TITLE: When is $t \mapsto \gamma (t) = \big( (1 + a \cos t) \cos t, (1 + a \cos t) \sin t \big)$ simple and closed?
QUESTION [2 upvotes]: Show that
$\gamma (t) = \big( (1 + a \cos t) \cos t, (1 + a \cos t) \sin t \big), t \in [0, 2\pi]$, where $a$ is a constant, is a simple closed curve if $|a| < 1$ , but that if $|a| > 1$  its complement is the disjoint union of three connected subsets of $\Bbb R^2$, two of which are bounded and one is unbounded. What happens if $a = \pm 1$?
Hopf's theorem states that the total signed curvature of a simple closed curve in $\Bbb R^2$ is $\pm 2 \pi$. But if a curve has total signed curvature of $\pm 2 \pi$ that does not mean it is a simple closed one. Right? So any simple way to solve this one? Have to prove that it has a period in order to be closed. And to be simple it must not have self-intersections..I try to solve $γ(t_1)=γ(t_2)$ but i get to
$$((1+acost_1)cost_1)=((1+acost_2)cost_2$$
cant solve for $t_1$ or $t_2$

REPLY [1 votes]: Partial answer.$\gamma (t)$ has period $2\pi$ and is continuous...(I)... If $|a|<1$ then $1+a\cos t$ is never $0$, so $\gamma (t)=\gamma (u)\implies (\cos t=\cos u\land \sin t=\sin u)\implies (t-u)/2\pi\in Z,$ so for $v\in [0,1]$ the function $g(v)=\gamma (2\pi v)$ is a homeomorphic embedding of the unit circle of $C$ into $R^2$. (II). If $|a|>1$ there are $2$ values of $t\in (0,2\pi)$ for which $\cos t=-1/a$ and $\gamma (t)=(0,0).$


TITLE: wave superposition of electrons and quarks
QUESTION [1 upvotes]: Is quantum wave superposition of electrons and quarks possible?
If not, can different types of elementary particles be mixed in wave superposition?

REPLY [0 votes]: Is quantum wave superposition of electrons and quarks possible?

It depends on what you define as "quantum wave",  and "quarks". In s-wave scattering of electrons off protons, since protons are made up of quarks, the s-wave electron is superimposed over the quarks in order to interact with the proton.

If not, can different types of elementary particles be mixed in wave superposition?

The wave we are talking of here is a probability wave, i.e the probability of finding a particle at (x,y,z,t) with fourmomentum (p_x,p_y,p_z, E). If one creates a mixed beam one has also the probability of being particle A or particle B ( which is true for many beams since they cannot be pure of one particle type).  So their probabilities are superposed.


TITLE: Algebraic numbers and fraction field.
QUESTION [2 upvotes]: If $a\in\mathbb{C}$ is algebraic, then $\mathbb{Q}(a)=\mathbb{Q}[a]$, and the converse holds too. I am having trouble proving that the same holds for several elements. Is it true that $a_1,\ldots, a_n$ are algebraic iff $\mathbb{Q}(a_1,\ldots, a_n)=\mathbb{Q}[a_1,\ldots, a_n]$?

REPLY [1 votes]: Try an inductive argument and observe that
$$\Bbb F(a_1,...,a_n)=\Bbb F(a_1,...,a_{n-1})(a_n)=\Bbb F(a_1,...,a_{n-1})[a_n]\Longleftrightarrow a_n$$
is algebraic over $\,\Bbb F(a_1,...,a_{n-1})\,$


TITLE: finding a matrix $A_2$ such that $A_2(\mbox{vect}(e_2,e_3)) \subseteq \mbox{vect}(e_2,e_3)$
QUESTION [3 upvotes]: Let us consider
$$A_1 = \left[\begin{array}{ccc} 2&-1&0 \\  0&1&0\\  0&0&0\end{array}\right].$$

I want to find a matrix $A_2$ such that $A_1A_2=A_2A_1$ and $A_2F\subseteq F$ with $F=\mbox{vect}(e_2,e_3)$

Thank you.

REPLY [3 votes]: You can take $A_2$ to be Identity matrix.


TITLE: Buildings associated to generalized $BN$ pairs
QUESTION [6 upvotes]: I'll begin by asking a general question, and then specializing to the situation I really care about.
Let $G$ be a group and let $(B, N)$ be a $BN$-pair in $G$ (see, for instance, page 39 of Tits' "Buildings of Spherical Type and Finite BN-pairs).  Then $(B,N)$ has associated to it a chamber complex $X$ with a number of nice properties; most specifically, $X$ is connected, each apartment $\Sigma$ is a Coxeter complex, and $G$ acts transitively on the set $\{(\Sigma,\, C)\}$ of pairs where $\Sigma$ is an apartment and $C \in \Sigma$ is a chamber.
When $G$ is a semisimple group over a $p$-adic field $K$, there is a $BN$-pair where $B$ is an Iwahori subgroup.
More generally, when $G$ is reductive, there is no such $BN$-pair; there is, however, an generalized $BN$-pair.
My general question is:

Is there a connected building $X$ associated to a generalized $BN$-pair; that is, a connected chamber complex $X$ with an action of $G$ that is transitive on the set $\{(\Sigma,\, C)\}$ as above, and where each apartment $\Sigma$ is a Coxeter complex?

