TITLE: Rainbowish phenomenon outside my home
QUESTION [0 upvotes]: I took a photo of an amazing sight I had never seen outside my home ( southern india ) but I could not tell if it was a rainbow type thing or an aurora borealis. Can someone tell what this is and it’s reason of occurrence

REPLY [2 votes]: That is due to ice particles high in the atmosphere.  It's a "rainbow type thing", but different from a rainbow caused by water droplets.   The angle between the sun and the rainbow is different from a droplet rainbow.  Also, the ice crystals tend to be oriented with one axis parallel or normal to the ground while suspended in the air,  so the shape of the rainbow is not necessarily an arc like the standard rainbow.  See https://www.crystalinks.com/sundogs.html.


TITLE: Convergence of Improper Integral for any polynomials
QUESTION [1 upvotes]: Let $f, g$ be real polynomials. How can I show that the improper integral $\int_{0}^{\infty}\frac{f(x)g(x)}{e^x}dx$ converges absolutely for any $f$ and $g$? I think some sort of bounding argument should be used where $f$ and $g$ can be replaced by explicit polynomials that bound $f$ and $g$, but I'm not really sure how to approach this.

REPLY [0 votes]: Note that, if $f$ is of degree $n$ and $g$ is of degree $m$, then $$\lim_{x\to\infty}\frac{|f(x)|}{x^{n+1}}=\lim_{x\to\infty}\frac{|g(x)|}{x^{m+1}}=0.$$ Therefore, there exists $M>0$ such that, if $x\geq M$, then $|f(x)|\leq x^{n+1}$ and $|g(x)|\leq x^{m+1}$. Also, $$\lim_{x\to\infty}\frac{x^{m+n+4}}{e^x}=0,$$ so there exists $M_0>M$ such that, if $x\geq M_0$, then $x^{m+n+4}\leq e^x$. Therefore, for $x\geq M_0$, $$\frac{|f(x)g(x)|}{e^x}\leq\frac{x^{m+1}x^{n+1}}{e^x}=\frac{1}{x^2}\frac{x^{m+n+4}}{e^x}\leq\frac{1}{x^2}.$$ But, the improper integral $\int_{M_0}^{\infty}\frac{1}{x^2}\,dx$ converges, so the original integral converges as well.


TITLE: Čech cocycles and monodromy
QUESTION [2 upvotes]: It is well known that over a topological space $X$ (and choosing an open cover $\mathfrak{U}$) every locally constant Cech cocycle $g$ on $\mathfrak{U}$ with coefficients in a group $G$ yields a $G$-covering space $X_g \rightarrow X$. As such, the monodromy action of this covering space gives a homomorphism from the fundamental group $\pi_1(X,x)$ on a point $x\in X$ to $G$.
I am trying to write this homomorphism explicitly in terms of the cocycle $g$. In his Bachelor Thesis, Lemma 5.5, M.P. Noordman claims that this can be done in the following way. You consider a loop $\sigma:[0,1]\rightarrow X$ and apply the Lebesgue number lemma to get a finite subcover $\{U_1,...,U_n\}$ of $\mathfrak{U}$ and a partition $t_0<t_1<...<t_n$ of the interval $[0,1]$ in such a way that $\sigma([t_{i-1},t_i])\subset U_i$. Now, you can define the homomorphism $f:\pi_1(X,x)\rightarrow G$ as
$$
f([\sigma])=g_{12} g_{23} g_{34} \cdots g_{(n-1) n}.
$$
However, it is not clear to me why this does not depend on the choice of the "Lebesgue subcover" $\{U_1,...,U_n\}$ or on the choice of the representative of the class $[\sigma]$.
For example, consider the case where $\mathfrak{U}=\{U,V,W\}$ consists on three open sets with $U\cap V \neq \varnothing$ and $U,V \subset W$. If we choose a path contained in $U\cup V$, we could choose the Lebesgue covering to be $\{U,V\}$, which would yield $f([\sigma])=g_{UV}$ or we could choose the covering to be simply $\{W\}$, which would yield $f([\sigma])=1$, and I do not see why these should coincide.

