small enough to get it inside that radius. Even a neutron star isn't massive enough.

But a star that becomes a black hole is. We don't actually know what happens to a star once it's become a black hole - the edges of the "hole" itself is simply the Schwarzschild radius - the point light can't escape. From outside, it doesn't matter whether the matter collapsed to the point that the neutrons started overlapping, whether it stopped just inside the radius, or whether it continued collapsing until it broke all known physical laws. The edges are still the same, because they're just a cutoff based on the escape velocity.

---

¹ I'm ignoring the red giant phase here, since it's just a delay in the "run out of fuel" step. Basically, the core is helium "ash", while the hydrogen fusion process takes place further and further out. Once that runs out, you get a nova and the collapse continues.

² Likewise, I'm ignoring the mass that stars shed in their various nova phases. All given masses are based on the remnants left behind.

³Every source I've found for Chandrasekhar mass, except Wikipedia, gives 1.44 or 1.4 solar masses (which are compatible). Wikipedia gives 1.39, and gives at least one source to back that number.

Fate of the Universe - Astronomy

Which scenario will be worse for the ultimate fate of the universe?

Well, that's a wee bit opinionated. It depends on your point of view - that is, how you take a liking to the following two scenarios. Everyone has their own opinion; from the details I give below, I'll let you answer that yourself.

To infinity - and . . . well, more infinity: The perpetually expanding universe

There are actually three types of universes that could fit this model: The accelerating expansion one, the constant expansion one, and the decelerating expansion one. At this point in time, we appear to be in a universe undergoing accelerated expansion, thanks to dark energy. However, all three scenarios give us a similar end result: A dead universe.

In the far future, as long as the universe does not curl back into a Big Crunch, slowly all processes as we know them will begin to end. Star formation will eventually cease, and black holes will slowly start to gobble up a lot of matter. In the even farther future, black holes are evaporate to decay via Hawking radiation, leaving the universe a collection of various subatomic particles. (If protons decay, then perhaps there will no longer be any atoms). The temperature of the universe will drop until everything is about thermodynamically equivalent. Not a great place to be (well, there won't be any life, so nobody will be around to see it, but still. . .).

Beyond this. . . well, there are a few different ideas regarding what could happen. The Big Freeze idea states that the universe will eventually just cool off to this low-temperature state. The Big Rip idea says that matter itself could be torn apart bye extreme expansion of space. Cosmological false vacuum theories suggest that quantum tunneling effects could make some interesting expansion possible. However, I emphasize that we do not know whether any of these long-term theories are correct. They are too far in the future, and there is too much uncertainty among the relevant factors.

In the short term, however, it appears our universe will expand forever.

Back to the beginning: The Big Crunch universe and all its friends

The Big Crunch scenario says that if the density of the universe is greater than the critical density, the universe will fall back onto itself into a singularity - just like the Big Bang. This leads to other theories, such as the epykrotic universe, or a Big Bounce, which postulate that our current universe is just undergoing one of infinitely many stages of expansion and collapse.

This isn't exactly a great outcome, either, but again, I'll leave you to judge.

co.combinatorics - Do all correlation coefficients induce a pseudometric?

The statement is false. Consider the Pearson sample correlation coefficient:

c(x,y) = (x-mean(x)).(y-mean(y))/sqrt(|x-mean(x)|^2*|y-mean(y)|^2)

Here is an example, where the triangle inequality for the distance defined in
the question is not satisfied

x = [0.5847 -0.3048 -0.4431 0.5032 -0.3401],

y = [0.2018 0.4547 -0.2230 0.3350 -0.7685],

z = [0.5226 -0.5159 -0.5439 0.3701 0.1671].

co.combinatorics - What can I say about the permutation $alphabeta$ if I know the permutation $betaalpha$?

I'm looking into a secret sharing scheme that has a secret permutation $theta$ which has the cycle structure (n/2)+(n/2) (i.e. two (n/2)-cycles).

The permutation $theta$ is decomposed into two permutations $alpha$ and $beta$, where $alpha$ is generated uniformly at random. So with knowledge of both $alpha$ and $beta$, we can find $theta$, while with knowledge of $alpha$ xor $beta$, we cannot find $theta$ (although, we could guess).

At this point, I want to make public $betaalpha(L)$ (L is actually a Latin square, but this is not too relevant for the question I want to ask). It is possible that an attacker could find $betaalpha$ from $betaalpha(L)$. However, I worry that knowledge of $betaalpha$ might give information about $theta$.