More specifically, if $G$ is a reductive group over a $p$-adic field, is there a canonical way to construct a chamber complex $X$ with the above properties, such that each Iwahori subgroup fixes a unique chamber?
For example, if $G = GL_2(\mathbb{Q}_p)$, then there is a generalized $BN$-pair where $B$ is the standard Iwahori subgroup, the Weyl group $W$ is generated by $w_1 = \pmatrix{0 & 1 \\ 1 & 0}$ and $w_2 = \pmatrix{0 & p \\ p^{-1} & 0}$, and $\Omega$ is generated by $s = \pmatrix{0 & p \\ 1 & 0}$ (here I am using the notation given at the beginning of Iwahori's Generalized Tits Systems on $p$-adic Semisimple Groups).  One way I can think to build a building is as follows:  $X$ is one-dimensional, where the set of vertices is $G/I_1 \cup G/I_2$, where $I_j$ is the parahoric generated by $Bw_jB$.  The set of edges corresponds to $G/B$, and there is the obvious $G$-action.  This, however, is disconnected since it is a countable disjoint union of trees.

REPLY [4 votes]: You may either work with an extended building as
L. Spice suggests in his answer, or consider the following based on the fact that a  generalized $BN$-pair contains a genuine $BN$-pair. With you notation, the group $N$ writes as a semidirect product $\Omega\ltimes N_0$ for some invariant subgroup $N_0$ of $N$. Then $G_0 = BN_0 B$ is  a group with a true $BN$-pair $(B,N_0 )$, and we have $G = \Omega \ltimes G_0$. We may form the building $X$ for $G_0$ with respect to the $BN$-pair $(B,N_0 )$. Recall that $X$ is a simplicial complex whose vertices are in $G_0$-equivariant bijection with the parahoric subgroups $P$ of $G_0$. The action of $G$ by conjugation permutes the parahoric subgroups of $G_0$, whence induces a simplicial action of $G$ on $X$.


TITLE: Proof check: if $x_n \to x$ prove $\sup(x_n) \to x$
QUESTION [1 upvotes]: My attempt,
if $x_n \to x$ then $x_n$ is bdd therefore there exists $\sup(x_n)$, by defn we have $x_n \leq \sup(x_n)$ however note that $\sup(x_n) - 1/n$ is not an upper bound hence, there exists a $n$ such that $\sup(x_n) - 1/n < x_n \leq \sup(x_n)$ and thus taking the limit as $n \to \infty$ we get by the squeezing property that $\lim_n \sup(x_n) < x \leq \lim_n \sup(x_n)$ so $\lim_n \sup(x_n) = x$?
Did I use the squeeze theorem properly? If not how can I do it using a similar method?

REPLY [0 votes]: This isn't correct. Let $S = \sup (x_n)$. For each $n$ there exists at least one term of the sequence in between $S-1/n$ and $S$.  This term probably isn't $x_n$, and in fact could be the same term for different $n$.
For instance, if $x_n = 2^{-n}$, then $S = 1/2$ but $x_n \to 0$.


TITLE: What is the cardinality of the set?
QUESTION [0 upvotes]: The set in question is $S=\Bbb{N}×A$, where $A$ is a set with infinitely many elements and $\Bbb{N}$ is the set of natural numbers.
I think cardinality of $S$ is the same as $A$ no matter whether $A$ is countable or not.
I have tried to give a bijection between them but no success.
Any help would be appreciated.

REPLY [3 votes]: This depends, as Steven Stadnicki said, on whether we assume the axiom of choice.
First, a quick observation: it's not hard to prove by transfinite induction that if $\alpha$ is an infinite ordinal, then $\vert\mathbb{N}\times\alpha\vert=\vert\alpha\vert$. Interestingly, more is true: for all infinite ordinals $\alpha$ we have $\vert\alpha\times\alpha\vert=\vert\alpha\vert$. This is harder to prove, but not much harder.
If we have the axiom of choice, then we're done: given $A$, the axiom of choice lets us prove that there is some ordinal in bijection with $A$; now just apply the previous paragraph.
If choice fails, however, things can get weird: it's consistent with ZF (= the usual axioms of set theory, but without choice) that there are very weird infinite sets! For example, there might be amorphous sets - these are infinite sets which can't be partitioned into two infinite disjoint subsets! It's easy to show that $\mathbb{N}\times A$ can't be amorphous, regardless of what $A$ is (HINT: think about evens vs. odds ...).

