general relativity - What is our universe expanding into?

It's a rather difficult concept to grab but, in short, the image of expansion you have in your mind is wrong. It's not your fault, it's everywhere out there, the inflatable balloon etc. And it's terrible :)

Another way to look at it is that the universe properties are changing. Whether it is infinite or finite, its size (as measured within itself) changes. Things are pulled appart. You could imagine that free vacuum keeps being generated between all objects but that would be a misrepresentation of the theory.

The most simple way I see it is actually what the equations say. It's lame but sometimes the math are the simplest way to understand a phenomenon. The equation say the metric is expanding. It means two point in the fabric of the universe are measured at increasing distances over time. As a result, unless objects are pulled to each other with sufficient force, they tend to drift appart. It's a property of the universe that has not been explained yet (at least not completely), and we still are on the fence as to what influences the expansion rate.

Ultimately, this concept is very easy to accept when you cease to represent the universe from outside of itself. The universe is everything that is so if any image in your mind represents the universe from outside, it won't give you a good picture. Balloons are especially bad :). If you need an image, draw a grid or take a notebook, draw some points here and there on it and imagine the grid keeps getting bigger and bigger. But don't look at the whole page, forget the page limits, focus on the grid. And keep in mind that the points do not expand, only the grid does...

Can impact craters on the moon act like giant radio telescopes?

Could large craters on the moon be used as reflective lenses for radio
signals?

You'd have to line the surface with something reflective to microwaves, like a metallic mesh, or similar materials.

Secondly, the shape of the crater is probably not quite ideal, so it would have to be adjusted a little, carved up a bit in various places. But it's a good start, and definitely better than starting with a flat ground.

There is also the question of stability - you need to make sure that whatever changes you make (carving a different shape, lining it with mesh) do not affect the stability of the crater, or else various parts may slide or collapse. This is an engineering problem.

Acting like a large radio telescope reflecting radio waves to a
satellite positioned over the crater.

Not possible unless the crater is exactly on the equator, and even then it would be tricky.

But a crater like the one in your picture is so strongly curved, the focal length is about the same as the diameter. In other words, if the diameter of the hole is X, the altitude of the receiver is pretty close to X - give or take something like 50% or so, depending on the exact curvature. It might be easier to just build a giant arch over the crater. Again, this is a matter of engineering.

reference request - Topology of function spaces?

Let $X,Y$ be finite-dimensional differentiable manifolds, and let's assume that they are connected. In fact, in applications I would like both $X$ and $Y$ to be riemannian manifolds.

Let $C^infty(X,Y)$ denote the space of smooth maps $f: X to Y$. I'm interested in, say, the connected components, fundamental group,... of this space, but I'm really not sure where to start looking. I realise that I first need to topologise this space. My only experience in this realm is an introductory point-set topology course (based on Munkres) I took as a graduate student. Munkres talks about the compact-open topology for the space $C^0(X,Y)$ of continuous maps between two topological spaces and shows, for instance, that if $X$ is locally compact Hausdorff then the evaluation map $X times C^0(X,Y) to Y$ is continuous. Later in the book he also applies this to give a slick proof of the existence of covering spaces with prescribed covering group.

Back to the differentiable category, ideally I'd like to be able to do calculus on $C^infty(X,Y)$, hence I'd like to think of $C^infty(X,Y)$ as an infinite-dimensional differentiable manifold and possibly even riemannian whenever so are $X$ and $Y$.

In case it helps to focus the question, let me say a few words of (physical, I fear) motivation.

When $X,Y$ are riemannian, $C^infty(X,Y)$ plays the rôle of the configuration space for a physical model known as the nonlinear sigma model, whose action functional, assigning to $sigma: X to Y$, the value of the integral (either take $X$ to be compact or else restrict the possible functions further to assure convergence)
$$ S[sigma] = int_X |dsigma|^2 operatorname{dvol}_X,$$
where I'd like to think of $dsigma$ as a one-form on $X$ with values in the pullback $sigma^*TY$ by $sigma$ of the tangent bundle to $Y$, and $|dsigma|^2$ involves the metric on the bundle $T^*X otimes sigma^*TY$ induced from the riemannian metrics on $X$ and on $Y$. The extrema of $S$ are then the harmonic maps.

We are often interested in the quantum theory (un)defined formally by a path integral. A mathematically conservative point of view is that the path integral simply gives a recipe for the perturbative treatment of the quantum theory, where we fix an extremum $sigma_0$ of $S$ and quantise the fluctuations around $sigma_0$. By definition, fluctuations around $sigma_0$ lie in the connected component of $sigma_0$ and as a first approximation, the path integral becomes a sum over the connected components of the space of maps. Hence the interest in determining the connected components of $C^infty(X,Y)$, which in this context are often called superselection sectors.

So in summary, a possible question would be this:

What can be said about the topology (e.g., homotopy type) of $C^infty(X,Y)$ in terms of $X$ and $Y$?

