TITLE: maximization of a quadratic function with a quadratic constraint
QUESTION [3 upvotes]: I have the following quadratic maximization problem
$\max_{\mathbf X} \quad tr(\mathbf A\mathbf X\mathbf B\mathbf X^H)+tr(\mathbf C\mathbf X)+tr(\mathbf C^H\mathbf X^H)$
subject to the quadratic constraint $tr(\mathbf X \mathbf X^H)\leq r$ where $\mathbf A, \mathbf B \succeq \mathbf 0$ (positive semi-definite) and $\mathbf C$ is any matrix (not necessarily hermitian).
I am looking for a closed-form solution/structure for $\mathbf X$ or even a nice upper bound. A simple upper bound of it can be found using inequality results, so we know that the problem has a finite maximum.
I would really appreciate any help/comment/solution :)
Thanks in advance,
Saeed

REPLY [1 votes]: Your problem is in the form
$\max_x f(x)$ subject to $g(x) < 0$, where $x$ is the set of components $x_i^j$ of the matrix $X$.
Use one Lagrange multiplier to write $f(x) + \lambda g(x)$. Both nonlinear terms are quadratic, so this is the most favorable case for the purpose of obtaining solutions. To take the derivative of the expression above with respect to the set of unknowns $x$, index notation as in tensor calculus could be helpful. Then you are in the standard form of constrained optimization.


TITLE: Embedding of general linear groups
QUESTION [1 upvotes]: Can we embedd $GL(2n,q)$ into $GL(n,q^2)$ for $n\in \mathbb{N}$ and $q=p^m$, $p$ a prime? If yes, how?

REPLY [4 votes]: As others have mentioned, this is clearly not true in special cases: GL(2,2) ≅ Sym(3) clearly does not embed in GL(1,4) ≅ Alt(3), and indeed the embedding should go the other way (in general even, not just for n=1, q=2).
If q is a prime power, and n is a positive integer, then GL(2n,q) never embeds in GL(n,qq).  One way to see this is the size of the Sylow p-subgroup.  GL(k,pi) has a Sylow p-subgroup of order p(ik(k−1)/2).  So in our cases, we have GL(2n,q) has a Sylow p-subgroup of order qn(2n−1) and GL(n,qq) has a Sylow p-subgroup of order qn(n−1)/2.  Since n(n−1)/2 < n(2n−1) whenever n ≥ 1, we never have an embedding.
In fact, a little more is true: homomorphisms from GL(n,q) into GL(k,qj) are actually conjugate to homomorphisms from GL(n,q) into GL(k,q).  You don't need to make the field bigger in order to map a GL(n) into a GL(k), at least not for finite fields.
Of course there are inclusions of GL(n,qq) into GL(2n,q), since an n-dimensional vector space over GF(qq) is also a 2n-dimensional vector space over GF(q).  You can even have inclusions of GL(n,q) into GL(k,q) for k ≥ n that are not conjugate to the obvious embedding.  In all cases though, the dimension cannot go down, and there is no sense making the field bigger, only occasionally making it smaller.


TITLE: Need clarification from an already answered question.
QUESTION [0 upvotes]: I was reading this question (Proving $E[X^4]=3σ^4$). It would be really helpfull to know why $X=\sigma U$ (Please, see the comments in the question). Where does $\sigma$ come from? I have no problems on why $E(U^4)=3$, tho.

REPLY [1 votes]: Define $U$ as $\frac 1 {\sigma } X$. Then $var (U)=EU^{2}-(EU)^{2}=\frac 1 {\sigma^{2} } EX^{2}-(EX)^{2}=\frac 1 {\sigma^{2} } var (X)=1$. Hence $U$ has $N(0,1)$ distribution and we can calculate $EU^{4}$ by the method described in your link.


TITLE: Every 3 disjoint closed subsets of normal space can be separated with 3 disjoint open subsets
QUESTION [0 upvotes]: Let X be a normal space ($T_4$),and let $F_1,F_2,F_3$ be closed disjoint subsets of X.
Prove that there are $U_1,U_2,U_3$ open disjoint subsets,such that $F_i$ is subset of $U_i$ for each $1\le i \le 3$.
I'm not even sure how to begin this..any ideas?

REPLY [4 votes]: HINT: First use normality to get disjoint open nbhds $U_1$ and $V$ of $F_1$ and $F_2\cup F_3$, respectively. Then use it to get $U_2$ and $U_3$ inside $V$.


TITLE: To which function does the lambda function refere in this context
QUESTION [1 upvotes]: In a tech report from Microsoft there is an equation on page 11 (or actually 16) to calculate the entropy of a d-variate Gaussian density:
$$H(S)={1\over2}log((2\pi e)^d|\Lambda(S)|)$$
I don't know what the $\Lambda(S)$ part of the equation means and what the $\Lambda$ function is. Can someone explain this to me please?

REPLY [0 votes]: It's an unusual notation, but presumably it's representing the differential entropy of a multivariate Gaussian random variate S with covariance matrix S.
$\lambda$ in this case represents the determinate.
So your original equation in standard form looks like this:
$$H(S)={1\over2}log((2\pi e)^d|det(S)|)$$
Recommended further reading on the topic: Elements of Information Theory (Wiley Series in Telecommunications and Signal Processing) 99th Edition


TITLE: Prime divides $n^2 + 1 \Rightarrow$ prime doesn't divide $n$
QUESTION [5 upvotes]: How can I show that if a prime $p$ divides $$n^2 + 1$$ then it doesn't divide $n$?

REPLY [1 votes]: If $p=2
 $ and $2\mid n
 $ we have that $n^{2}
 $ is even and so $n^{2}+1
 $ is odd. Now assume that $p$ is odd. We can use the Legendre symbol. If we assume that $n\equiv0\mod p
 $ we have $n^{2}\equiv0 \mod p
 $. So $$\left(\frac{n^{2}}{p}\right)=0
 $$ but since $n^{2}\equiv-1 \mod p$ we also have, by the law of quadratic reciprocity $$\left(\frac{n^{2}}{p}\right)=\left(\frac{-1}{p}\right)=1^{\frac{p-1}{2}}=\begin{cases}
1 & p\equiv1\,\mod\,4\\
-1 & p\equiv3\,\mod\,4
\end{cases}$$ and this is absurd.


TITLE: Equating Coefficients of Cos and Sin
QUESTION [0 upvotes]: I've got a nonlinear system
\begin{align}
x'&=\frac{1}{2}x-y-\frac{1}{2}(x^3+y^2x)\\
y'&=x+\frac{1}{2}y-\frac{1}{2}(y^3+x^2y)
\end{align}
I am to analyse the system when the system is changed to polar coordinates. The writer of the book states that he computes the following
\begin{align}
r'\cos\theta-r(\sin\theta)\theta'&=x'=\frac{1}{2}(r-r^3)\cos\theta-r\sin\theta\\
r'\sin\theta+r(\cos\theta)\theta'&=y'=\frac{1}{2}(r-r^3)\sin\theta+r\cos\theta
\end{align}
And by equating the coefficients of $\cos\theta,\sin\theta$ he gets the following system
\begin{align}
r'&=r(1-r^2)/2\\
\theta'&=1
\end{align}
My issues is then:
I'm not sure where he gets the following identities
\begin{align}
r'\cos\theta-r(\sin\theta)\theta'&=x'\\
r'\sin\theta+r(\cos\theta)\theta'&=y'
\end{align}
And I'm not that familiar with the equating coefficients method, so I've read a bit about it, and I can't seem to compute the presented system.
If I were to make a phase portrait of the system, would the axis be $r'$ and $\theta'$?
Source: [c. 8, p. 162], Differential Equations, Dynamical Systems & An Introduction to Chaos (2nd Ed) by Morris W. Hirsch, Stephen Smale and Robert L. Devaney.

