TITLE: Product of Primes
QUESTION [3 upvotes]: Let $\mathbb{P}$ denote the set of prime numbers. How would one evaluate
$$\prod_{p\in \mathbb{P}}\frac{p-1}{p}$$ I do not think that the fact that $$\prod_{n=2}^{\infty}\frac{n-1}{n}=\lim_{n\to\infty}\frac{1}{n}=0$$ can be applied, but it is worth noting. Additionally, if we take the log, we obtain
$$\log\left(\prod_{p\in \mathbb{P}}\frac{p-1}{p}\right)=\sum_{p\in\mathbb{P}}\log\left(\frac{p-1}{p}\right)=\sum_{p\in\mathbb{P}}\log(p-1)-\sum_{p\in\mathbb{P}}\log(p)$$

REPLY [0 votes]: You could also use Mertens' 3rd theorem which states that
$$
\lim_{n\to\infty}\log n\prod_{p\le n}\left(1-\frac1p\right)=e^{-\gamma}\;.
$$


TITLE: Approximating higher dimension step function
QUESTION [0 upvotes]: Let $s \in R^{n}$ (meaning $s$ is $n \times 1$ vector), where $n$ is the dimension of the vector. The ideal sliding term, $\nu$ is taken to be:
    \begin{equation}
    \nu = \frac{s}{\|s\|}
    \end{equation}
where $\| . \|$ is taken to be the Euclidean norm. When $n = 1$, the above equation represents a step function but with the value of at $s=0$ undefined.
So, my question is what other continuous approximations can I use to approximate the above equation such that all values are well-defined and the function is smooth? Notice that the approximation has to be able to hold true for vectors. Considering just the scalar case is not good enough for me. Ideally, I would the approximation to have the following characteristics:

differentiable with respect to all s
having a closed form expression
Isn't a piecewise function

I know this might sound a lot. I looking something like that so that I can use it to do Lyapunov analysis (stability theory). I have a feeling that there is something simple out there.
I have written a long description in PDF format. You can download it here. Any input is greatly appreciated. If there is anything unclear, feel free to ask me. Thank you!

REPLY [0 votes]: A first pass to get continuity: for $||s||\geq \epsilon$, $f(s) = s/||s||$.  For $||s||<\epsilon$,
$f(s) = \frac{s}{||s||} (e^{1/\epsilon^2-1/||s||^2})$
If you want differentiability you just need to fiddle with the $< \epsilon$ function to have a derivative of 0 in the increasing $||s||$ direction at $||s||=\epsilon$ (I'm not quite motivated enough to figure it out)
Alternatively if you're willing to accept your function never quite being correct
$f(s) = e^{-\epsilon/||s||^2} \frac{s}{||s||}$


TITLE: Computing $\displaystyle\int \frac{1}{x^{2}\sqrt{x^{2}+7}}$
QUESTION [0 upvotes]: Find the primitive function of $$\frac{1}{x^{2}\sqrt{x^{2}+7}}$$
Attempt. And my answer is
$$\arcsin \left( \frac{\sqrt{\left( x^{2}+7 \right)}}{\sqrt{7}} \right)$$
Why am I wrong? See link for whole calculation.
[Wolfram](http://www.wolframalpha.com/input/?i=integrate+1/(x%5E2sqrt(x%5E2+7)).

REPLY [0 votes]: Let, $x^2+7=t^2x^2\implies 2xdx=2xt^2dx+2tx^2dt\implies dx=\frac {xt dt}{1-t^2}$
Given integral, $I=\int\frac { \frac {xt dt}{1-t^2}}{\frac {7}{t^2-1}.xt}=-\int\frac {dt}{7}=- \frac t7+c=\sqrt {1+\frac {7}{x^2}}+c$
And Jacob obviously you are wrong since $\frac {\sqrt {x^2+7}}{\sqrt 7}=\sqrt {1+\frac{x^2}{7}}\geq 1$ So $\arcsin \frac {\sqrt {x^2+7}}{\sqrt 7}$ seems to be undefined !!


TITLE: What are the logical underpinnings of the epsilon- delta definition of limits?
QUESTION [3 upvotes]: I'm having trouble getting my head around the epsilon-delta defintion of limits. I learned about conditional statements and I know that in order for a conditional to be true , one of the following cases have to hold

both antecedent and consequent are true, or
both are false, or
the consequent is true and antecedent is false.

Now in proving the definition of a limit say
$$\lim_{x \rightarrow 4} x^2 + x -11 = 9 $$
which case do we prove? Is it the same for every limit or are different limits proved by different cases?
How does the concept of conditional truth-tables relate to this definition?
Thank you in advance.

REPLY [2 votes]: This is a good question!
Strictly speaking, there is a logical implication here. But once you understand that we only care about one case, your job becomes much easier (and less confusing). Why do we only care about one case?
Another way to look at the statement "$Q$, whenever $P$" (equivalently, "If $P$, then $Q$") is that the statement is true if

Both $P$ and $Q$ are true, or if
$P$ (the antecedent) is false.

The latter case, when the antecedent is false, we often say that the statement "$Q$, whenever $P$" is vacuously true (and I think of it as "true by a technicality", or "true by default"). It's similar to the statement "If I go to the store, I'll bring you something". You wouldn't consider me a liar if I don't go to the store (regardless of whether I've brought you anything), would you? It's essentially the same kind of "true by a technicality." If I don't go to the store, then you don't even need to ask "Have you brought me anything?" because I technically couldn't possibly have lied to you.

Aside: My thoughts are that the logic we use in calculus isn't really the same kind of logic that's done in a purely logical setting; at least, we don't have the same goals. We really don't care about implications with false antecedents because we're only interested in using implications (and properties defined with implications, like continuity) to deduce things. For that we need modus ponens, which requires our antecedents to be true.
For this reason, I think a truth table will generally be nearly useless in a calculus setting.

So, when confronted with the task of showing that

for all $\epsilon > 0$, there exists a $\delta >0$ such that $\lvert f(x) - 9 \rvert < \epsilon$ whenever $\lvert x - 4 \rvert < \delta$,

we do a few things:

Pick a generic $\epsilon > 0$.
Find some $\delta > 0$ so that when we assume $\lvert x - 4 \rvert < \delta$, we can show that $\lvert f(x) - 9 \rvert < \epsilon$.

Step 2. is where all the work happens. It usually involves using some unknown $\delta$, assuming $\lvert x - 4 \rvert < \delta$, and seeing what kind of relationship we can derive, algebraically, between $\epsilon$ and $\delta$.
(Notice that I said "we assume $\lvert x - 4 \rvert < \delta$". This goes back to the fact that, if $\lvert x - 4 \rvert \nless \delta$, then our antecedent is false, and the "$Q$, whenever $P$" statement involving the two inequalities is true by default -- there's literally nothing to show, so we don't even bother talking about it.)
Once we have our $\delta$, actually showing that $\lvert f(x) - 9 \rvert < \epsilon$ using our $\delta$ and $x$ such that $\lvert x - 4 \rvert < \delta$ can happen almost as if by magic: We generally don't even need to say how we found such a $\delta$ (although we had to go through the algebra in step 2. to do so), only that we've found one, and shown that it works.


TITLE: If an abelian category $\mathcal{A}$ has enough injectives then so does $\mathrm{Ch}^{\geq 0}(\mathcal{A})$
QUESTION [3 upvotes]: Well my question is as clear as its title suggests. So here I would like to clarify on the fact that an object $A^\cdot$ in $\mathrm{Ch}^{\geq 0}(\mathcal{A})$ is injective if and only if
$0\longrightarrow\mathrm{H}^0(A^\cdot)\longrightarrow A^0\longrightarrow A^1\longrightarrow\cdot\cdot\cdot$ is exact and $\mathrm{H}^0(A^\cdot)$ and all $A^n$'s are injectives in $\mathcal{A}$.
Many thanks in advance!

REPLY [6 votes]: If $X^{\bullet}$ is an object of $\mathrm{Ch}^{\geq0}(\mathcal{A})$ and for each $i\geq0$ $X^i\to I^i$ is an embedding into an injective, then $X^{\bullet}$ embeds in
$$I^{\bullet}:=\dots\to 0\to I^0\oplus I^1\to I^1\oplus I^2\to I^2\oplus I^3\to\dots$$
(with the obvious maps as differentials). $I^{\bullet}$ is the product (and coproduct) of cochain complexes $J_0^{\bullet}:=\dots0\to I^0\to0\to\dots$ and $J_i^{\bullet}:=\dots0\to I^i\to I^i\to0\to\dots$ for $i>0$. Since products of injective objects are injective, it suffices to prove that $J_i^{\bullet}$ is injective for all $i\geq0$, which is straightforward since for any object $Y^{\bullet}$ of $\mathrm{Ch}^{\geq0}(\mathcal{A})$, $\operatorname{Hom}(Y^{\bullet},J_i^{\bullet})=\operatorname{Hom}_{\mathcal{A}}(Y^i,I^i)$.


TITLE: Give all the nodes on a tree a positive number $v_i$ such that no adjacent nodes share the same number, what is the minimum $\sum v_i$?
QUESTION [0 upvotes]: Suppose you have to give all the nodes on a tree a positive number $v_i$ such that no adjacent nodes share the same number, what is the minimum $\sum v_i$?

I came across this problem on a programming platform. I had solved it but I had a few doubts concerning it.

When I was solving this problem, I thought that $\max(v_i) = 3$ were always enough but apparently there were some cases where $\max(v_i)$  had to be $4$. Could anyone provide an example of such a tree?
Is there a bound on $\max(v_i)$? If yes, why?

If this is a well-known problem (especially on a tree), please provide me with some reference links cuz I can't find any. Thanks in advance.
Example of $\max(v_i)$ being $3$:

REPLY [4 votes]: For every $n$, there exists a tree $T_n$ such that every sum-minimizing assignment assigns $n$ to the root.
This is clear for $n=1$. Assume we know this to be true for all $k<n$. Build a tree starting with a root $v$ and adding $n-k+1$ copies of each $T_k$, $k<n$, as children. Then any assignment of some $k<n$ to $v$ implies that the copies of $T_k$ cannot have $k$ at their roots, hence cannot be optimal, hence have their sum at least $1$ too big. Then the sum for the whole tree is at least $(n-k+1)\cdot 1+(k-n)=1$ more than what we could obtain by assigning $n$ to $v$ and minimizing assignments to all subtrees.
Following this construction, the folowing trees have $1,2,3,4$ at their root (top node) in every minimizing assignment:


TITLE: Is it necessary that if a limit exists at a point it should be also defined at that point?
QUESTION [9 upvotes]: Say there exists a limit $\lim_{x \to x_0}f(x) = L$. Is it necessary that $f$ be defined at the point $x_0$ itself?
Well, what I think of it is that it's OK to be undefined at that point because I guess that won't cease the limit of that function to exist there would it?

REPLY [28 votes]: "No", in a particularly strong sense: The very reason limits came into mathematics was to extract numerical values from difference quotients
$$
\frac{f(x) - f(x_{0})}{x - x_{0}}
$$
(which are, of course, algebraically indeterminate at $x_{0}$), in the limit as $x \to x_{0}$.
The formal definition of $L = \lim(f, x_{0})$ ("For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - x_{0}| < \delta$, then $|f(x) - L| < \varepsilon$") explicitly avoids evaluation of $f$ at $x_{0}$, precisely because $f$ might be undefined at $x_{0}$.


TITLE: If matrix $T$ has rank $1$ then it can be written as $T=\underline{uv}^T$
QUESTION [1 upvotes]: In my textbook there was the following statement:

If matrix $T$ has rank $1$ then it can be written as $T=\underline{{u}{v}}^T$.

I'm trying to figure out why it's true. For example:
$$
T=\begin{bmatrix}1 & 2 & 3\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
$$
The matrix $T$ has rank $1$ but how do I re-write it as $T=\underline{u}\underline{v}^T$? what is the general decomposition for any $T$?

REPLY [2 votes]: If $T:V\to W$ has rank $1$, then its image has dimension $1$, meaning that it is spanned by a single vector $u$. So $Tx=\lambda(x) u$, where $\lambda(x)$ is a scalar depending on $x$. And since $T$ is linear, $\lambda:V\to K$ ($K$ being the field we're working over) also has to be linear. But then it has a matrix representation $\lambda (x)=\Lambda x$, where $\Lambda$ is an $n\times 1$ matrix, $n$ being the dimension of $V$. In other words, it's a row vector, so $\Lambda=v^T$ for some $v\in V$. Then we have
$$\lambda(x)=\Lambda x=v^Tx.$$
Putting everything together:
$$Tx=\lambda(x)u=u\lambda(x)=uv^Tx.$$
So in the end, $T=uv^T$, where $u$ is a basis vector of $\operatorname{im} T$, and $v^T$ is chosen such that $uv^Tx=Tx$ for all $x\in V$. To find its components, you have to calculate $Tb_i$ for every basis vector $b_i$ of $V$. If the basis is orthonormal (like the standard basis of $\mathbb R^n$), then
$$Tb_i=uv^Tb_i=u v_i,$$
which will allow you to obtain the $i$-th component $v_i$ of $v$ in the basis $\{b_i~\vert~i=1,\dots,\dim V\}$.


text:
\begin{document}

\markboth{ALEXANDER I. NESTEROV } {SMOOTH LOOPS AND FIBER BUNDLES:
THEORY OF PRINCIPAL Q-BUNDLES}

\catchline{}{}{}{}{}

\title{SMOOTH LOOPS AND FIBER BUNDLES: THEORY OF PRINCIPAL Q-BUNDLES}

\author{ALEXANDER I. NESTEROV }

\address{ Departamento de F{\'\i}sica, CUCEI, Universidad de
Guadalajara, Av. Revoluci\'on 1500\\ Guadalajara, CP 44420, Jalisco,
M\'exico\\
\email{nesterov@cencar.udg.mx } }

\maketitle

\begin{history}
\received{(Day Month Year)} \revised{(Day Month Year)}
\end{history}
\begin{abstract}
A nonassociative generalization of the principal fiber bundles
with a smooth loop mapping on the fiber is presented. Our approach
allows us to construct a new kind of gauge theories that involve
higher ''nonassociative'' symmetries.
\end{abstract}

\keywords{Smooth loops; Quasigroups; Gauge loops; Nonassociative
Gauge Theories; Fiber bundles.}

Mathematical Subject Classification 2000: 81S99, 55R65, 53Z05, 57R20

\section{Introduction}

During the last few decades, nonassociative structures have been
employed in various fields of modern physics. Among others, one
may mention the rise of nonassociative objects such as
3-cocycles, which are linked with violations of the Jacobi identity in
anomalous quantum field theory, and quantum mechanics with the Dirac
monopole, the appearance of Lie groupoids and algebroids in the context
of Yang-Mills theories, and the application of nonassociative algebras
to gauge theories on commutative but nonassociative fuzzy spaces
\cite{Jac,G2,Gr,Gr1,G1,N1a,N3,NF,N1,NI,Strobl,Ram,MR}.