A side comment: if choice fails, then "$\vert A\times A\vert=\vert A\vert$" must fail for some infinite set $A$: the statement "for all infinite sets $A$, $\vert A\times A\vert=\vert A\vert$" is equivalent to the axiom of choice (this was proved by Zermelo). Meanwhile, the statement "$\vert A\times\mathbb{N}\vert=\vert A\vert$" is strictly weaker than the axiom of choice, although it isn't provable in ZF alone.


TITLE: The sum: $ 1+ {n \choose 1}\cos{\theta} + {n \choose 2}\cos{2\theta} + …. + {n \choose n}\cos{n\theta} $ equals
QUESTION [1 upvotes]: The sum:
$$ 1+ {n \choose 1}\cos{\theta} + {n \choose 2}\cos{2\theta} + …. + {n \choose n}\cos{n\theta} $$
equals

REPLY [3 votes]: It is $2^n cos^n(\theta/2) \cos(n\theta/2).$


TITLE: Are there any good Discrete Mathematics video online?
QUESTION [25 upvotes]: I want to learn discrete mathematics by reading book by myself but I find sometime it's very hard to understand what author trying to say.
I want to know, are there any good online video that teach discrete mathematic?

REPLY [1 votes]: Warning: Shameless self-promotion.
Discrete Math Playlist: https://youtube.com/playlist?list=PLtxJg53s2o0OcotIx1_uKY0gEjqq96sx6
Combinatorial Optimization Playlist: https://youtube.com/playlist?list=PLtxJg53s2o0PqIAEphtjIaqXTouaWYR0j
These are videos from two university courses in Discrete Math that I've taught in the past.


TITLE: Does aerosol particulate matter attenuate UVA more as compared to UVB?
QUESTION [3 upvotes]: I am trying to understand if winter aerosol particulate matter has a greater impact on UVB radiation as compared to UVA radiation. Would the attenuation be mainly due to scattering?
I am asking this question for latitudes less than 30 degrees north where UVB radiation is expected to reach ground even in winter.

REPLY [6 votes]: Firstly, a definition of what exactly is termed as particulate matter, according to the Australian Government Dept. of Environment and Energy page Particles. the are defined as being:

Airborne particles are sometimes referred to as 'particulate matter' or 'PM'. They include dust, dirt, soot, smoke, and liquid droplets. Some particles are large enough or dark enough to be seen as soot or smoke, while others are so small they can only be detected individually with a microscope

Particulate matter is further defined by their size (from the link above):

Particles can be classified on the basis of their size, referred to as
  their 'aerodynamic diameter'. 'Coarse particles' are those between 10
  and 2.5 micrometres (µm) in diameter; 'fine particles' are smaller
  than 2.5 µm; and 'ultrafine particles' are smaller than 0.1 µm. For
  comparison, the diameter of a human hair is 70 µm and this is seven
  times the diameter of the largest 'coarse particles'.

The classifications are often referred to as $PM_{10}$, $PM_{2.5}$ and $PM_{0.1}$ for 'course', 'fine' and 'ultrafine' particulate matter respectively.  Further classification can be made of the chemical toxicity etc - but that is beyond the scope of this answer.
The particulate matter species' sizes have been observed to be bimodal, as can be seen in the diagram from the Middle Tennessee State University (MTSU) webpage "Visibility and Light Scattering in the Atmosphere":

The point between the 2 peaks is approximately 2.5 µm
A comparison between the size fractions and other small entities is shown below, just to gain some perspective:

Image source: clinicalgate.com
Studying the effects of all aerosols, particularly particulate matter on incoming solar radiation has been of major interest to atmospheric sciences as it constitutes a major source of uncertainty in radiative models.  The research in how PM affects the UV (both UVA and UVB) has been, in part, a focus of many (like me) who are involved with photobiology.
There are several complications with PM, some species are emitted at a certain size, but due to chemical and other processes, they can grow into much larger secondary particulates. Additionally, some species have differing absorption of specific wavebands in the UV (and other wavelengths of light). So, this answer will concentrate on the size of the particles and their effect on the incoming UV radiation.
A key point made by the MTSU webpage (linked above) provides a general answer to one of your main questions ("Would the attenuation be mainly due to scattering?"):

Light scattering is caused by all atmospheric particles, but particles very near the same diameter as the wavelength of light scatter photons most effectively.

The MTSU webpage provides a comparison chart between electromagnetic radiation and some common types of aerosol/PM - of particular interest for this question is the solar radiation in the UV:

It is important to realise that scattering refers to light received by a particle and re-radiated at the same wavelength.
This relates to the main part of your question, whether aerosol particulate matter attenuates more in the UVA than in the UVB - (essentially wavelengths of about 295 nm and above and divided at 315 or 320 nm depending on what reference you refer to (IARC Working Group Monograph - Solar and Ultraviolet Radiation)).
Essentially, a general answer is - it depends on the size of the particulate aerosol present.  The process is summarised nicely by the MTSU webpage as:

Particles larger than the wavelength of light can scatter light through three processes: (1) diffraction, (2) refraction, and (3) phase shift. Photons of light can also be absorbed by particles and converted to internal molecular energy. The efficiency with which a particle can scatter light and the direction in which the incident light is redistributed are dependent on all four of these effects.