REPLY [0 votes]: While the method in the book is correct, it is often easier to derive the polar coordinates version using
\begin{align}
rr'&=xx'+yy'\\
r^2\theta'&=xy'-yx'
\end{align}
which is obtained by differentiating
$$
r^2=x^2+y^2,\quad\tan\theta=\frac{y}{x}.
$$
Indeed, the $r'$-equation is clear, and differentiating $\tan\theta=y/x$ gives
$$
(\tan\theta)'=\theta'\sec^2\theta=\theta'(1+\tan^2\theta)=\theta'\frac{x^2+y^2}{x^2}=r^2\theta'
$$
in the LHS and
$$
\left(\frac{y}{x}\right)'=\frac{xy'-yx'}{x^2}
$$
in the RHS.


TITLE: A divisor sum involving generalized Moebius function
QUESTION [1 upvotes]: I am trying to prove the following claim:

Let $n$ be a squarefree natural number. Denote by $\mu_k$ the generalized Moebius function: $\mu_k=\underbrace{\mu \ast \ldots \ast \mu}_{k}$ where $\ast$ is a Dirichlet convolution, and let $\omega(n)$ be the prime omega function. Then,
$$\displaystyle\sum_{d \mid n} |\mu_k(d)|=(k+1)^{\omega(n)} \quad \text{for} \quad k\ge 1$$

The SageMath cell that demonstrates this claim can be found here.
My attempt:
I tried to apply a mathematical induction:

Induction base (k=1): $\displaystyle\sum_{d \mid n} |\mu(d)|=2^{\omega(n)}$ - This is a well known identity.

Induction hypothesis (k=m): $\displaystyle\sum_{d \mid n} |\mu_m(d)|=(m+1)^{\omega(n)}$

Induction step (k=m+1): $\displaystyle\sum_{d \mid n} |\mu_{m+1}(d)|=(m+2)^{\omega(n)}$

Since $\mu_{m+1}=\mu_m \ast \mu$ we have:
$$\displaystyle\sum_{d \mid n} |(\mu_{m} \ast \mu)(d)|=(m+2)^{\omega(n)}$$
$$\displaystyle\sum_{d \mid n} \left|(\mu_{m}(d)\mu \left(\frac{n}{d}\right)\right|=(m+2)^{\omega(n)}$$
Now, I am stuck here. How should I proceed?

REPLY [2 votes]: First, your last goal is incorrectly derived from the penultimate one: it should read
$$
\sum_{d \mid n} \biggl| \sum_{c\mid d} \mu_{m}(c)\mu \biggl(\frac{d}{c}\biggr)\biggr|=(m+2)^{\omega(n)}.
$$
Now note that since $n$ is squarefree, $d$ and $d/c$ must also be squarefree, and so $|\mu(d/c)=1|$ always. Using the induction hypothesis and sorting the left-hand side according to the value of $\omega(d)$ then yields
\begin{align*}
\sum_{d \mid n} \biggl| \sum_{c\mid d} \mu_{m}(c)\mu \biggl(\frac{d}{c}\biggr)\biggr| &= \sum_{d \mid n} \biggl| \sum_{c\mid d} \mu_{m}(c) \biggr| \\
&= \sum_{d \mid n} (m+1)^{\omega(d)} \\
&= \sum_{k=0}^{\omega(n)} (m+1)^k \sum_{\substack{d \mid n \\ \omega(d)=k}} 1 \\
&= \sum_{k=0}^{\omega(n)} (m+1)^k \binom{\omega(n)}k = \big((m+1)+1)^{\omega(n)}=(m+2)^{\omega(n)}
\end{align*}
by the binomial theorem. (The fact that $\#\{d\mid n\colon \omega(d)=k\} = \binom{\omega(n)}k$ is because such divisors $d$ correspond exactly to $k$-element subsets of the set of $\omega(n)$ prime factors of $n$.)


TITLE: Polynomials for Bicubic Planar Dessin d'Enfants
QUESTION [11 upvotes]: A dessin d'enfant is a graph, with its vertices colored alternately black and white, embedded in ... a plane. For the coloring to exist, the graph must be bipartite. ... The ... embedding may be described combinatorially using a rotation system...
Any dessin can provide the surface it is embedded in with a structure as a Riemann surface. It is natural to ask which Riemann surfaces arise in this way.

The wikipage on dessin d'enfants, also gives an example:

The dessin d'enfant arising from the rational function $f = −(x − 1)^3(x − 9)/64x$. Not to scale.
Further a table, listing the degrees of vertices, is given. Later Shabat polynomials are mentioned, which are the corresponding ones for trees.
My question:

Is it possible to assign a rational function to any kind of bipartite planar cubic graph?
Or more specific: How to construct the polynomial for the chamfered cube?

REPLY [2 votes]: Some introductory words. My interest in dessins d'enfants (dessins for short from now on) was related some years ago to (my speculation concerning) some connections with the K-theory, where the object "$\Bbb P^1_{\Bbb Q}\setminus\{0,1,\infty\}$" is a central object. (Symbols, Galois Groups over $\Bbb Q$, regulators in K-theory, ... are related to this object.) There was also an Arbeitsgemeinschaft (AG) on the subject, MFO-2000 centered on Dessins d'Enfants, and some very specific applications. (The report of the AG from loc. cit. shows a wide spectrum of related domains.) Often, the starting point is a given more or less complicated Belyi function. (E.g. related to quotients of the upper half plane by modular groups, where the Belyi function is a $j$-invariant.)
Building the dessin gives a "compact information" on the situation.
In this case, the question of the OP is the converse one. We have the dessin, and need a realization. To obtain an answer to the question, it is natural to get a simplest situation of a map. Which is the situation of a rational map $f=\frac PQ:\Bbb P^1(\Bbb C)\to\Bbb P^1(\Bbb C)$, defined over $\bar{\Bbb Q}$, unramified above $\Bbb P^1(\Bbb C)-\{0,1,\infty\}$. If this is possible.
Existence theorems show it is possible. I found a good reference for this, the master thesis of Jeroen Sijsling.
Let us see the machinery at work for the case of the Chamfered Cube.

The planar graph for the ($1$-skeleton of the) cube is as follows:

To obtain its chamfered version, we draw inside of each (deformated) square a smaller one. For the outer square we must go outside. Every "old vertex" gets connected to the three new square vertices in its proximity.
The new picture is:

The circles (in black and/or blue) $\circ$ and $\color{blue}{\circ}$ are preimages of $0$ via the searched function $f$ (up to repositioning by hand, which does not destroy the dessin). The ramification index of each such point is $3$, i.e. locally around each such point, after a new choice of uniformizers, the map $f$ looks like $z\to z^3$. (Our graph is a cubic graph.)
The bullets (in black and/or blue) $\bullet$ and $\color{blue}{\bullet}$ are preimages of $1$ via the searched function $f$ (up to same repositioning).
The ramification index of each such point is $3$.
The edges correspond to preimages via $f$ of the interval $(0,1)\subset \Bbb C\subset \Bbb P^1(\Bbb C)$.
In each "region" delimited by such edges there is exactly one preimage of $\infty$, we build the preimages of the three distinct paths between $0,1,\infty$ that build the projective real line $\Bbb P^1(\Bbb R)$ inside $\Bbb P^1(C)=\Bbb H\cup\Bbb P^1(\Bbb R)\cup\Bbb H'$. Here $\Bbb H$ is the upper half plane, and $\Bbb H'$ is its reflection w.r.t. the real affine line $\Bbb R\subset \Bbb C\subset \Bbb P^1(C)$. Imagine that the three paths are building a "triangle", and both $\Bbb H$ and $\Bbb H'$ may be considered to fill it, we give them two different colors. Preimages are then colored in the same way. Then the six "squares" and the twelve "hexagons" are colored with two white and two gray triangles as follows:

My paint dried short before painting the two "bounded" regions (those with one arc as boundary) outside the outer most square, please imagine them also as painted in gray. (The "triangle region", and the "unbounded region" are painted in white.)
Each preimage of $0$ and/or $1$ becomes thus a vertex of three white and three gray triangles, the corresponding ramification index is three. For the pink bullets $\color{magenta}{\bullet}$, preimages of $\color{magenta}{\infty}$ we have two different situations, there are:

either two white and two gray "triangles" building a "square", and the corresponding ramification index is two,
or three white and three gray "triangles" building a "hexagon", and the corresponding ramification index is three.