Nonassociative algebraic structures such as quasigroups and loops have
considerable potential interests for mathematical physics, especially in view
of the appearance of nonassociative algebras, such as the Mal'cev algebra
\cite{Mal}, related to the problem of the chiral gauge anomalies, the emergence of a
nonassociative electric field algebra in a two-dimensional gauge theory, and so
on \cite{C,Jo,NSem,Sem,Jac}. Quasigroups and loops have recently been
employed in general relativity and for the description of the Thomas precession,
coherent states, geometric phases, and nonassociative gauge theories
\cite{NI,N2,N4,N5,NS3,NS3a,NS2,NS1,U1,U2,K}.

In this paper, we give a detailed account of nonassociative fiber bundles
leading to ``nonassociative'' gauge theories \cite{N1,NI,N2}. In general terms,
a consequence of nonassociativity is that the structure constants of the gauge
algebra have to be changed by structure functions.

The paper is organized as follows. In Section 2, we outline the basic
constructions of the smooth loops theory. In Section 3,
the theory of nonassociative principal fiber bundles ($Q$-bundles)
is introduced.

\section{Smooth Loops}

\subsection{Basic notations }

Here, we outline the main results on the algebraic theory of quasigroups
and loops. The details may be found in \cite{Bel,Bruck,Pfl,Ch,Sab1}.

Let $\langle Q,{\mathbf \cdot}\rangle $ be a groupoid, i.e. a set
with a binary operation $(a,b) \rightarrow{} {a \mathbf \cdot b}$. A
groupoid $\langle Q,{\mathbf \cdot}\rangle $ is called a {\it
quasigroup} if each of the equations $a{\mathbf \cdot}
x=b,~y{\mathbf \cdot} a=b$ has unique solutions:  $x=a\backslash b$,
$y=b/a$. A {\it loop} is a quasigroup with a two-sided identity
$a{\mathbf \cdot} e= e{\mathbf \cdot} a=a, \forall a \in Q$. A loop
$\langle Q,{\mathbf \cdot},e\rangle $ with a smooth operation
$\phi(a,b):=a{\mathbf \cdot} b$ is called the {\it smooth loop}. We
define
\begin{eqnarray}
&L_a b=R_b a=a{\mathbf \cdot} b,\quad
l_{(a,b)}=L^{-1}_{a{\mathbf \cdot} b}\circ L_a\circ L_b, \nonumber\\
& \hat l_{(a,b)}= L_a\circ L_b \circ L^{-1}_{a{\mathbf \cdot} b} ,
\quad r_{(b,c)} = R^{-1}_{b{\mathbf \cdot}c}\circ R_c \circ R_b,
\label{Ll}
\end{eqnarray}
where $L_a$ is a {\it  left translation}, $R_b$ is a {\it right
translation}, $l_{(a,b)}$ is a {\it left associator}, $\hat
l_{(a,b)}$ is an {\it adjoint associator} and $r_{(b,c)}$ is a {\it right
associator}.

Let $T_e(Q)$ be the tangent space of $Q$ at the neutral element $e$. Then
for each $X_e \in T_e(Q)$, we construct a smooth vector field on $Q$
\[
X_b = L_{b\mathbf  \ast } X_e, \quad b\in Q, \quad X_e \in T_e(Q),
\; X_b \in T_b(Q)
\]
where $L_{b\mathbf \ast }: T_e(Q)\rightarrow{} T_b(Q)$ denotes the
differential of the left translation. Notice that $X_b$ depends smoothly on both variables $b \in Q$ and $X_e \in T_e(Q)$, and satisfies
\begin{eqnarray}
L_{a\mathbf  \ast }X_b = \hat l_{(a,b){\mathbf \ast} } X_{a{\mathbf
\cdot} b}.
\end{eqnarray}

\begin{definition}
A vector field $X$ on $Q$ that satisfies the
relation $L_{a\mathbf  \ast }X_b =\hat l_{(a,b)\mathbf  \ast }
X_{a{\mathbf \cdot} b}$ for any $a,b \in Q$ is called the {\it left
fundamental} or {\it  left quasi-invariant} vector
field.
\end{definition}

Let ${V}$ denote the set of left fundamental vector
fields on $Q$. It becomes a vector space under the operations
\begin{eqnarray}
&&(X + Y)_a = X_a + Y_a, \quad X,Y \in {V},\; a\in Q\nonumber,\\
&&(\alpha X)_a = \alpha X_a , \quad \alpha \in {\mathbb R}.
\nonumber
\end{eqnarray}
The following lemma is evident.

{\begin{lemma} The vector space of left fundamental vector fields is
isomorphic to the tangent space $T_e(Q)$ at the neutral element. The
isomorphism is defined by the map $L^{-1}_{a\mathbf  \ast}$:
$X_a\rightarrow{} X_e, \; X_e\in T_e(Q), \; X_a\in
T_a(Q)$.
\end{lemma}}

The vector space $V$ can be equipped with a Lie commutator. This leads us
to the notion of {\it quasialgebra}.

{\begin{definition} We define a {\it quasialgebra} $\mathfrak q$ on
$Q$ as the vector space of left fundamental vector fields, under the
Lie commutator operation.\end{definition}}

Let $\Gamma_i =R^j_i\partial/\partial a^j, \; i=1,2, \dots, r$ be a basis of the space of  left fundamental vector fields. Then, we have
\begin{eqnarray}
[\Gamma_i, \Gamma_j] = C^p_{ij}(a) \Gamma_p
\end{eqnarray}
where $C^p_{ij}(a)$ are the {\em structure functions} satisfying the
modified Jacobi identity
\begin{eqnarray}
C^p_{ij,n}R^n_k + C^p_{jk,n}R^n_i + C^p_{ki,n}R^n_j
+ C^l_{ij}C^p_{kl} +  C^l_{jk}C^p_{il} + C^l_{ki}C^p_{jl}=0.
\label{eq2}
\end{eqnarray}

In view of the noncommutativity of the right and left translations,
there exists a problem in the definition of the `adjoint' map of
$Q$ on itself. Indeed, in the case of the group $G$, the adjoint
map is given by
\[
Ad_g g' = g g' g^{-1} \equiv L_g \circ R^{-1}_g g' \equiv
 R^{-1}_g \circ L_g g'.
\]

For quasigroups, the left and right translations do not commute, $L_a
\circ R_b  \neq R_b \circ L_a $, and it follows that the definitions
$Ad_g g' = L_g \circ R^{-1}_g g'$ and $\tilde Ad_g g' =R^{-1}_g
\circ L_g g'$ are not equivalent. We introduce a generalized
adjoint map of $Q$ on itself in the following way.
{\begin{definition} The map
\begin{eqnarray}
Ad_b(a)= L^{-1}_a\circ R^{-1}_b\circ L_{a{\mathbf \cdot}b}: \;
Q\rightarrow{} Q \label{A_1}
\end{eqnarray}
is called the {\it Ad-map.} \end{definition}}

{\begin{remark} \rm The Ad-map $Ad_b(a)$ leaves invariant the
neutral element $e$ and generates the map $T_e(Q) \rightarrow{}
T_e(Q)$ as follows:
\begin{eqnarray}
{\sf Ad}_b(a):=\bigl(Ad_b(a)\bigr)_{\mathbf \ast}= L^{-1}_{a\ast}
R^{-1}_{b\ast} L_{a{\mathbf \cdot}b \ast}.
\end{eqnarray}
It is easy to check that $Ad_b(e)= Ad_b:=R^{-1}_b\circ
L_b$.\end{remark}}

{\begin{definition} The vector-valued 1-form
$\mbox{\boldmath$\omega$}$ defined through the relation
\begin{eqnarray}
\mbox{\boldmath$\omega$}(V_a)=V_e \quad {\rm where}  \quad V_e
=L^{-1}_{a^{\mathbf \ast} } V_a, \quad V_e \in T_e(Q), \; V_a \in
T_a(Q)
\end{eqnarray}
where $V$ is a left fundamental vector field, is called the {\em
canonical} $ \sf Ad$-form. \end{definition}}

{\begin{theorem} The canonical $\sf Ad$-form ${\omega}$ is a left
fundamental form and is transformed under left (right) translations as follows:
\begin{eqnarray}
&(L^{\mathbf \ast}_b\omega)V_a=l_{(b,a)\mathbf \ast}
\omega (V_a) \label{AL} \\
&(R^{\mathbf \ast} _b\omega)V_a={\sf Ad}^{-1}_b(a) \omega(V_a).
\label{AR}
\end{eqnarray}
\label{Th1}\end{theorem}}

{\begin{proof} For the left translations, we obtain
\[ (L^{\mathbf \ast} _b\omega)V_a=\omega(L_{b\mathbf  \ast
}V_a)=L^{-1}_{b{\mathbf \cdot} a\ast}L_{b\mathbf  \ast } L_{a\mathbf
\ast } V_e = l_{(b,a)\mathbf  \ast }\omega (V_a),
\]
thus (\ref{AL}) is true.

Similarly, for the right translations, we find
\begin{eqnarray}
(R^{\mathbf \ast} _b\omega)V_a=&\omega(R_{b\mathbf  \ast }V_a)
=L^{-1}_{a{\mathbf \cdot} b\mathbf  \ast }R_{b\mathbf  \ast }
L_{a\mathbf
\ast }L^{-1}_{a\mathbf  \ast } V_a \nonumber  \\
= &L^{-1}_{a{\mathbf \cdot} b\mathbf  \ast } R_{b\mathbf  \ast }
L_{a\mathbf  \ast }\omega (V_a) ={\sf Ad}^{-1}_b(a) \omega(V_a)
\end{eqnarray}
and the proof is completed. \end{proof}}

\subsubsection{Examples}

The examples given below play an important role in the description of generalized
coherent states, Thomas precession and nonassociative geometry \cite{N3,NS3,NS3a,NS2,NS1,U1,U2}.

\begin{example}
Loop $\mathbf \langle {\mathbb R}/{\mathbb
Z},\mathbf \ast\rangle $. Let $\ast$ be the operation defined on
${\mathbb R}/{\mathbb Z}$ as follows:
\begin{eqnarray}
&&x\ast y = x+y + f(x) +f(y) - f(x+y)
\end{eqnarray}
where $f(x)= (1-\cos 2\pi x)/4 $. Then $\langle {\mathbb
R}/{\mathbb Z}, \ast\rangle $ is the analytical loop \cite{H1,H2}.
\end{example}

\begin{example}
Loop $\rm Q\mathbb C \label{ejemplo}$. Let
${\mathbb C}$ be a complex plane and $\zeta,\eta\in{\mathbb C}$.
Then, the loop $\rm Q\mathbb C $ is obtained by introducing on
${\mathbb C}$ the nonassociative multiplication defined as follows
\begin{eqnarray}
R_\eta^\zeta=L_\zeta \eta=\zeta\mathbf
\cdot\eta=\frac{\zeta+\eta}{1-\overline\zeta\eta}, \quad
\zeta,\eta\in{\mathbb C}, \label{eq5}
\end{eqnarray}
where a bar denotes complex conjugation. The inverse operation is is found to
be
\begin{eqnarray}
L^{-1}_\zeta \eta=\frac{\eta-\zeta}{1+\overline\zeta\eta}, \quad
\zeta,\eta\in{\rm Q\mathbb C}. \label{eq6}
\end{eqnarray}
The associator $l_{(\zeta,\eta)}$ is given by
\begin{eqnarray}
l_{(\zeta,\eta)}\xi=\frac{1-\zeta\overline\eta}{1-\eta\overline\zeta}\xi
\end{eqnarray}
and can also be written as $l_{(\zeta,\eta)}= \exp({\rm i}\alpha),$ where $
\alpha =2\arg(1-\zeta\overline\eta)$.

The basis of left quasi-invariant vectors and the dual basis of 1-forms
are found to be
\begin{eqnarray}
\Gamma_1 = (1 + |\eta|^2)\partial_\eta,
\quad \Gamma_2 = (1 + |\eta|^2)\partial_{\bar\eta}, \\ \theta^1 =
\frac{d\eta}{1 + |\eta|^2}, \quad \theta^2 = \frac{d\bar\eta}{1 +
|\eta|^2}.
\end{eqnarray}
The computation of the commutator yields
\begin{eqnarray}
[\Gamma_1,\Gamma_2] = \bar\eta\Gamma_2 - \eta\Gamma_1,
\end{eqnarray}
And, for the structure functions, we find $C^1_{12}= -\eta, C^2_{12}=
\bar\eta$.

