TITLE: Fundamental group of annulus with relation
QUESTION [1 upvotes]: I'm trying to find the fundamental group of space $X$ obtained from an annulus $\{p \in \mathbb R^2 : 1 \leq |p|  \leq2 \}$ by identifying the point $(x,y)$ on the inner circle of radius $1$ with the point $(-2x,-2y)$ on the outer circle of radius $2$.
I think that we can use CW-complexes do that, but I'm not sure how to do it in this case. Moreover, I think I can apply van Kampen here too by taking the sets $A= X \smallsetminus \{\text{inner circle}\}$ and $B= X \smallsetminus \{\text{outer circle}\}$. Is that correct?
Thanks...

REPLY [2 votes]: The use of Van Kampen isn't quite correct - you can't remove the inner circle from $X$ without also removing the outer circle. Try Van Kampening with two half-annuli intersecting at the $y$-axis (or a thickened neighborhood of it). Note that the intersection is actually connected, due to the unusual gluing map.


TITLE: limit problems involving infinity
QUESTION [0 upvotes]: find $$lim_{x \to \infty}\frac{20+2\sqrt{x}+3x^{\frac{1}{3}}}{2+\sqrt{4x-3}+(8x-4)^{\frac{1}{3}}}$$
i confused on seeing the cube root.please give me a hint.Thanks

REPLY [0 votes]: Generally,

$$\lim _{ x\rightarrow \infty  }{ \frac { a{ x }^{ m }+... }{ { b }{ x }^{ n }+... }  } =\begin{cases} +\infty ,\quad if\quad m>n\quad  \\ \frac { a }{ b } ,\quad if\quad m=n \\ 0,\quad \quad \quad if\quad m<n \end{cases}$$


TITLE: Generalization of $\lim_{n \rightarrow \infty} \prod_{i=1}^{n}\frac{2i-1}{2i}$ for a character $\chi:\mathbb{Z}/s \mathbb{Z} \rightarrow \mathbb{C}^*$
QUESTION [4 upvotes]: Playing with some infinite products I came up with this problem, that I'm not able to figure it out by myself. Moreover in the internet it doesn't seem to appear anywhere.
Maybe it is just an easy consequence of properties of characters that I'm not aware of, anyway thank you in advance for any help/answers/suggestions.
All of us know (it is fairly easy to see) that $$\lim_{n \rightarrow \infty} \frac{1 \cdot 3\cdot \dots \cdot (2n-1)}{2 \cdot 4 \cdot \dots \cdot (2n)}=0$$
Now this fact could be reformulated in this fashion: let $$\chi:\mathbb{Z}/2 \mathbb{Z} \rightarrow \mathbb{C}^*$$
the only non trivial character of $\mathbb{Z}/2 \mathbb{Z}$, then the expression $$\lim_{n \rightarrow \infty}(\prod_{i=1}^{n}[2(i-1)+1]^{\chi(2(i-1)+1 \: \text{mod} 2\mathbb{Z})}\cdot [2i]^{\chi(2i \: \text{mod} s\mathbb{Z})})^{-1}$$
More generally, we can perform this construction for every $s \in \mathbb{N}$.
Indeed all we have to do is to consider a non trivial character $$\chi:\mathbb{Z}/s \mathbb{Z} \rightarrow \mathbb{C}^*$$
and consider the limit
$$\lim_{n \rightarrow \infty}(\prod_{i=1}^{n}[s(i-1)+1]^{\chi(s(i-1)+1 \: \text{mod} s\mathbb{Z})}\cdot \dots \cdot [si]^{\chi(si \: \text{mod} s\mathbb{Z})})^{-1}$$
Now it is true that:

The value of the limit is finite for every $s$ and every non trivial character $\chi$?
If so the value of the limit depends only on $s$ or also on $\chi$?
The limit is always a real number? (Possibly $0$?)

REPLY [8 votes]: Summary: I consider the limit $\lim_{n\to\infty}\prod_{i=1}^n i^{\chi(i)}$ (let me drop the $\mod s\mathbb Z$ for brevity). If we restrict to $n$ divisible by $s$, then the limit will always be equal to zero. If we consider the limit over all $n$, the limit will never exist. [The question was edited to only ask about the first of those cases since I've started writing this, so parts of this answer are now irrelevant. For interest I've decided to keep them though, they are enclosed in square brackets]
[The limit you claim is zero, $\lim_{n\to\infty}\prod_{i=1}^n i^{\chi(i)}$ with $\chi:\mathbb Z/2\mathbb Z\to\{\pm 1\}$, actually does not exist. It only exists and is equal to $0$ if you group the terms into pairs, which amounts to only looking at $n$ even. If you restrict to $n$ odd, then the product will tend to infinity by a very similar argument.]
For the products with arbitrary $s$, let me first consider the sum up to $n=ms$ for some $m$. The logarithm of your product is $-\sum_{k=1}^\infty\chi(k)\log k$. Splitting into residue classes modulo $s$, we have for $a=1,\dots,s$ a sum $\chi(a)\sum_{i=0}^{m-1}\log(is+a)$. Finding asymptotics for those sums can be done using the Stirling formula for the gamma function:
$$\prod_{i=0}^{m-1}(is+a)=s^m\prod_{i=0}^m(\frac{a}{s}+i)=s^m\frac{\Gamma(\frac{a}{s}+m)}{\Gamma(\frac{a}{s})},$$
so by Stirling the logarithm is
$$\sum_{i=0}^{m-1}\log(is+a)=m\log s+\log\Gamma(\frac{a}{s}+m)-\log\Gamma(\frac{a}{s})\\
=m\log s+(\frac{a}{s}+m)\log(\frac{a}{s}+m)-(\frac{a}{s}+m)-\frac{1}{2}\log(\frac{a}{s}+m)+O(1)\\
=m\log(\frac{a}{s}+m)+m(\log s-1)+(\frac{a}{s}-\frac{1}{2})\log(\frac{a}{s}+m)+O(1).$$
Note we have $\log(\frac{a}{s}+m)=\log m+\log(1+\frac{a}{sm})=\log m+O(\frac{1}{m})$, so this sum is further equal to
$$m\log m+m(\log s-1)+(\frac{a}{s}-\frac{1}{2})\log(m)+O(1).$$
Now multiply by $\chi(a)$ and sum over all $a$. Since $\sum_{a=1}^s\chi(a)=0$, all terms not involving $a$ cancel out and we are left with
$$\sum_{n=1}^{ms}\chi(n)\log n=\frac{1}{s}\log m\cdot\sum_{a=1}^s a\chi(a)+O(1).$$
Therefore the behavior of your product depends on what the sum $\sum_{a=1}^s a\chi(a)$. If this sum turns out to be a complex number of positive real part [which it does in the case of interest, see below], then the product will tend to zero. For $\chi$ as at the start this is indeed what happens: $1\chi(1)+2\chi(2)=-1+2=1$. If this turned out to have negative real part, then the sum would tend to infinity. If the real part of the sum is zero, then the exact behavior will depend on the $O(1)$ term. We could very well make that computation too, it would be similar but more messy.
[Now, what happens if we take $n$ to not be divisible by $n$, but instead say of the form $n=ms+c$ for some $c$? In that case, the easiest thing to do given computations already done above is separate some of the original terms:
$$\prod_{i=1}^n i^{\chi(i)}=\prod_{i=1}^c i^{\chi(i)}\prod_{i=c+1}^{ms+c}i^{\chi(i)}.$$
The first factor is a nonzero constant, so doesn't impact the convergence properties. The same calculation as above will establish the same kind of result as for $c=0$, except it will instead depend on the sum $\sum_{a=c+1}^{c+s} a\chi(a)$. For your original $\chi$ and $c=1$ this gives $2\chi(2)+3\chi(3)=2-3=-1$, which is indeed what we observe with the limit over odd $n$ being infinite.]
All that said, the behavior of the sum revolves around sums of the form $\sum_{a=c+1}^{c+s} a\chi(a)$. Let me indicate how to find (and study) them. First consider the case of $c=0$ [which after the edit is the only relevant one]. Let $z=\chi(1)$. Since $\chi$ is a character, $\chi(a)=z^a$, and so our sum becomes $\sum_{a=1}^sa z^a$. This sum is equal to
$$\frac{z(sz^{s+1}-(s+1)z^s+1)}{(z-1)^2}=\frac{z(sz-(s+1)+1)}{(z-1)^2}=\frac{sz(z-1)}{(z-1)^2}=\frac{sz}{z-1}$$
since $z^s=\chi(s)=1$ and $z\neq 1$ since $\chi$ is nontrivial. This value will always have positive real part - there is a simple geometric argument: the argument of this value is the angle formed by points $z,0,z-1$. But since $|z-0|=|z-(z-1)|=1$, the triangle they form will be isosceles, so the angle has to be acute. Therefore, in light of all that was said above, the product over $n$ divisible by $s$ will always tend to zero.
[If we permit other values of $c$, then we can just make the transformation $$\sum_{a=c+1}^{c+s} a\chi(a)=\sum_{a=1}^s(a+c)\chi(a+c)=\chi(c)\sum_{a=1}^sa\chi(a)+c\chi(c)\sum_{a=1}^s\chi(a)=\chi(c)\sum_{a=1}^sa\chi(a)=\frac{sz^{c+1}}{z-1}.$$
This sum can now have real part positive, negative, or zero, so its behavior can vary. For $c=s-1$ this term will become $\frac{s}{z-1}$ and will always have negative real part (since $z-1$ does), so the product will tend to infinity. In particular we get that the product will never exist if we take it over all values of $n$]


TITLE: How to prove equality of sets?
QUESTION [0 upvotes]: Prove that $[x \in \mathbb{Z}:6|x] = [x \in \mathbb{Z}:2|x]\cap[x\in\mathbb{Z}:3|x]. $
I understand that I have to prove this from multiple angles, but this is my first time working with proofs using sets. How do I go about tackling a problem like this?
Any input is greatly appreciated!