We can now make an Ansatz for $f=P/Q$:

there are $16$ points $a_1,a_2, \dots a_{16}$ building $f^{-1}(0)$, each ramification index is three,
there are $12$ points $b_1,b_2, \dots b_{12}$ with ramification index three in $f^{-1}(\infty)$, they correspond to the points $\color{magenta}{\bullet}$  in the "hexagons",
there are $6$ points $b'_1,b'_2, \dots b'_6$ with ramification index two in  $f^{-1}(\infty)$, they correspond to the points $\color{magenta}{\bullet}$  in the "squares",
there are $16$ points $c_1,c_2, \dots c_{16}$ building $f^{-1}(1)$, each ramification index is three.

We need to exhibit the above constants, so that:
Ansatz:
$$
A\underbrace{\left(\prod_{1\le i\le 16}(x-a_i)\right)^3}_{P=p^3}
-
B\underbrace{\left(
\prod_{1\le j\le 12}(x-b_j)
\right)^3
\cdot\left(
\prod_{1\le j\le 6}(x-b'_j)
\right)^2
}_{Q=q^3s^2}
=
C\underbrace{
\left(\prod_{1\le k\le 16}(x-c_k)
\right)^3
}_{R=r^3}
\ .
$$
(Each term has degree $48$, which is $3\cdot 16=3\cdot 12+2\cdot 6$.)
There is of course a homogenous version of the above polynomial relation, obtained by setting $x=X/Y$. The factors from above are $(X-a_iY)$, $(X-b_jY)$, etc. - and
using Möbius isomorphisms, we may change the above, so that three points among the many  points $(a_i)$, $(b_j)$, $(c_k)$ become respectively $\infty,0,1$. So according to our future needs, if $\rho$ is one such root among the three, we may arrange that $(X-\rho Y)$ becomes $X$ if we want to set $\rho=0$, then $(X-\rho Y)$ becomes $Y$ if we move $\rho$ to $\infty$, and $(X-\rho Y)$ becomes $(X-Y)$, when $\rho$ is moved to one. For this point, we only arrange that $b_1=\infty$, and still have the chance to move two points... We dehomogenize, so that we work further with
$$
\begin{aligned}
p(x) &=(x-a_1)(x-a_2)\dots(x-a_{16})\ ,\\
q(x) &=(x-b_2)\dots(x-b_{12})\ ,\qquad\text{($b_1$ was moved to $\infty$)}\\
s(x) &=(x-b'_1)\dots(x-b'_6)\ ,\\
r(x) &=(x-c_1)(x-c_2)\dots(x-c_{16})\ .
\end{aligned}
$$
All above polynomials are monic, and the degrees are now $16$ for $p,r$, $12-1=11$ for $q$, and $6$ for $s$.
We may and do take now $A=C=1$. It remains to realize in the factorial / UFD ring $\bar{\Bbb Q}[x]$ - best even with $p,q,r,s$ having coefficients in $\Bbb Q$ or some simple
extension of $\Bbb Q$- a relation of the shape:
$$
(p-r)(p-\omega r)(p-\omega^2 r)=p^3 - r^3=B\; q^3s^2\ .
$$
We have denoted by $\omega$ a primitive thirs root of unity.
Let $F=\Bbb Q(\omega)=\Bbb Q(\sqrt{-3})$ be the corresponding cyclotomic field.
Here, $p,q,s,r$ have different roots, so we use UFD to get the following arithmetic information:
$$
\left\{
\begin{aligned}
p-r &= B\; q_1^3\; s_1^2\ ,\\
p-\omega r &= (1-\omega)\; q_2^3\; s_2^2\ ,\\
p-\omega^2 r &= (1-\omega^2)\; q_3^3\; s_3^2\ .
\end{aligned}
\right.
$$
The pieces $q_1,q_2,q_3;s_1s_2,s_3$ are monic in the further search.
One of the factors in $(p-r)(\dots)=p^3-r^3$ must go down in degree.
We have chosen the first one to do this.
So the degree for $q_2^3\; s_2^2$, $q_3^3\; s_3^2$
is $16$. In the product $p^3-r^2=(p-r)(\dots)$ the degree goes down by three, so  this happens in $(p-r)$, whose degree is thus $13$.
We may even search for $q_2,q_3$, respectively $s_2,s_3$ which are mapped in each other by the conjugation in $F$,
$\omega\to\omega^2$, which is also the complex conjugation after some choice of an embedding $F\subset \mathbb C$, e.g. via $\omega=(-1+i\sqrt 3)/2$.
(So no such polynomial has rational roots.)
We may now extract $p,r$ from the first two equations above,
and plug in in the third one. We obtain an equation only in the $q_1,q_2,q_3;s_1,s_2,s_3$ polynomials.
This equation may be also obtained by observing that $(p-r)+\omega(p-\omega r)+\omega^2(p-\omega^2 r)=0$.
So we would like to realize:
$$
B\; q_1^3s_1^2
=
\omega  (1-\omega)\; q_2^3s_2^2 +
\omega^2(1-\omega^2)\; q_3^3s_3^2 \ .
$$
Note that $\omega  (1-\omega)=-\omega^2(1-\omega^2)$, so we may indeed get cancellations with monic polynomials on the R.H.S. above.
I have to make a decision, so i am taking $\deg q_1=3$, $\deg q_2=\deg q_3=4$, $\deg s_1=\deg s_2=\deg s_2=2$.
It is time to remember that we may further norm two roots now, so $q_1$ is a good candidate to be taken in the form
$$
q_1=x(x-1)(x-b_4)\ .
$$

I have to stop here, but i hope it is clear how the search goes on.
Some simple sage program did not found an explicit solution, but i will come back here in the eventuality that i manage to find
an explicit solution.


TITLE: Multivariable Chain Rule / Partial Derivatives
QUESTION [2 upvotes]: If $z = f(x − y)$ and $g(x, y) = x − y$, so that $z =f∘g$
Why/How does the chain rule imply this?
$$\frac{∂z}{∂x} =\frac{∂f}{∂g}\frac{∂g}{∂x} =\frac{∂f}{∂g}$$
and
$$\frac{∂z}{∂y}=\frac{∂f}{∂g}\frac{∂g}{∂y} =-\frac{∂f}{∂g}$$

REPLY [3 votes]: Just note that $z = f(x-y)$ is one variable function depend on $g = x-y$. Because $z = f (g(x,y))$,
$$
\frac{\partial z}{\partial x} = \frac{\partial f(g(x,y))}{\partial x} = \frac{\partial f}{\partial g} \frac{\partial g}{\partial x} = \frac{\partial f}{\partial g} \cdot 1 = \frac{\partial f}{\partial g}
$$
and the same thing for $\frac{\partial z}{\partial y}$.