The loop thus obtained is related to the two-sphere $S^2$, which admits a natural quasigroup
structure. Namely, $S^2$ is the local two-parametric Bol loop $QS^2$
\cite{N3,N1,NS2,NS1}. The isomorphism between points of the sphere
and the complex plane ${\mathbb C}$ is established by the
stereographic projection from the south pole of the sphere:
$\zeta=e^{\rm i\varphi}\tan(\theta/2)$.
\end{example}

\begin{example}
Loop $QSU(2)$. Let us consider a set of the unitary
matrices $QSU(2)\subset$ $SU(2)$, where an arbitrary element of
$QSU(2)$ has the form
\begin{displaymath}
U=
\left ( \begin{array}{cc}
\alpha & \beta    \\
- \bar\beta & \bar\alpha
\end{array}
\right ), \quad |\alpha|^2 +  |\beta|^2 =1, \quad \alpha =\bar\alpha.
\end{displaymath}

It is convenient to set
\[
\alpha = \frac{1}{\sqrt{1+ |\eta|^2}}, \quad
\beta  = \frac{\eta}{\sqrt{1+ |\eta|^2}},
\]
and write $U$ as
\begin{eqnarray}
U_{\eta} = \frac{1}{\sqrt{1+ |\eta|^2}}
\left ( \begin{array}{cc}
1 & \eta    \\
- \bar\eta & 1
\end{array}
\right ).
\label{QS}
\end{eqnarray}

For arbitrary elements of $QSU(2)$, we define the nonassociative binary operation
as follows:
\begin{eqnarray}
U_{\eta}\ast U_{\zeta} = U_{\eta} U_{\zeta} \Lambda (\eta, \zeta)
\label{U},
\end{eqnarray}
where, on the right-hand side, the matrices are multiplied in the usual way, and
\begin{displaymath}
\Lambda (\eta, \zeta) =
\left ( \begin{array}{cc}
e^{i\varphi} & 0    \\
0           & e^{-i\varphi}
\end{array}
\right ),
\quad \varphi = 2\arg(1 - \bar\eta \zeta).
\end{displaymath}
The set of matrices (\ref{QS}) with the operation $\ast$ is called the loop
$QSU(2)$.

The loop $QSU(2)$ forms the so-called nonassociative representation of
$QS^2$. Indeed, writing  Eq. (\ref{U}) as
\begin{eqnarray}
U_{\eta}\ast U_{\zeta} = \frac{1}{\sqrt{1+ |\eta{\mathbf
\cdot}\zeta|^2}} \left ( \begin{array}{cc}
1 & \eta {\mathbf \cdot}\zeta   \\
- \overline{\eta {\mathbf \cdot}\zeta}& 1
\end{array}
\right ),
\label{QS1}
\end{eqnarray}
we find (see Ex. 9)
\begin{eqnarray}
L_{\eta}\zeta \equiv\eta{\mathbf
\cdot}\zeta=\frac{\eta+\zeta}{1-\overline\eta\zeta}, \quad
l_{(\eta,\zeta)}\xi =\frac{1- \eta\overline\zeta}
{1-\overline\eta\zeta}\xi, \quad \zeta,\eta,\xi\in{\mathbb C}.
\end{eqnarray}
\end{example}
\begin{example}
Loop $QH^2$. This loop is associated with the group
$SU(1,1)$ and its action on the two-dimensional unit hyperboloid
$H^2$. Let $D\subset\mathbb C$ be the open unit disk,
$D=\{\zeta\in\mathbb C:\left|\zeta\right|<1 \}$. We define the
nonasociative binary operation $\ast$ as
\begin{eqnarray}
\zeta\ast\eta=\frac{\zeta+\eta}{1+\bar\zeta\eta}, \quad\zeta,\eta\in D.
\end{eqnarray}
The associator $l_{(\zeta,\eta)}$ on $QH^2$ is determined by
\begin{eqnarray}
l_{(\zeta,\eta)}\xi=\frac{1+\zeta\overline\eta}{1+\eta\overline\zeta}\xi,
\end{eqnarray}
and can also be written as $l_{(\zeta,\eta)}=\exp({\rm i}  \alpha), \;
\alpha =2\arg(1+\zeta\overline\eta)$. Inside the disk $D$, the set of
complex numbers with the operation $\ast$ forms a two-sided loop $QH^2$, which
is isomorphic to the geodesic loop of two-dimensional Lobachevsky space
realized as the upper part of the two-sheeted unit hyperboloid. The
isomorphism is established by $\zeta=e^{\rm i\varphi}\tanh(\theta/2)$,
where $(\theta,\varphi)$ are inner coordinates on $H^2$.

The basis of left quasi-invariant vectors and the dual basis of 1-forms are
found to be
\begin{eqnarray}
\Gamma_1 = (1 - |\eta|^2)\partial_\eta, \quad
\Gamma_2 = (1 - |\eta|^2)\partial_{\bar\eta}, \\
\theta^1 = \frac{d\eta}{1 - |\eta|^2}, \quad
\theta^2 = \frac{d\bar\eta}{1 - |\eta|^2},
\end{eqnarray}
and the computation of the commutator gives
\begin{eqnarray}
[\Gamma_1,\Gamma_2] = \eta\Gamma_1 - \bar \eta\Gamma_2.
\end{eqnarray}
\end{example}

\begin{example} Loop $\rm Q{\sf H}_{\mathbb R}$. Let us consider
the quaternionic algebra over the complex field ${\mathbb C}(1,\rm i)$
\[ {\sf H}_{\mathcal C} = \{ \alpha + \beta i+ \gamma j + \delta k\;
\mid \; \alpha , \beta,\gamma,\delta \in
 {\mathbb C} \} \] with the multiplication operation defined by the
property of bilinearity and following rules for $i, \; j, \; k$:
\begin{eqnarray}
&&i^{2}=j^{2}=k^{2} =-1,\;\; jk=-kj = i, \nonumber\\
&&ki=-ik=j,\;\; ij=-ji= k.\nonumber
\end{eqnarray}
For $q=\alpha + \beta i+ \gamma j + \delta k$, the quaternionic conjugation
is defined as follows:
\begin{equation}
q^+=\alpha - \beta i -\gamma j - \delta k.
\end{equation}

Furthermore, we restrict ourselves to the set of quaternions
 ${\sf H}_{\mathbb R}\subset {\sf H}_{\mathcal C}$:
\begin{eqnarray*}
{\sf H}_{\mathbb R} =\{\zeta=\zeta^0 + {\rm i}(\zeta^1 i+ \zeta^2 j+
\zeta^3 k):\;{\rm i}^2=-1,\;\rm i\in{\mathbb C},\;
\zeta^0,\zeta^1,\zeta^2,\zeta^3  \in {\mathbb R} \},
\end{eqnarray*}
with the norm $\|\zeta\|^2$ given by
\begin{equation}
\|\zeta\|^2=\zeta {\zeta}^+=(\zeta^0)^2 - (\zeta^1)^2-(\zeta^3)^2 -
(\zeta^4)^2.
\end{equation}

Introducing a binary operation
\begin{equation}
\zeta\ast\eta=\big(\zeta+\eta\big)\big/\big(1 + \frac{K}{4}\zeta^+\eta \big),
\quad \zeta,\eta \in {\sf H}_{\mathbb R}, \label{QL}
\end{equation}
where $K$ is constant and $/$ denotes the right division, we find that the
set of quaternions ${\sf H}_{\mathbb R}$ with the binary operation $\ast$
forms a loop, which we denote by Q${\sf H}_{\mathbb R}$. The associator is found
to be
\begin{equation}
l_{(\zeta,\eta)}\xi= \big(1+\frac{K}{4}\zeta\eta^+ \big)\xi
\big/\big(1+\frac{K}{4}\zeta^+\eta \big).
\end{equation}

The loop Q${\sf H}_{\mathbb R}$ is related to the spacetime of constant
positive curvature $K>0$ ({\em de Sitter spacetime}), which is locally
characterized by the condition \cite{NS3}
\begin{equation}
R_{\mu\nu\lambda\sigma}= K(g_{\mu\lambda}g_{\nu\sigma} -
g_{\mu\sigma}g_{\nu\lambda}). \label{R}
\end{equation}
\end{example}

\subsection{Infinitesimal theory of smooth loops in brief}

It is well known that the infinitesimal theory of Lie groups arises from the
associativity of the operation
\[
a{\mathbf \cdot}(b{\mathbf \cdot} c)=(a{\mathbf \cdot} b){\mathbf
\cdot} c.
\]
In what follows, we will show that the quasiassociativity identities
\begin{equation}
R_c\circ R_b a=R_{R^a_c b}a, \quad L_a \circ L_b c = L_{(a{\mathbf
\cdot})}\circ l_{(a,b)}c
\end{equation}
lead to the infinitesimal theory of smooth loops. We start with the definition of a quasigroup of transformations introduced by Batalin \cite{Bat}.

\begin{definition} Let $\mathfrak M$ be an n-dimensional manifold and let the
continuous law of transformation be given by $x'=T_a x , \quad  x\in
{\mathfrak M}$, where \{$a^i$\} is the set of real parameters,
$i=1,2,\dots,r$. The set of transformations $\{T_a\}$ forms a
$r$-parametric quasigroup of
transformations (with right map on ${\mathfrak M}$), if \\
1) there exists a unit element that is common for all $x^i$ and corresponds to
$a^i=0:\;T_a x|_{a=0}=x$;\\
2) the modified composition law holds:
\[
T_aT_b x =T_{\varphi(b,a;x)} x;
\]
3) the left and right units coincide:
\[
\varphi(a,0;x)=a, \quad \varphi(0,b;x)=b;
\]
4) the modified law of associativity is satisfied:
\[
 \varphi(\varphi(a,b;x),c;x)=\varphi(a,\varphi(b,c;T_a x);x);
\]
4) the inverse transformation of $T_a$ exists: $x=T^{-1}_a x'$.
\end{definition}

The generators of infinitesimal transformations
\[
\Gamma_i=(\partial(T_ax)^i/\partial a^i)|_{a=0}
\partial/\partial x^\alpha \equiv R^\alpha_i\partial/\partial x^\alpha
\]
form a {\it quasialgebra} and obey the commutation relations
\begin{equation}
[\Gamma_i, \Gamma_j] =C^p_{ij}(x)\Gamma_p, \label{eq1}
\end{equation}
where $C^p_{ij}(x)$ are the structure functions satisfying the modified Jacobi
identity
\begin{eqnarray}
C^p_{ij,\alpha}R^\alpha_k + C^p_{jk,\alpha}R^\alpha_i
+C^p_{ki,\alpha}R^\alpha_j

+ C^l_{ij}C^p_{kl} + C^l_{jk}C^p_{il} + C^l_{ki}C^p_{jl}=0 . \label{eq2a}
\end{eqnarray}

{\begin{theorem}
Let the given functions $R^\alpha_i, \;C^p_{kj}$
obey the equations (\ref{eq1}), (\ref{eq2}). Then, locally the
quasigroup of transformations can be reconstructed as the solution of
the set of differential equations
\begin{eqnarray}
\frac{\partial\tilde x^\alpha}{\partial a^i} = R^\alpha_j(\tilde
x)\lambda^j_i(a;x),\quad
\tilde x^\alpha(0)=x^\alpha, \label{eq3} \\
\frac{\partial\lambda^i_j}{\partial a^p} -\frac{\partial\lambda^i_p}{\partial
a^j} + C^i_{mn}(\tilde x)\lambda^m_p\lambda^n_j=0, \quad
\lambda^i_j(0;x)=\delta^i_j \label{eq4}.
\end{eqnarray}
\end{theorem}}

Eq. (\ref{eq3}) is an analog of the Lie equation, and Eq.
(\ref{eq4}) is the generalized Maurer-Cartan equation.

The Batalin approach can be easily extended to the case of smooth loops
if we consider the action of the loop on itself. Indeed, for smooth loops, the modified associativity law can be written as follows:
\begin{equation} \varphi^i(\varphi (a,b),c)=\varphi^i(a,\tilde \varphi (b,c;a)), \quad
\varphi^i(a,0)=a^i, \quad \varphi^i(0,b)=b^i. \label{ID_1}
\end{equation}
The derivation of Eq.(\ref{ID_1}) in $e$ with respect to $c^j$  yields
\begin{equation}
\left.\frac{\partial\varphi^i(\varphi (a,b),c)}{\partial c^j} \right| _e =
\left.\frac{\partial\varphi^i(a,b)}{\partial b^k}
\frac{\partial\tilde\varphi^k(b,c;a)}{\partial c^j}\right| _e \label{Lie1}
\end{equation}
Let us introduce
\begin{equation}
\alpha^k_j(b;a)=\left.\frac{\partial \tilde\varphi^k(b,c;a)}{\partial c^j}
\right|_e,\quad \omega^i_p \alpha^p_j=\delta^i_j, \quad  q^i_p=
\left.\frac{\partial \varphi^i(a,b)}{\partial b^p} \right|_e. \label{Def}
\end{equation}
Substituting (\ref{Def}) in (\ref{Lie1}), we obtain an analog of Lie equation
(see Eq.(\ref{eq3}))
\begin{equation}
\frac{\partial\varphi^i(a,b)}{\partial b^j} =
q^i_p(\varphi(a,b))\omega^p_q(b;a). \label{Lie_1}
\end{equation}
Actually, the parametric operation $\varphi(b,c;a)$ is not independent, and it can be written in the terms of left translations as follows:
\[
\varphi(b,c;a) = L_b \circ l^{-1}_{(a,b)}c.
\]
This yields
\[
\omega^p_q(b;a) = [l_{(a,b),\ast,e}]^p_r \omega^r_q(b;e).
\]
Taking into account that $\omega^r_q(b;e)= \omega^r_q(b)$, we obtain the
generalized Lie equation deduced by Sabinin \cite{Sab1}
\begin{equation}
\frac{\partial\varphi^i(a,b)}{\partial b^j}  =
q^i_p(\varphi(a,b))[l_{(a,b),\ast,e}]^p_r \omega^r_q(b).
\end{equation}

\section{Principal Q-bundles}

\subsection{Basic definitions and examples}

Let $\mathfrak M$ be a manifold and $\langle Q, {\mathbf \cdot},
e\rangle$ be a smooth two-sided loop.  A {\em principal loop Q-bundle}
(a principal bundle with structure loop Q) is a triple $(P,\pi,
M)$, where $P$ is a manifolds and the following conditions hold:

(1) $Q$ acts freely on $P$ by the right map: $(p,a)\in P\times Q
\mapsto{} \tilde R_a p \equiv pa \in P$.

(2) $\mathfrak M$ is the quotient space of $P$ by the equivalence relation
induced by $Q$, $\mathfrak M=P/Q$, and the canonical projection  $\pi:
P\rightarrow M$ is a smooth map.