REPLY [0 votes]: Proof writing. Let $A=\{x\in \mathbb Z: 6|x\}.$ Let $B=\{x\in Z:2|x\}.$ Let $C=\{x\in Z:3|x\}.$
To prove $$A\subset (B\cap C)\iff$$ $$\iff \forall x\;(x\in A\implies (x\in B\land x\in C)\iff$$ $$\iff \forall x\in \mathbb Z\;(6|x\implies (2|x\land 3|x)).$$ At this point you just present a proof that $2|x$ and $3|x$ whenever $x$ is an integer divisible by $6.$
To prove $(B\cap C)\subset A ,$ : $$(B\cap C)\subset A\iff $$ $$\iff\forall x\;(x\in b\cap C\implies x\in A)\iff$$ $$\iff \forall x\in \mathbb Z\;((2|x\land 3|x)\implies 6|x).$$ At this point you just present  a proof that $6|x$ whenever $x$ is an integer divisible by $2$ and by $3.$
This method of proving equality of sets is often useful, especially in more complicated problems.


TITLE: What is the remainder when $4^{96}$ is divided by 6
QUESTION [3 upvotes]: What is the remainder when $4^{96}$ is divided by 6
My approach :
When 4 is divided by 6 gives 4 remainder, when $4^2$ is divided by 6 gives remainder 4 ,... is it for the complete series.. that means when $4^{96}$ is divided by 6 gives 4 as remainder or is there any short cut or other answer of this question.
Please suggest if there is any sort of short cut to such problems.
Thanks.

REPLY [0 votes]: In all cases,
n-odd$\rightarrow 2^n\equiv 2 \mod6$
n-even$\rightarrow 2^n\equiv 4\mod6$
$\implies 2^{96}\equiv 4\mod 6$


TITLE: Why the 'low' part of the sky looks lite blue?
QUESTION [2 upvotes]: When looking on the sky its lower part looks lite blue instead blue like the rest of the sky. Why is it so?

REPLY [4 votes]: As you probably already know, the sky is blue due to scattering of light. Blue gets scattered the most1. The light which "comes" from the horizon has to travel the largest distance (compared to the light which comes from points in the sky closer to the sun), so the light is scattered more. As a consequence, some of the blue light is "removed" by scattering, making the light closer to the horizon appear more white.

(Source)

1 It is actually purple that gets scattered the most, see Why is the sky not purple?

REPLY [3 votes]: When you look in different directions you look through different amounts of air (and different layers): the total distance scales as $\approx 1+\tan^2(\theta)$ where $\theta$ is the zenith angle.
That means that besides the Rayleigh light scattering that makes the sky blue, there will also be more Mie scattering along the path. This is amplified by the larger amount of water droplets, dust, pollen and other scattering particles in the lower layers of the atmosphere. They all scatter sunlight, and this adds to the light along the path. This scattering tends to be more whitish since it scatters all wavelengths roughly equally, unlike the Rayleigh scattering.


TITLE: Non-canonical examples of divergent sequences that are square summable?
QUESTION [0 upvotes]: The canonical example of a divergent sequence that is square summable,  $\sum_{n = 1}^\infty a_n$ is finite, is the harmonic sequence: $\sum_{n = 1}^\infty 1/n$.
Are there examples of sequences that also satisfy this property outside of the harmonic sequence that also satisfies these properties (divergent + square convergent)?

REPLY [0 votes]: Yes, of course. See stuff like $\{b_n\}_{n\ge 10}$ where, to name a few examples \begin{align}b_n&=\frac1{\sqrt n\log n}\\ b_n&=\frac1{\sqrt{n\log n}\log\log n}\\ b_n&=\frac1{\sqrt{n\log n}(\log\log n)^{2/3}}\\ b_n&=\frac1{\sqrt{n\log n\log\log n}\log\log\log n}\end{align}


text:
\begin{document}

\maketitle

\begin{abstract}

Meeks, Pérez and Ros conjectured that a closed Riemannian $3$-manifold which does not admit any closed embedded minimal surface whose two-sided covering is stable must be diffeomorphic to a quotient of the $3$-sphere. We give a counterexample to this conjecture.\\
Also, we show that if we consider immersed surfaces instead of embedded ones, then the corresponding statement is true.

\end{abstract}

\providecommand{\abs}[1]{\lvert#1\rvert}

\linespread{1}

\section{Introduction}

We say that a closed minimal surface $\si$ immersed in a Riemannian $3$-manifold $(M,g)$ is stable if the second variation of area is non-negative for all smooth variations of $\si$. It is well known that if $M$ contains some non-trivial (homology, homotopy or isotopy) class of surfaces, then one can minimize area on the given class and obtain a stable minimal surface in $(M,g)$. On the other hand, quotients of the $3$-sphere endowed with a metric of positive Ricci curvature does not admit two-sided, closed, stable, minimal surfaces. These facts suggest that the existence of closed stable minimal surfaces is related to the topology of the ambient space. In \cite{M.P.R}, Meeks, P\'erez and Ros stated the following conjecture concerning this relation.

\begin{conj}[Meeks, P\'erez, Ros]
Let $(M,g)$ be a closed Riemannian $3$-manifold. If $(M,g)$ does not admit any closed embedded minimal surface whose two-sided covering is stable, then $M$ is finitely covered by the $3$-sphere.
\end{conj}

\begin{rem}
When the surface is one-sided, we consider the two-sided immersed surface associated to it, otherwise we consider the surface itself.
\end{rem}

In this work we will give a family of counter-examples to this conjecture. The examples are locally homogeneous Riemannian manifolds modelled on the {\it Thurston geometries} $\nil$ and $\psl(\re)$. The construction uses a topological characterization of these spaces and the classification of its stable minimal surfaces (due to Pitts-Rubinstein \cite{P.R}). The main result is (for a more precise version see {\it the main example} in Section \ref{SFS} and the Theorem \ref{coex}):

\begin{thm}\label{cx}
Let $(\tilde{M},\tilde{h})$ be $\nil$ or $\psl(\re)$ endowed with a homogeneous metric $\tilde{h}$ whose isometry group has dimension $4$. There are closed Riemannian $3$-manifolds $(M,h)$ obtained as quotients of $(\tilde{M},\tilde{h})$ and which do not contain any closed, embedded, minimal surface whose two-sided cover is stable.
\end{thm}

One can wonder about what is the correct statement to the conjecture of Meeks, P\'erez and Ros. In this direction, we first prove that if we change the class of surfaces considered to immersed instead of embedded, then the corresponding statement is true, i.e, we have:

\begin{thm}\label{ims}
Let $(M,g)$ be a closed Riemannian $3$-manifold. If $(M,g)$ does not admit any closed immersed minimal surface whose two-sided covering is stable, then $M$ is finitely covered by the $3$-sphere.
\end{thm}

The proof of this theorem uses theorems of existence of area-minimizing surfaces and some of the developments in the theory of $3$-manifolds of the last fifteen years, in particular the work of Perelman on the {\it Geometrization conjecture} \cite{Pe}, and the work of Kahn-Markovic on the {Surface subgroup conjecture} \cite{K.M}. However, if one keep considering embedded surfaces, we will show in section \ref{last} that our counterexample and the proof of theorem \ref{ims} reduces the problem to the following question:\\

\noindent
{\bf Question}: Let $M$ be a non-Haken closed $3$-manifold, which is homeomorphic to a hyperbolic manifold. Let $g$ be a Riemannian metric on $M$. Does $(M,g)$ admit a closed, embedded, minimal surface whose two-sided covering is stable?\\

Using the following theorem we can obtain interesting consequences about the geometry of the manifolds in theorem \ref{cx}.

\begin{teo}
Let $(M,g)$ be a closed Riemannian $3$-manifold which does not admit any closed embedded minimal surface whose two-sided covering is stable. Then:
\begin{enumerate}
\item (Meeks, P\'erez, Ros, \cite{M.P.R}) Any two closed, immersed minimal surfaces in
$(M,g)$ intersect (Frankel property).\\
\item (Meeks, P\'erez, Ros, \cite{M.P.R}) Every closed, two-sided minimal surface embedded in $(M,g)$ is a Heegaard splitting.\\
\item (Marques-Neves, \cite{M.N1}) $(M,g)$ contains an infinite number of distinct closed, embedded, minimal surfaces.\\
\item (Song, \cite{S1}) If $M$ is orientable, a strongly irreducible Heegaard surface in $M$ is isotopic to a minimal surface of index at most $1$ .\\
\item (Mazet-Rosenberg, \cite{M.R}) The embedded minimal surface of least area in $(M,g)$ is orientable, has index $1$ and is a min-max surface that realizes the Almgren-Pitts width $W(M,g)$.
\end{enumerate}

\end{teo}

Concerning item $(3)$, the {\it Yau conjecture} says that any closed Riemannian $3$-manifold contains an infinite number of distinct closed, immersed, minimal surfaces. This conjecture was proved very recently by Song \cite{S2}, after successive advancements by many authors. The proof uses the recent developments of the {\it Min-max theory of minimal surfaces} carried out by by Marques, Neves and their collaborators.

