expansion - Is dark energy evenly spaced throughout Universe?

If dark energy varied by location, then plots of 1a supernova brightness vs redshift should vary depending on which direction you look in the sky. AFAIK, that's not the case. For example, although coords are not accounted for, there's not a lot of scatter in this plot:

In the relationship between the distance and redshift of Type 1a supernovae, the data (points) agree with the equation in which light propagates through the expanding universe on the least-time path (solid line). Image credit: Annila. ©2011 Royal Astronomical Society

Read more at: http://phys.org/news/2011-10-supernovae-universe-expansion-understood-dark.html#jCp

computer science - post correspondence problem

As Tsuyoshi said, it doesn’t make sense to search for an undecidable instance of a problem. It’s only the problem itself that can be undecidable.

In particular, for every instance of PCP (or any other problem for that matter) there trivially exists an algorithm that gives the correct answer for that particular instance. If we’re dealing with the decision version of the problem, it’s either the algorithm that always answers “yes”, or the algorithm that always answers “no” (granted, this is not a constructive proof).

On the other hand, you might find specific instances of PCP without a known answer, for example by exploiting any open problem of mathematics and the fact that the halting problem reduces to PCP, say via a many-one reduction R.

Consider the Turing machine M that searches for a proof of the Riemann hypothesis by enumerating all proofs, and halts when it finds it. If RH is provable, this machine will halt in a finite amount of time, otherwise it will run forever. You can use the reduction from the halting problem to construct a PCP instance R(M) = x. Now, by deciding whether x is a positive or negative instance of PCP, you also decide RH. But that’s an open problem, and so the status of x also is.

ag.algebraic geometry - Flatness of modules via Tor

As far as I understand, this is false. Here is an example (familiar to $D$-module people):
$A=k[x,y]$; $M=k[a,b]$ on which $x$ (resp. $y$) acts as $frac{d}{da}$ (resp. $frac{d}{db}$).
Since the action of both $x$ and $y$ is locally nilpotent, $M$ is supported at the origin of
$Spec(A)$. Therefore, the only non-zero Tor's of the kind you consider are $Tor_i(M,k)$, where both $x$ and $y$ act on $k$ by zero. These Tor's are easy to compute (they amount to computing de Rham cohomology of affine plane with coordinates $a$ and $b$), and they are non-zero precisely when $i=2$. (Essentially, the calculation repeats the proof of Kashiwara's Lemma.)

complex geometry - question about kahler cone of a compact kahler manifold

Hi to all!

I'm studying complex geometry from Huybrechts book "Complex Geometry"
and i have problems with an exercise, please can anyone help me?

I define the kahler cone of a compact kahler manifold X as the set

$K_X subseteq H^{(1,1)}(X)cap H^2(X,mathbb{R})$
of kahler classes. I have to prove that $K_X$ doesn't contain any line
of the form $alpha + t beta$ with $alpha , betain H^{(1,1)}(X)cap H^2(X,mathbb{R})$
and $betaneq 0$ (i identify classes with representatives).

This is what i thought: i know that a form $omega in H^{(1,1)}(X)cap H^2(X,mathbb{R})$
that is positive definite (locally of the form $frac{i}{2}sum_{i,j} h_{ij}(x)dz^iwedge d overline{z}^{j}$ and $(h_{ij}(x))$ is a positive definite hermitian matrix $forall xin X$) is the kahler form associated to a kahler structure. Supposing $alpha$ a kahler class i want to show that there is a $tinmathbb{R}$ such that $alpha + t beta$ is not a kahler class. Since $betaneq0$ i can find a $tinmathbb{R}$ such that $alpha + t beta$ is not positive definite any more, now i want to prove that there is no form $omega in H^{(1,1)}(X)cap H^2(X,mathbb{R})$ such that $omega=dlambda$ with $lambda$ a real 1-form and $omega=overline{partial}mu$ with $mu$ a complex (1,0)-form (what i'd like to prove is: correcting representatives of cohomology classes with an exact form i don't get a kahler class). From $partialoverline{partial}$-lemma and a little work i know that $omega=ipartialoverline{partial}f$ with f a real function. And now (and here i can't go on) i want to prove that i can't have a function f such that $alpha + t beta+ipartialoverline{partial}f$ is positive definite.

Please, if i made mistakes, or you know how to go on, or another way to solve this, tell me.

Thank you in advance.

nt.number theory - Proof of no rational point on Selmer's Curve 3x^3+4y^3+5z^3=0

The "standard" technique for killing the Hasse priniciple for elliptic curves is to show that the Tate-Shafarevich group has a copy of (Z/mZ)^2 for some m - see chapter X in Silverman's the arithmetic of Eliptic curves, both for the theory and examples. All the examples which Silverman presents ar with m = 2. Selmers example requires m = 3, which requires (much) more computations. Poonen has an example
on his web page of a family of elliptic curves violating the Hasse principle, and containing Selmers example, but you'd have to dive through a labirinth of references.

co.combinatorics - Is there a combinatorial reason that the (-1)st Catalan number is -1/2?

The $n$th Catalan number can be written in terms of factorials as
$$ C_n = {(2n)! over (n+1)! n!}. $$
We can rewrite this in terms of gamma functions to define the Catalan numbers for complex $z$:
$$ C(z) = {Gamma(2z+1) over Gamma(z+2) Gamma(z+1)}. $$
This function is analytic except where $2n+1, n+2$, or $n+1$ is a nonpositive integer -- that is, at $n = -1/2, -1, -3/2, -2, ldots$.

At $z = -2, -3, -4, ldots$, the numerator of the expression for $C(z)$ has a pole of order 1, but the denominator has a pole of order $2$, so $lim_{z to n} C(z) = 0$.

At $z = -1/2, -3/2, -5/2, ldots$, the denominator is just some real number and the numerator has a pole of order 1, so $C(z)$ has a pole of order $1$.

But at $z = -1$:
- $Gamma(2z+1)$ has a pole of order $1$ with residue $1/2$;
- $Gamma(z+2) = 1$;
- $Gamma(z+1)$ has a pole of order $1$ with residue $1$.
Therefore $lim_{z to -1} C(z) = 1/2$, so we might say that the $-1$st Catalan number is $-1/2$.

Is there an interpretation of this fact in terms of any of the countless combinatorial objects counted by the Catalan numbers?

amateur observing - Can you see city lights on the Moon from Earth?

This is the opposite of another question. That question is about whether you could see cities on Earth if you were standing on the Moon.

Let's there are cities on the Moon and you're standing on the Earth on a clear night. Could you see the city lights?

If you're looking at Earth cities from the Moon, your line of sight is not affected by atmospheric turbulence because the Moon is airless. But if you are on Earth, then you have to look through the turbulent atmosphere to see faint lights. So you won't have to put up with the turbulence that makes stars look blurry or twinkley from the Earth.

But if you're on the Earth, any city lights on the Moon might look blurry or twinkley from our atmospheric turbulence. If it helps, imagine yourself at the top of a mountain so there is less atmosphere to look through and hence less turbulence.

In this picture of the Moon (and Venus in the background), imagine cities on dark part of the Moon's surface facing Earth. Those city lights should be easier to see than cities within the lit crescent.