If I know $theta=alphabeta$, and I'm given the permutation $betaalpha$, what can I say about $theta$? (without a priori knowledge of $alpha$, $beta$ or $theta$)

gn.general topology - Finite versus infinite on non-Hausdorff topologies

Seeing as the comment thread to the original question is running out of control, let me just record some attempts to formulate a question which might (a) be related to what Ian Durham is asking, and (b) is more palatable to some of the people, myself included, who find the original question hard to answer meaningfully.

First of all: I guess we are taking as a working principle

... it is impossible to simultaneously have an infinite number of physical objects of non-zero size in the universe

The example given of an object in the original question is something like "a quantum channel" - now since I'm a physics ignoramus I don't know what the ontological status is of such a beast, but let's suppose for sake of discussion that it does have "size" and that therefore only finitely many of such can exist in a given physical system. This is presumably some argument about physical observables being quantized, but someone else is welcome to correct me on this.

Secondly: there are constructions in mathematical physics which seem to be of an infinite nature. The example given seems to be a "potential infinity", i.e. what are we approaching if we tensor a channel with itself repeatedly.

Now, my interpretation of what Ian may be trying to ask -- and I have to say, in my personal opinion I've not found it at all easy to discern what his underlying question is -- goes like this:

(i) are there contexts in "mainstream abstract mathematics" where an implicitly defined "object" -- such as, the solution space of some differential equation, the solution set of some algebraic equation, the set of accumulation points of some sequence -- which depends on some outside flavour (choice of ground-field for an algebraic equation; choice of topology on some ambient space which reasonably admits more than one topology; an ambient topos in which the construction is supposed to live), might have finite cardinality for one choice of flavour, but infinite cardinality for other choices?

(ii) does this have anything to do with whether we equip a given space, broadly and vaguely conceived, with a Hausdorff or a non-Hausdorff topology?

(iii) do either of these have any connection to the original subject, namely that certain mathematical constructions appear to have physical meaning yet be defined in terms of unphysical infinities?

The answer to (i) is in my view "yes, but so what?" and the answer to (ii) is in my view "I don't really think so". Moreover, I don't think (iii) is really dependent on (ii), and so my overall impression is that the "Hausdorff discussion" is a red herring.

Lastly, I am having difficulty making sense of the reasoning behind this sentence in the original question:

Now suppose that one of the various branching spacetime interpretations of quantum mechanics (MWI, MMI, etc.) is correct (personal aside: I am agnostic on this issue). The topology of the multiverse would thus be non-Hausdorff and, given these interpretations of QM, there ought to be an infinite number of branches. Given that, an infinite physical realization becomes possible.

graph theory - Szemeredi's Regularity Lemma

Theorem: (SRL)
For every $epsilon>0$ and integer $mgeq 1$ there is an $M$ such that every graph $G$, with $|G|geq m$ has an $epsilon$-regular partition $V(G)=V_0cupldotscup V_k$ for some $mleq kleq M$.

Can someone explain to me why this statement is not trivial? For instance, what stops me choosing $M$ larger than $|G|$ and picking $k=|G|$, so I can split $G$ up into singletons, which is trivally $epsilon$-regular for any $epsilon>0$.

cosmology - Is there a paper on galaxy mergers in clusters of galaxies?

Galaxy mergers in clusters explain the large central galaxies in clusters. In fact, the centers of clusters often host a special type of galaxy called a cD which is usually extremely massive because of galactic cannibalism. Dynamical friction makes galaxies spiral down toward the center where they get eaten by the central galaxy and that in turn makes the central galaxy more massive and more efficient at this process.

In the rest of the cluster, the velocity dispersion is significantly higher than the escape velocities from individual galaxies. So a) galaxies cannot gravitationally attract other galaxies because they go by too fast and b) even if there is a head on collision by sheer coincidence, they would pass through and not capture one another. There may well be huge consequences to both galaxies, such as stellar rings and shells formed and tidal tails extracted, but the two remnants would separate and perhaps never meet again.

This is an oversimplification though because clusters often have subgroups of galaxies which have lower velocity dispersion wrt each other and these could merge much as galaxies in groups merge. But the tidal field of the cluster actually works against this and so the rate would be lower.