(3) $P$ is {\em locally trivial}, i.e. any $x\in  M$ has a
neighborhood $U$ and a diffeomorphism $\Phi$ of
$\pi^{-1}(U)\mapsto{} U\times Q$ such that, for any point $p\in
\pi^{-1}(U)$, it has the form $\Phi(p)=(\pi(p),\varphi(p))$, where
$\varphi$ is the map from $\pi^{-1}(U)$ to $Q$ satisfying
$\varphi(\tilde R_a p) = R_a \varphi(p)$.

We say that $P$ is a {\em total space} or {\em bundle space}, $Q$ is
the {\em structure loop}, $M$ is the {\em base} or {\em base
space} and $\pi$ is the {\em projection}. For any $x\in M$, the
inverse image $\pi^{-1}(x)$ is the {\em fiber} $F_x$ over $x$. Any
fiber is diffeomorphic to $Q$, and a transitive map of the loop $Q$
is defined on each fiber so that
\[
\tilde R_{b}\circ\tilde R_{a} p= \tilde R_{a{\mathbf \cdot}b}\tilde
r_{(a,b)}p,\quad
\]
where the map $\tilde r_{(a,b)}$ satisfies $\varphi(\tilde r_{(a,b)}p)=
r_{(a,b)}\varphi(p)$. The neutral element acts as the identity map.

Let us consider a covering $\{ U_\alpha\}_{\alpha \in J}$ of $ \mathfrak  M$
with $\bigcup_{\alpha \in J} U_\alpha =  \mathfrak M $, which can be chosen in
such a way that the restriction of the fibration to each open set $U_\alpha$ is
{\em trivializable }. This means that there exists a diffeomorphism
$\Phi_\alpha$: $\pi^{-1}(U_\alpha) \rightarrow{} U_\alpha\times Q$. The sets
$\{U_\alpha,\Phi_\alpha\}$ are called a family of {\em local trivializations}.
It is easy to prove the following \cite{N1,N2}.

{\begin{proposition} The right map on the fiber does not depend on
the trivialization: $\Phi^{-1}_\alpha
\big(\pi(p),\varphi_\alpha(\tilde R_q p) \big ) = \Phi^{-1}_\beta
\big(\pi(p),\varphi_\beta (\tilde R_q p) \big )$. \end{proposition}
}

{\begin{definition} Let $\{U_\alpha, \Phi_\alpha\}$ be a local
trivialization and $e\in Q$ a neutral element; then,
$\sigma_\alpha=\Phi^{-1}_\alpha(x,e),\; x\in U_\alpha$ is called the
{\em canonical local section} associated with this
trivialization.\end{definition}}

Let $u\in P$; then, one can describe the local section as follows:
$\sigma_\alpha = \tilde R^{-1}_{q_\alpha(u)}u$, where we set $q_\alpha(u)
= \varphi_\alpha (u).$

{\begin{proposition} The section $\sigma_\alpha$ does not depend on the choice of the point in the fiber.\end{proposition}}

{\begin{proof} Let $u,p\in P$ and $u = \tilde
R_{q_\alpha(u)}\sigma_\alpha$, $p = \tilde
R_{q_\alpha(p)}\sigma'_\alpha$. There exists an element $a\in Q$
such that $p=\tilde R_a u$. Then we have
\begin{eqnarray}
p= \tilde R_{q_{\alpha}(p)}\sigma'_{\alpha}= \tilde R_a u =\tilde
R_a \circ \tilde R_{q_\alpha(u)}\sigma_\alpha = &
\varphi^{-1}_\alpha\bigl(R_a \circ R_{q_\alpha(u)}
\varphi_\alpha(\sigma_\alpha) \bigr) \nonumber \\
= &\varphi^{-1}_\alpha\bigl(R_a\circ R_{q_\alpha(u)}e \bigr)= \tilde
R_{q_\alpha {\mathbf \cdot} a}\sigma_\alpha.
\end{eqnarray}
Taking into account that $q_{\alpha(p)} = q_{\alpha(u)}{\mathbf
\cdot}a$, we obtain
$R_{q_{\alpha}(p)}\sigma'_{\alpha}=R_{q_{\alpha}(p)}\sigma_{\alpha}$.
This yields $\sigma_\alpha=\sigma'_\alpha$, and, hence, the section
$\sigma_\alpha$ does not depend on the point of the fiber, while the
dependence on the base, $\sigma_\alpha=\sigma_\alpha(x)$, may still
hold. \end{proof}}

{\begin{proposition} Let the point $x$ lie in the intersection of the
neighborhoods $U_\alpha$ and $U_\beta$, $x\in U_\alpha\cap U_\beta$.
Then, the following formula holds:
\begin{eqnarray}
\sigma_\alpha(x)=\tilde R_{q_{\beta\alpha}}\sigma_\beta(x).
\label{sigma}
\end{eqnarray}\end{proposition}}

{\begin{proof} Let $u=\tilde R_{q_\alpha}\sigma_\alpha$, rewriting
this formula in the chart $U_\beta$, we obtain
\begin{eqnarray}
\Phi_\beta\bigl(\tilde R_{q_\alpha}\sigma_\alpha\bigr) =&\Phi_\beta\bigl(
\tilde R_{q_\beta}\sigma_\beta\bigr) =\bigl(\pi(u),R_{q_\beta}\varphi_\beta
(\sigma_\beta) \bigr)
\nonumber \\
=&\bigl(\pi(u),R_{q_{\beta\alpha}{\mathbf \cdot}q_\alpha}
\varphi_\beta(\sigma_\beta) \bigr)=  \bigl(\pi(u),R_{q_\alpha}\circ
R_{q_{\beta\alpha}}e \bigr) \nonumber \\
= &\bigl(\pi(u),R_{q_\alpha}\circ R_{q_{\beta\alpha}} \varphi_\beta(\sigma_\beta
)\bigr) =\Phi_\beta\bigl(\tilde R_{q_\alpha}\circ \tilde
R_{q_{\beta\alpha}}\sigma_\beta\bigr), \nonumber
\end{eqnarray}
where the relation $\varphi_\beta(\sigma_\beta)=e$ has been used.
Finally, we obtain $\sigma_\alpha(x)=\tilde
R_{q_{\beta\alpha}}\sigma_\beta(x)$ in the intersection
$U_\alpha\cap U_\beta$ . \end{proof}}

{\begin{definition} The family of maps $\{q_{\beta\alpha}(p) =
R^{-1}_{q_\alpha}q_\beta: \;\pi^{-1}( U_\alpha \cap
U_\beta)\rightarrow {} Q\}$, where $q_\alpha: =
{\varphi_\alpha(p)},\; q_\beta: = \varphi_\beta(p),\; p\in \pi^{-1}(
U_\alpha \cap U_\beta)$, is called the family of {\em transition
functions} of the bundle ${P({ M},Q)}$ corresponding to the
trivializing covering $\{U_\alpha,\Phi_\alpha\}_{\alpha \in J}$ of
$\mathfrak M$.\end{definition}}

{\begin{proposition} The transition functions $q_{\beta\alpha}(p)$
change under right translations as
\[
R_a q_{\beta\alpha}(p) = r_{(q_\alpha,a)}q_{\beta\alpha}(p)
\]
where $r_{(q_\alpha,a)}= R^{-1}_{q_\alpha{\mathbf \cdot}a}\circ R_a
\circ R_{q_\alpha}$ is the right associator.\end{proposition}  }

{\begin{proof} According to the definition, we have
\begin{eqnarray}
R_a q_{\beta\alpha}(p) = q_{\beta\alpha}(pa)
=R^{-1}_{q_\alpha{\mathbf \cdot} a} \circ R_a q_\beta
=R^{-1}_{q_\alpha{\mathbf \cdot} a} \circ R_a \circ
R_{q_\alpha}\circ R^{-1}_{q_\alpha}q_\beta=
r_{(q_\alpha,a)}q_{\beta\alpha}(p). \nonumber
\end{eqnarray}
\end{proof}}

We find that $q_{\beta\alpha}$ is changed under the right
translations, and, therefore, it depends on the points of the fiber:
$q_{\beta\alpha}=q_{\beta\alpha}(p),\; p\in F_x ,\;x \in U_\alpha
\cap U_\beta$. The transition functions obey the so-called {\it cocycle condition}:
\begin{eqnarray}
&&q_{\beta\alpha}(p){\mathbf \cdot}q_\alpha(p)= q_{\beta\gamma}(p)
{\mathbf \cdot} \bigl(q_{\gamma\alpha}(p) {\mathbf \cdot}
q_\alpha(p)\bigr)     \\
&&\forall x \in U_\alpha \cap U_\beta \cap U_\gamma \; {\rm and}\; \forall
p \in F_x , \nonumber
\end{eqnarray}
which may also be written as
\begin{eqnarray}
&&q_{\beta\alpha}{\mathbf \cdot}q_\alpha= q_{\beta\gamma} {\mathbf
\cdot} q_{\gamma\alpha}{\mathbf \cdot} \bigl(l_{(q_{\beta\gamma},
q_{\gamma\alpha})} q_\alpha \bigr). \nonumber
\end{eqnarray}

{\begin{definition} Let $P({  M},Q,\pi)$ be a principal loop
$Q$-bundle and $\{U_\alpha\}$ some covering of the base $\mathfrak
M$. Then we say that the system $\{ { \mathfrak
M},Q,\{U_\alpha\},q_{\alpha\beta} \}$ is a {\em principal coordinate
$Q$-bundle}\end{definition}}

Assume that there are two families of local sections, $\sigma_\alpha$ and
$\sigma'_\alpha$. These sections are linked by the right translations. This
implies that $\exists q_\alpha \in Q: \sigma'_\alpha =
R_{q_\alpha}\sigma_\alpha.$ Using the set $\{\sigma_\alpha\}$, let us
introduce the new transition functions $q'_{\alpha\beta}$. Then, the
following transformation law holds:
\[
q'_{\alpha\beta}=L^{-1}_{q_{\alpha}}\circ R_{q_\beta}q_{\alpha\beta}.
\]

\begin{example}
Principal $QS^2$ bundle over $S^1$. Here, we will
show that $S^3$ is the $QS^2$ bundle over $S^1$. We consider the
unit-three sphere $S^3$ as a subset of $\mathbb C^2$:
\begin{eqnarray}\label{eq7}
S^3 =\left\{(z_1,z_2)\in  \mathbb C^2:\; |z_1|^2+|z_2|^2=1 \right\}.
\end{eqnarray}
We define the map $\pi: S^3 \rightarrow S^1 $  by
\begin{eqnarray}\label{eq8}
\pi(z_1,z_2) =\bigg (\frac{\bar z_1 + z_1}{2\sqrt{1-|z_2|^2}}, \frac{ z_1 -
\bar z_1}{2i\sqrt{1-|z_2|^2}},0\bigg ).
\end{eqnarray}
Parameterizing $S^3$ by
\begin{eqnarray*}
z_1 = \cos\frac{\theta}{2}\,e^{i\psi_1}, \quad z_2
=\sin\frac{\theta}{2}\,e^{i\psi_2}
\end{eqnarray*}
where $0\leq \theta \leq \pi$ and $\psi_1,\psi_2 \in \mathbb R$, we get
\begin{eqnarray}\label{eq9}
\pi\big(\cos\frac{\theta}{2}\,e^{i\psi_1},
\sin\frac{\theta}{2}\,e^{i\psi_2}\big)= (\cos\psi_1,\sin\psi_1,0),
\end{eqnarray}
and one can see that $\pi$ indeed maps $S^3$ to $S^1$. By this means, one can see that locally
$S^3 \approx S^1 \times S^2$; however, this is not true globally since the spaces $S^1 \times S^2$ and $S^3$ have different second homotopy groups \cite{G3}:
\begin{eqnarray}\label{eq9a}
  \pi_2(S^1 \times S^2) =\mathbb Z, \quad  \pi_2(S^3) = 0.
\end{eqnarray}

Let $U_{-}=(-\pi,\pi)$ and $U_{+}=(0,2\pi)$ be an open covering of $S^1$. We define the
local trivializations, $ \Phi_{\pm}:  \pi^{-1}(U_{\pm})\rightarrow U_{\pm}\times QS^2 $,
as follows:
\begin{eqnarray}\label{eq10}
\Phi_{-}\big(\cos\frac{\theta}{2}\,e^{i\psi_1},
\sin\frac{\theta}{2}\,e^{i\psi_2}\big)
= \big(e^{i\psi_1}, \tan\frac{\theta }{2}\,e^{i(\psi_2-\psi_1)}\big) \\
\Phi_{+}\big(\cos\frac{\theta}{2}\,e^{i\psi_1},
\sin\frac{\theta}{2}\,e^{i\psi_2}\big) = \big(e^{i\psi_1},
\tan\frac{\theta +\pi/2}{2}\,e^{i(\psi_2-\psi_1)}\big).
\end{eqnarray}

Let us set
  \begin{eqnarray}\label{eq11}
  \zeta_{-} = \tan\frac{\theta }{2}\,e^{i(\psi_2-\psi_1)} ,\quad
  \zeta_{+} = \tan\frac{\theta +\pi/2}{2}\,e^{i(\psi_2-\psi_1)},
  \end{eqnarray}
then the transition function relative to our trivialization determined by
\begin{eqnarray}
   &q_{+-}: U_{-}\cap U_{+} \rightarrow QS^2
\end{eqnarray}
and
\begin{eqnarray} \label{eq12}
      & \zeta_{+}= L_{q_{+-}} \zeta_{-} = \displaystyle\frac{q_{+-} + \zeta_{-}}{1 - \bar q_{+-}\zeta_{-}},
\end{eqnarray}
is given by $q_{+-} = e^{i(\psi_2-\psi_1)}$, and we have $(\cos\psi_1,\sin\psi_1,0)\rightarrow\, e^{i(\psi_2-\psi_1)}\in QS^2$.