TITLE: Cycle lengths with prime denominators
QUESTION [2 upvotes]: For a math course I've been playing with the following piecewise recurrence relation,
$$f(x_t) = x_{t+1} =
\begin{cases}
      3x_t & 0 \leq x_t < \frac{1}{3} \\
      3x_t - 1 &  \frac{1}{3} \leq x_t < \frac{2}{3} \\
      3 - 3x_t &  \frac{2}{3} \leq x_t \leq 1.
\end{cases}
$$
When the initial condition $x_0$ = $\frac{1}{p}$, where $p$ is prime, successive iterations lie on a cycle. By brute force checking many primes it appears that all cycles have at most $\frac{p-1}{2}$ elements (rarely some have fewer). With my skills I haven't been able to show (or really get an intuition for) why this is the case.
Any help would be appreciated.

REPLY [2 votes]: Let $\{x\} = x-\lfloor x\rfloor$, for $p$ prime $\ge 5$ and $x_0=  \frac1p$ then by induction $$x_t =\left\{  \frac{(-1)^{e_t} 3^t}p\right\}, \qquad e_t=\pm 1$$
Let $r$ be the least positive integer such that $3^r \equiv \pm 1 \bmod p$, since $\Bbb{Z/pZ}^\times / \pm 1$ is a group with $\frac{p-1}{2}$ elements then

$r$ divides $\frac{p-1}{2}$

Either $x_r = \frac1p=x_0$ and we have a cycle of length $r$,
Or $x_r=- \frac1p$ which means $x_{r+1} =  \frac3p=x_1$ and we have a cycle of length $r$.


TITLE: Virtual particles/quantum tunneling - conservation of energy?
QUESTION [5 upvotes]: I'm confused as to how the above phenomena can take place since arent they breaking the law of conservation of energy (even, if temporarily)?

REPLY [3 votes]: Of course, one usual intepretation of quantum tunneling is that the particle will borrow some energy from the vaccum in order to pass an unsurmountable barrier otherwise and then restitute it asap after crossing the barrier.
As many others have said, this is a valid interpretation. I am not sure it is necessary though.
In fact, in quantum tunneling, what really differs from classical physics per se is not energy conservation (after all the energy is supposed to be conserved and quantum tunneling in the NH3 molecule is at fixed energy for instance) but barrier crossing.
The only thing quantum tunneling tells us is that it is not the case that a quantum particle with energy $E$ cannot overcome a barrier of height $V > E$, period. All the rest is a matter of interpretation and taste I think.
Now, how can a particle pass a barrier if it doesn't have enough energy to jump over it? The answer lies, in my opinion, in the fact that a quantum particle has rarely a definite position even when its energy is perfectly defined. Therefore, assuming that such an uncatchable object can remain always on one side of a finite barrier would be the strange thing to observe and not the other way around.

REPLY [2 votes]: First of all what you need to understand about quantum physics is that it's a theory of probability, not realist classical probability, but it is still a probability theory. The second thing you need to understand is that realism is wrong.
Conservation of energy in quantum physics simply means that the Hamiltonian is not time dependent. That's it. From this follows that the expectation value of the energy does not change in time. You shouldn't picture a particle moving around either and then some energy magically appears and the particle can jump across some potential. This is a wrong way of thinking about quantum physics and Bell's theorem is clear proof of that. Particles do not have well defined classical properties before measurement. The sooner you realize this the better.


TITLE: How to derive the cosets of $A_4$?
QUESTION [0 upvotes]: In the alternating group $A_4$ for the subgroup of $H\le A_4$ find all of its left and right cosets where $$H=\{\text{Id},(1,2,3),(1,3,2)\}.$$
We have by Lagrange's theorem that $$|A_4:H|=\frac{|A_4|}{|H|}=\frac{24/2}{3}=4$$ which means that $H$ has $4$ left and $4$ right cosets.
By definition a left coset of $H$ in $A_4$ determined by $x\in A_4$ is the set $\{xh:h\in H\}$.
So, the left cosets will be $x\text{Id},x(1,2,3),x(1,3,2)$, right?
How can I find the $4$th coset?

REPLY [2 votes]: Obviously, the 100% way to do this is get every element of $A_4$ multiply it to left and right of $H$ to get your $aH$ and $Hb$. I will answer to answer how to come up with the selection of elements to get 4 left cosets and 4 right cosets.
I am copying the listing of $A_4$ by johnny09 posted in his or her answer, but with a rearrangement for emphasizing certain elements: $$A_4=\{\text{Id},$$
$$(1,2,3),(1,3,2),(2,3,4),(2,4,3),(1,2,4),(1,4,2),(1,3,4),(1,4,3)$$
$$(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$$
Looking at $A_4$ this way, we have 3 to cosets derive because $H$ is its own coset, and we also have 3 elements other than the identity that are not like the others (We can also say we have 4 cosets to derive and 4 elements that are not like the others). I am going to base my selection of elements on these 3 elements.
A permutation property is: $$(a,b)(c,d)=(adc)(abc)$$
From this, we can write:
$$(1,2)(3,4) = (143)(123) \tag{1}$$
$$(1,3)(2,4) = (142)(132) \tag{2}$$
$$(1,4)(2,3) = (132)(142) \tag{3}$$
A permutation property is: $$(a,b)(c,d)=(dba)(dca)$$
From this, we can write:
$$(1,2)(3,4) = (421)(431) \tag{4}$$
$$(1,3)(2,4) = (431)(421) \tag{5}$$
$$(1,4)(2,3) = (341)(321) \tag{6}$$
My selection for $a$ to derive left cosets $aH$ will be based on $(1),(2),(6)$
$$a = 1, (143), (142), (341)$$
My selection for $b$ to derive right cosets $Hb$ is the reverse of $(1),(2),(6)$:
$$(1,2)(3,4) = (143)(123) \tag{1}$$
$$(1,3)(2,4) = (142)(132) \tag{2}$$
$$(1,4)(2,3) = (341)(321) \tag{6}$$
$$(1,2)(3,4) = (321)(341) \tag{1(rev)}$$
$$(1,3)(2,4) = (231)(241) \tag{2(rev)}$$
$$(1,4)(2,3) = (123)(143) \tag{6(rev)}$$
I select $$b = 1, (341), (241), (143)$$
To prove that my choices for a work, I used the property: $(a,b,c)(c,b,e)=(abe)$. Other permutation properties are: $(a,b)(c,d)=(cda)(cba)=(acd)(abd)=(dab)(dcb)$. For b, see: Why did this reversal from the left cosets of $\langle (1, 2, 3) \rangle$ in $A_4$ give me the right cosets?

For a systematic approach (but still aiming for the products of disjoint transposition), we could try to solve $(1,2)(3,4)=(123)(abc)$, $(1,2)(3,4)=(132)(abc)$ and also for the others.


text:
\section{Teichm\"uller space and the deformation space of complex projective structures}\label{TSCPS}

\subsection{\texorpdfstring{$\TS$ and $\CPS$}{{T(S)} and {CP(S)}}}\label{defts}

Let $S$ be a surface. Unless otherwise stated, we will assume that $S$ is connected\footnote{In some
sections (e.g. \ref{cphyp}), we will allow $S$ to be disconnected in order to be able to consider the case where $S$
is the boundary
of a compact $3$-manifold. This does not cause any issue in the exposition above.}, oriented, smooth,
closed and with genus $g \geqslant 2$.