REPLY [3 votes]: First, one can pull back the Čech cocycle to S^1 and work directly with S^1
instead of X.
Any two open covers have a common refinement,
so it suffices to show that the monodromy map does not change
under passing to refinements.
As already pointed out in the comments, the open cover must be cyclic: $U_0=U_n$.
By induction, it suffices to show invariance under adding a single new point
$t_{i-1}<t_{i-1/2}<t_i$ with $[t_{i-1},t_{i-1/2}]⊂σ^*U_{i-1/2}$ and $[t_{i-1/2},t_i]⊂σ^*U_i$.
But this invariance is precisely what the Čech cocycle condition says for the open
sets $U_{i-1}$, $U_i$, and $U_{i-1/2}$: $$g_{i-1,i}=g_{i-1/2,i}g_{i-1,i-1/2}.$$


TITLE: Abstract algebra. splitting field
QUESTION [0 upvotes]: Find the degree of the splitting field of the polynomial $x^6-7$ over $F_{3}$ (the field with $3$ elements).
Here, $x^6-7$|${x^6}^2-7^2 = x^{12}-1$  where $7^2 = 1\mod 3$
After this step, I couldn't make anything. How can I continue for solution? or if this step is wrong, what is the way to solve this question?
Thank you

REPLY [1 votes]: Hint: $x^6-7=(x+1)^3(x+2)^3$ over $\mathbb{F}_3$. So the degree is $1$.


TITLE: Maclaurin Series for $(\sec(x))^2$
QUESTION [0 upvotes]: Question 2a ii
So I know I have to use the given tangent MacLaurin Series for solve for 2aii, but how did
$(\sec(x))^2 = 1+ (a_1x + a_3x^3 + a_5x^5 +....)^2 $
turn into
$ (\sec(x))^2= 1 + a_1^2x^2 + 2a_1a_3x^4 +....$ (Real Answer)
I thought it would be $a_1^2x^2 + a_3^2x^6$

REPLY [0 votes]: The term $2a_1 a_3 x^4$ in (Real Answer) comes from $$(a_1x + a_3x^3 + ...)^2=(a_1x + a_3x^3 + ...)(a_1x + a_3x^3 + ...)=a_1^2 x^2+a_1x a_3x^3+a_3x^3a_1x+a_3^2 x^6=a_1^2 x^2+2a_1a_3x^4 +a_3^2 x^6$$
So what you computed is also a term in the Maclaurin series for $sec^2(x)$.


text:
\section{Results and Discussions}\label{sec:experiments}
We begin this section by stating the metrics for evaluating performance of our single individual
hapolotyping algorithm.
\subsection{Performance Metrics}\label{sec:perf_metric}

A widely used metric for characterizing the quality of single individual haplotyping is the minimum
error correction (MEC) score. This metric captures the smallest number of entries of $P_{\Omega}(\R)$
which need to be changed from $1$ to $-1$ and vice versa so that $P_{\Omega}(\R)$ can be
interpreted as a noiseless version of $P_{\Omega}(\M)$. Essentially, it is the most likely number of
sequencing errors, defined for diploids as
\begin{align}
\text{MEC} &= \sum\limits_{i=1}^n~\min \left( D(r_i,\hat{h}_1), D(r_i,\hat{h}_{-1})\right),
\label{eq:MEC_defn}
\end{align}
where $D(r_i,\hat{h}_j)$ denotes the generalized Hamming distance between read $r_i$ (regarded as $m$ length vector in $\{1,-1,\text{`-'}\}$) and the estimated parent haplotypes $\hat{h}_k, ~k=1, -1$. This, in turn, is defined as
$D(r_i,\hat{h}_j) = \sum_{j=1}^m~d(r_{i,j}, \hat{h}_{k,j})$, where
\begin{align}
d(x,y) = \begin{cases} 1, & ~\text{if}~x\neq \text{`-' ~and ~} y\neq \text{`-' ~and ~} x\neq y \\
0, &~\text{otherwise},
\end{cases}
\label{eq:ham_dist_defn}
\end{align} and
$r_{i,j}$ and $\hat{h}_{k,j}$ denote the $j^{th}$ entries of $r_i$ and $\hat{h}_k$, respectively.