Since the structure loop $QS^2$ acts on the standard fiber $S^2$ by left
translations (\ref{eq12}), we indeed have a $QS^2$ principal bundle over $S^1$.
For the right translatons of $QS^2$ on $S^3$, we find
\begin{eqnarray}\label{eq14}
    \tilde R_{\eta}(z_1,z_2)  = \Bigg(\frac{z_1|1-\bar \eta z_2/z_1|}{\sqrt{1+ |\eta|^2}},
    \frac{z_2(1+ \eta z_1/z_2)|1-\bar \eta z_2/z_1|}{(1-\bar \eta z_2/z_1)\sqrt{1+
    |\eta|^2}}\Bigg), \quad \eta\in QS^2.
\end{eqnarray}
\end{example}

\begin{example}
Principal $QS^2$ bundle over $S^2$. We construct
the principal $QS^2$ bundle over $S^2$ by taking
\begin{tabbing}
\qquad Base ${\mathfrak  M} = S^2$ \qquad \= with coordinates
$0 \leq \theta \leq \pi, \; 0 \leq \varphi < 2\pi$, \\ \kill \\
\qquad Fiber $QS^2$ \> with coordinates $ \zeta \in {\mathbb
C}$.
\end{tabbing}
We split $S^2$ into two hemispheres $H_{\pm}$ with
$H_{+}\cap H_{-}$ being  a thin strip parameterized by the
equatorial angle $\varphi$. Thus, locally the bundle can be described
as
\begin{tabbing}
\qquad $H_{-}\times QS^2$  \qquad \= with coordinates
$(\theta,\varphi,
\zeta_{-})$,\\ \kill \\
\qquad $H_{+}\times QS^2$ \qquad \= with coordinates
$(\theta,\varphi,
\zeta_{+})$. \\
\end{tabbing}
The transition functions must be elements of the loop $QS^2$ in order to give a
principal $Q$-bundle. We choose to relate the $H_{+}$ and $H_{-}$  fiber
coordinates in $H_{+}\cap H_{-}$  by the left map (\ref{eq12})
\begin{eqnarray}\label{eq15}
L_{q_{+-}} \zeta_{-} = \frac{q_{+-} + \zeta_{-}}{1 - \bar q_{+-}\zeta_{-}}.
\end{eqnarray}
Since $q_{+-}$ maps $S^2$ to $S^2$, it is classified by $\pi_2(S^2)= \mathbb Z$.
This implies that
\begin{eqnarray}
\zeta_{+} =\big( L_{q_{+-}}\big)^n\zeta_{-},
\label{eq16}
\end{eqnarray}
and the power $n$ of $L_{q_{+-}}$ must be an integer to give a well-defined
manifold. Let us define $q^n_{+-}$ by the following relation:
$\big( L_{q_{+-}}\big)^n = L_{q^n_{+-}}$; then, we get
\begin{eqnarray}
q^n_{+-} = \frac{\zeta_{+}(1- \zeta_{-}\overline\zeta_{+})
-\zeta_{-}(1- \overline\zeta_{-}\zeta_{+})}
{1 - |\zeta_{-}|^2|\zeta_{+}|^2}.
\end{eqnarray}
Setting $q_{+-}=e^{i\gamma}\tan(\theta/2)$ and using Eq.(\ref{eq16}) ,
we obtain $q^n_{+-}=e^{i\gamma}\tan(n\theta/2)$.
\end{example}

\subsubsection{Associated bundles}

In this section, we will discuss a method of constructing {\em quasigroup}
fiber bundles that are associated with some principle $Q$-bundle. Let $F$ be a
space on which $Q$ acts (on the left) as the loop of transformations. The basic
idea is to construct for a particular principal $Q$-bundle $P({\mathfrak M},Q)$
a fiber bundle with fiber $F$. First, we need the concept of a $Q$-{\em product}
of a pair of spaces on which $Q$ acts by the right mappings.

{\begin{definition} Let $X$ and $Y$ be any pair of right $Q$-spaces.
Then, the $Q$-{\em product} of $X$ and $Y$ is the space of orbits
of the $Q$-mappping on the Cartesian product $X\times Y. $
\end{definition}}

This definition implies that there is defined an equivalence relation on
$X\times Y$ in which $(x,y) \equiv (x',y')$ iff $\exists q\in Q$ such that $x'=
R_{\displaystyle q} x$ and $y' = R_q y$. The $Q$-product is denoted $X\times_Q
Y$.

{\begin{definition} Let $P({\mathfrak M},\pi,Q)$ be a principal
$Q$-bundle and let $F$ be a manifold on which $Q$ acts by left maps.
Define $P_F :=P\times_Q F$ and the map $Q$ on $P_F$ as follows
\[
(p,\xi)\rightarrow (R_a p, \tilde L^{-1}_a \xi), \quad p\in P,\; \xi \in F,
\]
where $\tilde L_a$ denotes the left map on $F$. Then, $E({\mathfrak
M},Q,F,P) =P\times_Q F/Q$ is the {\em associated} fiber bundle over
$\mathfrak M$ with fiber $F$.
\end{definition}
The differential structure over $E({\mathfrak M},Q,F,P)$ is defined
as follows. Let $\pi_E$ be the projection map $E\rightarrow
{\mathfrak M}: (u,\xi) \rightarrow \pi(u)$, that is induced by the
projection $\pi: P\times {\mathfrak M}\rightarrow \mathfrak M$. For
any $x\in {\mathfrak M}$, the set $\pi^{-1}_E (x)$ is called the {\em fiber}
in $E$ over $x$ and the $Q$-map on $\pi^{-1}_E (U)\times F$ is
defined by $(x,a,\xi)\rightarrow (x,ab,\tilde L^{-1}_b \xi)$, where
$(x,a,\xi)\in U\times Q\times F, \; b\in Q, U\subset {\mathfrak M}$
and $\pi^{-1}_E (U)$ is the open manifold in $E$: $\pi^{-1}_E (U)
\approx U\times F$. Then, the projection $\pi_E^{-1}$ is a differentiable
mapping $E\rightarrow {\mathfrak M}$. A map $\sigma: {\mathfrak
M}\rightarrow E$, such that $\pi_E\circ \sigma= {\rm Id}$ is the identity map ${\mathfrak M}\rightarrow {\mathfrak M}$, is called a
section of the fiber bundle $E({\mathfrak M},Q,F,P)$. }

\subsection{Connection, curvature and Bianchi identities}

\subsubsection{Connection on principal $Q$-bundles}

Let $P({  \mathfrak M},Q)$ be a principal $Q$-bundle over the manifold $
\mathfrak M$. For any $u\in P$, we denote a tangent space at $u$ by $T_u(P)$ (or
$T_u$), and the tangent to the fiber passing through $u$ by ${\mathcal V}_u$. We
call ${\mathcal V}_u$ a {\em vertical subspace}. It is generated by the right
translations on the fiber: $ u \rightarrow{}  \tilde R_a u, \; a\in Q, \; u\in
P$:
\begin{eqnarray}
X_u=\left.\frac{d}{dt}\tilde R_{a(t)} u\right |_{t=0}, \quad X_u\in
{\mathcal V}_u, \; a\in Q, \; u\in P. \label{ver0}
\end{eqnarray}

Let $\gamma(t)\in   \mathfrak M$ be a smooth curve. A {\em horizontal lift} of
$\gamma$ is a curve $\tilde\gamma(t)\in P$ such that
$\pi(\tilde\gamma(t))=\gamma(t)$. Evidently, to determine $\tilde\gamma(t)$,
it is sufficient to define at any point a tangent vector $\tilde X:
\pi_\ast \tilde X = X$, where $X$ is the tangent vector to $\gamma(t)$. A set
$\{ X\}$ is called a {\em horizontal subspace} ${\mathcal H}_u$.

{\begin{definition} A {\em connection} on a principal $Q$-bundle is
a smooth assignment to each point $p\in P$ of a horizontal subspace
${\mathcal H}_p$ of $T_p(P)$ such that:

(i) $ T_p={\mathcal V}_p \oplus{\mathcal H}_p$ (a direct sum).

(ii) The family of horizontal subspaces is invariant under the right
map of the loop $Q$; i.e., for any $a\in Q, \;p\in P$
\[
{\mathcal H}_{\tilde R_a p}= (\tilde R_a)_\ast {\mathcal H}_ p,
\]
where $\tilde R_{a\ast}$ denotes the right map by $a$ on a tangent
vector. \end{definition}}

The connection allows us to decompose any vector $Z\in T_p(P)$ in
the form $Z=X+Y$, where $X=hor Z \in {\mathcal H}_p$ is the
horizontal component of the vector $Z$ and $Y=ver Z \in {\mathcal
V}_p$ is the vertical one. The map $\varphi$ induces the map of the
vertical subspace ${\mathcal V}_p$ onto the tangent space to $Q$,
${\mathcal V}_p \stackrel{\varphi_\ast}\longrightarrow T_q(Q),\;
q=\varphi(p)$:
\begin{eqnarray}
{\tilde V}_p=\left.\frac{d}{dt}\tilde R_{a(t)} p\right |_{t=0}
\stackrel{\varphi_\ast}{\longrightarrow}
{\hat V}_q=\left.\frac{d}{dt}R_{ a(t)} q\right |_{t=0} .
\label{ver1}
\end{eqnarray}
Notice that the vector field $\hat V$ is the left quasi-invariant
vector field. Indeed, it can be written as $\hat V_q=(L_{q})_\ast
V_e$, where
\begin{eqnarray}
V_e=\left. \frac{da(t)}{dt}\right|_{t=0} \in T_e(Q).
\label{ver2}
\end{eqnarray}

\begin{definition} Let $ V_e \in T_e(Q)$. The vector field $\tilde
V$ connecting with ${V}_e$ by means of (\ref{ver1}), (\ref{ver2}) is
called a {\em fundamental} vector field.
\end{definition}

The notion of connection may be reformulated in the following way:

{\begin{definition} A connection form on a principal $Q$-bundle is a
vector-valued 1-form taking values at $T_e(Q)$, which satisfies:

(i) $\omega(X_p)= X_e$, where $X_p\in {\mathcal V}_p, \; X_e\in
T_e(Q)$ are determined according to {\rm (\ref{ver1}), (\ref{ver2})}.

(ii)$ \bigl(\tilde R^\ast_a \omega\bigr)X_p =
{\sf Ad}^{-1}_a(q) \omega(X_p)$, where $q=\varphi(p)$.

(iii) The horizontal subspace ${\mathcal H}_p$ is defined as a
kernel of $\omega$:
\[
{\mathcal H}_p=\{X_p \in T_p(P): \omega(X_p)=0 \}.
\]
\label{conn} \end{definition}}

A local 1-form taking values in $T_e(Q)$ can be
associated with the given connection form as follows. Let $\sigma: U\subset { \mathfrak M}\rightarrow{}
\sigma(U) \subset P,\; \pi\circ \sigma ={\rm id}$ be a local section of a
$Q$-bundle $Q\rightarrow{} P \rightarrow{} { \mathfrak  M}$ that is equipped
with a connection 1-form $\omega$. Define the {\em local
$\sigma$-representative} of $\omega$ to be the vector-valued 1-form $\omega^U$ (taking
values at $T_e(Q)$) on the open set $U\subset { \mathfrak  M}$ given
by $\omega^U:=\sigma^\ast \omega.$

{\begin{theorem} {\rm (On reconstruction of the connection form)}
For a given canonical {\sf Ad}-form $\tilde\omega$ defined on
$U\subset {  M}$ with values in $T_e(Q)$ and a given section $\sigma: U
\rightarrow{} \pi^{-1}(U)$, there exists one and only one connection
1-form $\omega$ on $\pi^{-1}(U)$ such that $\sigma^\ast
\omega=\tilde\omega$. \label{Rec}\end{theorem} }

{\begin{proof} Let $p_0=\sigma(x)$ and $Z\in T_{p_0}(P)$. We have
$Z=X_1 +X_2$, where $X_1:= (\sigma_\ast\circ\pi_\ast)Z$  and $X_2\in
{\mathcal V}_{p_0}, \; \pi_\ast X_2 = 0.$  Define at $p_0$ the
1-form $\omega$ to be the vector-valued 1-form given by
$\omega_{p_0} =\tilde\omega_x (\pi_\ast X) + \hat X_2$. The continuation
of the 1-form $\omega$ onto all points of the fiber is realized by
means of right translations, namely, $\forall p\in P,\;\exists a\in
Q: p=\tilde R_a p_0.$ This yields
\[
\omega_p\bigl((\tilde
R_a)_\ast X\bigr) = {\sf Ad}^{-1}_a(q_0)\omega_{p_0}(X), \quad
q_0=\varphi(p_0).
\]
It is easy to see that the 1-form so obtained satisfies all the conditions
of Definition \ref{conn}. \end{proof}}

This construction can be generalized to the whole base $\mathfrak M$ and leads
to the following theorem.

{\begin{theorem} Any smooth principal $Q$-bundle $(P,\pi,\mathfrak
M)$ has a connection. \end{theorem} }

{\begin{proof} Let $\tilde\omega$ be a differentiable 1-form on
$\mathfrak M$ with values in $T_e(Q)$ and $\{ U_\alpha,
\Phi_\alpha\}$ a family of local trivializations associated with the
covering $\{ U_\alpha\}$ of $\mathfrak M$. For each local
trivialization, there exists the local connection form
$\omega_\alpha$ on $\pi^{-1}(U_\alpha)$ obtained from $\tilde\omega$
by a choice of a section $\sigma_\alpha$ over $U_\alpha$ (see
Theorem \ref{Rec}). Let $\lambda_\alpha$ be a partition of unity
subordinate to the covering $\{U_\alpha\}$, $\lambda = 0$ outside of
$U_\alpha$ and $\sum_\alpha \lambda_\alpha (x) =1$ at an arbitrary
point  $x \in  \mathfrak M$. We form
\[
\omega = \sum_\alpha (\lambda_\alpha \circ \pi) \omega_\alpha.
\]
This is a 1-form on $P$ with values in $T_e(Q)$ and for each $\alpha$
we have $\omega\left|_{{}_{U_\alpha}}\right. =
(\lambda_\alpha\circ\pi) \omega_\alpha$. It is easily seen that
$\omega$ satisfies the conditions of the Definition \ref{conn} and
thus this is a connection form on $P$. \end{proof}}

\subsubsection{Connection in the local trivialization }

We consider a principal Q-bundle $(P,\pi, M)$ over $\mathfrak M$ and fix a
trivializating covering $\{U_\alpha,\Phi_\alpha\}_{\alpha \in J}$. Let
$\overline{\rm Id}:  U_\alpha\rightarrow{} U_\alpha\times Q$ by $x
\rightarrow{} (x,e)$. A trivialization $\Phi_\alpha$ defines a {\em canonical}
section $\sigma_\alpha$ by the equation

\[
\sigma_\alpha =
\Phi^{-1}_\alpha \circ\overline{\rm Id},
\]
and vice versa. Further, we denote by $\omega_{C}$ a canonical $\sf Ad$-form.