The type of scattering is summarised in the diagram below (from Pennsylvania State University):

Mie Scattering

"Mie scattering" suggests situations where the size of the scattering particles is comparable to the wavelength of the light, rather than much smaller or much larger.

Ratyleigh Scattering

is the (dominantly) elastic scattering of light or other electromagnetic radiation by particles much smaller than the wavelength of the radiation.

The TL:DR of this is that, in general, UVB tends to be scattered by particulate matter more than UVA.  - As emphasised by the graph (from MTSU) below:

Absorption is also a major factor when it comes to attenuation of UV, particularly soot or 'black carbon' which is every present, not only in urban environments.  However, there are many species that absorb ultraviolet radiation - this is largely dependent on the properties of the PM species itself - but in general, absorption affects UVB more, as can be seen in the SCHIMACHY graph below:

Image Source: Global Aerosol Direct Effect Retrieval from Passive Hyperspectral Measurements
So summing all this up:
The amount of attenuation due to scattering and absorption of UV (UVA and UVB) by particulate matter depends on the amount. size and the type of particulate matter present AND the intensity of incoming UV radiation (seasons and cloud cover etc).  In general, UVB - which is already filtered by the atmosphere, is subjected to more scattering and absorption (depending on the PM species) than UVA.


TITLE: If $H \subseteq S_n$ is a normal subgroup, and H contains a cycle $\alpha$ of length $r$, then prove $H$ contains every cycle of length $r$.
QUESTION [1 upvotes]: Textbook: A First Course in Abstract Algebra with Applications 3rd Rotman
If $H \subseteq S_n \text{ (the symmetric group) }$ is a normal subgroup, and H contains a cycle $\alpha$ of length $r$, then prove $H$ contains every cycle of length $r$.
Every example I have seen of a normal subgroup of $S_n$ have been alternating groups. So for my problem, can I assume that $H=A_n$?

REPLY [1 votes]: Here is an example with a 3-cycle to show how conjugating a 3-cycle actually works. It's just the same for an $r$-cycle, but there's more subscripts, and it's messier, but the idea is the same.
We want to compute $\sigma(a\ b\ c)\sigma^{-1}$.
First we suppose $\sigma(a) = k,\ \sigma(b) = m,\ \sigma(c) = n$.
Clearly, then, $\sigma^{-1}(k) = a,\ \sigma^{-1}(m) = b,\ \sigma^{-1}(n) = c$.
Case $1$: $x \not\in \{k,m,n\}$. Whatever $\sigma^{-1}(x)$ may be, it is neither $a,\ b$ nor $c$, because permutations are one-to-one.
Therefore, viewing $(a\ b\ c)$ as a function, we have:
$(a\ b\ c)(\sigma^{-1}(x)) = \sigma^{-1}(x)$, since $a,b,c$ are the only elements permuted by $(a\ b\ c)$.
Thus $\sigma(a\ b\ c)\sigma^{-1}(x) = \sigma((a\ b\ c)(\sigma^{-1}(x)) = \sigma(\sigma^{-1}(x)) = x$ for $x \not\in \{k,m,n\}$.
Case $2$: $x \in \{k,m,n\}$. First we apply $\sigma^{-1}$:
$k \mapsto a$
$m \mapsto b$
$n \mapsto c$ (we are doing "all three at once").
Next we apply $(a\ b\ c)$ to get $(a b c)\sigma^{-1}$:
$k \mapsto a \mapsto b$
$m \mapsto b \mapsto c$
$n \mapsto c \mapsto a$
Finally, we apply $\sigma$, to get $\sigma(a\ b\ c)\sigma^{-1}$:
$k \mapsto a \mapsto b \mapsto m$
$m \mapsto b \mapsto c \mapsto n$
$n \mapsto c \mapsto a \mapsto k$
So for all $x\not\in \{k,m,n\}$, we have $\sigma(a\ b\ c)\sigma^{-1}$ fixes $x$, and on the set $\{k,m,n\}$ we have that $\sigma(a\ b\ c)\sigma^{-1}$ acts as: $(k\ m\ n)$, that is:
$\sigma(a\ b\ c)\sigma^{-1} = (\sigma(a)\ \sigma(b)\ \sigma (c))$
So any conjugate of a $3$-cycle is another $3$-cycle (as I pointed out before, the "same idea" works for any $r$-cycle).
The question then becomes: given a particular $3$-cycle, say $(a\ b\ c)$, can I write any OTHER $3$-cycle, say $(k\ m\ n)$, as a conjugate of my first $3$-cycle? That is, can we find an appropriate $\sigma$? And the answer is yes, ANY $\sigma$ with $\sigma(a) = k, \sigma(b) = m$ and $\sigma(c) = m$ will do (choose the images of $\{1,2,\dots,n\} - \{a,b,c\}$ any way you like).
The upshot of all this is: a normal subgroup $H$ of $S_n$ contains ALL examples of any given cycle-type it contains-it is a union of conjugacy classes (for example, $A_n$ is the union of all cycles types that are even).