A complex structure on $S$ is a maximal atlas of charts mapping open sets of $S$ into the complex line $\C$ such that
the transition maps are holomorphic.
The atlas is required to be compatible with the orientation and smooth structure on $S$.
A Riemann surface $X$ is a surface $S$ equipped with a complex structure.

The group $\mathrm{Diff}^+(S)$ of orientation preserving diffeomorphisms of $S$ acts on the set
of all complex structures on $S$ in a natural way:
a compatible complex atlas on $S$ is pulled back to another one by such diffeomorphisms.
Denote by $\mathrm{Diff}^+_0(S)$ the identity component of $\mathrm{Diff}^+(S)$, its elements are the
orientation preserving diffeomorphisms of $S$ that are homotopic to the identity.
The quotient $\TS$ of the set of all complex structures on $S$ by $\mathrm{Diff}^+_0(S)$
is called the \emph{Teichm\"uller space} of $S$, its elements are called \emph{marked Riemann surfaces}.

In a similar fashion, define a \emph{complex projective structure} on $S$ as a maximal atlas of charts
mapping open sets of $S$ into the complex
projective line $\CP$ such that the transition maps are (restrictions of) projective linear transformations
(\textit{i.e.} Möbius transformations of the Riemann sphere). The atlas is also required to be compatible
with the orientation and smooth structure on $S$. A complex projective surface $Z$ is a surface $S$ equipped with a complex
projective structure.
In terms of geometric structures (see e.g. \cite{thurston}),
a complex projective structure is a $(\CP,\PSL_2(\C))$-structure.

Again, $\mathrm{Diff}^+(S)$ naturally acts on the set of all complex projective structures on $S$. The quotient $\CPS$ by the subgroup
$\mathrm{Diff}^+_0(S)$ is called the \emph{deformation space of complex projective structures} on $S$, its
elements are \emph{marked complex projective surfaces}.

\subsubsection*{\texorpdfstring{$\TS$ and $\CPS$ are complex manifolds}{{T(S)} and {CP(S)} are complex manifolds}}\label{pif}

Kodaira-Spencer deformation theory (see \cite{KD}, also \cite{earleeells}) applies and it shows that
$\TS$ is naturally a complex manifold with holomorphic
tangent space given by $T^{1,0}_X \TS = \check{H}^1(X,\Theta_X)$,
where $\Theta_X$ is the sheaf of holomorphic vector fields on $X$.
Denote by $K$ the canonical bundle over $X$ (the holomorphic cotangent bundle of $X$).
By Dolbeault's theorem, $\check{H}^1(X,\Theta_X)$ is isomorphic to the Dolbeault cohomology space $H^{-1,1}(X)$.
Elements of $H^{-1,1}(X)$ are Dolbeault classes of smooth sections of $K^{-1} \otimes \bar{K}$,
which are called \emph{Beltrami differentials}.
In a complex chart $z : U \subset S \rightarrow \C$, a Beltrami differential $\mu$
has an expression of the form $\mu = u(z) \frac{d\overline{z}}{dz}$ where $u$ is a smooth function.
The fact that we only consider (Dolbeault) classes of Beltrami differentials can be expressed as follows :
if $V$ is a vector field on $X$ of type $(1,0)$, then the Beltrami differential $\overline{\partial} V$ induces
a trivial (infinitesimal) deformation of the complex structure $X$.
Recall that $X$ carries a unique hyperbolic metric within its conformal class (called the Poincar\'e metric)
by the uniformization theorem.
By Hodge theory, every Dolbeault cohomology class has a unique harmonic representative $\mu$.
The tangent space $T_X \TS$ is thus also identified with the space $\HB(X)$ of harmonic Beltrami differentials.

We can also derive a nice description of the Teichm\"uller cotangent space using cohomology machinery.
$\check{H}^1(X,\Theta_X) = H^1(X,K^{-1})$ because $\mathrm{dim}_\C X= 1$ and this space is dual to $H^0(X,K^{2})$ by Serre duality.
An element $\varphi \in Q(X) := H^0(X,K^{2})$ \label{qx} is called a \emph{holomorphic quadratic differential}.
In a complex chart $z : U \subset S \rightarrow \C$, $\varphi$
has an expression of the form $\varphi = \phi(z) dz^{2}$, where $\phi$ is a holomorphic function.
The holomorphic cotangent space ${T}^*_X \TS$ is thus identified with the space $Q(X)$ of holomorphic quadratic differentials.
The duality pairing $Q(X) \times H^{-1,1}(X) \to \C$ is just given by $(\varphi,\mu) \mapsto \int_S \varphi \cdot \mu$.
Note that we systematically use tensor contraction (when dealing with line bundles over $X$) : $\varphi \cdot \mu$
is a section of $K \otimes \bar{K} \approx |K|^2$, so it defines a conformal density and can be integrated over $S$.
With the notations above, $\varphi \cdot \mu$ has local expression $\phi(z) u(z) |dz|^2$.

An easy consequence of the Riemann-Roch theorem is that $\mathrm{dim}_\C Q(X) = 3g-3$, so that $\TS$ is a complex manifold
of dimension $3g-3$.

Similarly, Kodaira-Spencer deformation theory applies to show that $\CPS$ is naturally a complex manifold with
tangent space $T_Z \CPS = \check{H}^1(Z,\Xi_Z)$, where $\Xi_Z$ is the sheaf of projective vector fields on $Z$ (see also \cite{hubbard}).
It follows that $\CPS$ is a complex manifold of dimension $6g-6$.

Unlike Teichm\"uller tangent vectors, there is no immediate way to describe tangent vectors to $\CPS$ in a more tangible way.
However, note that a complex projective atlas is in particular a holomorphic atlas,
so that a complex projective surface $Z$ has an underlying structure of a
Riemann surface $X$. This yields a forgetful map
\begin{equation} p : \CPS \to \TS \label{pfor} \end{equation} which is easily seen to be holomorphic.
We will see in section \ref{affinebundle} that the fiber $p^{-1}(X)$ is naturally a complex affine space whose underlying
vector space is $Q(X)$. In particular $\dim_\C \CPS = \dim_\C \TS \times \dim_\C Q(X) = 6g-6$ as expected.

\subsubsection*{\texorpdfstring{The Weil-Petersson K\"ahler metric on $\TS$}{The Weil-Petersson K\"ahler metric on {T(S)}}}

The \emph{Weil-Petersson product} of two holomorphic quadratic differentials $\Phi$ and $\Psi$ is given by
\begin{equation}
 \left<\varphi,\psi\right>_{WP} = -\frac{1}{4} \int_X \varphi \cdot {\sigma}^{-1} \cdot \overline{\psi}
\end{equation}
where ${\sigma}^{-1}$ is the dual current of the area form $\sigma$ for the Poincar\'e metric\footnote{In
a complex chart with values in the upper half-plane $z = x + i y : U \subset X \to \HH^{2}$,
the tensor product $-\frac{1}{4} \varphi \cdot {\sigma}^{-1} \cdot \overline{\psi}$ reduces to the classical expression
$\begin{displaystyle}
y^2 \varphi(z) \overline{\psi(z)} dx \wedge dy
\end{displaystyle}$.}. It is a Hermitian
inner product on the complex vector space $Q(X)$.

By duality, this gives a Hermitian product also denoted by $\left<\cdot,\cdot\right>_{WP}$ on $H^{-1,1}(X)$ and globally a Hermitian
metric $\left<\cdot,\cdot\right>_{WP}$ on the manifold $\TS$.
It was first shown to be K\"ahler by Ahlfors \cite{ahlfors2}
and Weil.