There is a lot of literature on cD or central galaxies since it is one of the standard types of galaxies. Here is a recent one:

"Growth of brightest cluster galaxies via mergers since z=1", Claire Burke and Chris A. Collins, 2014, MNRAS, v. 434, p. 2856.
http://mnras.oxfordjournals.org/content/434/4/2856

A study of merger rates in the outer parts of clusters may take a major effort of going through the results of N-body simulations on your own and your answer may depend on which simulation you use.

saturn - How long do planetary rings last?

I'm surprised that this question hasn't been asked before (here or on Physics), to the best of my knowledge. It's one that I might have asked when I was a bit younger, and one that I think other people will ask.

Anyway, it's clear that Saturn's rings won't form a moon, and the same is likely to be true for other ring systems. However, I'm guessing that they won't last forever (it's just a guess).

How long do planetary rings in general last? What mechanisms could cause them to dissipate/fall apart/end? I'm guessing the Poynting-Robertson effect could come into play, but I'm not sure.

And for anyone curious, yes, I checked just for the fun of it, and Yahoo Answers had a bunch of really, really bad, unsourced and most likely inaccurate answers (given that there was no consensus), ranging from '3 million years' to '13-18 billion years' to 'forever'.

homotopy theory - Are injective Omega-spectra the S-local objects of symmetric spectra for some class S?

You have to realize it has been a long time since we wrote that paper. But I'll give it my best shot.

I think we intentionally chose the injective Omega-spectra because they are "extra fibrant", so to speak. That is, I think S-local spectra don't have to be injective, just Omega-spectra.

The injective Omega-spectra should be the fibrant objects in a different model structure. There should be an injective level structure, which I guess we did not construct, where the cofibrations are monomorphisms and the weak equivalences are level equivalences. The fibrant objects would then be the injective spectra. The injective Omega-spectra are then the fibrant objects in the left Bousfield localization of this category with respect to the stable equivalences.

co.combinatorics - Undecidable graph problems?

From a MathSciNet search:

---

Földes, Stéphane; Steinberg, Richard
A topological space for which graph embeddability is undecidable.
J. Combin. Theory Ser. B 29 (1980), no. 3, 342--344.

From the introduction: ``From Edmonds' permutation theorem and a generalization due to Stahl, it follows that graph embeddability is decidable for all surfaces, orientable as well as nonorientable. We show the existence of a topological space $hat G$ such that there is no algorithm to decide whether a finite graph is embeddable in $hat G$. In fact, $hat G$ will be a path-connected subspace of the real plane.''

---

ag.algebraic geometry - Question on an exercise in Hartshorne: Equivalence of categories

This is a slight reformulation of exercise II.5.9.(c) in Hartshorne's "Algebraic Geometry" which I don't understand.

Let $K$ be a field and $S=K[X_0,ldots,X_n]$ a graded ring. Set $X=Proj(S)$ and let $M$ be a graded $S$-module. The functors $Gamma_*$ definied by
$$
Gamma_*(mathcal{F})=bigoplus_{ninmathbb{Z}} (mathcal{F}(n))(X)
$$
and $~widetilde{phantom{cdot}}~$ (the "graded associated sheaf functor", see Hartshorne II.5. page 116 for a definition)
induce an equivalence of categories between the category $mathcal{A}$ of quasi-finitely generated (i.e. in relation to a finitely generated module) graded $S$-modules modulo a certain equivalence relation $approx$ and the category $mathcal{B}$ of coherent $mathcal{O}_X$-modules. The equivalence relation is: $Mapprox N$ if there is an integer $d$ such that $oplus_{kgeq d}M_kcongoplus_{kgeq d}N_k$.

I don't know what an "equivalence of categories" is in this context. Formally an "equivalence of categories" means in particular that there are isomorphisms $$hom_mathcal{A}(M,N)cong hom_mathcal{B}(widetilde{M},widetilde{N})$$ and $$hom_mathcal{B}(Y,Z)cong hom_mathcal{A}(Gamma_*(Y),Gamma_*(Z))$$
of sets. This is my problem: How is the sheaf $mathcal{H}om_mathcal{B}(Y,Z)$ considered as a set? Perhaps it should be $Gamma_*(mathcal{H}om_mathcal{B}(Y,Z))cong hom_mathcal{A}(Gamma_*(Y),Gamma_*(Z))$?

ag.algebraic geometry - Dual of Torsion free/Reflexive Coherent Sheaf

If $E$ is a coherent sheaf on a noetherian scheme, the dual $E^*=Hom_{O_X}(E, O_X)$ is always coherent. If $A$ is an affine open subset, then $E^*$ is the sheaf associated to the $A$-module $Hom_A( Gamma(A, E), Gamma(A, O_X))$. More generally, sheaf hom of any two sheaves preserves coherence.