N.B.: If by $\sigma\tau$ one means "apply $\sigma$, then $\tau$", one uses $\sigma^{-1}(a\ b\ c)\sigma$ instead (the argument is the same, though).


TITLE: Algorithms to find irreducible polynomials of a given degree
QUESTION [8 upvotes]: I need to know what are the efficient algorithms to find all the irreducible polynomials of a given degree, say $d$ over a given finite field, say $\mathbb{F}_{p^n}.$
One way is to factorize the polynomial $x^{p^{dn}}-x$, which is the product of all irreducible polynomials whose degree divides $d$, using factorization algorithms and collect all the degree $d$ factors. But I guess we are doing some extra job here. Are there better algorithms to find all irreducible polynomials of degree $d$ ?
I also want to know about the algorithms to find one irreducible polynomial of a given degree over a given finite field.

REPLY [5 votes]: The last word on the second question is this paper of Couveignes and Lercier. The question is highly nontrivial.


TITLE: Prove that the arithmetic progressions intersect if and only if $\gcd(a, b) \mid (s − r)$.
QUESTION [0 upvotes]: Let $a,b,r,s$ be given constants. Prove that the arithmetic progressions
  $\{ax + r : x \in \Bbb Z\}$ and $\{by + s : y \in \Bbb Z\}$
  intersect if and only if $\gcd(a, b) \mid (s − r)$.

I supposed that the two intersect. Which gives $ax+r=by+s$. I believe that there's a point where they're equal, thus, some $x$ and $y$ exist such that $ax+r=by+s$ is true. Rearranging the equation yields you with $ax-by=s-r$. I know that $ax-by=\gcd(a,b)$. Now, I am just lost on how to compute the gcd(a,b).
Thank you for your time.

REPLY [2 votes]: You've pretty much done it. You just can't say that "$ax-by=\gcd(a,b)$" — that's not necessarily true. But it is always true $\gcd(a,b)$ divides any expression of the form $ap+bq$ (with integer $p$ and $q$, of course). Therefore, you've obtained that $\gcd(a,b)\mid ax-by=s-r$, finishing your proof in this direction. Note that you still have to prove the other direction of "if and only if".


TITLE: Existence of analytic function on a unit dsic (Converse of Schwaraz Pick Lemma )
QUESTION [0 upvotes]: (Schwarz - Pick Lemma) Suppose that $f$ is analytic on the Unit Disk $\triangle$ and satisfies the following two conditions :
(1) $|f(z) \leq 1$ for all z $\in \triangle$
(2) $f (a) = b $ for some $a , b \in \triangle$
Then
$$ |f'(a) | \leq \frac{1 - |f(a)|^2}{1 - |a|^2}$$
My question is.
$\bullet$  Is the converse of the Schwarz - Pick Lemma exist.
$\bullet$ Show that there exist an analytic function $ f : \triangle \rightarrow \triangle$ such that $f(1/2) = 0$ and $|f' (1/2)| \leq 4/3$ which satisfies the
$$ |f'(a) | \leq \frac{1 - |f(a)|^2}{1 - |a|^2}$$

REPLY [1 votes]: Please correct me if I misunderstand your question. But here it is: It seems like if you require $a$ to be $1/2$, then the condition on the estimate of $|f'(a)|$ is redundant with the condition $|f'(1/2)|\leq 4/3$. Also, just from the fact that if such an analytic function $f$ satisfying $f(1/2)=0$ exists, it is totally possible for us to derive an estimate for $f'(1/2)$. To see this, consider the map $\phi:$ $z\mapsto \frac{z+1/2}{1+z/2} \in \text{Aut}(\mathbb{D})$. Then the composition $$g:=f\circ\phi: \mathbb{D} \longrightarrow \mathbb {D}$$ is a holomorphic map that fixes $0$. So by Schwarz' lemma and chain rule, we have: $$|\phi'(0)||f'(\phi(0))|=|g'(0)|\leq 1.(*)$$ Note that $\phi(0)=1/2$. Hence, $|f'(1/2)|\leq 4/3$.
Now the question is, does such function $f$ exist? Sure, I will give two examples:

$f'(1/2)=4/3$ when the equality of $(*)$ takes place. That is when $g(z)=e^{i\theta}z$, from some $\theta$. Then $f=g\circ\phi^{-1}$, which means $f(z)=e^{i\theta}\frac{z-1/2}{1-z/2}$.
If you want strict inequality, just take any analytic function on the unit disk that isn't an automorphism, and takes $1/2$ to $0$. For example, $z\mapsto 2z/3-1/3$ would certainly do the job.