The MEC score
is a relevant and most commonly studied performance metric for single individual haplotyping
\cite{schwartz2010theory}, and critically important for experimental data where the ground truth is not
known in advance. It is also a proxy for the most meaningful haplotype assembly metric referred to as

\textit{reconstruction rate}.
Recall that $\mathcal{H} = \{h_1,h_{-1}\}$ denotes the set of true haplotypes.
Then the reconstruction rate of $\hat{\mathcal{H}}= \{\hat{h}_1,\hat{h}_{-1}\}$ with respect to $\mathcal{H}$ is defined as \cite{geraci}
\begin{eqnarray}
\mathcal{R}_{\mathcal{H},\hat{\mathcal{H}}} = 1- \frac{\min \{ D(h_1,\hat{h}_1+D(h_{-1},\hat{h}_{-1}), D(h_1,\hat{h}_{-1})+D(h_{-1},\hat{h}_1) \}}{2m},
\end{eqnarray}
where $D(h_i,h_j)$ denotes the generalized Hamming distance between the haplotype pair $h_i$ and $h_j$.

\subsection{Experiments}\label{sec:exp_results}

In this section, for convenience we refer to our algorithm for single individual haplotyping as \textit{HapAltMin} .

All of the methods described here were run on a Linux OS desktop with 3.07 GHz CPU and 8 GB RAM on an
Intel Core i7 880 Processor.

We first tested our algorithm on the experimental dataset containing Fosmid pool-based NGS
data for HapMap trio child $NA12878$ \cite{duitama}. The Fosmid dataset is characterized
by very long fragments, high SNP to read ratio, and low sequencing coverage of about 3X, consisting of around $1,342,091$ SNPs spread across $22$ chromosomes. We compare the performance of HapAltMin with
the structurally-constrained gradient descent (SCGD) algorithm of \cite{cai_vikalo} and one
more recent SIH software ProbHap \cite{probhap}, which was shown to be superior to several
prior methods on this dataset \cite{bansal_hapcut,duitama,mixsih}.  Table~\ref{table:exp} shows
the MEC rate (average number of mismatches per SNP position across the reads) and runtimes
for all $22$ chromosomes. As seen there, our HapAltMin outperforms other methods for majority
of the chromosomes shown; it is second best in terms of runtime (behind SCGD).

\begin{table}[h]

\caption{MEC rates and runtimes on HapMap sample NA12878 dataset.}
\label{table:exp}
\centering
\begin{tabular}{|C{0.7cm}|C{1.5cm}|C{1.4cm}|C{1.5cm}|C{1.4cm}|C{1.5cm}|C{1.4cm}|}

\hline\hline
{\bf Chr} &  \multicolumn{2}{|c|}{{\bf HapAltMin}} & \multicolumn{2}{|c|}{{\bf SCGD}}  & \multicolumn{2}{|c|}{{\bf ProbHap}} \\ \hline
& {\bf MEC}  & {\bf time(s)} & {\bf MEC}  & {\bf time(s)} & {\bf MEC}  & {\bf time(s)} \\ \hline
\hline
1&  0.034 & 65.0 & 0.04 & 44.2& 0.058 & 87.7 \\\hline
2&  0.035  & 71.6 & 0.035 & 49.5 & 0.055 & 88.9   \\ \hline
3& 0.034 & 61.1 & 0.036 & 41.5 & 0.057 & 84.3 \\ \hline
4& 0.029 & 60.7 & 0.034 & 41.8 & 0.053 & 67.1\\ \hline
5&  0.032  & 52.9 & 0.036 & 39.9 & 0.054 & 64.6 \\ \hline
6 & 0.038 &  34.7 &  0.037 & 27.9&  0.050 & 53.4 \\ \hline
7 & 0.038 &26.4&  0.035 &25.05& 0.055 & 40.8\\ \hline
8 & 0.033 &24.3&  0.034 &23.9& 0.05 & 42.8\\ \hline
9 & 0.036 &21.9&  0.037 &17.6& 0.052& 45.2\\ \hline
10 & 0.036 & 24.7&  0.037 &21.0& 0.053& 44.4 \\ \hline
11 & 0.034 & 24.7&  0.038 &20.8& 0.055& 39.5\\ \hline
12 & 0.037 &23.5&  0.037 & 20.2& 0.057& 38.9\\ \hline
13 & 0.039 &14.6&  0.035 &15.6& 0.053& 26.4\\ \hline
14 & 0.035 &16.6&  0.039 &13.7& 0.055& 27.4\\ \hline
15 & 0.038 & 14.1&  0.041 &11.9& 0.056& 26.5\\ \hline
16 & 0.046 &20.3&  0.0405 & 12.2& 0.051& 36.5\\ \hline
17 & 0.048 &15.3&  0.046 &11.1& 0.061& 27.4\\ \hline
18 & 0.033 &12.2&  0.037 &11.8& 0.053& 24.4\\ \hline
19 & 0.052 &12.8&  0.046 &9.0& 0.063& 19.8\\ \hline
20& 0.044 & 18.1 & 0.044 & 13.0 & 0.055 & 30.9\\ \hline
21&  0.035 & 11.5& 0.041 & 8.5 & 0.051 & 15.6\\ \hline
22&  0.054 & 11.7& 0.055 & 8.6 & 0.061 & 31.4 \\ \hline
\end{tabular}
\vspace{-0.05in}
\end{table}