I'm not asking for a tutorial, just for some orientation to the available literature.

Thanks in advance.

Added

In response to the helpful answers I have received already, I'd like to point out that my interest is really on the homotopy type of the space of maps. I only mentioned the analytic aspects in case that narrows down the topology one would put on the space. As pointed out in the answers, most reasonable topologies are equivalent, so this is a relief.

As for concrete examples, I am particularly interested in the case where $X$ is a compact Riemann surface and $Y$ a compact Lie group. I'm happy to put the constant curvature metric on $X$ and a bi-invariant metric on $Y$.

pulsar - Brightest Radio Source In the Universe

I would have assumed that the brightest radio source in the Universe is a quasar of some kind (perhaps 3C 273) given that the average pulsar has luminosity of $~10^{40} text{Watts}$ and this is the brightest (optically) and is radio loud.

My professor seems to be of the opinion that the Arecibo message is louder, but that does not seem to pass the straight face test, given that it was only transmitted at a power of only $1text{MW}$, and only a very small fraction of the pulsars output (a mere $1/10^{34}$th) would be sufficient to outshine the message.

Is one of these the brightest radio source in the universe (excluding the Big Bang if it matters), or is it something else entirely?

Additionally, is Watts even the right unit to be measuring this in?

Update: Here is the text of the question (it was a true/false, and the answer is true):

In 1975, Frank Drake and Carl Sagan sent a message in to the Universe, directed to extraterrestrials, from the radio telescope at Arecibo in Puerto Rico that, for a brief time, made the Earth the brightest radio object in the Universe.

ac.commutative algebra - Proving that two local PIDs, one inside the other, with the same field of fractions are equal.

I believe that the OP meant to include the condition that each of the local PIDs is not a field. In this case the result is true, and as several people have said, is a rather standard exercise.

At this moment it seems to me that if we get asked a rather standard question that has not been asked on MO before, it would be nice to use it as an opportunity to explain something a little deeper / slightly less standard related to the question. In this regard, let me mention a generalization:

A local PID $R$ (which is not a field!) with fraction field $K$ is precisely a discrete valuation ring, i.e., is the valuation ring $R = {x in K | |x| leq 1 }$ of a norm $| |: K rightarrow mathbb{R}^{geq 0}$ such that $|K^{times}|$ is a discrete subgroup of $mathbb{R}^{times}$. Now for nontrivial norms $| |_1$, $| |_2$ (Archimedean or not) on a field $K$, there is the following result:

Theorem: The following are equivalent:
(i) There exists $alpha > 0$ such that $| |_2 = | |_1^{alpha}$.
(ii) For all $x in K$, $|x|_1 < 1 implies |x|_2 < 1$.
(iii) For all $x in K$, $|x|_1 leq 1 implies |x|_2 leq 1$.

(See e.g. http://math.uga.edu/~pete/8410Chapter1.pdf, p. 4, for a proof.)

Now the implication (iii) $implies$ (i) shows that there can be no proper containments among DVRs with the same fraction field. The same holds for all rank one valuation ring because, by definition, a rank one valuation ring is one whose value group is a subgroup of $mathbb{R}$; therefore the data of a rank one valuation is equivalent to that of a non-Archimedean norm. (Note that if $| |$ is a non-Archimedean norm, then $v = - log | |$ is a valuation, and conversely if $v$ is a rank one valuation, then $| | = e^{-v}$ is a non-Archimedean norm.) It is not true for valuation rings of higher rank.

Visibility of earth from moon during day-time of moon

Inspired by this question. I am curious whether earth, besides being nearly fixed on one place on the moon's sky, is it visible during the day-time on moon too?

My understanding is that earth should be visible as moon has no atmosphere. Also, if the NASA didn't edit the following photograph, it suggests that the day-time sky of moon is all black and earth should be visible in it.

I think, Stellarium don't take into account the atmosphere once you are on other planet.

rt.representation theory - Is every finite group a group of "symmetries"?

The permutohedron may have additional symmetries. For example, the order 3 permutohedron {(1,2,3),(1,3,2),(2,1,3),(3,1,2),(3,2,1)} is a regular hexagon contained in the plane x+y+z=6, which has more than 6 symmetries.

I think we can solve it as follows:

Let G be a group with finite order n thought via Cayley's representation as a subgroup of S_n.

Let S={A_1,...,A_n} be the set of vertices of a regular simplex centered at the origin in a (n-1)-dimensional real inner product space V. Let r be the distance between the origin and A_1. The set of vertices S is an affine basis for V.

First unproven claim: If a closed ball that has radius r contains S, then it is centered at the origin. Let B be this ball.

The group of isometries that fix S hence contains only isometries that fix the origin and permute the vertices, which can be identified with S_n in the obvious way. The same is true if we replace S by its convex hull.

Now G can be thought of as a group containing some of the symmetries of S.