Hope this helps.


text:
\section{Some isomorphisms of combinatorial quantum groups}\label{section_isomorphisms}

In this section we will consider combinatorial quantum groups $\Astarn$ for $* \in \{b,b',s',c,p\}$. They are free products  or direct sums of known quantum groups. For $* \in \{b',s',c,p\}$ it turns out that their fusion rings are not free.

\begin{theorem}\label{theorem_isomorphisms}
We have the following isomorphisms of compact quantum groups (not necessarily preserving the fundamental corepresentation).
\begin{enumerate}
  \item $\Abn$ is isomorphic to $\textup{A}_\textup{o}(n - 1)$.
  \item $\Asprimen$ is isomorphic to the direct sum $\Asn \otimes C^*(\ZZ/{2\ZZ})$.
  \item $\Abprimen$ is isomorphic to the free product $\Abn * C^*(\ZZ/{2\ZZ})$.
  \item $\Apn$ is isomorphic to the free product $\Asn * \cont(\textup{S}^1)$.
  \item $\Acn$ is isomorphic to the free product $\Abn * \cont(\textup{S}^1)$.
\end{enumerate}
\end{theorem}

\begin{remark}
 Note that in the case $n \leq 3$ we have the isomorphisms  $\Asn \cong \cont{(\textup{S}_\textup{n})}$ and $\textup{A}_\textup{o}(1) \cong \cont(\{-1,1\})$. So the given descriptions can be further simplified.
\end{remark}

Theorem \ref{theorem_isomorphisms}(1) is proven by the following remark. Let $U \in \mat{n}{A}$ be an orthogonal matrix, i.e. $\overline{U} = U$ unitary, where $A$ is any unital $\Cstar$-algebra. Then $U$ is bistochastic if and only if the vector $(1,1, \dotsc,1)^\textup{t}$ is a right eigenvector and $(1,1, \dotsc,1)$ is a left eigenvector of $U$. If $T \in \Cmat{n}$ denotes any orthogonal matrix such that $T(1,0, \dotsc,0)^\textup{t} = (1/{\sqrt{n}},\dotsc,1/{\sqrt{n}})^\textup{t}$ , then an orthogonal matrix $U$ is bistochastic if and only if $T^\textup{t}UT$ is of block form with $1$ in the upper left corner and an orthogonal $(n - 1) \times (n - 1)$ matrix in the lower right corner. \\
The key observation for the rest of \ref{theorem_isomorphisms} is the following lemma.

\begin{lemma}
Let $* \in \{b', s', c, p\}$. The fundamental corepresentation of $\Astarn$ contains a one dimensional non-trivial corepresentation $U_z$ which fulfils $U_z \boxtimes \overline{U_z} \simeq 1$. If $* \in \{b', s'\}$ then $U_z \simeq \overline{U_z}$.
\begin{proof}
  Consider $* = b', s'$ first. The element $z = \sum_i{u_{ij}}$ is easily seen to be a unitary group-like element, so it corresponds to a one dimensional unitary corepresentation of $\Astarn$. Consider the group $\textup{S}_n \oplus \ZZ/{2\ZZ} \subset U_n$ as permutation matrices with entries $+1$ and $-1$. Let $U_{\textup{S}_n \oplus \ZZ/{2\ZZ}}$ be the canonical fundamental corepresentation of $\cont(\textup{S}_n \oplus \ZZ/{2\ZZ})$. Then the image of $z$ under the map $(\Astarn, U_*) \rightarrow (\cont(\textup{S}_n \oplus \ZZ/{2\ZZ}), U_{\textup{S}_n \oplus \ZZ/{2\ZZ}})$ is $-1$, so $z$ is non-trivial. \\
  For $* = p,c$ consider $z := \id_{\textup{S}^1}$ as coming from the copy of $\cont(\textup{S}^1)$. This copy is contained in $\Astarn$, since the trivial corepresentation is contained in the fundamental corepresentation of $\Abn$ and $\Asn$. \\
  Using the relations of $\Astarn$ we can check the rest of the claim by simple calculations.
\end{proof}
\end{lemma}

\begin{remark}
The last lemma shows, that the fusion rules of neither of the quantum groups $\Astarn$ for $\* \in \{b', s', c, p\}$ can be described by a free fusion ring. Actually in a free fusion ring any element $a \neq 1$ satisfies $a \cdot a^* \neq 1$. This gives two counterexamples to the conjecture that for $n \geq 4$ the fusion rules of all free orthogonal quantum groups  can be described by a free fusion ring, which was stated in \cite{banicavergnioux09}.
\end{remark}