{\begin{definition} Let $\omega_\alpha = \sigma_\alpha^\ast \omega$,
where $\omega$ is the connection form. The form $\omega_\alpha$ on
$U_\alpha$ is called the {\em connection form in the local
trivialization} $\{U_\alpha,\Phi_\alpha\}.$ \end{definition}}

{\begin{theorem} Let $\{U_\alpha, \Phi_\alpha\}_{\alpha \in J}$ be a
family of local trivializations for $P$ with $\bigcup_{\alpha\in J}
U_\alpha = M$; then, on $U_\alpha\cap U_\beta$, the local connection forms
$\omega_\alpha$ and $\omega_\beta$, corresponding to the same
connection $\omega$ on $P$, are related by
\begin{eqnarray}
\omega_\beta = {\sf Ad}^{-1}_{q_{\alpha\beta}}(q_{\beta\alpha})
\omega_\alpha + l_{(q_{\beta\alpha},q_{\alpha\beta})\mathbf \ast}
\theta_{\alpha\beta}, \label{local}
\end{eqnarray}
where the $q_{\alpha\beta}$ are the transition functions, and
$\theta_{\alpha\beta}=q^{\ast}_{\alpha\beta}\omega_{C}$ denotes the
pullback on $U_\alpha\cap U_\beta$ of the canonical 1-form
$\omega_{C}$ on $Q$. Vice versa, for any set of the local forms $\{
\omega_\alpha\}_{\alpha \in J}$ satisfying {\rm(\ref{local})}, there
exists the unique connection form $\omega$ on $P$ generating this
family of local forms, namely
$\omega_\alpha=\sigma^\ast_\alpha\omega,\; \forall\alpha \in J$.
\end{theorem}}

{\begin{proof}
1. The direct theorem. Let $x\in U_\alpha\cap
U_\beta$. Applying (\ref{sigma}), we obtain $\sigma_\beta(x) =
\tilde R_{q_{\alpha\beta}}\sigma_\alpha(x),\; \forall x\in
U_\alpha\cap U_\beta$. The map $(\sigma_\beta(x))_\ast$ transforms
any vector $X\in T_x(U_\alpha\cap U_\beta)$ into
$(\sigma_\beta)_\ast X \in T_{\sigma_\beta(x)}(P)$. Using the
Leibniz formula (see \cite{KN1}, Ch. I), we get
\[
(\sigma_\beta)_\ast = (\tilde  R_{q_{\alpha\beta}})_\ast
(\sigma_\alpha)_\ast + (L_{q_{\beta\alpha}})_\ast (q_{\alpha\beta})_\ast
\]
where $q_{\beta\alpha}:=\varphi_\beta(\sigma_\alpha)$. Applying $\omega$ to
both sides of this relation, we find
\begin{eqnarray}
\omega_\beta(X):=\omega((\sigma_\beta)_\ast X)&& =
\omega\bigl((\tilde R_{q_{\alpha\beta}})_\ast (\sigma_\alpha)_\ast
X\bigr)
+\omega\bigl((L_{q_{\beta\alpha}})_\ast (q_{\alpha\beta})_\ast X\bigr) \nonumber\\
&&={\sf
Ad}^{-1}_{q_{\alpha\beta}}(q_{\beta\alpha})\,\omega_{\alpha}(X) +
l_{(q_{\beta\alpha},q_{\alpha\beta})\ast}
(q^\ast_{\alpha\beta}\,\omega_{C})(X)\nonumber \\
&&={\sf
Ad}^{-1}_{q_{\alpha\beta}}(q_{\beta\alpha})\,\omega_{\alpha}(X) +
l_{(q_{\beta\alpha},q_{\alpha\beta})\ast}
\theta_{\alpha\beta}\,\omega_{C}(X).
\end{eqnarray}

2. The inverse theorem. Let us define the 1-form $\tilde\omega$ as
follows:
\begin{eqnarray}
\tilde\omega={\sf Ad}^{-1}_{q_\alpha}(e)(\pi^\ast \omega_\alpha)
+ q^\ast_\alpha\, \omega_{C}, \quad q_\alpha:= \varphi_\alpha(p).
\label{L1}
\end{eqnarray}
Let $X\in T_u(P)$ be an arbitrary vector and $u=\sigma_\alpha(\pi(x))$.
Decompose $X$ into horizontal $Y$ and vertical $Z$ components:
\[
X=Y+Z, \quad Y=(\sigma_\alpha)_\ast (\pi_\ast X), \quad \pi_\ast Z =0.
\]
This yields
\begin{eqnarray}
\tilde\omega(X)&&= {\sf Ad}^{-1}_{q_\alpha}(e)\omega_\alpha(\pi_\ast
X ) + (q^\ast_\alpha\, \omega_{C})(X) ={\sf
Ad}^{-1}_{q_\alpha}(e)\omega((\sigma_\alpha)_\ast \pi_\ast X )
+\omega_{C} ((q_\alpha)_\ast X)\nonumber\\
&&= (\tilde R^\ast_{q_\alpha} \sigma^\ast_\alpha) \omega(\pi_\ast Y)
+ \omega_{C}((q_\alpha)_\ast Z) =
\omega((\sigma_\alpha)_\ast \pi_\ast Y )  + \omega_{C} ((q_\alpha)_\ast Z)\nonumber\\
&&=\omega(Y) + \hat Z = \omega(Y)+ \omega(Z) = \omega(X),
\end{eqnarray}
and one sees that $\tilde\omega = \omega$ at any point of the
section $\sigma_\alpha$. So, these forms are transformed in the same
way under the right translations and therefore coincide on
$\pi^{-1}(U_\alpha)$. \end{proof}}

{\begin{corollary} For arbitrary sections $\sigma_1$ and $\sigma_2$
such that $\sigma_2 = R_q \sigma_1$ and $\omega_1 = \sigma^\ast_1
\omega, \;\omega_2 = \sigma^\ast_2 \omega$, the following relation
holds:
\[
\omega_2 = {\sf Ad}^{-1}_q(q_1)\omega_1 + l_{(q_1,q))\ast} (q^\ast
\omega_{C}),
\]
where $q_1:=\varphi(\sigma_1)$. \end{corollary}}

\subsection{Covariant derivative. Curvature form}

Let $\{x^\mu, y^i\}$ be a local coordinate system in the neighborhood
$\pi^{-1}(U_\alpha)$, where the $x^\mu$ are coordinates in $U_\alpha \in {M}$ and
the $y^i$ are coordinates in the fiber. Locally $\pi^{-1}(U_\alpha)$ can be
presented as a direct product $U_\alpha\times Q$. The connection form
can be written in the form: $\omega=\omega^i L_i$, with $\{L_i\}$ being the
basis of left fundamental fields and $\{\omega^i\}$ the basis of 1-forms.
Taking into account Eq. (\ref{L1}), we find that in the coordinates
$\{x^\mu, y^i\}$,
\begin{eqnarray}
\omega^i= \bigl({\sf Ad}^{-1}_{\displaystyle y}(e)\bigr)^i_j
A^j_\mu(x)dx^\mu + \omega^i{}_j(y) dy^j
\label{om}
\end{eqnarray}
where $A^i_\mu(x)dx^\mu = \pi^{\ast}(\omega^i) $ and $\omega^i{}_j(y) dy^j
= \bigl(L^{-1}_{*}\bigr)^i_j dy^j$.

{\begin{definition} A covariant derivative $D_\mu$ in the principal
$Q$-bundle is defined as follows:
\[
D_\mu = \partial_\mu - A^i_\mu (x)\bar L_i,
\]
where the $\bar L_i=(R_\ast)^j_i \partial/\partial y^j$ are generators of the
left translations (right quasi-invariant vector
fields).\end{definition} }

Now, we show that $\omega(D_\mu) = 0$. Indeed, Eq. (\ref{om}) implies
\[
\omega(D_\mu)= [\bigl({\sf Ad}^{-1}_{\displaystyle y}(e)\bigr)^i_j
A^j_\mu(x) - \omega^i{}_p A^j_\mu (R_\ast)^p_j]L_i.
\]
Noting that $\omega^i{}_p= (L^{-1}_\ast)^i_ p$, we obtain
$\omega^i{}_p(R_\ast)^p_j = ({\sf Ad}^{-1}_q (e))^i_j$, and hence
$\omega(D_\mu) =0$.

The computation of the commutator $[D_\mu,D_\nu]$ yields
\[
[D_\mu,D_\nu]=(\partial_\nu A^i_\mu - \partial_\mu A^i_\nu)\bar L_i
+ A^i_\mu A^j_\nu [\bar L_i, \bar L_j],
\]
and introducing $[\bar L_i, \bar L_j] = C^p_{ij}(y)\bar L_p$, we get
\begin{eqnarray}
[D_\mu,D_\nu]=- F^i_{\mu\nu}\bar L_i,\\
F^i_{\mu\nu}: = \partial_\mu A^i_\nu - \partial_\nu A^i_\mu
- A^j_\mu A^p_\nu C^i_{jp}.
\end{eqnarray}

{\begin{definition} Let $\Psi$ be a vector-valued $r$-form in the
principal $Q$-bundle. A $(r+1)$-form $D\Psi$ defined by
\begin{eqnarray}
D\Psi(X_1,X_2,\dots,X_{r+1})=d\Psi(hor X_1,\dots,hor X_{r+1})
\end{eqnarray}
is called a {\em covariant differential} of the form $\Psi$.
\end{definition}}

{\begin{definition} A vector-valued 2-form $\Omega(X,Y)$ defined as
\begin{eqnarray}
\Omega(X,Y) = D\omega (X,Y) =d\omega (hor X, hor Y)
\end{eqnarray}
where $\omega$ is a connection form, is called a {\em curvature form}.
\end{definition}}

{\begin{lemma} Let $X,Y$ be horizontal fields, then the following
relation holds:
\[
\omega([X,Y]) = -2\Omega(X,Y).
\]
\label{lem2}\end{lemma} }

{\begin{proof} Applying the exterior differentiation to the 1-form
$\omega$, we obtain
\[
2d\omega(X,Y)=X\omega(Y) - Y\omega(X) -\omega([X,Y]).
\]
As $X,Y \in {\mathcal H}_p$, then $\omega(X)=\omega(Y)=0$. This
yields $\omega([X,Y]) = -2\Omega(X,Y)$. \end{proof}}

{\begin{corollary} The curvature form can be defined as follows:
\[
\Omega(X,Y)=-\frac{1}{2}\omega([hor\bar Y,hor\bar Y])
\]
where $\bar X, \bar Y$ are any continuations of  the vectors $X,Y
\in T_p(P)$, respectively.\end{corollary} }

\begin{corollary}
The 2-form $\Omega$ is $\sf Ad$-form and is
transformed under the right translations as
\begin{eqnarray}
(\tilde R^\ast_a \Omega)(X,Y)= {\sf Ad}^{-1}_a(q)\,\Omega \vert_p
(X,Y), \; {\rm where} \; q=\varphi(p).
\end{eqnarray}
\end{corollary}

{\begin{theorem} The curvature form $\Omega$ satisfies the structure
equation
\begin{eqnarray}
\Omega = d \omega + \frac{1}{2}{[\omega,\omega]}.
\label{str}
\end{eqnarray}
\end{theorem}}

{\begin{proof} We are going to prove (\ref{str}), considering all
possible pairs $X,Y$.

1. Let $X,Y$ be any two horizontal vectors then $\omega(X)=\omega(Y)=0$ and
we find
\[
d\omega(X,Y) + \frac{1}{2}{[\omega(X),\omega(Y)]} = d\omega(hor X, hor Y) =
\Omega(X,Y)
\]

2. Let $X,Y$ be any two vertical vectors. Without loss of generality, one can
assume that $X,Y$ are fundamental vector fields. This implies $\omega(X) =
\hat X,\; \omega (Y) =\hat Y, \; \hat X, \hat Y \in T_e(Q)$.
The computation yields
\begin{eqnarray}
2 d \omega(X,Y)= X\omega(Y) - Y\omega(X) -\omega([X,Y])
= X(\hat Y) - Y(\hat X) -\omega([X,Y])   \\
= -[\hat X, \hat Y] =- [\omega(X),\omega(Y)] \Longrightarrow
d \omega(X,Y) + \frac{1}{2}[\omega(X),\omega(Y)] =0.
\end{eqnarray}
As $X,Y$ are the vertical fields, then $\Omega(X,Y)= d\omega(hor X,
hor Y) =0$.

3. Let $X$ be a horizontal vector field and $Y$ be a vertical one (or
fundamental vector field). From the one side, we have
\[
\Omega(X,Y)=d\omega(hor X, hor Y) =0,
\]
and from the other side,
\[
d\omega(X,Y)=X\omega(Y) - Y\omega(x) -\omega([X,Y]) =0,
\]
since the commutator $[X,Y]$ is the horizontal vector field.
\end{proof}}

The last statement arises from the following lemma.