In \cite{M.N1} Marques and Neves, proved the conjecture for manifolds which satisfies the embedded Frankel property (in particular manifolds with $\ric(g) > 0$). In \cite{I.M.N}, Irie, Marques and Neves proved the strong fact that for generic metrics (in the Baire sense) the union of all closed, embedded minimal hypersurfaces is a dense subset of a closed manifold of dimension $3 \leq n \leq 7$. Also, in \cite{M.N.S} Marques, Neves and Song obtained a quantitative version of the previous result. The main ingredient in the proof of these results is the Weyl law for the volume spectrum by Liokumovich, Marques and Neves \cite{L.M.N}.

Later, working on the Allen-Cahn version of min-max theory developed by Guaraco and Gaspar in \cite{G,G.G1}, Chodosh and Mantoulidis \cite{C.M} proved a version of the {\it Multiplicity One Conjecture} of Marques-Neves in dimension $3$, and following a strategy of these authors proved the Yau conjecture for bumpy metrics and metrics with $\ric(g) > 0$ (the nolvety of their statement in the generic case, is that there is a sequence of two-sided embedded minimal surfaces $\{\si_{k}\}$, with $\ind(\si_k) = k$, and $|\si_k| \thicksim k^{\frac{1}{3}})$. Also, in \cite{G.G2}, Gaspar and Guaraco obtained variations of the results in \cite{I.M.N,M.N.S}, and using the work in \cite{C.M} proved that in dimension $3$ the surfaces can be chosen to be separating in the density statement. Finally, building in the ideas of \cite{M.N1}, Song settled the full conjecture \cite{S2}, proving that in dimensions $3 \leq n \leq 7$ there exists infinitely many embedded closed minimal hypersurfaces for any metric.
\\

\noindent
{\bf Ackowledgements}: It is a pleasure to thank Lucas Ambrozio and Harold Rosenberg for discussions and suggestions on the manuscript. I also thank, Laurent Mazet for a discussion about the main results, and Joaquin P\'erez for personal correspondence about this work.

\section{Seifert Fibered Spaces}\label{SFS}

In this section we recall some facts about Seifert Fibered Spaces. The main references used here are \cite{Ma,S}. Let $D$ denote the unit disc of the complex plane centered at the origin. Let $p,q$ be two coprime integers with $p>0$. A \emph{standard fibered solid torus} with coefficients $(p,q)$ is a solid torus
$$\frac{D \times [0,1]}{\bigl\{(z,0) \sim (\psi_{p,q}(z),1)\bigr\} }$$
where $\psi_{p,q}\colon D  \to D $ is given by $\psi_{p,q}(z) = e^{(2\pi iq/p)}z$. The fibration by \emph{vertical} segments $\{x\} \times [0,1]$ extends to a fibration into circles of the solid torus. The central fibre obtained by identifying the endpoints of $\{0\} \times [0,1]$ is the core of the solid torus, and every non-central fiber winds $p$ times around the core of $M$. The positive number $p$ is the \emph{multiplicity} of the central fibre. If $p=1$ the fibered solid torus is diffeomorphic to the usual product fibration $D\times S^1$ and the central fibre is called \emph{regular}. If $p > 1$, the central fibre is called \emph{singular}.

A \emph{Seifert Fibered Space} is an oriented 3-manifold $M$ foliated by circles, such that every circle has a fibered neighbourhood diffeomorphic to a standard fibered solid torus. Let $S$ be the topological space obtained from $M$ by quotienting the circles to points, then $S$ is a compact connected surface and has a natural orbifold structure: if the preimage of $x\in S$ is a fibre of order $p$, we see $x$ as a cone point of angle $\frac{2\pi}{p}$. The map $M \to S$ is called a Seifert fibration, (one should think of this fibration as a circle bundle over the orbifold $S$). A Seifert fibration without singular fibres is a circle bundle over a surface, in the usual sense.

Let $\si \subset M$ be a closed embedded surface. We say $\si$ is \emph{vertical} if it is the union of regular fibers. In this case we have two possibilities: $\si$ is a torus whose projection to the base surface is an orientation-preserving simple closed curve contained in the complement of the cone points of $S$; or $\si$ is a Klein bottle, which projects to an orientation-reversing simple closed curve, again inside the regular part of $S$ (on this case, the orbifold is necessarily non-orientable). On the other hand, we say $\si$ is \emph{horizontal} if is everywhere transverse to the fibers. On a closed Seifert manifold $M$ there is an invariant associated to it, called the Euler number and denoted by $e(M)$, which detects the existence of such surfaces: $M$ contains a \emph{horizontal} surface if, and only if, $e(M) = 0$.
\\

\noindent
{\bf Geometric metrics}:
Let $\mathbb{E}$ denote one of the fibered {\it Thurston geometries}, i.e, $\rt, \st,\sd\times\re, \hp\times\re, \nil, \psl(\re)$ (we will denote $\psl(\re)$ by $\psl$). Each one of these spaces has a homogeneous metric with isometry group of dimension at least $4$ and these metrics satisfy the following properties: there is a Riemannian submersion
$\Pi: \mathbb{E} \rightarrow \mathbb{M}^{2}(\kappa),$
where $\mathbb{M}^{2}(\kappa)$ is the complete simply-connected Riemannian surface with constant curvature $\kappa$, and the fibers of $\Pi$ are the integral curves of a unit Killing vector field $\xi$. We have $\kappa = 0$ in the case of $\rt$ and $\nil$, $\kappa = 1$ in the case of $\sd\times\re$ and $\st$, and $\kappa = -1$ in the case of $\hp\times\re$ and $\psl$. Moreover each geometry has the structure of a line or circle bundle over $\mathbb{M}^{2}(\kappa)$, such that the bundle projection coincides with the submersion $\Pi$, and all the isometries of these metrics preserve this bundle structure.

Every Seifert Fibered Space is diffeomorphic to a quotient of some $\mathbb{E}$, and conversely every quotient of $\mathbb{E}$ has a Seifert Fibered structure. In view of this, we see that each Seifert fibered space has a locally homogeneous metric, and its Seifert fibration is induced by the bundle structure on the universal cover and the projection $\Pi$. Thus, outside the singular fibers, we have an $SO(2)$-action on the fibers by isometries and the projection on the base is a Riemannian submersion, where the base orbifold is endowed with a constant curvature metric with isolated cone singularities. We call such a metric a {\it geometric metric}.

\begin{rem}
Two closed $3$-manifolds with different Thurston geometries are not diffeomorphic (see theorem $5.2$ in \cite{S}). This fact will be crucial to our counterexample in the next section.
\end{rem}

The appropriate geometry for a closed \emph{Seifert Fibered Space} is determined in terms of the Euler number $e(M)$ and the orbifold Euler characteristic of the base $\bigl($denoted $\chi(S)\bigr)$ by the following table:

\vspace{0.2cm}

\begin{tabular}[c]{lcccccr}
\hline
 & & $\chi < 0$ & & $\chi = 0$ & & $\chi > 0$ \\\\
$e = 0$ & & $\hp\times\re$ & & $\rt$ & &$\sd\times\re$ \\\\
$e \neq 0$ & & $\psl$ & & $\nil$ & & $\st$ \\
\hline
\label{The geometry of closed Seifert manifolds}
\end{tabular}

\vspace{0.1cm}
\noindent
\label{main}{\bf Main example}: Let $p_1,p_2,p_3 \geqslant 2$ be natural numbers and $\Delta$ be a triangle in $\rp$, or $\hp$, with inner angles $\frac{\pi}{p_1}$, $\frac{\pi}{p_2}$, $\frac{\pi}{p_3}$. By reflecting iteratively $\Delta$ along its sides we get a tessellation $T$. The \emph{triangle group} $\Gamma (p_1,p_2,p_3)$ is the group of isometries of $\rp$, or $\hp$, generated by the reflections along the three sides of $\Delta$. This group acts freely and transitively on the triangles of the tessellation $T$, hence it is discrete and $\Delta$ is a fundamental domain for $\Gamma$. Consider the index-two subgroup $\Gamma = \Gamma^{\rm or}(p_1,p_2,p_3) \triangleleft \Gamma(p_1,p_2,p_3)$ consisting of orientation-preserving isometries. Taking the quotient of the model space by $\Gamma$ we obtain an orbifold $(S,g)$, which is a sphere with exactly $3$ cone points of orders $p_1$, $p_2$ and $p_3$. $(S,g)$ is hyperbolic, or flat, according to whether $\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3}$ is smaller than $1$, or equal to $1$, respectively.

Assume that $\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} < 1$. Recall that the geometry of $\psl$ can be constructed in the following way: consider the unit tangent bundle $U\hp$ of $\hp$ endowed with the {\it Sasaki metric}, and then take the universal cover of $U\hp$ endowed with the pullback metric.