The K\"ahler form of the Weil-Petersson metric on $\TS$ is the real symplectic form
\begin{equation}
 \omega_{WP} = - \mathrm{Im} \left<\cdot,\cdot\right>_{WP}~. \label{WP}
\end{equation}

\subsection{Fuchsian and quasi-Fuchsian projective structures}\label{fqf}

Note that whenever a Kleinian group $\Gamma$ ({\it i.e.} a discrete subgroup of $\PSL_2(\C)$) acts freely and properly
on some open subset $U$
of the complex projective line $\CP$, the quotient surface $U/\Gamma$ inherits a complex projective structure.
This gives a variety (but not all) of complex projective surfaces, called embedded projective structures.

Fuchsian projective structures are a fundamental example of embedded projective structures.
Given a marked complex structure $X$, the uniformization theorem provides a representation $\rho : \pi_1(S) \to \PSL_2(\R)$
such that $X$ is isomorphic to $\HH^2/\rho(\pi_1(S))$ as a Riemann surface (where $\HH^2$ is the upper half-plane).
$\HH^2$ can be seen as an open set (a disk) in $\CP$ and the Fuchsian group $\rho(\pi_1(S)) \subset \PSL_2(\R)$ is in particular a
Kleinian group, so the quotient
$X \approx \HH^2/\rho(\pi_1(S))$ inherits a complex projective structure $Z$. This defines a section
\begin{equation}\label{fufuch}
 \sigma_{\cal F} : \TS \to \CPS
\end{equation}
to $p$, called the \emph{Fuchsian section}.
It shows in particular that the projection $p$ is surjective.
We call $\FS:=\sigma_{\cal F}(\TS)$ the (deformation) space of (standard) Fuchsian (projective) structures on $S$,
it is an embedded copy of $\TS$ in $\CPS$.

\emph{Quasi-Fuchsian structures} are another important class of embedded projective structures.
Given two marked complex structures $(X^+,X^-) \in \TS \times {\cal T}(\overline{S})$\footnote{
Note that ${\cal T}(\overline{S})$ is canonically identified with $\overline{\TS}$,
which denotes the manifold $\TS$ equipped with the opposite complex structure.
The same remark holds for ${\cal CP}(\overline{S})$ and $\overline{\CPS}$.}
 (where $\overline{S}$ is the surface $S$ with reversed orientation),
Bers' simultaneous uniformization theorem states that there exists a
unique representation $\rho : \pi_1(S) \stackrel{\sim}{\to} \Gamma \subset \PSL_2(\C)$ up to conjugation such that:
\begin{itemize}
 \item[$\bullet$] The limit set\footnote{The \emph{limit set} $\Lambda = \Lambda(\Gamma)$
 is defined as the complement in $\CP$ of the domain
of discontinuity $\Omega$, which is the maximal open set on which $\Gamma$ acts freely and properly. Alternatively,
$\Lambda$ is described as the closure in $\CP$ of the set of fixed points of elements of $\Gamma$.} $\Lambda$ is a Jordan curve.
 The domain of discontinuity $\Omega$ is then
 the disjoint union of two simply connected domains $\Omega^+$ and $\Omega^-$. A such $\Gamma$ is called a
 quasi-Fuchsian group.
 \item[$\bullet$] As marked Riemann surfaces, $X^+ \approx \Omega^+/\Gamma$ and $X^- \approx \Omega^-/\Gamma$.
\end{itemize}
Again, both Riemann surfaces $X^+$ and $X^-$ inherit embedded complex projective structures $Z^+$ and $Z^-$ by this construction.
This defines a map
$\beta  = (\beta^+,\beta^-) :
\TS \times  {\cal T}(\overline{S}) \to \CPS \times  {\cal CP}(\overline{S})$
which is a holomorphic section to $p \times p : \CPS \times  {\cal CP}(\overline{S}) \to  \TS \times  {\cal T}(\overline{S})$
by Bers' theorem.
The map $\beta$ has the obvious symmetry property: $\overline{\beta^-(X^+,X^-)} = \beta^+(\overline{X^-},\overline{X^+})$.

In particular, when $X^- \in {\cal T}(\overline{S})$ is fixed, the map
$\sigma_{X^-} := \beta^+(\cdot,X^-) : \TS \to \CPS \label{bbss}$
is a holomorphic section to $p$, called a \emph{Bers section},
and its image $\sigma_{X^-}(\TS)$ in $\CPS$ is called a \emph{Bers slice}. \label{berssection}
On the other hand, when $X^+ \in\TS$ is fixed, the map
$f_{X^+} = \beta^+(X^+,\cdot) $
 is an embedding of ${\cal T}(\overline{S})$
 in the fiber $P(X^+) := p^{-1}(X^+) \subset \CPS$\footnote{which has the structure of a complex affine space as we
 will see in section \ref{schwarzian}.},
 $f_{X^+}$ is called a \emph{Bers embedding}.
Also, note that $\sigma_{\cal F}(X) = \beta^+(X,\bar{X}) = \overline{\beta^-(X,\bar{X})} = \sigma_{\bar{X}}(X)$. This shows that the Fuchsian section
$\sigma_{\cal F}$ is real analytic but not holomorphic, in fact it is a maximal totally real analytic embedding, see section \ref{ancont}.

$\QFS:=\beta^+(\TS \times  {\cal T}(\overline{S})) \subset \CPS$ \label{qfdef} is called the (deformation) space of
(standard) quasi-Fuchsian (projective) structures on $S$. It is an open neighborhood of $\FS$ in $\CPS$
(this is a consequence of general arguments mentioned in the next paragraph),
and it follows from the discussion above that Bers slices and Bers embeddings
define two transverse foliations of $\QFS$ by holomorphic copies of $\TS$.

\subsection{\texorpdfstring{Complex projective structures and hyperbolic $3$-manifolds}{Complex projective structures and hyperbolic 3-manifolds}} \label{cphyp}

In this paragraph, we briefly review the relation between complex projective structures on the boundary of a compact $3$-manifold $\hat{M}$
and hyperbolic structures
on its interior. The quasi-Fuchsian projective structures presented in the
previous section occur as a particular case of this discussion.
We then define \emph{generalized Bers sections} and \emph{generalized Bers embeddings}, and
fix a few notations for later sections.

Let $M$ be a connected complete hyperbolic $3$-manifold. The universal cover of $M$ is isometric to hyperbolic $3$-space $\HH^3$,
this defines a unique faithful representation $\rho : \pi_1(M) \to \mathrm{Isom}^+(\HH^3) \approx \PSL_2(\C)$ up to conjugation such that
$\Gamma:=\rho(\pi_1(M))$
acts freely and properly on $\HH^3$ and $M \approx \HH^3/\Gamma$. Let $\Omega \subset \CP$ be the domain of discontinuity of the Kleinian group $\Gamma$, it is
the maximal open set on which $\Gamma$ acts freely and properly.
Here $\CP$ is seen as the ``ideal boundary'' of $\HH^3$, also denoted $\partial_\infty \HH^3$.
The possibly disconnected surface $\partial_\infty M := \Omega / \Gamma$ is called the \emph{ideal boundary}
\label{idboundef} of $M$ and it inherits an embedded complex
projective structure as the quotient of $\Omega \subset \CP$ by the Kleinian group $\Gamma$.
Conversely, any torsion-free Kleinian group $\Gamma$ acts freely and properly on $\HH^3 \sqcup \Omega$
(where $\Omega$ is the domain of discontinuity
of $\Gamma$), and the quotient consists of a $3$-manifold
$\hat{M} = M \sqcup \partial_\infty M$, where $M = \HH^3 / \Gamma$ is a complete hyperbolic $3$-manifold
and $\partial_\infty M = \Omega/\Gamma$ is its ideal boundary. In general, the manifold $\hat{M}$ is not compact, if it is then $\hat{M}$
is topologically the end compactification of $M$. In that case we say that the hyperbolic structure on $M$ is \emph{convex cocompact}.
The \emph{convex core} of $M$ is the quotient of the convex hull of the limit set $\Lambda$
in $\HH^3$ by $\Gamma$.
It is well-known that $M$ is convex cocompact if and only if its convex core is a compact deformation retract of $M$.