\begin{remark}
The fundamental corepresentation of any matrix quantum group that has $(\Asn, U_\textup{s})$ as a sub quantum group cannot be the sum of more than two irreducible corepresentations. In particular the last lemma already gives a decomposition $U \simeq U_z \boxplus V$ with $U_z$ non-trivial and one dimensional and $V$ irreducible, where $U$ is the fundamental corepresentation of $\Astarn$.
\end{remark}

\begin{proof}[Proof of Theorem \ref{theorem_isomorphisms}]
 The isomorphism of (2) is given by $\Asn \otimes C^*(\ZZ/{2\ZZ}) \rightarrow \Asprimen: \: u^\textup{s}_{ij} \otimes 1 \mapsto u^{\textup{s}'}_{ij} \cdot z, \, 1 \otimes u_{\overline{1}} \mapsto z$. This map exists since $z$ is central in $\Asprimen$ as an easy calculation shows. The inverse map is given by \[\Asprimen \rightarrow \Asn \otimes C^*(\ZZ/{2\ZZ}): \: u^{\textup{s'}}_{ij} \rightarrow u^{\textup{s}}_{ij} \otimes u_{\overline{1}}.\]
 In order to prove (3) we use again an orthogonal matrix $T \in \Cmat{n}$ such that $T(1,0,...,0)^\textup{t} = (1/{\sqrt{n}},...,1/{\sqrt{n}})^\textup{t}$. Then a matrix $U \in \mat{n}{A}$ for some $\Cstar$-algebra $A$ satisfies the relations of $U_{\textup{b'}}$ if and only if $T^{\textup{t}}UT$ is a block matrix with a self-adjoint unitary in the upper left corner and an orthogonal $(n - 1) \times (n - 1)$ matrix in the lower right corner. This proves $\Abprimen \cong \textup{A}_\textup{o}(n - 1) * C^*(\ZZ/{2\ZZ}) \cong \Abn * C^*(\ZZ/{2\ZZ})$. \\
 The isomorphism of (4) is given by  \[\Asn * \cont(\textup{S}^1) \rightarrow \Apn: \: u^{\textup{s}}_{ij} \mapsto u^\textup{p}_{ij} \cdot z^*, \, \id_{\textup{S}^1} \mapsto z.\]
 The isomorphism of (5) is given by \[\Abn * \cont(\textup{S}^1) \rightarrow \Acn: \: u^{\textup{b}}_{ij} \mapsto u^\textup{c}_{ij} \cdot z^*, \, \id_{\textup{S}^1} \mapsto z.\]
 All the isomorphisms respect the comultiplication, since $z$ is group-like. Hence, they are isomorphisms of quantum groups.
\end{proof}

config: arxiv


TITLE: What is the lower bound for highly composite numbers?
QUESTION [10 upvotes]: if $x=d(n)$ is the number of divisors of $n$, what is the tightest lower-bound for $n$ only given $x$?
http://en.wikipedia.org/wiki/Highly_composite_number

REPLY [2 votes]: It would be nice to have an inequality $n \ge f(x)$. If the poser wants numerical results, here are two:
The least number having exactly x divisors is given by OEIS sequence http://www.oeis.org/A005179. It is a pretty wild function. The nice paper by Grost is recommended.
The least number having x (or more) divisors is given by the OEIS sequence http://www.oeis.org/A061799.


TITLE: Find/Prove Compact Subspace that is Non-Closed
QUESTION [0 upvotes]: Let $\tau = \{\varnothing\} \cup \{G \subseteq \mathbb R: \mathbb R-G \ $is finite $\} $ and  $(\tau,\mathbb R)$ is the topology of the cofinite space.
Find a compact set $K \subseteq \mathbb R $ that is not closed.
We have just started topology and I am a bit stumped.
My ideas: As I need to find a set $K \subseteq\mathbb R $ whereby $K^{C}$ is finite. I would naturally suggest $\mathbb R-\{0\}$ or excluding some other finite subset, but then I get trouble in proving that it is compact. Any hints?

REPLY [1 votes]: Take any countable subset $C$ of $\mathbb R$. Since it is neither $\mathbb R$ nor a finite set, it is not closed.
However, it is compact. Let $(A_\lambda)_{\lambda\in\Lambda}$ is an open cover of $C$. Take $c\in C$. Then there is a $\lambda_0\in\Lambda$ such that $c\in A_{\lambda_0}$. In particular, $A_{\lambda_0}\neq\emptyset$. SInce it is an open set, $\mathbb{R}\setminus A_{\lambda_0}$ is finite. So, only finitely many elements $c_1,\ldots,c_n$ of $C$ don't belong to $a_{\lambda_0}$. For each $k\in\{1,2,\ldots,n\}$, take $\lambda_k\in\Lambda$ such that $c_k\in A_{\lambda_k}$. Then $\bigl\{A_{\lambda_k}\,|\,k\in\{0,1,\ldots,n\}\bigr\}$ is a finite subcover of $(A_\lambda)_{\lambda\in\Lambda}$.