For each isometry $f$ of $\hp$, the differential $f_{*}$ is an isometry of $U\hp$, and therefore using the covering $\psl \to U\hp$, $f_{*}$ lifts to an isometry of $\psl$. So, there is a group of isometries of $\psl$, denoted by $\tilde{\Gamma}$, which is an extension of $\Gamma$ by the central $\z$ subgroup of $\iso(\psl)$: we have the exact sequence
$$0 \to \z \to \tilde{\Gamma} \to \Gamma \to 0.$$
We will verify that the action of $\tilde\Gamma$ has no fixed point. It suffices to check this to $f_{*}$, where $f$ is any isometry of $\hp$. We have
$$f_{*}(x,v) = (x,v) \ \textrm{if, and only if,} \ f(x) = x \ \textrm{and} \ (f_{*})_{x}(v) = v.$$
Since $f \in \Gamma$, $f$ is either conjugate to a rotation or to an hyperbolic translation. On this last case there is no fixed point and on the former case an calculation shows that $f_{*}$ has no fixed points. Now is easy to prove that $M = \psl/\tilde{\Gamma}$ is a closed Seifert Fibered Space projecting on $(S,g)$, and a calculation shows that $e(M) \neq 0$ and $M$ has $3$ singular fibers whose fibered torus have coefficients $(p_i,1)$, $i=1,2,3$. Geometrically we can see $M$ as the unit tangent bundle of the orbifold $(S,g)$. In general, it is possible to construct other circle bundles over $(S,g)$ where $e \neq 0$ and at least one of the singular fibers has coefficients $(p_i,q_i)$ satisfying $q_i \neq 1$. Similar constructions in the case $\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} = 1$ produces quotients of $\nil$.

\section{The counterexample}

The following lemma describes the stable minimal surfaces in Seifert manifolds, and is crucial to our work.

\begin{lem}[Pitts, Rubinstein, \cite{P.R}]\label{PR}
Let $\si$ be an closed two-sided stable minimal surface in a Seifert Fibered Space endowed with a geometric metric. Then $\si$ is vertical or horizontal.
\end{lem}

\begin{proof}
The properties of being {\it vertical} or everywhere transverse to the fibers are local, moreover, the lift of a two-sided stable minimal surface under a covering is still two-sided and stable, so we can pass to a covering if necessary and suppose that $M$ is a $\mathbb{S}^1$-bundle over a closed surface (any closed Seifert manifold admits such a covering). In this case, we have a well-defined $SO(2)$-action on the fibers by isometries of the metric. Let $\xi$ be the vector field associated to the action on the fibers and let $N_{\si}$ be a unit normal to $\si$. Suppose $\si$ is neither vertical nor is everywhere transverse to the fibers. Then, the function $\phi = \langle N_{\si},\xi\rangle$ is not identically zero, but there is a least one point $x \in \si$ such that $\phi(x) = 0$.

On the other hand, since the action on the fibers is by isometries we have
\begin{equation}\label{jacobi}
L(\phi) = 0,
\end{equation}
where $L$ is the Jacobi operator of $\si$. Since $\si$ is stable and equation \eqref{jacobi} is satisfied, by standard elliptic theory we have that $\phi$ is everywhere positive or negative, or is identically zero. But this contradicts the last line of the previous paragraph. Therefore $\si$ is \emph{vertical} or \emph{horizontal}.
\end{proof}

\begin{prop}\label{geod}
Let $(S,g)$ be a two-sphere endowed with a flat or hyperbolic metric $g$ with three cone points. Then $(S,g)$ does not contain a simple closed geodesic inside its regular part.
\end{prop}

\begin{proof}
Suppose there is such a geodesic, and denote it by $\gamma$. Consider disks $D_i$ around the cone points $p_i$, $i = 1, 2, 3$, such that $\gamma \subset \tilde{S} = S\backslash(\cup D_i)$. We can choose $D_i$ as the quotient of a geodesic disc $\tilde{D}_i$ in $\rp$ or $\hp$ by a rotation of angle $2\pi/q_i$ (for some $q_i \in \mathbb{Z}$), where the center of this disc projects on the cone point. So, the boundary $\rd D_i$ is the projection of $\rd\tilde{D}_i$. By this construction the boundary components of $\tilde{S}$ have negative geodesic curvature with respect to the unit normal vector pointing into $\tilde{S}$.

Since the underlying surface is a sphere, the curve $\gamma$ must separate $S$. If $\gamma$ were the boundary of a disc $U$ on $\tilde{S}$, then by the Gauss-Bonnet formula and the hypothesis on the curvature of the orbifold we would have
\begin{equation}
2\pi = \int_{U} K_{S} \ dA + \int_{\gamma} \kappa_{g} \ dL = \left\{
\begin{array}{rl}
0, & \text{if } S \ \textrm{is flat},\\\\
-|U|, & \text{if } S \ \textrm{is hyperbolic};
\end{array} \right.
\end{equation}
which is a contradiction.

Thus, $\gamma$ must be homotopic to one of the boundary components of $\tilde{S}$. We denote this component by $\tilde{\gamma}$. Consider the annulus $\mathcal{A}$ with boundary components $\gamma$ and $\tilde{\gamma}$. Since $\tilde\gamma$ has negative geodesic curvature we obtain
\begin{equation}
0 = \int_{\mathcal{A}} K_{S} \ dA + \int_{\gamma} \kappa_{g} + \int_{\tilde\gamma} \kappa_{g} \ dL < 0,
\end{equation}
which is again a contradiction.
\end{proof}

Let us recall some definitions concerning $3$-manifolds.
\begin{defin}
$1)$ We say that a $3$-manifold $M$ is irreducible, if every $2$-sphere embedded in $M$ bounds an embedded $3$-ball in $M$. $M$ is $P^2$-irreducible if it is irreducible and contains no embedded two-sided projective plane.\\\\
$2)$ We say that a closed oriented $3$-manifold $M$ is Haken, if is irreducible and it contains an embedded two-sided incompressible surface $\si$ (i.e, the map $\pi_1(\si) \to \pi_1(M)$ induced by the embedding is one-to-one) of genus $g \geq 1$. Otherwise, we say $M$ is non-Haken.
\end{defin}

\begin{thm}\label{coex}
Let $(M,h)$ be a closed, irreducible, non-Haken Seifert Fibered Space with infinite fundamental group, endowed with a geometric metric $h$. Then $(M,h)$ does not contain any closed, embedded, minimal surface whose two-sided cover is stable.
\end{thm}

\begin{proof}
It is well-known that a closed, irreducible, non-Haken Seifert Fibered Space $M$ with infinite fundamental group has $e(M) \neq 0$ and the base orbifold is a sphere with three cone points (see for example proposition $2$ in \cite{H}). The possible geometric structures for $M$ are $\nil$ and $\psl$, and for the geometric metrics on $M$, the base has a flat or hyperbolic metric. So, $M$ belong to the class of manifolds discussed on the main example \ref{main}.

Suppose there is $\si \subset M$, a closed embedded minimal surface whose two-sided covering is stable.  On the case $\si$ is two-sided, by lemma \ref{PR}, $\si$ is either \emph{vertical} or \emph{horizontal}. If $\si$ is \emph{vertical} then its projection $\gamma$ on the base surface is a simple closed curve inside the regular part. Since the metric is geometric, the following calculation shows that $\gamma$ is a geodesic: \\
\noindent
$\bullet$ Let $T$ and $\eta$ be respectively a unit tangent field and a unit normal field to $\gamma$. Given a point $x \in \si$ consider an small neighbourhood $U$ of $\Pi(x)$ in $S$, such that the projection $\Pi: \Pi^{-1}(U) \to U$ is a Riemannian submersion, and let $\widetilde{T}$ and $\widetilde{\eta}$ be the corresponding horizontal lifts to $\Pi^{-1}(U)$ of $T$ and $\eta$. Then $\{\widetilde{T},\xi\}$ is an orthonormal basis on $T_{x}\si$ and $\widetilde{\eta}$ a unit normal to $\si$, where $\xi$ is the Killing vertical vector field of $\Pi^{-1}(U)$. Also, denote by $\bar{\nabla}$ and $\nabla$ the connections on $M$ and $S$ respectively. The mean curvature of $\si$ at a $x$ is then calculated as
\begin{align*}
0&=-\Big\langle\bar{\nabla}_{\widetilde{T}}\widetilde{\eta},\widetilde{T}\Big\rangle-\Big\langle \bar{\nabla}_{\xi}\widetilde{\eta},\xi\Big\rangle\\
&=\Big\langle \widetilde{\nabla_TT},\widetilde{\eta}\Big\rangle=\Big\langle \nabla_TT,\eta\Big\rangle\\
&=\rm{the\:geodesic\;curvature\;of\;\gamma\;at\;the\;point\;\Pi(x)}.
\end{align*}

However, the previous proposition shows that $S$ does not admit simple closed geodesics in its regular part. On the other hand, as we mentioned in the last section closed manifolds with the geometries of $\nil$ and $\psl$, do not contain \emph{horizontal} surfaces. So, in any case we obtain a contradiction.

Finally, if $\si$ is one-sided we can pass to a double covering $\tilde{M}$ of $M$ so that the lift $\tilde\si$ of $\si$ is a connected closed two-sided minimal surface (see proposition $3.7$ in \cite{Z}), which is stable by the hypothesis. The manifold $\tilde{M}$ is also a Seifert Fibered space, endowed with the geometric metric $\tilde{h} = \Pi^{*}h$, so again by lemma \ref{PR}, $\tilde\si$ is either \emph{vertical} or \emph{horizontal}, thus this also holds for $\si$, and the contradiction follows as before.