Consider now a smooth $3$-manifold with boundary $\hat{M}$ with the following topological restrictions:
$\hat{M}$ is connected, oriented, compact, irreducible\footnote{meaning that every embedded $2$-sphere
bounds a ball.}, atoroidal\footnote{meaning that it does not contain any embedded, non-boundary parallel, incompressible tori.}
and with infinite fundamental group.
Let $M = \hat{M} \setminus \partial \hat{M}$
denote the interior of $\hat{M}$. For simplicity, we also assume that the boundary $\partial \hat{M}$ is incompressible\footnote{
meaning that the map $\iota_* : \pi_1(\partial \hat{M}) \to \pi_1(\hat{M})$ induced by the inclusion map $\iota$ is injective.} and contains no tori,
so that it consists of a finite number of surfaces $S_1, \dots, S_N$ of genera at least $2$.
The Teichm\"uller space ${\cal T}(\partial \hat{M})$ is described as the direct product
${\cal T}(\partial \hat{M}) = {\cal T}(S_1) \times \dots \times {\cal T}(S_N)$,
similarly ${\cal CP}(\partial \hat{M}) = {\cal CP}(S_1) \times \dots \times {\cal CP}(S_N)$
and there is a holomorphic ``forgetful'' projection
$p = p_1 \times \dots \times p_N : {\cal CP}(\partial \hat{M}) \rightarrow {\cal T}(\partial \hat{M})$.
Let  $\pr_k : {\cal CP}(\partial \hat{M}) \rightarrow {\cal CP}(S_k)$ denote the
$k^{\textrm{th}}$ \label{defprmap} projection map. Let us consider the space ${\cal HC}(M)$ of convex cocompact
hyperbolic structures on $M$ up to homotopy. In other words, we define ${\cal HC}(M)$ as the quotient of the set of convex cocompact hyperbolic metrics on $M$
by the group of orientation-preserving diffeomorphisms of $M$ that are homotopic to the identity. Let us mention that Marden \cite{mardeng}
and Sullivan \cite{sullivang} showed that
${\cal HC}(M)$ is a connected component of the interior of the subset of discrete and faithful representations in the
character variety ${\cal X}(M,PSL_2(\C))$. By the discussion above, any element of ${\cal HC}(M)$ determines a marked
complex projective structure $Z \in {\cal CP}(\partial \hat{M})$. We thus have a map $\varphi : {\cal HC}(M) \to {\cal CP}(\partial \hat{M})$, and it is shown to be
holomorphic, this is a straightforward consequence of the fact
that the holonomy map is holomorphic (see section
\ref{holonomy}). Considering the induced conformal structure on $\partial \hat{M}$, define the map $\psi = p \circ \varphi$ as in the following diagram:
$$\xymatrix{{\cal HC}(M) \ar[r]^\varphi \ar[rd]^{\psi} & {\cal CP}(\partial \hat{M}) \ar[d]^p \\
& {\cal T}(\partial \hat{M})~.}$$
The powerful theorem mainly due to Ahlfors, Bers, Kra, Marden, Maskit, Sullivan and Thurson\footnote{see \cite{CanaryMC} chapter 7. for a detailed exposition
of this theorem,
containing in particular the description of the different contributions of the several authors.
A non-exhaustive list of references includes \cite{abg},
\cite{ahlforsg}, \cite{bersg}, \cite{krag}, \cite{mardeng}, \cite{maskitg}, \cite{sullivang}.}
says in this context that:
\begin{theorem}[Ahlfors, Bers, Kra, Marden, Maskit, Sullivan, Thurston]
The map $\psi : {\cal HC}(M) \rightarrow {\cal T}(\partial \hat{M})$ is bijective.
\end{theorem}
Let us mention that this statement has to be slightly modified if $\hat{M}$ has compressible boundary.
As a consequence of this theorem, we get
\begin{propo}
The map \begin{equation}\beta = \varphi \circ \psi^{-1} : {\cal T}(\partial \hat{M}) \to {\cal CP}(\partial \hat{M})\end{equation} is a
holomorphic section to $p : {\cal CP}(\partial \hat{M}) \to {\cal T}(\partial \hat{M})$.
\end{propo}
We call $\beta$ the \emph{(generalized) simultaneous uniformization section}. \label{betasimul}
This map allows us to define ``generalized Bers sections'' and ``generalized Bers embeddings'' by letting only one of the boundary components' conformal
structure vary and by looking at the resulting complex projective structure on some other (or the same) boundary component.
This idea is made precise as follows.
If an index $j \in \{1,\dots,N\}$ and marked complex structures $X_i \in {\cal T}(S_i)$ are fixed for all $i \neq j$,
we denote by $\iota_{(X_i)}$ the
injection
\begin{equation} \label{caninj}
\iota_{(X_i)} :
\begin{array}{ccl}
{\cal T}(S_j) & \rightarrow & {\cal T}(\partial \hat{M})\\
  X & \mapsto & (X_1, \dots, X_{j-1},X,X_{j+1}, \dots, X_N)~.
\end{array}
\end{equation}
Let $f_{(X_i),k} = \pr_k \circ \beta \circ \iota_{(X_i)}$ as in the following diagram:
\begin{equation}
\xymatrix{
{\cal T}(\partial \hat{M}) \ar[rr]^\beta & & {\cal CP}(\partial \hat{M}) \ar[d]^{\pr_k}\\
{\cal T}(S_j) \ar[u]^{\iota_{(X_i)}}\ar[rr]^{f_{(X_i),k}} & & {\cal CP}(S_k)~.
}\end{equation}
If $j = k$, then $\sigma_{(X_i)}:=f_{(X_i),j}$ is a holomorphic section to $p_j : {\cal CP}(S_j) \to {\cal T}(S_j)$, that we call a \emph{generalized Bers
section}. On the other hand, if $j \neq k$, then $f_{(X_i),k}$ maps ${\cal T}(S_j)$ in the affine\footnote{see section \ref{affinebundle}.}
fiber $P(X_k) \subset {\cal CP}(S_k)$, we call a $f_{(X_i),k}$ a \emph{generalized Bers embedding}. We apologize for this misleading
terminology: a ``generalized Bers embedding'' is \emph{not} an embedding in general.\label{genbers}

Note that quasi-Fuchsian structures discussed in the previous paragraph just correspond to the case where $M = S \times \R$.
Let us also mention that this discussion is easily
adapted when $\partial M$ contains tori or is no longer assumed incompressible, with a few precautions. When $\hat{M}$ only has one boundary component,
this gives the notion of a \emph{Schottky section}.

config: arxiv


TITLE: How do you calculate the width of the Poset Lattice of Divisors?
QUESTION [2 upvotes]: Let $n = 10800 = 2^43^35^2$
I can find a set of eleven divisors of $n$ such that none divides another:
$$\begin{array}{ccccc}
  &   &   & 2 3^3 & 3^35\\
  &   & 2^23^2 & 23^25 & 3^25^2 \\
  & 2^33  & 2^235 & 235^2 &   \\
 2^4 & 2^35  & 2^25^2 &   &
\end{array}$$
I have written them in this way to show some structure. The factors of two decrease to the right so no term may divide any term to the right. The factors of three decrease and the factors of five increase going down. So the only chance for a term to divide another is to the left somehow, but since the total number of prime divisors is constantly four this is impossible.
Question number 1) Is it possible to find a larger set of such divisors with no element dividing another?
Question number 2) (the answer implies an answer to #1)
Given a prime decomposition of $n$ how do I calculate the size of a largest set of divisors such that no element divides another?
$$n = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$$
Clearly this must be a function of the $\alpha_i$ only, but I don't see how do to it for the general case.