TITLE: Prove that $x_n$ converge if subsequences $(x_{2n})$, $(x_{2n+1})$ and $(x_{3n})$ converge
QUESTION [0 upvotes]: Let $x_n$ be a sequence of real numbers such that the subsequences subsequences $(x_{2n})$, $(x_{2n+1})$ and $(x_{3n})$ converge. Prove that $x_n$ converge. I want to prove that all the subsequences that converge, converge to the same point. In other words, I want to prove that their limit is the same. (If not, $x_n$ doesn't converge) Though, I don't know how to go about to proving $\lim(x_{2n})=\lim(x_{2n+1})$.

REPLY [0 votes]: $(x_{3n})$ has a subsequence which is a subsequence of $(x_{2n})$. $(x_{3n})$ also has another subsequence which is a subsequence of $(x_{2n+1})$, you are ready to claim that they all converge to the same value.


TITLE: Solve $e^{-x} < \frac{1}{x^2 + 1}$
QUESTION [1 upvotes]: I want to solve $e^{-x} < \frac{1}{x^2 + 1}$, assuming $x \in \mathbb{R}^+$. This can be solved via finding the solutions of $e^x>x^2+1$. This seems less troublesome, but I've found this inequality is very hard as well. I've tried everything I could think of, from logarithmic approaches to remove the exponential function to expressing $e^x$ as $\sum_{k=0}^\infty \frac{x^k}{k!}$. This last approach renders the inequality $-x^2 +\sum_{k=3}^\infty \frac{x^k}{k!} > 0$, which isn't any less tricky.
A last, desperate approach was to prove the last inequality via induction for $x \in \mathbb{N}$ and see if that could be extended to $\mathbb{R}$ somehow. Needless to say such long shot failed. How would one go about solving this inequality?
PD: This inequality is just a particular case of a more general inequality $e^{-\frac{|x|}{\alpha}}<\frac{\alpha}{x^2+\alpha}$, with $\alpha \in \mathbb{R}^+$.  The problem presented above is this inequality under the assumptions $\alpha = 1, x \in \mathbb{R}^+$. Any thoughts on this more general, and far more complicated, problem are also welcomed -though are not my main interest.
Thanks in advance.

REPLY [5 votes]: Let us define $f(x) := e^{x} - x^{2} - 1$. Then $f(0) = 0$ and $f'(x) = e^{x} - 2x > 0$ for every real number (due to this result). In particular, $f(x) > 0$ for every $x > 0$ (because $f$ is strictly increasing on $\mathbb{R}_{>0}$).
Hopefully this helps!


TITLE: Bounds of system of closed intervals
QUESTION [0 upvotes]: From Vladimir Zorich Analysis I:

if $I$ is any system of nested closed intervals, then
$$\sup \{a \in \mathbb{R} \mid [a, b] \in I \} = \alpha \leq \beta =
 \inf \{b \in \mathbb{R} \mid [a, b] \in I\}$$
and
$$[\alpha, \beta] = \bigcap\limits_{[a, b] \in I}[a, b]$$

By visualizing finite nested intervals on a line, the assertion above seems trivial. However, in order to mathematically show it, Cantor's intersection theorem was the first thing that came to my mind:
$$\exists c \in \mathbb{R}, c \in \bigcap\limits_{[a, b] \in I}[a, b]$$
In other words:
$$\exists c \in \mathbb{R}, a \leq c \leq b, \forall [a, b] \in I$$
Now if $I$ is finite (and by definition closed and bounded), then $\alpha$ and $\beta$ belong to their respective sets as maximum and minimum elements respectively and by Cantor's intersection theorem $\alpha \leq c \leq \beta \Leftrightarrow \alpha \leq \beta$. Let $c \in [\alpha, \beta]$ any point of our interval and $I_k = [a_k, b_k] \in I$ be any interval, then by definition, $a_k \geq \alpha \geq c \geq \beta \geq b_k \Leftrightarrow c \in I_k$, hence all intervals in $I$ contain all points of $[\alpha, \beta]$.
Question:
What happens when $I$ is infinitely nested?
I know by monotone convergence theorem sets $\{a \in \mathbb{R}\, \mid [a, b] \in I\}$ and $\{b \in \mathbb{R}\ \mid [a, b] \in I\}$ both converge to $\alpha$ and $\beta$ respectively. However how can I show that it still applies that $\exists c \in \mathbb{R} \alpha \leq c \leq \beta$ ?
Thank you!

REPLY [1 votes]: If no $c\in\mathbb R$ exists with $\alpha\leq c\leq\beta$ then we must have $\beta<\alpha$.
Taking $a_0\in A$ close enough to $\alpha$ and $b_1\in B$ close enough to $\beta$ we find $b_1<a_0$.
Then $b_0\in B$ and $a_1\in A$ must exist with $[a_i,b_i]\in I$ for $i=1,2$.
But then $[a_1,b_1]\cap [a_0,b_0]=\varnothing$ and a contradiction is found.