Therefore there is no such $\si$.
\end{proof}

\section{Further discussion}\label{last}

\begin{thm}
Let $(M,g)$ be a closed Riemannian $3$-manifold. If $(M,g)$ does not admit any closed immersed minimal surface whose two-sided covering is stable, then $M$ is finitely covered by the $3$-sphere.
\end{thm}

\begin{proof}
Suppose $M$ is not a quotient of the $3$-sphere, we will prove that $(M,g)$ contains a two-sided, immersed, stable minimal surface, for any Riemannian metric $g$ on $M$. Under the assumption, it follows from the work of Perelman that $M$ has infinite fundamental group, thus there are four possibilities:\\
\noindent
{\bf 1) $M$ is not irreducible}: In this case, there is an embedded sphere $S \subset M$, which represents a non zero element of $\pi_{2}(M)$. So, combining the results in \cite{S.U1} and \cite{M.Y}, there is an immersed stable minimal sphere $\si$ on $M$, and either $\si$ is embedded or double covers an embedded projective plane.\\
\noindent
{\bf 2) $M$ is orientable, irreducible and Haken}: Combining the results in \cite{S.Y,S.U2} with that of \cite{F.H.S}, M admits an incompressible, orientable, stable minimal surface of genus $g \geq 1$, moreover this surface is embedded or double-covers an embedded non-orientable surface (We could exclude this case, since the arguments of the next case also work if $M$ is Haken. However it is relevant to the question of when the surface obtained is embedded.) \\
\noindent
{\bf 3) $M$ is orientable, irreducible and non-Haken}: We have two sub-cases. If $\pi_1(M)$ contains $\mathbb{Z}\oplus\mathbb{Z}$ as a subgroup, then there an immersion of the 2-torus in $M$. Thus it follows from the results in \cite{S.Y,S.U2} that $M$ contains an immersed stable minimal torus; if $\pi_1(M)$ does not contain any subgroup isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$, then by the work of Perelman, $M$ admits a hyperbolic metric. It follows from the work of Kahn-Markovic \cite{K.M} that $\pi_1(M)$ contain a subgroup isomorphic to the fundamental group of a orientable closed surface of genus $g \geq 2$. So using again \cite{S.Y,S.U2} we conclude that $M$ contains an immersed stable minimal surface of genus $g \geq 2$.\\
\noindent
{\bf 4) $M$ is non-orientable and irreducible}: If $M$ contains an embedded $2$-sided projective plane $P$, then by proposition $2.3$ \cite{BBEN}, $M$ contains a embedded, $2$-sided, stable minimal surface homeomorphic to $P$. If $M$ is $P^{2}$-irreducible, by lemmas $6.6$ and $6.7$ in \cite{He}, $M$ contains a embedded, $2$-sided nonseparating incompressible surface $S$. Using the theorems $3.1$ and $5.1$ of \cite{F.H.S} we obtain a stable minimal surface $\si$ homotopic to $S$, moreover, $\si$ is embedded or double covers an embedded one-sided surface.
\end{proof}

Analysing the previous proof, we see that if $M$ is either non-orientable, or not irreducible, or irreducible and Haken, then it contains an embedded minimal surface whose two-sided covering is stable. If $M$ is irreducible, non-Haken and $\pi_1(M)$ contains $\mathbb{Z}\oplus\mathbb{Z}$ as a subgroup, it follows from the {\it Seifert Fibered Space theorem} (see section $2.3.2$ in \cite{Pr}) that $M$ is a non-Haken Seifert Fibered Space with infinite $\pi_1$, and as we saw in the theorem \ref{coex}, these spaces admit a metric which do not contain the desired surfaces. So the only case remaining to study is that when $M$ is a hyperbolic manifold. There are examples of non-Haken hyperbolic manifolds which does not contain any non-orientable embedded surface, so in this case the surfaces produced by the last theorem are not embedded and do not cover any non-orientable surface.\\

\noindent
{\bf Question}: Let $M$ be a non-Haken closed hyperbolic $3$-manifold. Let $g$ be a Riemannian metric on $M$. Does $(M,g)$ admit a closed, embedded, minimal surface whose two-sided covering is stable?\\

It is important to highlight that this question is open even when the metric $g$ is the hyperbolic metric.

config: arxiv


TITLE: "psi-epistemic theories" in 3 or more dimensions
QUESTION [31 upvotes]: In their recent paper The Quantum State Can Be Interpreted Statistically, Lewis et al. end with a very nice mathematical question, one whose answer (either way) would have interesting implications for the foundations of quantum mechanics.  In the hopes of getting some non-quantum math folks interested in their question---and maybe even finding someone to say "the answer is trivial for the following reason..." :-)---I decided to state the question for the MO community, shorn of all the physics and philosophy.
Let Hd be the set of unit vectors in $\mathbb{C}^d$.  A ψ-epistemic theory in d dimensions consists of the following:

A measurable space Λ (called the "space of ontic states").
A function mapping each unit vector ψ∈Hd to a probability measure Dψ over Λ.
A function f(λ,M,i)∈[0,1], which takes as input an ontic state λ∈Λ, an ordered orthonormal basis M=(v1,...,vd) for $\mathbb{C}^d$, and an index i∈{1,...,d}.

f must satisfy the following two conditions:
(i) $\sum_{i=1}^{d}f(\lambda,M,i)=1$ for all λ and M.  (Intuitively, f must give rise to a probability distribution over the "measurement outcomes" v1,...,vd in M.)
(ii) $\int_{\lambda \sim D_{\psi}} f(\lambda,M,i) d\lambda = |v_{i}^{*}\psi|^{2}$ for all ψ, M, and i.  (Intuitively, the probability of the measurement outcome vi, averaged over all λ drawn from Dψ, must equal the squared projection of ψ onto vi.)
Note that we can trivially satisfy conditions (i) and (ii) as follows:

Λ=Hd
Dψ assigns probability 1 to λ=ψ, and probability 0 to all other states in Λ
f(ψ,M,i) = |vi*ψ|2

Thus, let Supp(D)⊆Λ be the support of D, and call a ψ-epistemic theory nontrivial if there exist ψ≠ϕ such that $Supp(D_{\psi})\cap Supp(D_{\phi}) \ne \emptyset$.
Observe that, if ψ and ϕ are orthogonal, then Supp(Dψ) and Supp(Dϕ) must be disjoint.  This is because, if we set v1=ψ and v2=ϕ, then $v_{1}^{*}\psi = v_{2}^{*}\phi = 1$ and $v_{1}^{*}\phi = v_{2}^{*}\psi = 0$, which is not possible if Dψ and Dϕ have any nonzero overlap.  Motivated by this observation, call a theory maximally nontrivial if $Supp(D_{\psi})\cap Supp(D_{\phi}) \ne \emptyset$ whenever ψ and ϕ are not orthogonal.
I can now state Lewis et al.'s open problem:
Does there exist a maximally-nontrivial ψ-epistemic theory in dimensions d≥3?
Update: See the comments for an extremely nice solution by George Lowther, plus my followup questions.
I know of two results directly relevant to this problem.
First, there exists a maximally-nontrivial theory in dimension d=2, which was found by Kochen and Specker in 1967.  See this paper by Rudolph for more details, including why the obvious generalizations to 3 or more dimensions seem to fail.  Briefly, the Kochen-Specker theory is defined as follows:

Λ=H2.
Dψ assigns probability measure $2 | \psi^{\*} \phi|^{2} - 1$ to ϕ if $| \psi^{\*} \phi|^{2} \geq 1/2$, and probability measure 0 to ϕ otherwise.
f(ψ,M,i) = 1 if $|v_{i}^{\*} \psi|^{2} \geq 1/2$, and f(ψ,M,i) = 0 otherwise.

(Warning: I converted from a different representation, and can't promise I didn't get a factor of 2 wrong or something like that.)
The second result is that, for all finite d, there exists a nontrivial ψ-epistemic theory (though it's far from being maximally nontrivial).  This is the main result of Lewis et al.
My own guess is that maximally-nontrivial theories don't exist for d≥3, but I'd only give it 60% confidence.
To anticipate some questions:

Yes, I'd also be interested in this problem with $\mathbb{R}$ in place of $\mathbb{C}$ (though I suspect the two cases are pretty similar).
Yes, I'd be interested in negative results for restricted classes of theories.  Here are a few examples of restrictions one could look at, in various combinations: Λ=Hd, f∈{0,1}, f is continuous, symmetry under unitary transformations, symmetry under relabeling of the vi's.
No, I don't know how to rule out that the answer could depend on the Axiom of Choice or something crazy like that (but I doubt it).

Update (March 20, 2013): Adam Bouland, Lynn Chua, George Lowther, and myself now have a paper on ψ-epistemic theories originating with this MO post.  The paper contains the construction below, but also proves impossibility results for ψ-epistemic theories when an additional symmetry condition is imposed.