This is a corollary of the fact that if $M,N$ are finitely presented $A$-modules, then for any multiplicative subset $S$, $S^{-1}Hom_A(M,N) = Hom_{S^{-1}A}(S^{-1}M, S^{-1}N)$, which can be found in any commutative algebra textbook (e.g. Eisenbud).

cosmology - What are the concrete technical arguments supporting the idea that the wave function of the universe can be written as partition function?

In this talk, the not yet settled down idea that the wave function of the universe could potentially be written as the partition function of a scale invariant statistical field theory is mentioned:

$$
Psi[g] = Z[g] = int D(fields) e^{-S[g,fields]}
$$

If our universe were AdS, this relation could already be well enough explained by the AdS/CFT correspondence, but as our expanding universe correspond to a dS geometry, things are less clear.

What are the concrete technical arguments, ideas, or hints that this relationship should hold for our dS universe too? What work has already been done on this?

Is sundial time entirely dependent on solar azimuth?

Any sundial that gives the same result as this is correct and any other is wrong (but sometimes close enough):
_ /############
/| /#############
skewer / /############## (central) v / /###############
| north /################
| (S in S. /#################
| hmsphre) /##################
________|________ /###################
hoop | /##### LEVEL #######
(from the | side) /###### GROUND #######
| /####### (SOIL) #######
latitude--|- /#######################
(use || /########################
protractor)| /#########################
|V /##########################
| /###########################
|/############################
j#############################
,|#############################
/#|#############################
/##|#############################
/###|#############################
/####|#############################
/#####|#############################
/######|#############################
/#######|#############################
/########V#############################
/#######################################

#

write noon on hoop's inside closest to ground, midnight opposite, 6 pm on the east side, 9 pm midway between the last two, and so on (hours only occurring in darkness optional).

If your sundial reads 6 pm at due west all the time then you're doing it wrong. Let's say I put a vertical stick in the ground, draw a 24 hour clock face around it, put noon poleward and think it's a sundial. In New York City, it could literally be saying 6 am when a genuine sundial says 9 am. That's just middle latitudes. At the equator on the equinox, it would read 6 am all morning and 6pm all afternoon at the equator. If you go 1 mile south of where the next noon, summer solstice and Tropic of Cancer coincide, it would say about 4:30am at sunrise, go forwards at first, then backwards, finally showing the middle 6 hours of the night passing in 7 seconds. Backwards. At noon. Then it will run forwards again until it reaches 7:30pm at sunset. If you place the stick right, you can even make it stay between 4:30a and 6a all morning, stop at 6 am at the instant of noon then instantly become midnight, run infinity years per second backwards for an infinitely short amount of time, go almost 6 hours backwards in seconds, then later forwards again very slowly until it shows about 7:30pm at sunset. This is why sundials cannot be made that way.

(of course, this is theoretical, there are no infinitely thin, vertical, and straight sticks, shadows are fuzzy, they can be too short to see, the Earth wobbles a bit, the speed of light is not infinite, this would only be true if only the Sun and Earth existed, even a flea jumping in Russia moves the Earth etc.)

And yes, the sundial time can be up to 16 minutes away from mean solar time (the equation of time), easily noticeable, but if you wanted correct local clock time instead of correct sundial time then you could put as many dates as needed on another dial and rotate the hour scale until the arrows point at the current time of year. The shadow then shows mean solar time.

That should be close enough to mean solar time that you wouldn't care for a number of centuries, certainly a century if you're real picky. As for clock time, what sundial disagreements are possible is limited only by the whims of man. The sundial is several hours wrong in West China. Cause they use the zone that's good for Shanghai (or Tokyo when daylight savings).

And all sundials without moving parts are latitude specific. Some are adjustible, though. Some designs are more suited for some latitudes or even become impossible in some places, like the kind with a wedge or rod on a level face. They will also not work on days with polar night. Though you could use a moondial if it's also not polar moon's below the horizon for days or weeks. Yes, moondials exist! You need to correct for moon phase and time of year or they're useless.

Click edit to see the drawing. Don't click save of course.