{\begin{lemma}  Let $Z$ be a fundamental vector field and $X$ be a
horizontal vector field; then, the commutator $[X,Z]$ is a horizontal
vector field. \end{lemma}  }

{\begin{proof} The fundamental vector field is induced by the right
translations $\tilde R_{a(t)}$. The commutator can be defined (see,
e.g., \cite{KN1}) by
\[
[X,Z] = \lim_{t\rightarrow{} 0}\frac{1}{t}((\tilde R_{a(t)})_\ast X - X).
\]
If $X$ is a horizontal vector field, then $(\tilde R_{a(t)})_\ast X$
is a horizontal vector field, and, hence, $[X,Z]$ is also
horizontal. \end{proof}}

Let us compute $\Omega_{\mu\nu} = 2\Omega(D_\mu,D_\nu)$. The result is given by
\begin{eqnarray}
\Omega_{\mu\nu}= -\omega([D_\mu,D_\nu])
= F^i_{\mu\nu}\omega(\bar L_i)= F^i_{\mu\nu}
\bigl({\sf Ad}^{-1}_q(e)\bigr)^j_i \hat L_j \nonumber
\label{Om}
\end{eqnarray}
where $\{\hat L_j =\omega(L_j)\}$ is the basis of the left quasi-invariant
vector fields at $T_e(Q),\; q=\varphi(p),\; p\in P$. Choosing the family of
 local sections $\sigma_\alpha$ associated with the trivialization
$U_\alpha, \Phi_\alpha$ and taking into account that
$\varphi_\alpha(\sigma_\alpha)=e$ where $\varphi_\alpha$ is the restriction of
$\Phi_\alpha$ on $\pi^{-1}(U_\alpha)$, we obtain
\begin{eqnarray}
\pi^\ast \Omega\vert_\alpha =\frac{1}{2}F_{\mu\nu}dx^\mu \wedge dx^\nu.
\end{eqnarray}
introducing $F_{\mu\nu}= F^i_{\mu\nu}L_$.

Since $\Omega$ is a $\sf Ad$-form, the following transformation law is true:
\begin{eqnarray}
\Omega\vert_\beta = {\sf Ad}^{-1}_{q_{\alpha\beta }}
(q_{\beta\alpha})\Omega\vert_\alpha.
\end{eqnarray}

{\begin{theorem} Bianchi identity: $D\Omega =0$. \end{theorem}}

{\begin{proof} It is sufficient to show that $d\Omega(X,Y,Z) =0$, if
$X,Y,Z$ are horizontal vector fields. Applying the exterior
derivative  to (\ref{str}), we obtain $d\Omega(X,Y,Z) =0$, if
$X,Y,Z$ are horizontal vector fields. \end{proof}}

\section*{Acknowledgements}

This work was partly supported by SEP-PROMEP (Grant No.
103.5/04/1911).

config: arxiv


TITLE: Empty lattice simplex or White's theorem
QUESTION [4 upvotes]: White has proved (White, G. K. Lattice tetrahedra -- Canad. J. Math. 16 1964 389–396.) the following theorem:
If $T$ is a closed tetrahedron and $\Lambda$ is a lattice which contains the vertices of $T$, then the two following conditions are equivalent:
(1) The only points of $\Lambda$ in $T$ other than the vertices lie on a pair of opposite edges of $T$;
(2) There is a pair of parallel lattice planes of $\Lambda$ through a pair of opposite edges of $T$ such that no points of $\Lambda$ lie between these planes.
His proof is rather complicated. Has anybody proved this theorem in a simpler way? What proof is the simplest one?

REPLY [1 votes]: I have not looked at the literature, so the following may fail for some geometric reason of which I am unaware; this reflects my thinking, and I do not yet see how else to get the results.
It should be clear that (2) implies (1), so let's think about not (2) implying not (1).  Then for each pair of parallel planes going through opposite edges of T, if they are lattice planes then there is another lattice plane between
them that goes through T and has "as many" points of the lattice as do the parallel planes.  I am hoping there is enough regularity so that the middles plane looks like one of the parallel planes shifted; perhaps I am wrong.  Now we have for each pair of opposite edges of T a lattice plane going through T between the pair.
Following Sidney Harris, "And then a miracle occurs...", which means I do not know how to take the three planes intersecting the middle of T and produce a point lying in T and on the three planes, and make it a lattice point.  If I were looking for a simple
(and natural) proof of the result, however, this is what I would try.  Perhaps I would need a map to pull the argument into Z^3 and use some analytic geometry to finish things, but I hope not.
Gerhard "Not Copying Someone Else's Catchphrase" Paseman, 2012.08.14


TITLE: Probability of including one person and not including an other one when creating a 4 person group
QUESTION [2 upvotes]: In a classroom there are 24 students, including Joe, Robert and Oliver.
We randomly assemble a four-student group (The order does not matter).
What is the probability that the 4 student group will include Oliver but not Robert.
I have tried removing one student (Robert) and calculating the possibilities with Oliver. I did: ${23 \choose 4}$ - ${22 \choose 4}$ = 1540, which I divided by ${23 \choose 4}$ = 1540/8855 (but it doesn't seem right)

REPLY [2 votes]: Let Oliver in four-student group. Now it will be selecting three students out of $22$, since Robert is not in the group.
P(including Oliver but not Robert)$=\frac{\binom{22}{3}}{\binom{24}{4}}=\frac{10}{69}$


TITLE: Showing that the irreducible factors of the cyclotomic polynomial in $\mathbb{Q}[x]$ have the same degree.
QUESTION [2 upvotes]: For each $n\in \mathbb{N},\, \,$ $\phi_{n}(x):=\displaystyle \prod_{\substack{1\leq k\leq n\\ (k,n)=1}}(x-e^{\frac{2\pi i}{n}k})$
Show that the irreducible factors of $\phi_n$ in $\mathbb{Q}[x]$ have the same degree.
I have a confusion with this exercise. I understand that the cyclotomic polynomial $\phi_n$ is irreducible in $\mathbb{Q}[x]$ so $\phi_n$ has a unique irreducible factor, the same. Is this correct or am I confusing something?

REPLY [2 votes]: The argument you gave is correct, but proving irreducibility of the cyclotomic polynomials over $\mathbb Q$ takes some work. You can solve the question at hand without citing this result, which I'm guessing is the intent of the exercise. I will, however, be using some field theory.
Let $\phi_n = \prod f_i$ with each $f_i$ monic and irreducible. The $f_i$ are distinct as $\phi_n$ has no repeated roots. Let $\alpha_i$ be a root of $f_i$. These are therefore roots of $\phi_n$ so they are primitive $n^{th}$ roots of unity. Let's consider the field $\mathbb Q(\alpha_i)$. The degree of this field is the degree of the minimal polynomial of $\alpha_i$, which is $f_i$. Furthermore, as $\alpha_i$ is a primitive $n^{th}$ root of unity, the set $\{\alpha_i^k : k \in \mathbb Z\}$ is precisely the set of $n^{th}$ roots of unity. But since $\mathbb Q(\alpha_i)$ is a field, each of these powers lies in $\mathbb Q(\alpha_i)$. Hence, $\mathbb Q(\alpha_i)$ contains all the $n^{th}$ roots of unity. Furthermore, because the above result held for all $i$, we have $\alpha_j \in \mathbb Q(\alpha_i)$ for all pairs $i,j$. The definition of $\mathbb Q(\alpha_j)$ is the smallest field containing $\mathbb Q$ and $\alpha_j$, so $\mathbb Q(\alpha_j) \subseteq \mathbb Q(\alpha_i)$ for all pairs $i,j$. Thus, all of these fields $\mathbb Q(\alpha_i)$ are the same. In particular, they have the same degree over $\mathbb Q$. We have that $[\mathbb Q(\alpha_i) : \mathbb Q] = \deg f_i$, so for all $i,j$ $\deg f_i = \deg f_j$ as desired.


TITLE: Probability distribution of faulty buses made during one week
QUESTION [2 upvotes]: I've been trying to figure out the following question:

I modelled the discrete random variable X with a binomial distribution such that:
X ~ Bin(3, 3/5)
I took the number of trials to be 3 because there can be between zero and three faulty buses (successful trials) on the Monday in question.
I took the probability of having one faulty bus on a given day to be 3/5. I calculated this by taking the probability of one bus on one day to be 1/5 (because there are five days) and multiplying by three (because there are three faulty buses made that week). This gave me Pr(success) = 3/5
When I used the equation V(X) = npq (n=trials, p=success probability, q=failure probability) I found the variance of X to be equal to 18/25. (npq = 3 x 3/5 x 2/5 = 18/25)
In the answers for this question though, the variance is listed as being 12/25:

I don't know what I've done wrong here. Have I missed out some fundamental bit of information? Is the answer sheet wrong?
Any help is massively appreciated. Thank you for taking the time to read all of this :)

REPLY [1 votes]: There are $3$ independent experiments, each of them represented by a faulty bus, and with equal probability to succeed.
We speak of a success if a faulty bus is made on Monday.
Then we are dealing with binomial distribution with parameters $n=3$ and $p=\frac15$.
That gives variance $np(1-p)=3\frac15\frac45=\frac{12}{25}$


TITLE: Convergence of $a_{n+1}=\sqrt{2-a_n}$
QUESTION [5 upvotes]: I'm attempting to prove the convergence (or divergence, though I strongly suspect it converges) of a sequence defined as  $a_{n+1}=\sqrt{2-a_n}$ with $a_1=\sqrt{2}$.
I cannot use the monotonic sequence theorem as the sequence is not monotonically increasing. In fact, the first few values of the sequence are:
$a_1 =\sqrt{2}\approx 1.4142$
$a_2 =\sqrt{2-\sqrt{2}}\approx .7653$
$a_3 =\sqrt{2-\sqrt{2-\sqrt{2}}}\approx 1.1111$
Thus, it seems that $a_{n \to \infty} \to 1$
It seems that the sequence is behaving similarly to $\frac{\sin x}{x}$, leading me to think that the squeeze theorem may be useful. Still, I cannot seem to make any progress besides numerical computation of successive terms.

REPLY [1 votes]: Here is a relatively simple argument that shows the convergence of the sequence $(a_n)_{n\geq 1}$.
First, we show by induction that  $1/3<a_n<5/3$ for $n\geq 1$.  This is clearly the case for $n=1$, since $a_1=\sqrt{2}$.  Assuming the inductive hypothesis, we have:
\begin{eqnarray*}
1/3&<&a_n<5/3 \\
5/3&>&2-a_n>1/3 \\
5/3>\sqrt{5/3}&>&a_{n+1}>\sqrt{1/3}>1/3\,.
\end{eqnarray*}
QED
Next, use the recurrence to get
$$a_{n+1}^2-1=(2-a_n)-1=-(a_n-1)\,.$$ Hence
$|a_{n}-1|=|a_{n+1}+1|\,|a_{n+1}-1|>4/3\,|a_{n+1}-1|$ and $|a_{n+1}-1|<3/4 \, |a_{n}-1|$ (for $n\geq 0$).  Consequently,
$$|a_{n}-1|<(3/4)^n$$
(for $n\geq 1$).  It follows that $\lim_{n\to \infty}a_n=1$.


TITLE: What is a continuous representation space for 3D rotations under cube symmetries?
QUESTION [3 upvotes]: I'm not sure if I'm using the right terminology on my title but I'll explain here. By "continuous representation space" I mean that I want to represent all physical rotations (call this the set $X$) in some manifold in $R = \mathbb{R}^n$ where I'm not so concerned about what the $n$ is. The conditions on this space would be:

I have a mapping $g: X \rightarrow R$ and a mapping $f: R \rightarrow X$ and one mapping is the inverse of another. I think this is called a bijection.
Small nudges to some $\mathbf{r} \in R$ cause proportionally small nudges to $\mathbf{x} = f(\mathbf{r}) \in X$. Same goes for the inverse. This is true everywhere.

It's known that the group of orthonormal matrices in $SO(3)$ are such a representation. Quaternions are not because they are a double cover on $SO(3)$ and even if you restrict them to one half space they still have a discontinuity with angles of $\pi$.
So far so good. But what if I wish to represent physical orientations of a perfect cube where there are 24 symmetries? This doesn't work with the $SO(3)$ matrices. For example a rotation of $\pi/4$ about one of the edges of the cube gives the same results as a rotation of $-\pi/4$ about that same edge. This situatuion satisfies nor 1 nor 2. To satisfy 1 I could squish $SO(3)$ up to only include those rotations that have the smallest geodesic distance to one of the 24 cube symmetries but this still wouldn't satisfy 2 (and I couldn't think of a way to write that squishing operation into closed form).
For reference/visualization, this GIF shows what happens when a take a small arrow and apply a uniform(ish) grid of all possible rotations to it. Then what happens if I remap all arrows via my squishing algorithm.