REPLY [12 votes]: Since George Lowther seems to have a lot of late nights, I decided to express my gratitude to him by writing up his lovely answer myself and thereby saving him the trouble.
The answer to my (and Lewis et al.'s) question is that yes, maximally-nontrivial ψ-epistemic theories do exist for every finite dimension $d$.
The first realization is that we can "mix" small ε-balls around any two non-orthogonal vectors.
Lemma 1: Given any two non-orthogonal unit vectors $\psi,\phi \in \mathbb{C}^d$, there exists a ψ-epistemic theory $T = T(\psi,\phi)$ such that $Supp(D_{\psi})$ and $Supp(D_{\phi})$ have nonempty intersection.  Moreover, for this $T$, there exists an $\epsilon \gt 0$ (for example, $\epsilon = |\psi^{*} \phi|/2d$) such that $Supp(D_{\psi'})$ and $Supp(D_{\phi'})$ have nonempty intersection for all $\psi',\phi'$ such that $||\psi - \psi'||,||\phi - \phi'||\lt \epsilon$.
Proof: Our ontic state space will be $\Lambda = H_{d} \times [0,1]$.  Given an orthonormal basis $M=(v_1,...,v_d)$, first sort the $v_i$'s in decreasing order of $min(|v_{i}^{∗}\psi|,|v_{i}^{∗}\phi|)$. Then the outcome of measurement M on ontic state $(w,p)\in \Lambda$ will equal the smallest positive integer $i$ such that
$|v_1^* w|^2+...+|v_{i−1}^*w|^2 \le p \le |v_1^* w|^2+...+|v_i^*w|^2$.
In other words, $f((w,p),M,i)$ will equal $1$ if $i$ satisfies the above and no $j \lt i$ does, and $0$ otherwise.  For now, assume that $D_{w}$ is simply the uniform distribution over all states $(w,p)$ with $p \in [0,1]$.  It is easy to check that this yields a valid ψ-epistemic theory, albeit so far a trivial one.
Since $|\psi^{*} \phi|\gt 0$, I claim that there exists an $\epsilon \gt 0$ such that for all orthonormal bases $M=(v_1,...,v_d)$, there exists an $i$ such that $|v_{i}^{∗}\psi|\ge \epsilon$ and $|v_{i}^{∗}\phi|\ge \epsilon$.  Indeed, by the triangle inequality, setting $\epsilon := |\psi^{*} \phi| / d$ will work.
Now, the above means that, for all measurements $M$ and all $p \in [0,\epsilon]$, the outcome is always $i=1$ when $M$ is applied to either of the ontic states $(\psi,p)$ or $(\phi,p)$.  Following Lewis et al., this implies that we can "mix" the corresponding distributions $D_{\psi}$ and $D_{\phi}$―i.e., have them intersect each other in the region $p \in [0,\epsilon]$―without affecting any outcome of any measurement $M$.
Furthermore, suppose $\psi'$ and $\phi'$ are $\epsilon /2$-close to $\psi$ and $\phi$ respectively, in some standard metric such as trace distance.  Then by continuity, we can similarly mix the distributions $D_{\psi'}$ and $D_{\phi'}$―i.e., have them intersect each other in the region $p\in [0,\epsilon /2]$―without affecting any measurement outcome.   (One subtlety is that, as we vary $M$, the sorting procedure can make $v_1$ "jump" discontinuously from one basis vector of $M$ to another.  However, this jumping is not a problem, since it depends only on the fixed vectors $\psi$ and $\phi$, not on $\psi'$ or $\phi'$.  So it happens the same way everywhere in the $\epsilon /2$-balls.) QED
The second realization is that we can take "convex combinations" of ψ-epistemic theories.  Given two ψ-epistemic theories $T=(\Lambda,D,f)$ and $T'=(\Lambda',D',f')$ (where $D,D'$ are the functions that map vectors $\psi \in H^d$ onto ontic distributions), and a constant $c \in (0,1)$, define the new theory $c T + (1-c)T' =(\Lambda_c,D_c,f_c)$ as follows:

$\Lambda_c := \Lambda \cup \Lambda'$.
$D_c := c D + (1-c) D'$.
$f_c : \Lambda_c \rightarrow [0,1]$ equals $f$ on $\Lambda$ and $f'$ on $\Lambda'$.

Lemma 2: $c T + (1-c)T'$ is a ψ-epistemic theory.  Furthermore, if $T$ mixes the ontic distributions of two vectors $\psi$ and $\phi$, and $T'$ mixes the ontic distributions of two other vectors $\psi'$ and $\phi'$, then $c T + (1-c)T'$ mixes both pairs of distributions.
Proof: Immediate.
Using Lemmas 1 and 2, we now construct a maximally-nontrivial ψ-epistemic theory.  Let $T(\psi,\phi)$ be the theory returned by Lemma 1 given vectors $\psi,\phi\in H^d$.  Also, for all positive integers $n$, let $A_n$ be a $1/n$-net for $H^d$: that is, a finite subset $A_n \subseteq H^d$ such that for all $v\in H^d$, there exists a $w \in A_n$ satisfying $|| w - v || \lt 1/n$.  By making small perturbations, we can easily ensure the property that $u^{*}v \neq 0$ for all $u,v\in A_n$.  Then our theory $T$ is defined as follows:
$$T = \frac{6}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2} \left( \frac{1}{|A_n|^2} \sum_{u,v \in A_n} T(u,v) \right). $$
One can check that $T$ mixes the distributions $D_{\psi}$ and $D_{\phi}$ for all non-orthogonal $\psi$ and $\phi$.
Comment: As often in math, I'd say the true value of knowing this answer is that it points us toward the questions we (or rather Lewis et al.) "really" meant to ask!  In the above construction, the overlap between $D_{\psi}$ and $D_{\phi}$ is indeed nonzero for any non-orthogonal $\psi,\phi$, but the amount of overlap falls off (by my crude estimate) as $ ( |\psi^{*} \phi| / d) ^{ \Theta(d) } $ .
A skeptic of ψ-epistemic theories might argue that for large $d$ (and of course, $d$ can be huge in quantum mechanics), such an overlap is physically irrelevant.  So one obvious followup question is how large the overlap can be―for example, whether it can fall off only as $(|\psi^{*} \phi| / d)^{O(1)}$.  I'd better stop here, though, since I know MO is not for open-ended research discussions.  The question, as I stated it, has been answered.


TITLE: How many of all cube's edges 3-colorings have exactly 4 edges for each color?
QUESTION [5 upvotes]: I've found the number of different cube's edges  colorings with three colors available.
 (We say that the two colorings are the same if one can obtain a second by turning cube and permuting colors)
My resualts credit are given to the following solution :
Edge coloring of the cube .
[solution summery : First of all define a group G, a set X and group action of G on X. We want to find the number of different cube edges coloring while we know that if we can obtain some coloring by turning another one – it is the same coloring. So we actually need to find the number of orbits of X (the orbit of an element x in X is the set of elements of X to which x can be moved by the elements of G) and then use Burnside's lemma . ]
So,  Now I want calculate how many of the 3-colorings I found have exactly 4 edges for each color ?

this question asked in Combinatorics course as a bonus question.

Thanks!
Bar

REPLY [5 votes]: First we construct the generating function of the edge colorings before the symmetric group acts on the colors. This requires the cycle index of the edge permutation group $E$ of the cube. We enumerate the permutations in this group by their cycle structure.
First, there is the identity, which contributes $$a_1^{12}.$$
There are rotations about an axis (three of them) passing through the center of two opposite faces, which gives $$3\times (a_2^6 + 2 a_4^3).$$
Rotations about a diagonal passing through opposite vertices give
$$4\times 2 a_3^4.$$
Finally there are 180 degree rotations about an axis passing through the centers of opposite edges, which gives $$6\times a_1^2 a_2^5.$$
This gives the cycle index
$$Z(E) = \frac{1}{24}
\left(a_1^{12} + 3 a_2^6 + 6 a_4^3 + 8 a_3^4 + 6 a_1^2 a_2^5\right).$$
Substituting into $Z(E)$ we obtain the following formula for the number of $N$-colorings of the edges of a cube the value
$$\frac{1}{24}(N^{12} + 3 N^6 + 6 N^3 + 8 N^4 + 6 N^7).$$
This gives the following sequence
$$1, 218, 22815, 703760, 10194250, 90775566, 576941778, 2863870080, 11769161895,\ldots$$
which is OEIS A060530.
The substituted cycle index for 3-colorings as asked for by the OP is
$$Z(E)(R+G+B)=
1/24\, \left( R+G+B \right) ^{12}+1/8\, \left( {R}^{2}+{G}^{2}+{B}^{2}
 \right) ^{6}\\+1/4\, \left( {R}^{4}+{G}^{4}+{B}^{4} \right) ^{3}+1/3\, \left(
{R}^{3}+{G}^{3}+{B}^{3} \right) ^{4}+1/4\, \left( R+G+B \right) ^{2} \left( {
R}^{2}+{G}^{2}+{B}^{2} \right) ^{5}$$
which yields
$$[R^4 G^4 B^4]Z(E)(R+G+B)=1479.$$
Addendum as of Tue Dec 17 22:50:41 CET 2013. The OP also mentioned in his query that as an extra twist we may consider edge colorings of the cube where the symmetric group acts on the colors, so that two colorings are not only equivalent when there is a rotation that transforms one into the other, but also a rotation combined with a permutation of the colors from the symmetric group. The technique for this kind of problem is called Power Group Enumeration Theorem.
There are at least two possible approaches here. There is the classic approach from the book Graphical Enumeration by Frank Harary and Edgar Palmer, a landmark text that continues to be relevant today. They have a chapter on this namely chapter six, section four "Power Group Enumeration / Graphs with colored lines." I will not typeset their formulas here as I would not add anything new and the text is available at any major library. Let me just say that they treat the following scenario: enumerate under the power group the colorings of various graphs, where an edge maybe on or off and has $N$ possible colors in the on state. I implemented their formula in Maple and the result was the following sequence for edge colorings of the cube under the symmetric group acting on the colors using at most $N$ colors:
$$1, 114, 3891, 29854, 87981, 143797, 170335, 177160, 178153, 178243,\ldots$$
This is the Maple code for the Harary/Palmer computation.