\begin{table}[!t]

\caption{MEC rate comparison of Algorithm~\ref{algo:mod_altmin} and Algorithm~\ref{algo:sigm_altmin}.}
\label{table:comparison}
\centering
\begin{tabular}{|C{2cm}|C{3.2cm}|C{3.2cm}|}

\hline\hline
{\bf Chr} &  {\bf Algorithm~\ref{algo:mod_altmin} (``Hard'' updates)} &  {\bf Algorithm~\ref{algo:sigm_altmin} (``Soft'' updates)}\\ \hline
\hline
1    & 0.0369 & 0.0339 \\ \hline
2  & 0.0357 & 0.035\\ \hline
3  & 0.0345& 0.0335 \\ \hline
20  & 0.0447 & 0.0446\\ \hline
21  & 0.0367 & 0.035\\ \hline
22  & 0.0568& 0.0535 \\ \hline
\end{tabular}
\vspace{-0.05in}
\end{table}

We further compare the MEC performance of Algorithm \ref{algo:sigm_altmin} with that of the discrete version (Algorithm~\ref{algo:mod_altmin}) in Table~\ref{table:comparison} for first $3$ and last $3$ of the $22$ chromosomes from the Fosmid dataset. As can be seen from Table~\ref{table:comparison}, the modified algorithm performs better than the discrete alternating minimization algorithm, as indicated in the discussion in Section~\ref{sec:algo}.

Next, we focus on the evaluation of performance on simulated dataset using reconstruction rate metric. For this purpose, we use

a widely popular standard benchmarking dataset from \cite{geraci} which also provides the true haplotypes used to generate the read data.  The dataset contains reads at a sequencing error rate with values in the set $\{0.0, 0.1, 0.2, 0.3\}$, and a depth of coverage in the set $\{3X, 5X, 8X, 10X\}$.
Reconstruction rate of our algorithm is
compared with that of SCGD \cite{cai_vikalo} and two more recent SIH methods known as HGHap \cite{hghap} and MixSIH \cite{mixsih}. In particular, \cite{hghap} is chosen for performance comparison since it has been shown to outperform a number of existing SIH methods such as \cite{levy,chen2008linear,bansal_hapcut,genovese2007fast,panconesi2004fast}.
A comparison with ProbHap is not shown for this data since it reconstructed haplotypes with a large fraction of SNPs missing (and therefore has inferior performance compared to the methods
used in the comparisons).
The results, shown in Table~\ref{table:simu}, are obtained by averaging over $100$ simulation runs for each combination of sequencing error rate and sequencing coverage, for a haplotype length of $700$ base pairs and pairwise hamming
distance $0.7$.  As evident from the results, our method is either the best or the second best in all of the scenarios. HapAltMin performs particularly well in the more realistic error range of $0.0-0.2$ and is marginally inferior to MixSIH only for higher sequencing error values.

\begin{table}[!t]

\caption{Reconstruction rate comparison on simulated data. Boldface values indicate best performance.}
\label{table:simu}
\centering
\begin{tabular}{|C{1.7cm}|C{1.2cm}|C{2.3cm}|C{2.1cm}|C{2.1cm}|C{2.1cm}|}