Let C=k(A_1+2A_2+3A_3+...+nA_n)/(1+2+...+n), with k a positive real that makes the distance between C and the origin a number r' slightly smaller than r.

Let GC={g(C) / g in G}. It has n distinct points, as a consequence of being S an affine basis of V.

Let P be the convex hull of the points of S union GC.

Remark: A closed ball of radius r contains P iff it is B. The intersection of the border of B and P is S.

Second unproven claim: The extremal points of P are the elements of S union GC.

Claim: G is the group of symmetries of P.

If g is in G, g is a symmetry of GB and of S, and it is therefore a symmetry of P.
If T is a symmetry of P, then T(P)=P, and in particular, T(P) is contained in B, and hence T(0)=0 (i.e. T is also a symmetry of B). T must also fix the intersection of P and the border of B, so T permutes the points of S, and it can be thought of as an element s of S_n sending A_i to A_s(i). And since T fixes the set of extremal points of P, T also permutes GC. Let's see that s is in G.

Since T(C) must be an element of g(C) of GC, we have T(C)=g(C). But since T is linear, T(C/k)=g(C/k). Expanding,

(A_s(1)+2A_s(2)+...+nA_s(n)/(1+...+n)=(A_g(1)+2A_g(2)+...+nA_g(n))/(1+...+n).

For each i in {1,...,n} the coefficient that multiplyes A_i is s⁻1(i)/(1+...+n) in the left hand side and g⁻1(i) in the right hand side. It follows that s=g.

I think that, taking n into account, the ratio r'/r can be set to substantiate the second unproven claim. The first unproven claim may be a consequence of Jung's inequality.

EDIT: With the previous argument, we can represent a finite group of order n as the group of linear isometries of a certain polytope in an n-1 dimensional real inner product space.

Now, if a finite group G of linear isometries of an (n-1)-dimensional inner product space V is given, can we define a polytope that has G as its group of symmetries? Yes. I'll give a somehow informal proof.

Let G={g_1,...,g_m}. Let A={a_1,...,a_n} be the set of vertices of a regular n-simplex centered at the origin of V. Let S be the sphere centered at the origin that contains A, and let C be the closed ball. Notice that C is the only minimun closed ball contaiing A.

(Remark: The set A need not be a regular simplex. It may be any finite subset of S that intersects all the possible hemispheres of S. C will then still be only minimum closed ball
containing it.)

Remark: An isometry of V is linear iff it fixes the origin.

Before proceeding, we need to be sure that the m copies of A obtained by making G act on it are disjoint. If that is not the case, our set A is useless but we can find a linear isometry T such that TA does de job. We consider the set M of all linear isometries with the usual operator metric, and look into it for an isometry T such that for all (g,a) and (h,b) distinct elements of GxA the equation g(Ta)=h(Tb) does not hold. Because each of the n*m(n*m-1) equations spoils a closed subset of M with empty interior(*), most of the choices of T will do.

Let K={ga/g in G, a in A}. We know that it has n*m points, which are contained in the sphere S. Now let e be a distance that is smaller than a quarter of any of the distances between different points of K. Now, around each vertex a=a_i of A make a drawing D_i. The drawing consists of a finite set of points of the sphere S, located near a (at a distance smaller than e). One of the points must be a itself, and the others (if any) should be apart from a and very near each other, so that a can be easily distinguished. Furthermore, for i=1 the drawing D_i must have no symmetries, i.e, there must be no linear isometries fixing D_1 other than the identity. For other values of i, we set D_i={a_i}. The union F of all the drawings contains A, but has no symmetries. Notice that each drawing has diameter less than 2*e,

Now let G act on F and let Q be the union of the m copies obtained. Q is a union of n*m drawings. Points of different drawings are separated by a distance larger than 2*e. Hence the drawings can be identified as the maximal subsets of Q having diameter less than 2*e. Also, the ball C can be identified as the only sphere with radius r containing Q. S can be identified as the border of C.

Let's prove that the set of symmetries of Q is G. It is obvious that each element of G is a symmetry. Let T be an isometry that fixes Q. It must fix S, so it must be linear. Also, it must permute the drawings. It must therefore send D_1 to some gD_i with g in G and 1<=i<=n. But i must be 1, because for other values of i, gD_i is a singleton. So we have TD_1=gD_1. Since D_1 has no nontrivial symmetries, T=g.

We have constructed a finite set Q with group of symmetries G. Q is not a polytope, but its convex hull is a polytope, and Q is the set of its extremal points.

(*) To show that for any (g,a) and (h,b) distinct elements of GxA the set of isometries T satisfying equation g(Ta)=h(Tb) has empty interior, we notice that if an isometry T satisfies the equation, any isometry T' with T'a=Ta and T'b=/=Tb must do (since h is injective). Such T' may be found very near T, provided dimV>2. The proof doesn't work for n=1 or 2, but these are just the easy cases.