with(numtheory);
with(group):
with(combinat):

pet_cycleind_symm :=
proc(n)
        option remember;

        if n=0 then return 1; fi;

        expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;

pet_varinto_power :=
proc(N, ZA)
           local beta, res, indvars, v, t, d, m, pot, ZB;

           ZB := pet_cycleind_symm(N);
           if N=1 then ZB := [ZB]; fi;

           indvars := indets(ZA);

           res := 0;
           for beta in ZB do
               t := [];
               for v in indvars do
                   pot := op(1, v);

                   m := 0;
                   for d in divisors(pot) do
                       m := m +
                       d*degree(beta, op(0,v)[d]);
                   od;

                   t := [op(t), v=1+m*z^pot];
               od;

               res := res+lcoeff(beta)*subs(t, ZA);
           od;

           res;
end;

cube_edge_ind :=
1/24*(a[1]^12+3*a[2]^6+6*a[4]^3+8*a[3]^4+6*a[1]^2*a[2]^5);

vgf :=
proc(N)
         option remember;
         local gf;

         gf := pet_varinto_power(N, cube_edge_ind);
         expand(gf);
end;

v := N -> coeff(vgf(N), z, 12);

The second approach uses a different Power Group Enumeration (PGET) formula as presented by Harald Fripertinger in his brilliant paper "Enumeration in Musical Theory" from 1993. Again I will not typeset the formulas here as I would not be adding anything new. The clinch is that this paper presents a so-called polynomial form of the PGET (in the first chapter) which makes it possible not only to calculate the total number of colorings but also to classify them according to the distribution of colors used. I have implemented this in Maple and I obtain for the sequence of colorings once more the values (that is as it ought to be):
$$1, 114, 3891, 29854, 87981, 143797, 170335,\ldots$$
Now, however, we also get the distributions of colors. For two colors we have
$$29\,{\it Q6\_Q6}+{\it Q12}+5\,{\it Q2\_Q10}+27\,{\it Q4\_Q8}+{\it Q1\_Q11}+13\,{\it
Q3\_Q9}+38\,{\it Q5\_Q7}  $$
and for three colors
$$ {\it Q12}+38\,{\it Q5\_Q7}+27\,{\it Q4\_Q8}+{\it Q1\_Q11}+5\,{\it Q2\_Q10}+13\,{\it
Q3\_Q9}+370\,{\it Q2\_Q5\_Q5}\\+29\,{\it Q6\_Q6}+600\,{\it Q2\_Q4\_Q6}+236\,{\it
Q1\_Q5\_Q6}+85\,{\it Q1\_Q3\_Q8}+77\,{\it Q2\_Q2\_Q8}\\+30\,{\it Q1\_Q2\_Q9}+340\,{\it
Q2\_Q3\_Q7}+5\,{\it Q1\_Q1\_Q10}+170\,{\it Q1\_Q4\_Q7}\\+412\,{\it Q3\_Q3\_Q6}+282\,{\it
Q4\_Q4\_Q4}+1170\,{\it Q3\_Q4\_Q5}$$
In particular we are now in a position to answer the query that prompted the post which is how many colorings there are using each of the three colors four times and the answer is
$$282$$
Here are the colorings using at most four colors:
$${\it Q12}+38\,{\it Q5\_Q7}+27\,{\it Q4\_Q8}+{\it Q1\_Q11}+5\,{\it Q2\_Q10}+13\,{\it
Q3\_Q9}+370\,{\it Q2\_Q5\_Q5}\\+698\,{\it Q3\_Q3\_Q3\_Q3}+29\,{\it Q6\_Q6}+3480\,{\it
Q1\_Q2\_Q4\_Q5}+3510\,{\it Q2\_Q2\_Q3\_Q5}\\+600\,{\it Q2\_Q4\_Q6}+2330\,{\it
Q1\_Q3\_Q3\_Q5}+2915\,{\it Q1\_Q3\_Q4\_Q4}+600\,{\it Q1\_Q1\_Q4\_Q6}\\+510\,{\it
Q1\_Q2\_Q2\_Q7}+2266\,{\it Q2\_Q2\_Q4\_Q4}+236\,{\it Q1\_Q5\_Q6}+85\,{\it Q1\_Q3\_Q8}\\+
77\,{\it Q2\_Q2\_Q8}+30\,{\it Q1\_Q2\_Q9}+2320\,{\it Q1\_Q2\_Q3\_Q6}+340\,{\it
Q1\_Q1\_Q3\_Q7}\\+13\,{\it Q1\_Q1\_Q1\_Q9}+626\,{\it Q2\_Q2\_Q2\_Q6}+340\,{\it Q2\_Q3\_Q7
}+5\,{\it Q1\_Q1\_Q10}\\+5850\,{\it Q2\_Q3\_Q3\_Q4}+135\,{\it Q1\_Q1\_Q2\_Q8}+170\,{\it
Q1\_Q4\_Q7}+412\,{\it Q3\_Q3\_Q6}\\+282\,{\it Q4\_Q4\_Q4}+370\,{\it Q1\_Q1\_Q5\_Q5}+1170
\,{\it Q3\_Q4\_Q5} $$
The code for the Fripertinger computation follows.

with(numtheory);
with(group):
with(combinat):

pet_disjcyc :=
proc(p)
        local dc, pos;

        dc := convert(p, 'disjcyc');

        for pos to nops(p) do
            if p[pos] = pos then
                dc := [op(dc), [pos]];
            fi;
        od;

        dc;
end;

pet_varinto_power :=
proc(N, ZA)
        local indvars, delta, v, pot, s, t, poly, k, degs,
                res, cyc;

        indvars := indets(ZA);

        res := 0;
        for delta in permute(N) do
            t := [];
            for v in indvars do
                pot := op(1, v);

                s := 0;
                for cyc in pet_disjcyc(delta) do
                    if pot mod nops(cyc) = 0 then
                        poly :=
                        mul(cat(`X`, cyc[k])^(pot/nops(cyc)),
                            k=1..nops(cyc));
                        s := s + nops(cyc)*poly;
                    fi;
                od;

                t := [op(t), v=s];
            od;

            res := res+subs(t, ZA);
        od;

        degs :=
        proc(t) local q;
            local vs, v, dl, d, sym;

            vs := indets(t);
            dl := [];

            for v in vs do
                dl := [op(dl), degree(t, v)];
            od;

            sym := [];
            for d in sort(dl) do
                if nops(sym) = 0 then
                    sym := cat(`Q`, d);
                else
                    sym := cat(sym, `_`, `Q`, d);
                fi;
            od;

            return lcoeff(t)*sym;
        end;

        if N=1 then
            return cat(`Q`, degree(res, `X1`));
        fi;

        map(degs, expand(res))/N!;
end;

cube_edge_ind :=
1/24*(a[1]^12+3*a[2]^6+6*a[4]^3+8*a[3]^4+6*a[1]^2*a[2]^5);

v :=
proc(N)
        option remember;
        local dist;

        dist := pet_varinto_power(N, cube_edge_ind);
        map(lcoeff, dist);
end;

Thanks go to the OP for asking such a fascinating question. Here is a link to a chain of Polya Counting Computations.
Addendum as of Sat Dec 21 03:20:27 CET 2013. The above algorithm can also be used on generic graphs rather than just cubes. With two colors, we get the sequence
$$1, 2, 6, 18, 78, 522, 6178, 137352, 6002584, 509498932, 82545586656, 25251015686776,\ldots$$
which is OEIS A007869.
Three interchangeable colors produces the sequence
$$1, 3, 15, 142, 4300, 384199, 98654374, 70130880569, \\136638863494089, 730439999032117301,\ldots$$
which does not yet have an OEIS entry.
With four interchangeable colors we get
$$1, 3, 22, 513, 67685, 37205801, 74992370359, 543437207831908, 14224090440652751128,\ldots $$
The computation of the relevant cycle indices is show at this MSE link. This MSE link II shows Power Group Enumeration with the flip group acting on the slots.
Addendum Dec  13 2016. The reader  may be interested to  know that
these generating functions can also  be computed from first principles
namely  the  Burnside  lemma  as   was  done  at  the  following  MSE
link. The algorithm
is  documented  there and  in  the  linked-to  pages. I  present  this
algorithm here for reference. The output  up to sort order is the same
as from the Fripertinger computation.

with(combinat);

pet_flatten_term :=
proc(varp)
        local terml, d, cf, v;

        terml := [];

        cf := varp;
        for v in indets(varp) do
            d := degree(varp, v);
            terml := [op(terml), seq(v, k=1..d)];
            cf := cf/v^d;
        od;