\hline\hline
{\bf Error Rate} & {\bf Cov.}  & {\bf HapAltMin} & {\bf SCGD}  & {\bf HGHap} & {\bf MixSIH} \\
\hline\hline
0.0 & 3X & {\bf 1}  & 0.983 & 0.934 	&
0.776
\\ \hline
0.0 & 5X & {\bf 1}  & 0.976 &  0.989	& 0.923
\\\hline
0.0 & 8X &  {\bf 1}   &  0.999 & 0.994		& 0.995
\\\hline
0.0 & 10X &  {\bf 1} &  0.999 & 0.999 	& {\bf 1}
\\\hline
0.1 & 3X & {\bf 0.935}  & 0.869 & 0.934 	&
0.775
\\ \hline
0.1 & 5X & 0.979  & 0.951 & {\bf 0.990}		& 0.942
\\\hline
0.1 & 8X &  {\bf 0.996}   & {\bf 0.996} & 0.987		& 0.972
\\\hline
0.1 & 10X & {\bf 0.999}  & {\bf 0.999} & 0.997 	& 0.993
\\\hline
0.2 & 3X & {\bf 0.735}   & 0.677 & 0.677		& 0.68
\\\hline
0.2 & 5X & 0.864   & 0.785 & {\bf 0.91}		& 0.774
\\\hline
0.2 & 8X & {\bf 0.943}   & 0.899 & 0.884	& 0.932
\\\hline
0.2 & 10X & 0.966  & 0.934 & 0.894	& {\bf 0.969}
\\\hline
0.3 & 3X & 0.555   & 0.527 & 0.592		& {\bf 0.65}
\\\hline
0.3 & 5X & 0.595   & 0.524 &  0.621		& {\bf 0.667}
\\\hline
0.3 & 8X & 0.68   & 0.518 & 0.646	& {\bf 0.714}
\\\hline
0.3 & 10X & 0.723  & 0.58 & 0.696 & {\bf 0.751}
\\\hline
\end{tabular}
\vspace{-0.05in}
\end{table}

config: arxiv


TITLE: Sen's $\gamma$ and $\alpha$ imply rationalizability of choice
QUESTION [1 upvotes]: I need the proof or a reference for the proof that Sen's $\gamma$ and $\alpha$ imply rationalizability of choice (iff?).
Moreover, on Rubinstein's notes, rationalizability can be shown by Sen's $\alpha$ and the condition that the domain of the choice function contains at least all subsets of X of cardinality 3 or less. I cannot understand this last condition and how it "replaces" Sen's $\gamma$
A brief discussion on the extra assumption implied by Sen's $\beta$, which makes the choice rationalizable under preference relation would be useful.
[I have already posted the question of economics stack, but it seems it is a rather technical question, I didn't get much success]
EDIT
Def If $\forall A,B  \subseteq X; x \in C(A) \cap C(B) \Rightarrow x \in C(A \cup B)$, then the choice rule $C(\cdot)$ satisfies Sen’s $\gamma$ Axiom.

REPLY [0 votes]: The proof that the Weak Axiom of Revealed Preference (WARP) implies Sen's properties alpha and gamma can be found in "Incompleteness and Intransitivity"
by Hector Lopez. Rationalizability is equivalent to the Weak Axiom of Revealed Preference (WARP) (proof).


TITLE: dividing an octave to $7$ instead of $12$
QUESTION [1 upvotes]: Usually an octave is divided into $12$ parts based on the harmonic series(basic zeta function).

how can I calculate the frequency of a note if I divide the octave into $7$ parts?

$N_1=A_4(440Hz)$
$N_8=N_1*2=A_5(880Hz)$
$N_2....N_7???$
Thanks guys!

REPLY [3 votes]: If you want equal musical intervals, then you need to multiply each frequency by a factor of $2^{1/7}$ to get the next higher frequency. Thus, your frequencies will be $440, 440\times 2^{1/7}, 440\times 2^{2/7}, \dots, 440\times 2^{7/7}=880$.
Edit:
The reasoning goes like this:
$$N_1 = 440 \\ N_2 = 440r \\ N_3 = 440r^2 \\ \ldots \\ N_8 = 440r^7 = 880$$
So that $$r^7 = 2$$
I warn you that the harmonies will sound terrible!


TITLE: Queries about rotational groups $\mathrm{SO}(3)$ and $\mathrm{SU}(2)$ in QM
QUESTION [6 upvotes]: In a QM text I am using (Sakurai 2nd edition 'Modern Quantum Mechanics'), he describes two rotation groups, namely the $\mathrm{SO}(3)$ rotation group and $\mathrm{SU}(2)$ rotation group (unitary unimodular group).
He defines $\mathrm{SO}(3)$ as a group with matrix multiplication on a set of orthogonal matrices (which are matrices which satisfy $R^TR = 1 = RR^T$), he then states that this group only includes rotational operators (and not also inverse operators which would be the group $\mathrm{O}(3)$). He does not ever rigorously define 'rotational  operation'.