REPLY [1 votes]: I think I see at least how to calculate this at all. here is my answer.
Start with the tuple $$( \alpha_1, \cdots, \alpha_k )$$
Then form a set of the $k$ possible tuples by subtracting one from each entry.
Note the size of this set.
Continue making new sets by forming all possible tuples by subtracting one from each possible entry--every element of the list should have the same tuple sum at each stage. (These are sets: do not count duplicate elements!)
The maximum length of this list is your answer.
With $n = 10800$ we proceed as follows:
Step 0:$$\{(4,3,2)\}$$
Step 1:$$\{(3,3,2),(4,2,2),(4,3,1)\}$$
Step 2:$$\{(2,3,2),(3,2,2),(3,3,1),(4,1,2),(4,2,1),(4,3,0)\}$$
Step 3:$$\{(1,3,2),(2,2,2),(2,3,1),(3,1,2),(3,2,1),(3,3,0),(4,0,2),(4,1,1),(4,2,0)\}$$
Step 4:$$\{(0,3,2),(1,2,2),(1,3,1),(2,1,2),(2,2,1),(2,3,0),(3,0,2)(3,1,1),(3,2,0),(4,0,1),(4,1,0)\}$$
At step five you have the same number of elements (11) and then it decreases after that until you get down to $(0,0,0)$ at step 10.
This algorithm will work in the general case, although I'm still very interested if there is a better, number theoretic way.


TITLE: If two charts are smoothly compatible with the atlas, then they are naturally compatible
QUESTION [2 upvotes]: I have been enrolled in an course on Riemannian Geometry and was struck on this exercise which was given as homework.

Question: If two charts are compatible with the atlas, then show that they are naturally compatible.

Attempt: Two charts $( U, \phi)$ and $(V, \psi)$ are said to be smoothly compatible if either $U\cap  V = \varnothing $  or $ \psi \circ \phi^{-1} $ is a diffeo.
Let $( U, \phi)$ be a chart and {$(W_{\alpha} ,X_{\alpha})$} be an atlas. Let $(V,\phi)$be another chart.
I couldn't understand what natural means, to me it's not a mathematical term, so, I decided to omit it anyway and try moving towards a rigorious proof.
Case 1: $U\cap W_{\alpha} = \varnothing$ and $ V \cap W_{\alpha} = \varnothing$. To show that either $U\cap V=\varnothing$ or $ \psi \circ \phi^{-1} $ is  a diffeo.
I can't prove $U\cap V=\varnothing$ ( It is easy to find a counterexample). So, I must show that $\psi \circ \phi^{-1}$ is a diffeo. ( But I don't have any ideas on how to do it.) I know that $\phi$ and $\psi$ are homeomorphisms but I am not able to relate them to diffeomorphisms.
(b) $U \cap W_{\alpha} =\varnothing $ and $X_{\alpha} \circ \psi^{-1}$ is diffeo. I have to prove that either $U \cap V =\varnothing$ or $\phi \circ \psi^{-1}$ is diffeo.
Let $ U \cap V \neq \varnothing$  because if the previous expression is $\phi$ then there is nothing to prove. It is given that $ \psi \circ X_{\alpha}^{-1} $ is $C^{\infty}$. To prove that  $  \phi \circ \psi^{-1} $ is a diffeo. But again , I am not able to relate to what is given to what is proved.
(iii) is similar to (ii).
(iv) Let $U \cap W_{\alpha}\neq \varnothing$ and $ V \cap W_{\alpha} \neq \varnothing$. In this case I have, $X_{\alpha} \circ \phi^{-1} $ is diffeo. and $X_{\alpha} \circ \psi^{-1}$ is diffeo. But again , I am not able to prove $\phi \circ \psi^{-1} $ is diffeo.

The problem I am having can be boiled down to how to use the given diffeomorphisms to prove a function is diffeo. but I am unable to. I am struck on it for the past 6 hrs.

Kindly help me by giving hints.
Thanks!

REPLY [2 votes]: Let $(U_1,\varphi_1)$ and $(U_2,\varphi_2)$ be two charts compatible with some atlas $\mathcal{A}=\{(W_{\alpha},\psi_{\alpha})\}_{\alpha}$.
Let $U:=U_1\cap U_2$.
Case 1.
Suppose that $U = \varnothing$.
Then, by definition, $(U_1,\varphi_1)$ and $(U_2,\varphi_2)$ are compatible.
Case 2.
Suppose that $U\neq \varnothing$, and let us show that $(\varphi_1|_U)\circ (\varphi_2|_U)^{-1}$ is a diffeomorphism.
First of all, we know that both $\varphi_1\colon U_1 \to \varphi_1(U_1)$ and $\varphi_2\colon U_2\to \varphi_2(U_2)$ are homeomorphisms (they are charts).
Then so are $\varphi_1|_{U}\colon U\to \varphi_1(U)$ and $\varphi_2|_U\colon U \to \varphi_2(U)$.
To conclude, it suffices to show that both $(\varphi_1|_U)\circ (\varphi_2|_U)^{-1}$ and $(\varphi_2|_U)\circ (\varphi_1|_U)^{-1}$ are smooth.
Let $y\in \varphi_2(U)$ and let $x\in U$ such that $\varphi_2(x)=y$.
Let $(W,\psi)$ be any chart of the atlas $\mathcal{A}$ with $x\in W$.
Consider $V=U\cap W\neq$, which is non empty since $x\in V$.
It is clear that $\left((\varphi_1|_U)\circ(\varphi_2|_U)^{-1}\right)|_{\varphi_2(V)}= (\varphi_1|_V)\circ (\varphi_2|_V)^{-1}$, and moreover, we have
$$
(\varphi_1|_V)\circ (\varphi_2|_V)^{-1} = (\varphi_1|_V)\circ \operatorname{Id}_V\circ (\varphi_2|_V)^{-1} = (\varphi_1|_V)\circ(\psi|_V)^{-1}\circ (\psi|_V)\circ  (\varphi_2|_V)^{-1}.
$$
Since $\varphi_1$ is compatible with $\psi$, as well as $\varphi_2$, the functions $(\varphi_1|_V)\circ (\psi|_V)^{-1}$ and $(\psi|_V)\circ (\varphi_2|_V)^{-1}$ are smooth (between open subsets of $\Bbb R^n$).
It follows that their composition is smooth, and we have already shown that it is $\left((\varphi_1|_U)\circ(\varphi_2|_U)^{-1}\right)|_{\varphi(V)}$.
Finally, $(\varphi_1|_U)\circ(\varphi_2|_U)^{-1}$ is smooth when restricted to $\varphi_2(V)$, so in particular, it is smooth at the point $y$.
This being true for all $y\in \varphi_2(U)$, we have shown that $(\varphi_1|U)\circ (\varphi_2|_U)^{-1}\colon \varphi_2(U)\to \varphi_1(U)$ is smooth.
Note that there was nothing particular with picking $\varphi_1$ first, so that the same proof shows that $(\varphi_2|_U)\circ (\varphi_1|_U)^{-1}$ is also smooth.
We have then shown that the charts $(U_1,\varphi_1)$ and $(U_2,\varphi_2)$ are compatible.