        [cf, terml];
end;

pet_autom2cyclesA :=
proc(src, aut)
local numa, numsubs;
local marks, pos, data, item, cpos, clen;

    numsubs := [seq(src[k]=k, k=1..nops(src))];
    numa := subs(numsubs, aut);

    marks := Array([seq(true, pos=1..nops(aut))]);

    pos := 1; data := table();

    while pos <= nops(aut) do
        if marks[pos] then
            clen := 0; item := []; cpos := pos;

            while marks[cpos] do
                marks[cpos] := false;
                item := [op(item), aut[cpos]];

                clen := clen+1;
                cpos := numa[cpos];
            od;

            if type(data[clen], 'list') then
                data[clen] :=
                [op(data[clen]), item];
            else
                data[clen] := [item];
            fi;
        fi;

        pos := pos+1;
    od;

    return data;
end;

pet_flatten_automcycs :=
proc(tbl)
local res, sz, cyc;
    res := [];

    for sz in [indices(tbl, 'nolist')] do
        for cyc in tbl[sz] do
            res := [op(res), cyc];
        od;
    od;

    res;
end;

cube_edge_cind :=
1/24*(a[1]^12 + 8*a[3]^4 + 6*a[4]^3 + 3*a[2]^6 + 6*a[1]^2*a[2]^5);

cube_edge_colorings :=
proc(n)
option remember;
local idx_colors, res, a, b,
    flat_a, flat_b, cyc_a, cyc_b, len_a, len_b, p, q,
    src, perm, acycs, gf, msetgf, term, deg, rep;

    res := 0; src := [seq(el, el=1..n)];

    for a in cube_edge_cind do
        flat_a := pet_flatten_term(a);

        perm := firstperm(n);
        while type(perm, `list`) do
            acycs := pet_autom2cyclesA(src, perm);
            flat_b := pet_flatten_automcycs(acycs);

            p := 1;
            for cyc_a in flat_a[2] do
                len_a := op(1, cyc_a);
                q := 0;

                for cyc_b in flat_b do
                    len_b := nops(cyc_b);

                    if len_a mod len_b = 0 then
                        q := q +
                        len_b*mul(Q[el], el in cyc_b)
                        ^(len_a/len_b);
                    fi;
                od;

                p := p*q;
            od;

            res := res + p*flat_a[1];

            perm := nextperm(perm);
        od;
    od;

    gf := expand(res/n!);
    msetgf := 0;

    for term in gf do
        rep := 0;

        for deg in
        sort(map(Q-> degree(term, Q),
                 convert(indets(term), `list`)))
        do
            if type(rep, `integer`) then
                rep := cat(`Q`, deg);
            else
                rep := cat(rep, `_Q`, deg);
            fi;
        od;

        msetgf := msetgf + lcoeff(term)*rep;
    od;

    msetgf;
end;


TITLE: A monotonicity property for ratios of power means
QUESTION [2 upvotes]: Let $Z$ be any non-degenerate positive random variable with pdf $%
g(z)$. Let $a>0$ and $r\neq 0$ denote arbitrary real numbers. Define the "$r$
-mean" of $Z$ shifted by the constant $a$ as follows. (It can be assumed that the
necessary moments are finite.):
$M_{r}(a)=E \left[ (a+Z)^{r}\right] ^\frac{1}{r}=\left[ \int\limits_{0}^{\infty
}(a+z)^{r}g(z)dz\right] ^{1/r}$
Furthermore, for some given $s>0$, define
$k(a)=\dfrac{M_{-1}(a)}{M_{s}(a)}=\dfrac{\left[ \int\limits_{0}^{\infty
}(a+z)^{-1}g(z)dz\right] ^{-1}}{\left[ \int\limits_{0}^{\infty
}(a+z)^{s}g(z)dz\right] ^{1/s}}$
I was able to show that $0<k(a)<1$. However, it needs to be shown
that $k(a)$ is non-decreasing, so if $a_{1}<a_{2}$, then $%
k(a_{1})\leq k(a_{2})$. (In other words, $k'(a)\geq 0$. )
I played with specific distributions (lognormal,
uniform, exponential) and it seems to work, but it would be nice to have
some general proof. I'd appreciate any suggestions. (Note: I am just an
economist, not a mathematician. I apologize if I used wrong terminology or I
am unaware of some well-known result.)

REPLY [2 votes]: Indeed the function $k$ is non-decreasing, here is a proof.
The derivative of $E((a+Z)^x)$ with respect to $a$ being $xE((a+Z)^{x-1})$, for every $x$, it suffices to show that, for $s\gt0$,
$$
E((a+Z)^{-2})\,E((a+Z)^{s})\geqslant E((a+Z)^{-1})\,E((a+Z)^{s-1}).
$$
More generally, it happens that, for every positive $X$ such that every moment involved exists,
$$
E(X^{-2})\,E(X^s)\geqslant E(X^{-1})\,E(X^{s-1}).
$$
To prove this, consider a random variable $Y$ independent of $X$ with the same distribution, and integrate the pointwise inequality
$$
(X^{s+1}-Y^{s+1})\,(X-Y)\,X^{-2}\,Y^{-2}\geqslant0.
$$
The expectation of each product $X^aY^b$ being $E(X^a)E(Y^b)=E(X^a)E(X^b)$, one is left with the result that twice the difference between the LHS and the RHS of the inequality one wants to prove, is nonnegative.


TITLE: When two determinantal ideals together generate a power of the maximal ideal?
QUESTION [8 upvotes]: (A somewhat technical question, but maybe it is well known.)
Consider matrices over the ring $k[[x_1,\dots,x_n]]$, whose entries vanish at the origin (i.e. belong to the maximal ideal $\mathfrak{m}$). Denote by $J(A_{k,l})$ the ideal of maximal minors of the matrix  $A_{k,l}\in Mat(k,l,\mathfrak{m})$.
Given two generic enough (and mutually generic!) matrices, consider the ideal generated by all the maximal minors: $\Big( J(A_{k_1,l_1}),J(B_{k_2,l_2})  \Big)$. (Assume $k_1\le l_1$ and $k_2\le l_2$.)
'{\bf Q.}': When does this ideal contain $\mathfrak{m}^{k_1+k_2-1}$?
Of course, an obvious necessary condition is: $(l_1-k_1+1)+(l_2-k_2+1)\ge n$, but is this condition sufficient?
example 1. Suppose $k_1=l_1$, $k_2=l_2$, then for the generic case we have: det(A)
 is of order $k_1$ and $det(B)$ is of order $k_2$ and the lowest order parts of the two polynomials form a regular sequence. Thus, if $n\le 2$ we get: $(det(A),det(B))\supset\mathfrak{m}^{k_1+k_2-1}$.
example 2. Suppose $k_1=1$, take $A_{1,l}$ with generic entries of first order. Again, one can show that the needed property holds.
What about the general case? How to address such questions?

REPLY [3 votes]: Let me  discuss the graded case, that is the ring is the polynomial ring and the matrices have general homogeneous entries of degree $1$. The local version should follows by taking "lowest order part" as you do in example $1$.
Consider first of the case of the polynomial ring $S$ in variables $x_{ij}$ and $y_{ij}$ and matrices $X=(x_{ij})$ of size $a\times b$ and $Y=(y_{ij})$ of size $c\times d$ with $a\leq b$ and $c\leq d$. Then take the ideal $I$ of minors of size $a$ of $X$ and the ideal $J$ of minors of size $c$ of $Y$.
Set $A=S/I+J$.
The minimal free resolution of $A$ is obtained by taking the tensor product of the resolution of $S/I$ with that of $S/J$ because the ideals are generated by polynomials in different variables. We may use this fact, in combination with the fact that $I$ and $J$ are resolved by the Eagon-Northcot complex, to compute the Castelnuovo-Mumford regularity $\operatorname{reg}(A)$ of $A$, its projective dimension and its dimension. We have:
$\operatorname{reg}(A)=\operatorname{reg}(S/I)+\operatorname{reg}(S/J)=(a-1)+(b-1)$
$\dim(A) = ab+cd-(b-a+1)-(d-c+1)$
and $A$ is Cohen-Macaulay.
Now we specialize generically the $x_{ij}$'s and the $y_{ij}$'s  generically to linear forms in variables $z_1,..,z_n$.  The ring you want to understand gets identified with  $A/L$ where $L$ is generated  by $ab+cd-n$ general linear forms in the $x_{ij}$ and $y_{ij}$.
Now if $n\leq (b-a+1)+(d-c+1)$ then $ab+cd-n\geq ab+cd-(b-a+1)-(d-c+1)$ and $ab+cd-(b-a+1)-(d-c+1)$ of the general linear forms generating $L$ form a maximal regular sequence in $A$. Let $U$ be the ideal generated by $ab+cd-(b-a+1)-(d-c+1)$ of the general linear forms generating $L$.
Then $\operatorname{reg}(A/U)=\operatorname{reg}(A)=(a-1)+(b-1)$ and $\dim(A/U)=0$. The regularity for a $0$-dimensional module $M$ is the largest index $i$ such that $M_i\neq 0$. It follows that $(A/U)_i=0$ for $i>(a-1)+(b-1)$. And the same is true for $A/L$ (because it is a quotient of $A/U$). Hence we have $(A/L)_i=0$ for $i>(a-1)+(b-1)$.
On the size of the $z$'s it says that the ideal $(z_1,..,z_n)$ to the power $(a-1)+(b-1)+1$ is contained in ideal of definition, which is exactly what you wanted to prove.