How you would distinguish between rotational operators and inverse
operators, would a sufficient definition be that rotational
operators is a transformation with one fixed point?

He also defines the the group $\mathrm{SU}(2)$ which consists of unitary unimodular matrices, and states that the most general unitary matrix in two dimensions has four independent parameters and it is defined as $$U = e^{i \gamma}
  \left( {\begin{array}{cc}
   a & b \\
   -b^* & a^* \\
  \end{array} } \right)
$$
where $|a|^2 + |b|^2 = 1,~~~\gamma^* = \gamma.$

Am I right to assume that the $\mathrm{SO}(3)$ rotation group does
not have much of application in quantum mechanics but is rather used
more in classical mechanics whereas $\mathrm{SU}(2)$ is used more in
quantum mechanics, particularly for $s =\frac{1}{2}$ spin systems
where we work in a two dimensional Hilbert space?
How does it follow that there are four independent parameters for the
general unitary matrix, the way I see it there are three independent
parameters, namely, $a$, $b$ and $\gamma$?

REPLY [5 votes]: When classifying representations of a group in QM, it is necessary to allow for projective representations, because states are actually rays (equivalence classes) in the Hilbert space. This means that in order to study the rotational symmetry of a system, you want the projective representations of $\mathrm{SO}(3)$, which are standard representations of $\mathrm{SU}(2)$, because the latter is the universal cover of the former. This is the reason $\mathrm{SU}(2)$ is important in QM.

REPLY [5 votes]: The already present answers have covered the difference between $\mathrm{O}(3)$ and $\mathrm{SO}(3)$ at length, so I will not repeat that. Let me instead explain the point about the "usage" of $\mathrm{SO}(3)$ vs. the "usage" of $\mathrm{SU}(2)$, which I think has not yet been made clear:

$\mathrm{SU}(2)$ is a double cover of $\mathrm{SO}(3)$, meaning there is a two-to-one group homomorphism $\mathrm{SU}(2)\overset{2:1}{\to}\mathrm{SO}(3)$, or, equivalently, $\mathrm{SO}(3)\cong\mathrm{SU}(2)/\mathbb{Z}_2$. It is also simply connected, meaning it is the universal cover. The Lie algebras of both of these Lie groups are the same, i.e. $\mathfrak{so}(3)\cong\mathfrak{su}(2)$. A representation of a Lie algebra always induces a linear representation of the simply connected Lie group associated to it, but it does not always induce a representation of the other groups. More specifically, the spin-1/2 representation is a linear representation of $\mathfrak{so}(3)$, but not of $\mathrm{SO}(3)$, only of $\mathrm{SU}(2)$.
The spin-1/2 representation, is, however, a so-called projective representation of $\mathrm{SO}(3)$. Quantum mechanics actually does not require ordinary linear representations of symmetry groups, but projective ones. For the general reason this is the case, see this Q&A of mine. In this case, it turns out that projective representations of $\mathrm{SO}(3)$ are equivalent to linear representations of $\mathfrak{so}(3)$, or equivalently linear representations of $\mathrm{SU}(2)$. This is the reason $\mathrm{SU}(2)$ appears in quantum mechanics, but not in classical mechanics, when representing the symmetry group of rotations  on our space of states.
The spin-1/2 representation is given by the "standard" representation of $\mathrm{SU}(2)$, i.e. just by the 2-by-2 special unitary matrices. But it is still also a representation of $\mathfrak{so}(3)\cong\mathfrak{su}(2)$ and a projective representation of $\mathrm{SO}(3)$. The spin-1 representation is given by the "standard" representation of $\mathrm{SO}(3)$ as 3-by-3 special orthogonal matrices, but it is still also a representation of $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ and a representation of $\mathrm{SU}(2)$ via the 2-to-1 map.