REPLY [2 votes]: It sounds to me like you're looking for a description of the quotient space of $SO(3)$ by the subgroup of rotations of a cube, specifically an embedding of this space into Euclidean space. This is a homogeneous space for a Lie group (either $SO(3)$ or $SU(2)$) and lots of stuff is known about these.
This particular homogeneous space is, among other things, a closed smooth $3$-manifold. By the Whitney embedding theorem this implies that it can be smoothly embedded into $\mathbb{R}^6$. Hirsch, Wall, and Rokhlin apparently proved that a smooth $3$-manifold in fact embeds into $\mathbb{R}^5$. Unfortunately these proofs are, as far as I know, pretty non-explicit.
Interestingly this question has been asked before (with no answers), $6$ years ago, but that time the asker wanted a stronger condition of an isometric embedding. The space you're interested in can be described as a quotient of the $3$-sphere $S^3 \cong SU(2)$ by the binary octahedral group, which reveals it to be a spherical $3$-manifold. In particular it inherits (up to scale) a metric of constant positive curvature from the $3$-sphere and one can ask for an embedding which respects this metric (which, for example, is true of the embedding of $SO(3)$ into $\mathbb{R}^9$ given by the usual matrix representation). For a closed Riemannian $3$-manifold the Nash embedding theorem guarantees such an embedding into $\mathbb{R}^{12}$ but is again, as far as I know, pretty non-explicit.
Here's an idea for writing down a (terrible) explicit embedding which doesn't quite work, which is a variation of Yuval Peres' answer to your other question. We'll think of your space as the quotient of $S^3$ by the binary octahedral group which we'll just write as $\Gamma$. Write $x_1, \dots x_4$ for the $4$ coordinates of the usual embedding $S^3 \to \mathbb{R}^4$. These coordinates, thought of as functions on $S^3$, are acted on linearly by $\Gamma$. Now consider the $48$ non-leading coefficients of the $4$ polynomials
$$f_i(t) = \prod_{g \in \Gamma} (t - g x_i).$$
The coefficients are by construction smooth functions invariant under $\Gamma$, and moreover by construction knowing these coefficients is enough to recover the orbit $\{ gx_i \}$ of each coordinate function $x_i$ independently. That means if we know the values of all of these coefficients we know, generically, that they came from $48^4$ possible points, where each factor of $24$ comes from using one of each of the possible values of $\{ gx_i \}$ in each orbit. This is bad; what we'd like to do is to recover a single orbit of $48$ points. However, we can hope that most of the resulting $4$-tuples of vectors are not unit vectors (which they would need to be to lie on the $3$-sphere $S^3$) and so we can hope that in practice actually most of them are ruled out. I don't know if this happens but even if it does the resulting map takes values in $\mathbb{R}^{192}$ and is annoying to invert. We can probably remove the ambiguity entirely by adding a few more functions but the result would still be annoying to invert.
Generally speaking, pure mathematicians prefer to avoid working with embeddings at all and will just work directly on a manifold itself, independent of a choice of embedding, partly to avoid annoying issues like this. If you could say more about what you need computationally from such an embedding it might be easier to suggest alternative approaches.


text:
The landmark paper by Lighthill and Whitham (1955) set the tone for many researchers' investigations into the theory of traffic flow. Introduced for traffic flows on a single, long, and rather idealized road, the LWR model proposes that their dynamics is described by the following PDE:
\bqn
\r_t+f(\r)_x&=&0, \label{lwr:eqn1}
\eqn
where subscript $t$ means the partial derivative with respect to time $t$, and subscript $x$ means the partial derivative with respect to location $x$.
In \refe{lwr:eqn1}, the function $f(\r)=\r \vs(\r)$ is called the fundamental diagram of traffic flow, in which $\vs(\r)$ reflects the equilibrium relationship between travel speed and traffic density. It's generally assumed that $f(\r)$ is a concave function, i.e., $f_{\r\r}(\r)<0$. The characteristic wave speed $\lambda(\r)=f_{\r}(\r)=\vs(\r)+\r \vs'(\r)$. $\lambda(\r)$ can be positive or negative.

Weak solutions (Smoller, 1983\nocite{smoller1983}) for \refe{lwr:eqn1} satisfy the integral form of the conservation law:
\bqn
\frac{\partial}{\partial t}\int_{x_1}^{x_2} \r(x,t)\m{dx}&=&f(x_1,t)-f(x_2,t).
\eqn
First, we discuss the Riemann problem for \refe{lwr:eqn1} theoretically and then give the numerical solutions.

\section{The Riemann problem}
We consider the Riemann problem for the LWR model with the following jump initial condition:
\bqn
\r(x,t=0)&=&\cas{{ll}\rl, & x<0\\\rr, &x> 0}. \label{lwr:ini}
\eqn

There are two types of wave solutions to the Riemann problem. The discontinuous solution is a shock wave
\bqn
\r(x,t>0)&=&\cas{{ll}\rl, & x/t<s\\\rr, &x/t> s},
\eqn
where $s$ is the shock wave speed. The shock wave speed is determined by the Rankine-Hugoniot jump condition,
\bqn
s&=&\frac {[f(\r)]}{[\r]}.
\eqn

The wave speed of a valid shock wave solution has to satisfy the entropy condition:
\bqn
\lambda(\rl)>s>\lambda(\rr).
\eqn
Specifically, for a concave fundamental diagram, a shock wave is a solution to the Riemann problem for \refe {lwr:eqn1} with initial conditions \refe{lwr:ini}  when $\rl<\rr$; i.e., the upstream traffic density is lower.

When the upstream traffic density is higher, solution to the LWR model with initial data \refe{lwr:ini} is a continuous rarefaction wave.  The rarefaction wave is given by $\r=\r(\xi), \xi=x/t$, where $\r$ satisfies the ordinary differential equation
\bqn
-\xi \r_{\xi}+f(\r)_{\xi}&=&0.
\eqn
If $\r(\xi)\neq 0$, we obtain
\bqn
\lambda(\r(\xi))=\xi,
\eqn
from which we can find $\r(\xi)$ uniquely. On any characteristic $x/t=\xi$, $\r$ is constant.

\section {Computation of boundary fluxes}
Given a wave solution to the Riemann problem, we can compute the average $\r^{\ast}$ at $x=0$ and therefore the flux $f(\r^{\ast})$ through the boundary. There are the following five cases:
\bi
\item[Case 1] When the solution to the Riemann problem is a shock wave with wave speed $s>0$, $\r^{\ast}=\rl$.
\item[Case 2] When the solution to the Riemann problem is a shock wave with wave speed $s\leq0$, $\r^{\ast}=\rr$.
\item[Case 3] When the solution to the Riemann problem is a rarefaction wave, and $\lambda(\rl)>0$, $\r^{\ast}=\rl$.
\item[Case 4] When the solution to the Riemann problem is a rarefaction wave, and $\lambda(\rr)<0$, $\r^{\ast}=\rr$.
\item[Case 5] When the solution to the Riemann problem is a rarefaction wave, $\lambda(\rl)<0$ and $\lambda(\rr)>0$, $\r^{\ast}$ is the solution of $\lambda(\r^{\ast})=0$.
\ei

Given initial and boundary conditions, we can use a first-order Godunov method to calculate the traffic conditions. The numerical solutions are given in next section.

\section {Numerical solutions to the LWR model}
In this section, we solve the Riemann problem for \refe{conse} numerically. We use Newell's equilibrium model,
\bqn
v_{\ast}(\rho)&=&v_f\left(1-\exp\{\frac {|c_j|}{v_f}(1-\rho_j/\rho\}\right)
\eqn
Without loss of generality, we set $v_f=1, c_j=1, \rho_j=1$ to obtain
\bqn
v_{\ast}(\rho)&=&1-\exp{(1-\frac 1{\rho})}
\eqn
The valid range for $\r$ and $v$ is $0<\r, v \leq 1$.  The corresponding flow rate, $f_{\ast}=\rho v_{\ast}$, is a normalized fundamental diagram.

\ben
\item Given the initial conditions:
\bqn
\rho(x,0)&=&\cas {{ll} 0.65& x\in [0l, 200l]\\0.4& \m{otherwise}}\\
v(x,0)&=&v_{\ast}(\rho(x,0)), \qquad \forall x\in [0l, 800l]
\eqn
we obtain the solutions shown in  \reff{LWRRiemann3_3d},\reff{LWRRiemann3_2d}.

\item Given the initial conditions:
\bqn
\rho(x,0)&=&\cas {{ll} 0.65& x\in [0l, 200l]\\0.9& \m{otherwise}}\\
v(x,0)&=&v_{\ast}(\rho(x,0)), \qquad \forall x\in [0l, 800l]
\eqn
we obtain the solutions shown in \reff{LWRRiemann4_3d},\reff{LWRRiemann4_2d}.

\een

From these numerical solutions we can see the formation of shock waves and rarefaction waves. These solutions are only an approximation of real solutions. For example, the solutions after $t=0$ in \reff{LWRRiemann4_2d} are not exact jumps, while the theoretical solution to the LWR model with initial conditions \refe{lwr:ini} is still a jump at any time. However as $\dx\to0$ and $\dt\to 0$, the solutions given by Godunov's method converge to the exact solutions.

\bfg
\bc\includegraphics [height=8cm] {LWRRiemann3_3d}\ec
\caption {Rarefaction wave solution of the LWR model with initial data \refe{lwr:ini}} \label {LWRRiemann3_3d}
\efg
\bfg
\bc\includegraphics [height=8cm] {LWRRiemann3_2d}\ec
\caption {Solutions from \reff{LWRRiemann3_3d} at selected times} \label {LWRRiemann3_2d}
\efg

\bfg
\bc\includegraphics [height=8cm]{LWRRiemann4_3d}\ec
\caption {Shock wave solution of the LWR model with initial data \refe{lwr:ini}} \label {LWRRiemann4_3d}
\efg
\bfg
\bc\includegraphics [height=8cm] {LWRRiemann4_2d}\ec
\caption {Solutions from \reff{LWRRiemann4_3d} at selected times} \label {LWRRiemann4_2d}
\efg


TITLE: Is there a natural homorphism $\pi_1(X/G) \rightarrow G$?
QUESTION [1 upvotes]: In Basic Topology (Armstrong, M.A., 1983) we have the following:

(5.13) Theorem. If $G$ acts as a group of homeomorphisms on a simply connected space $X$, and if each point $x \in X$ has a neighbourhood $U$ which satisfies $U \cap g(U) = \emptyset$ for all $g \in G - \{e\}$, then $\pi_1(X/G)$ is isomorphic to $G$.

Probably a very naive question, but is there a version of this without the assumption of simply-connectedness that gives us a (mere) homomorphism $\pi_1(X/G) \rightarrow G$? If so, how is this homomorphism defined?
I think that deck transformations are relevant here, but my understanding of them is a bit sketchy and I'm having trouble pulling my intuitions together into a coherent thought.
This question is related, but seems to go the other way.

REPLY [1 votes]: I'm assuming that $G$ is a discrete group. Then given the existence of neighbourhoods $U$ at each point $x\in X$ satisfying $U\cap g(U)=\emptyset$, a standard result is that $X\rightarrow X/G$ is a principal $G$-covering (this is given as Proposition 14.1.12 in tom Dieck's "Algebraic Topology", for instance).
This means that $G\hookrightarrow X\xrightarrow{p}X/G$ is a fibration sequence, where $p$ is the quotient map. This gives a long exact homotopy sequence
$$0\rightarrow\pi_1X\xrightarrow{p_*} \pi_1(X/G)\xrightarrow{\Delta} \pi_0G\rightarrow \pi_0X\rightarrow\dots$$
where $\Delta$ is the fibration connecting map.
Since we assumed that $G$ was discrete we have that $\pi_0G\cong G$, not only as a set, but also as a group. This follows since the fibration is principal, which means that it classifies by some map $\delta:X/G\rightarrow BG\simeq K(G,1)$ which in turn allows us to identify $\pi_0G\cong \pi_1BG\cong G$ and the map $\Delta$ as the homomorphism induced by $\delta_*:\pi_1(X/G)\rightarrow \pi_1BG$.
Hence, identifying $\pi_0G\cong G$ we have a homomorphism
$$\Delta:\pi_1(X/G)\rightarrow G.$$
Of course, to get this homomorphism we may equally well construct the map $\delta:X/G\rightarrow BG$, identify $\pi_1BG\cong G$ and then identify the homomorphism $\delta_*:\pi_1(X/G)\rightarrow G$.
Of course in either case, the kernel of the homomorphism is simply $\pi_1X$.


TITLE: Is there $B \in M_4({\mathbb{R}})$ such that $B^2$ is similar to $A$ ? If it is possible, find it.
QUESTION [1 upvotes]: $A =\left ( \begin{matrix}
    0 & 0 & 1 & 0\\
    1 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
   \end{matrix} \right )$
Is there $B \in M_4({\mathbb{R}})$ such that $B^2$ is similar to $A$ ? If it is possible, find it.
The Jordan normal form of $A$ is $J=\begin{pmatrix}
    0 & 0 & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1\\
    0 & 0 & 0 & 0\\
   \end{pmatrix}$, $P=\begin{pmatrix}
    0 & 0 & 1 & 0\\
    0 & 1 & 0 & 0\\
    0 & 0 & 0 & 0\\
    1 & 0 & 0 & 1\\
   \end{pmatrix}$ such that $P^{-1}AP=J$
Then, $J=B^2$ but how can I find B? How do I know if it is possible to find $B$?

REPLY [2 votes]: Your initial step of finding the Jordan form of $A$ is a helpful. With that, we can see that $A^3 = 0$, but $A^2 \neq 0$. This is enough for us to figure out that there is no $B$ such that $B^2$ is similar to $A$.
Indeed, if $B^2$ were similar to $A$, then it would follow that $B^6 = (B^2)^3$ similar to $A^3 = 0$, which means that $B$ is nilpotent, which means that the characteristic polynomial of $B$ must be $p(x) = x^4$. From this, it follows that $p(B) = B^4 = 0$. On the other hand, $B^4 = (B^2)^2$ must be similar to $A^2 \neq 0$. Thus, $B^4$ is similar to both the zero matrix and $A^2 \neq 0$, which means that $A^2$ is a non-zero matrix that is similar to the zero matrix, which is impossible.

The question of finding a general rule that determines whether $A$ has a square root is a bit trickier. As it turns out, this is true iff the blocks of the Jordan form of $A$ with length greater than $1$ can be paired off such that the sizes of the blocks in each pair are at most one apart.


TITLE: Prove that $(\frac{13}{16})^{12}=\sum_{k=0}^{12}{12 \choose k}\left(\frac{1}{4}\right)^{2k}\left(\frac{3}{4}\right)^{12-k}$
QUESTION [0 upvotes]: I tried to solve a probability question where I got $\sum_{k=0}^{12}{12 \choose k}\left(\frac{1}{4}\right)^{2k}\left(\frac{3}{4}\right)^{12-k}$ but did not succeed in evaluating that to $(\frac{13}{16})^{12}$ by hand. There is an answer that requires defining a specific probability space, but my question is whether there's a more "straightforward" way of doing this proof.

REPLY [2 votes]: This is nothing but the binomial expansion.
$$\frac{13}{16} = \frac{1}{16} + \frac{12}{16}$$
hence
$$\left(\frac{13}{16}\right)^{12} = \left(\frac{1}{16} + \frac{12}{16}\right)^{12} = \sum_{k =0}^{12} \binom{12}{k}\left(\frac{1}{16}\right)^k \left(\frac{12}{16}\right)^{12-k}$$
Now just notice that
$$\frac{12}{16} = \frac{3}{4}$$
and
$$\frac{1}{16} = \left(\frac{1}{4}\right)^2$$
whence the identity.
Binomial Expansion Review
$$(A + B)^n = \sum_{k =0}^n \binom{n}{k} A^n B^{n-k}$$