TITLE: A proof involving Fermat's Little Theorem.
QUESTION [0 upvotes]: Let $p$ be a prime and n be an integer such that $p$ does not divide $n$. Suppose $d$ is the smallest natural number such that $n^d$ is congruent to $1 \mod p$. Prove that $d$ divides $(p-1)$.
So far, I've done:
$n^{p-1}$ is congruent to $1 \mod p$.
Moreover, $n^d$ is congruent to $n^{p-1} \mod p$.
I'm not sure how to continue.

REPLY [2 votes]: We can write $p-1=a\cdot d+r$ where $0\le r<d$ and $a,r$ are nonnegative integers
Now $n^d\equiv1\pmod p, n^{p-1}\equiv1\implies n^r\cdot(n^d)^a\equiv1\implies n^r\equiv1$
But $d$ is the smallest positive integer such that $\implies n^d\equiv1$
So, $r$ must be $0\implies p-1=?$

REPLY [0 votes]: The multiplicative group of the field $Z_p$ is  of order $p-1$, since $n$ is an element of that group its order is a divisor of the order of the group.


TITLE: Does symplectic morphism after geometric quantization induce Hilbert spaces morphism?
QUESTION [5 upvotes]: Let us have two symplectic manifolds $(M, \,\omega)$ and $(N, \,\omega')$ and morphism between them:
$$ \varphi \ :\ M \to N.$$
Then we geometrically quantize these systems:

we add a prequantum line bundle, $ L_M \to M$ and $L_N \to N$ correspondingly;
choose some real polarizations $P_M$ and $P_N$ such that $(\varphi_*P_M)(x) \neq P_N(x), \, \text{for }x \in \varphi(M)$;
let the quotient maps $\pi_{M}\ :\ M \to M/P_M$ and $\pi_N\ :\ N \to N/P_N$ be fibration.
Then the space of compactly supported sections of $L_M/P_M \otimes |T(M/P_M)|^{1/2} $ after completion realise Hilbert space $H_M$. The same procedure for $N$ gives $H_N$.

I don't understand how to build morphism between $H_M$ and $H_N$ from $\varphi$. Is there any natural way to do it?
I'm not sure whether it's a trivial question or not. I don't have deep understanding of geometric quantization, I've only read this review by Lerman. So if some comprehensible text about the question exists, I'll gladly accept a reference.
P.S. I'm reading this paper by Nekrasov which describes morphism between classical Calogero-Moser and Calogero-Sutherland models, and derives from it morphism between their quantum versions. Nekrasov briefly describes how he gets quantum morphism from classical one in the beginning of 4th section. But I don't understand it. So I'm interested whether there is a general way to do it. If it is possible only in this particular case, I'd like to get a more thoroughly explanation of Nekrasov's procedure.

REPLY [2 votes]: The slogan to keep in mind is that "quantization is not a functor". For details on this slogan, see the (decade old!) Mathoverflow question What does “quantization is not a functor” really mean?. Basically, "is not a functor" means that there is no canonical way to build morphisms of Hilbert spaces from morphisms of symplectic manifolds.
Of course, in many practical cases there may be some "best" morphism of Hilbert spaces for a given "nice" morphism of symplectic manifolds, or some family of "best" morphisms. For instance, perhaps in your practical case you have some control over how your $\varphi$ interacts with the prequantum and polarization data.


TITLE: An inequality in Hilbert spaces
QUESTION [0 upvotes]: Suppse $A$ is a self-adjoint linear and bounded operator in a complex Hilbert space. Let $x,y$ be two elements in the Hilbert space such that $\|x\|=\|y\|=1$. How can I prove that
$$ |\langle Ax,y\rangle| \leq |Re \{ \langle Ax,y\rangle \}|$$

REPLY [2 votes]: Your inequality is not true. For instance let $A$ be the identity and let $x$ and $y$ be any two unit vectors whose inner product $\langle x,y\rangle$ is not real. Then, the magnitude of $\langle x,y\rangle$ will strictly exceed the magnitude of the real part of $\langle x,y\rangle$.
For an explicit example, let $e_1$ and $e_2$ be orthonormal basis vectors. Let $x=e_1$ and let $$y=\frac{ie_1+e_2}{\sqrt{2}}.$$
Then $$\langle x,y\rangle=\frac{i}{\sqrt{2}},$$
hence the real part of $\langle Ax,y\rangle$ is zero, whereas $\langle Ax,y\rangle$ has magnitude $1/\sqrt{2}$.